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| author | Trupti Kini | 2016-05-12 23:30:31 +0600 |
|---|---|---|
| committer | Trupti Kini | 2016-05-12 23:30:31 +0600 |
| commit | 5218db5dbae9c0890958f57af323c8d296ac9b07 (patch) | |
| tree | 23dfd4b8c171a9f4ca5e20525a4f92a216484ce3 | |
| parent | da40306957548bcc4a1a8d5e2cdc8928017fe5da (diff) | |
| download | Python-Textbook-Companions-5218db5dbae9c0890958f57af323c8d296ac9b07.tar.gz Python-Textbook-Companions-5218db5dbae9c0890958f57af323c8d296ac9b07.tar.bz2 Python-Textbook-Companions-5218db5dbae9c0890958f57af323c8d296ac9b07.zip | |
Added(A)/Deleted(D) following books
A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter25_2.ipynb
A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter26_2.ipynb
A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter27_2.ipynb
A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter28_2.ipynb
A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter29_2.ipynb
A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter30_2.ipynb
A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter31_2.ipynb
A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter32_2.ipynb
A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter33_2.ipynb
A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter34_2.ipynb
A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter35_2.ipynb
A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter36_2.ipynb
A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter37_2.ipynb
A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter38_2.ipynb
A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter39_2.ipynb
A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/screenshots/chapter29example32_2.png
A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/screenshots/chapter29example33_2.png
A A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/screenshots/chapter32example30_2.png
A Advance_Semiconductor_Devices_by_K._C._Nandi/chapter1_1.ipynb
A Advance_Semiconductor_Devices_by_K._C._Nandi/chapter2_1.ipynb
A Advance_Semiconductor_Devices_by_K._C._Nandi/chapter5_1.ipynb
A Advance_Semiconductor_Devices_by_K._C._Nandi/screenshots/KC_econ1_chapter1_1.png
A Advance_Semiconductor_Devices_by_K._C._Nandi/screenshots/KC_econ_ch1_1.png
A Advance_Semiconductor_Devices_by_K._C._Nandi/screenshots/KC_percentChangeinDiodeCurrent_chapter2_1.png
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch15_1.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch15_2.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch15_3.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch16_1.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch16_2.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch16_3.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch17_1.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch17_2.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch17_3.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch18_1.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch18_2.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch18_3.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch19_1.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch19_2.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch19_3.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch20_1.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch20_2.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch20_3.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch21_1.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch21_2.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch21_3.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch22_1.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch22_2.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch22_3.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch23_1.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch23_2.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch23_3.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch24_1.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch24_2.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch24_3.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch25_1.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch25_2.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch25_3.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch26_1.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch26_2.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch26_3.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch27_1.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch27_2.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch27_3.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch28_1.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch28_2.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch28_3.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch29_1.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch29_2.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch29_3.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch30_1.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch30_2.ipynb
D College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch30_3.ipynb
A Electronic_Devices_and_Circuits_by_J._Paul/Ch10_1.ipynb
A Electronic_Devices_and_Circuits_by_J._Paul/Ch11_1.ipynb
A Electronic_Devices_and_Circuits_by_J._Paul/Ch12_1.ipynb
A Electronic_Devices_and_Circuits_by_J._Paul/Ch13_1.ipynb
A Electronic_Devices_and_Circuits_by_J._Paul/Ch1_1.ipynb
A Electronic_Devices_and_Circuits_by_J._Paul/Ch2_1.ipynb
A Electronic_Devices_and_Circuits_by_J._Paul/Ch3_1.ipynb
A Electronic_Devices_and_Circuits_by_J._Paul/Ch4_1.ipynb
A Electronic_Devices_and_Circuits_by_J._Paul/Ch5_1.ipynb
A Electronic_Devices_and_Circuits_by_J._Paul/Ch6_1.ipynb
A Electronic_Devices_and_Circuits_by_J._Paul/Ch7_1.ipynb
A Electronic_Devices_and_Circuits_by_J._Paul/Ch8_1.ipynb
A Electronic_Devices_and_Circuits_by_J._Paul/Ch9_1.ipynb
A Electronic_Devices_and_Circuits_by_J._Paul/screenshots/4CollCurr_1.png
A Electronic_Devices_and_Circuits_by_J._Paul/screenshots/4MaxNBasRes_1.png
A Electronic_Devices_and_Circuits_by_J._Paul/screenshots/4saturationMode_1.png
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch1.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch1_1.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch1_2.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch1_3.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch1_4.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch1_5.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch2.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch2_1.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch2_2.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch2_3.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch2_4.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch2_5.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch3.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch3_1.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch3_2.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch3_3.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch3_4.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch3_5.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch4.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch4_1.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch4_2.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch4_3.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch4_4.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch4_5.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch6.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch6_1.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch6_2.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch6_3.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch6_4.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch6_5.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch7.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch7_1.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch7_2.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch7_3.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch7_4.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch7_5.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch8.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch8_1.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch8_2.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch8_3.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch8_4.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch8_5.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter1_1.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter2_1.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter3_1.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter4_1.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter6_1.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter7_1.ipynb
D Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter8_1.ipynb
A Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch11_1.ipynb
A Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch12_1.ipynb
A Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch13_1.ipynb
A Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch14_1.ipynb
A Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch3_1.ipynb
A Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch4_1.ipynb
A Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch5_1.ipynb
A Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch6_1.ipynb
A Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch7_1.ipynb
A Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch8_1.ipynb
A Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch9_1.ipynb
A Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/BiasingVoltage_1.png
A Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/InOutCrrnt_1.png
A Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/inAndOutCurrnt_1.png
_by_A._K._Theraja_B.L_Thereja/screenshots/chapter29example32_2.png?id=5218db5dbae9c0890958f57af323c8d296ac9b07){kind=link}
_by_A._K._Theraja_B.L_Thereja/screenshots/chapter29example33_2.png?id=5218db5dbae9c0890958f57af323c8d296ac9b07){kind=link}
_by_A._K._Theraja_B.L_Thereja/screenshots/chapter32example30_2.png?id=5218db5dbae9c0890958f57af323c8d296ac9b07){kind=link}
151 files changed, 43630 insertions, 45726 deletions
diff --git a/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter25_2.ipynb b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter25_2.ipynb new file mode 100644 index 00000000..884c7e96 --- /dev/null +++ b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter25_2.ipynb @@ -0,0 +1,173 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:0a9697b2451ba5bc5f24eb67c66ef466539d8d3c214c7c35bb64d3c339daf3f9" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 25: Elements of Electro-Mechanical Energy Conversion" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 25.1, Page Number:876" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#variable declaration\n", + "sod=15#stator-core outer diameter\n", + "sid=10.05#stator-core inner diameter\n", + "rod=10.00#rotor-core outer diameter\n", + "rid=5#rotor-core inner diameter\n", + "a=8#axial lenght of the machine\n", + "b=1.20\n", + "ur=1000\n", + "#calculations\n", + "vs=(3.14/4)*((sod*sod)-(sid*sid))*a#volume of stator-core\n", + "vr=(3.14/4)*((rod*rod)-(rid*rid))*a#volume of rotor-core\n", + "va=(3.14/4)*((sid*sid)-(rod*rod))*a#volume of air-gap in the machine\n", + "ed=(.5*b*b)/(4*3.14*math.pow(10,-7))\n", + "e=ed*va*math.pow(10,-6)\n", + "edm=(.5*b*b)/(4*3.14*math.pow(10,-7)*ur)\n", + "es=edm*vs*math.pow(10,-6)\n", + "er=edm*vr*math.pow(10,-6)\n", + "kr=(vs+vr)/vs\n", + "ke=(es+er)/e\n", + "ratio=kr/ke\n", + "eratio=e/(es+er)\n", + "\n", + "#result\n", + "print \"Energy stored in air gap= \",e,\" Joules\"\n", + "print \"Energy stored in stator-core= \",round(es,2),\" Joules\"\n", + "print \"Energy stored in rotor core= \",er,\" Joules\"\n", + "print \"Ratio of energy dtored in air-gap to that stored in the cores=\",round(eratio)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy stored in air gap= 3.609 Joules\n", + "Energy stored in stator-core= 0.45 Joules\n", + "Energy stored in rotor core= 0.27 Joules\n", + "Ratio of energy dtored in air-gap to that stored in the cores= 5.0\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 25.2, Page Number:877" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#variable declaration\n", + "n=800#turns\n", + "area=5*5#cross sectional area\n", + "i=1.25#amp\n", + "x=0.25#cm\n", + "l=0.402\n", + "#calculations\n", + "p=4*3.14*10**(-7)*area*10**(-4)/(0.5*10**(-2))\n", + "l=n**2*p\n", + "em=.5*i*i*l\n", + "W=-1*0.5*n**2*4*3.14*10**(-7)*area*10**(-4)*i**2/(0.5*10**(-2))**2\n", + "\n", + "#result\n", + "print \"a)i)coil inductance=\",l,\"H\"\n", + "print \" ii)field energy stored=\",em,\"J\"\n", + "print \"b)mechanical energy output=\",W,\"NW\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a)i)coil inductance= 0.40192 H\n", + " ii)field energy stored= 0.314 J\n", + "b)mechanical energy output= -62.8 NW\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 25.4, Page Number:882" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "lo=50#mH\n", + "xo=0.05#cm\n", + "r=0.5#ohm\n", + "x=0.075#cm\n", + "i2=3#A\n", + "x2=0.15#cm\n", + "\n", + "#calculation\n", + "l1=2*lo/(1+(x/xo))\n", + "lambda1=l1*i2*10**(-3)\n", + "W=0.5*l1*i2**2*10**(-3)\n", + "l2=2*lo/(1+(x2/xo))\n", + "lambda2=l2*i2*10**(-3)\n", + "w2=0.5*i2*(lambda1-lambda2)\n", + "\n", + "#result\n", + "print \"a)magnetic stored energy=\",W,\"J\"\n", + "print \"b)change in magnetic stored energy=\",w2,\"J\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a)magnetic stored energy= 0.18 J\n", + "b)change in magnetic stored energy= 0.0675 J\n" + ] + } + ], + "prompt_number": 19 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter26_2.ipynb b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter26_2.ipynb new file mode 100644 index 00000000..1af9bb80 --- /dev/null +++ b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter26_2.ipynb @@ -0,0 +1,1600 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:fbc29937443ef7eae8e50df5118b16ddcc8ed6efb4b30db1cb412240bf7eac02" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 26: D.C. Generators" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.3, Page Number:912" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "i=450#A\n", + "v=230#v\n", + "rs=50#ohm\n", + "ra=.03#ohm\n", + "\n", + "#calculations\n", + "ish=v/rs\n", + "ia=i+ish\n", + "va=ia*ra\n", + "E=v+va\n", + "\n", + "#result\n", + "print \"e.m.f. generated in the armature= \",E,\" V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "e.m.f. generated in the armature= 243.62 V\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.4, Page Number:913" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "i=50#A\n", + "v=500#v\n", + "rs=250#ohm\n", + "ra=.05#ohm\n", + "rseries=0.03#ohm\n", + "b=1#V\n", + "\n", + "#calculations\n", + "ish=v/rs\n", + "ia=i+ish\n", + "vs=ia*rseries\n", + "va=ia*ra\n", + "vb=ish*b\n", + "E=v+va+vs+vb\n", + "\n", + "#result\n", + "print \"generated voltage in the armature= \",E,\" V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "generated voltage in the armature= 506.16 V\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.5, Page Number:913" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "i=30#A\n", + "v=220#v\n", + "rs=200#ohm\n", + "ra=.05#ohm\n", + "rseries=0.30#ohm\n", + "b=1#V\n", + "\n", + "#calculations\n", + "vs=i*rseries\n", + "vshunt=v+vs\n", + "ish=vshunt/v\n", + "ia=i+ish\n", + "vb=b*2\n", + "E=v+vs+vb+(ia*ra)\n", + "\n", + "#result\n", + "print \"generated voltage in the armature= \",E,\" V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "generated voltage in the armature= 232.552045455 V\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.6, Page Number:913" + ] + }, + { + "cell_type": "code", + "collapsed": true, + "input": [ + "#variable declaration\n", + "v=230.0#v\n", + "i=150.0#A\n", + "rs=92.0#ohm\n", + "rseries=0.015#ohm\n", + "rd=0.03#ohm(divertor)\n", + "ra=0.032#ohm\n", + "\n", + "#calculations\n", + "ish=v/rs\n", + "ia=i+ish\n", + "sdr=(rd*rseries)/(rd+rseries)\n", + "tr=ra+sdr\n", + "vd=ia*tr\n", + "Eg=v+vd\n", + "tp=Eg*ia\n", + "pl=(ia*ia*ra)+(ia*ia*sdr)+(v*ish)+(v*i)\n", + "\n", + "#resuts\n", + "print \"i) Induced e.m.f.= \",Eg,\" V\"\n", + "print \"ii)Total power generated= \",tp,\" W\"\n", + "print \"iii)Distribution of the total power:\"\n", + "print \" power lost in armature= \", ia*ia*ra\n", + "print \"power lost in series field and divider= \", ia*ia*sdr\n", + "print \"power dissipated in shunt winding= \", v*ish\n", + "print \"power delivered to load= \", v*i\n", + "print \" ------------\"\n", + "print \"Total= \", pl" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i) Induced e.m.f.= 236.405 V\n", + "ii)Total power generated= 36051.7625 W\n", + "iii)Distribution of the total power:\n", + " power lost in armature= 744.2\n", + "power lost in series field and divider= 232.5625\n", + "power dissipated in shunt winding= 575.0\n", + "power delivered to load= 34500.0\n", + " ------------\n", + "Total= 36051.7625\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.7, Page Number:914" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=300000.0#w\n", + "v=600.0#v\n", + "sr=75.0#ohm\n", + "abr=0.03#ohm\n", + "cr=0.011#ohm\n", + "rseries=0.012#ohm\n", + "dr=0.036#ohm\n", + "\n", + "#calculatons\n", + "io=p/v#output current\n", + "ish=v/sr\n", + "ia=io+ish\n", + "sdr=(rseries*dr)/(rseries+dr)\n", + "tr=abr+cr+sdr\n", + "vd=ia*tr\n", + "va=v+vd\n", + "pg=va*ia\n", + "W=pg/1000\n", + "\n", + "#result\n", + "print \"Voltage generatedby the armature= \",va,\" V\"\n", + "print \"Power generated by the armature= \",W, \"kW\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage generatedby the armature= 625.4 V\n", + "Power generated by the armature= 317.7032 kW\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.8, Page Number:915" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "phi=7*math.pow(10,-3)\n", + "z=51*20\n", + "a=p=4\n", + "n=1500#r.p.m\n", + "\n", + "#calculations\n", + "Eg=(phi*z*n*p)/(a*60)\n", + "\n", + "#result\n", + "print \"Voltage generated= \",Eg,\" V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage generated= 178.5 V\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.9, Page Number:916" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=a=8\n", + "phi=0.05#Wb\n", + "n=1200#rpm\n", + "N=500#armature conductor\n", + "\n", + "#calculations\n", + "E=phi*(n/60)*(p/a)*N\n", + "\n", + "#result\n", + "print \"e.m.f generated= \",E,\" V\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "e.m.f generated= 500.0 V\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.10, Page Number:916" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=127#v\n", + "vt=120#v(terminal voltage)\n", + "r=15#ohms\n", + "i1=8.47#A\n", + "ra=0.02#ohms\n", + "fi=8#A\n", + "\n", + "#calculations\n", + "Eg=v+(i1*ra)\n", + "ia=(Eg-vt)/ra\n", + "il=ia-fi\n", + "\n", + "#result\n", + "print \"Load current \",il,\" A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Load current 350.47 A\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.11(a), Page Number:917" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=8\n", + "z=778\n", + "n=500\n", + "ra=0.24\n", + "rl=12.5\n", + "r=250\n", + "v=250\n", + "a=2\n", + "#calculations\n", + "il=v/rl\n", + "si=v/r\n", + "ai=il+si\n", + "emf=v+(ai*ra)\n", + "phi=(emf*60*a)/(p*z*n)\n", + "\n", + "#result\n", + "print \"armature current= \",ai,\" A\"\n", + "print \"induced e.m.f.= \",emf,\" V\"\n", + "print \"flux per pole= \",round(phi*1000,2),\" mWb\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "armature current= 21.0 A\n", + "induced e.m.f.= 255.04 V\n", + "flux per pole= 9.83 mWb\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.11(b), Page Number:916" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=a=4\n", + "P=5000.0#w\n", + "P2=2500.0#W\n", + "v=250.0#v\n", + "ra=0.2#ohm\n", + "r=250.0#ohm\n", + "z=120\n", + "N=1000#rpm\n", + "\n", + "#calculations\n", + "gc=P/v\n", + "li=P2/v\n", + "ti=gc+li\n", + "fc=1\n", + "ai=ti+fc\n", + "ard=ai*ra\n", + "emf=v+ard+2\n", + "phi=(emf*60*a)/(p*z*N)\n", + "ac_perparralelpath=ai/p\n", + "\n", + "#result\n", + "print \"Flux per pole= \",phi*1000,\" mWb\"\n", + "print \"Armature current per parallel path= \",ac_perparralelpath,\" A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Flux per pole= 129.1 mWb\n", + "Armature current per parallel path= 7.75 A\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.12, Page Number:918" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "i=200.0#A\n", + "v=125.0#V\n", + "n1=1000#rpm\n", + "n2=800#rpm\n", + "ra=0.04#ohm\n", + "bd=2.0#V(brush drop)\n", + "\n", + "#calculations\n", + "R=v/i\n", + "E1=v+(i*ra)+bd\n", + "E2=(E1*n2)/n1\n", + "il=(E2-bd)/0.675\n", + "\n", + "#result\n", + "print \"Load current when speed drops to 800 r.p.m.= \",round(il,2),\" A\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Load current when speed drops to 800 r.p.m.= 157.04 A\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.13, Page Number:918" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "p=4\n", + "n=900 #rpm\n", + "V=220#V\n", + "E=240#V\n", + "ra=0.2#ohm\n", + "phi=10#mWb\n", + "N=8\n", + "\n", + "#calculations\n", + "ia=(E-V)/ra\n", + "Z=(E*600*2)/(phi*math.pow(10,-3)*n*p)\n", + "#since there ae 8 turns in a coil,it means there are 16 active conductor\n", + "number_of_coils=Z/16\n", + "\n", + "#result\n", + "print \"armature current= \",ia,\" A\"\n", + "print \"number of coils= \",number_of_coils" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "armature current= 100.0 A\n", + "number of coils= 500.0\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.14, Page Number:919" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "V=120.0#V\n", + "ra=0.06#ohm\n", + "rs=25#ohm\n", + "rsw=0.04#ohm(series winding)\n", + "il=100.0#A\n", + "#i)Long shunt\n", + "ish=V/rs\n", + "ia=il+ish\n", + "vd=ia*rsw\n", + "vda=ia*ra\n", + "E=V+vd+vda\n", + "\n", + "print \"Induced e.m.f. when the machine is connected to long shunt= \",E,\" V\"\n", + "print \"Armature current when the machine is connected to long shunt=\",ia,\" A\"\n", + "\n", + "#i)Short shunt\n", + "vds=il*rsw\n", + "vs=V+vds\n", + "ish=vs/rs\n", + "ia=il+ish\n", + "vd=ia*rsw\n", + "vda=ia*ra\n", + "E=V+vd+vda\n", + "\n", + "print \"Induced e.m.f. when the machine is connected to short shunt= \",E,\" V\"\n", + "print \"Armature current when the machine is connected to short shunt=\",ia,\" A\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Induced e.m.f. when the machine is connected to long shunt= 130.48 V\n", + "Armature current when the machine is connected to long shunt= 104.8 A\n", + "Induced e.m.f. when the machine is connected to short shunt= 130.496 V\n", + "Armature current when the machine is connected to short shunt= 104.96 A\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.15, Page Number:920" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=25000.0#W\n", + "V=500.0#V\n", + "ra=0.03#ohm\n", + "rs=200.0#ohm\n", + "rseries=0.04#ohm\n", + "vb=1.0#V\n", + "n=1200#rpm\n", + "phi=0.02#Wb\n", + "\n", + "#calculations\n", + "i=p/V\n", + "ish=V/rs\n", + "ia=i+ish\n", + "p=4\n", + "vds=ia*rseries\n", + "vda=ia*ra\n", + "vdb=vb*2\n", + "E=V+vds+vda+vdb\n", + "Z=(E*60*4)/(phi*n*p)\n", + "\n", + "#result\n", + "print \"The e.m.f. generated= \",E,\" V\"\n", + "print \"The number of conductors=\",Z" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The e.m.f. generated= 505.675 V\n", + "The number of conductors= 1264.1875\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.16, Page Number:920" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=4\n", + "n=750#rpm\n", + "e=240.0#V\n", + "z=792\n", + "phi=0.0145#Wb\n", + "\n", + "#calculations\n", + "phi_working=(e*60*2)/(n*z*p)\n", + "lambda_=phi/phi_working\n", + "\n", + "#results\n", + "print \"Leakage coefficient= \",round(lambda_,1)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Leakage coefficient= 1.2\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.17, Page Number:920" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=a=4\n", + "phi=0.07#Wb\n", + "t=220\n", + "rt=0.004#ohm\n", + "n=900#rpm\n", + "ia=50.0#A\n", + "\n", + "#calculations\n", + "z=2*t\n", + "E=(phi*z*n*p)/(60*a)\n", + "rtotal=t*rt\n", + "r_eachpath=rtotal/p\n", + "ra=r_eachpath/a\n", + "vda=ia*ra\n", + "V=E-vda\n", + "\n", + "#result\n", + "print \"Terminal Voltage= \",V, \" V\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Terminal Voltage= 459.25 V\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.18, Page Number:920" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=a=4\n", + "phi=0.07#Wb\n", + "t=220\n", + "rturn=0.004#ohm\n", + "rs=100.0#ohm\n", + "rsc=0.02#ohm\n", + "n=900#rpm\n", + "ia=50.0#A\n", + "\n", + "#calculations\n", + "z=2*t\n", + "E=(phi*z*n*p)/(60*a)\n", + "ra=0.055#ohm\n", + "ra=ra+rsc\n", + "va=ia*ra\n", + "v=E-va\n", + "ish=v/rs\n", + "i=ia-ish\n", + "output=v*i\n", + "\n", + "#result\n", + "print \"Output= \",round(output/1000,3),\" kW\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Output= 20.813 kW\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.19, Page Number:921" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n1=1200#rpm\n", + "ia=200#A\n", + "v=125#V\n", + "n2=1000#rpm\n", + "ra=0.04#ohm\n", + "vb=2#V\n", + "\n", + "#calculations\n", + "E1=v+vb+(ia*ra)\n", + "E2=E1*n2/n1*0.8\n", + "\n", + "#results\n", + "print \"Generated e.m.f. when field current is reduced to 80%=\",E2,\" V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Generated e.m.f. when field current is reduced to 80%= 90.0 V\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.20(a), Page Number:921" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=4\n", + "rs=100.0#ohm\n", + "ra=1.0#ohm\n", + "z=378\n", + "phi=0.02#Wb\n", + "rl=10.0#ohm\n", + "n=1000#rpm\n", + "a=2\n", + "\n", + "#calculations\n", + "E=(phi*z*n*p)/(60*a)\n", + "V=(100.0/111.0)*E\n", + "il=V/rl\n", + "P=il*V\n", + "\n", + "#result\n", + "print \"Power absorbed by the load is= \",P,\" W\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Power absorbed by the load is= 5154.12710007 W\n" + ] + } + ], + "prompt_number": 50 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.20(b), Page Number:921" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=a=4\n", + "z=300\n", + "phi=0.1#Wb\n", + "n=1000#rpm\n", + "ra=0.2#rpm\n", + "rf=125#ohm\n", + "il=90#A\n", + "\n", + "#calculations\n", + "E=(phi*z*n*p)/(60*a)\n", + "ifield=E/rf\n", + "ia=ifield+il\n", + "V=E-(ia*ra)\n", + "\n", + "#result\n", + "print \"Terminal voltage= \",V,\" V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Terminal voltage= 481.2 V\n" + ] + } + ], + "prompt_number": 51 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.21(a), Page Number:922" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=6\n", + "n=1200#rpm\n", + "e=250.0#V\n", + "d=350.0#mm\n", + "air_gap=3.0#mm\n", + "al=260.0#mm\n", + "fringing=0.8\n", + "coils=96\n", + "t=3\n", + "\n", + "#calculations\n", + "z=t*coils*2\n", + "a=p*2\n", + "phi=(e*60*a)/(n*z*p)\n", + "di=d+air_gap\n", + "pole_arc=(3.14*di*fringing)/6\n", + "B=phi/(pole_arc*0.000001*al)\n", + "\n", + "#result\n", + "print \"flux per pole= \",phi,\" Wb\"\n", + "print \"effective pole arc lenght= \",pole_arc*0.001,\" m\"\n", + "print \"flux density= \",B,\" T\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "flux per pole= 0.0434027777778 Wb\n", + "effective pole arc lenght= 0.147789333333 m\n", + "flux density= 1.12953862717 T\n" + ] + } + ], + "prompt_number": 57 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.21(b), Page Number:922" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "p=a=4\n", + "z=1200\n", + "e=250.0#v\n", + "n=500#rpm\n", + "b=35.0#cm\n", + "ratio=0.7\n", + "lpole=20.0#cm\n", + "\n", + "#calculations\n", + "pole_pitch=(b*3.14)/p\n", + "polearc=ratio*pole_pitch\n", + "pole_area=polearc*lpole\n", + "phi=(e*60*a)/(n*z*p)\n", + "mean_flux=phi/(pole_area*math.pow(10,-4))\n", + " \n", + "#result\n", + "print \"Mean flux density= \",mean_flux,\" Wb/m2\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mean flux density= 0.649941505265 Wb/m2\n" + ] + } + ], + "prompt_number": 67 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.21(d), Page Number:923" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "i=200.0#A\n", + "v=100.0#V\n", + "ra=0.04#ohm\n", + "rseries=0.03#ohm\n", + "rs=60.0#ohm\n", + "\n", + "#calculations\n", + "va=v+(i*rseries)\n", + "ish=va/rs\n", + "ia=i+ish\n", + "e=va+(ia*ra)\n", + "\n", + "#long shunt\n", + "ishunt=v/rs\n", + "vd=ia*(ra+rseries)\n", + "e2=v+vd\n", + "\n", + "#result\n", + "print \"emf generated(short shunt)\",e,\" V\"\n", + "print \"emf generated(long shunt)\",e2,\" V\"\n", + "\n", + "\n", + "#result\n", + "print " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "emf generated(short shunt) 114.070666667 V\n", + "emf generated(long shunt) 114.123666667 V\n", + "\n" + ] + } + ], + "prompt_number": 73 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.22, Page Number:923" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n=1000#rpm\n", + "w=20000.0#W\n", + "v=220.0#v\n", + "ra=0.04#ohm\n", + "rs=110.0#ohm\n", + "rseries=0.05#ohm\n", + "efficiency=.85\n", + "\n", + "#calculations\n", + "il=w/v\n", + "i_f=v/rs\n", + "ia=il+i_f\n", + "ip=w/efficiency#input power\n", + "total_loss=ip-w\n", + "copper_loss=(ia*ia*(ra+rseries))+(i_f*i_f*rs)\n", + "ironloss=total_loss-copper_loss\n", + "omega=2*3.14*n/60\n", + "T=ip/omega\n", + "\n", + "#omega\n", + "print \"Copper loss= \",copper_loss,\" W\"\n", + "print \"Iron and friction loss= \",ironloss,\" W\"\n", + "print \"Torque developed by the prime mover= \",T,\"Nw-m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Copper loss= 1216.88892562 W\n", + "Iron and friction loss= 2312.52283909 W\n", + "Torque developed by the prime mover= 224.803297115 Nw-m\n" + ] + } + ], + "prompt_number": 75 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.23, Page Number:928" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declartaion\n", + "power=10000.0#W\n", + "v=250.0#V\n", + "p=a=6\n", + "n=1000.0#rpm\n", + "z=534\n", + "cu_loss=0.64*1000#W\n", + "vbd=1.0#V\n", + "\n", + "#calculations\n", + "ia=power/v\n", + "ra=cu_loss/(ia*ia)\n", + "E=v+(ia*ra)+vbd\n", + "phi=(E*60*a)/(n*z*p)\n", + "\n", + "#result\n", + "print \"flux per pole= \",phi*1000,\" mWb\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "flux per pole= 30.0 mWb\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.24(a), Page Number:928" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "i=195#A\n", + "pd=250#V\n", + "ra=0.02#ohm\n", + "rsh=50#ohm\n", + "p=250#W\n", + "strayloss=950#W\n", + "#calculations\n", + "ish=pd/rsh\n", + "ia=i+ish\n", + "vda=ia*ra\n", + "E=pd+vda\n", + "cu_loss=(ia*ia*ra)+(pd*ish)\n", + "output_prime=(pd*i)+strayloss+cu_loss\n", + "power_a=output_prime-strayloss\n", + "neu_m=(power_a/output_prime)\n", + "neu_e=(pd*i)/((pd*i)+cu_loss)\n", + "neu_c=(pd*i)/output_prime\n", + "\n", + "#result\n", + "print \"a)e.m.f. generated= \",E,\" V\"\n", + "print \" b)Cu losses= \",cu_loss,\" W\"\n", + "print \" c)output of prime mover= \",output_prime,\" W\"\n", + "print \" d)mechanical efficiency= \",neu_m*100,\" %\"\n", + "print \" electrical efficiency= \",neu_e*100,\" %\"\n", + "print \" commercial efficiency= \",neu_c*100,\" %\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a)e.m.f. generated= 254.0 V\n", + " b)Cu losses= 2050.0 W\n", + " c)output of prime mover= 51750.0 W\n", + " d)mechanical efficiency= 98.1642512077 %\n", + " electrical efficiency= 95.9645669291 %\n", + " commercial efficiency= 94.2028985507 %\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.24(b), Page Number:929" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=500.0#V\n", + "i=5.0#A\n", + "ra=0.15#ohm\n", + "rf=200.0#ohm\n", + "il=40.0#A\n", + "\n", + "#calculations\n", + "output=v*il\n", + "total_loss=(v*i*0.5)+((il+i*0.5)*(il+i*0.5)*ra)+(v*i*0.5)\n", + "efficiency=output/(output+total_loss)\n", + "\n", + "#result\n", + "print \"Efficiency= \",efficiency*100,\" %\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Efficiency= 87.8312542029 %\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.25, Page Number:929" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "i=196#A\n", + "v=220#V\n", + "stray_loss=720#W\n", + "rsh=55#ohm\n", + "e=0.88\n", + "\n", + "#calculations\n", + "output=v*i\n", + "inpute=output/e\n", + "total_loss=inpute-output\n", + "ish=v/rsh\n", + "ia=i+ish\n", + "cu_loss=v*ish\n", + "constant_loss=cu_loss+stray_loss\n", + "culoss_a=total_loss-constant_loss\n", + "ra=culoss_a/(ia*ia)\n", + "I=math.sqrt(constant_loss/ra)\n", + "\n", + "#result\n", + "print \"Load curent corresponding to maximum efficiency\",I,\" A\" " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Load curent corresponding to maximum efficiency 122.283568103 A\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.26, Page Number:929" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n=1000#rpm\n", + "p=22*1000#w\n", + "v=220#V\n", + "ra=0.05#ohm\n", + "rsh=110#ohm\n", + "rseries=0.06#ohm\n", + "efficiency=.88\n", + "\n", + "#calculations\n", + "ish=v/rsh\n", + "I=p/v\n", + "ia=ish+I\n", + "vdseries=ia*rseries\n", + "cu_loss=(ia*ia*ra)+(ia*ia*rseries)+(rsh*ish*ish)\n", + "total_loss=(p/efficiency)-p\n", + "strayloss=total_loss-cu_loss\n", + "T=(p/efficiency*60)/(2*3.14*n)\n", + "\n", + "#result\n", + "print \"a)cu losses= \",cu_loss,\" W\"\n", + "print \"b)iron and friction loss= \",strayloss,\" W\"\n", + "print \"c)Torque exerted by the prime mover= \",T,\" N-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a)cu losses= 1584.44 W\n", + "b)iron and friction loss= 1415.56 W\n", + "c)Torque exerted by the prime mover= 238.853503185 N-m\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.27, Page Number:930" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=4\n", + "i=20#A\n", + "r=10#ohm\n", + "ra=0.5#ohm\n", + "rsh=50#ohm\n", + "vdb=1#V(voltage drop per brush)\n", + "\n", + "#calculations\n", + "v=i*r\n", + "ish=v/rsh\n", + "ia=i+ish\n", + "E=v+(ia*ra)+(2*vdb)\n", + "totalpower=E*ia\n", + "output=v*i\n", + "efficiency=output/totalpower\n", + "\n", + "#result\n", + "print \"induced e.m.f.= \",E,\" V\"\n", + "print \"efficiency= \",efficiency*100,\" %\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "induced e.m.f.= 214.0 V\n", + "efficiency= 77.8816199377 %\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.28, Page Number:930" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=240#V\n", + "i=100#A\n", + "ra=0.1#ohm\n", + "rseries=0.02#ohm\n", + "ri=0.025#ohm\n", + "rsh=100#ohm\n", + "ironloss=1000#W\n", + "frictionloss=500#W\n", + "\n", + "#calculations\n", + "output=v*i\n", + "totalra=ra+rseries+ri\n", + "ish=v/rsh\n", + "ia=i+ish\n", + "copperloss=ia*ia*totalra\n", + "shculoss=ish*v\n", + "total_loss=copperloss+ironloss+frictionloss+shculoss\n", + "efficiency=output/(output+total_loss)\n", + "\n", + "#result\n", + "print \"F.L. efficiency of the machine= \",efficiency*100,\" %\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "F.L. efficiency of the machine= 87.3089843128 %\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.31, Page Number:931" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "output=10.0*1000#W\n", + "v=240.0#V\n", + "ra=0.6#ohm\n", + "rsh=160.0#ohm\n", + "mechcoreloss=500.0#W\n", + "culoss=360.0#W\n", + "\n", + "#calculations\n", + "ish=v/rsh\n", + "i=output/v\n", + "ia=ish+i\n", + "culossa=ia*ia*ra\n", + "totalloss=culoss+mechcoreloss+culossa\n", + "inputp=output+totalloss\n", + "efficiency=output/inputp\n", + "\n", + "#result\n", + "print \"Power required= \",inputp*0.001,\" kW\"\n", + "print \"efficinecy= \",efficiency*100,\" %\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Power required= 11.9780166667 kW\n", + "efficinecy= 83.486275552 %\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.32, Page Number:932" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=110*1000#W\n", + "v=220#V\n", + "ra=0.01#ohm\n", + "rse=0.002#ohm\n", + "rsh=110#ohm\n", + "\n", + "#calculations\n", + "il=p/v\n", + "ish=v/rsh\n", + "ia=il+ish\n", + "E=v+ia*(ra+rse)\n", + "\n", + "#result\n", + "print \"induced emf= \",E,\" V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "induced emf= 226.024 V\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 26.33 Page Number:932" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=4\n", + "E=216.0#V\n", + "n=600.0#rpm\n", + "slots=144\n", + "con=6\n", + "n2=500.0#rpm\n", + "\n", + "#calculations\n", + "z=con*slots\n", + "a=p\n", + "phi=(E*60*a)/(n*z*p)\n", + "a=2\n", + "armatureE=(phi*z*n2*p)/(60*a)\n", + "\n", + "#result\n", + "print \"the armature emf= \",armatureE,\" V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the armature emf= 360.0 V\n" + ] + } + ], + "prompt_number": 34 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter27_2.ipynb b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter27_2.ipynb new file mode 100644 index 00000000..638b15f1 --- /dev/null +++ b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter27_2.ipynb @@ -0,0 +1,730 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:02f2208937b2d82cdc7150d6d9062a1310b3e2fcf2346b8c885c3f6fe2fe5405" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 27: Armature Reaction and Commutation" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 27.1, Page Number:943" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=4\n", + "z=722\n", + "ia=100.0#A\n", + "theta_m=8.0#degrees\n", + "\n", + "#calculatons\n", + "i=ia/2\n", + "atd_perpole=z*i*theta_m/360\n", + "atc_perpole=z*i*((1/(2.0*p))-(theta_m/360.0))\n", + "\n", + "#result\n", + "print \"armature demagnetization=\",atd_perpole\n", + "print \"cross-magnetization=\",atc_perpole" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "armature demagnetization= 802.222222222\n", + "cross-magnetization= 3710.27777778\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 27.2, Page Number:943" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=8\n", + "z=1280\n", + "v=500#V\n", + "ia=200.0#A\n", + "commuter=160\n", + "advanced_segments=4\n", + "\n", + "#calculatons\n", + "i=ia/8\n", + "theta_m=advanced_segments*360/commuter\n", + "atd_perpole=z*i*theta_m/360\n", + "atc_perpole=z*i*((1/(2.0*p))-(theta_m/360.0))\n", + "\n", + "#result\n", + "print \"armature demagnetization=\",atd_perpole\n", + "print \"cross-magnetization=\",atc_perpole" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "armature demagnetization= 800.0\n", + "cross-magnetization= 1200.0\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 27.3(a), Page Number:943" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=4\n", + "z=880\n", + "ia=120.0#A\n", + "theta_m=3.0#degrees\n", + "n=1100#tturns/pole\n", + "#calculatons\n", + "i=ia/2\n", + "atd_perpole=z*i*theta_m/360\n", + "atc_perpole=z*i*((1/(2.0*p))-(theta_m/360.0))\n", + "iadditional=(atd_perpole/n)\n", + "\n", + "\n", + "#result\n", + "print \"a)armature demagnetization=\",atd_perpole,\"AT\"\n", + "print \"b)cross-magnetization=\",atc_perpole,\"AT\"\n", + "print \"c)additional field current=\",iadditional,\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a)armature demagnetization= 440.0 AT\n", + "b)cross-magnetization= 6160.0 AT\n", + "c)additional field current= 0.4 A\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 27.3(b), Page Number:943" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=4\n", + "z=480\n", + "ia=150.0#A\n", + "theta_m=10.0*2#degrees\n", + "\n", + "#calculatons\n", + "i=ia/4\n", + "total=(z*i)/(2*p)\n", + "atd_perpole=total*(2*theta_m/180)\n", + "atc_perpole=total*(1-(2*theta_m/180))\n", + "\n", + "#result\n", + "print \"armature demagnetization=\",atd_perpole\n", + "print \"cross-magnetization=\",atc_perpole" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "armature demagnetization= 500.0\n", + "cross-magnetization= 1750.0\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 27.4, Page Number:944" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "z=492\n", + "theta_m=10.0\n", + "ia=143.0+10.0\n", + "\n", + "#calculations\n", + "i1=ia/2#wave wound\n", + "i2=ia/4#lap wound\n", + "atd_perpole1=z*i1*theta_m/360#wave wound\n", + "extra_shunt1=atd_perpole1/theta_m\n", + "atd_perpole2=z*i2*(theta_m/360.0)#lap wound\n", + "extra_shunt2=atd_perpole2/theta_m\n", + "#result\n", + "print \"wave wound:\"\n", + "print \"demagnetization per pole=\",atd_perpole1,\"AT\"\n", + "print \"extra shunt field turns=\",int(extra_shunt1)\n", + "print \"lap wound:\"\n", + "print \"demagnetization per pole=\",atd_perpole2,\"AT\"\n", + "print \"extra shunt field turns=\",int(extra_shunt2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "wave wound:\n", + "demagnetization per pole= 1045.5 AT\n", + "extra shunt field turns= 104\n", + "lap wound:\n", + "demagnetization per pole= 522.75 AT\n", + "extra shunt field turns= 52\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 27.5, Page Number:944" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "pole=4\n", + "p=50*1000.0#W\n", + "v=250.0#V\n", + "z=400\n", + "commuter=4\n", + "rsh=50.0#ohm\n", + "a=2\n", + "\n", + "#calculations\n", + "i=p/v\n", + "ish=v/rsh\n", + "ia=i+ish\n", + "i=ia/2\n", + "segments=z/a\n", + "theta=pole*360.0/segments\n", + "atd=z*i*(theta/360)\n", + "extra=atd/ish\n", + "\n", + "#result\n", + "print \"demagnetisation=\",atd,\"AT\"\n", + "print \"extra shunt turns/poles\",extra" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "demagnetisation= 820.0 AT\n", + "extra shunt turns/poles 164.0\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 27.6, Page Number:943" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "z=500\n", + "ia=200.0#A\n", + "p=6\n", + "theta=10.0#degrees\n", + "lambda_=1.3\n", + "\n", + "#calculations\n", + "i=ia/2\n", + "atc=((1/(2.0*p))-(theta/360.0))*z*i\n", + "atd=z*i*theta/360\n", + "extra=lambda_*atd/ia\n", + "\n", + "#result\n", + "print \"i)cross magnetization ampere-turns=\",atc\n", + "print \"ii)back ampere-turns\",atd\n", + "print \"iii)series turns required to balance the demagnetising ampere turns\",int(extra)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)cross magnetization ampere-turns= 2777.77777778\n", + "ii)back ampere-turns 1388.88888889\n", + "iii)series turns required to balance the demagnetising ampere turns 9\n" + ] + } + ], + "prompt_number": 45 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 27.7, Page Number:945" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=22.38#kW\n", + "v=440.0#V\n", + "pole=4\n", + "z=840\n", + "commutator=140\n", + "efficiency=0.88\n", + "ish=1.8#A\n", + "back=1.5\n", + "\n", + "#calculations\n", + "motor_input=p*1000.0/efficiency\n", + "input_i=motor_input/v\n", + "ia=input_i-ish\n", + "i=ia/2.0\n", + "theta=back*360/commutator\n", + "atd=z*i*(theta/360.0)\n", + "atc=((1/(2.0*pole))-(theta/360.0))*z*i\n", + "#result\n", + "print \"armature demagnetization amp-turns/pole=\",atd\n", + "print \"distorting amp-turns/pole=\",atc" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "armature demagnetization amp-turns/pole= 251.998140496\n", + "distorting amp-turns/pole= 2687.98016529\n" + ] + } + ], + "prompt_number": 59 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 27.8, Page Number:945" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=400#V\n", + "ia=1000#A\n", + "p=10\n", + "z=860\n", + "per=0.7\n", + "\n", + "#calculations\n", + "i=ia/p\n", + "at=per/p*z*(i/2)\n", + "\n", + "#result\n", + "print \"AT/pole for compensation winding=\",at" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "AT/pole for compensation winding= 3010.0\n" + ] + } + ], + "prompt_number": 62 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 27.9, Page Number:948" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n=800.0#rpm\n", + "segment=123\n", + "wb=3\n", + "#calculations\n", + "v=n/60.0*segment\n", + "commutation=wb/v\n", + "\n", + "#result\n", + "print \"commutation time=\",commutation*1000,\"millisecond\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "commutation time= 1.82926829268 millisecond\n" + ] + } + ], + "prompt_number": 64 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 27.10, Page Number:948" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=4\n", + "n=1500#rpm\n", + "d=30#cm\n", + "ia=150#A\n", + "wb=1.25#cm\n", + "L=0.07*0.001#H\n", + "\n", + "#calculation\n", + "i=ia/2\n", + "v=3.14*d*(n/60)\n", + "tc=wb/v\n", + "E=L*2*i/tc\n", + "\n", + "#result\n", + "print \"average emf=\",E,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "average emf= 19.782 V\n" + ] + } + ], + "prompt_number": 65 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 27.11, Page Number:949" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "segments=55\n", + "n=900\n", + "wb=1.74\n", + "L=153*math.pow(10,-6)#H\n", + "i=27#A\n", + "\n", + "#calculations\n", + "v=segments*n/60\n", + "Tc=wb/v\n", + "E=L*2*i/Tc\n", + "\n", + "#result\n", + "print \"average emf=\",E,\"V\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "average emf= 3.91732758621 V\n" + ] + } + ], + "prompt_number": 67 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 27.12, Page Number:949" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=4\n", + "n=1500.0#rpm\n", + "ia=150.0#A\n", + "z=64\n", + "wb=1.2\n", + "L=0.05#mH\n", + "\n", + "#calculations\n", + "L=L*0.001\n", + "v=n/60*z\n", + "tc=wb/v\n", + "i=ia/p\n", + "#i.linear\n", + "E1=L*2*i/tc\n", + "#ii.sinusoidal\n", + "E2=1.11*E1\n", + "\n", + "#result\n", + "print \"Linear commutation,E=\",E1,\"V\"\n", + "print \"Sinosoidal commutation,E=\",E2,\"V\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Linear commutation,E= 5.0 V\n", + "Sinosoidal commutation,E= 5.55 V\n" + ] + } + ], + "prompt_number": 68 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 27.13, Page Number:951" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "p=6\n", + "B=0.5#Wb/m2\n", + "Ig=4.0#mm\n", + "ia=500.0#A\n", + "z=540\n", + "\n", + "#calculations\n", + "arm_mmf=z*(ia/p)/(2*p)\n", + "compole=int(B*Ig*0.001/(4*3.14*math.pow(10,-7)))\n", + "mag=0.1*compole\n", + "total_compole=int(compole+mag)\n", + "total_mmf=arm_mmf+total_compole\n", + "Ncp=total_mmf/ia\n", + "\n", + "#result\n", + "print \"Number of turns on each commutating pole=\",int(Ncp)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of turns on each commutating pole= 11\n" + ] + } + ], + "prompt_number": 89 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 27.14, Page Number:957" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p1=100.0#kW\n", + "V1=250#V\n", + "p2=300.0#kW\n", + "V2=250#V\n", + "i1=200#A\n", + "i2=500#A\n", + "il=600#A\n", + "\n", + "#calculations\n", + "delI1=p1/(p1+p2)*il\n", + "delI2=p2/(p1+p2)*il\n", + "\n", + "#result\n", + "print \"Current supplied by generator 1 with additional load=\",delI1,\"A\"\n", + "print \"Current supplied by generator 2 with additional load=\",delI2,\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current supplied by generator 1 with additional load= 150.0 A\n", + "Current supplied by generator 2 with additional load= 450.0 A\n" + ] + } + ], + "prompt_number": 92 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 27.23, Page Number:963" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "va=400#V\n", + "ra=0.25#ohm\n", + "vb=410#V\n", + "rb=0.4#ohm\n", + "V=390#V\n", + "\n", + "#calculations\n", + "loada=(va-V)/ra\n", + "loadb=(vb-V)/rb\n", + "pa=loada*V\n", + "pb=loadb*V\n", + "net_v=vb-va\n", + "total_r=ra+rb\n", + "i=net_v/total_r\n", + "terminal_v=va+(i*ra)\n", + "power_AtoB=terminal_v*i\n", + "\n", + "#result\n", + "print \"Current=\",i,\"A\"\n", + "print \"Voltage=\",terminal_v,\"V\"\n", + "print \"Power=\",power_AtoB,\"W\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current= 15.3846153846 A\n", + "Voltage= 403.846153846 V\n", + "Power= 6213.01775148 W\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter28_2.ipynb b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter28_2.ipynb new file mode 100644 index 00000000..447ef8ab --- /dev/null +++ b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter28_2.ipynb @@ -0,0 +1,388 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:6743417a1c79c6197a7cd49755318e10828c09b3cb248c5af8d5364367840700" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 28: Generator Characteristics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 28.13, Page Number:984" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=220#V\n", + "#emf increases by 1 V for every increase of 6 A\n", + "ra=0.02#ohm\n", + "i=96#A\n", + "\n", + "#calculations\n", + "voltageincrease=i/6\n", + "vd=i*ra\n", + "voltage_rise=voltageincrease-vd\n", + "vconsumer=v+voltage_rise\n", + "power_supplied=voltage_rise*i\n", + "\n", + "#result\n", + "print \"voltage supplied ot consumer= \",vconsumer,\" V\"\n", + "print \"power supplied by the booster itself= \",power_supplied/1000,\" kW\" " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "voltage supplied ot consumer= 234.08 V\n", + "power supplied by the booster itself= 1.35168 kW\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 28.14, Page Number:985" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=50.0#V\n", + "i=200.0#A\n", + "r=0.3#ohm\n", + "i1=200.0#A\n", + "i2=50.0#A\n", + "\n", + "#calculations\n", + "vd=i*r\n", + "voltage_decrease=v-vd\n", + "feeder_drop=v*r\n", + "booster_voltage=v*v/i1\n", + "voltage_net=feeder_drop-booster_voltage\n", + "\n", + "#result\n", + "print \"Net decrease in voltage= \",voltage_net,\" V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Net decrease in voltage= 2.5 V\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 28.15, Page Number:986" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "inl=5.0#A\n", + "v=440.0#V\n", + "il=6.0#A\n", + "i_full=200.0#A(full load)\n", + "turns=1600\n", + "\n", + "#calcuations\n", + "shunt_turns1=turns*inl\n", + "shunt_turns2=turns*il\n", + "increase=shunt_turns2-shunt_turns1\n", + "n=increase/i_full#number of series turns required\n", + "\n", + "#result\n", + "print \"Number of series turns required= \",n,\" tunrs/pole\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of series turns required= 8.0 tunrs/pole\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 28.16, Page Number:987" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n=1000#turns/pole\n", + "series_winding=4#turns/pole\n", + "r=0.05#ohm\n", + "increase_i=0.2#A\n", + "ia=80#A\n", + "\n", + "#calculations\n", + "additional_at=n*increase_i\n", + "current_required=additional_at/series_winding\n", + "R=(current_required*r)/(ia-current_required)\n", + "\n", + "#result\n", + "print \"Divertor resistance= \",R,\" ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Divertor resistance= 0.0833333333333 ohm\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 28.17, Page Number:987" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=220.0#V\n", + "i=100.0#A\n", + "ra=0.1#ohm\n", + "rsh=50.0#ohm\n", + "rse=0.06#ohm\n", + "divertor=0.14#ohm\n", + "\n", + "#calculations\n", + "#short shunt\n", + "vd=i*rse\n", + "ish=v/rsh\n", + "ia=i+ish\n", + "armature_drop=ia*ra\n", + "E=v+vd+armature_drop\n", + "#long shunt\n", + "vd=ia*(ra+rse)\n", + "print vd\n", + "E2=v+vd\n", + "current_divertor=(ia*divertor)/(divertor+rse)\n", + "change=(current_divertor/ia)*100\n", + "\n", + "#result\n", + "print \"a)emf induced using short shunt= \",E\n", + "print \"b)emf induced using long shunt= \",E2\n", + "print \"c)series amp-turns are reduced to \",change,\" %\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "16.704\n", + "a)emf induced using short shunt= 236.44\n", + "b)emf induced using long shunt= 236.704\n", + "c)series amp-turns are reduced to 70.0 %\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 28.18, Page Number:988" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=250*1000#W\n", + "v=240#V\n", + "v2=220#V\n", + "i=7#A\n", + "inl=12#A\n", + "shunt=650#turns/pole\n", + "series=4#turns/pole\n", + "rse=0.006#ohm\n", + "\n", + "#calculations\n", + "i_fulload=p/v\n", + "shunt_increase=shunt*(inl-i)\n", + "ise=shunt_increase/series\n", + "i_d=i_fulload-ise\n", + "Rd=(ise*rse)/i_d\n", + "\n", + "#results\n", + "print \"resistance of the series amp-turns at no-load\",Rd,\"ohm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resistance of the series amp-turns at no-load 0.0212751091703 ohm\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 28.19, Page Number:988" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "p=60.0*1000#W\n", + "n=1600.0#turns/pole\n", + "inl=1.25#A\n", + "vnl=125#V\n", + "il=1.75#A\n", + "vl=150.0#V\n", + "\n", + "#calculations\n", + "extra_excitation=n*(il-inl)\n", + "ise=p/vl\n", + "series=extra_excitation/ise\n", + "ise2=extra_excitation/3\n", + "i_d=ise-ise2\n", + "rd=(ise2*0.02)/i_d\n", + "reg=(vnl-vl)*100/vl\n", + "\n", + "#result\n", + "print \"i)minimum number of series turns/pole= \",series\n", + "print \"ii)divertor resistance= \",rd\n", + "print \"iii)voltage regulation= \",reg,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)minimum number of series turns/pole= 2.0\n", + "ii)divertor resistance= 0.04\n", + "iii)voltage regulation= -16.6666666667 %\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 28.20, Page Number:989" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=50.0#v\n", + "i=200.0#A\n", + "r=0.3#ohm\n", + "i1=160.0#A\n", + "i2=50.0#A\n", + "\n", + "#calculations\n", + "#160 A\n", + "vd=i1*(r-(v/i))\n", + "#50 A\n", + "vd2=i2*(r-(v/i))\n", + "\n", + "#result\n", + "print \"voltage drop at 160 A=\",vd,\"V\"\n", + "print \"voltage drop at 50 A=\",vd2,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "voltage drop at 160 A= 8.0 V\n", + "voltage drop at 50 A= 2.5 V\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter29_2.ipynb b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter29_2.ipynb new file mode 100644 index 00000000..f3eda54f --- /dev/null +++ b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter29_2.ipynb @@ -0,0 +1,2343 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:f1e5688d45c7bb285838d2aad7b4c0c08dc93f4afbba4c253d97655938545a41" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 29: D.C. Motor" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.1, Page Number:999" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=220#V\n", + "r=0.5#ohm\n", + "i=20#A\n", + "\n", + "#calculation\n", + "#as generator \n", + "eg=v+i*r\n", + "#as motor\n", + "eb=v-i*r\n", + "\n", + "#result\n", + "print \"as generator:eg=\",eg,\"V\"\n", + "print \"as motor:eb=\",eb,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "as generator:eg= 230.0 V\n", + "as motor:eb= 210.0 V\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.2, Page Number:999" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "ia=Symbol('ia')\n", + "r=0.1#ohm\n", + "brush_drop=2#V\n", + "n=1000#rpm\n", + "i=100#A\n", + "v=250#V\n", + "n2=700#rpm\n", + "\n", + "#calculations\n", + "rl=v/i\n", + "eg1=v+i*r+brush_drop\n", + "eg2=eg1*n2/n\n", + "ia=solve(eg2-2-ia*r-2.5*ia,ia)\n", + "\n", + "#result\n", + "print \"current delivered to the load=\",ia[0],\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "current delivered to the load= 69.7692307692308 A\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.3, Page Number:999" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=440#V\n", + "ra=0.8#ohm\n", + "rf=200#ohm\n", + "output=7.46#kW\n", + "efficiency=0.85\n", + "\n", + "#calculations\n", + "input_m=output*1000/efficiency\n", + "im=output*1000/(efficiency*v)\n", + "ish=v/rf\n", + "ia=im-ish\n", + "eb=v-ia*ra\n", + "\n", + "#results\n", + "print \"back emf=\",eb,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "back emf= 425.642780749 V\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.4, Page Number:1000" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=25#kW\n", + "v=250#V\n", + "ra=0.06#ohm\n", + "rf=100#ohm\n", + "\n", + "#calculations\n", + "#as generator\n", + "i=load*1000/v\n", + "ish=v/rf\n", + "ia=i+ish\n", + "eb=v+ia*ra\n", + "power=eb*ia/1000\n", + "\n", + "print \"As generator: power=\",power,\"kW\"\n", + "\n", + "#as motor\n", + "i=load*1000/v\n", + "ish=v/rf\n", + "ia=i-ish\n", + "eb=v-ia*ra\n", + "power=eb*ia/1000\n", + "\n", + "print \"As generator: power=\",power,\"kW\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "As generator: power= 26.12424 kW\n", + "As generator: power= 23.92376 kW\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.5, Page Number:1000" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "p=a=4\n", + "z=32\n", + "v=200.0#V\n", + "i=12.0#A\n", + "ra=2.0#ohm\n", + "rf=200.0#ohm\n", + "n=1000.0#rpm\n", + "i2=5.0#A\n", + "#calculations\n", + "ia=i+v/rf\n", + "eg=v+ia*ra\n", + "phi=eg*a*60/(z*n*p)\n", + "#as motor\n", + "ia=i2-v/rf\n", + "eb=v-ia*ra\n", + "n=60*eb/(phi*z)\n", + "\n", + "#result\n", + "print \"flux per pole=\",phi,\"wb\"\n", + "print \"speed of the machine=\",math.ceil(n),\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "flux per pole= 0.42375 wb\n", + "speed of the machine= 850.0 rpm\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.6, Page Number:1002" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "ia=110#A\n", + "v=480#V\n", + "ra=0.2#ohm\n", + "z=864\n", + "p=a=6\n", + "phi=0.05#Wb\n", + "\n", + "#calculations\n", + "eb=v-ia*ra\n", + "n=60*eb/(phi*z)\n", + "ta=0.159*phi*z*ia*p/a\n", + "\n", + "#result\n", + "print \"the speed=\",math.floor(n),\"rpm\"\n", + "print \"the gross torque=\",ta,\"N-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the speed= 636.0 rpm\n", + "the gross torque= 755.568 N-m\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.7, Page Number:1003" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=250#V\n", + "z=782\n", + "ra=rf=0.5#ohm\n", + "ia=40#A\n", + "phi=25*0.001#Wb\n", + "p=4\n", + "a=2\n", + "#calculation\n", + "eb=v-ia*ra\n", + "n=60*eb/(phi*z)\n", + "ta=0.159*phi*z*ia*p/a\n", + "\n", + "print \"the speed=\",math.floor(n),\"rpm\"\n", + "print \"the gross torque=\",ta,\"N-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the speed= 705.0 rpm\n", + "the gross torque= 248.676 N-m\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.8, Page Number:1003" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "eb=250.0#V\n", + "n=1500.0#rpm\n", + "ia=50.0#A\n", + "\n", + "#calculations\n", + "pm=eb*ia\n", + "ta=9.55*eb*ia/n\n", + "\n", + "#result\n", + "print \"torque=\",ta,\"N-m\"\n", + "print \"machanical power=\",pm,\"W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "torque= 79.5833333333 N-m\n", + "machanical power= 12500.0 W\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.9, Page Number:1003" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=220#V\n", + "p=4\n", + "z=800\n", + "load=8.2#kW\n", + "ia=45#A\n", + "phi=25*0.001#Wb\n", + "ra=0.6#ohm\n", + "a=p/2\n", + "\n", + "#calculation\n", + "ta=0.159*phi*z*ia*p/a\n", + "eb=v-ia*ra\n", + "n=eb*a/(phi*z*p)\n", + "tsh=load*1000/(2*3.14*n)\n", + "\n", + "#result\n", + "print \"developed torque=\",ta,\"N-m\"\n", + "print \"shaft torque=\",tsh,\"N-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "developed torque= 286.2 N-m\n", + "shaft torque= 270.618131415 N-m\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.10, Page Number:1003" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=220.0#V\n", + "n=500.0#rpm\n", + "i=50.0#A\n", + "ra=0.2#ohm\n", + "\n", + "#calculation\n", + "ia2=2*i\n", + "fb1=v-(i*ra)\n", + "eb2=v-(ia2*ra)\n", + "n2=eb2*n/fb1\n", + "#result\n", + "print \"speed when torque is doubled=\",n2,\"N-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed when torque is doubled= 476.19047619 N-m\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.11, Page Number:1003" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "r=Symbol('r')\n", + "v=500#V\n", + "load=37.3#kW\n", + "n=1000#rpm\n", + "efficiency=0.90\n", + "ra=0.24#ohm\n", + "vd=2#v\n", + "i=1.8#A\n", + "ratio=1.5\n", + "\n", + "#calculation\n", + "input_m=load*1000/efficiency\n", + "il=input_m/v\n", + "tsh=9.55*load*1000/n\n", + "il=ratio*il\n", + "ia=il-i\n", + "r=solve(ia*(r+ra)+vd-v,r)\n", + "\n", + "#result\n", + "print \"full-load line current=\",il,\"A\"\n", + "print \"full-load shaft torque\",tsh,\"N-m\"\n", + "print \"total resistance=\",r[0],\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "full-load line current= 124.333333333 A\n", + "full-load shaft torque 356.215 N-m\n", + "total resistance= 3.82420021762787 ohm\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.12, Page Number:1004" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=a=4\n", + "v=220#V\n", + "z=540\n", + "i=32#A\n", + "output=5.595#kW\n", + "ra=0.09#ohm\n", + "i_f=1#A\n", + "phi=30*0.001#Wb\n", + "\n", + "#calculation\n", + "ia=i-i_f\n", + "eb=v-ia*ra\n", + "n=eb*a*60/(phi*z*p)\n", + "tsh=9.55*output/n\n", + "\n", + "#result\n", + "print \"speed=\",n,\"rpm\"\n", + "print \"torque developed=\",tsh*1000,\"N-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed= 804.481481481 rpm\n", + "torque developed= 66.4182473183 N-m\n" + ] + } + ], + "prompt_number": 43 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.13(a), Page Number:1004" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=220.0#V\n", + "load=20.0#kW\n", + "i=5.0#A\n", + "ra=0.04#ohm\n", + "phi=0.04#Wb\n", + "z=160\n", + "il=95.0#A\n", + "inl=9.0#A\n", + "p=4\n", + "a=2\n", + "#calculation\n", + "#no load\n", + "ea0=v-(inl-i)*ra\n", + "n0=ea0*a*60/(phi*z*p)\n", + "#load\n", + "ea=v-(il-i)*ra\n", + "n=ea*n0/ea0\n", + "\n", + "#result\n", + "print \"no-load speed=\",n0,\"rpm\"\n", + "print \"load speed=\",n,\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "no-load speed= 1030.5 rpm\n", + "load speed= 1014.375 rpm\n" + ] + } + ], + "prompt_number": 58 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.13(b), Page Number:1004" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=a=6\n", + "i=400#A\n", + "n=350#rpm\n", + "phi=80*0.001#Wb\n", + "z=600*2\n", + "loss=0.03#percentage\n", + "\n", + "#calculation\n", + "e=phi*z*n*p/(60*a)\n", + "pa=e*i\n", + "t=pa/(2*3.14*n/60)\n", + "t_net=0.97*t\n", + "bhp=t_net*36.67*0.001/0.746\n", + "#result\n", + "print \"brake-horse-power\",bhp,\"HP\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "brake-horse-power 291.551578696 HP\n" + ] + } + ], + "prompt_number": 66 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.13(c), Page Number:1004" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=4\n", + "z=774\n", + "phi=24*0.001#Wb\n", + "ia=50#A\n", + "a=2\n", + "#calculations\n", + "t=0.159*phi*z*ia*p/a\n", + "\n", + "#result\n", + "print \"torque=\",t,\"N-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "torque= 295.3584 N-m\n" + ] + } + ], + "prompt_number": 67 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.13(d), Page Number:1005" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=500.0#V\n", + "i=5.0#A\n", + "ra=0.15#ohm\n", + "rf=200.0#ohm\n", + "il=40.0#A\n", + "\n", + "#calculations\n", + "ih=v/rf\n", + "pi=v*i\n", + "cu_loss_f=cu_loss=v*ih\n", + "output=v*il\n", + "cu_loss_a=(il+ih)**2*ra\n", + "total_loss=cu_loss+cu_loss_a+cu_loss_f\n", + "efficiency=output/(output+total_loss)\n", + "#result\n", + "print \"efficiency=\",efficiency*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "efficiency= 87.8312542029 %\n" + ] + } + ], + "prompt_number": 81 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.13(e), Page Number:1006" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable delcration\n", + "ia=40#A\n", + "v=220#V\n", + "n=800#rpm\n", + "ra=0.2#ohm\n", + "rf=0.1#ohm\n", + "loss=0.5#kW\n", + "\n", + "#calculations\n", + "eb=v-ia*(ra+rf)\n", + "ta=9.55*eb*ia/n\n", + "cu_loss=ia**2*(ra+rf)\n", + "total_loss=cu_loss+loss*1000\n", + "input_m=v*ia\n", + "output=input_m-total_loss\n", + "\n", + "#result\n", + "print \"output of the motor=\",output/1000,\"kW\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "output of the motor= 7.82 kW\n" + ] + } + ], + "prompt_number": 88 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.14, Page Number:1006" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "f=400.0#N\n", + "d=10.0#cm\n", + "n=840#rpm\n", + "v=220.0#V\n", + "n1=1800#rpm\n", + "efficiency=.80\n", + "d2=24.0#cm\n", + "\n", + "#calculations\n", + "tsh=f*d*0.01/2\n", + "output=tsh*2*3.14*n/60\n", + "input_m=output/efficiency\n", + "i=input_m/v\n", + "d1=n*d2/n1\n", + "\n", + "#calculation\n", + "print \"current taken by the motor=\",round(i),\"A\"\n", + "print \"size of motor pulley=\",d1,\"cm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "current taken by the motor= 10.0 A\n", + "size of motor pulley= 11.2 cm\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.15, Page Number:1006" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=200.0#V\n", + "p=4\n", + "z=280\n", + "ia=45.0#A\n", + "phi=18*0.001#Wb\n", + "ra=0.5+0.3#ohm\n", + "loss=800.0#W\n", + "d=0.41\n", + "a=4\n", + "#calculation\n", + "eb=v-ia*ra\n", + "n=eb*60*a/(phi*z*p*4)\n", + "inpt=v*ia\n", + "cu_loss=ia**2*ra\n", + "total_loss=loss+cu_loss\n", + "output=inpt-total_loss\n", + "tsh=9.55*output/n\n", + "f=tsh*2/d\n", + "\n", + "#result\n", + "print \"pull at the rim of the pulley=\",f,\"N-m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "pull at the rim of the pulley= 628.016180845 N-m\n" + ] + } + ], + "prompt_number": 102 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.16, Page Number:1007" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=4\n", + "v=240#V\n", + "output=11.19#kW\n", + "n=1000#rpm\n", + "ia=50#A\n", + "i=1#A\n", + "z=540\n", + "ra=0.1#ohm\n", + "vd=1#V\n", + "a=2\n", + "#calculation\n", + "eb=v-ia*ra\n", + "ta=9.55*eb*ia/n\n", + "tsh=9.55*output*1000/n\n", + "phi=eb*60*a*1000/(z*n*p)\n", + "input_a=v*ia\n", + "cu_loss=ia**2*ra\n", + "brush_loss=ia*2\n", + "power=input_a-(cu_loss+brush_loss)\n", + "rotational_loss=power-output*1000\n", + "input_m=v*(ia+i)\n", + "efficiency=output*1000/input_m\n", + "\n", + "#result\n", + "print \"total torque=\",ta,\"N-m\"\n", + "print \"useful torque=\",tsh,\"N-m\"\n", + "print \"flux/pole=\",phi,\"mWb\"\n", + "print \"rotational losses=\",rotational_loss,\"W\"\n", + "print \"efficiency=\",efficiency*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "total torque= 112.2125 N-m\n", + "useful torque= 106.8645 N-m\n", + "flux/pole= 13.0555555556 mWb\n", + "rotational losses= 460.0 W\n", + "efficiency= 91.4215686275 %\n" + ] + } + ], + "prompt_number": 106 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.17, Page Number:1007" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=460.0#v\n", + "n=500.0#rpm\n", + "i=40.0#A\n", + "i2=30.0#A\n", + "ra=0.8#ohm\n", + "\n", + "#calculation\n", + "t2_by_t1=i2**2/i**2\n", + "change=(1-t2_by_t1)*100#percentage\n", + "eb1=v-i*ra\n", + "eb2=v-i2*ra\n", + "n2=eb2*i*n/(eb1*i2)\n", + "#result\n", + "print \"speed=\",n2,\"rpm\"\n", + "print \"percentage change in torque=\",change,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed= 679.127725857 rpm\n", + "percentage change in torque= 43.75 %\n" + ] + } + ], + "prompt_number": 111 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.18, Page Number:1008" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=460.0#V\n", + "output=55.95#kW\n", + "n=750#rpm\n", + "I=252.8#kg-m2\n", + "ia1=1.4\n", + "ia2=1.8\n", + "\n", + "#calculations\n", + "ia=(ia1+ia2)/2\n", + "n=n/60.0\n", + "tsh=output*1000/(2*3.14*n)\n", + "torque_avg=(ia-1)*tsh\n", + "dt=(I*2*3.14*n)/torque_avg\n", + "\n", + "#result\n", + "print \"approximate time to attain full speed=\",dt,\"s\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "approximate time to attain full speed= 46.4050282991 s\n" + ] + } + ], + "prompt_number": 129 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.19, Page Number:1008" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "output=14.92#kW\n", + "v=400.0#V\n", + "n=400.0#rpm\n", + "i=40.0#A\n", + "I=7.5#kg-m2\n", + "ratio=1.2\n", + "\n", + "#calculations\n", + "n=n/60\n", + "t=output*1000/(2*3.14*n)\n", + "torque=(ratio-1)*t\n", + "dt=(I*2*3.14*n)/torque\n", + "\n", + "print \"time to attain full speed=\",dt,\"s\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "time to attain full speed= 4.4055406613 s\n" + ] + } + ], + "prompt_number": 138 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.20, Page Number:1009" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=4\n", + "z=944\n", + "phi=34.6*0.001#Wb\n", + "ta=209.0#N-m\n", + "v=500.0#V\n", + "ra=3.0#ohm\n", + "a=2\n", + "#calculation\n", + "ia=ta/(0.159*phi*z*(p/a))\n", + "ea=v-ia*ra\n", + "n=ea/(phi*z*(p/a))\n", + "\n", + "#result\n", + "print \"line current=\",ia,\"A\"\n", + "print \"speed=\",n*60,\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "line current= 20.1219966813 A\n", + "speed= 403.798260345 rpm\n" + ] + } + ], + "prompt_number": 143 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.21, Page Number:1010" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=250#v\n", + "n=1000#rpm\n", + "ia=8#A\n", + "ra=0.2#ohm\n", + "rf=250#ohm\n", + "i2=50#A\n", + "\n", + "#calculation\n", + "ish=v/rf\n", + "eb0=v-(ia-ish)*ra\n", + "eb=v-(i2-ish)*ra\n", + "n=eb*n/eb0\n", + "\n", + "#result\n", + "print \"speed when loaded=\",n,\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed when loaded= 966.21078037 rpm\n" + ] + } + ], + "prompt_number": 144 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.22, Page Number:1010" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n=800#rpm\n", + "ia=100#A\n", + "v=230#V\n", + "ra=0.15#ohm\n", + "rf=0.1#ohm\n", + "ia2=25#A\n", + "ratio=0.45\n", + "\n", + "#calculation\n", + "eb1=v-(ra+rf)*ia\n", + "eb2=v-ia2*(ra+rf)\n", + "n2=eb2*n/(eb1*ratio)\n", + "\n", + "#result\n", + "print \"speed at which motor runs=\",n2,\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed at which motor runs= 1940.37940379 rpm\n" + ] + } + ], + "prompt_number": 148 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.23, Page Number:1010" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "ia2=Symbol('ia2')\n", + "#variable declaration\n", + "v=230.0#V\n", + "ra=0.5#ohm\n", + "rf=115.0#ohm\n", + "n1=1200#rpm\n", + "ia=2.5#A\n", + "n2=1120#rpm\n", + "\n", + "#calculation\n", + "eb1=v-ra*ia\n", + "x=n2*eb1/n1\n", + "ia2=solve((v-ra*ia2)-x,ia2)\n", + "ia=ia2[0]+(v/rf)\n", + "input_m=v*ia\n", + "\n", + "#result\n", + "print \"line current=\",round(ia,1),\"A\"\n", + "print \"power input=\",round(input_m,1),\"W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "line current= 35.0 A\n", + "power input= 8050.0 W\n" + ] + } + ], + "prompt_number": 158 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.24, Page Number:1010" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "power=100.0#kW\n", + "n1=300#rpm\n", + "v=220.0#V\n", + "load=10.0#kW\n", + "ra=0.025#ohm\n", + "rf=60.0#ohm\n", + "vd=1.0#V\n", + "\n", + "#calculation\n", + "i=power*1000/v\n", + "ish=v/rf\n", + "ia=i+ish\n", + "eb=v+ia*ra+2*vd\n", + "i=load*1000/v\n", + "ia2=i-ish\n", + "eb2=v-ia2*ra-2*vd\n", + "n2=eb2*n1/eb\n", + "\n", + "#result\n", + "print \"speed=\",n2,\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed= 278.796797778 rpm\n" + ] + } + ], + "prompt_number": 174 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.25, Page Number:1011" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=250.0#V\n", + "n=1000.0#rpm\n", + "ra=0.5#ohm\n", + "rf=250.0#ohm\n", + "ia=4.0#A\n", + "i=40.0#A\n", + "ratio=0.04#percentage by whih armature reaction weakens field\n", + "\n", + "#calculations\n", + "ish=v/rf\n", + "ia2=ia-ish\n", + "eb0=v-ia2*ra\n", + "n0=n*eb0/v\n", + "ia=i-ish\n", + "eb=v-ia*ra\n", + "n=eb*n0/(eb0*(1-ratio))\n", + "\n", + "#result\n", + "print \"speed of machine=\",math.floor(n),\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed of machine= 960.0 rpm\n" + ] + } + ], + "prompt_number": 190 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.26, Page Number:1011" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=250#V\n", + "ooutput=14.92#kW\n", + "n=1000#rpm\n", + "i=75#A\n", + "ra=0.25#ohm\n", + "ratio=0.20\n", + "\n", + "#calculation\n", + "eb1=v-i*ra\n", + "eb_inst=eb1*(1-ratio)\n", + "ia_inst=(v-eb_inst)/ra\n", + "t_inst=9.55*eb_inst*ia_inst/n\n", + "ia2=i/(1-ratio)\n", + "eb2=v-ia2*ra\n", + "n2=eb2*n/(eb1*(1-ratio))\n", + "\n", + "#result\n", + "print \"armature current=\",ia2,\"A\"\n", + "print \"speed=\",n2,\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "armature current= 93.75 A\n", + "speed= 1224.66216216 rpm\n" + ] + } + ], + "prompt_number": 191 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.27, Page Number:1012" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=200.0#V\n", + "i=4.0#A\n", + "n=700.0#rpm\n", + "rf=100.0#A\n", + "v2=6.0#V\n", + "i2=10.0#A\n", + "input_m=8.0#kW\n", + "\n", + "#calculation\n", + "ish=v/rf\n", + "il=input_m*1000/v\n", + "ia=il-ish\n", + "ra=v2/i2\n", + "eb0=v-ish*ra\n", + "eb=v-ia*ra\n", + "n=eb*n/eb0\n", + "ta=9.55*eb*ia/n\n", + "inpt=v*i\n", + "cu_loss=ish**2*ra\n", + "constant_loss=inpt-cu_loss\n", + "cu_loss_arm=ia**2*ra\n", + "total_loss=constant_loss+cu_loss_arm\n", + "output=input_m*1000-total_loss\n", + "efficiency=output/(input_m*1000)\n", + "print \n", + "#result\n", + "print \"speed on load=\",n,\"rpm\"\n", + "print \"torque=\",ta,\"N-m\"\n", + "print \"efficiency=\",efficiency*100,\"%\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "speed on load= 623.943661972 rpm\n", + "torque= 103.0636 N-m\n", + "efficiency= 79.2 %\n" + ] + } + ], + "prompt_number": 197 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.28, Page Number:1012" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variabe declaration\n", + "v=220#V\n", + "load=11#kW\n", + "inl=5#A\n", + "n_nl=1150#rpm\n", + "ra=0.5#ohm\n", + "rsh=110#ohm\n", + "\n", + "#calculations\n", + "input_nl=v*inl\n", + "ish=v/rsh\n", + "ia0=inl-ish\n", + "cu_loss_nl=ia1**2*ra\n", + "constant_loss=input_nl-cu_loss_nl\n", + "i=load*1000/v\n", + "ia=i-ish\n", + "cu_loss_a=ia**2*ra\n", + "total_loss=cu_loss_a+constant_loss\n", + "output=load*1000-total_loss\n", + "efficiency=output*100/(load*1000)\n", + "eb_nl=v-(ia0*ra)\n", + "eb=v-ia*ra\n", + "n=n_nl*eb/eb_nl\n", + "ta=9.55*eb*ia/n\n", + "\n", + "#result\n", + "print \"torque developed=\",ta,\"N-m\"\n", + "print \"efficiency=\",efficiency,\"%\"\n", + "print \"the speed=\",n,\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "torque developed= 87.096 N-m\n", + "efficiency= 79.5361818182 %\n", + "the speed= 1031.57894737 rpm\n" + ] + } + ], + "prompt_number": 200 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.29, Page Number:1013" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=18.65#kW\n", + "v=250.0#V\n", + "ra=0.1#ohm\n", + "vb=3#V\n", + "rf=0.05#ohm\n", + "ia=80.0#A\n", + "n=600.0#rpm\n", + "i2=100.0#A\n", + "\n", + "#calculation\n", + "eb1=v-ia*(ra+rf)\n", + "eb2=v-i2*(ra+rf)\n", + "n2=eb2*ia*n/(eb1*i2)\n", + "\n", + "#result\n", + "print \"speed when current is 100 A=\",n2,\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed when current is 100 A= 473.949579832 rpm\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.30, Page Number:1013" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=220.0#V\n", + "n=800.0#rpm\n", + "i=100.0#A\n", + "ra=0.1\n", + "ratio=1.0/2.0\n", + "#calculation\n", + "ia1=i*math.sqrt(ratio)\n", + "eb1=v-i*ra\n", + "eb2=v-ia1*ra\n", + "n2=eb2*i*n/(eb1*ia1)\n", + "#result\n", + "print \"speed when motor will run when developing half the torque=\",round(n2,0),\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed when motor will run when developing half the torque= 1147.0 rpm\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.31, Page Number:1013" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "p=a=4\n", + "n=600#rpm\n", + "ia=25#A\n", + "v=450#V\n", + "z=500\n", + "phi=1.7*0.01*math.pow(ia,0.5)\n", + "\n", + "#calculation\n", + "eb=n*phi*z*p/(60*a)\n", + "iara=v-eb\n", + "ra=iara/ia\n", + "i=math.pow((phi*ia*math.sqrt(ia)/(phi*2)),2.0/3.0)\n", + "eb2=v/2-i*ra\n", + "phi2=1.7*0.01*math.pow(i,0.5)\n", + "n2=eb2*phi*n/(eb*phi2)\n", + "\n", + "#result\n", + "print \"speed at which motor will run=\",round(n2,0),\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed at which motor will run= 372.0 rpm\n" + ] + } + ], + "prompt_number": 224 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.32, Page Number:1017" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%pylab\n", + "import math\n", + "#variable declaration\n", + "v=460.0#V\n", + "ra=0.5#ohm\n", + "\n", + "def f(ia,t):\n", + " n=(v*ia-ia**2*ra)*60/(2*3.14*t)\n", + " return(n)\n", + "\n", + "n1=f(20.0,128.8)\n", + "n2=f(30.0,230.5)\n", + "n3=f(40.0,349.8)\n", + "n4=f(50.0,469.2)\n", + "T=[128.8,230.5,349.8,469.2]\n", + "N=[n1,n2,n3,n4]\n", + "a=plot(T,N)\n", + "xlabel(\"Torque(NM.m)\") \n", + "ylabel(\"Speed(rpm)\") \n", + "plt.xlim((0,500))\n", + "plt.ylim((0,800))\n", + "show(a)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Using matplotlib backend: TkAgg\n", + "Populating the interactive namespace from numpy and matplotlib\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.33, Page Number:1017" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%pylab\n", + "import math\n", + "#variable declaration\n", + "output=5.968#kW\n", + "n=700#rpm\n", + "v1=500#V\n", + "n2=600#rpm\n", + "ra=3.5#ohm\n", + "loss=450#W\n", + "\n", + "#calculation\n", + "\n", + "def fp(i,v):\n", + " p=5.968*((n2*(v1-i*ra)/(v*n))**2)\n", + " return(p)\n", + "\n", + "def fm(i,v):\n", + " m=((v1-i*ra)*i-loss)/1000\n", + " return(m)\n", + "\n", + "p1=fp(7.0,347.0)\n", + "p2=fp(10.5,393.0)\n", + "p3=fp(14.0,434.0)\n", + "p4=fp(27.5,468.0)\n", + "\n", + "m1=fm(7.0,347.8)\n", + "m2=fm(10.5,393.0)\n", + "m3=fm(14.0,434.0)\n", + "m4=fm(27.5,468.0)\n", + "\n", + "#plot\n", + "I=[7,10.5,14,27.5]\n", + "P=[p1,p2,p3,p4]\n", + "M=[m1,m2,m3,m4]\n", + "a=plot(I,P)\n", + "a=plot(I,M)\n", + "xlabel(\"Current\") \n", + "ylabel(\"Power(kW)\") \n", + "plt.xlim((0,30))\n", + "plt.ylim((0,12))\n", + "show(a)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.34, Page Number:1022" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=500#V\n", + "i=3#A\n", + "ia=3.5#A\n", + "ib=4.5#A\n", + "\n", + "#calculation\n", + "loss=v*i\n", + "#B unexcited\n", + "loss1=v*(ia-i)\n", + "#B excited\n", + "loss2=v*(ib-i)\n", + "loss=loss2-loss1\n", + "\n", + "#result\n", + "print \"iron losses of B=\",loss,\"W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "iron losses of B= 500.0 W\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.35, Page Number:1023" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=220.0#V\n", + "ra=0.2#ohm\n", + "rf=110.0#ohm\n", + "ia=5.0#A\n", + "n=1500#rpm\n", + "i2=52.0#A\n", + "\n", + "#calculation\n", + "ish=v/rf\n", + "ia1=ia-ish\n", + "ia2=i2-ish\n", + "eb1=v-ia1*ra\n", + "eb2=v-ia2*ra\n", + "n2=round(eb2*n/eb1,0)\n", + "input_nl=v*ia\n", + "cu_loss_nl=ia1**2*ra\n", + "constant_loss=input_nl-cu_loss_nl\n", + "cu_loss_l=ia2**2*ra\n", + "total_loss=constant_loss+cu_loss_l\n", + "input_l=v*i2\n", + "output=input_l-total_loss\n", + "tsh=9.55*output/n2\n", + "\n", + "#result\n", + "print \"speed=\",n2,\"rpm\"\n", + "print \"shaft torque=\",tsh,\"N-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.36, Page Number:1023" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=250#V\n", + "n=1000#rpm\n", + "ia=5#A\n", + "ra=0.2#ohm\n", + "rf=250#ohm\n", + "i=50#A\n", + "ratio=0.03#percentage by which armature reaction weakens field\n", + "\n", + "#calculations\n", + "ish=v/rf\n", + "ia1=ia-ish\n", + "ia2=i-ish\n", + "eb1=v-ia1*ra\n", + "eb2=v-ia2*ra\n", + "n2=eb2*n/(eb1*(1-ratio))\n", + "\n", + "#result\n", + "print \"speed=\",round(n2,0),\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed= 994.0 rpm\n" + ] + } + ], + "prompt_number": 241 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.37, Page Number:1023" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=500#V\n", + "ia=5#A\n", + "ra=0.22#A\n", + "rf=250#ohm\n", + "i=100#A\n", + "\n", + "#calculations\n", + "ish=v/rf\n", + "ia0=ia-ish\n", + "eb0=v-ia0*ra\n", + "cu_loss=ia0**2*ra\n", + "input_m=v*ia\n", + "constant_loss=input_m-cu_loss\n", + "ia=i-ish\n", + "eb=v-ia*ra\n", + "cu_loss=ia**2*ra\n", + "total_loss=cu_loss+constant_loss\n", + "input_m=v*i\n", + "output=input_m-total_loss\n", + "efficiency=output*100/input_m\n", + "per=(eb-eb0)*100/eb0\n", + "\n", + "#result\n", + "print \"efficiency=\",round(efficiency,1),\"%\"\n", + "print \"percentage change in speed=\",round(per,2),\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "efficiency= 90.8 %\n", + "percentage change in speed= -4.19 %\n" + ] + } + ], + "prompt_number": 244 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.38, Page Number:1024" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=250#V\n", + "n=1000#rpm\n", + "i=25#A\n", + "i2=50#A\n", + "ratio=0.03#percentage by which the armature reaction weakens field\n", + "ra=0.2#ohm\n", + "rf=250#ohm\n", + "vd=1\n", + "#calculation\n", + "ish=v/rf\n", + "ia1=i-ish\n", + "ebh=v-ia1*ra-2*vd\n", + "ia2=i2-ish\n", + "eb2=v-ia2*ra-2*vd\n", + "n2=eb2*n/(ebh*(1-ratio))\n", + "ta1=9.55*eb1*ia1/n\n", + "ta2=9.55*eb2*ia2/n2\n", + "\n", + "#result\n", + "print \"speed=\",round(n2,0),\"rpm\"\n", + "print \"torque in first case=\",ta1,\"N-m\"\n", + "print \"torque in second case=\",ta2,\"N-m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed= 1010.0 rpm\n", + "torque in first case= 57.11664 N-m\n", + "torque in second case= 110.3912768 N-m\n" + ] + } + ], + "prompt_number": 247 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.39, Page Number:1024" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=250.0#V\n", + "n1=1000.0#rpm\n", + "ra=0.5#ohm\n", + "rf=250.0#ohm\n", + "ia=4.0#A\n", + "i=40.0#A\n", + "ratio=0.04#percentage by which the armature reaction weakens field\n", + "eb1=250.0#V\n", + "\n", + "#calculation\n", + "ish=v/rf\n", + "eb2=v-(i-ish)*ra\n", + "n2=eb2*n/(eb1*(1-ratio))\n", + "cu_loss=(ia-ish)**2*ra\n", + "input_m=v*ia\n", + "constant_loss=input_m-cu_loss\n", + "cu_loss_a=(i-ish)**2*ra\n", + "total_loss=constant_loss+cu_loss_a\n", + "inpt=v*i\n", + "output=inpt-total_loss\n", + "efficiency=output*100/inpt\n", + "\n", + "#result\n", + "print \"speed=\",round(n2,0),\"rpm\"\n", + "print \"efficiency=\",efficiency,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed= 960.0 rpm\n", + "efficiency= 82.44 %\n" + ] + } + ], + "prompt_number": 254 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.40, Page Number:1025" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=4\n", + "v=250#V\n", + "z=120*8\n", + "a=4\n", + "phi=20*0.001#Wb\n", + "i=25#A\n", + "ra=0.1#ohm\n", + "rf=125#ohm\n", + "loss=810#W\n", + "\n", + "#calculations\n", + "ish=v/rf\n", + "ia=i-ish\n", + "eb=v-ia*ra\n", + "n=eb*a*60/(p*z*phi)\n", + "ta=9.55*eb*ia/n\n", + "cu_loss=ia**2*ra\n", + "cu_loss_shunt=v*ish\n", + "total_loss=loss+cu_loss+cu_loss_shunt\n", + "input_m=v*i\n", + "output=input_m-total_loss\n", + "tsh=9.55*output/n\n", + "efficiency=output*100/input_m\n", + "\n", + "#result\n", + "print \"gross torque=\",ta,\"N-m\"\n", + "print \"useful torque=\",tsh,\"N-m\"\n", + "print \"efficiency=\",efficiency,\"%\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "gross torque= 70.288 N-m\n", + "useful torque= 60.2946209124 N-m\n", + "efficiency= 78.1936 %\n" + ] + } + ], + "prompt_number": 256 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.41, Page Number:1025" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "output=14.92#kW\n", + "n=1150#rpm\n", + "p=4\n", + "a=2\n", + "z=620\n", + "ra=0.2#ohm\n", + "i=74.8#A\n", + "i2=3#A\n", + "v=230#V\n", + "#calculation\n", + "ia=i-i2\n", + "eb=v-ia*ra\n", + "phi=eb*a*60/(p*z*n)\n", + "ta=9.55*eb*ia/n\n", + "power=eb*ia\n", + "loss_rot=power-output*1000\n", + "input_m=v*i\n", + "total_loss=input_m-output*1000\n", + "per=total_loss*100/input_m\n", + "\n", + "#result\n", + "print \"flux per pole=\",phi*1000,\"mWb\"\n", + "print \"torque developed=\",ta,\"N-m\"\n", + "print \"rotational losses=\",loss_rot,\"W\"\n", + "print \"total losses expressed as a percentage of power=\",per,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "flux per pole= 9.07321178121 mWb\n", + "torque developed= 128.575818783 N-m\n", + "rotational losses= 562.952 W\n", + "total losses expressed as a percentage of power= 13.2759823297 %\n" + ] + } + ], + "prompt_number": 263 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.42, Page Number:1025" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "ia1=Symbol('ia1')\n", + "output=7.46#kW\n", + "v=250#V\n", + "i=5#A\n", + "ra=0.5#ohm\n", + "rf=250#ohm\n", + "\n", + "#calculation\n", + "input_m=v*i\n", + "ish=v/rf\n", + "ia=i-ish\n", + "cu_loss=v*ish\n", + "cu_loss_a=ra*ia**2\n", + "loss=input_m-cu_loss\n", + "ia1=solve(ra*ia1**2-v*ia1+output*1000+loss,ia1)\n", + "i2=ia1[0]+ish\n", + "input_m1=v*i2\n", + "efficiency=output*100000/input_m1\n", + "ia=math.sqrt((input_m-cu_loss_a)/ra)\n", + "input_a=v*ia\n", + "cu_loss=ia**2*ra\n", + "output_a=input_a-(cu_loss+loss)\n", + "\n", + "#result\n", + "print \"efficiency=\",efficiency,\"%\"\n", + "print \"output power at which efficiency is maximum=\",output_a/1000,\"kW\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "efficiency= 79.5621535016683 %\n", + "output power at which efficiency is maximum= 10.2179357944 kW\n" + ] + } + ], + "prompt_number": 271 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.43, Page Number:1026" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n2_by_n1=1.0/2.0\n", + "ia2_by_ia1=phi1_by_phi2=1.0/2.0\n", + "v2_by_v1=n2_by_n1*phi1_by_phi2\n", + "reduction_v=(1-v2_by_v1)*100\n", + "reduction_i=(1-ia2_by_ia1)*100\n", + "\n", + "#result\n", + "print \"percentage reduction in the motor terminal voltage=\",reduction_v,\"%\"\n", + "print \"percentage fall in the motor current=\",reduction_i,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "percentage reduction in the motor terminal voltage= 75.0 %\n", + "percentage fall in the motor current= 50.0 %\n" + ] + } + ], + "prompt_number": 272 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.44, Page Number:1026" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=6\n", + "v=500#V\n", + "z=1200\n", + "phi=20*0.001#Wb\n", + "ra=0.5#ohm\n", + "rf=250#ohm\n", + "i=20#A\n", + "loss=900#W\n", + "a=2\n", + "#calculation\n", + "ish=v/rf\n", + "ia=i-ish\n", + "eb=v-ia*ra\n", + "n=eb*a*60/(p*z*phi)\n", + "ta=9.55*eb*ia/n\n", + "cu_loss=ia**2*ra\n", + "cu_loss_f=v*ish\n", + "total_loss=cu_loss+cu_loss_f+loss\n", + "input_m=v*i\n", + "output=input_m-total_loss\n", + "tsh=9.55*output/n\n", + "efficiency=output*100/input_m\n", + "\n", + "#result\n", + "print \"useful torque=\",ta,\"N-m\"\n", + "print \"output=\",output/1000,\"Kw\"\n", + "print \"efficiency==\",efficiency,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "useful torque= 206.28 N-m\n", + "output= 7.938 Kw\n", + "efficiency== 79.38 %\n" + ] + } + ], + "prompt_number": 275 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 29.45, Page Number:1027" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "ia1=Symbol('ia1')\n", + "output=37.3*1000#W\n", + "v=460#V\n", + "i=4#A\n", + "n=660#rpm\n", + "ra=0.3#ohm\n", + "rf=270#ohm\n", + "\n", + "#calculations\n", + "ish=v/rf\n", + "cu_loss=v*ish\n", + "ia=i-ish\n", + "cu_loss_a=ia**2*ra\n", + "input_a=loss=v*ia\n", + "ia1=solve(ra*ia1**2-v*ia1+output+loss,ia1)\n", + "i=ia1[0]+ish\n", + "eb1=v-(ia*ra)\n", + "eb2=v-(ia1[0]*ra)\n", + "n2=n*eb2/eb1\n", + "ia=math.sqrt((cu_loss+input_a)/ra)\n", + "\n", + "#result\n", + "print \"the current input=\",i,\"A\"\n", + "print \"speed=\",round(n2,0),\"rpm\"\n", + "print \"armature current at which efficiency is maximum=\",ia,\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the current input= 90.2860908863713 A\n", + "speed= 623.0 rpm\n", + "armature current at which efficiency is maximum= 78.3156008298 A\n" + ] + } + ], + "prompt_number": 280 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter30_2.ipynb b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter30_2.ipynb new file mode 100644 index 00000000..ce13ea95 --- /dev/null +++ b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter30_2.ipynb @@ -0,0 +1,2629 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:072a977ff7e7f41108f647b699866e16f58bf91b148a03cefc5a07bc1eeda05b" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 30:Speed Control of D.C. Motors" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.1, Page Number:1032" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=500#V\n", + "n=250#rpm\n", + "ia=200#A\n", + "ra=0.12#ohm\n", + "ratio=0.80\n", + "ia2=100#A\n", + "\n", + "#calculations\n", + "eb1=v-ia*ra\n", + "eb2=v-ia2*ra\n", + "n2=eb2*n/(eb1*ratio)\n", + "\n", + "#result\n", + "print \"speed=\",round(n2),\"rpm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed= 320.0 rpm\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.2, Page Number:1032" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=250#V\n", + "ra=0.25#ohm\n", + "ia=50#A\n", + "n=750#rpm\n", + "ratio=1-0.10\n", + "\n", + "#calculation\n", + "ia2=ia/ratio\n", + "eb1=v-ia*ra\n", + "eb2=v-ia2*ra\n", + "n2=eb2*n/(eb1*ratio)\n", + "\n", + "#result\n", + "print \"speed=\",round(n2),\"rpm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed= 828.0 rpm\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.3, Page Number:1032" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=230.0#V\n", + "n=800#rpm\n", + "ia=50.0#A\n", + "n2=1000#rpm\n", + "ia2=80.0#A\n", + "ra=0.15#ohm\n", + "rf=250.0#ohm\n", + "\n", + "#calculation\n", + "eb1=v-ia*ra\n", + "eb2=v-ia2*ra\n", + "ish1=v/rf\n", + "r1=(n2*eb1*v)/(n*eb2*ish1)\n", + "r=r1-rf\n", + "ish2=v/r1\n", + "torque_ratio=ish2*ia2/(ish1*ia)\n", + "\n", + "#result\n", + "print \"resistance to be added=\",r,\"ohm\"\n", + "print \"ratio of torque=\",torque_ratio" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resistance to be added= 68.9506880734 ohm\n", + "ratio of torque= 1.25411235955\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.3, Page Number:1033" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=250.0#V\n", + "rf=250.0#ohm\n", + "ra=0.25#ohm\n", + "n=1500#rpm\n", + "ia=20.0#A\n", + "r=250.0#ohm\n", + "\n", + "#calculations\n", + "ish=v/rf\n", + "ish2=v/(rf+r)\n", + "ia2=ia*1/ish2\n", + "eb2=v-ia2*ra\n", + "eb1=v-ia*ra\n", + "n2=eb2*n/(eb1*ish2)\n", + "\n", + "#result\n", + "print \"new speed=\",round(n2),\"rpm\"\n", + "print \"new armature current=\",ia2,\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "new speed= 2939.0 rpm\n", + "new armature current= 40.0 A\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.5, Page Number:1033" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "rt=Symbol('rt')\n", + "v=250.0#V\n", + "ra=0.5#ohm\n", + "rf=250.0#ohm\n", + "n=600.0#rpm\n", + "ia=20.0#A\n", + "n2=800.0#rpm\n", + "\n", + "#calculation\n", + "ish1=v/rf\n", + "eb1=v-ia*ra\n", + "rt=solve(((n2*eb1*(v/rt))/(n*(v-(ia*ra/(v/rt)))))-1,rt)\n", + "r=rt[0]-rf\n", + "\n", + "#result\n", + "print \"resistance to be inserted=\",r,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resistance to be inserted= 88.3128987990058 ohm\n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.6, Page Number:1034" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "x=Symbol('x')\n", + "v=220#V\n", + "ra=0.5#ohm\n", + "ia=40#A\n", + "ratio=1+0.50\n", + "\n", + "#calculation\n", + "eb1=v-ia*ra\n", + "x=solve((ratio*eb1/((v-ia*ra*x)*x))-1,x)\n", + "per=1-1/x[0]\n", + "\n", + "#result\n", + "print\"main flux has to be reduced by=\",per*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "main flux has to be reduced by= 37.2991677469778 %\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.7, Page Number:1034" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=220#V\n", + "load=10#kW\n", + "i=41#A\n", + "ra=0.2#ohm\n", + "rw=0.05#ohm\n", + "ri=0.1#ohm\n", + "rf=110#ohm\n", + "ratio=1-0.25\n", + "r=1#ohm\n", + "ratio1=1-0.50\n", + "n=2500\n", + "#calculation\n", + "ish=v/rf\n", + "ia1=i-ish\n", + "ia2=ratio1*ia1/ratio\n", + "eb1=v-ia1*(ra+ri+rw)\n", + "eb2=v-ia2*(r+ra+ri+rw)\n", + "n2=eb2*n/(eb1*ratio)\n", + "\n", + "#result\n", + "print \"armature current=\",ia2,\"A\"\n", + "print \"motor speed=\",round(n2),\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "armature current= 26.0 A\n", + "motor speed= 2987.0 rpm\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.8, Page Number:1035" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=220#V\n", + "load=15#kW\n", + "n=850#rpm\n", + "ia=72.2#A\n", + "ra=0.25#ohm\n", + "rf=100#ohm\n", + "n2=1650#rpm\n", + "ia2=40#A\n", + "\n", + "#calculation\n", + "ish=v/rf\n", + "ia1=ia-ish\n", + "eb1=v-ia1*ra\n", + "eb2=v-ia2*ra\n", + "ratio=(n*eb2)/(n2*eb1)\n", + "per=1-ratio\n", + "#result\n", + "print \"percentage reduction=\",per*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "percentage reduction= 46.5636857585 %\n" + ] + } + ], + "prompt_number": 46 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.9, Page Number:1035" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "ia2=Symbol('ia2')\n", + "v=220#V\n", + "ra=0.5#ohm\n", + "ia=40#A\n", + "ratio=0.50+1\n", + "\n", + "#calculation\n", + "eb1=v-ia*ra\n", + "ia2=solve((((v-ra*ia2)*ia2)/(eb1*ratio*ia))-1,ia2)\n", + "per=ia/ia2[0]\n", + "\n", + "#result\n", + "print \"mail flux should be reduced by=\",round(per,4)*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "mail flux should be reduced by= 62.7 %\n" + ] + } + ], + "prompt_number": 49 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.10, Page Number:1035" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "ia=20.0#A\n", + "v=220.0#V\n", + "ra=0.5#ohm\n", + "ratio=0.50\n", + "\n", + "#calculation\n", + "eb1=v-ia*ra\n", + "eb2=ratio*(v-ia*ra)\n", + "r=(v-eb2)/ia-ra\n", + "\n", + "#result\n", + "print \"resistance required in the series=\",r,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resistance required in the series= 5.25 ohm\n" + ] + } + ], + "prompt_number": 53 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.11, Page Number:1036" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=250#V\n", + "n=1000#rpm\n", + "ia=8#A\n", + "i_f=1#A\n", + "ra=0.2#ohm\n", + "rf=250#ohm\n", + "i=50#A\n", + "\n", + "#calculations\n", + "eb0=v-(ia-i_f)*ra\n", + "kpsi=eb0/1000\n", + "ia=i-i_f\n", + "eb1=v-ia*ra\n", + "n1=eb1/kpsi\n", + "\n", + "#result\n", + "print \"speed=\",round(n1,1),\"rpm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed= 966.2 rpm\n" + ] + } + ], + "prompt_number": 55 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.12, Page Number:1037" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=240#V\n", + "ra=0.25#ohm\n", + "n=1000#rpm\n", + "ia=40#A\n", + "n2=800#rpm\n", + "i2=20#A\n", + "#calculation\n", + "eb=v-ia*ra\n", + "eb2=n2*eb/n\n", + "r=(v-eb2)/(ia)-ra\n", + "eb3=v-i2*(r+ra)\n", + "n3=eb3*n/eb\n", + "\n", + "#result\n", + "print \"additional resistance=\",r,\"ohm\"\n", + "print \"speed=\",round(n3),\"rpm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "additional resistance= 1.15 ohm\n", + "speed= 922.0 rpm\n" + ] + } + ], + "prompt_number": 61 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.13, Page Number:1037" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=7.48#kW\n", + "v=220#V\n", + "n=990#rpm\n", + "efficiency=0.88\n", + "ra=0.08#ohm\n", + "ish=2#A\n", + "n2=450#rpm\n", + "\n", + "#calculation\n", + "input_p=load*1000/efficiency\n", + "losses=input_p-load*1000\n", + "i=input_p/v\n", + "ia=i-ish\n", + "loss=v*ish\n", + "cu_loss=ia**2*ra\n", + "loss_nl=losses-cu_loss-loss\n", + "eb1=v-20-(ia*ra)\n", + "eb2=n2*eb1/n\n", + "r=(eb1-eb2)/ia\n", + "total_loss=ia**2*(r+ra)+loss+loss_nl\n", + "output=input_p-total_loss\n", + "efficiency=output/(input_p)\n", + "\n", + "#result\n", + "print \"motor input=\",input_p/1000,\"kW\"\n", + "print \"armature current=\",ia,\"A\"\n", + "print \"external resistance=\",r,\"ohm\"\n", + "print \"efficiency=\",efficiency*100,\"%\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "motor input= 8.5 kW\n", + "armature current= 36.6363636364 A\n", + "external resistance= 2.93403113016 ohm\n", + "efficiency= 41.6691237902 %\n" + ] + } + ], + "prompt_number": 81 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.14, Page Number:1038" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "eb1=230.0#V\n", + "n=990.0#rpm\n", + "n2=500.0#rpm\n", + "ia=25.0#A\n", + "\n", + "#calculation\n", + "eb2=eb1*n2/n\n", + "r=(eb1-eb2)/ia\n", + "\n", + "#result\n", + "print \"resistance required in series=\",r,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resistance required in series= 4.55353535354 ohm\n" + ] + } + ], + "prompt_number": 83 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.15, Page Number:1038" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=220.0#V\n", + "ra=0.4#ohm\n", + "rf=200.0#ohm\n", + "ia=20.0#A\n", + "n=600.0#rpm\n", + "n2=900.0#rpm\n", + "\n", + "#calculation\n", + "if1=v/rf\n", + "eb1=v-ia*ra\n", + "k2=eb1/(if1*n)\n", + "if2=n*if1/n2\n", + "rf1=v/if1\n", + "rf2=v/if2\n", + "r=rf2-rf1\n", + "\n", + "#result\n", + "print \"resistance to be added=\",r,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resistance to be added= 100.0 ohm\n" + ] + } + ], + "prompt_number": 90 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.16, Page Number:1039" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "ia2=Symbol('ia2')\n", + "v=220.0#V\n", + "ra=0.4#ohm\n", + "rf=200.0#ohm\n", + "ia=22.0#A\n", + "n=600.0#rpm\n", + "n2=900.0#rpm\n", + "\n", + "#calculation\n", + "if1=v/rf\n", + "eb1=v-ia*ra\n", + "k1=eb1/(if1*n)\n", + "if2=n*if1/n2\n", + "if2=n2*ia/n\n", + "ia2=solve(v-ra*ia2-(k1*ia*if1*n2)/ia2,ia2)\n", + "if2=ia*if1/ia2[0]\n", + "r=v/if2\n", + "\n", + "#result\n", + "print \"new field resistance to be added=\",r,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "new field resistance to be added= 306.828780053869 ohm\n" + ] + } + ], + "prompt_number": 103 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.17, Page Number:1040" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=250#V\n", + "output=25#kW\n", + "efficiency=0.85\n", + "n=1000#rpm\n", + "ra=0.1#ohm\n", + "rf=125#ohm\n", + "ratio=1.50\n", + "\n", + "#calculation\n", + "input_p=output*1000/efficiency\n", + "i=input_p/v\n", + "if1=v/rf\n", + "ia=i-if1\n", + "il=ratio*ia\n", + "r=v/il\n", + "r_ext=r-ra\n", + "\n", + "#result\n", + "print \"starting resistance=\",round(r_ext,3),\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "starting resistance= 1.341 ohm\n" + ] + } + ], + "prompt_number": 105 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.18, Page Number:1042" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=200.0#V\n", + "n=1000.0#rpm\n", + "ia=17.5#A\n", + "n2=600.0#rpm\n", + "ra=0.4#ohm\n", + "\n", + "#calculation\n", + "eb1=v-ia*ra\n", + "rt=(v-(n2*eb1/n))/ia\n", + "r=rt-ra\n", + "#result\n", + "print \"resistance to be inserted=\",round(r,1),\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resistance to be inserted= 4.4 ohm\n" + ] + } + ], + "prompt_number": 111 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.19, Page Number:1042" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=500#V\n", + "ra=1.2#ohm\n", + "rf=500#ohm\n", + "ia=4#A\n", + "n=1000#rpm\n", + "i=26#A\n", + "r=2.3#ohm\n", + "ratio=0.15\n", + "\n", + "#calculation\n", + "ish=v/rf\n", + "ia1=ia-ish\n", + "eb1=v-ia1*ra\n", + "ia2=i-ish\n", + "eb2=v-ia2*ra\n", + "n2=n*eb2/eb1\n", + "eb2=v-ia2*(r+ra)\n", + "n2_=n*eb2/eb1\n", + "n2__=n*eb2/(eb1*(1-ratio))\n", + "\n", + "#result\n", + "print \"speed when resistance 2.3 ohm is connected=\",round(n2_),\"rpm\"\n", + "print \"speed when shunt field is reduced by 15%=\",round(n2__),\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed when resistance 2.3 ohm is connected= 831.0 rpm\n", + "speed when shunt field is reduced by 15%= 978.0 rpm\n" + ] + } + ], + "prompt_number": 113 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.20, Page Number:1043" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=250.0#V\n", + "ia1=ia2=20.0#A\n", + "n=1000.0#rpm\n", + "ra=0.5#ohm\n", + "n2=500.0#ohm\n", + "\n", + "#calculation\n", + "eb1=v-ia1*ra\n", + "rt=(v-((n2/n)*eb1))/ia2\n", + "r=rt-ra\n", + "ia3=ia2/2\n", + "n3=n*(v-ia3*rt)/eb1\n", + "#result\n", + "print \"speed=\",round(n3),\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed= 771.0 rpm\n" + ] + } + ], + "prompt_number": 117 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.21, Page Number:1043" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=250.0#V\n", + "ra1=0.5#ohm\n", + "n=600.0#rpm\n", + "ia2=ia1=20#A\n", + "r=1.0#ohm\n", + "\n", + "#calculations\n", + "eb1=v-ia1*ra1\n", + "ra2=r+ra1\n", + "eb2=v-ia2*ra2\n", + "n2=eb2*n/eb1\n", + "#torque is half the full-load torque\n", + "ia2=1.0/2.0*ia1\n", + "eb22=v-ia2*ra2\n", + "n2_=eb22*n/eb1\n", + "#result\n", + "print \"speed at full load torque=\",round(n2),\"rpm\"\n", + "print \"speed at half full-load torque=\",round(n2_),\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed at full load torque= 550.0 rpm\n", + "speed at half full-load torque= 588.0 rpm\n" + ] + } + ], + "prompt_number": 137 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.22, Page Number:1044" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=220.0#V\n", + "ra1=0.5#ohm\n", + "n=500.0#rpm\n", + "ia2=ia1=30.0#A\n", + "r=1.0#ohm\n", + "\n", + "#calculations\n", + "eb1=v-ia1*ra1\n", + "ra2=r+ra1\n", + "eb2=v-ia2*ra2\n", + "n2=eb2*n/eb1\n", + "\n", + "#torque is half the full-load torque\n", + "ia2=2.0*ia1\n", + "eb22=v-ia2*ra2\n", + "n2_=eb22*n/eb1\n", + "#result\n", + "print \"speed at full load torque=\",round(n2),\"rpm\"\n", + "print \"speed at double full-load torque=\",round(n2_),\"rpm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed at full load torque= 427.0 rpm\n", + "speed at double full-load torque= 317.0 rpm\n" + ] + } + ], + "prompt_number": 142 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.23, Page Number:1044" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=37.3*1000#W\n", + "v=500.0#V\n", + "n=750.0#rpm\n", + "efficiency=0.90\n", + "t2=250.0#N-m\n", + "r=5.0#ohm\n", + "ra=0.5#ohm\n", + "\n", + "#calculation\n", + "t1=load/(2*3.14*(n/60))\n", + "ia1=load/(efficiency*v)\n", + "ia2=ia1*math.sqrt(t2/t1)\n", + "eb1=v-ia1*ra\n", + "eb2=v-ia2*(r+ra)\n", + "n2=eb2*ia1*n/(eb1*ia2)\n", + "\n", + "#result\n", + "print \"speed at which machine will run=\",round(n2),\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed at which machine will run= 381.789716486 rpm\n" + ] + } + ], + "prompt_number": 157 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.24, Page Number:1044" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "output=7.46*1000#W\n", + "v=220.0#V\n", + "n=900.0#rpm\n", + "efficiency=0.88\n", + "ra=0.08#ohm\n", + "ish=2.0#A\n", + "n2=450.0#rpm\n", + "#calculation\n", + "i=output/(efficiency*v)\n", + "ia2=ia1=i-ish\n", + "eb1=v-ia2*ra\n", + "rt=(v-20-((n2/n)*eb1))/ia2\n", + "r=rt-ra\n", + "input_m=(v)*(ia2+ish)\n", + "total_loss=input_m-output\n", + "cu_loss=ia2**2*ra\n", + "cu_loss_f=v*ish\n", + "total_cu_loss=cu_loss+cu_loss_f\n", + "stray_loss=total_loss-total_cu_loss\n", + "stray_loss2=stray_loss*n2/n\n", + "cu_loss_a=ia1**2*rt\n", + "total_loss2=stray_loss2+cu_loss_f+cu_loss_a\n", + "output2=input_m-total_loss2\n", + "efficiency=output2*100/input_m\n", + "\n", + "#result\n", + "print \"motor output=\",output2,\"W\"\n", + "print \"armature current=\",ia2,\"A\"\n", + "print \"external resistance=\",r,\"ohm\"\n", + "print \"overall efficiency=\",efficiency,\"%\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "motor output= 4460.66115702 W\n", + "armature current= 36.5330578512 A\n", + "external resistance= 2.42352222599 ohm\n", + "overall efficiency= 52.619059225 %\n" + ] + } + ], + "prompt_number": 175 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.25, Page Number:1044" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=240.0#V\n", + "ia=15.0#A\n", + "n=800.0#rpm\n", + "ra=0.6#ohm\n", + "n2=400.0#rpm\n", + "\n", + "#calculation\n", + "eb1=v-ia*ra\n", + "r=((v-(n2*eb1/n))/ia)-ra\n", + "ia3=ia/2\n", + "eb3=v-ia3*(r+ra)\n", + "n3=eb3*n/eb1\n", + "\n", + "#result\n", + "print \"speed=\",n3,\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed= 615.584415584 rpm\n" + ] + } + ], + "prompt_number": 187 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.26, Page Number:1045" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "r=Symbol('r')\n", + "v=400.0#V\n", + "inl=3.5#A\n", + "il=59.5#A\n", + "rf=267.0#ohm\n", + "ra=0.2#ohm\n", + "vd=2.0#V\n", + "ratio=0.02\n", + "speed_ratio=0.50\n", + "\n", + "#calculations\n", + "ish=v/rf\n", + "ia1=inl-ish\n", + "eb1=v-ia1*ra-vd\n", + "ia2=il-ish\n", + "eb2=v-ia2*ra-vd\n", + "n1_by_n2=eb1*(1-ratio)/eb2\n", + "per_change=(1-1/n1_by_n2)*100\n", + "r=solve(eb2*speed_ratio/(eb2-ia2*r)-1,r)\n", + "#result\n", + "print \"change in speed=\",per_change,\"%\"\n", + "print \"resistance to be added=\",r[0],\"ohm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "change in speed= 0.83357557339 %\n", + "resistance to be added= 3.33092370774547 ohm\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.27, Page Number:1046" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaraion\n", + "v=200.0#V\n", + "i=50.0#A\n", + "n=1000.0#rpm\n", + "n2=800.0#rpm\n", + "ra=0.1#ohm\n", + "rf=100.0#ohm\n", + "\n", + "#calculations\n", + "ish=v/rf\n", + "ia1=i-ish\n", + "ia2=ia1*(n2/n)**2\n", + "eb1=v-ia1*ra\n", + "eb2=v-ia2*ra\n", + "rt=(v-(n2*eb1/n))/ia2\n", + "r=rt-ra\n", + "#result\n", + "print \"resustance that must be added=\",r,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resustance that must be added= 1.32708333333 ohm\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.28, Page Number:1047" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=250#V\n", + "load=37.3#kW\n", + "efficiency=0.90\n", + "n=1000#rpm\n", + "ra=0.1#ohm\n", + "rf=115#ohm\n", + "ratio=1.5\n", + "\n", + "#calculation\n", + "tsh=9.55*load*1000/n\n", + "i=load*1000/(v*efficiency)\n", + "ish=v/rf\n", + "ia=i-ish\n", + "eb=v-ia*ra\n", + "ta=9.55*eb*ia/n\n", + "i_permissible=i*ratio\n", + "ia_per=i_permissible-ish\n", + "ra_total=v/ia_per\n", + "r_required=ra_total-ra\n", + "torque=ratio*ta\n", + "#result\n", + "print \"net torque=\",ta,\"N-m\"\n", + "print \"starting resistance=\",r_required,\"ohm\"\n", + "print \"torque developed at starting=\",torque,\"N-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "net torque= 365.403326173 N-m\n", + "starting resistance= 0.913513513514 ohm\n", + "torque developed at starting= 548.104989259 N-m\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.29, Page Number:1047" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "I=Symbol('I')\n", + "v=200.0#V\n", + "rf=40.0#ohm\n", + "ra=0.02#ohm\n", + "i=55.0#A\n", + "n=595.0#rpm\n", + "r=0.58#ohm\n", + "n2=630.0#rpm\n", + "ia_=15.0#A\n", + "rd=5.0#ohm\n", + "ia2=50.0#A\n", + "\n", + "#calculation\n", + "ish=v/rf\n", + "ia1=i-ish\n", + "ra1=r+ra\n", + "eb1=v-ra1*ia1\n", + "ia2=ia1\n", + "eb2=eb1*(n2/n)\n", + "r=(v-eb2)/ia1\n", + "eb2_=v-ia_*ra1\n", + "n2=eb2_*n/eb1\n", + "eb3=eb1\n", + "IR=v-eb3-ia2*ra\n", + "pd=v-IR\n", + "i_d=pd/rd\n", + "i=ia2+i_d\n", + "R=IR/i\n", + "I=solve(rd*(I-ia_)-v+R*I,I)\n", + "eb4=v-R*I[0]-ia_*ra\n", + "n4=n*(eb4/eb1)\n", + "\n", + "#result\n", + "print \"armature circuit resistance should be reduced by=\",ra1-r,\"ohm\"\n", + "print \"speed when Ia=\",n2,\"rpm\"\n", + "print \"value of series resistance=\",R,\"ohm\"\n", + "print \"speed when motor current falls to 15A=\",n4,\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "armature circuit resistance should be reduced by= 0.2 ohm\n", + "speed when Ia= 668.5 rpm\n", + "value of series resistance= 0.344418052257 ohm\n", + "speed when motor current falls to 15A= 636.922222222222 rpm\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.31, Page Number:1051" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "i=15#A\n", + "n=600#rpm\n", + "\n", + "#calculation\n", + "ia2=math.sqrt(2*2**0.5*i**2)\n", + "n2=n*2*i/ia2\n", + "\n", + "#result\n", + "print \"speed=\",n2,\"rpm\"\n", + "print \"current=\",ia2,\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed= 713.524269002 rpm\n", + "current= 25.2268924576 A\n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.32, Page Number:1052" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n=707#rpm\n", + "ia1=100#A\n", + "v=85#V\n", + "rf=0.03#ohm\n", + "ra=0.04#ohm\n", + "\n", + "#calculation\n", + "ra_total=ra+(2*rf)\n", + "eb1=v-ia1*ra_total\n", + "ia2=ia1*2**0.5\n", + "rf=rf/2\n", + "eb2=v-ia2*(ra+rf)\n", + "n2=n*(eb2/eb1)*(2*ia1/ia2)\n", + "rt=(v-((n/n2)*eb2))/ia2\n", + "r=rt-ra-rf\n", + "\n", + "#result\n", + "print \"speed=\",n2,\"rpm\"\n", + "print \"additional resistance=\",r,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed= 1029.46885374 rpm\n", + "additional resistance= 0.171040764009 ohm\n" + ] + } + ], + "prompt_number": 44 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.33, Page Number:1052" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#varable declaration\n", + "v=240.0#V\n", + "ia=40.0#A\n", + "ra=0.3#ohm\n", + "n=1500.0#rpm\n", + "n2=1000.0#rpm\n", + "#calculation\n", + "R=v/ia-ra\n", + "eb1=v-ia*ra\n", + "r=(v-((n2/n)*eb1))/ia-ra\n", + "\n", + "#result\n", + "print \"resistance to be added at starting=\",R,\"ohm\"\n", + "print \"resistance to be added at 1000 rpm\",r,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resistance to be added at starting= 5.7 ohm\n", + "resistance to be added at 1000 rpm 1.9 ohm\n" + ] + } + ], + "prompt_number": 49 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.34, Page Number:1053" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n=600.0#rpm\n", + "v=250.0#V\n", + "ia1=20.0#A\n", + "ratio=2.0\n", + "\n", + "#calculations\n", + "ia2=ia1*2**(3.0/4.0)\n", + "n2=n*ratio*ia1/ia2\n", + "\n", + "#result\n", + "print \"current=\",ia2,\"A\"\n", + "print \"speed=\",n2,\"rpm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "current= 33.6358566101 A\n", + "speed= 713.524269002 rpm\n" + ] + } + ], + "prompt_number": 50 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.35, Page Number:1053" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "V=Symbol('V')\n", + "ra=1.0#ohm\n", + "v=220.0#V\n", + "n=350.0#rpm\n", + "ia=25.0#A\n", + "n2=500.0#rpm\n", + "\n", + "#calculation\n", + "ia2=ia*(n2/n)\n", + "eb1=v-ia*ra\n", + "V=solve((n2*eb1*ia2/(n*ia))+ia2-V,V)\n", + "\n", + "#result\n", + "print \" current=\",ia2,\"A\"\n", + "print \"voltage=\",V[0],\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " current= 35.7142857143 A\n", + "voltage= 433.673469387755 V\n" + ] + } + ], + "prompt_number": 58 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.36, Page Number:1053" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n=1000.0#rpm\n", + "ia=20.0#A\n", + "v=200.0#V\n", + "ra=0.5#ohm\n", + "rf=0.2#ohm\n", + "i=20.0#A\n", + "rd=0.2#ohm\n", + "i_f=10.0#A\n", + "ratio=0.70\n", + "\n", + "#calculation\n", + "eb1=v-(ra+rf)*ia\n", + "r_total=ra+rf/2\n", + "eb2=v-r_total*ia\n", + "n2=(eb2*n/(eb1*ratio))\n", + " \n", + "#result\n", + "print \"speed=\",round(n2),\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed= 1444.0 rpm\n" + ] + } + ], + "prompt_number": 61 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.37, Page Number:1054" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=200.0#V\n", + "ia=40.0#A\n", + "n=700.0#rpm\n", + "ratio=0.50+1\n", + "ra=0.15#ohm\n", + "rf=0.1#ohm\n", + "\n", + "#calculations\n", + "ia2=(ratio*2*ia**2)**0.5\n", + "eb1=v-ia*(ra+rf)\n", + "eb2=v-ia2*(ra+rf)\n", + "n2=(eb2/eb1)*(ia*2/ia2)*n\n", + "\n", + "#result\n", + "print \"speed=\",n2,\"rpm\"\n", + "print \"speed=\",ia2,\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed= 777.147765122 rpm\n", + "speed= 69.2820323028 A\n" + ] + } + ], + "prompt_number": 63 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.38, Page Number:1055" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=250#V\n", + "ia=20#A\n", + "n=900#rpm\n", + "r=0.025#ohm\n", + "ra=0.1#ohm\n", + "rd=0.2#ohm\n", + "\n", + "#calculation\n", + "#when divertor is added\n", + "eb1=v-ia*(ra+4*r)\n", + "ia2=(ia**2*(ra+rd)/rd)**0.5\n", + "ra_=rd*ra/(ra+rd)\n", + "eb2=v-ia2*ra_\n", + "n2=(eb2/eb1)*(ia*3/(2*ia2))*n\n", + "\n", + "#rearranged field coils in two series and parallel group\n", + "ia2=(ia**2*2)**0.5\n", + "r=ra+r\n", + "eb2=v-ia2*r\n", + "n2_=(eb2/eb1)*(ia*2/(ia2))*n\n", + "\n", + "#result\n", + "print \"speed when divertor was added=\",n2,\"rpm\"\n", + "print \"speed when field coils are rearranged=\",n2_,\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed when divertor was added= 1112.87640676 rpm\n", + "speed when field coils are rearranged= 1275.19533144 rpm\n" + ] + } + ], + "prompt_number": 74 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.39, Page Number:1055" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=230.0#V\n", + "n=1000.0#rpm\n", + "i=12.0#A\n", + "rf=0.8#ohm\n", + "ra=1.0#ohm\n", + "il=20#A\n", + "ratio=0.15\n", + "\n", + "#calculation\n", + "eb1=v-i*(ra+rf)\n", + "eb2=v-il*(ra+rf/4)\n", + "n2=(eb2/eb1)*(1/(1-ratio))*n\n", + "\n", + "#result\n", + "print \"speed=\",n2,\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed= 1162.92198261 rpm\n" + ] + } + ], + "prompt_number": 75 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.40, Page Number:1056" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "i2=Symbol('i2')\n", + "v=200.0#v\n", + "n=500.0#rpm\n", + "i=25.0#A\n", + "ra=0.2#ohm\n", + "rf=0.6#ohm\n", + "rd=10.0#ohm\n", + "\n", + "#calculation\n", + "r=ra+rf\n", + "eb1=v-i*r\n", + "i2=solve(((rd+rf)*i2**2)-(v*i2)-(i**2*rd),i2)\n", + "pd=v-i2[1]*rf\n", + "ia2=((rd+rf)*i2[1]-v)/rd\n", + "eb2=pd-ia2*ra\n", + "n2=(eb2/eb1)*(i/i2[1])*n\n", + "#result\n", + "print \"speed=\",n2,\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed= 342.848235418389 rpm\n" + ] + } + ], + "prompt_number": 97 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.41, Page Number:1056" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=440#V\n", + "ra=0.3#ohm\n", + "i=20#A\n", + "n=1200#rpm\n", + "r=3#ohm\n", + "i2=15#A\n", + "ratio=0.80\n", + "\n", + "#calculation\n", + "eb1=v-i*ra\n", + "eb2=v-(r+ra)*i2\n", + "n2=n*(eb2/eb1)/ratio\n", + "power_ratio=(n*i)/(n2*i2*ratio)\n", + "\n", + "#result\n", + "print \"new speed=\",n2,\"rpm\"\n", + "print \"ratio of power outputs=\",power_ratio" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "new speed= 1349.65437788 rpm\n", + "ratio of power outputs= 1.48186086214\n" + ] + } + ], + "prompt_number": 99 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.42, Page Number:1057" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "i=50#A\n", + "v=460#V\n", + "ratio=1-0.25\n", + "\n", + "#calculation\n", + "I=(i**2*ratio**3)**0.5\n", + "eb2=I*ratio*v/i\n", + "R=(v-eb2)/I\n", + "pa=v*i/1000\n", + "power_n=pa*ratio**4\n", + "pa=eb2*I\n", + "\n", + "#result\n", + "print \"Resistance required=\",R,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Resistance required= 7.26432660412 ohm\n" + ] + } + ], + "prompt_number": 103 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.44, Page Number:1060" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n=500#rpm\n", + "n2=550#rpm\n", + "i=50#A\n", + "v=500#V\n", + "r=0.5#ohm\n", + "\n", + "#calculation\n", + "eb1=v-i*r\n", + "kphi1=eb1/n\n", + "eb2=v-i*r\n", + "kphi2=eb2/n2\n", + "eb_=v-i*2*r\n", + "n=eb_/((eb1/n2)+(eb2/n))\n", + "#result\n", + "print \"speed=\",n,\"rpm\"\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed= 248.120300752 rpm\n" + ] + } + ], + "prompt_number": 109 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.45, Page Number:1061" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=14.92#kW\n", + "v=250#V\n", + "n=1000#rpm\n", + "ratio1=5.0\n", + "ratio2=4.0\n", + "t=882#N-m\n", + "\n", + "#calculation\n", + "i=load*1000/v\n", + "k=v/(n*i/60)\n", + "I=(t/((ratio1+ratio2)*0.159*k))**0.5\n", + "nsh=v/((ratio1+ratio2)*k*I)\n", + "eb1=ratio1*k*I*nsh\n", + "eb2=ratio2*k*I*nsh\n", + "\n", + "#result\n", + "print \"current=\",I,\"A\"\n", + "print \"speed of shaft=\",round(nsh*60),\"rpm\"\n", + "print \"voltage across the motors=\",round(eb1),\"V,\",round(eb2),\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "current= 49.5202984449 A\n", + "speed of shaft= 134.0 rpm\n", + "voltage across the motors= 139.0 V, 111.0 V\n" + ] + } + ], + "prompt_number": 117 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.46, Page Number:1063" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=220#V\n", + "t=700#N-m\n", + "n=1200#rpm\n", + "ra=0.008#ohm\n", + "rf=55#ohm\n", + "efficiency=0.90\n", + "t2=375#N-m\n", + "n2=1050#rpm\n", + "\n", + "#calculation\n", + "output=2*3.14*n*t/60\n", + "power_m=output/efficiency\n", + "im=power_m/v\n", + "ish=v/rf\n", + "ia1=im-ish\n", + "eb1=v-ia1*ra\n", + "ia2=ia1*t2/t\n", + "eb2=eb1*n2/n\n", + "r=eb2/ia2-ra\n", + "\n", + "#result\n", + "print \"dynamic break resistance=\",r,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "dynamic break resistance= 0.795525014538 ohm\n" + ] + } + ], + "prompt_number": 118 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.47, Page Number:1064" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=400.0#V\n", + "load=18.65#kW\n", + "n=450.0#rpm\n", + "efficiency=0.746\n", + "ra=0.2#ohm\n", + "\n", + "#calculations\n", + "I=load*1000/(efficiency*v)\n", + "eb=v-I*ra\n", + "vt=v+eb\n", + "i_max=2*I\n", + "r=vt/i_max\n", + "R=r-ra\n", + "N=n/60\n", + "phizp_by_a=eb/N\n", + "k4=phizp_by_a*v/(2*3.14*r)\n", + "k3=phizp_by_a**2/(2*3.14*r)\n", + "tb=k4+k3*N\n", + "tb0=k4\n", + "#result\n", + "print \"breaking resistance=\",R,\"ohm\"\n", + "print \"maximum breaking torque=\",tb,\"N-m\"\n", + "print \"maximum breaking torque when N=0 =\",tb0,\"N-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "breaking resistance= 6.1 ohm\n", + "maximum breaking torque= 1028.3970276 N-m\n", + "maximum breaking torque when N=0 = 522.360394972 N-m\n" + ] + } + ], + "prompt_number": 122 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.48, Page Number:1069" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=120#V\n", + "ra=0.5#ohm\n", + "l=20*0.001#H\n", + "ka=0.05#V/rpm motor constant\n", + "ia=20#A\n", + "\n", + "#calculations\n", + "vt=ia*ra\n", + "alpha=vt/v\n", + "#when alpha=1\n", + "eb=v-ia*ra\n", + "N=eb/ka\n", + "\n", + "#result\n", + "print \"range of speed control=\",0,\"to\",N,\"rpm\"\n", + "print \"range of duty cycle=\",(alpha),\"to\",1" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " range of speed control= 0 to 2200.0 rpm\n", + "range of duty cycle= 0.0833333333333 to 1\n" + ] + } + ], + "prompt_number": 124 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.49, Page Number:1080" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=7.46#kW\n", + "v=200#V\n", + "efficiency=0.85\n", + "ra=0.25#ohm\n", + "ratio=1.5\n", + "\n", + "#calculation\n", + "i=load*1000/(v*efficiency)\n", + "i1=ratio*i\n", + "r1=v/i1\n", + "r_start=r1-ra\n", + "eb1=v-i*r1\n", + "\n", + "#result\n", + "print \"starting resistance=\",r_start,\"ohm\"\n", + "print \"back emf=\",eb1,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "starting resistance= 2.78842716711 ohm\n", + "back emf= 66.6666666667 V\n" + ] + } + ], + "prompt_number": 125 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.50, Page Number:1080" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=220.0#V\n", + "ra=0.5#ohm\n", + "ia=40.0#A\n", + "n=7\n", + "\n", + "#calculations\n", + "r1=v/ia\n", + "k=(r1/ra)**(1.0/(n-1))\n", + "r2=r1/k\n", + "r3=r2/k\n", + "r4=r3/k\n", + "r5=r4/k\n", + "r6=r5/k\n", + "p1=r1-r2\n", + "p2=r2-r3\n", + "p3=r3-r4\n", + "p4=r4-r5\n", + "p5=r5-r6\n", + "p6=r6-ra\n", + "\n", + "#result\n", + "print \"resistance of 1st section=\",round(p1,3),\"ohm\"\n", + "print \"resistance of 2nd section=\",round(p2,3),\"ohm\"\n", + "print \"resistance of 3rd section=\",round(p3,3),\"ohm\"\n", + "print \"resistance of 4th section=\",round(p4,3),\"ohm\"\n", + "print \"resistance of 5th section=\",round(p5,3),\"ohm\"\n", + "print \"resistance of 6th section=\",round(p6,3),\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resistance of 1st section= 1.812 ohm\n", + "resistance of 2nd section= 1.215 ohm\n", + "resistance of 3rd section= 0.815 ohm\n", + "resistance of 4th section= 0.546 ohm\n", + "resistance of 5th section= 0.366 ohm\n", + "resistance of 6th section= 0.246 ohm\n" + ] + } + ], + "prompt_number": 132 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.51, Page Number:1081" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n=6\n", + "load=3.73#kW\n", + "v=200#V\n", + "ratio=0.50\n", + "i1=0.6#A\n", + "efficiency=0.88\n", + "\n", + "#calculation\n", + "output=load/efficiency\n", + "total_loss=output-load\n", + "cu_loss=total_loss*ratio\n", + "i=output*1000/v\n", + "ia=i-i1\n", + "ra=cu_loss*1000/ia**2\n", + "i_per=i*2\n", + "ia_per=i_per-i1\n", + "r1=v/ia_per\n", + "k=(r1/ra)**(1.0/(n-1))\n", + "r2=r1/k\n", + "r3=r2/k\n", + "r4=r3/k\n", + "r5=r4/k\n", + "p1=r1-r2\n", + "p2=r2-r3\n", + "p3=r3-r4\n", + "p4=r4-r5\n", + "p5=r5-ra\n", + "\n", + "\n", + "#result\n", + "print \"resistance of 1st section=\",round(p1,3),\"ohm\"\n", + "print \"resistance of 2nd section=\",round(p2,3),\"ohm\"\n", + "print \"resistance of 3rd section=\",round(p3,3),\"ohm\"\n", + "print \"resistance of 4th section=\",round(p4,3),\"ohm\"\n", + "print \"resistance of 5th section=\",round(p5,3),\"ohm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resistance of 1st section= 1.627 ohm\n", + "resistance of 2nd section= 1.074 ohm\n", + "resistance of 3rd section= 0.709 ohm\n", + "resistance of 4th section= 0.468 ohm\n", + "resistance of 5th section= 0.309 ohm\n" + ] + } + ], + "prompt_number": 146 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.52, Page Number:1081" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n=7\n", + "load=36.775#kW\n", + "v=400#V\n", + "ratio=0.05\n", + "rsh=200#ohm\n", + "efficiency=0.92\n", + "\n", + "#calculation\n", + "input_m=load*1000/efficiency\n", + "cu_loss=input_m*ratio\n", + "cu_loss_sh=v**2/rsh\n", + "cu_loss_a=cu_loss-cu_loss_sh\n", + "i=input_m/v\n", + "ish=v/rsh\n", + "ia=i-ish\n", + "ra=cu_loss_a/ia**2\n", + "k=(v/(ia*ra))**(1.0/(n))\n", + "i1=k*ia\n", + "r1=v/i1\n", + "r2=r1/k\n", + "r3=r2/k\n", + "r4=r3/k\n", + "r5=r4/k\n", + "r6=r5/k\n", + "r7=r5/k\n", + "p1=r1-r2\n", + "p2=r2-r3\n", + "p3=r3-r4\n", + "p4=r4-r5\n", + "p5=r5-r6\n", + "p6=r6-r7\n", + "p7=r7-ra\n", + "\n", + "#result\n", + "print \"resistance of 1st section=\",round(p1,3),\"ohm\"\n", + "print \"resistance of 2nd section=\",round(p2,3),\"ohm\"\n", + "print \"resistance of 3rd section=\",round(p3,3),\"ohm\"\n", + "print \"resistance of 4th section=\",round(p4,3),\"ohm\"\n", + "print \"resistance of 5th section=\",round(p5,3),\"ohm\"\n", + "print \"resistance of 6th section=\",round(p6,3),\"ohm\"\n", + "print \"resistance of 7th section=\",round(p7,3),\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resistance of 1st section= 0.974 ohm\n", + "resistance of 2nd section= 0.592 ohm\n", + "resistance of 3rd section= 0.36 ohm\n", + "resistance of 4th section= 0.219 ohm\n", + "resistance of 5th section= 0.133 ohm\n", + "resistance of 6th section= 0.0 ohm\n", + "resistance of 7th section= 0.081 ohm\n" + ] + } + ], + "prompt_number": 157 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.53, Page Number:1082" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "n=Symbol('n')\n", + "v=250.0#V\n", + "ra=0.125#ohm\n", + "i2=150.0#A\n", + "i1=200.0#A\n", + "\n", + "#calculation\n", + "r1=v/i1\n", + "n=solve((i1/i2)**(n-1)-(r1/ra),n)\n", + "k=i1/i2\n", + "r2=r1/k\n", + "r3=r2/k\n", + "r4=r3/k\n", + "r5=r4/k\n", + "r6=r5/k\n", + "r7=r6/k\n", + "r8=r7/k\n", + "p1=r1-r2\n", + "p2=r2-r3\n", + "p3=r3-r4\n", + "p4=r4-r5\n", + "p5=r5-r6\n", + "p6=r6-r7\n", + "p7=r7-r8\n", + "p8=r8-ra\n", + "#result\n", + "print \"resistance of 1st section=\",round(p1,3),\"ohm\"\n", + "print \"resistance of 2nd section=\",round(p2,3),\"ohm\"\n", + "print \"resistance of 3rd section=\",round(p3,3),\"ohm\"\n", + "print \"resistance of 4th section=\",round(p4,3),\"ohm\"\n", + "print \"resistance of 5th section=\",round(p5,3),\"ohm\"\n", + "print \"resistance of 6th section=\",round(p6,3),\"ohm\"\n", + "print \"resistance of 7th section=\",round(p7,3),\"ohm\"\n", + "print \"resistance of 8th section=\",round(p8,3),\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resistance of 1st section= 0.313 ohm\n", + "resistance of 2nd section= 0.234 ohm\n", + "resistance of 3rd section= 0.176 ohm\n", + "resistance of 4th section= 0.132 ohm\n", + "resistance of 5th section= 0.099 ohm\n", + "resistance of 6th section= 0.074 ohm\n", + "resistance of 7th section= 0.056 ohm\n", + "resistance of 8th section= 0.042 ohm\n" + ] + } + ], + "prompt_number": 163 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.54, Page Number:1083" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "n=Symbol('n')\n", + "v=500#V\n", + "z=20\n", + "ra=1.31#ohm\n", + "t=218#N-m\n", + "ratio=1.5\n", + "slot=60\n", + "phi=23*0.001#Wb\n", + "\n", + "#calculation\n", + "ia=t/(0.159*phi*slot*z)\n", + "i1=ia*ratio\n", + "i2=ia\n", + "k=i1/i2\n", + "r1=v/i1\n", + "n=solve(k**(n-1)-(r1/ra),n)\n", + "r2=r1/k\n", + "r3=r2/k\n", + "r4=r3/k\n", + "p1=r1-r2\n", + "p2=r2-r3\n", + "p3=r3-r4\n", + "p4=r4-ra\n", + "\n", + "#result\n", + "print \"resistance of 1st section=\",round(p1,3),\"ohm\"\n", + "print \"resistance of 2nd section=\",round(p2,3),\"ohm\"\n", + "print \"resistance of 3rd section=\",round(p3,3),\"ohm\"\n", + "print \"resistance of 4th section=\",round(p4,3),\"ohm\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resistance of 1st section= 2.237 ohm\n", + "resistance of 2nd section= 1.491 ohm\n", + "resistance of 3rd section= 0.994 ohm\n", + "resistance of 4th section= 0.678 ohm\n" + ] + } + ], + "prompt_number": 164 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.55, Page Number:1084" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=37.3#kW\n", + "v=440#V\n", + "drop=0.02\n", + "efficiency=0.95\n", + "i_per=1.30\n", + "\n", + "#calculation\n", + "il=load*1000/(v*efficiency)\n", + "i1=i_per*il\n", + "vd=drop*v\n", + "rm=vd/il\n", + "r1=v/i1\n", + "r=(r1-rm)/6\n", + "\n", + "#result\n", + "print \"resistance of each rheostat=\",r,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resistance of each rheostat= 0.615721729566 ohm\n" + ] + } + ], + "prompt_number": 165 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 30.56, Page Number:1085" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=55.95#kW\n", + "v=650.0#V\n", + "r=0.51#ohm\n", + "i1=140.0#A\n", + "i2=100.0#A\n", + "per=0.20\n", + "\n", + "#calculation\n", + "ratio=i1/i2\n", + "r1=v/i1\n", + "r2=((per+1)/ratio-per)*r1\n", + "r3=(per+1)*r2/ratio-per*r1\n", + "r4=((per+1)*r3/ratio)-per*r1\n", + "\n", + "p1=r1-r2\n", + "p2=r2-r3\n", + "p3=r3-r4\n", + "\n", + "#result\n", + "print \"number of steps=\",3\n", + "print \"resistance of 1st section=\",round(p1,3),\"ohm\"\n", + "print \"resistance of 2nd section=\",round(p2,3),\"ohm\"\n", + "print \"resistance of 3rd section=\",round(p3,3),\"ohm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "number of steps= 3\n", + "resistance of 1st section= 1.592 ohm\n", + "resistance of 2nd section= 1.364 ohm\n", + "resistance of 3rd section= 1.17 ohm\n" + ] + } + ], + "prompt_number": 170 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter31_2.ipynb b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter31_2.ipynb new file mode 100644 index 00000000..88c66f5b --- /dev/null +++ b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter31_2.ipynb @@ -0,0 +1,935 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:02fdabadd118404eca71c942f203b8c36bfc89b9baf1e3f2f8e7065ab9807edb" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 31: Testing of DC Machines" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 31.1, Page Number:1092" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "l=38.1#kg\n", + "d=63.53*0.01#cm\n", + "v=12#rps\n", + "i=49#A\n", + "V=220#V\n", + "\n", + "#calculations\n", + "r=d/2\n", + "torque=l*r*9.81\n", + "power=torque*2*3.14*v\n", + "motor_input=i*V\n", + "efficiency=power*100/motor_input\n", + "\n", + "#result\n", + "print \"Output power=\",round(power),\"W\"\n", + "print \"Efficiency=\",round(efficiency),\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Output power= 8947.0 W\n", + "Efficiency= 83.0 %\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 31.2(a), Page Number:1093" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "spring_b1=10.0#kg\n", + "spring_b2=35.0#kg\n", + "d=40*0.01#m\n", + "v=950.0#rpm\n", + "V=200.0#V\n", + "i=30.0#A\n", + "\n", + "#calculations\n", + "F=(spring_b2-spring_b1)*9.81\n", + "N=v/60\n", + "R=d/2\n", + "tsh=F*R\n", + "omega=2*3.14*N\n", + "output=tsh*omega\n", + "motor_input=V*i\n", + "efficiency=output/motor_input\n", + "\n", + "#result\n", + "print \"output power=\",output,\"W\"\n", + "print \"efficiency=\",efficiency*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "output power= 4877.205 W\n", + "efficiency= 81.28675 %\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 31.2(b), Page Number:1093" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "t1=2.9#kg\n", + "t2=0.17#kg\n", + "r=7*0.01#m\n", + "i=2.0#A\n", + "V=230.0#V\n", + "n=1500.0#rpm\n", + "\n", + "#calculations\n", + "force=(t1-t2)*9.81\n", + "torque=force*r\n", + "output=torque*2*3.14*n/60\n", + "efficiency=output/(V*i)\n", + "\n", + "#result\n", + "print \"torque=\",torque,\"N-m\"\n", + "print \"output\",output,\"W\"\n", + "print \"efficiency\",efficiency*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "torque= 1.874691 N-m\n", + "output 294.326487 W\n", + "efficiency 63.984018913 %\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 31.3, Page Number:1095" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "V=220.0#V\n", + "i=2.5#A\n", + "ra=0.8#ohm\n", + "rsh=200.0#ohm\n", + "I=20.0#A\n", + "\n", + "#calculations\n", + "input_noload=V*i\n", + "ish=V/rsh\n", + "ia0=i-ish\n", + "culoss=ia0**2*ra\n", + "constant_loss=input_noload-culoss\n", + "ia=32-ish\n", + "cu_lossa=ia**2*ra\n", + "total_loss=cu_lossa+constant_loss\n", + "input_=V*I\n", + "output=input_-total_loss\n", + "efficiency=(output/input_)*100\n", + "\n", + "#result\n", + "print \"Efficiency=\",efficiency,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Efficiency= 70.1754545455 %\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 31.4, Page Number:1096" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "V=400.0#V\n", + "i=5.0#A\n", + "ra=0.5#ohm\n", + "r=200.0#ohm\n", + "I=50.0#A\n", + "\n", + "#calculations\n", + "input_nl=V*i\n", + "ish=V/r\n", + "ia=i-ish\n", + "cu_loss=ia**2*ra\n", + "constant_loss=input_nl-cu_loss\n", + "Ia=I-ish\n", + "cu_lossa=Ia**2*ra\n", + "total_loss=constant_loss+cu_lossa\n", + "input_nl1=V*I\n", + "output=input_nl1-total_loss\n", + "efficiency=output/input_nl\n", + "Eb1=V-(ia*ra)\n", + "Eb2=V-(Ia*ra)\n", + "change=math.fabs((Eb1-Eb2)/Eb1)\n", + "\n", + "#result\n", + "print \"output=\",output,\"W\"\n", + "print \"efficiency=\",efficiency*10,\"%\"\n", + "print \"percentage change in speed=\",change*100,\"%\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "output= 16852.5 W\n", + "efficiency= 84.2625 %\n", + "percentage change in speed= 5.64617314931 %\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 31.8, Page Number:1098" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=200*1000.0#W\n", + "v=250.0#V\n", + "i1=36.0#A\n", + "I1=12.0#A\n", + "v1=250.0#V\n", + "pd=6.0#V\n", + "i2=400.0#A\n", + "\n", + "#calculations\n", + "#no load\n", + "ia=i1-I1\n", + "ra=pd/i2\n", + "cu_loss=ia**2*ra\n", + "input_nl=v*i1\n", + "constant_loss=input_nl-cu_loss\n", + "\n", + "#full load\n", + "output_i=p/v\n", + "ia=output_i+I1\n", + "cu_lossa=ia**2*ra\n", + "total_loss=cu_lossa+constant_loss\n", + "efficiency=p/(p+total_loss)\n", + "#result\n", + "print \"efficiency at full load=\",efficiency*100,\"%\"\n", + "\n", + "#half load\n", + "output_i=p/(2*v)\n", + "ia=output_i+I1\n", + "cu_lossa=ia**2*ra\n", + "total_loss=cu_lossa+constant_loss\n", + "efficiency=p/((p/2+total_loss)*2)\n", + "\n", + "#result\n", + "print \"efficiency at half load=\",efficiency*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "efficiency at full load= 91.3736344667 %\n", + "efficiency at half load= 89.6559292335 %\n" + ] + } + ], + "prompt_number": 42 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 31.9, Page Number:1098" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=250.0#V\n", + "p=14.92*1000#W\n", + "e=0.88\n", + "n=700.0#rpn\n", + "rsh=100.0#ohm\n", + "i=78.0#A\n", + "\n", + "#calculations\n", + "input_=0.8*p/e\n", + "total_loss=input_-0.8*p\n", + "input_i=input_/v\n", + "ish=v/rsh\n", + "ia=input_i-ish\n", + "ra=total_loss/(2*(ia**2))\n", + "Ia=i-ish\n", + "total_loss2=Ia**2*ra+total_loss/2\n", + "input__=v*i\n", + "efficiency=(input__-total_loss2)*100/input__\n", + "Eb1=v-(ia*ra)\n", + "Eb2=v-(Ia*ra)\n", + "n2=(n*Eb2)/Eb1\n", + "\n", + "#result\n", + "print \"efficiency=\",efficiency,\"%\"\n", + "print \"speed=\",n2,\"r.p.m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "efficiency= 86.9450046554 %\n", + "speed= 678.443304738 r.p.m\n" + ] + } + ], + "prompt_number": 48 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 31.10(a), Page Number:1101" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=220.0#V\n", + "p=100*1000.0#W\n", + "i2=90.0#A\n", + "\n", + "#calculations\n", + "i1=p/v\n", + "efficiency=math.sqrt(i1/(i1+i2))*100\n", + "\n", + "#result\n", + "print \"efficiency=\",round(efficiency,1),\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "efficiency= 91.4 %\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 31.11, Page Number:1102" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "i=15#A\n", + "v=200#V\n", + "motor_i=100#A\n", + "shunt_i1=3#A\n", + "shunt_i2=2.5#A\n", + "ra=0.05#ohm\n", + "cu_loss=500#W\n", + "cu_lossa=361#W\n", + "ia=85#A\n", + "#calculations\n", + "mech_core_stray_loss=0.5*((v*i)-(motor_i**2*ra)-(ia**2*ra))\n", + "cu_motor=v*shunt_i1\n", + "generator_motor=v*shunt_i2\n", + "total_loss=mech_core_stray_loss+cu_motor+generator_motor\n", + "input_=v*i+cu_motor\n", + "output=v*ia*10**(-3)\n", + "loss=cu_loss*10**(-3)+1.07+0.36\n", + "efficiency=output*100/(output+loss)\n", + "\n", + "#result\n", + "print \"eficiency=\",efficiency,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "eficiency= 89.8045430534 %\n" + ] + } + ], + "prompt_number": 52 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 31.12, Page Number:1103" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=110#V\n", + "i=48#A\n", + "i1=3#a\n", + "i2=3.5#A\n", + "motor_i=230#A\n", + "ra=0.035#ohm\n", + "\n", + "#calculations\n", + "#motor\n", + "cu_loss=motor_i**2*ra\n", + "brush_loss=motor_i*2\n", + "totalarm_culoss=cu_loss+brush_loss\n", + "shunt_cu=v*i1\n", + "total_cu_lossm=totalarm_culoss+shunt_cu\n", + "#generator\n", + "arm_i=233-i+i2\n", + "cu_loss=arm_i**2*ra\n", + "brush_loss=arm_i*2\n", + "totalarm_culoss=cu_loss+brush_loss\n", + "shunt_cu=v*i2\n", + "total_cu_lossg=totalarm_culoss+shunt_cu\n", + "#set\n", + "totalcu_loss=total_cu_lossm+total_cu_lossg\n", + "total_input=v*i\n", + "stray_loss=total_input-totalcu_loss\n", + "strayloss_per=stray_loss/2\n", + "#motor efficiency\n", + "input_=233*v\n", + "output=input_-(total_cu_lossm+strayloss_per)\n", + "e=output/input_*100\n", + "print \"motor efficiency=\",e,\"%\"\n", + "#generator efficiency\n", + "input_=110*185\n", + "output=input_-(total_cu_lossg+strayloss_per)\n", + "e=output/input_*100\n", + "100\n", + "print \"generator efficiency=\",e,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "motor efficiency= 88.4590884705 %\n", + "generator efficiency= 88.5893642506 %\n" + ] + } + ], + "prompt_number": 56 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 31.13, Page Number:1103" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable series\n", + "v=500.0#A\n", + "p=100*1000.0#w\n", + "auxiliary_i=30.0#A\n", + "output_i=200.0#A\n", + "i1=3.5#A\n", + "i2=1.8#A\n", + "ra=0.075#ohm\n", + "vdb=2.0#V\n", + "\n", + "#calculations\n", + "motor_arm=output_i+auxiliary_i\n", + "motorarm_culoss=(motor_arm**2*ra)+(motor_arm*2)\n", + "motorfield_culoss=v*i2\n", + "generatorarm_culoss=(output_i**2*ra)+(output_i*2)\n", + "generatoefield_culoss=v*i1\n", + "total_culoss=motorarm_culoss+motorfield_culoss+generatorarm_culoss+generatoefield_culoss\n", + "power=v*auxiliary_i\n", + "stray_loss=power-total_culoss\n", + "permachine=stray_loss/2\n", + "total_loss=generatorarm_culoss+generatoefield_culoss+permachine\n", + "output=v*output_i\n", + "e=output/(output+total_loss)\n", + "\n", + "#result\n", + "print \"efficiency=\",e*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "efficiency= 93.1001175389 %\n" + ] + } + ], + "prompt_number": 58 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 31.14, Page Number:1104" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=250.0#V\n", + "i=50.0#A\n", + "motor_i=400.0#A\n", + "i1=6.0#A\n", + "i2=5.0#A\n", + "ra=0.015#ohm\n", + "\n", + "#calculations\n", + "motora_culoss=motor_i**2*ra\n", + "generatora_culoss=(motor_i-i)**2*ra\n", + "power=v*i\n", + "stray_loss=power-(motora_culoss+generatora_culoss)\n", + "permachine=stray_loss/2\n", + "#motor\n", + "total_motor_loss=motora_culoss+(v*i2)+permachine\n", + "motor_input=(v*motor_i)+v*i2\n", + "motor_e=(motor_input-total_motor_loss)/motor_input\n", + "\n", + "#generator\n", + "total_gen_loss=generatora_culoss+(v*i1)+permachine\n", + "gen_output=v*(motor_i-i)\n", + "gen_e=(gen_output-total_gen_loss)/gen_output\n", + "\n", + "#result\n", + "print \"motor efficiency=\",motor_e*100,\"%\"\n", + "print \"generator efficiency\",gen_e*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "motor efficiency= 92.3148148148 %\n", + "generator efficiency 91.4642857143 %\n" + ] + } + ], + "prompt_number": 77 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 31.15, Page Number:1105" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=250.0#V\n", + "i=50.0#A\n", + "ia=380.0#A\n", + "i1=5.0#A\n", + "i2=4.2#A\n", + "ra=0.2#ohm\n", + "\n", + "#calculations\n", + "motora_culoss=ia**2*ra\n", + "generatora_culoss=(ia-i)**2*ra\n", + "power=v*i\n", + "stray_loss=power-(motora_culoss+generatora_culoss)\n", + "permachine=stray_loss/2\n", + "#motor\n", + "total_motor_loss=motora_culoss+(v*i2)+permachine\n", + "motor_input=(v*ia)+v*i2\n", + "motor_e=(motor_input-total_motor_loss)/motor_input\n", + "\n", + "#generator\n", + "total_gen_loss=generatora_culoss+(v*i1)+permachine\n", + "gen_output=v*(ia-i)\n", + "gen_e=(gen_output-total_gen_loss)/gen_output\n", + "\n", + "#result\n", + "print \"motor efficiency=\",motor_e*100,\"%\"\n", + "print \"generator efficiency\",gen_e*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "motor efficiency= 88.7038001041 %\n", + "generator efficiency 95.2121212121 %\n" + ] + } + ], + "prompt_number": 81 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 31.16, Page Number:1107" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=220.0#V\n", + "v2=190.0#V\n", + "t=30#sec\n", + "t2=20#sec\n", + "i=20.0#A\n", + "\n", + "#calculations\n", + "avg_v=(v+v2)/2\n", + "avg_i=i/2\n", + "power=avg_v*avg_i\n", + "W=power*(t2/(t-t2))\n", + "\n", + "#result\n", + "print \"Stray loss=\",W,\"W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Stray loss= 4100.0 W\n" + ] + } + ], + "prompt_number": 85 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 31.17, Page Number:1107" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variabledeclaration\n", + "n1=1525.0#rpm\n", + "n2=1475.0#ohm\n", + "dt=25.0#sec\n", + "p=1000.0#W\n", + "t2=20.0#sec\n", + "\n", + "#calculations\n", + "N=(n1+n2)/2\n", + "w=p*(t2/(dt-t2))\n", + "dN=n1-n2\n", + "I=(w*dt)/((2*3.14/60)**2*N*dN)\n", + "\n", + "#result\n", + "print \"Moment of Inertia=\",I,\"kg-m2\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Moment of Inertia= 121.708791432 kg-m2\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 31.18, Page Number:1108" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=240.0#V\n", + "v2=225.0#V\n", + "dt=25.0#sec\n", + "t2=6.0#ohm\n", + "iavg=10.0#A\n", + "i2=25.0#A\n", + "v3=250.0#V\n", + "ra=0.4#ohm\n", + "r=250.0#ohm\n", + "\n", + "#calculations\n", + "avg_v=(v+v2)/2\n", + "w_=avg_v*iavg\n", + "W=w_*(t2/(dt-t2))\n", + "ish=v3/r\n", + "ia=i2-ish\n", + "cu_loss=ia**2*ra\n", + "cu_shunt=v3*ia\n", + "total_loss=W+cu_loss+v3\n", + "e=((v*i2)-total_loss)/(v*i2)\n", + "\n", + "#result\n", + "print \"efficiency=\",e*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "734.210526316\n", + "efficiency= 79.7564912281 %\n" + ] + } + ], + "prompt_number": 97 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 31.19, Page Number:1108" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n=1000#rpm\n", + "n1=1030#rpm\n", + "n2=970#rpm\n", + "t1=36#sec\n", + "t2=15#sec\n", + "t3=9#sec\n", + "i=10#A\n", + "v=219#V\n", + "\n", + "#calculations\n", + "W=v*i*(t2/(dt-t2))\n", + "dN=n1-n2\n", + "I=(W*t2)/((2*3.14/60)**2*n*dN)\n", + "Wm=W*t2/t1\n", + "iron_loss=W-Wm\n", + "\n", + "#result\n", + "print \"i)moment of inertia=\",I,\"kg.m2\"\n", + "print \"ii)iron loss=\",iron_loss,\"W\"\n", + "print \"iii)mechanical losses=\",Wm,\"W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)moment of inertia= 74.9650087225 kg.m2\n", + "ii)iron loss= 1916.25 W\n", + "iii)mechanical losses= 1368.75 W\n" + ] + } + ], + "prompt_number": 99 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 31.20, Page Number:1110" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "iam=56.0#A\n", + "vam=590.0#V\n", + "vdm=40.0#V\n", + "iag=44.0#A\n", + "vag=400.0#V\n", + "vdg=40.0#V\n", + "r=0.3#ohm\n", + "\n", + "#calculations\n", + "input_total=(vdm+vam)*iam\n", + "output=vag*iag\n", + "total_loss=input_total-output\n", + "rse=vdg/iam\n", + "cu_loss=((r+2*rse)*iam**2)+(iag**2*r)\n", + "strayloss=total_loss-cu_loss\n", + "permachine=strayloss/2\n", + "#motor\n", + "inputm=vam*iam\n", + "culossm=(r+rse)*iam**2\n", + "totallossm=culossm+permachine\n", + "output=inputm-totallossm\n", + "em=output*100/inputm\n", + "#generator\n", + "inputg=vag*iag\n", + "culossg=(r)*iag**2\n", + "totalloss=culossg+permachine+(vdm*iam)\n", + "output=vag*iag\n", + "eg=output*100/(output+totalloss)\n", + "\n", + "print \n", + "#result\n", + "print \"motor efficiency=\",em,\"%\"\n", + "print \"generator efficiency=\",eg,\"%\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "motor efficiency= 72.6997578692 %\n", + "generator efficiency= 67.0220868241 %\n" + ] + } + ], + "prompt_number": 115 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter32_2.ipynb b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter32_2.ipynb new file mode 100644 index 00000000..a29de087 --- /dev/null +++ b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter32_2.ipynb @@ -0,0 +1,5311 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:69b299b5398cdb7b833f53d6a7d05a19c0a433537449ffb871db80e61817fe5c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 32: Transformer" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.1, Page Number:1123" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v1=250.0#V\n", + "v2=3000.0#V\n", + "f=50.0#Hz\n", + "phi=1.2#Wb-m2\n", + "e=8.0#V\n", + "\n", + "#calculations\n", + "n1=v1/e\n", + "n2=v2/e\n", + "a=v2/(4.44*f*n2*phi)\n", + "\n", + "#result\n", + "print \"primary turns=\",n1\n", + "print \"secondary turns=\",n2\n", + "print \"area of core=\",round(a,2),\"m2\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "primary turns= 31.25\n", + "secondary turns= 375.0\n", + "area of core= 0.03 m2\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.2, Page Number:1123" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=100#KVA\n", + "v1=11000#V\n", + "v2=550#V\n", + "f=50#Hz\n", + "bm=1.3#Tesla\n", + "sf=0.9\n", + "per=10#%\n", + "a=20*20*sf/10000#m2\n", + "\n", + "#calculation\n", + "n1=v1/(4.44*f*bm*a)\n", + "n2=v2/(4.44*f*bm*a)\n", + "e_per_turn=v1/n1\n", + "\n", + "#result\n", + "print \"HV TURNS=\",round(n1)\n", + "print \"LV TURNS=\",round(n2)\n", + "print \"EMF per turns=\",round(e_per_turn,1),\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "HV TURNS= 1059.0\n", + "LV TURNS= 53.0\n", + "EMF per turns= 10.4 V\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.3, Page Number:1123" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n1=400.0\n", + "n2=1000.0\n", + "a=60.0/10000.0#cm2\n", + "f=50.0#Hz\n", + "e1=520.0#V\n", + "\n", + "#calculations\n", + "k=n2/n1\n", + "e2=k*e1\n", + "bm=e1/(4.44*f*n1*a)\n", + "\n", + "#result\n", + "print \"peak value of flux density=\",bm,\"WB/m2\"\n", + "print \"voltage induced in the secondary winding=\",e2,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "peak value of flux density= 0.975975975976 WB/m2\n", + "voltage induced in the secondary winding= 1300.0 V\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.4, Page Number:1124" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=25.0#kVA\n", + "n1=500.0\n", + "n2=50.0\n", + "v=3000.0#V\n", + "f=50.0#Hz\n", + "\n", + "#calculations\n", + "k=n2/n1\n", + "i1=load*1000/v\n", + "i2=i1/k\n", + "e1=v/n1\n", + "e2=e1*n2\n", + "phim=v/(4.44*f*n1)\n", + "\n", + "#result\n", + "print \"primary and secondary currents=\",i1,\"A\", i2,\"A\"\n", + "print \"secondary emf=\",e2,\"V\"\n", + "print \"flux=\",phim*1000,\"mWB\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "primary and secondary currents= 8.33333333333 A 83.3333333333 A\n", + "secondary emf= 300.0 V\n", + "flux= 27.027027027 mWB\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.5, Page Number:1123" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "f=50#Hz\n", + "v1=11000#V\n", + "v2=550#V\n", + "load=300#kVA\n", + "phim=0.05#Wb\n", + "\n", + "#calculation\n", + "e=4.44*f*phim\n", + "e2=v2/1.732\n", + "t1=v1/e\n", + "t2=e2/e\n", + "output=load/3\n", + "HV=100*1000/v1\n", + "LV=100*1000/e2\n", + "\n", + "#result\n", + "print \"HV turns=\",t1\n", + "print \"LV turns=\",t2\n", + "print \"emf per turn=\",e2\n", + "print \"full load HV=\",HV\n", + "print \"full load LV=\",LV" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "HV turns= 990.990990991\n", + "LV turns= 28.6082849593\n", + "emf per turn= 317.551963048\n", + "full load HV= 9\n", + "full load LV= 314.909090909\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.6, Page Number:1124" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n1=500.0\n", + "n2=1200.0\n", + "a=80.0/10000.0#m2\n", + "f=50.0#Hz\n", + "v=500.0#V\n", + "\n", + "#calculation\n", + "phim=n1/(4.44*f*n1)\n", + "bm=phim/a\n", + "v2=n2*v/n1\n", + "\n", + "#result\n", + "print \"peak flux-density=\",bm,\"Wb\"\n", + "print \"voltage induced in the secondary=\",v2,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "peak flux-density= 0.563063063063 Wb\n", + "voltage induced in the secondary= 1200.0 V\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.7, Page Number:1125" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#varible declaration\n", + "load=25.0#kVA\n", + "n1=250.0\n", + "n2=40.0\n", + "v=1500.0#V\n", + "f=50.0#Hz\n", + "\n", + "#calculation\n", + "v2=n2*v/n1\n", + "i1=load*1000/v\n", + "i2=load*1000/v2\n", + "phim=v/(4.44*f*n1)\n", + "\n", + "#result\n", + "print \"i)primary current an secondary current=\",i1,\"A\",i2,\"A\"\n", + "print \"ii)seconary emf=\",v2,\"V\"\n", + "print \"iii)maximum flux=\",phim*1000,\"mWb\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)primary current an secondary current= 16.6666666667 A 104.166666667 A\n", + "ii)seconary emf= 240.0 V\n", + "iii)maximum flux= 27.027027027 mWb\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.8, Page Number:1125" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "f=50.0#Hz\n", + "a=20.0*20.0/10000#m2\n", + "phim=1.0#Wbm2\n", + "v1=3000.0#V\n", + "v2=220.0#V\n", + "\n", + "#calculation\n", + "t2=v2/(4.44*f*phim*a)\n", + "t1=t2*v1/v2\n", + "n1=t1/2\n", + "n2=t2/2\n", + "\n", + "#result\n", + "print \"HV turns=\",n1\n", + "print \"LV turns=\",n2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "HV turns= 168.918918919\n", + "LV turns= 12.3873873874\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.9, Page Number:1126" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v1=2200.0#V\n", + "v2=200.0#V\n", + "i1=0.6#A\n", + "p=400.0#W\n", + "v3=250.0#V\n", + "i0=0.5#A\n", + "pf=0.3\n", + "\n", + "#calculation\n", + "il=p/v1\n", + "imu=(i1**2-il**2)**0.5\n", + "iw=i0*pf\n", + "imu2=(i0**2-iw**2)**0.5\n", + "\n", + "#result\n", + "print \"magnetising currents=\",imu,\"A\"\n", + "print \"iron loss current=\",il,\"A\"\n", + "print \"magnetising components of no load primary current=\",imu2,\"A\"\n", + "print \"working components of no-load primary current=\",iw,\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "magnetising currents= 0.571788552492 A\n", + "iron loss current= 0.181818181818 A\n", + "magnetising components of no load primary current= 0.476969600708 A\n", + "working components of no-load primary current= 0.15 A\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.10, Page Number:1127" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n1=500.0\n", + "n2=40.0\n", + "l=150.0#cm\n", + "airgap=0.1#mm\n", + "e1=3000.0#V\n", + "phim=1.2#Wb/m2\n", + "f=50.0#Hz\n", + "d=7.8#grma/cm3\n", + "loss=2.0#watt/kg\n", + "\n", + "#calculation\n", + "a=e1/(4.44*f*n1*phim)\n", + "k=n2/n1\n", + "v2=k*e1\n", + "iron=l*5\n", + "air=phim*airgap/(1000*4*3.14*10**(-7))\n", + "bmax=iron+air\n", + "imu=bmax/(n1*2**0.5)\n", + "volume=l*a\n", + "im=volume*d*10\n", + "total_i=im*2\n", + "iw=total_i/(e1)\n", + "i0=(imu**2+iw**2)**0.5\n", + "pf=iw/i0\n", + "\n", + "#result\n", + "print \"a)cross sectional area=\",a*10000,\"cm2\"\n", + "print \"b)no load secondary voltage=\",v2,\"V\"\n", + "print \"c)no load current=\",imu,\"A\"\n", + "print \"d)power factor=\",pf" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a)cross sectional area= 225.225225225 cm2\n", + "b)no load secondary voltage= 240.0 V\n", + "c)no load current= 1.19577611723 A\n", + "d)power factor= 0.145353269536\n" + ] + } + ], + "prompt_number": 42 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.11, Page Number:1127" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "n1=1000\n", + "n2=200\n", + "i=3#A\n", + "pf=0.2\n", + "i2=280#A\n", + "pf2=0.8\n", + "\n", + "#calculations\n", + "phi1=math.acos(pf2)\n", + "i2_=i2/5\n", + "phi2=math.acos(pf)\n", + "sinphi=math.sin(phi2)\n", + "sinphi2=math.sin(math.acos(phi1))\n", + "i1=i*complex(pf,-sinphi)+i2_*complex(pf2,-sinphi2)\n", + "\n", + "#result\n", + "print \"primary current=\",abs(i1),\"/_\",math.degrees(phi1),\"degrees\"\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "primary current= 64.4918252531 /_ 36.8698976458 degrees\n" + ] + } + ], + "prompt_number": 51 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.12, Page Number:1130" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v1=440.0#v\n", + "v2=110.0#V\n", + "i0=5.0#A\n", + "pf=0.2\n", + "i2=120.0#A\n", + "pf2=0.8\n", + "\n", + "#calculation\n", + "phi2=math.acos(pf2)\n", + "phi0=math.acos(pf)\n", + "k=v2/v1\n", + "i2_=k*i2\n", + "angle=phi2-phi0\n", + "i1=(i0**2+i2_**2+(2*i0*i2_*math.cos(angle)))**0.5\n", + "\n", + "#result\n", + "print \"current taken by the primary=\",i1,\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "current taken by the primary= 33.9022604184 A\n" + ] + } + ], + "prompt_number": 53 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.13, Page Number:1130" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n1=800.0\n", + "n2=200.0\n", + "pf=0.8\n", + "i1=25.0#A\n", + "pf2=0.707\n", + "i2=80.0#A\n", + "#calculations\n", + "k=n2/n1\n", + "i2_=i2*k\n", + "phi2=math.acos(pf)\n", + "phi1=math.acos(pf2)\n", + "i0pf2=i1*pf2-i2_*pf\n", + "i0sinphi=i1*pf2-i2_*math.sin(math.acos(pf))\n", + "phi0=math.atan(i0sinphi/i0pf2)\n", + "i0=i0sinphi/math.sin(phi0)\n", + "\n", + "#result\n", + "print \"no load current=\",i0,\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "no load current= 5.91703050525 A\n" + ] + } + ], + "prompt_number": 59 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.14, Page Number:1131" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "i=10#A\n", + "pf=0.2\n", + "ratio=4\n", + "i2=200#A\n", + "pf=0.85\n", + "\n", + "#calculations\n", + "phi0=math.acos(pf)\n", + "phil=math.acos(pf)\n", + "i0=complex(2,-9.8)\n", + "i2_=complex(42.5,-26.35)\n", + "i1=i0+i2_\n", + "phi=math.acos(i1.real/57.333)\n", + "\n", + "#result\n", + "print \"primary current=\",i1,\"A\"\n", + "print \"power factor=\",math.degrees(phi),\"degrees\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "primary current= (44.5-36.15j) A\n", + "power factor= 39.0890154959 degrees\n" + ] + } + ], + "prompt_number": 60 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.15, Page Number:1136" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable decaration\n", + "load=30.0#KVA\n", + "v1=2400.0#V\n", + "v2=120.0#V\n", + "f=50.0#Hz\n", + "r1=0.1#ohm\n", + "x1=0.22#ohm\n", + "r2=0.034#ohm\n", + "x2=0.012#ohm\n", + "\n", + "#calculations\n", + "k=v2/v1\n", + "r01=r1+r2/k**2\n", + "x01=x1+x2/k**2\n", + "z01=(r01**2+x01**2)**0.5\n", + "r02=r2+r1*k**2\n", + "x02=x2+x1*k**2\n", + "z02=(r02**2+x02**2)**0.5\n", + "\n", + "#result\n", + "print \"high voltage side:\"\n", + "print \"equivalent winding resistance=\",r01,\"ohm\"\n", + "print \"reactance=\",x01,\"ohm\"\n", + "print \"impedence=\",z01,\"ohm\"\n", + "print \"low voltage side:\"\n", + "print \"equivalent winding resistance=\",r02,\"ohm\"\n", + "print \"reactance=\",x02,\"ohm\"\n", + "print \"impedence=\",z02,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "high voltage side:\n", + "equivalent winding resistance= 13.7 ohm\n", + "reactance= 5.02 ohm\n", + "impedence= 14.5907642021 ohm\n", + "low voltage side:\n", + "equivalent winding resistance= 0.03425 ohm\n", + "reactance= 0.01255 ohm\n", + "impedence= 0.0364769105051 ohm\n" + ] + } + ], + "prompt_number": 64 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.16, Page Number:1136" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=50.0#KVA\n", + "v1=4400.0#V\n", + "v2=220.0#V\n", + "r1=3.45#ohm\n", + "r2=0.009#ohm\n", + "x1=5.2#ohm\n", + "x2=0.015#ohm\n", + "\n", + "#calculations\n", + "i1=load*1000/v1\n", + "i2=load*1000/v2\n", + "k=v2/v1\n", + "r01=r1+r2/k**2\n", + "r02=r2+k**2*r1\n", + "x01=x1+x2/k**2\n", + "x02=x2+x1*k**2\n", + "z01=(r01**2+x01**2)**0.5\n", + "z02=(r02**2+x02**2)**0.5\n", + "cu_loss=i1**2*r01\n", + "\n", + "#result\n", + "print \"i)resistance=\"\n", + "print \"primary=\",r01,\"ohm\"\n", + "print \"secondary=\",r02,\"ohm\"\n", + "print \"iii)reactance=\"\n", + "print \"primary=\",x01,\"ohm\"\n", + "print \"secondary=\",x02,\"ohm\"\n", + "print \"iv)impedence=\"\n", + "print \"primary=\",z01,\"ohm\"\n", + "print \"secondary=\",z02,\"ohm\"\n", + "print \"v)copper loss=\",cu_loss,\"W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resistance=\n", + "primary= 7.05 ohm\n", + "secondary= 0.017625 ohm\n", + "reactance=\n", + "primary= 11.2 ohm\n", + "secondary= 0.028 ohm\n", + "impedence=\n", + "primary= 13.2341414531 ohm\n", + "secondary= 0.0330853536327 ohm\n", + "copper loss= 910.382231405 W\n" + ] + } + ], + "prompt_number": 68 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.17, Page Number:1137" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "ratio=10.0\n", + "load=50.0#KVA\n", + "v1=2400.0#V\n", + "v2=240.0#V\n", + "f=50.0#Hz\n", + "v=240.0#V\n", + "\n", + "#calculation\n", + "i2=load*1000/v\n", + "z2=v/(i2)\n", + "k=v2/v1\n", + "z2_=z2/k**2\n", + "i2_=k*i2\n", + "\n", + "#result\n", + "print \"a)load impedence=\",z2,\"ohm\"\n", + "print \"b)impedence referred to high tension side=\",z2_,\"ohm\"\n", + "print \"c)the value of current referred to the high tension side=\",i2_,\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a)load impedence= 1.152 ohm\n", + "b)impedence referred to high tension side= 115.2 ohm\n", + "c)the value of current referred to the high tension side= 20.8333333333 A\n" + ] + } + ], + "prompt_number": 70 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.18, Page Number:1137" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=100.0#kVA\n", + "v1=11000.0#V\n", + "v2=317.0#V\n", + "load2=0.62#kW\n", + "lvload=0.48#kW\n", + "\n", + "#calculations\n", + "k=v1/v2\n", + "i1=load*1000/v1\n", + "i2=load*1000/v2\n", + "r1=load2*1000/i**2\n", + "r2=lvload*1000/i2**2\n", + "r2_=r2*k**2\n", + "x01=4*v1/(i1*100)\n", + "x2_=x01*r2_/(r1+r2_)\n", + "x1=x01-x2_\n", + "x2=x2_*10/k**2\n", + "\n", + "#result\n", + "print \"i)r1=\",r1,\"ohm\"\n", + "print \"r2=\",r2,\"ohm\"\n", + "print \"r2_=\",r2_,\"ohm\"\n", + "print \"ii)reactance=\",x01,\"ohm\"\n", + "print \"x1=\",x1,\"ohm\"\n", + "print \"x2=\",x2,\"ohm\"\n", + "print \"x2_=\",x2_,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)r1= 7.502 ohm\n", + "r2= 0.004823472 ohm\n", + "r2_= 5.808 ohm\n", + "ii)reactance= 48.4 ohm\n", + "x1= 27.28 ohm\n", + "x2= 0.175398981818 ohm\n", + "x2_= 21.12 ohm\n" + ] + } + ], + "prompt_number": 76 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.19, Page Number:1137" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declarations\n", + "k=19.5\n", + "r1=25.0#ohm\n", + "x1=100.0#ohm\n", + "r2=0.06#ohm\n", + "x2=0.25#ohm\n", + "i=1.25#A\n", + "angle=30#degrees\n", + "i2=200#A\n", + "v=50#V\n", + "pf2=0.8\n", + "\n", + "#calculations\n", + "v2=complex(500,0)\n", + "i2=i2*complex(0.8,-0.6)\n", + "z2=complex(r2,x2)\n", + "e2=v2+i2*z2\n", + "beta=math.atan(e2.imag/e2.real)\n", + "e1=e2*k\n", + "i2_=i2/k\n", + "angle=beta+math.radians(90)+math.radians(angle)\n", + "i0=i*complex(math.cos(angle),math.sin(angle))\n", + "i1=-i2_+i0\n", + "v2=-e1+i1*complex(r1,x1)\n", + "phi=math.atan(v2.imag/v2.real)-math.atan(i1.imag/i1.real)\n", + "pf=math.cos(phi)\n", + "power=abs(v2)*i*math.cos(math.radians(60))\n", + "r02=r2+r1/k**2\n", + "cu_loss=abs(i2)**2*r02\n", + "output=500*abs(i2)*pf2\n", + "loss=cu_loss+power\n", + "inpt=output+loss\n", + "efficiency=output*100/inpt\n", + "\n", + "#result\n", + "print \"primary applied voltage=\",v2,\"V\"\n", + "print \"primary pf=\",pf\n", + "print \"efficiency=\",efficiency,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "primary applied voltage= (-11464.2126901-1349.15424294j) V\n", + "primary pf= 0.698572087114\n", + "efficiency= 86.7261056254 %\n" + ] + } + ], + "prompt_number": 94 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.20, Page Number:1138" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable description\n", + "load=100#KVA\n", + "v1=1100#V\n", + "v2=220#V\n", + "f=50#Hz\n", + "zh=complex(0.1,0.4)\n", + "zl=complex(0.006,0.015)\n", + "\n", + "#calculations\n", + "k=v1/v2\n", + "#HV \n", + "r1=zh.real+zl.real*k**2\n", + "x1=zh.imag+zl.imag*k**2\n", + "z1=(r1**2+x1**2)**0.5\n", + "#LV\n", + "r2=r1/k**2\n", + "x2=x1/k**2\n", + "z2=z1/k**2\n", + "\n", + "#result\n", + "print \"HV:\"\n", + "print \"resistance=\",r1,\"ohm\"\n", + "print \"reactance=\",x1,\"ohm\"\n", + "print \"impedence=\",z1,\"ohm\"\n", + "print \"LV:\"\n", + "print \"resistance=\",r2,\"ohm\"\n", + "print \"reactance=\",x2,\"ohm\"\n", + "print \"impedence=\",z2,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "HV:\n", + "resistance= 0.25 ohm\n", + "reactance= 0.775 ohm\n", + "impedence= 0.814324873745 ohm\n", + "LV:\n", + "resistance= 0.01 ohm\n", + "reactance= 0.031 ohm\n", + "impedence= 0.0325729949498 ohm\n" + ] + } + ], + "prompt_number": 96 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.21, Page Number:1141" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v1=230#V\n", + "v2=460#V\n", + "r1=0.2#ohm\n", + "x1=0.5#ohm\n", + "r2=0.75#ohm\n", + "x2=1.8#ohm\n", + "i=10#A\n", + "pf=0.8\n", + "\n", + "#calculation\n", + "k=v2/v1\n", + "r02=r2+k**2*r1\n", + "x02=x2+k**2*x1\n", + "vd=i*(r02*pf+x02*math.sin(math.acos(pf)))\n", + "vt2=v2-vd\n", + "\n", + "#result\n", + "print \"secondary terminal voltage=\",vt2,\"V\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "secondary terminal voltage= 424.8 V\n" + ] + } + ], + "prompt_number": 97 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.22, Page Number:1141" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "r=1.0#%\n", + "x=5.0#%\n", + "pf=0.8\n", + "\n", + "#calculation\n", + "mu=r*pf+x*math.sin(math.acos(pf))\n", + "mu2=r**2+x*0\n", + "mu3=r*pf-x*math.sin(math.acos(pf))\n", + "\n", + "#result\n", + "print \"regulation at pf=0.8 lag:\",mu,\"%\"\n", + "print \"regulation at pf=1:\",mu2,\"%\"\n", + "print \"regulation at pf=0.8 lead:\",mu3,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "regulation at pf=0.8 lag: 3.8 %\n", + "regulation at pf=1: 1.0 %\n", + "regulation at pf=0.8 lead: -2.2 %\n" + ] + } + ], + "prompt_number": 98 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.23, Page Number:1141" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "x=5#%\n", + "r=2.5#%\n", + "\n", + "#calculation\n", + "phi=math.atan(x/r)\n", + "cosphi=math.cos(phi)\n", + "sinphi=math.sin(phi)\n", + "regn=r*cosphi+x*sinphi\n", + "\n", + "#result\n", + "print \"regulation=\",regn,\"%\"\n", + "print \"pf=\",cosphi" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "regulation= 5.59016994375 %\n", + "pf= 0.4472135955\n" + ] + } + ], + "prompt_number": 100 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.24, Page Number:1142" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "r=2.5#%\n", + "x=5#%\n", + "load1=500#KVA\n", + "load2=400#KVA\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "kw=load2*pf\n", + "kvar=load2*math.sin(math.acos(pf))\n", + "drop=(r*kw/load1)+(x*kvar/load1)\n", + "\n", + "#result\n", + "print \"percentage voltage drop=\",drop,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "percentage voltage drop= 4.0 %\n" + ] + } + ], + "prompt_number": 102 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.26, Page Number:1145" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v1=600#V\n", + "v2=1080#V\n", + "v=720#V\n", + "load=8#W\n", + "load2=10#kVA\n", + "\n", + "#calculation\n", + "ir2=load*1000/v2\n", + "il2=load*1000/v\n", + "ir2_=ir2*v2/v1\n", + "il2_=il2*v/v1\n", + "ir2=math.sqrt(ir2_**2+il2_**2)\n", + "s=complex(load,load2)\n", + "s=abs(s)\n", + "pf=load/s\n", + "i=s*load2*100/v1\n", + "\n", + "#result\n", + "print \"primary current=\",i,\"A\"\n", + "print \"power factor=\",pf" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "primary current= 21.3437474581 A\n", + "power factor= 0.624695047554\n" + ] + } + ], + "prompt_number": 103 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.27, Page Number:1046" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=220#V\n", + "v1=110#V\n", + "i=0.5#A\n", + "p=30#W\n", + "r=0.6#ohm\n", + "\n", + "#calculation\n", + "ratio=v/v1\n", + "pf=p/(i*v)\n", + "sinphi=math.sqrt(1-pf**2)\n", + "ip=i*sinphi\n", + "iw=i*pf\n", + "cu_loss=i**2*r\n", + "iron_loss=p-cu_loss\n", + "\n", + "#result\n", + "print \"i)turns ratio=\",ratio\n", + "print \"ii)magnetising component of no-load current=\",ip,\"A\"\n", + "print \"iii)working component of no-load current=\",iw,\"A\"\n", + "print \"iv)the iron loss=\",iron_loss,\"W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)turns ratio= 2\n", + "ii)magnetising component of no-load current= 0.481045692921 A\n", + "iii)working component of no-load current= 0.136363636364 A\n", + "iv)the iron loss= 29.85 W\n" + ] + } + ], + "prompt_number": 104 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.28, Page Number:1047" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=5.0#kVA\n", + "v1=200.0#V\n", + "v2=1000.0#V\n", + "f=50.0#Hz\n", + "vo=2000.0#V\n", + "io=1.2#A\n", + "po=90.0#W\n", + "vs=50.0#V\n", + "i_s=5.0#A\n", + "ps=110.0#W\n", + "p=3.0#kW\n", + "pf=0.8\n", + "v=200.0#V\n", + "\n", + "#calculation\n", + "r0=v**2/po\n", + "ia0=v/r0\n", + "ip=math.sqrt(io**2-ia0**2)\n", + "xm=v/ip\n", + "z=vs/i_s\n", + "r=ps/25\n", + "x=math.sqrt(z**2-r**2)\n", + "r1=r*(v1/v2)**2\n", + "x1=x*(v1/v2)**2\n", + "i_lv1=load*1000/v\n", + "i_lv=(p*1000/pf)/v\n", + "sinphi=math.sin(math.acos(pf))\n", + "reg=i_lv*(r1*pf+x1*sinphi)/v\n", + "vt=v2-reg*1000/v\n", + "\n", + "#result\n", + "print \"LV crrent at rated load=\",i_lv1,\"A\"\n", + "print \"LV current at 3kW at 0.8 lagging pf\",i_lv,\"A\"\n", + "print \"output secondary voltage=\",vt,\"V\"\n", + "print \"percentage regulation=\",reg*100,\"%\"\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "LV crrent at rated load= 25.0 A\n", + "LV current at 3kW at 0.8 lagging pf 18.75 A\n", + "output secondary voltage= 999.832975251 V\n", + "percentage regulation= 3.34049498886 %\n" + ] + } + ], + "prompt_number": 105 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.29, Page Number:1048" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "A=Symbol('A')\n", + "B=Symbol('B')\n", + "loss1=52.0#W\n", + "f1=40.0#Hz\n", + "loss2=90.0#W\n", + "f2=60.0#Hz\n", + "f=50.0#Hz\n", + "\n", + "#calculation\n", + "ans=solve([(loss1/f1)-(A+f1*B),(loss2/f2)-(A+f2*B)],[A,B])\n", + "wh=ans[A]*f\n", + "we=ans[B]*f**2\n", + "\n", + "#result\n", + "print \"hysteresis=\",round(wh),\"W\"\n", + "print \"eddy current=\",round(we),\"W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "hysteresis= 45.0 W\n", + "eddy current= 25.0 W\n" + ] + } + ], + "prompt_number": 107 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.30, Page Number:1048" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%pylab\n", + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "A=Symbol('A')\n", + "B=Symbol('B')\n", + "m=10#kg\n", + "f=50.0#Hz\n", + "f1=25.0\n", + "f2=40.0\n", + "f3=50.0\n", + "f4=60.0\n", + "f5=80.0\n", + "l1=18.5/f1\n", + "l2=36.0/f2\n", + "l3=50.0/f3\n", + "l4=66.0/f4\n", + "l5=104.0/f5\n", + "#calculation\n", + "ans=solve([l1/f1-(A+f1*B),l2/f2-(A+f2*B)],[A,B])\n", + "eddy_loss_per_kg=ans[B]*f**2/m\n", + "\n", + "#result\n", + "print\"eddy current loss per kg at 50 Hz=\",eddy_loss_per_kg,\"W\"\n", + "\n", + "#plot\n", + "F=[f1,f2,f3,f4,f5]\n", + "L=[l1,l2,l3,l4,l5]\n", + "a=plot(F,L)\n", + "xlabel(\"f -->\") \n", + "ylabel(\"Wi/f\") \n", + "plt.xlim((0,100))\n", + "plt.ylim((0.74,2))\n", + "show(a)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Using matplotlib backend: TkAgg\n", + "Populating the interactive namespace from numpy and matplotlib\n", + "eddy current loss per kg at 50 Hz=" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " -0.118333333333333 W\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.31, Page Number:1148" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "A=Symbol('A')\n", + "B=Symbol('B')\n", + "v1=440#V\n", + "f1=50#Hz\n", + "p1=2500#W\n", + "v2=220#V\n", + "f2=25#Hz\n", + "p2=850#z\n", + "\n", + "#calculation\n", + "ans=solve([(p1/f1)-(A+f1*B),(p2/f2)-(A+f2*B)],[A,B])\n", + "wh=ans[A]*f\n", + "we=ans[B]*f**2\n", + "\n", + "#result\n", + "print \"hysteresis=\",round(wh),\"W\"\n", + "print \"eddy current=\",round(we),\"W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "hysteresis= 900.0 W\n", + "eddy current= 1600.0 W\n" + ] + } + ], + "prompt_number": 109 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.32, Page Number:1149" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v1=1000.0#V\n", + "f1=50.0#Hz\n", + "core=1000.0#W\n", + "wh=650.0#W\n", + "we=350.0#W\n", + "v2=2000.0#V\n", + "f2=100.0#Hz\n", + "\n", + "#calculation\n", + "a=wh/f1\n", + "b=we/f1**2\n", + "wh=a*f2\n", + "we=b*f2**2\n", + "new_core=wh+we\n", + "\n", + "#result\n", + "print \"new core loss=\",new_core,\"W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " new core loss= 2700.0 W\n" + ] + } + ], + "prompt_number": 111 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.33, Page Number:1149" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "phi=1.4#Wb/m2\n", + "we=1000.0#W\n", + "wh=3000.0#W\n", + "per=10.0#%\n", + "\n", + "#calculation\n", + "wh1=wh*1.1**1.6\n", + "we1=we*1.1**2\n", + "wh2=wh*0.9**(-0.6)\n", + "wh3=wh*1.1**1.6*1.1**(-0.6)\n", + "#result\n", + "print \"a)wh and we when applied voltage is increased by 10%=\",wh1,\"W\",\"and\",we1,\"W\"\n", + "print \"b)wh when frequency is reduced by 10%=\",wh2,\"W\"\n", + "print \"c)wh and we when both voltage and frequency are increased y 10%=\",wh3,\"W\",\"and\",we1,\"W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a)wh and we when applied voltage is increased by 10%= 3494.21441464 W and 1210.0 W\n", + "b)wh when frequency is reduced by 10%= 3195.77171838 W\n", + "c)wh and we when both voltage and frequency are increased y 10%= 3300.0 W and 1210.0 W\n" + ] + } + ], + "prompt_number": 119 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.34, Page Number:1150" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=2200.0#V\n", + "f=40.0#Hz\n", + "loss=800.0#W\n", + "wh=600.0#W\n", + "we=loss-wh\n", + "v2=3300.0#V\n", + "f2=60.0#Hz\n", + "\n", + "#calculations\n", + "a=wh/f\n", + "b=we/f**2\n", + "core_loss=a*f2+b*f2**2\n", + "\n", + "#result\n", + "print \"core loss at 60 Hz=\",core_loss,\"W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "core loss at 60 Hz= 1350.0 W\n" + ] + } + ], + "prompt_number": 122 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.35, Page Number:1151" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=30.0#KvA\n", + "v1=6000.0#V\n", + "v2=230.0#V\n", + "r1=10.0#ohm\n", + "r2=0.016#ohm\n", + "x01=34.0#ohm\n", + "\n", + "#calculations\n", + "k=v2/v1\n", + "r01=r1+r2/k**2\n", + "z01=(r01**2+x01**2)**0.5\n", + "i1=load*1000/v1\n", + "vsc=i1*z01\n", + "pf=r01/z01\n", + "\n", + "#result\n", + "print \"primary voltage=\",vsc,\"V\"\n", + "print \"pf=\",pf" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "primary voltage= 199.519931911 V\n", + "pf= 0.523468222173\n" + ] + } + ], + "prompt_number": 124 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.36, Page Number:1152" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v1=200.0#V\n", + "v2=400.0#V\n", + "f=50.0#Hz\n", + "vo=200.0#V\n", + "io=0.7#A\n", + "po=70.0#W\n", + "vs=15.0#v\n", + "i_s=10.0#A\n", + "ps=85.0#W\n", + "load=5.0#kW\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "cosphi0=po/(vo*io)\n", + "sinphi0=math.sin(math.acos(cosphi0))\n", + "iw=io*cosphi0\n", + "imu=io*sinphi0\n", + "r0=v1/iw\n", + "x0=v1/imu\n", + "z02=vs/i_s\n", + "k=v2/v1\n", + "z01=z02/k**2\n", + "r02=ps/i_s**2\n", + "r01=r02/k**2\n", + "x01=(z01**2-r01**2)**0.5\n", + "output=load/pf\n", + "i2=output*1000/v2\n", + "x02=(z02**2-r02**2)**0.5\n", + "drop=i2*(r02*pf+x02*math.sin(math.acos(pf)))\n", + "v2=v2-drop\n", + "print z02\n", + "#result\n", + "print \"secondary voltage=\",v2,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1.5\n", + "secondary voltage= 377.788243349 V\n" + ] + } + ], + "prompt_number": 130 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.37, Page Number:1152" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "k=1.0/6\n", + "r1=0.9#ohm\n", + "x1=5.0#ohm\n", + "r2=0.03#ohm\n", + "x2=0.13#ohm\n", + "vsc=330.0#V\n", + "f=50.0#Hz\n", + "\n", + "#calculations\n", + "r01=r1+r2/k**2\n", + "x01=x1+x2/k**2\n", + "z01=(r01**2+x01**2)**0.5\n", + "i1=vsc/z01\n", + "i2=i1/k\n", + "cosphisc=i1**2*r01/(vsc*i1)\n", + "\n", + "#result\n", + "print \"current in low voltage winding=\",i2,\"A\"\n", + "print \"pf=\",round(cosphisc,1)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "current in low voltage winding= 200.396236149 A\n", + "pf= 0.2\n" + ] + } + ], + "prompt_number": 132 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.38, Page Number:1153" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=10.0#kVA\n", + "v1=500.0#V\n", + "v2=250.0#V\n", + "f=50.0#Hz\n", + "r1=0.2#ohm\n", + "x1=0.4#ohm\n", + "r2=0.5#ohm\n", + "x2=0.1#ohm\n", + "r0=1500.0#ohm\n", + "x0=750.0#ohm\n", + "\n", + "#calculation\n", + "k=v2/v1\n", + "imu=v1/x0\n", + "iw=v1/r0\n", + "i0=(iw**2+imu**2)**0.5\n", + "pi=v1*iw\n", + "r01=r1+r2/k**2\n", + "x01=x1+x2/k**2\n", + "z01=(r01**2+x01**2)**0.5\n", + "i1=load*1000/v1\n", + "vsc=i1*z01\n", + "power=i1**2*r01\n", + "\n", + "#result\n", + "print \"reading of instruments=\",vsc,\"V,\",i1,\"A,\",power,\"W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "reading of instruments= 46.8187996429 V, 20.0 A, 880.0 W\n" + ] + } + ], + "prompt_number": 140 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.39, Page Number:1153" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "x=Symbol('x')\n", + "y=Symbol('y')\n", + "load=1000#kVA\n", + "v1=110#V\n", + "v2=220#V\n", + "f=50#Hz\n", + "per1=98.5#%\n", + "pf=0.8\n", + "per2=98.8#%\n", + "\n", + "#calculaions\n", + "output=load*1\n", + "inpt=output*100/per2\n", + "loss=inpt-output\n", + "inpt_half=(load/2)*pf*100/per1\n", + "loss2=inpt_half-400\n", + "ans=solve([x+y-loss,(x/4)+y-loss2],[x,y])\n", + "kva=load*(ans[y]/ans[x])*0.5\n", + "output=kva*1\n", + "cu_loss=ans[y]\n", + "total_loss=2*cu_loss\n", + "efficiency=output/(output+total_loss)\n", + "#result\n", + "print \"full load copper loss=\",cu_loss,\"kW\"\n", + "print \"maximum efficiency=\",efficiency,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "full load copper loss= 4.07324441521606 kW\n", + "maximum efficiency= 0.968720013059872 %\n" + ] + } + ], + "prompt_number": 148 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.40, Page Number:1154" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v1=200.0#v\n", + "v2=400.0#V\n", + "r01=0.15#ohm\n", + "x01=0.37#ohm\n", + "r0=600.0#ohm\n", + "x0=300.0#ohm\n", + "i2=10.0#A\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "imu=v1/x0\n", + "iw=v1/r0\n", + "i0=(imu**2+iw**2)**0.5\n", + "tantheta=iw/imu\n", + "theta=math.atan(tantheta)\n", + "theta0=math.radians(90)-theta\n", + "angle=theta0-math.acos(pf)\n", + "k=v2/v1\n", + "i2_=i2*k\n", + "i1=(i0**2+i2_**2+2*i0*i2_*math.cos(angle))**0.5\n", + "r02=k**2*r01\n", + "x02=x01*k**2\n", + "vd=i2*(r02*pf+x02*math.sin(math.acos(pf)))\n", + "v2=v2-vd\n", + "\n", + "#result\n", + "print \"i)primary current=\",i1,\"A\"\n", + "print \"ii)secondary terminal voltage=\",v2,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)primary current= 20.6693546639 A\n", + "ii)secondary terminal voltage= 386.32 V\n" + ] + } + ], + "prompt_number": 149 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.43, Page Number:1158" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=100.0#kVA\n", + "n1=400.0\n", + "n2=80.0\n", + "r1=0.3#ohm\n", + "r2=0.01#ohm\n", + "x1=1.1#ohm\n", + "x2=0.035#ohm\n", + "v1=2200.0#V\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "k=n2/n1\n", + "r01=r1+r2/k**2\n", + "x01=x1+x2/k**2\n", + "z01=complex(r01,x01)\n", + "z02=k**2*z01\n", + "v2=k*v1\n", + "i2=load*1000/v2\n", + "vd=i2*(z02.real*pf-z02.imag*math.sin(math.acos(pf)))\n", + "regn=vd*100/v2\n", + "v2=v2-vd\n", + "\n", + "#result\n", + "print \"i)equivalent impedence=\",z02,\"ohm\"\n", + "print \"ii)voltage regulation=\",regn,\"%\"\n", + "print \"secondary terminal voltage=\",v2,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)equivalent impedence= (0.022+0.079j) ohm\n", + "ii)voltage regulation= -1.53925619835 %\n", + "secondary terminal voltage= 446.772727273 V\n" + ] + } + ], + "prompt_number": 158 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.44, Page Number:1158" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=10.0#kVA\n", + "va=450.0#V\n", + "vb=120.0#V\n", + "v1=120.0#V\n", + "i1=4.2#A\n", + "w1=80.0#W\n", + "v2=9.65#V\n", + "i2=22.2#A\n", + "w2=120.0#W\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "k=vb/va\n", + "i0=i1*k\n", + "cosphi0=w1/(va*i0)\n", + "phi0=math.acos(cosphi0)\n", + "sinphi0=math.sin(phi0)\n", + "iw=i0*cosphi0\n", + "imu=i0*sinphi0\n", + "r0=va/iw\n", + "x0=va/imu\n", + "z01=v2/i2\n", + "r01=vb/i2**2\n", + "x01=(z01**2-r01**2)**0.5\n", + "i1=load*1000/va\n", + "drop=i1*(r01*pf+x01*math.sin(math.acos(pf)))\n", + "regn=drop*100/va\n", + "loss=w1+w2\n", + "output=load*1000*pf\n", + "efficiency=output/(output+loss)\n", + "iron_loss=w1\n", + "cu_loss=(0.5**2)*w2\n", + "total_loss=iron_loss+cu_loss\n", + "output=load*1000*pf/2\n", + "efficiency2=output/(output+total_loss)\n", + "\n", + "#result\n", + "print \"i)equivalent circuit constants=\"\n", + "print \"z01=\",z01,\"ohm\"\n", + "print \"x01=\",x01,\"ohm\"\n", + "print \"r01=\",r01,\"ohm\"\n", + "print \"ii)efficiency and voltage regulation at pf=0.8=\",efficiency*100,\"%\",regn,\"%\"\n", + "print \"iii)efficiency at half load and pf=0.8=\",efficiency2*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)equivalent circuit constants=\n", + "z01= 0.434684684685 ohm\n", + "x01= 0.360090249002 ohm\n", + "r01= 0.243486729973 ohm\n", + "ii)efficiency and voltage regulation at pf=0.8= 97.5609756098 % 2.02885695496 %\n", + "iii)efficiency at half load and pf=0.8= 97.3236009732 %\n" + ] + } + ], + "prompt_number": 162 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.45, Page Number:1159" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=20.0#kVA\n", + "va=2200.0#V\n", + "vb=220.0#V\n", + "f=50.0#Hz\n", + "v1=220.0#V\n", + "i1=4.2#A\n", + "w1=148.0#W\n", + "v2=86.0#V\n", + "i2=10.5#A\n", + "w2=360.0#W\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "z01=v2/i2\n", + "r01=w2/i2**2\n", + "x01=(z01**2-r01**2)**0.5\n", + "i1=load*1000/va\n", + "drop=i1*(r01*pf+x01*math.sin(math.acos(pf)))\n", + "regn=drop*100/va\n", + "pf=r01/z01\n", + "\n", + "#result\n", + "print \"regulation=\",regn,\"%\"\n", + "print \"pf=\",round(pf,1),\"lag\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "regulation= 2.94177963326 %\n", + "pf= 0.4 lag\n" + ] + } + ], + "prompt_number": 172 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.46, Page Number:1159" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=10.0#kVA\n", + "v1=2000.0#V\n", + "v2=400.0#V\n", + "v=60.0#V\n", + "i=4.0#A\n", + "w=100.0#W\n", + "pf=0.8\n", + "v_=400.0#V\n", + "\n", + "#calculations\n", + "z01=v/i\n", + "r01=w/i**2\n", + "x01=(z01**2-r01**2)**0.5\n", + "i1=load*1000/v1\n", + "vd=i1*(r01*pf+x01*math.sin(math.acos(pf)))\n", + "\n", + "#result\n", + "print \"voltage applied to hv side=\",v1+vd,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "voltage applied to hv side= 2065.90767043 V\n" + ] + } + ], + "prompt_number": 182 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.47, Page Number:1159" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v1=250.0#V\n", + "v2=500.0#V\n", + "vs=20.0#V\n", + "i_s=12.0#A\n", + "ws=100.0#W\n", + "vo=250.0#V\n", + "io=1.0#A\n", + "wo=80.0#W\n", + "i2=10#A\n", + "v2=500#V\n", + "pg=0.8\n", + "\n", + "#calculation\n", + "cosphi0=wo/(vo*io)\n", + "iw=io*cosphi0\n", + "imu=(1-iw**2)**0.5\n", + "r0=v1/iw\n", + "x0=v1/imu\n", + "r02=ws/i_s**2\n", + "z02=vs/i_s\n", + "x02=(z02**2-r02**2)**0.5\n", + "k=v2/v1\n", + "r01=r02/k**2\n", + "x01=x02/k**2\n", + "z01=z02/k**2\n", + "cu_loss=i2**2*r02\n", + "iron_loss=wo\n", + "total_loss=iron_loss+cu_loss\n", + "efficiency=i2*v2*pf/(i2*v2*pf+total_loss)\n", + "v1_=((vo*pf+x01)**2+(vo*math.sin(math.acos(pf))+i1*x01)**2)**0.5\n", + "\n", + "#result\n", + "print \"applied voltage=\",v1_,\"V\"\n", + "print \"efficiency=\",efficiency*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "applied voltage= 251.442641983 V\n", + "efficiency= 96.3984469139 %\n" + ] + } + ], + "prompt_number": 190 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.48, Page Number:1160" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v1=230.0#V\n", + "v2=230.0#V\n", + "load=3.0#kVA\n", + "vo=230.0#V\n", + "io=2.0#A\n", + "wo=100.0#W\n", + "vs=15.0#V\n", + "i_s=13.0#A\n", + "ws=120.0#W\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "i=load*1000/v1\n", + "cu_loss=ws\n", + "core_loss=wo\n", + "output=load*1000*pf\n", + "efficiency=output*100/(output+cu_loss+core_loss)\n", + "z=vs/i_s\n", + "r=ws/(vs**2)\n", + "x=(z**2-r**2)**0.5\n", + "regn=i*(r*pf+x*math.sin(math.acos(pf)))*100/v1\n", + "\n", + "#result\n", + "print \"regulation=\",regn,\"%\"\n", + "print \"efficiency=\",efficiency,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "regulation= 5.90121149256 %\n", + "efficiency= 91.6030534351 %\n" + ] + } + ], + "prompt_number": 194 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.49, Page Number:1161" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=10.0#kVA\n", + "v1=500.0#V\n", + "v2=250.0#V\n", + "efficiency=0.94\n", + "per=0.90\n", + "pf=0.8\n", + "\n", + "#calculation\n", + "output=per*load*1000\n", + "inpt=output/efficiency\n", + "loss=inpt-output\n", + "core_loss=loss/2\n", + "pc=core_loss/per**2\n", + "output=load*1000*pf\n", + "cu_loss=pc\n", + "efficiency=output/(output+cu_loss+core_loss)\n", + "\n", + "#result\n", + "print \"efficiency=\",efficiency*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "efficiency= 92.5728354534 %\n" + ] + } + ], + "prompt_number": 196 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.50, Page Number:1161" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=10.0#kVA\n", + "f=50.0#Hz\n", + "v1=2300.0#V\n", + "v2=230.0#V\n", + "r1=3.96#ohm\n", + "r2=0.0396#ohm\n", + "x1=15.8#ohm\n", + "x2=0.158#ohm\n", + "pf=0.8\n", + "v=230.0#V\n", + "\n", + "#calculations\n", + "i=load*1000/v\n", + "r=r2+r1*(v2/v1)**2\n", + "x=x1*(v2/v1)**2+x2\n", + "v1_=v2+i*(r*pf+x*math.sin(math.acos(pf)))\n", + "v1=v1_*(v1/v2)\n", + "phi=math.atan(r/x)\n", + "pf=math.cos(phi)\n", + "#result\n", + "print \"a)HV side voltage necessary=\",v1,\"V\"\n", + "print \"b)pf=\",round(pf,2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a)HV side voltage necessary= 2409.9826087 V\n", + "b)pf= 0.97\n" + ] + } + ], + "prompt_number": 199 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.51, Page Number:1162" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=5.0#kVA\n", + "v1=2200.0#V\n", + "v2=220.0#v\n", + "r1=3.4#ohm\n", + "x1=7.2#ohm\n", + "r2=0.028#ohm\n", + "x2=0.060#ohm\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "i=load*1000/v2\n", + "r=r1*(v2/v1)**2+r2\n", + "x=x1*(v2/v1)**2+x2\n", + "ad=i*r*pf\n", + "dc=i*x*math.sin(math.acos(pf))\n", + "oc=v2+ad+dc\n", + "bd=i*r*math.sin(math.acos(pf))\n", + "b_f=x*pf\n", + "cf=b_f-bd\n", + "v1_=(oc**2+cf**2)**0.5\n", + "v1=v1_*(v1/v2)\n", + "\n", + "#result\n", + "print \"terminal voltage on hv side=\",v1,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "terminal voltage on hv side= 2229.28500444 V\n" + ] + } + ], + "prompt_number": 200 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.52, Page Number:1163" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=4.0#kVA\n", + "v1=200.0#V\n", + "v2=400.0#V\n", + "i1=0.7#A\n", + "w1=65.0#W\n", + "v=15.0#V\n", + "i2=10.0#A\n", + "w2=75.0#W\n", + "pf=0.80\n", + "#calculation\n", + "il=load*1000/v1\n", + "ih=load*1000/v2\n", + "cu_loss=w2\n", + "constant_loss=w1\n", + "z=v/i2\n", + "r=w2/i2**2\n", + "x=(z**2-r**2)**0.5\n", + "efficiency=load*100000/(load*1000+cu_loss+constant_loss)\n", + "regn=i2*(r*pf+x*math.sin(math.acos(pf)))\n", + "\n", + "#result\n", + "print \"full load efficiency=\",efficiency,\"%\"\n", + "print \"full load regulation=\",regn,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "full load efficiency= 96.6183574879 %\n", + "full load regulation= 13.7942286341 V\n" + ] + } + ], + "prompt_number": 209 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.53, Page Number:1164" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v1=3300.0#V\n", + "v2=230.0#V\n", + "load=50.0#kVA\n", + "z=4\n", + "cu_loss=1.8\n", + "\n", + "#calculations\n", + "x=(z**2-cu_loss**2)**0.5\n", + "i1=load*1000/v1\n", + "r01=cu_loss*v1/(100*i1)\n", + "x01=x*v1/(100*i1)\n", + "z01=z*v1/(100*i1)\n", + "isc=i1*100/z\n", + "print \n", + "#result\n", + "print \"%x=\",x,\"%\"\n", + "print \"resistance=\",r01,\"ohm\"\n", + "print \"reactance=\",x01,\"ohm\"\n", + "print \"impedence=\",z01,\"ohm\"\n", + "print \"primary sc current=\",isc,\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "%x= 3.5721142199 %\n", + "resistance= 3.9204 ohm\n", + "reactance= 7.78006477094 ohm\n", + "impedence= 8.712 ohm\n", + "primary sc current= 378.787878788 A\n" + ] + } + ], + "prompt_number": 214 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.54, Page Number:1164" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=20.0#kVA\n", + "v1=2200.0#V\n", + "v2=220.0#V\n", + "f=50.0#Hz\n", + "vo=220.0#V\n", + "i_o=4.2#A\n", + "wo=148.0#W\n", + "vs=86.0#V\n", + "i_s=10.5#A\n", + "ws=360.0#W\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "k=v2/v1\n", + "r01=ws/i_s**2\n", + "r02=k**2*r01\n", + "z10=vs/i_s\n", + "x01=(z10**2-r01**2)**0.5\n", + "x02=k**2*x01\n", + "i1=load*1000/v1\n", + "v1_=((v1*pf+i1*r01)**2+(v1*math.sin(math.acos(pf))+i1*x01)**2)**0.5\n", + "regn1=(v1_-v1)/v1\n", + "i2=i1/k\n", + "core_loss=wo\n", + "cu_loss=i1**2*r01\n", + "cu_loss_half=(i1/2)**2*r01\n", + "efficiency=load*1000*pf*100/(load*1000*pf+core_loss+cu_loss)\n", + "efficiency_half=(load/2)*1000*pf*100/((load/2)*1000*pf+core_loss+cu_loss)\n", + "print v1_ \n", + "#result\n", + "print \"a)core loss=\",wo,\"W\"\n", + "print \"b)equivalent resistance primary=\",r01,\"ohm\"\n", + "print \"c)equivalent resistance secondary=\",r02,\"ohm\"\n", + "print \"d)equivalent reactance primary=\",x01,\"ohm\"\n", + "print \"e)equivalent reactance secondary=\",x02,\"ohm\"\n", + "print \"f)regulation=\",regn1*100,\"%\"\n", + "print \"g)efficiency at full load=\",efficiency,\"%\"\n", + "print \"h)efficiency at half load=\",efficiency_half,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "2265.01840886\n", + "a)core loss= 148.0 W\n", + "b)equivalent resistance primary= 3.26530612245 ohm\n", + "c)equivalent resistance secondary= 0.0326530612245 ohm\n", + "d)equivalent reactance primary= 7.51143635755 ohm\n", + "e)equivalent reactance secondary= 0.0751143635755 ohm\n", + "f)regulation= 2.95538222101 %\n", + "g)efficiency at full load= 97.4548448466 %\n", + "h)efficiency at half load= 95.0360304208 %\n" + ] + } + ], + "prompt_number": 222 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.55, Page Number:1165" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "er=1.0/100\n", + "ex=5.0/100\n", + "pf=0.8\n", + "\n", + "#calculation\n", + "regn=er*pf+ex*math.sin(math.acos(pf))\n", + "regn2=er*1\n", + "regn3=er*pf-ex*math.sin(math.acos(pf))\n", + "\n", + "#result\n", + "print \"i)regulation with pf=0.8 lag=\",regn*100,\"%\"\n", + "print \"ii)regulation with pf=1=\",regn2*100,\"%\"\n", + "print \"iii)regulation with pf=0.8 lead=\",regn3*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)regulation with pf=0.8 lag= 3.8 %\n", + "ii)regulation with pf=1= 1.0 %\n", + "iii)regulation with pf=0.8 lead= -2.2 %\n" + ] + } + ], + "prompt_number": 223 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.56, Page Number:1165" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=500#kVA\n", + "v1=3300#V\n", + "v2=500#V\n", + "f=50#Hz\n", + "per=0.97\n", + "ratio=3.0/4\n", + "zper=0.10\n", + "pf=0.8\n", + "\n", + "#calculation\n", + "output=load*ratio*1\n", + "x=0.75\n", + "pi=0.5*(output*(1/per-1))\n", + "pc=pi/x**2\n", + "i1=load*1000/v1\n", + "r=pc*1000/i1**2\n", + "er=i1*r/v1\n", + "ez=zper\n", + "ex=(ez**2-er**2)**0.5\n", + "regn=er*pf+ex*math.sin(math.acos(pf))\n", + "\n", + "#result\n", + "print \"regulation=\",regn*100,\"%\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "regulation= 7.52529846012 %\n" + ] + } + ], + "prompt_number": 225 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.57, Page Number:1166" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "cu_loss=1.5#%\n", + "xdrop=3.5#%\n", + "pf=0.8\n", + "\n", + "#calculation\n", + "pur=cu_loss/100\n", + "pux=xdrop/100\n", + "regn2=pur*pf+pux*math.sin(math.acos(pf))\n", + "regn1=pur*1\n", + "regn3=pur*pf-pux*math.sin(math.acos(pf))\n", + "\n", + "#result\n", + "print \"i)regulation at unity pf=\",regn1*100,\"%\"\n", + "print \"ii)regulation at 0.8 lag=\",regn2*100,\"%\"\n", + "print \"iii)regulation at 0.8 lead=\",regn3*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)regulation at unity pf= 1.5 %\n", + "ii)regulation at 0.8 lag= 3.3 %\n", + "iii)regulation at 0.8 lead= -0.9 %\n" + ] + } + ], + "prompt_number": 226 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.58, Page Number:1168" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=250#KVA\n", + "w1=5.0#kW\n", + "w2=7.5#kW\n", + "efficiency=0.75\n", + "pf=0.8\n", + "\n", + "#calculation\n", + "total_loss=w1+w2\n", + "loss=total_loss/2\n", + "cu_loss=efficiency**2*w2/2\n", + "output=load*efficiency*pf\n", + "efficiency=output*100/(output+cu_loss+2.5)\n", + "\n", + "#result\n", + "print \"efficiency=\",efficiency,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "efficiency= 97.0186963113 %\n" + ] + } + ], + "prompt_number": 229 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.59, Page Number:1170" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=25.0#kVA\n", + "v1=2000.0#V\n", + "v2=200.0#V\n", + "w1=350.0#W\n", + "w2=400.0#W\n", + "\n", + "#calculation\n", + "total_loss=w1+w2\n", + "output=load*1000*1\n", + "efficiency=output/(output+total_loss)\n", + "cu_loss=w2*(0.5)**2\n", + "total_loss=cu_loss+w1\n", + "efficiency2=(load*1000/2)/((load*1000/2)+total_loss)\n", + "\n", + "#result\n", + "print \"i)efficiency at full load=\",efficiency*100,\"%\"\n", + "print \"ii)efficiency at half load=\",efficiency2*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)efficiency at full load= 97.0873786408 %\n", + "ii)efficiency at half load= 96.5250965251 %\n" + ] + } + ], + "prompt_number": 232 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.60, Page Number:1170" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "efficiency=0.75\n", + "\n", + "#calculation\n", + "ratio=efficiency**2\n", + "\n", + "#result\n", + "print \"ratio of P1 and P2=\",ratio" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "ratio of P1 and P2= 0.5625\n" + ] + } + ], + "prompt_number": 233 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.61, Page Number:1170" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v1=11000.0#V\n", + "v2=230.0#V\n", + "load1=150.0#KVA\n", + "f=50.0#Hz\n", + "loss=1.4#kW\n", + "cu_loss=1.6#kW\n", + "pf=0.8\n", + "\n", + "#calculation\n", + "load=load1*(cu_loss/loss)**0.5\n", + "total_loss=loss*2\n", + "output=load*1\n", + "efficiency=output/(output+total_loss)\n", + "cu_loss=cu_loss*(0.5)**2\n", + "total_loss=total_loss+cu_loss\n", + "output2=(load/2)*pf\n", + "efficiency2=output2/(output2+total_loss)\n", + "\n", + "#result\n", + "print \"i)kVA load for max efficiency=\",load1,\"kVA\"\n", + "print \"max efficiency=\",efficiency*100,\"%\"\n", + "print \"ii)efficiency at half load=\",efficiency2*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)kVA load for max efficiency= 150.0 kVA\n", + "max efficiency= 98.283858876 %\n", + "ii)efficiency at half load= 95.2481856352 %\n" + ] + } + ], + "prompt_number": 237 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.62, Page Number:1171" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%pylab\n", + "#variable declaration\n", + "load=5#kVA\n", + "v1=2300#V\n", + "v2=230#V\n", + "f=50#Hz\n", + "iron_loss=40#W\n", + "cu_loss=112#W\n", + "pf=0.8\n", + "#calculations\n", + "def e(k):\n", + " e=k*pf*1000*100/(k*pf*1000+(cu_loss*(k/5)**2+40))\n", + " return(e)\n", + "\n", + "e1=e(1.25)\n", + "e2=e(2.5)\n", + "e3=e(3.75)\n", + "e4=e(5.0)\n", + "e5=e(6.25)\n", + "e6=e(7.5)\n", + "\n", + "K=[1.25,2.5,3.75,5.0,6.25,7.5]\n", + "E=[e1,e2,e3,e4,e5,e6]\n", + "a=plot(K,E)\n", + "xlabel(\"load,kVA\") \n", + "ylabel(\"Efficiency\") \n", + "plt.xlim((0,8))\n", + "plt.ylim((92,98))\n", + "show(a)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.63, Page Number:1171" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=200.0#kVA\n", + "efficiency=0.98\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "output=load*pf\n", + "inpt=output/efficiency\n", + "loss=inpt-output\n", + "x=loss*1000/(1+9.0/16)\n", + "y=(9.0/16)*x\n", + "cu_loss=x*(1.0/2)**2\n", + "total_loss=cu_loss+y\n", + "output=load*pf*0.5\n", + "efficiency=output/(output+total_loss/1000)\n", + "\n", + "#result\n", + "print \"efficiency at hald load=\",efficiency*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "efficiency at hald load= 97.9216626699 %\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.64, Page Number:1172" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=25.0#kVA\n", + "v1=2200.0#V\n", + "v2=220.0#V\n", + "r1=1.0#ohm\n", + "r2=0.01#ohm\n", + "pf=0.8\n", + "loss=0.80\n", + "\n", + "#calculations\n", + "k=v2/v1\n", + "r02=r2+k**2*r1\n", + "i2=load*1000/v2\n", + "cu_loss=i2**2*r02\n", + "iron_loss=loss*cu_loss\n", + "total_loss=cu_loss+iron_loss\n", + "output=load*pf*1000\n", + "efficiency=output/(output+total_loss)\n", + "\n", + "#result\n", + "print \"secondary resistance=\",r02,\"ohm\"\n", + "print \"efficiency=\",efficiency*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "secondary resistance= 0.02 ohm\n", + "efficiency= 97.7284199899 %\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.65, Page Number:1172" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=4.0#kVA\n", + "v1=200.0#V\n", + "v2=400.0#V\n", + "r01=0.5#ohm\n", + "x01=1.5#ohm\n", + "ratio=3.0/4\n", + "pf=0.8\n", + "v=220.0#V\n", + "loss=100.0#W\n", + "\n", + "#calculations\n", + "k=v2/v1\n", + "r02=k**2*r01\n", + "x02=k**2*x01\n", + "i2=1000*load*ratio/v2\n", + "drop=i2*(r02*pf+x02*math.sin(math.acos(pf)))\n", + "v2=v2-drop\n", + "cu_loss=i2**2*r02\n", + "total_loss=loss+cu_loss\n", + "output=load*ratio*pf\n", + "inpt=output*1000+total_loss\n", + "efficiency=output*1000/(inpt)\n", + "#result\n", + "print \"output=\",output,\"w\"\n", + "print \"efficiency=\",efficiency*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "output= 2.4 w\n", + "efficiency= 91.8660287081 %\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.66, Page Number:1172" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=20.0#KVA\n", + "v1=440.0#V\n", + "v2=220.0#V\n", + "f=50.0#Hz\n", + "loss=324.0#W\n", + "cu_loss=100.0#W\n", + "pf=0.8\n", + "#calculations\n", + "cu_loss=4*cu_loss\n", + "efficiency=load*pf/(load*pf+cu_loss/1000+loss/1000)\n", + "per=(loss/cu_loss)**0.5\n", + "\n", + "#result\n", + "print \"i)efficiency=\",efficiency*100,\"%\"\n", + "print \"ii)percent of full-load=\",per*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)efficiency= 95.6708921311 %\n", + "ii)percent of full-load= 90.0 %\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.67, Page Number:1173" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "load=4.0#kVA\n", + "v1=200.0#V\n", + "v2=400.0#V\n", + "pf=0.8\n", + "vo=200.0#V\n", + "io=0.8#A\n", + "wo=70.0#W\n", + "vs=20.0#V\n", + "i_s=10.0#A\n", + "ws=60.0#W\n", + "\n", + "#calculation\n", + "i2=load*1000/v2\n", + "loss=ws+wo\n", + "output=load*pf\n", + "efficiency=output/(output+loss/1000)\n", + "z02=vs/i_s\n", + "r02=ws/i2**2\n", + "x02=(z02**2-r02**2)**0.5\n", + "drop=i2*(r02*pf+x02*math.sin(math.acos(pf)))\n", + "v2=v2-drop\n", + "i1=load*1000/v1\n", + "load=load*(wo/ws)**0.5\n", + "load=load*1\n", + "\n", + "#result\n", + "print \"efficiency=\",efficiency*100,\"%\"\n", + "print \"secondary voltage=\",v2,\"V\"\n", + "print \"current=\",i1,\"A\"\n", + "print \"load at unity pf=\",load,\"kW\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "efficiency= 96.0960960961 %\n", + "secondary voltage= 383.752729583 V\n", + "current= 20.0 A\n", + "load at unity pf= 4.32049379894 kW\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.69, Page Number:1174" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "x=Symbol('x')\n", + "y=Symbol('y')\n", + "load=600.0#KVA\n", + "efficiency=0.92\n", + "per=0.60\n", + "\n", + "#calculation\n", + "inpt=load/efficiency\n", + "loss1=inpt-load\n", + "inpt2=load/(2*efficiency)\n", + "loss2=inpt2-load/2\n", + "ans=solve([x+y-loss1,x+y/4-loss2],[x,y])\n", + "cu_loss=ans[y]*0.36\n", + "loss=cu_loss+ans[x]\n", + "output=load*per\n", + "efficiency=output/(output+loss)\n", + "\n", + "#result\n", + "print \"efficiency=\",efficiency*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "389.913043478261\n", + "efficiency= 92.3282783229260 %\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.70, Page Number:1174" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=100#kVA\n", + "e1=0.98\n", + "e2=0.80\n", + "pf=8\n", + "z=0.05\n", + "pf1=0.8\n", + "\n", + "#calculations\n", + "output=load*pf1*e2\n", + "inpt=output/e1\n", + "loss=-output+inpt\n", + "cu_loss=loss/2\n", + "cu_loss_full=cu_loss/pf1**2\n", + "r=round(cu_loss_full*100/load)\n", + "sin=math.sin(math.acos(pf1))\n", + "regn=(r*pf1+5*sin)+(1.0/200)*(5*pf1-r*sin)**2\n", + "#result\n", + "print \"voltage regulation=\",regn,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "voltage regulation= 3.8578 %\n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.71, Page Number:1174" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=10.0#KVA\n", + "v1=5000.0#V\n", + "v2=440.0#V\n", + "f=25.0#Hz\n", + "cu_loss=1.5\n", + "we=0.5\n", + "wh=0.6\n", + "v2=10000.0\n", + "#calculations\n", + "cu_loss1=cu_loss*load/100\n", + "we1=we*load/100\n", + "wh1=wh*load/100\n", + "cu_loss2=cu_loss1\n", + "we2=(we1*(50.0/25.0)**2)\n", + "wh2=(wh1*(50.0/25))\n", + "e1=load*100/(load+cu_loss1+we1+wh1)\n", + "e2=load*2*100/(load*2+cu_loss2+we2+wh2)\n", + "\n", + "#result\n", + "print \"full load efficiency in first case=\",e1,\"%\"\n", + "print \"full load efficiency in second case=\",e2,\"%\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "20.47 0.06 0.05\n", + "full load efficiency in first case= 97.4658869396 %\n", + "full load efficiency in second case= 97.7039570103 %\n" + ] + } + ], + "prompt_number": 47 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.72, Page Number:1175" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=300#KVA\n", + "r=1.5#%\n", + "load1=173.2#kVA\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "cu_loss=r*load*1000/100\n", + "iron_loss=(load1/load)**2*cu_loss\n", + "total_loss=cu_loss+iron_loss\n", + "efficiency=(load*pf)*100/((load*pf)+(total_loss/1000))\n", + "\n", + "#result\n", + "print \"efficiency=\",efficiency,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "efficiency= 97.5610105096 %\n" + ] + } + ], + "prompt_number": 53 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.73, Page Number:1175" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=100#kVA\n", + "v1=2300#V\n", + "v2=230.0#V\n", + "f=50#Hz\n", + "phim=1.2#Wb/m2\n", + "a=0.04#m2\n", + "l=2.5#m\n", + "bm=1200\n", + "inpt=1200#W\n", + "pi=400#W\n", + "efficiency=0.75\n", + "pf=0.8\n", + "f2=100#Hz\n", + "\n", + "#calculation\n", + "n1=v1/(4.44*f*phim*a)\n", + "k=v2/v1\n", + "n2=k*n1\n", + "i=1989/n1\n", + "cu_loss=efficiency**2*inpt\n", + "total_loss=pi+cu_loss\n", + "output=load*efficiency*pf\n", + "efficiency=output*100/(output+total_loss/1000)\n", + "\n", + "#result\n", + "print \"a)n1=\",round(n1)\n", + "print \" n2=\",round(n2)\n", + "print \"b)magnetising current=\",i,\"A\"\n", + "print \"c)efficiency=\",efficiency,\"%\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "0.00643416423287\n", + "a)n1= 216.0\n", + " n2= 22.0\n", + "b)magnetising current= 9.21512347826 A\n", + "c)efficiency= 98.2398690135 %\n" + ] + } + ], + "prompt_number": 58 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.74, Page Number:1176" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "r=1.8\n", + "x=5.4\n", + "\n", + "#calculation\n", + "pf=r/x\n", + "phi=math.atan(pf)\n", + "phi2=math.atan(x/r)\n", + "regn=r*math.cos(phi2)+x*math.sin(phi2)\n", + "efficiency=100/(100+r*2)\n", + "\n", + "#result\n", + "print \"a)i)phi=\",math.degrees(phi),\"degrees\"\n", + "print \" ii)regulation=\",regn,\"%\"\n", + "print \"b)efficiency=\",efficiency*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a)i)phi= 18.4349488229 degrees\n", + " ii)regulation= 5.6920997883 %\n", + "b)efficiency= 96.5250965251 %\n" + ] + } + ], + "prompt_number": 60 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.75, Page Number:1176" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=10.0#kVA\n", + "f=50.0#Hz\n", + "v1=500.0#V\n", + "v2=250.0#V\n", + "vo=250.0#V\n", + "io=3.0#A\n", + "wo=200.0#W\n", + "vsc=15.0#V\n", + "isc=30.0#A\n", + "wsc=300.0#W\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "i=load*1000/v2\n", + "cu_loss=(i/isc)**2*wsc\n", + "output=load*1000*pf\n", + "efficiency=output*100/(output+cu_loss+wo)\n", + "z=vsc/isc\n", + "r=wsc/isc**2\n", + "x=(z**2-r**2)**0.5\n", + "regn=(i/v2)*(r*pf-x*math.sin(math.acos(pf)))*v2\n", + "\n", + "#result\n", + "print \"efficiency=\",efficiency,\"%\"\n", + "print \"regulation=\",regn,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "efficiency= 91.6030534351 %\n", + "regulation= 1.72239475667 %\n" + ] + } + ], + "prompt_number": 64 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.76, Page Number:1177" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=40.0#kVA\n", + "loss=400.0#W\n", + "cu_loss=800.0#W\n", + "\n", + "#calculation\n", + "x=(loss/cu_loss)**0.5\n", + "output=load*x*1\n", + "efficiency=output/(output+load*2/100)\n", + "\n", + "#result\n", + "print \"efficiency=\",efficiency*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "efficiency= 97.2493723732 %\n" + ] + } + ], + "prompt_number": 71 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.77, Page Number:1178" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=10#kVA\n", + "v1=500#V\n", + "v2=250#V\n", + "vsc=60#V\n", + "isc=20#A\n", + "wsc=150#W\n", + "per=1.2\n", + "pf=0.8\n", + "\n", + "#calculation\n", + "i=load*1000/v1\n", + "cu_loss=per**2*wsc\n", + "output=per*load*1.0\n", + "efficiency=output*100/(output+cu_loss*2/1000)\n", + "output=load*1000*pf\n", + "e2=output*100/(output+cu_loss+wsc)\n", + "\n", + "#result\n", + "print \"maximum efficiency=\",efficiency,\"%\"\n", + "print \"full-load efficiency=\",e2,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "maximum efficiency= 96.5250965251 %\n", + "full-load efficiency= 95.6251494143 %\n" + ] + } + ], + "prompt_number": 75 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.78, Page Number:1181" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=500.0#kVA\n", + "cu_loss=4.5#kW\n", + "iron_loss=3.5#kW\n", + "t1=6.0#hrs\n", + "t2=10.0#hrs\n", + "t3=4.0#hrs\n", + "t4=4.0#hrs\n", + "load1_=400.0#kW\n", + "load2_=300.0#kW\n", + "load3_=100.0#kW\n", + "pf1=0.8\n", + "pf2=0.75\n", + "pf3=0.8\n", + "\n", + "#calculations\n", + "load1=load1_/pf1\n", + "load2=load2_/pf2\n", + "load3=load3_/pf3\n", + "wc1=cu_loss\n", + "wc2=cu_loss*(load2/load1)**2\n", + "wc3=cu_loss*(load3/load1)**2\n", + "twc=(t1*wc1)+(t2*wc2)+(t3*wc3)+(t4*0)\n", + "iron_loss=24*iron_loss\n", + "total_loss=twc+iron_loss\n", + "output=(t1*load1_)+(t2*load2_)+(t3*load3_)\n", + "efficiency=output*100/(output+total_loss)\n", + "\n", + "#result\n", + "print \"efficiency=\",round(efficiency,1),\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "efficiency= 97.6 %\n" + ] + } + ], + "prompt_number": 86 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.79, Page Number:1182" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=100.0#kVA\n", + "loss=3.0#kW\n", + "tf=3.0#hrs\n", + "th=4.0#hrs\n", + "\n", + "#calculation\n", + "iron_loss=loss*24/2\n", + "wcf=loss*tf/2\n", + "wch=loss/8\n", + "wch=wch*4\n", + "total_loss=iron_loss+wch+wcf\n", + "output=load*tf+load*th/2\n", + "efficiency=output*100/(output+total_loss)\n", + "\n", + "#result\n", + "print \"efficiency=\",efficiency,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "efficiency= 92.2509225092 %\n" + ] + } + ], + "prompt_number": 89 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.80, Page Number:1182" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=100.0#KW\n", + "efficiency=0.98\n", + "tf=4.0#hrs\n", + "th=6.0#hrs\n", + "t10=14.0#hrs\n", + "\n", + "#calculations\n", + "#1st transformer\n", + "inpt=load/efficiency\n", + "tloss=inpt-load\n", + "y=tloss/2\n", + "x=y\n", + "iron_loss=x*24\n", + "cu_loss=x*tf+th*(x/2**2)+t10*(x/10**2)\n", + "loss=iron_loss+cu_loss\n", + "output=tf*load+th*load/2+t10*10\n", + "e1=output/(output+loss)\n", + "#2nd transformer\n", + "y=tloss/(1+1.0/4)\n", + "x=(tloss-y)\n", + "iron_loss=x*24\n", + "wc=tf*y+th*(y/2**2)+t10*(y/10**2)\n", + "loss=iron_loss+wc\n", + "e2=output/(output+loss)\n", + "\n", + "#result\n", + "print \"efficiency of forst transformer=\",e1*100,\"%\"\n", + "print \"efficiency ofsecond transformer=\",e2*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "0.408163265306 1.63265306122\n", + "efficiency of forst transformer= 96.5245532574 %\n", + "efficiency ofsecond transformer= 97.7876610788 %\n" + ] + } + ], + "prompt_number": 96 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.81, Page Number:1183" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=5.0#kVA\n", + "efficiency=0.95\n", + "nl=10.0#hrs\n", + "ql=7.0#hrs\n", + "hl=5.0#hrs\n", + "fl=2.0#hrs\n", + "\n", + "#calculations\n", + "inpt=load/efficiency\n", + "loss=inpt-load\n", + "wc_fl=loss/2\n", + "iron_loss=loss/2\n", + "wc_fl_4=(1.0/4)**2*wc_fl\n", + "wc_fl_2=(1.0/2)**2*wc_fl\n", + "wc_ql=ql*wc_fl_4\n", + "wc_hl=hl*wc_fl_2\n", + "wc_fl_2=fl*wc_fl\n", + "wc=wc_ql+wc_hl+wc_fl_2\n", + "wh=wc\n", + "loss=wh+24*iron_loss\n", + "output=load*1\n", + "half_output=(output/2)\n", + "q_load=(load/4)\n", + "output=ql*q_load+hl*half_output+fl*output\n", + "e=output*100/(output+loss)\n", + "\n", + "#result\n", + "print \"efficiency=\",e,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "efficiency= 89.5592740985 %\n" + ] + } + ], + "prompt_number": 115 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.82, Page Number:1183" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "efficiency=0.98\n", + "load=15#kVA\n", + "t1=12.0#hrs\n", + "t2=6.0#hrs\n", + "t3=6.0#hrs\n", + "pf1=0.5\n", + "pf2=0.8\n", + "k1=2#kW\n", + "k2=12#kW\n", + "\n", + "#calculations\n", + "output=load*1\n", + "inpt=output/efficiency\n", + "loss=inpt-output\n", + "wc=loss/2\n", + "wi=loss/2\n", + "w1=k1/pf1\n", + "w2=k2/pf2\n", + "wc1=wc*(4/load)\n", + "wc2=wc\n", + "wc12=t1*wc1\n", + "wc6=t2*wc2\n", + "wc=(wc12+wc6)\n", + "wi=24*wi\n", + "output=(k1*t1)+(t2*k2)\n", + "inpt=output+wc+wi\n", + "e=output*100/inpt\n", + "\n", + "#result\n", + "print \"efficiency=\",e,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "0.918367346939 3.67346938776\n", + "efficiency= 95.4351795496 %\n" + ] + } + ], + "prompt_number": 120 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.83, Page Number:1184" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=150.0#KVA\n", + "l1_=100.0#kVA\n", + "t=3.0#hrs\n", + "loss=1.0#KW\n", + "\n", + "#calculations\n", + "l1=l1_/2\n", + "l2=l1_\n", + "output=load*1\n", + "loss=loss*2\n", + "e1=output/(output+loss)\n", + "wc1=t*(1.0/3)**2*1\n", + "wc2=8*(2.0/3)**2*1\n", + "wc=wc1+wc2\n", + "wi=24*1\n", + "loss=wc+wi\n", + "output=3*(l1*1)+8*(l2*1)\n", + "e2=(output*100)/(output+loss)\n", + "\n", + "#result\n", + "print \"ordinary efficiency=\",e1*100,\"%\"\n", + "print \"all day efficiency=\",e2,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "ordinary efficiency= 98.6842105263 %\n", + "all day efficiency= 97.1480513578 %\n" + ] + } + ], + "prompt_number": 127 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.84, Page Number:1184" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=50#KVA\n", + "efficiency=0.94#%\n", + "nl=10\n", + "hl=5.0\n", + "ql=6.0\n", + "fl=3.0\n", + "\n", + "#calculations\n", + "pi=0.5*(load*1000)*(1-efficiency)/efficiency\n", + "wch=(0.5)**2*pi\n", + "eh=wch*hl/1000\n", + "wcq=(0.25)**2*pi\n", + "eq=ql*wcq/1000\n", + "e3=pi*3/1000\n", + "e2=pi*24/1000\n", + "e=25*hl+12.5*ql+50*fl\n", + "efficiency=e/(e+e2+eh+eq+e3)\n", + "\n", + "#result\n", + "print \"efficiency=\",efficiency*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "efficiency= 88.4557217274 %\n" + ] + } + ], + "prompt_number": 129 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.85, Page Number:1185" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=10.0#kVA\n", + "t1=7.0#hrs\n", + "t2=4.0#hrs\n", + "t3=8.0#hrs\n", + "t4=5.0#hrs\n", + "k1=3.0#kW\n", + "k2=8.0#kW\n", + "pf1=0.6\n", + "pf2=0.8\n", + "\n", + "#calculations\n", + "x1=k1/(pf1*load)\n", + "x2=k2/(pf2*load)\n", + "x3=load/(1*load)\n", + "pc1=(0.5)**2*0.1\n", + "pc2=pc3=0.10\n", + "o1=k1*t1\n", + "o2=k2*t2\n", + "o3=k2*load\n", + "output=o1+o2+o3\n", + "wc1=pc1*t1\n", + "wc2=pc2*t2\n", + "wc3=pc3*t3\n", + "cu_loss=wc1+wc2+wc3\n", + "loss=400.0*24/10000\n", + "efficiency=output/(output+loss+cu_loss)\n", + "\n", + "#result\n", + "print \"efficency=\",efficiency*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "efficency= 98.27465179 %\n" + ] + } + ], + "prompt_number": 142 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.86, Page Number:1185" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "efficiency=.98\n", + "load=15.0#kVA\n", + "t1=12.0\n", + "t2=6.0\n", + "t3=6.0\n", + "pf1=0.8\n", + "pf2=0.8\n", + "pf3=0.9\n", + "k1=2.0\n", + "k2=12.0\n", + "k3=18.0\n", + "#calculations\n", + "output=load*1000\n", + "inpt=output/efficiency\n", + "loss=inpt-output\n", + "cu_loss=loss/2\n", + "x1=k1/(0.5*load)\n", + "x2=k2/(pf2*load)\n", + "x3=k3/(pf3*load)\n", + "wc1=0.131\n", + "wc2=0.918\n", + "wc3=1.632\n", + "o1=t1*k1\n", + "o2=t2*k2\n", + "o3=t3*k3\n", + "output=o1+o2+o3\n", + "loss=wc1+wc2+wc3+0.153*24\n", + "efficiency=(output*100)/(output+loss)\n", + "\n", + "#result\n", + "print \"efficiency=\",efficiency,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "efficiency= 96.9798386522 %\n" + ] + } + ], + "prompt_number": 143 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.87, Page Number:1188" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=3.0#kW\n", + "v1=115.0#V\n", + "v2=230.0#V\n", + "\n", + "#calculation\n", + "k=v1/v2\n", + "power=load*(1-k)\n", + "power2=k*load\n", + "\n", + "#result\n", + "print \"a)power transferred inductively=\",power,\"kW\"\n", + "print \"b)power transferred conductively=\",power2,\"kW\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a)power transferred inductively= 1.5 kW\n", + "b)power transferred conductively= 1.5 kW\n" + ] + } + ], + "prompt_number": 145 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.88, Page Number:1188" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v1=500.0#V\n", + "v2=400.0#V\n", + "i=100.0#A\n", + "\n", + "#calculations\n", + "k=v2/v1\n", + "i1=k*i\n", + "saving=k*100\n", + "\n", + "#result\n", + "print \"economy of cu=\",saving" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "economy of cu= 80.0\n" + ] + } + ], + "prompt_number": 147 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.89, Page Number:1188" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=500.0#KVA\n", + "f=50.0#Hz\n", + "v1=6600.0#V\n", + "v2=5000.0#V\n", + "e=8.0#V\n", + "phim1=1.3#Wb/m2\n", + "\n", + "#calculations\n", + "phim=e/(4.44*f)\n", + "area=phim/phim1\n", + "n1=v1/e\n", + "n2=v2/e\n", + "\n", + "#result\n", + "print \"core area=\",area*10000,\"m2\"\n", + "print \"number of turns on the hv side=\",n1\n", + "print \"number of turns on the lv side=\",n2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "core area= 277.2002772 m2\n", + "number of turns on the hv side= 825.0\n", + "number of turns on the lv side= 625.0\n" + ] + } + ], + "prompt_number": 150 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.90, Page Number:1189" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=20.0#KVA\n", + "v1=2400.0#V\n", + "v2=240.0#V\n", + "\n", + "#calculation\n", + "i1=round(load*1000/v1,1)\n", + "k=v2/v1\n", + "i2=i1/k\n", + "kva=2640*i2*0.001\n", + "kva_per=kva*100/load\n", + "i1_=kva*1000/v1\n", + "ic=i1_-i2\n", + "over=ic*100/i1\n", + "\n", + "#result\n", + "print \"i)i1=\",i1,\"A\"\n", + "print \"ii)i2=\",i2,\"A\"\n", + "print \"iii)kVA rating=\",kva,\"kVA\"\n", + "print \"iv)per cent increase in kVA=\",kva_per,\"%\"\n", + "print \"v)I1=\",i1_,\"A\"\n", + "print \" Ic=\",ic,\"A\"\n", + "print \"vi)per cent overload=\",over,\"%\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)i1= 8.3 A\n", + "ii)i2= 83.0 A\n", + "iii)kVA rating= 219.12 kVA\n", + "iv)per cent increase in kVA= 1095.6 %\n", + "v)I1= 91.3 A\n", + " Ic= 8.3 A\n", + "vi)per cent overload= 100.0 %\n" + ] + } + ], + "prompt_number": 159 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.91, Page Number:1190" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=20.0#KVA\n", + "v1=2400.0#V\n", + "v2=240.0#V\n", + "\n", + "#calculation\n", + "i1=round(load*1000/v1,1)\n", + "k=v2/v1\n", + "i2=i1/k\n", + "kva=2160*i2*0.001\n", + "kva_per=kva*100/load\n", + "i1_=kva*1000/v1\n", + "ic=i2-i1_\n", + "over=ic*100/i1\n", + "\n", + "#result\n", + "print \"i)i1=\",i1,\"A\"\n", + "print \"ii)i2=\",i2,\"A\"\n", + "print \"iii)kVA rating=\",kva,\"kVA\"\n", + "print \"iv)per cent increase in kVA=\",kva_per,\"%\"\n", + "print \"v)I1=\",i1_,\"A\"\n", + "print \" Ic=\",ic,\"A\"\n", + "print \"vi)per cent overload=\",over,\"%\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)i1= 8.3 A\n", + "ii)i2= 83.0 A\n", + "iii)kVA rating= 179.28 kVA\n", + "iv)per cent increase in kVA= 896.4 %\n", + "v)I1= 74.7 A\n", + " Ic= 8.3 A\n", + "vi)per cent overload= 100.0 %\n" + ] + } + ], + "prompt_number": 160 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.92, Page Number:1190" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=5.0#kVA\n", + "v1=110.0#V\n", + "v2=110.0#V\n", + "f=50.0#Hz\n", + "efficiency=0.95\n", + "iron_loss=50.0#W\n", + "v=220.0#V\n", + "\n", + "#calculations\n", + "cu_loss=load*1000/efficiency-load*1000-iron_loss\n", + "efficiency=load*1000/(load*1000+cu_loss/4+iron_loss)\n", + "i2=(load*1000+cu_loss/4+iron_loss)/v\n", + "\n", + "#result\n", + "print \"efficiency=\",efficiency*100,\"%\"\n", + "print \"current drawn on hv side=\",i2,\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "efficiency= 97.9760216579 %\n", + "current drawn on hv side= 23.1967703349 A\n" + ] + } + ], + "prompt_number": 163 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.93, Page Number:1191" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v1=11500#V\n", + "v2=2300#V\n", + "\n", + "#calculations\n", + "kva=(v1+v2)*50*0.001\n", + "\n", + "#result\n", + "print \"voltage output=\",v1+v2,\"V\"\n", + "print \"kVA rating of auto transformer=\",kva,\"kVA\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "voltage output= 13800 V\n", + "kVA rating of auto transformer= 690.0 kVA\n" + ] + } + ], + "prompt_number": 164 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.94, Page Number:1191" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v1=11500.0#V\n", + "v2=2300.0#V\n", + "load=100.0#KVA\n", + "\n", + "#calculations\n", + "i1=load*100/v1\n", + "i2=load*100/v2\n", + "kva1=(v1+v2)*i1/(100)\n", + "kva2=(v1+v2)*i2/(100)\n", + "#result\n", + "print \"voltage ratios=\",(v1+v2)/v1,\"or\",(v1+v2)/v2\n", + "print \"kVA rating in first case=\",kva1\n", + "print \"kVA rating in second case=\",kva2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "voltage ratios= 1.2 or 6.0\n", + "kVA rating in first case= 120.0\n", + "kVA rating in second case= 600.0\n" + ] + } + ], + "prompt_number": 167 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.95, Page Number:1192" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v1=2400.0#v\n", + "v2=240.0#V\n", + "load=50.0#kVA\n", + "\n", + "#calculations\n", + "i1=load*1000/v1\n", + "i2=load*1000/v2\n", + "output=2640*i2\n", + "i=i2*2640/v1\n", + "k=2640/v1\n", + "poweri=v1*i1*0.001\n", + "power=output/1000-poweri\n", + "\n", + "#result\n", + "print \"rating of the auto-transformer=\",output/1000,\"kVA\"\n", + "print \"inductively transferred powers=\",poweri,\"kW\"\n", + "print \"conductively transferred powers=\",power,\"kW\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "rating of the auto-transformer= 550.0 kVA\n", + "inductively transferred powers= 50.0 kW\n", + "conductively transferred powers= 500.0 kW\n" + ] + } + ], + "prompt_number": 169 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.96, Page Number:1196" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "za=complex(0.5,3)\n", + "zb=complex(0.,10)\n", + "load=100#KW\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "s=load/pf*complex(pf,math.sin(math.acos(pf)))\n", + "sa=s*zb/(za+zb)\n", + "sb=s*za/(za+zb)\n", + "\n", + "#result\n", + "print \"SA=\",abs(sa)*math.cos(math.atan(sa.imag/sa.real)),\"kW\"\n", + "print \"SB=\",abs(sb)*math.cos(math.atan(sb.imag/sb.real)),\"kW\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "96.082805253\n", + "SA= 74.5937961595 kW\n", + "SB= 25.4062038405 kW\n" + ] + } + ], + "prompt_number": 174 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.97, Page Number:1197" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "r1=0.005#ohm\n", + "r2=0.01#ohm\n", + "x1=0.05#ohm\n", + "x2=0.04#ohm\n", + "pf=0.8\n", + "\n", + "#calculation\n", + "za=complex(r1,x1)\n", + "zb=complex(r2,x2)\n", + "pf=math.cos(math.degrees((-1)*math.acos(pf))*math.degrees(math.atan((za/zb).imag/(za/zb).real)))\n", + "\n", + "#result\n", + "print \"load of B=\",abs(za/zb)\n", + "print \"pf of B=\",pf" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "load of B= 1.21872643265\n", + "pf of B= 0.613584256393\n" + ] + } + ], + "prompt_number": 202 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.98, Page Number:1197" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=250#kVA\n", + "za=complex(1,6)\n", + "zb=complex(1.2,4.8)\n", + "load1=500#kVA\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "s=load1*complex(-pf,math.sin(math.acos(pf)))\n", + "sa=s*zb/(za+zb)\n", + "sb=s*za/(za+zb)\n", + "\n", + "#result\n", + "print \"SA=\",abs(sa),math.degrees(math.atan(sa.imag/sa.real)),\"degrees\"\n", + "print \"SB=\",abs(sb),math.degrees(math.atan(sb.imag/sb.real)),\"degrees\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "SA= 224.451917244 -39.3923099293\n", + "SB= 275.942423833 -34.8183886694\n" + ] + } + ], + "prompt_number": 205 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.99, Page Number:1197" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variabledeclaration\n", + "load=100.0#KW\n", + "r1=0.5\n", + "x1=8.0\n", + "r2=0.75\n", + "x2=4.0\n", + "load1=180.0#kW\n", + "pf=0.9\n", + "\n", + "#calculations\n", + "load=load1/pf\n", + "s=load*complex(pf,-math.sin(math.acos(pf)))\n", + "z1=complex(r1,x1)\n", + "z2=complex(r2,x2)\n", + "s1=s*z2/(z1+z2)\n", + "s2=s*z1/(z1+z2)\n", + "kw1=abs(s1)*math.cos(math.atan(s1.imag/s1.real))\n", + "kw2=abs(s2)*math.cos(math.atan(s2.imag/s2.real))\n", + "\n", + "#result\n", + "print \"kW1=\",kw1,\"kW\"\n", + "print \"kW2=\",kw2,\"kW\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(1.25+12j)\n", + "kW1= 58.119626171 kW\n", + "kW2= 121.880373829 kW\n" + ] + } + ], + "prompt_number": 214 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.100, Page Number:1197" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=200.0#kW\n", + "pf=0.85\n", + "za=complex(1,5)\n", + "zb=complex(2,6)\n", + "\n", + "#calculations\n", + "s=load/pf*complex(0.85,-0.527)\n", + "sa=s*zb/(za+zb)\n", + "sb=s*za/(za+zb)\n", + "\n", + "#result\n", + "print \"kVA for A=\",abs(sa),math.cos(math.atan(sa.imag/sa.real)),\"lag\"\n", + "print \"kVA for B=\",abs(sb),math.cos(math.atan(sb.imag/sb.real)),\"lag\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "kVA for A= 130.53263665 0.819364787986 lag\n", + "kVA for B= 105.238776124 0.884143252833 lag\n" + ] + } + ], + "prompt_number": 216 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.101, Page Number:1198" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v1=2200.0#V\n", + "v2=110.0#V\n", + "load=125.0#kVA\n", + "pf=0.8\n", + "za=complex(0.9,10)\n", + "zb=(100/50)*complex(1.0,5)\n", + "\n", + "#calculation\n", + "s=load*complex(pf,-math.sin(math.acos(pf)))\n", + "sa=s*zb/(za+zb)\n", + "sb=s*za/(za+zb)\n", + "\n", + "#result\n", + "print \"SA=\",abs(sa),math.degrees(math.atan(sa.imag/sa.real)),\"degrees\"\n", + "print \"SB=\",abs(sb),math.degrees(math.atan(sb.imag/sb.real)),\"degrees\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "SA= 63.0780848499 -39.929442891 degrees\n", + "SB= 62.1031510961 -33.7622749748 degrees\n" + ] + } + ], + "prompt_number": 218 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.102, Page Number:1199" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load1=500#kVA\n", + "za=complex(1,5)\n", + "load2=250#kVA\n", + "zb=complex(1.5,4)\n", + "v2=400#V\n", + "load=750#kVA\n", + "pf=0.8\n", + "\n", + "#calculation\n", + "zb=(500/load2)*zb\n", + "s=load*complex(pf,-math.sin(math.acos(pf)))\n", + "sa=s*zb/(za+zb)\n", + "sb=s*za/(za+zb)\n", + "\n", + "#result\n", + "print \"SA=\",abs(sa),math.degrees(math.atan(sa.imag/sa.real)),\"degrees\"\n", + "print \"SB=\",abs(sb),math.degrees(math.atan(sb.imag/sb.real)),\"degrees\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "SA= 471.125736359 -40.3232138964 degrees\n", + "SB= 281.165527855 -31.0771011508 degrees\n" + ] + } + ], + "prompt_number": 219 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.103, Page Number:1199" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "i=1000#A\n", + "pf=0.8\n", + "za=complex(2,3)\n", + "zb=complex(2.5,5)\n", + "\n", + "#calculations\n", + "i=i*complex(pf,-math.sin(math.acos(pf)))\n", + "ratio=zb/za\n", + "ib=i/(1+ratio)\n", + "ia=i-ib\n", + "ratio=ia.real/ib.real\n", + "\n", + "#result\n", + "print \"IA=\",ia\n", + "print \"IB=\",ib\n", + "print \"ratio of output=\",ratio" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "IA= (504.451038576-341.246290801j)\n", + "IB= (295.548961424-258.753709199j)\n", + "ratio of output= 1.70682730924\n" + ] + } + ], + "prompt_number": 220 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.104, Page Number:1200" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v1=1000.0#V\n", + "v2=500.0#V\n", + "load=100.0#kVA\n", + "za=complex(1.0,5.0)\n", + "zb=complex(2.0,2.0)\n", + "load1=300.0#kVA\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "zb=(100.0/250)*zb\n", + "s=load1*complex(pf,-math.sin(math.acos(pf)))\n", + "sa=s*zb/(za+zb)\n", + "sb=s*za/(za+zb)\n", + "zab=za*zb/(za+zb)\n", + "drop=zab.real*240/100+zab.imag*180/100\n", + "v2=v2-v2*drop/100\n", + "\n", + "#result\n", + "print \"SA=\",abs(sa),math.degrees(math.atan(sa.imag/sa.real)),\"degrees\"\n", + "print \"SB=\",abs(sb),math.degrees(math.atan(sb.imag/sb.real)),\"degrees\"\n", + "print \"secondary voltage=\",v2,\"V\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "SA= 55.8895719399 -64.6284382469 degrees\n", + "SB= 251.890896741 -30.9383707209 degrees\n", + "secondary voltage= 486.177874187 V\n" + ] + } + ], + "prompt_number": 223 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.105, Page Number:1200" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n11=5000.0\n", + "n12=440.0\n", + "load1=200#kVA\n", + "n21=5000.0\n", + "n22=480.0\n", + "load2=350#kVA\n", + "x=3.5\n", + "\n", + "#calculation\n", + "i1=load1*1000/n12\n", + "i2=load2*1000/n22\n", + "x1=x*n12/(100*i1)\n", + "x2=x*n22/(100*i2)\n", + "ic=(n22-n12)/0.057\n", + "\n", + "#result\n", + "print \"no-load circulation current=\",ic/i1,\"times the normal current of 200 kVA unit\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "no-load circulation current= 1.54385964912 times the normal current of 200 kVA unit\n" + ] + } + ], + "prompt_number": 225 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.106, Page Number:1203" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variabe declaration\n", + "ea=6600#V\n", + "eb=6400#V\n", + "za=complex(0.3,3)\n", + "zb=complex(0.2,1)\n", + "zl=complex(8.0,6.0)\n", + "ia=(ea*zb+(ea-eb)*zl)/(za*zb+zl*(za+zb))\n", + "ib=(eb*za-(ea-eb)*zl)/(za*zb+zl*(za+zb))\n", + "\n", + "#result\n", + "print \"IA=\",abs(ia),\"A\"\n", + "print \"IB=\",abs(ib),\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "IA= 195.492387533 A\n", + "IB= 422.567795916 A\n" + ] + } + ], + "prompt_number": 227 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.107, Page Number:1204" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load1=100.0#kVA\n", + "load2=50.0#kVA\n", + "v1=1000.0#V\n", + "v2=950.0#V\n", + "r1=2.0\n", + "r2=2.5\n", + "x1=8.0\n", + "x2=6.0\n", + "\n", + "#calculations\n", + "ia=load1*1000/v1\n", + "ra=v1*r1/(100*ia)\n", + "xa=v1*x1/(100*ia)\n", + "ib=load2*1000/v2\n", + "rb=v2*r2/(100*ib)\n", + "xb=v2*x2/(100*ib)\n", + "z=((ra+rb)**2+(xa+xb)**2)**0.5\n", + "ic=(v1-v2)/z\n", + "alpha=math.atan((xa+xb)/(ra+rb))\n", + "\n", + "#result\n", + "print \"no load circulating current=\",ic,\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "no load circulating current= 25.0948635944 A\n" + ] + } + ], + "prompt_number": 231 + }, + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Example Number 32.108, Page Number:1204" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load1=1000.0#KVA\n", + "load2=500.0#kVA\n", + "v1=500.0#V\n", + "v2=510.0#V\n", + "z1=3.0\n", + "z2=5.0\n", + "r=0.4\n", + "\n", + "#calculation\n", + "ia=load1*1000/480\n", + "ib=load2*1000/480\n", + "za=z1*v1/(100*ia)\n", + "zb=z2*v2/(100*ib)\n", + "ic=(v2-v1)/(za+zb)\n", + "\n", + "#result\n", + "print \"cross current=\",ic,\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "cross current= 315.656565657 A\n" + ] + } + ], + "prompt_number": 233 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.109, Page Number:1204" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "loada=500.0#KVA\n", + "loadb=250.0#kVA\n", + "load=750.0#KVA\n", + "pf=0.8\n", + "v1=405.0#V\n", + "v2=415.0#V\n", + "ra=1.0\n", + "rb=1.5\n", + "xa=5.0\n", + "xb=4.0\n", + "\n", + "#calculations\n", + "ia=loada*1000/400\n", + "ra=400/(100*ia)\n", + "xa=xa*400/(100*ia)\n", + "ib=loadb*1000/400\n", + "rb=rb*400/(100*ib)\n", + "xb=xb*400/(100*ib)\n", + "za=complex(ra,xa)\n", + "zb=complex(rb,xb)\n", + "zl=400**2*0.001/load*complex(pf,math.sin(math.acos(pf)))\n", + "ic=(v1-v2)/(za+zb)\n", + "ia=(v1*zb+(v1-v2)*zl)/(za*zb+zl*(za+zb))\n", + "ib=(v2*za-(v1-v2)*zl)/(za*zb+zl*(za+zb))\n", + "sa=400*ia/1000\n", + "sb=400*ib/1000\n", + "pf1=math.cos(math.atan(sa.imag/sa.real))\n", + "pf2=math.cos(math.atan(sb.imag/sb.real))\n", + "\n", + "#result\n", + "print \"a)cross current=\",-abs(ic),math.degrees(math.atan(ic.imag/ic.real))\n", + "print \"b)SA=\",abs(sa),pf1,\"lag\"\n", + "print \" SB=\",abs(sb),pf2,\"lag\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a)cross current= -229.754569404 -72.8972710309\n", + "b)SA= 387.844943528 0.820048560714 lag\n", + " SB= 351.964386212 0.738709225528 lag\n" + ] + } + ], + "prompt_number": 243 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.110, Page Number:1205" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "zl=complex(2.0,1.5)\n", + "za=complex(0.15,0.5)\n", + "zb=complex(0.1,0.6)\n", + "ea=207#V\n", + "eb=205#V\n", + "\n", + "#calculations\n", + "ia=(ea*zb+(ea-eb)*zl)/(za*zb+zl*(za+zb))\n", + "ib=(eb*za-(ea-eb)*zl)/(za*zb+zl*(za+zb))\n", + "v2_=(ia+ib)*zl\n", + "angle=math.atan(v2_.imag/v2_.real)-math.atan(ia.imag/ia.real)\n", + "pfa=math.cos(angle)\n", + "angle=math.atan(v2_.imag/v2_.real)-math.atan(ib.imag/ib.real)\n", + "pfb=math.cos(angle)\n", + "pa=abs(v2_)*abs(ia)*pfa\n", + "pb=abs(v2_)*abs(ib)*pfb\n", + "\n", + "#result\n", + "print \"power output:\"\n", + "print \" A:\",pa,\"W\"\n", + "print \" B:\",pb,\"W\"\n", + "print \"power factor:\"\n", + "print \" A:\",pfa\n", + "print \" B:\",pfb\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "power output:\n", + " A: 6535.37583042 W\n", + " B: 4925.36941503 W\n", + "power factor:\n", + " A: 0.818428780129\n", + " B: 0.775705655277\n" + ] + } + ], + "prompt_number": 248 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 32.111, Page Number:1206" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "ia=200.0#A\n", + "ib=600.0#A\n", + "ra=0.02#ohm\n", + "rb=0.025#ohm\n", + "xa=0.05#ohm\n", + "xb=0.06#ohm\n", + "ea=245.0#V\n", + "eb=240.0#V\n", + "zl=complex(0.25,0.1)\n", + "\n", + "#calculation\n", + "za=(ea/ia)*complex(ra,xa)\n", + "zb=(eb/ib)*complex(rb,xb)\n", + "i=(ea*zb+eb*za)/(za*zb+zl*(za+zb))\n", + "v2=i*zl\n", + "\n", + "#result\n", + "print \"terminal voltage=\",round(abs(v2)),round(math.degrees(math.atan(v2.imag/v2.real))),\"degrees\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "terminal voltage= 230.0 -3.0 degrees\n" + ] + } + ], + "prompt_number": 251 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter33_2.ipynb b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter33_2.ipynb new file mode 100644 index 00000000..495cee05 --- /dev/null +++ b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter33_2.ipynb @@ -0,0 +1,1433 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:62e227cc38186a0706017dd159987c82bd21be1d7e8602e20c55cf079ab30efe" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 33: Transformer:Three Phase" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.1, Page Number:1216" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "p=3\n", + "f=50.0#Hz\n", + "vd=22000.0#V\n", + "vs=400.0#V\n", + "phi=0.8\n", + "i=5.0#A\n", + "\n", + "#calcuations\n", + "v_phase_secondary=vs/math.sqrt(3)\n", + "K=(vs/vd)/math.sqrt(3)\n", + "i_primary=i/math.sqrt(3)\n", + "i_secondary=i_primary/K\n", + "il=i_secondary\n", + "output=math.sqrt(3)*il*vs*phi\n", + "\n", + "#result\n", + "print \"Output=\",output/10000,\"kW\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Output= 15.2420471066 kW\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.2, Page Number:1217" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "w=500.0#kVA\n", + "f=50.0#Hz\n", + "vls=11.0#kV\n", + "vld=33.0#kV\n", + "rh=35.0#ohm\n", + "rl=0.876#ohm\n", + "iron_loss=3050.0#W\n", + "phi1=1.0\n", + "phi2=0.8\n", + "\n", + "#calculations\n", + "\n", + "K=(vls*1000)/(math.sqrt(3)*vld*1000)\n", + "r02=rl+K**2*rh\n", + "i_Secondary=(w*1000)/(math.sqrt(3)*vls*1000)\n", + "#full load\n", + "fl_culoss=3*((w/(vls*math.sqrt(3)))**2)*r02\n", + "fl_totalloss=fl_culoss+iron_loss\n", + "fl_efficiency1=w*1000/(w*1000+fl_totalloss)\n", + "fl_efficiency2=(phi2*w*1000)/(w*phi2*1000+fl_totalloss)\n", + "#half load\n", + "cu_loss=.5**2*fl_culoss\n", + "totalloss=cu_loss+iron_loss\n", + "efficiency1=(w*1000/2)/((w*1000/2)+totalloss)\n", + "efficiency2=(w*1000*phi2/2)/((phi2*w*1000/2)+totalloss)\n", + "#result\n", + "print \"full load efficiency at p.f. 1=\",fl_efficiency1*100,\"%\"\n", + "print \"full load efficiency at p.f. 0.8=\",fl_efficiency2*100,\"%\"\n", + "print \"half load efficiency at p.f. 1=\",efficiency1*100,\"%\"\n", + "print \"half load efficiency at p.f. 0.8=\",round(efficiency2*100),\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "full load efficiency at p.f. 1= 98.5147491838 %\n", + "full load efficiency at p.f. 0.8= 98.1503046336 %\n", + "half load efficiency at p.f. 1= 98.3585709725 %\n", + "half load efficiency at p.f. 0.8= 98.0 %\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.3, Page Number:1218" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "r=0.02\n", + "va=2000\n", + "reactance=0.1\n", + "pf=0.8\n", + "phi=math.acos(pf)\n", + "#calculation\n", + "cu_loss=r*100*va/100\n", + "regn=r*100*math.cos(phi)+reactance*100*math.sin(phi)\n", + "\n", + "#result\n", + "print \"Cu loss=\",cu_loss,\"kW\"\n", + "print \"Regulation=\",regn,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Cu loss= 40.0 kW\n", + "Regulation= 7.6 %\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.4, Page Number:1218" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "w=120.0#kVA\n", + "v1=6000.0\n", + "v2=400.0\n", + "f=50.0#Hz\n", + "iron_loss=1600.0#W\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "cu_loss_fl=iron_loss*((4/3)**2)\n", + "fl_output=w*pf*1000\n", + "total_loss=iron_loss+cu_loss_fl\n", + "efficiency1=fl_output/(fl_output+total_loss)\n", + "cu_loss_hl=0.5**2*cu_loss_fl\n", + "total_loss2=cu_loss_hl+iron_loss\n", + "efficiency2=(w*1000/2)/((w*1000/2)+total_loss2)\n", + "total_loss3=2*iron_loss\n", + "output=(3.0/4)*w*1000\n", + "inpt=output+total_loss3\n", + "efficiency=output/inpt\n", + "\n", + "\n", + "#result\n", + "print \"full load efficiency=\",efficiency1*100,\"%\"\n", + "print \"half load efficiency=\",efficiency2*100,\"%\"\n", + "print \"3/4 load efficiency=\",efficiency*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "full load efficiency= 96.7741935484 %\n", + "half load efficiency= 96.7741935484 %\n", + "3/4 load efficiency= 96.5665236052 %\n" + ] + } + ], + "prompt_number": 46 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.5, Page Number:1218" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "rp=8.0#ohm\n", + "rs=0.08#ohm\n", + "z=0.07\n", + "pf=0.75\n", + "v1=33.0\n", + "v2=6.6\n", + "w=2*10.0**6\n", + "phi=math.acos(pf)\n", + "#calculations\n", + "fl_i=w/(math.sqrt(3)*v2*10**3)\n", + "K=v2/(math.sqrt(3)*v1)\n", + "r02=rs+(rp*(K*K))\n", + "z_drop=z*v2*1000/math.sqrt(3)\n", + "z02=z_drop/fl_i\n", + "x02=math.sqrt((z02*z02)-(r02*r02))\n", + "drop=fl_i*(r02*math.cos(phi)+x02*math.sin(phi))\n", + "secondary_v=v2*1000/math.sqrt(3)\n", + "V2=secondary_v-drop\n", + "line_v=V2*math.sqrt(3)\n", + "regn=drop*100/secondary_v\n", + "\n", + "#result\n", + "print \"secondary voltage\",line_v,\"V\"\n", + "print \"regulation=\",regn,\"%\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "secondary voltage 6254.29059005 V\n", + "regulation= 5.23802136291 %\n" + ] + } + ], + "prompt_number": 59 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.6, Page Number:1219" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "w=100.0#kWA\n", + "f=50.0#Hz\n", + "v1=3300.0#V\n", + "v2=400.0#V\n", + "rh=3.5#ohm\n", + "rl=0.02#ohm\n", + "pf=0.8\n", + "efficiency=0.958\n", + "\n", + "#calculations\n", + "output=0.8*100\n", + "inpt=output/efficiency\n", + "total_loss=(inpt-output)*1000\n", + "K=v2/(math.sqrt(3)*v1)\n", + "r02=rl+K**2*rh\n", + "i2=((w*1000)/math.sqrt(3))/v2\n", + "cu_loss=3*i2**2*r02\n", + "iron_loss=total_loss-cu_loss\n", + "#result\n", + "print \"ironloss=\",iron_loss,\"W\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "0.0371411080502\n", + "2321.31925314\n", + "ironloss= 1185.98763622 W\n" + ] + } + ], + "prompt_number": 75 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.7, Page Number:1219" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "w=5000.0#kVA\n", + "v1=6.6#kV\n", + "v2=33.0#kV\n", + "nl=15.0#kW\n", + "fl=50.0#kW\n", + "drop=0.07\n", + "load=3200.0#kw\n", + "pf=0.8\n", + "phi=math.acos(pf)\n", + "#calculations\n", + "i2=w*1000/(math.sqrt(3)*v2*1000)\n", + "impedence_drop=drop*(v2/math.sqrt(3))*1000\n", + "z02=impedence_drop/i2\n", + "cu_loss=fl-nl\n", + "r02=cu_loss*1000/(3*i2**2)\n", + "x02=math.sqrt(z02**2-r02**2)\n", + "print \"full-load x02:\",x02\n", + "\n", + "#when load=3200#kW\n", + "i2=load/(math.sqrt(3)*v2*0.8)\n", + "drop_=drop*1000*(r02*math.cos(phi)+z02*math.sin(phi))\n", + "regn=(drop_*100)/(v2*1000/math.sqrt(3))\n", + "vp=v1+regn/100*v1\n", + "print \"Primary voltage=\",vp*1000,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "full-load x02: 15.1695784661\n", + "Primary voltage= 6851.39317975 V\n" + ] + } + ], + "prompt_number": 95 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.8, Page Number:1219" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "r=1\n", + "x=6\n", + "v=6600#V\n", + "v2=4800#V\n", + "pf=0.8\n", + "phi=math.acos(pf)\n", + "#calculations\n", + "regn=(r*math.cos(phi)+z*math.sin(phi))\n", + "secondary_v=v2+regn/100*v2\n", + "secondary_vp=secondary_v/math.sqrt(3)\n", + "K=secondary_vp/v\n", + "\n", + "#result\n", + "print \"Transformation Ratio=\",K" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Transformation Ratio= 0.423426587968\n" + ] + } + ], + "prompt_number": 96 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.9, Page Number:1220" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "w=2000#kVA\n", + "v1=6600#V\n", + "v2=400#V\n", + "pf=0.8\n", + "scv=400#V\n", + "sci=175#A\n", + "scw=17#kW\n", + "ocv=400#V\n", + "oci=150#A\n", + "ocw=15#kW\n", + "phi=math.acos(pf)\n", + "#calculations\n", + "i1=sci/math.sqrt(3)\n", + "z01=scv/i1\n", + "r01=scw*1000/(3*i1*i1)\n", + "x01=math.sqrt(z01**2-r01**2)\n", + "r=i1*r01*100/v1\n", + "x=i1*x01*100/v1\n", + "regn=(r*math.cos(phi)-x*math.sin(phi))\n", + "I1=w*1000/(math.sqrt(3)*v1)\n", + "total_loss=scw+ocw\n", + "fl_output=w*pf\n", + "efficiency=fl_output/(fl_output+total_loss)\n", + "\n", + "#result\n", + "print \"% resistance=\",r,\"%\"\n", + "print \"% reactance=\",x,\"%\"\n", + "print \"% efficiency=\",efficiency*100,\"%\"\n", + "print \"%regulation=\",regn,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "% resistance= 0.849779616989 %\n", + "% reactance= 6.00073499035 %\n", + "% efficiency= 98.0392156863 %\n", + "%regulation= -2.92061730062 %\n" + ] + } + ], + "prompt_number": 109 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.10, Page Number:1220" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v1=11000.0#V\n", + "v2=440.0#V\n", + "i=5.0#A\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "secondary_rating=v2/math.sqrt(3)\n", + "primary_i=i/math.sqrt(3)\n", + "voltsamps=v1*5/math.sqrt(3)\n", + "i2=voltsamps/secondary_rating\n", + "output=pf*voltsamps/1000\n", + "\n", + "#result\n", + "print \"Each coil current=\",i2,\"A\"\n", + "print \"Total output=\",output,\"kW\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Each coil current= 125.0 A\n", + "Total output= 25.4034118443 kW\n" + ] + } + ], + "prompt_number": 116 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.12, Page Number:1224" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=40#kVA\n", + "\n", + "#calculations\n", + "kVA_per_transformer=load/2*1.15\n", + "delta_delta_rating=kVA_per_transformer*3\n", + "increase=(delta_delta_rating-load)*100/load\n", + "\n", + "#result\n", + "print \"increase=\",increase,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "increase= 72.5 %\n" + ] + } + ], + "prompt_number": 126 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.13, Page Number:1224" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "w=20#kVA\n", + "v1=2300#v\n", + "v2=230#V\n", + "load=40#kVA\n", + "\n", + "#calculations\n", + "kva_load=load/math.sqrt(3)\n", + "percent_rated=kva_load*100/w\n", + "kvarating_vv=2*w*0.866\n", + "vv_delta=kvarating_vv*100/60\n", + "percentage_increase=kva_load/(load/3)\n", + "\n", + "#result\n", + "print \"i)kVA load of each transformer=\",kva_load,\"kVA\"\n", + "print \"ii)per cent of rated load carried by each transformer=\",percent_rated,\"%\"\n", + "print \"iii)total kVA rating of the V-V bank\",kvarating_vv,\"kVA\"\n", + "print \"iv)ratio of the v-v bank to delta-delta bank\",vv_delta,\"%\"\n", + "print \"v)percent increase in load=\",percentage_increase*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)kVA load of each transformer= 23.0940107676 kVA\n", + "ii)per cent of rated load carried by each transformer= 115.470053838 %\n", + "iii)total kVA rating of the V-V bank 34.64 kVA\n", + "iv)ratio of the v-v bank to delta-delta bank 57.7333333333 %\n", + "v)percent increase in load= 177.646236674 %\n" + ] + } + ], + "prompt_number": 130 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.14, Page Number:1225" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "load=150.0#kW\n", + "v1=1000.0#V\n", + "pf=0.866\n", + "v=2000.0#V\n", + "\n", + "#calculations\n", + "il=load*1000/(pf*math.sqrt(3)*1000)\n", + "ip=il/math.sqrt(3)\n", + "ratio=v1/v\n", + "ip=ip*ratio\n", + "I=il\n", + "Ip=I*ratio\n", + "pf=86.6/100*pf\n", + "\n", + "#result\n", + "print \"delta-delta:current in the windings=\",ip,\"A\"\n", + "print \"v-v:current in the windings=\",Ip,\"A\"\n", + "print \"Power factor\",pf" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "delta-delta:current in the windings= 28.8683602771 A\n", + "v-v:current in the windings= 50.0014667312 A\n", + "Power factor 0.749956\n" + ] + } + ], + "prompt_number": 133 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.15, Page Number:1225" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "load=3000#kW\n", + "v=11#kV\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "I=load*1000/(math.sqrt(3)*v*1000*pf)\n", + "transformer_pf=86.6/100*pf\n", + "additional_load=72.5/100*load\n", + "total_load=additional_load+load\n", + "il=total_load*1000/(math.sqrt(3)*v*1000*pf)\n", + "\n", + "#result\n", + "print \"Il=\",il,\"A\"\n", + "print \"phase current=\",il/math.sqrt(3),\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Il= 339.521323075 A\n", + "phase current= 196.022727273 A\n" + ] + } + ], + "prompt_number": 134 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.16, Page Number:1225" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "load=400#kVA\n", + "pf=0.866\n", + "v=440#V\n", + "\n", + "#calculations\n", + "kVA_each=(load/2)/pf\n", + "phi=math.acos(pf)\n", + "p1=kVA_each*math.cos(math.radians(30-phi))\n", + "p2=kVA_each*math.cos(math.radians(30+phi))\n", + "p=p1+p2\n", + "\n", + "#result\n", + "print \"kVA supplied by each transformer=\",kVA_each,\"kVA\"\n", + "print \"kW supplied by each transformer=\",p,\"kW\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "kVA supplied by each transformer= 230.946882217 kVA\n", + "kW supplied by each transformer= 399.995027715 kW\n" + ] + } + ], + "prompt_number": 136 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.17, Page Number:1228" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=400.0#V\n", + "load=33.0#kVA\n", + "v2=3300.0#V\n", + "\n", + "#calculations\n", + "vl=0.866*v2\n", + "ilp=load*1000/(math.sqrt(3)*v2)\n", + "ils=ilp/(440/v2)\n", + "main_kva=v2*ilp*0.001\n", + "teaser_kva=0.866*main_kva\n", + "\n", + "#result\n", + "print \"voltage rating of each coil=\",vl\n", + "print \"current rating of each coil=\",ils\n", + "print \"main kVA=\",main_kva,\"kVA\"\n", + "print \"teaser kVA=\",teaser_kva,\"kVA\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "voltage rating of each coil= 2857.8\n", + "current rating of each coil= 43.3012701892\n", + "main kVA= 19.0525588833 kVA\n", + "teaser kVA= 16.4995159929 kVA\n" + ] + } + ], + "prompt_number": 139 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.18, Page Number:1231" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=440.0#V\n", + "v2=200.0#V\n", + "output=150.0#kVA\n", + "\n", + "#calculations\n", + "ratio=v2/v\n", + "i2=output*1000/(2*v2)\n", + "i1=i2*ratio\n", + "primary_volts=(math.sqrt(3)*v)/2\n", + "ratio=v2/primary_volts\n", + "\n", + "#result\n", + "print \"primary current=\",i1,\"A\"\n", + "print \"turns ratio\",ratio" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "primary current= 170.454545455 A\n", + "turns ratio 0.524863881081\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.19, Page Number:1231" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=100.0#V\n", + "v2=3300.0#V\n", + "p=400.0#kW\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "K=v/v2\n", + "i2=p*1000/(pf*v)\n", + "ip=1.15*K*i2\n", + "I2m=K*i2\n", + "i2=ip/2\n", + "i1m=math.sqrt(I2m**2+i2**2)\n", + "\n", + "#reslult\n", + "print \"Current=\",i1m,\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current= 174.77684841 A\n" + ] + } + ], + "prompt_number": 150 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.20, Page Number:1232" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "w1=300#kW\n", + "w2=450#kW\n", + "v1=100#V\n", + "pf=0.707\n", + "v2=3300#V\n", + "\n", + "#calculations\n", + "K=v/v2\n", + "i2t=(w2*1000)/(100*pf)\n", + "i1t=1.15*K*i2t\n", + "I2m=(K*w1*1000)/(100*pf)\n", + "i2=i1t/2\n", + "i1m=math.sqrt(I2m**2+i2**2)\n", + "\n", + "#result\n", + "print \"Current=\",i1m,\"A\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current= 169.804606659 A\n" + ] + } + ], + "prompt_number": 163 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.21, Page Number:1233" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v1=80.0#V\n", + "v2=11000.0#V\n", + "w1=500.0#kW\n", + "w2=800.0#kW\n", + "pf=0.5\n", + "\n", + "#calculations\n", + "K=v1/v2\n", + "#unity pf\n", + "i2t=w1*1000/v1\n", + "i1t=1.15*K*i2t\n", + "i2m=K*w2*1000/v1\n", + "i1t_half=i1t/2\n", + "ip=math.sqrt(i2m**2+i1t_half**2)\n", + "\n", + "print \"unity pf\"\n", + "print \"one 3 phase line carries\",i1t,\"A whereas the other 2 carry\",ip,\"A each\"\n", + "#0.5 pf\n", + "i2t=w1*1000/(v1*pf)\n", + "i1t=1.15*K*i2t\n", + "i2m=K*w2*1000/(v1*pf)\n", + "i1t_half=i1t/2\n", + "ip=math.sqrt(i2m**2+i1t_half**2)\n", + "print \"0.5 pf\"\n", + "print \"one 3 phase line carries\",i1t,\"A whereas the other 2 carry\",ip,\"A each\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "unity pf\n", + "one 3 phase line carries 52.2727272727 A whereas the other 2 carry 77.281082436 A each\n", + "0.5 pf\n", + "one 3 phase line carries 104.545454545 A whereas the other 2 carry 154.562164872 A each\n" + ] + } + ], + "prompt_number": 171 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.22, Page Number:1234" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v1=50#V\n", + "v2=4.6*1000#V\n", + "load=350#kW\n", + "w=200#kW\n", + "pf=0.8\n", + "\n", + "#calculation\n", + "K=v1/v2\n", + "i2t=w*1000/(v1*pf)\n", + "i1t=1.15*K*i2t\n", + "i2m=load*1000/(v1*pf)\n", + "Ki2m=K*i2m\n", + "i1t_half=i1t/2\n", + "i1m=math.sqrt(Ki2m**2+i1t_half**2)\n", + "\n", + "#result\n", + "print \"current in line A=\",i1t\n", + "print \"current in line B=\",i1m\n", + "print \"current in line C=\",i1m" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "current in line A= 62.5\n", + "current in line B= 100.11107076\n", + "current in line C= 100.11107076\n" + ] + } + ], + "prompt_number": 173 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.23, Page Number:1234" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=231#V\n", + "v2=6600#v\n", + "volt_induced=8#v\n", + "\n", + "#calculations\n", + "hv=v2/volt_induced\n", + "vl=v*math.sqrt(3)\n", + "n_lv1=vl/volt_induced\n", + "n_lv2=math.sqrt(3)*n_lv1/2\n", + "n=2*n_lv2/3\n", + "\n", + "#result\n", + "print \"neutral point is located on the\",math.ceil(n),\"th turn from A downwards\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "neutral point is located on the 29.0 th turn from A downwards\n" + ] + } + ], + "prompt_number": 176 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.24, Page Number:1235" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=6000.0#V\n", + "v2=440.0#V\n", + "f=50.0#Hz\n", + "area=300.0#cm2\n", + "flux=1.2#Wb/m2\n", + "\n", + "#calculations\n", + "n1=v/(4.44*f*flux*area*0.0001*0.9)\n", + "K=v2/v\n", + "n2=n1*K\n", + "n_lv=math.sqrt(3)*n2/2\n", + "turns=n_lv*2/3\n", + "\n", + "#result\n", + "print \"NUmber of turns in AN=\",math.floor(turns)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " NUmber of turns in AN= 35.0\n" + ] + } + ], + "prompt_number": 183 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.25, Page Number:1235" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=250.0#V\n", + "load=30.0#kVA\n", + "v2=250.0#V\n", + "\n", + "#calculations\n", + "il=load*1000/(math.sqrt(3)*v2)\n", + "vl=0.866*v2\n", + "kva=il*vl*(0.001)\n", + "\n", + "#result\n", + "print \"Voltage=\",vl,\"V\"\n", + "print \"kVA rating\",kva,\"kVA\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage= 216.5 V\n", + "kVA rating 14.9995599935 kVA\n" + ] + } + ], + "prompt_number": 185 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.26, Page Number:1237" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import cmath\n", + "#vaiable declaration\n", + "load=500#kVA\n", + "pf=0.8\n", + "za=complex(2,6)\n", + "zb=complex(2,5)\n", + "phi=math.acos(pf)\n", + "#calculations\n", + "s=load*complex(math.cos(phi),math.sin(phi))\n", + "z1=za/zb\n", + "z2=zb/za\n", + "sa=s/(1+z1)\n", + "sb=s/(1+z2)\n", + "pfa=cmath.phase(sa)\n", + "pfb=cmath.phase(sb)\n", + "#result\n", + "print \"sa=\",abs(sa)\n", + "print \"sb=\",abs(sb)\n", + "print \"cos phi_a=\",pfa\n", + "print \"cos phi_b=\",pfb" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "sa= 230.042839552\n", + "sb= 270.171613479\n", + "cos phi_a= 0.611765735265\n", + "cos phi_b= 0.670521557981\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.27, Page Number:1237" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import cmath\n", + "#variable declaration\n", + "w=2000#kVA\n", + "w1=4000#kVA\n", + "w2=5000#kVA\n", + "pf=0.8\n", + "za=complex(2,8)\n", + "zb=complex(1.6,3)\n", + "\n", + "#calculations\n", + "za_per=(w1/w)*za\n", + "zb_per=zb\n", + "z=za_per+zb_per\n", + "s=complex(w1,w-w2)\n", + "sb=s*(za/z)\n", + "sa=s-sb\n", + "\n", + "#result\n", + "print \"sa=\",sa\n", + "print \"sb=\",sb" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "sa= (2284.2287695-1821.49046794j)\n", + "sb= (1715.7712305-1178.50953206j)\n" + ] + } + ], + "prompt_number": 211 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.28, Page Number:1237" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import cmath\n", + "#variable declaration\n", + "load=1400#kVA\n", + "pf=0.866\n", + "w1=1000#kVA\n", + "w2=500#kVA\n", + "v1=6600\n", + "v2=400\n", + "za=complex(0.001,0.003)\n", + "zb=complex(0.0028,0.005)\n", + "phi=math.acos(pf)\n", + "#calculations\n", + "zb=(w1/w2)*zb\n", + "z=za/(za+zb)\n", + "x=math.cos(-phi)\n", + "y=math.sin(-phi)*1j\n", + "s=load*(x+y)\n", + "sb=s*z\n", + "sa=s-sb\n", + "\n", + "#result\n", + "print \"sa=\",sa\n", + "print \"sb=\",sb" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "sa= (929.911014012-588.664867724j)\n", + "sb= (282.488985988-111.396729565j)\n" + ] + } + ], + "prompt_number": 240 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.29, Page Number:1238" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import cmath\n", + "#variable declaration\n", + "load=750#kVA\n", + "pf=0.707\n", + "w1=500#kVA\n", + "w2=250#kVA\n", + "v1=3300\n", + "v2=400\n", + "za=complex(2,3)\n", + "zb=complex(1.5,4)\n", + "phi=math.acos(pf)\n", + "#calculations\n", + "zb=(w1/w2)*zb\n", + "z=za/(za+zb)\n", + "x=math.cos(-phi)\n", + "y=math.sin(-phi)*1j\n", + "s=load*(x+y)\n", + "sb=s*z\n", + "sa=s-sb\n", + "per_r=za.real*(sa.real)/w1\n", + "per_x=(za.imag)*(sa.imag)/w1\n", + "total_per=per_r+per_x\n", + "vl=v2-(total_per*4)\n", + "#result\n", + "print \"sa=\",sa\n", + "print \"sb=\",sb" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "sa= (399.511103547-348.770523615j)\n", + "sb= (130.738896453-181.639636072j)\n" + ] + } + ], + "prompt_number": 242 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.30, Page Number:1240" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "ratio=100/5\n", + "i=5#A\n", + "i1=3.5#A\n", + "\n", + "#calculations\n", + "il=i1*ratio\n", + "\n", + "#result\n", + "print \"Line current=\",il,\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Line current= 70.0 A\n" + ] + } + ], + "prompt_number": 214 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 33.31, Page Number:1240" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "i1=2000#A\n", + "i2=2500#A\n", + "i=5#A\n", + "\n", + "#calculations\n", + "ratio1=i1/i\n", + "ratio2=i2/i\n", + "\n", + "#result\n", + "print \"ratio in first case=\",ratio1\n", + "print \"ratio in second case=\",ratio2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "ratio in first case= 400\n", + "ratio in second case= 500\n" + ] + } + ], + "prompt_number": 216 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter34_2.ipynb b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter34_2.ipynb new file mode 100644 index 00000000..d05f1eeb --- /dev/null +++ b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter34_2.ipynb @@ -0,0 +1,3065 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:0f43ef5b4c05930620c5e3871d199970ead64e15a20629e8e926abd11e2e9167" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 34:Induction Motors" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.1, Page Number:1255" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n=290.0#rpm\n", + "f=50.0#Hz\n", + "Ns=300.0#rpm(considered)\n", + "#calculation\n", + "P=120*f/Ns\n", + "s=(Ns-n)/Ns\n", + "\n", + "#result\n", + "print \"no. of poles=\",P\n", + "print \"slip=\",s*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "no. of poles= 20.0\n", + "slip= 3.33333333333 %\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.2, Page Number:1255" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n=3\n", + "slot=3\n", + "f=50#Hz\n", + "\n", + "#calculation\n", + "P=2*n\n", + "slots_total=slot*P*n\n", + "Ns=120*f/P\n", + "\n", + "#result\n", + "print \"No. of stator poles=\",P\n", + "print \"Total number of slots=\",slots_total\n", + "print \"Speed=\",Ns,\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " No. of stator poles= 6\n", + "Total number of slots= 54\n", + "Speed= 1000 rpm\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.3, Page Number:1255" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=4\n", + "n=3\n", + "f=50#Hz\n", + "slip1=0.04\n", + "slip2=0.03\n", + "\n", + "#calculation\n", + "Ns=120*f/p\n", + "N=Ns*(1-slip1)\n", + "f1=slip2*f*60\n", + "#at standstill s=1\n", + "f2=1*f\n", + "\n", + "#calculation\n", + "print \"speed at which magnetic field of the stator is rotating=\",Ns,\"rpm\"\n", + "print \"speed of the rotor when the slip is 0.04=\",N\n", + "print \"frequency of rotor current=\",f1,\"rpm\"\n", + "print \"frequency of the rotor current at standstill=\",f2,\"Hz\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed at which magnetic field of the stator is rotating= 1500 rpm\n", + "speed of the rotor when the slip is 0.04= 1440.0\n", + "frequency of rotor current= 90.0 rpm\n", + "frequency of the rotor current at standstill= 50 Hz\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.4, Page Number:1255" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n=3.0\n", + "p=4.0\n", + "f=50.0#Hz\n", + "slip=0.04\n", + "n=600.0#rpm\n", + "\n", + "#calculations\n", + "Ns=120*f/p\n", + "N=Ns*(1-slip)\n", + "s=(Ns-n)/Ns\n", + "f1=s*f\n", + "\n", + "#result\n", + "print \"the synchronous speed=\",Ns,\"rpm\"\n", + "print \"the rotor speed=\",N,\"rpm\"\n", + "print \"the rotor frequency when n=600 rpm=\",f1,\"Hz\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the synchronous speed= 1500.0 rpm\n", + "the rotor speed= 1440.0 rpm\n", + "the rotor frequency when n=600 rpm= 30.0 Hz\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.5, Page Number:1256" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=12\n", + "n=3\n", + "N=500#rpm\n", + "p2=8\n", + "slip=0.03\n", + "\n", + "#calculation\n", + "f=p*N/120\n", + "Ns=120*f/p2\n", + "N=Ns-slip*Ns\n", + "\n", + "#result\n", + "print \"full load speed of the motor=\",N,\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "full load speed of the motor= 727.5 rpm\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.6, Page Number:1258" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "e=80#V\n", + "r=1#ohm\n", + "x=4#ohm\n", + "rheo=3#ohm\n", + "\n", + "#calculation\n", + "E=e/(3)**0.5\n", + "z=(r**2+x**2)**0.5\n", + "i=E/z\n", + "pf=r/z\n", + "R=rheo+r\n", + "z2=(R**2+x**2)**0.5\n", + "i2=E/z2\n", + "\n", + "pf2=R/z2\n", + "\n", + "#result\n", + "print \"slip rings are short circuited:\"\n", + "print \"current/phase\",i,\"A\"\n", + "print \"pf=\",pf\n", + "print \"slip rings are onnected to a star-connected rheostat of 3 ohm\",\n", + "print \"current/phase\",i2,\"A\"\n", + "print \"pf=\",pf2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "slip rings are short circuited:\n", + "current/phase 11.2022406722 A\n", + "pf= 0.242535625036\n", + "slip rings are onnected to a star-connected rheostat of 3 ohm current/phase 8.16496580928 A\n", + "pf= 0.707106781187\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.7, Page Number:1258" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n=3\n", + "v=400#V\n", + "ratio=6.5\n", + "r=0.05#ohm\n", + "x=0.25#ohm\n", + "\n", + "#calculations\n", + "k=1/ratio\n", + "e2=v*k/(3**0.5)\n", + "R=x-r\n", + "r2=x\n", + "z=(x**2+r2**2)**0.5\n", + "i2=e2/z\n", + "\n", + "#result\n", + "print \"external resistance=\",R,\"ohm\"\n", + "print \"starting current=\",i2,\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "external resistance= 0.2 ohm\n", + "starting current= 100.491886883 A\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.8, Page Number:1259" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=1100#V\n", + "f=50#Hz\n", + "ratio=3.8\n", + "r=0.012#ohm\n", + "x=0.25#ohm\n", + "s=0.04\n", + "#calculation\n", + "e=v/ratio\n", + "z=(r**2+x**2)**0.5\n", + "i=e/z\n", + "pf=r/z\n", + "xr=s*x\n", + "zr=(r**2+xr**2)**0.5\n", + "er=s*e\n", + "i2=er/zr\n", + "pf2=r/zr\n", + "i2=100*ratio\n", + "z2=e/i2\n", + "r2=(z2**2-x**2)**0.5\n", + "R=r2-r\n", + "\n", + "#result\n", + "print \"current with slip rings shorted=\",i,\"A\"\n", + "print \"pf with slip rings shorted=\",pf\n", + "print \"current with slip=4% and slip rings shorted=\",i2\n", + "print \"pf withslip=4% and slip rings shorted=\",pf2\n", + "print \"external resistance=\",R,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "current with slip rings shorted= 1156.56314266 A\n", + "pf with slip rings shorted= 0.0479447993684\n", + "current with slip=4% and slip rings shorted= 380.0\n", + "pf withslip=4% and slip rings shorted= 0.768221279597\n", + "external resistance= 0.70758173952 ohm\n" + ] + } + ], + "prompt_number": 41 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.9, Page Number:1259" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=15#kW\n", + "v=3000#V\n", + "f=50#Hz\n", + "p=6\n", + "ratio=3.6\n", + "r=0.13#ohm\n", + "l=3.61*0.001#H\n", + "\n", + "#calculation\n", + "v=v/3**0.5\n", + "x2=2*3.14*l*f\n", + "k=1/ratio\n", + "r2_=0.1/k**2\n", + "x2_=ratio**2*x2\n", + "is1=v/((r**2+x2_**2)**0.5)\n", + "ns=120*f/p\n", + "ts=(3*3/(2*3.14*f))*((v**2)*r2_)/(r2_**2+x2_**2)\n", + "\n", + "#result\n", + "print \"starting current=\",is1,\"A\"\n", + "print \"ts=\",ts,\"N-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "starting current= 117.896733436 A\n", + "ts= 512.375725888 N-m\n" + ] + } + ], + "prompt_number": 49 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.10, Page Number:1261" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "zs=complex(0.4,4)\n", + "zr=complex(6,2)\n", + "v=80#V\n", + "s=0.03\n", + "\n", + "#calculation\n", + "e2=v/3**0.5\n", + "i=e2/abs(zr+zs)\n", + "er=s*e2\n", + "xr=s*zs.imag\n", + "ir=er/abs(complex(zs.real,xr))\n", + "\n", + "#result\n", + "print \"rotor current at standstill=\",i,\"A\"\n", + "print \"rotor current when slip-rings are short-circuited=\",ir,\"A\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "rotor current at standstill= 5.26498126493 A\n", + "rotor current when slip-rings are short-circuited= 3.31800758166 A\n" + ] + } + ], + "prompt_number": 51 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.11, Page Number:1261" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n=3\n", + "e=120#V\n", + "r2=0.3#ohm\n", + "x2=1.5#ohm\n", + "s=0.04\n", + "\n", + "#calculations\n", + "e2=e/3**0.5\n", + "er=s*e2\n", + "xr=s*x2\n", + "zr=(r2**2+xr**2)**0.5\n", + "i=er/zr\n", + "s=r2/x2\n", + "xr=s*x2\n", + "zr=(xr**2+r2**2)**0.5\n", + "er=s*e2\n", + "i2=er/zr\n", + "\n", + "#result\n", + "print \"rotor when running short-circuited=\",i,\"A\"\n", + "print \"slip=\",s\n", + "print \"current when torque is maximum=\",i2,\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "rotor when running short-circuited= 9.05821627316 A\n", + "slip= 0.2\n", + "current when torque is maximum= 32.6598632371 A\n" + ] + } + ], + "prompt_number": 54 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.12, Page Number:1264" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=8\n", + "f=50.0#Hz\n", + "s=0.04\n", + "tb=150.0#kg-m\n", + "n=660.0#rpm\n", + "r=0.5#ohm\n", + "\n", + "#calculation\n", + "ns=120*f/p\n", + "sb=(ns-n)/ns\n", + "x2=r/sb\n", + "t=tb*(2/((sb/s)+s/sb))\n", + "\n", + "#result\n", + "print \"torque=\",t,\"kg-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "torque= 90.0 kg-m\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.13(a), Page Number:1266" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variablde declaration\n", + "n=3\n", + "vd=0.90\n", + "\n", + "#calculation\n", + "ratio_s=(1/vd)**2\n", + "ratio_i=ratio_s*vd\n", + "cu_loss_increase=ratio_i**2\n", + "\n", + "#result\n", + "print \"increase in motor copper losses=\",cu_loss_increase" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "increase in motor copper losses= 1.23456790123\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.13(b), Page Number:1264" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=230.0#V\n", + "p=6\n", + "f=50.0#Hz\n", + "p1=15.0#kW\n", + "n=980.0#rpm\n", + "efficiency=0.93\n", + "vd=0.10\n", + "fd=0.05\n", + "\n", + "#calculation\n", + "v2=(1-vd)*v\n", + "f2=(1-fd)*f\n", + "n1=120*f/p\n", + "n2=120*f2/p\n", + "s1=(n1-n)/n1\n", + "ratio_f=s1*(v*(1-vd)/v)**2*f2/f\n", + "n2=n2*(1-ratio_f)\n", + "p2=p1*n2/n1\n", + "#result\n", + "print \"the new operating speed=\",n2,\"rpm\"\n", + "print \"the new output power=\",p2,\"kW\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the new operating speed= 935.3795 rpm\n", + "the new output power= 14.0306925 kW\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.14(a), Page Number:1267" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=3\n", + "v1=400#V\n", + "v2=200#V\n", + "r=0.06#ohm\n", + "x=0.3#ohm\n", + "a=1\n", + "#calculations\n", + "r=x-r\n", + "\n", + "#result\n", + "print \"additional resistance=\",r,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "additional resistance= 0.24 ohm\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.14(b), Page Number:1267" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "n=3\n", + "f=50#Hz\n", + "p=8\n", + "s=0.02\n", + "r=0.001#ohm\n", + "x=0.005#ohm\n", + "\n", + "#calculation\n", + "ns=120*f/p\n", + "a=r/x\n", + "n2=(1-s)*ns\n", + "ratio=2*s**2*a/(a**2+s**2)\n", + "\n", + "#result\n", + "print \"ratio of the maximum to full-load torque=\",ratio*1000,\"10^-3\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "ratio of the maximum to full-load torque= 3.9603960396 10^-3\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.14(c), Page Number:1267" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=12\n", + "v=600#V\n", + "f=50#Hz\n", + "r=0.03#ohm\n", + "x=0.5#ohm\n", + "n=495#rpm\n", + "s=0.01\n", + "#calculation\n", + "Ns=120*f/p\n", + "a=r/x\n", + "n=Ns*(1-a)\n", + "ratio=2*a*s/(a**2+s**2)\n", + "\n", + "#result\n", + "print \"speed of max torque=\",n,\"rpm\"\n", + "print \"ratio of torques=\",ratio" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed of max torque= 470.0 rpm\n", + "ratio of torques= 0.324324324324\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.15, Page Number:1267" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=746.0#kW\n", + "f=50.0#Hz\n", + "p=16\n", + "zr=complex(0.02,0.15)\n", + "n=360.0#rpm\n", + "\n", + "#calculation\n", + "ns=120*f/p\n", + "s=(ns-n)/ns\n", + "a=zr.real/zr.imag\n", + "ratio=2*a*s/(a**2+s**2)\n", + "N=ns*(1-a)\n", + "R=zr.imag-zr.real\n", + "\n", + "#result\n", + "print \"ratio of torques=\",ratio\n", + "print \"speed at maximum torque=\",N,\"rpm\"\n", + "print \"rotor resistance=\",R,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "ratio of torques= 0.550458715596\n", + "speed at maximum torque= 325.0 rpm\n", + "rotor resistance= 0.13 ohm\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.16, Page Number:1268" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "a=Symbol('a')\n", + "p=4\n", + "f=50.0#Hz\n", + "r=0.025#ohm\n", + "x=0.12#ohm\n", + "ratio=3.0/4.0\n", + "\n", + "#calculations\n", + "s=r/x\n", + "ns=120*f/p\n", + "n=ns*(1-s)\n", + "a=solve(ratio-(2*a/(1+a**2)),a)\n", + "r=a[0]*x-r\n", + "\n", + "#result\n", + "print \"speed at maximum torque=\",n,\"rpm\"\n", + "print \"additional resistance=\",r,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed at maximum torque= 1187.5 rpm\n", + "additional resistance= 0.0291699475574164 ohm\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.17, Page Number:1268" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "f=50#Hz\n", + "s=0.04\n", + "r=0.01#ohm\n", + "x=0.1#ohm\n", + "p=8\n", + "#calculation\n", + "a=r/x\n", + "t_ratio=2*a*s/(a**2+s**2)\n", + "ns=120*f/p\n", + "n=(1-a)*ns\n", + "\n", + "#result\n", + "print \"ratio of torques=\",1/t_ratio\n", + "print \"speed=\",n,\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "ratio of torques= 1.45\n", + "speed= 675.0 rpm\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.18, Page Number:1268" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "a=Symbol('a')\n", + "a2=Symbol('a2')\n", + "p=3\n", + "t_ratio=2.5\n", + "t_ratio2=1.5\n", + "s=0.03\n", + "\n", + "#calculation\n", + "t_ratio3=t_ratio2/t_ratio\n", + "a=solve(t_ratio3-(2*a/(1+a**2)),a)\n", + "a2=solve(a2**2-0.15*a2+0.0009,a2)\n", + "r_red=(a[0]-a2[1])/a[0]\n", + "#result\n", + "print \"percentage reduction in rotor circuit resistance=\",r_red*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "percentage reduction in rotor circuit resistance= 56.8784093726987 %\n" + ] + } + ], + "prompt_number": 46 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.19, Page Number:1269" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=8\n", + "f=50#Hz\n", + "r=0.08#ohm\n", + "n=650.0#rpm\n", + "\n", + "#calculation\n", + "ns=120*f/p\n", + "sb=(ns-n)/ns\n", + "x2=r/sb\n", + "a=1\n", + "r=a*x2-r\n", + "#result\n", + "print \"extra resistance=\",r,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "extra resistance= 0.52 ohm\n" + ] + } + ], + "prompt_number": 51 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.20, Page Number:1269" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "R=Symbol('R')\n", + "p=4\n", + "f=50.0#Hz\n", + "t=162.8#N-m\n", + "n=1365.0#rpm\n", + "r=0.2#ohm\n", + "\n", + "#calculations\n", + "ns=120*f/p\n", + "sb=(ns-n)/ns\n", + "x2=r/sb\n", + "R=solve(1.0/(4*x2)-((r+R)/((r+R)**2+x2**2)),R)\n", + "\n", + "#result\n", + "print \"resistance to be added=\",round(R[0],1),\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resistance to be added= 0.4 ohm\n" + ] + } + ], + "prompt_number": 56 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.21, Page Number:1270" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=4.0\n", + "f=50.0#Hz\n", + "load=7.46#kW\n", + "t_ratios=1.60\n", + "t_ratiom=2.0\n", + "\n", + "#calcualtion\n", + "t_ratio=t_ratios/t_ratiom\n", + "#0.8a2-2*a+0.8 a=0.04\n", + "#0.5=2*a*sf/a2+sf2 sf=0.01\n", + "a=0.04\n", + "sf=0.01\n", + "ns=120*f/p\n", + "n=ns-sf*ns\n", + "N=ns-a*ns\n", + "\n", + "#result\n", + "print \"full-load speed=\",n,\"rpm\"\n", + "print \"speed at maximum torque=\",N,\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "full-load speed= 1485.0 rpm\n", + "speed at maximum torque= 1440.0 rpm\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.22, Page Number:1270" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=6\n", + "v=240#V\n", + "f=50#Hz\n", + "r=0.12#ohm\n", + "x=0.85#ohm\n", + "ratio=1.8\n", + "s=0.04\n", + "\n", + "#calculations\n", + "k=1/ratio\n", + "e2=k*(v/3**0.5)\n", + "ns=120*f/p\n", + "tf=(3/(2*3.14*f/3))*(s*e2*e2*r/(r**2+(s*x)**2))\n", + "s=r/x\n", + "tmax=(3/(2*3.14*f/3))*(s*e2*e2*r/(r**2+(s*x)**2))\n", + "n=ns*(1-s)\n", + "\n", + "#result\n", + "print \"developed torque=\",tf,\"N-m\"\n", + "print \"maximum torque=\",tmax,\"N-m\"\n", + "print \"speed at maximum torque=\",n,\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "developed torque= 52.4097855621 N-m\n", + "maximum torque= 99.9125764956 N-m\n", + "speed at maximum torque= 858.823529412 rpm\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.23, Page Number:1270" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "r=0.015#ohm\n", + "x=0.09#ohm\n", + "s=0.03\n", + "\n", + "#calculation\n", + "ns=100#rpm considered\n", + "n=(1-s)*ns\n", + "n2=n/2\n", + "s2=(ns-n2)/ns\n", + "ratio=((s2/s)*(r**2+(s*x)**2)/(r**2+(s2*x)**2))**0.5\n", + "per=1-1/ratio\n", + "phi=math.atan(s2*x/r)\n", + "pf=math.cos(phi)\n", + "\n", + "#result\n", + "print \"percentage reduction=\",per*100,\"%\"\n", + "print \"pf=\",pf\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "percentage reduction= 22.8528060715 %\n", + "pf= 0.307902262948\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.26, Page Number:1272" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=440#V\n", + "f=50#Hz\n", + "p=4\n", + "t=100#N-m\n", + "n=1200#rpm\n", + "\n", + "#calculation\n", + "e2=v/2\n", + "ns=120*f/p\n", + "n=ns-n\n", + "n2=n+ns/2\n", + "\n", + "#result\n", + "print \"stator supply voltage=\",e2,\"V\"\n", + "print \"new speed=\",n2,\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "stator supply voltage= 220 V\n", + "new speed= 1050 rpm\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.24, Page Number:1274" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable delclaration\n", + "v=400.0#V\n", + "f=60.0#Hz\n", + "p=8.0\n", + "n=1140.0#rpm\n", + "e=440.0#V\n", + "e2=550.0#V\n", + "\n", + "#calculations\n", + "ns=120*f/p\n", + "s1=(ns-n)/ns\n", + "s2=s1*(e/e2)**2\n", + "n2=ns*(1-s2)\n", + "\n", + "#result\n", + "print \"speed=\",n2,\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed= 1053.6 rpm\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.25, Page Number:1274" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=450.0#V\n", + "f=60.0#Hz\n", + "p=8.0\n", + "n=873.0#rpm\n", + "t=23.0#degrees\n", + "n2=864.0#rpm\n", + "alpha=1.0/234.0#per degrees centrigrade\n", + "\n", + "#calculation\n", + "s1=(900-n)/900\n", + "s2=(900-n2)/900\n", + "ratio=s2/s1-1\n", + "t2=(s2/s1-1)/alpha+23 \n", + "\n", + "#result\n", + "print \"increase in rotor resistance=\",ratio*100,\"%\"\n", + "print \"approx temperature=\",t2,\"degrees centigrade\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "increase in rotor resistance= 33.3333333333 %\n", + "approx temperature= 101.0 degrees centigrade\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.27, Page Number:1283" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=440.0#V\n", + "f=500.0#Hz\n", + "p=6.0\n", + "load=80.0#kW\n", + "alt=100.0\n", + "ns=120.0*f/60.0\n", + "#calculation\n", + "s=alt/(60.0*f)\n", + "n=(1-s)*ns\n", + "cu_loss=(1.0/3.0)*load*1000/3.0\n", + "\n", + "#result\n", + "print \"slip=\",s*1000,\"%\"\n", + "print \"rotor speed=\",n,\"rpm\"\n", + "print \"rotor copper loss=\",cu_loss/10000,\"kW\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "slip= 3.33333333333 %\n", + "rotor speed= 996.666666667 rpm\n", + "rotor copper loss= 0.888888888889 kW\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.28, Page Number:1283" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=440.0#V\n", + "f=50.0#Hz\n", + "p=4.0\n", + "n=1425.0#rpm\n", + "z=complex(0.4,4)\n", + "ratio=0.8\n", + "loss=500.0#W\n", + "\n", + "#calculation\n", + "ns=120*f/p\n", + "s=75/ns\n", + "e1=v/3**0.5\n", + "tf=(3*2/(2*3.14*f))*(((e1*ratio)**2)*z.real*s)/(z.real**2+(s*z.imag)**2)\n", + "ir=s*ratio*e1/(z.real**2+(s*z.imag)**2)**0.5\n", + "cu_loss=3*ir**2*z.real\n", + "pm=2*3.4*(n/60)*tf\n", + "pout=pm-loss\n", + "s=z.real/z.imag\n", + "tmax=(3*2/(2*3.14*f))*(((e1*ratio)**2)*z.real*s)/(z.real**2+(s*z.imag)**2)\n", + "nmax=ns-s*ns\n", + "i=ratio*e1/abs(z)\n", + "tst=(3*2/(2*3.14*f))*(((e1*ratio)**2)*z.real)/(z.real**2+(z.imag)**2)\n", + "\n", + "#result\n", + "print \" full load torque=\",tf,\"N-m\"\n", + "print \"rotor current=\",ir,\"A\"\n", + "print \"cu_loss=\",cu_loss,\"W\"\n", + "print \"power output=\",pout,\"W\"\n", + "print \"max torque=\",tmax,\"N-m\"\n", + "print \"speed at max torque=\",nmax,\"rpm\"\n", + "print \"starting current=\",i,\"A\"\n", + "print \"starting torque=\",tst,\"N-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " full load torque= 78.9197452229 N-m\n", + "rotor current= 22.7215022978 A\n", + "cu_loss= 619.52 W\n", + "power output= 12245.5388535 W\n", + "max torque= 98.6496815287 N-m\n", + "speed at max torque= 1350.0 rpm\n", + "starting current= 50.5546790867 A\n", + "starting torque= 19.5345904017 N-m\n" + ] + } + ], + "prompt_number": 47 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.30, Page Number:1286" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=60#kW\n", + "loss=1#kW\n", + "s=0.03\n", + "\n", + "#calculations\n", + "p2=load-loss\n", + "pm=(1-s)*p2\n", + "cu_loss=s*p2\n", + "rotor_loss=cu_loss*1000/3\n", + "\n", + "#result\n", + "print \"mechanical power developed=\",pm,\"kW\"\n", + "print \"rotor copper loss=\",rotor_loss,\"W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "mechanical power developed= 57.23 kW\n", + "rotor copper loss= 590.0 W\n" + ] + } + ], + "prompt_number": 52 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.31, Page Number:1287" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=400#V\n", + "f=50#Hz\n", + "p=6\n", + "load=20#KW\n", + "s=0.03\n", + "i=60#A\n", + "\n", + "#calculation\n", + "fr=s*f\n", + "ns=120*f/p\n", + "n=ns*(1-s)\n", + "cu_loss=s*load*1000\n", + "r2=cu_loss/(3*i**2)\n", + "\n", + "#result\n", + "print \"frequency of rotor current=\",fr,\"Hz\"\n", + "print \"rotor copper loss=\",cu_loss,\"W\"\n", + "print \"rotor resistance=\",r2,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "frequency of rotor current= 1.5 Hz\n", + "rotor copper loss= 600.0 W\n", + "rotor resistance= 0.0555555555556 ohm\n" + ] + } + ], + "prompt_number": 54 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.32, Page Number:1287" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=6\n", + "f=50#Hz\n", + "load=3.73#KW\n", + "n=960#rpm\n", + "loss=280#W\n", + "\n", + "#calculation\n", + "ns=120*f/p\n", + "input_r=load*1000*ns/n\n", + "input_s=input_r+loss\n", + "\n", + "#result\n", + "print \"stator input=\",input_s,\"W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "stator input= 4165.41666667 W\n" + ] + } + ], + "prompt_number": 55 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.33, Page Number:1287" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=400.0#V\n", + "f=50.0#Hz\n", + "p=6.0\n", + "p2=75.0#KW\n", + "alt=100.0\n", + "\n", + "#calculations\n", + "f1=alt/60\n", + "s=f1/f\n", + "ns=120*f/p\n", + "n=ns*(1-s)\n", + "cu_loss_r_per_phase=s*p2/3\n", + "pm=(1-s)*p2\n", + "\n", + "#result\n", + "print \"slip=\",s*100,\"%\"\n", + "print \"rotor speed=\",n,\"rpm\"\n", + "print \"rotor copper loss per phase=\",cu_loss_r_per_phase,\"kW\"\n", + "print \"mechancal power=\",pm,\"kW\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "slip= 3.33333333333 %\n", + "rotor speed= 966.666666667 rpm\n", + "rotor copper loss per phase= 0.833333333333 kW\n", + "mechancal power= 72.5 kW\n" + ] + } + ], + "prompt_number": 57 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.34, Page Number:1287" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=500.0#V\n", + "f=50.0#Hz\n", + "p=6.0\n", + "n=975.0#rpm\n", + "p1=40.0#KW\n", + "loss_s=1.0#kW\n", + "loss=2.0#KW\n", + "\n", + "#calculation\n", + "ns=120*f/p\n", + "s=(ns-n)/ns\n", + "p2=p1-loss_s\n", + "cu_loss=s*p2\n", + "pm=p2-cu_loss\n", + "pout=pm-loss\n", + "efficiency=pout/p1\n", + "\n", + "#result\n", + "print \"slip=\",s*100,\"%\"\n", + "print \"rotor copper loss=\",cu_loss,\"kW\"\n", + "print \"shaft power=\",pout,\"kW\"\n", + "print \"efficiency=\",efficiency*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "slip= 2.5 %\n", + "rotor copper loss= 0.975 kW\n", + "shaft power= 36.025 kW\n", + "efficiency= 90.0625 %\n" + ] + } + ], + "prompt_number": 59 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.35, Page Number:1287" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "output=100#KW\n", + "v=3300#V\n", + "f=50#Hz\n", + "n=500#rpm\n", + "s=0.018\n", + "pf=0.85\n", + "cu_loss=2440#W\n", + "iron_loss=3500#W\n", + "rotational_loss=1200#W\n", + "\n", + "#calculations\n", + "pm=output+rotational_loss/1000\n", + "cu_loss_r=(s/(1-s))*pm\n", + "p2=pm+cu_loss_r\n", + "input_s=p2+cu_loss/1000+iron_loss/1000\n", + "il=input_s*1000/(3**0.5*v*pf)\n", + "efficiency=output/input_s\n", + "\n", + "#result\n", + "print \"rotor copper loss=\",cu_loss_r,\"kW\"\n", + "print \"line current=\",il,\"A\"\n", + "print \"efficiency=\",efficiency*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "rotor copper loss= 1.85132382892 kW\n", + "line current= 22.1989272175 A\n", + "efficiency= 92.7202341611 %\n" + ] + } + ], + "prompt_number": 62 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.36, Page Number:1288" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=440.0#V\n", + "f=50.0#Hz\n", + "p=6.0\n", + "p2=100.0#W\n", + "c=120.0\n", + "\n", + "#calculations\n", + "s=c/(f*60)\n", + "ns=120*f/p\n", + "n=ns*(1-s)\n", + "pm=(1-s)*p2\n", + "cu_loss=s*p2/3\n", + "n2=ns-n\n", + "\n", + "#result\n", + "print \"slip=\",s*100,\"%\"\n", + "print \"rotor speed=\",n,\"rpm\"\n", + "print \"mechanical power=\",pm,\"kW\"\n", + "print \"copper loss=\",cu_loss,\"kW\"\n", + "print \"speed of stator field with respect to rotor=\",n2,\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "slip= 4.0 %\n", + "rotor speed= 960.0 rpm\n", + "mechanical power= 96.0 kW\n", + "copper loss= 1.33333333333 kW\n", + "speed of stator field with respect to rotor= 40.0 rpm\n" + ] + } + ], + "prompt_number": 69 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.37, Page Number:1288" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "efficiency=0.9\n", + "output=37#kW\n", + "ratio=1.0/3.0\n", + "\n", + "#calculation\n", + "input_m=output*1000/efficiency\n", + "total_loss=input_m-output*1000\n", + "x=total_loss/(3+0.5)\n", + "input_r=output*1000+x/2+x\n", + "s=x/input_r\n", + "\n", + "#result\n", + "print \"slip=\",s*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "slip= 3.0303030303 %\n" + ] + } + ], + "prompt_number": 74 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.38, Page Number:1289" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=400#V\n", + "f=50#Hz\n", + "p=6\n", + "load=45#KW\n", + "i=75#A\n", + "s=0.03\n", + "iron_loss=1200#kW\n", + "loss=900#kW\n", + "r=0.12#ohm\n", + "\n", + "#calculations\n", + "pf=load*1000/(3**0.5*v*i)\n", + "r=r*3/2\n", + "cu_loss=3*(i/3**0.5)**2*r\n", + "cu_loss_r=s*42788\n", + "pm=42788-cu_loss_r\n", + "output_s=pm-loss\n", + "efficiency=output_s/(load*1000)\n", + "t=(output_s*60)/(2*3.14*970)\n", + "\n", + "#result\n", + "print \"pf=\",pf\n", + "print \"rotor cu loss=\",cu_loss_r,\"W\"\n", + "print \"p out=\",output_s,\"W\"\n", + "print \"efficiency=\",efficiency*100,\"%\"\n", + "print \"torque=\",t,\"N-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "pf= 0.866025403784\n", + "rotor cu loss= 1283.64 W\n", + "p out= 40604.36 W\n", + "efficiency= 90.2319111111 %\n", + "torque= 399.937881673 N-m\n" + ] + } + ], + "prompt_number": 78 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.39(a), Page Number:1287" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=4.0\n", + "v=220.0#V\n", + "f=50.0#Hz\n", + "r=0.1#ohm\n", + "x=0.9#ohm\n", + "ratio=1.75\n", + "s=0.05\n", + "\n", + "#calculations\n", + "k=1/ratio\n", + "e1=v/3**0.5\n", + "e2=k*e1\n", + "z=(r**2+(s*x)**2)**0.5\n", + "i2=s*e2/z\n", + "pcr=3*i2**2*r\n", + "pm=pcr*(1-s)/s\n", + "ns=120*f/p\n", + "n=ns*(1-s)\n", + "tg=9.55*pm/n\n", + "sm=r/x\n", + "n=ns*(1-sm)\n", + "e3=sm*e2\n", + "\n", + "#result\n", + "print \"load torque=\",tg/9.81,\"kg-m\"\n", + "print \"speed at maximum torque=\",n,\"rpm\"\n", + "print \"rotor emf at max torque=\",e3,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "load torque= 4.26478644041 kg-m\n", + "speed at maximum torque= 1333.33333333 rpm\n", + "rotor emf at max torque= 8.06457518868 V\n" + ] + } + ], + "prompt_number": 88 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.39(b), Page Number:1290" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=400#V\n", + "f=50#Hz\n", + "p=4\n", + "i=10#A\n", + "pf=0.86\n", + "loss=0.05\n", + "cu_r=0.04\n", + "m_loss=0.03\n", + "\n", + "#calculation\n", + "input_m=3**0.5*v*i*pf\n", + "loss_s=loss*input_m\n", + "input_r=input_m-loss_s\n", + "cu_lossr=cu_r*input_r\n", + "mec_loss=m_loss*input_r\n", + "output_shaft=input_r-cu_lossr-mec_loss\n", + "s=cu_lossr/input_r\n", + "ns=120*f/p\n", + "n=ns*(1-s)\n", + "wr=2*3.14*n/60\n", + "output_r=input_r-cu_lossr\n", + "tr=output_r/wr\n", + "tin=output_shaft/wr\n", + "\n", + "#result\n", + "print \"slip=\",s*100,\"%\"\n", + "print \"rotor speed=\",n,\"rpm\"\n", + "print \"torque developed in the rotor=\",tr,\"Nw-m\"\n", + "print \"shaft torque=\",tin,\"Nw-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "slip= 4.0 %\n", + "rotor speed= 1440.0 rpm\n", + "torque developed in the rotor= 36.0531340072 Nw-m\n", + "shaft torque= 34.9264735695 Nw-m\n" + ] + } + ], + "prompt_number": 91 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.40, Page Number:1291" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=440.0#V\n", + "p=40.0\n", + "f=50.0#Hz\n", + "r=0.1#ohm\n", + "x=0.9#ohm\n", + "ratio=3.5\n", + "s=0.05\n", + "\n", + "#calculation\n", + "e1=v/3**0.5\n", + "k=1/ratio\n", + "e2=k*e1\n", + "er=s*e2\n", + "z=(r**2+(s*x)**2)**0.5\n", + "i2=er/z\n", + "cu_loss=3*i2**2*r\n", + "output=cu_loss*(1-s)/s\n", + "sm=r/x\n", + "er=sm*e2\n", + "zr=(r**2+(x*sm)**2)**0.5\n", + "i2=er/zr\n", + "cu_loss=3*i2**2*r\n", + "input_r=cu_loss/sm\n", + "\n", + "#result\n", + "print \"gross output at 5% slip=\",output,\"W\"\n", + "print \"maximum torque=\",input_r,\"W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "gross output at 5% slip= 6242.77652849 W\n", + "maximum torque= 8780.04535147 W\n" + ] + } + ], + "prompt_number": 107 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.41, Page Number:1291" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "pout=18.65#kW\n", + "p=4.0\n", + "f=50.0#Hz\n", + "loss=0.025\n", + "s=0.04\n", + "\n", + "#calculations\n", + "pw=loss*pout*1000\n", + "pm=pout*1000+pw\n", + "cu_loss=s*pm/(1-s)\n", + "p2=cu_loss/s\n", + "ns=120*f/p\n", + "n=ns*(1-s)\n", + "tsh=9.55*pout*1000/n\n", + "tg=9.55*pm/n\n", + "\n", + "#result\n", + "print \"rotor cu loss=\",cu_loss,\"W\"\n", + "print \"rotor input=\",p2,\"W\"\n", + "print \"shaft torque=\",tsh,\"N-m\"\n", + "print \"gross electromagnetic torque=\",tg,\"N-m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "rotor cu loss= 796.510416667 W\n", + "rotor input= 19912.7604167 W\n", + "shaft torque= 123.685763889 N-m\n", + "gross electromagnetic torque= 126.777907986 N-m\n" + ] + } + ], + "prompt_number": 109 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.42, Page Number:1291" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=8\n", + "f=50.0#Hz\n", + "n=710#rpm\n", + "load=35#kW\n", + "loss=1200#W\n", + "loss_r=600#W\n", + "\n", + "#calculation\n", + "p2=load*1000-loss\n", + "ns=120*f/p\n", + "s=(ns-n)/ns\n", + "cu_loss=s*p2\n", + "pm=p2-cu_loss\n", + "tg=9.55*pm/n\n", + "pout=pm-loss_r\n", + "tsh=9.55*pout/n\n", + "\n", + "#result\n", + "print \"rotor copper loss=\",cu_loss/1000,\"kW\"\n", + "print \"gross torque=\",tg,\"N-m\"\n", + "print \"mechanical power=\",pm,\"W\"\n", + "print \"net torque=\",tsh,\"N-m\"\n", + "print \"mechanical power output=\",pout,\"W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "rotor copper loss= 1.80266666667 kW\n", + "gross torque= 430.386666667 N-m\n", + "mechanical power= 31997.3333333 W\n", + "net torque= 422.316244131 N-m\n", + "mechanical power output= 31397.3333333 W\n" + ] + } + ], + "prompt_number": 113 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.43, Page Number:1292" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=6\n", + "f=50.0#Hz\n", + "s=0.04\n", + "tsh=149.3#N-m\n", + "loss=200#W\n", + "cu_loss=1620#W\n", + "\n", + "#calculations\n", + "ns=120*f/p\n", + "n=ns*(1-s)\n", + "pout=tsh*2*3.14*(n/60)\n", + "output=pout+loss\n", + "p2=output*ns/n\n", + "cu_lossr=p2-output\n", + "p1=p2+cu_loss\n", + "efficiency=pout*100/p1\n", + "\n", + "#result\n", + "print \"output power=\",pout/1000,\"kW\"\n", + "print \"rotor cu loss=\",cu_lossr,\"W\"\n", + "print \"the efficiency=\",efficiency,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "output power= 15.001664 kW\n", + "rotor cu loss= 633.402666667 W\n", + "the efficiency= 85.9444669361 %\n" + ] + } + ], + "prompt_number": 116 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.44, Page Number:1291" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "pout=18.65#kW\n", + "p=6\n", + "f=50.0#Hz\n", + "n=960#rpm\n", + "i2=35#A\n", + "loss=1#kW\n", + "\n", + "#calculation\n", + "pm=pout+loss\n", + "ns=120*f/p\n", + "s=(ns-n)/ns\n", + "cu_lossr=pm*s*1000/(1-s)\n", + "r2=cu_lossr/(3*i2**2)\n", + "\n", + "#result\n", + "print \"resistane per phase=\",r2,\"ohm/phase\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resistane per phase= 0.222789115646 ohm/phase\n" + ] + } + ], + "prompt_number": 120 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.45, Page Number:1291" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "sf=Symbol('sf')\n", + "v=400#V\n", + "p=4\n", + "f=50#Hz\n", + "r=0.01#ohm\n", + "x=0.1#ohm\n", + "ratio=4\n", + "\n", + "#calculation\n", + "e1=v/3**0.5\n", + "e2=e1/ratio\n", + "sm=r/x\n", + "ns=120*f/p\n", + "tmax=(3/(2*3.14*25))*(e2**2/(2*x))\n", + "a=r/x\n", + "sf=solve(0.5*(a**2+sf**2)-2*a*sf,sf)\n", + "n=ns*(1-sf[0])\n", + "tf=tmax/2\n", + "output=2*3.14*n*tf/60\n", + "\n", + "#result\n", + "print \"maximum torque=\",tmax,\"N-m\"\n", + "print \"full load slip=\",sf[0]\n", + "print \"power output=\",output,\"W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "maximum torque= 318.47133758 N-m\n", + "full load slip= 0.0267949192431123\n", + "power output= 24330.1270189222 W\n" + ] + } + ], + "prompt_number": 129 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.46, Page Number:1291" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=4\n", + "f=50.0#Hz\n", + "v=200.0#V\n", + "r=0.1#ohm\n", + "x=0.9#ohm\n", + "k=0.67\n", + "s=0.04\n", + "#calculations\n", + "e1=v/3**0.5\n", + "e2=e1*k\n", + "z=(r**2+(s*x)**2)**0.5\n", + "i2=s*e2/z\n", + "cu_loss=3*i2**2*r\n", + "pm=cu_loss*(1-s)/s\n", + "ns=120*f/p\n", + "n=ns*(1-s)\n", + "tg=9.55*pm/n\n", + "sm=r/x\n", + "er=sm*e2\n", + "zr=(r**2+(sm*x)**2)**0.5\n", + "i2=er/zr\n", + "cu_lossr=3*i2**2*r\n", + "output=cu_lossr*(1-sm)/sm\n", + "n=(1-sm)*ns\n", + "tmax=9.55*output/n\n", + "\n", + "#result\n", + "print \"torque=\",tg,\"N-m\"\n", + "print \"maximum torque=\",tmax,\"N-m\"\n", + "print \"speed at max torque=\",n,\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "torque= 40.4815391879 N-m\n", + "maximum torque= 63.511037037 N-m\n", + "speed at max torque= 1333.33333333 rpm\n" + ] + } + ], + "prompt_number": 143 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.47, Page Number:1293" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "r=0.015#ohm\n", + "x=0.09#ohm\n", + "f=50#Hz\n", + "s=0.04\n", + "p=4\n", + "e2=110#V\n", + "\n", + "#calculations\n", + "z=(r**2+x**2)**0.5\n", + "pf=r/z\n", + "xr=s*x\n", + "zr=(r**2+xr**2)**0.5\n", + "pf2=r/zr\n", + "ns=120*f/p\n", + "n=ns*(1-s)\n", + "er=s*e2\n", + "i2=er/zr\n", + "cu_loss=3*i2**2*r\n", + "pm=cu_loss*(1-s)/s\n", + "tg=9.55*pm/n\n", + "\n", + "#result\n", + "print \"pf of motor at start=\",pf\n", + "print \"pf of motor at s=4%\",pf2\n", + "print \"full load torque=\",tg,\"N-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "pf of motor at start= 0.164398987305\n", + "pf of motor at s=4% 0.972387301981\n", + "full load torque= 582.728189612 N-m\n" + ] + } + ], + "prompt_number": 144 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.48, Page Number:1294" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=6.0\n", + "f=50.0#Hz\n", + "tsh=162.84#N-m\n", + "c=90.0\n", + "t=20.36#N-m\n", + "loss=830.0#W\n", + "\n", + "#calculation\n", + "ns=120*f/p\n", + "fr=c/60\n", + "s=fr/f\n", + "n=ns*(1-s)\n", + "output=2*3.14*n*tsh/60\n", + "tg=tsh+t\n", + "p2=tg*ns/9.55\n", + "cu_lossr=s*p2\n", + "p1=p2+cu_lossr\n", + "efficiency=output*100/p1\n", + "\n", + "#result\n", + "print \"motor output=\",output,\"W\"\n", + "print \"cu loss=\",cu_lossr,\"W\"\n", + "print \"motor input\",p1,\"W\"\n", + "print \"efficiency=\",efficiency,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "motor output= 16532.6024 W\n", + "cu loss= 575.497382199 W\n", + "motor input 19758.7434555 W\n", + "efficiency= 83.6723369441 %\n" + ] + } + ], + "prompt_number": 146 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.49, Page Number:1294" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=18.65#kW\n", + "v=420.0#V\n", + "p=6\n", + "f=50.0#Hz\n", + "r=1.0#ohm\n", + "z=complex(0.25,0.75)\n", + "zr=complex(0.173,0.52)\n", + "v1=420.0#V\n", + "v2=350.0#V\n", + "\n", + "#calculations\n", + "k=v2/v1\n", + "r02=zr.real+k**2*z.real\n", + "x02=zr.imag+k**2*z.imag\n", + "z02=((r+r02)**2+x02**2)**0.5\n", + "i2=v2/(3**0.5*z02)\n", + "cu_loss=i2**2*(r+zr.real)\n", + "p2=cu_loss*3\n", + "ns=120*f/p\n", + "tst=9.55*p2/(ns*9.81)\n", + "#result\n", + "print \"torque=\",tst,\"kg-m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "torque= 48.2909354778 kg-m\n" + ] + } + ], + "prompt_number": 157 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.50, Page Number:1295" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=8\n", + "load=37.3#ohm\n", + "v=280#V\n", + "f=50.0#Hz\n", + "i=200#A\n", + "pf=0.25\n", + "r=0.15#ohm\n", + "k=1.0/3\n", + "#calculation\n", + "wsc=2*v*i*pf\n", + "power_phase=v*i*pf\n", + "R=power_phase/i**2\n", + "r2_=R-r\n", + "r2=k**2*r2_\n", + "p2=3*i**2*r2_\n", + "ns=120*f/p\n", + "t=9.55*p2/ns\n", + "\n", + "#result\n", + "print \"resistance perphaseof therotor winding=\",r2,\"ohm\"\n", + "print \"startingtorque=\",t,\"N-m\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resistance perphaseof therotor winding= 0.0222222222222 ohm\n", + "startingtorque= 305.6 N-m\n" + ] + } + ], + "prompt_number": 158 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.51, Page Number:1295" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "ratios=1.6\n", + "ratiom=2.0\n", + "sf=0.01\n", + "sb=0.04\n", + "#calculation\n", + "i=(ratios/sf)**0.5\n", + "\n", + "#result\n", + "print \"slip at full load=\",sf\n", + "print \"slip at maximum torque=\",sb\n", + "print \"rotor current=\",i" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "slip at full load= 0.01\n", + "slip at maximum torque= 0.04\n", + "rotor current= 12.6491106407\n" + ] + } + ], + "prompt_number": 159 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.52, Page Number:1297" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=200#km/h\n", + "f=100#Hz\n", + "\n", + "#calculation\n", + "w=v*5.0/18/(2*f)\n", + "\n", + "#result\n", + "print \"pole pitch=\",w*1000,\"mm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "pole pitch= 277.777777778 mm\n" + ] + } + ], + "prompt_number": 162 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.53, Page Number:1297" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=4\n", + "w=6#mm\n", + "f=25#Hz\n", + "p=6#kW\n", + "loss=1.2#kW\n", + "v=2.4#m/s\n", + "\n", + "#calculation\n", + "vs=2*f*w/100\n", + "s=(vs-v)/vs\n", + "p2=p-loss\n", + "pcr=s*p2\n", + "pm=p2-pcr\n", + "f=p2*1000/vs\n", + "\n", + "#result\n", + "print \"synchronous speed=\",vs,\"m/s\"\n", + "print \"slip=\",s\n", + "print \"cu loss=\",pcr,\"kW\"\n", + "print \"mechanical power=\",pm,\"kW\"\n", + "print \"thrust=\",f/1000,\"kN\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "synchronous speed= 3 m/s\n", + "slip= 0.2\n", + "cu loss= 0.96 kW\n", + "mechanical power= 3.84 kW\n", + "thrust= 1.6 kN\n" + ] + } + ], + "prompt_number": 163 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.54, Page Number:1304" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "s=0.12\n", + "r=0.08#ohm/phase\n", + "pg=9000.0#W\n", + "\n", + "#calculations\n", + "rl=r*(1/s-1)\n", + "v=(pg*rl/3)**0.5\n", + "il=v/rl\n", + "\n", + "#result\n", + "print \"load resistance=\",rl,\"ohm\"\n", + "print \"load voltage=\",v,\"V\"\n", + "print \"load current=\",il,\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "load resistance= 0.586666666667 ohm\n", + "load voltage= 41.9523539268 V\n", + "load current= 71.5096941934 A\n" + ] + } + ], + "prompt_number": 166 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.55, Page Number:1305" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=400.0#V\n", + "f=50.0#Hz\n", + "p=4\n", + "r1=0.15#ohm\n", + "x1=0.45#ohm\n", + "r2_=0.12#ohm\n", + "x2_=0.45#ohm\n", + "xm=complex(0,28.5)#ohm\n", + "s=0.04\n", + "#calculations\n", + "rl_=r2_*(1/s-1)\n", + "i2_=(v/3**0.5)/complex(r1+rl_,x1)\n", + "i0=(v/3**0.5)/xm\n", + "i1=i0+i2_\n", + "pf=math.cos(math.atan(i1.imag/i1.real))\n", + "\n", + "#result\n", + "print \"stator current=\",i1,\"A\"\n", + "print \"power factor=\",pf" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "stator current= (74.5730253701-19.1783634605j) A\n", + "power factor= 0.968485280755\n" + ] + } + ], + "prompt_number": 177 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.56, Page Number:1305" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=220#V\n", + "p=4\n", + "f=50#Hz\n", + "power=3.73#kW\n", + "r1=0.45#ohm\n", + "x1=0.8#ohm\n", + "r2_=0.4#ohm\n", + "x2_=0.8#ohm\n", + "b0=-1.0/30\n", + "loss=50#W\n", + "lossr=150#W\n", + "s=0.04\n", + "\n", + "#calculations\n", + "zab=complex(30*complex(r2_/s,x2_))/complex(r2_/s,x2_-1/b0)\n", + "z01=complex(r1,x1)+zab\n", + "vph=v/3**0.5\n", + "i1=v1/z01\n", + "pf=math.cos(math.atan(i1.imag/i1.real))\n", + "p2=3*i1.real**2*zab.real\n", + "pm=(1-s)*p2\n", + "ns=120*f/p\n", + "n=ns*(1-s)\n", + "tg=9.55*pm/n\n", + "power_o=pm-lossr\n", + "cu_loss=3*i1.real**2*r1\n", + "cu_lossr=s*p2\n", + "total_loss=loss+cu_loss+cu_lossr+lossr\n", + "efficiency=power_o/(power_o+total_loss)\n", + "\n", + "#result\n", + "print \"input current=\",i1,\"A\"\n", + "print \"pf=\",pf\n", + "print \"air gap power=\",p2,\"W\"\n", + "print \"mechanical power=\",pm,\"W\"\n", + "print \"electro magnetic torque=\",tg,\"N-m\"\n", + "print \"output power=\",power_o,\"W\"\n", + "print \"efficiency=\",efficiency*100,\"%\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "input current= (21.9914486234+42.6194245913j) A\n", + "pf= 0.45854949826\n", + "air gap power= 5173.46132109 W\n", + "mechanical power= 4966.52286825 W\n", + "electro magnetic torque= 32.9377037443 N-m\n", + "output power= 4816.52286825 W\n", + "efficiency= 81.9644851937 %\n" + ] + } + ], + "prompt_number": 184 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.57, Page Number:1306" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=440#V\n", + "f=50#Hz\n", + "load=37.3#kW\n", + "r1=0.1#ohm\n", + "x1=0.4#ohm\n", + "r2_=0.15#ohm\n", + "x2_=0.44#ohm\n", + "loss=1250#W\n", + "lossr=1000#W\n", + "i=20#A\n", + "pf=0.09\n", + "s=0.03\n", + "\n", + "#calculation\n", + "v1=v/3**0.5\n", + "i2_=v1/complex(r1+r2_/s,x1+x2_)\n", + "i1=i2_+complex(1.78,19.9)\n", + "pf=math.cos(math.atan(i1.imag/i1.real))\n", + "p2=3*i2_.real**2*r2_/s\n", + "ns=120*f/p\n", + "tg=9.55*p2/ns\n", + "pm=p2*(1-s)\n", + "pout=pm-1000\n", + "cu_losss=3*i1.real**2*r1\n", + "cu_lossr=s*p2\n", + "total_loss=loss+cu_losss+cu_lossr+lossr\n", + "efficiency=pout/(pout+total_loss)\n", + "\n", + "#result\n", + "print \"line current=\",i1,\"A\"\n", + "print \"pf=\",pf\n", + "print \"electromagnetic torque=\",tg,\"N-m\"\n", + "print \"output=\",pout,\"W\"\n", + "print \"efficiency=\",efficiency*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "line current= (50.2750367599+11.9125821807j) A\n", + "pf= 0.973057118792\n", + "electromagnetic torque= 224.593900377 N-m\n", + "output= 33218.2329894 W\n", + "efficiency= 89.0932246577 %\n" + ] + } + ], + "prompt_number": 186 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.58, Page Number:1306" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=400#V\n", + "z=complex(0.06,0.2)\n", + "zr=complex(0.06,0.22)\n", + "\n", + "#calculation\n", + "r01=z.real+zr.real\n", + "x01=z.imag+zr.imag\n", + "z01=(r01**2+x01**2)**0.5\n", + "s=z.real/(z.real+z01)\n", + "v1=v/3**0.5\n", + "pmax=3*v1**2/(2*(r01+z01))\n", + "\n", + "#result\n", + "print \"maximum gross power=\",pmax,\"W\"\n", + "print \"slip=\",s" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "maximum gross power= 143676.459572 W\n", + "slip= 0.120771344025\n" + ] + } + ], + "prompt_number": 188 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.59, Page Number:1307" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v1=115#V\n", + "f=60.0#Hz\n", + "p=6\n", + "z=complex(0.07,0.3)\n", + "zr=complex(0.08,0.3)\n", + "gd=0.022#mho\n", + "bo=0.158#mho\n", + "s=0.02\n", + "\n", + "#calculation\n", + "rl_=1/bo*(1/s-1)\n", + "z=complex(z.real+zr.real+rl_,0.6)\n", + "v=v1/3**0.5\n", + "i2=complex(16,-2.36)\n", + "io=v*complex(gd,-bo)\n", + "i1=io+i2\n", + "pf=math.cos(math.atan(i1.imag/i1.real))\n", + "pg=3*abs(i2)**2*rl_/100\n", + "ns=120*f/p\n", + "n=(1-s)*ns\n", + "tg=9.55*pg/n\n", + "p2=3**0.5*v1*abs(i1)*pf\n", + "efficiency=pg*100/p2\n", + "\n", + "#result\n", + "print \"secondary current=\",i2,\"A\"\n", + "print \"primary current=\",i1,\"A\"\n", + "print \"pf=\",pf\n", + "print \"power output=\",pg,\"W\"\n", + "print \"torque=\",tg,\"N-m\"\n", + "print \"input=\",p2,\"W\"\n", + "print \"efficiency=\",efficiency,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "secondary current= (16-2.36j) A\n", + "primary current= (17.460696181-12.8504543912j) A\n", + "pf= 0.805393212665\n", + "power output= 2433.59058228 W\n", + "torque= 19.7625765823 N-m\n", + "input= 3477.92348593 W\n", + "efficiency= 69.9725164204 %\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 34.60, Page Number:1308" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=400#V\n", + "z=complex(0.4,1)\n", + "zr=complex(0.6,1)\n", + "zm=complex(10,50)\n", + "s=0.05\n", + "\n", + "#calculation\n", + "sm=zr.real/(z.real**2+(z.imag+zr.imag)**2)**0.5\n", + "v1=v/3**0.5\n", + "i2=v1/((z.real+zr.real)**2+(zr.imag+z.imag)**2)**0.5\n", + "tgmax=3*i2**2*z.real*60/(sm*2*3.14*1500)\n", + "#result\n", + "print \"maximum torque=\",tgmax,\"N-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "maximum torque= 277.144160399 N-m\n" + ] + } + ], + "prompt_number": 208 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter35_2.ipynb b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter35_2.ipynb new file mode 100644 index 00000000..1c89c3bd --- /dev/null +++ b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter35_2.ipynb @@ -0,0 +1,1220 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:87ef53401e46d15eef2e50d8ed392f8c9e3784abe371e55cb0923dbffffe7b33" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 35: Computations and Circle Diagrams" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.1, Page Number:1316" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "i=10#A\n", + "p=450#W\n", + "v=110#V\n", + "r=0.05#ohm\n", + "loss=135#w\n", + "\n", + "#calculations\n", + "cu_loss=3*i**2*r\n", + "core_loss=p-loss-cu_loss\n", + "volt=v/math.sqrt(3)\n", + "g=core_loss/(3*(v/math.sqrt(3))**2)\n", + "y=i*math.sqrt(3)/v\n", + "b=math.sqrt(y**2-g**2)\n", + "\n", + "#result\n", + "print \"exciting conductance=\",g,\"seimens/phase\"\n", + "print \"susceptance/phase=\",b,\"seimens/phase\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "exciting conductance= 0.0247933884298 seimens/phase\n", + "susceptance/phase= 0.155494939853 seimens/phase\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.2, Page Number:1317" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=110.0#V\n", + "i=25.0#A\n", + "v2=30.0#V\n", + "inpt=440.0#W\n", + "loss=40.0#W\n", + "r=0.1#ohm\n", + "ratio=1.6\n", + "\n", + "#calculations\n", + "vs=v2/math.sqrt(3)\n", + "z01=vs/i\n", + "losses=inpt-loss\n", + "r01=losses/(3*i**2)\n", + "x01=math.sqrt(z01**2-r01**2)\n", + "dc_r=r/2.0\n", + "ac_r=dc_r*ratio\n", + "effective_r=r01-ac_r\n", + "\n", + "#result\n", + "print \"x01=\",x01,\"ohm\"\n", + "print \"r1=\",ac_r,\"ohm\"\n", + "print \"r2=\",effective_r,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "x01= 0.659157711696 ohm\n", + "r1= 0.08 ohm\n", + "r2= 0.133333333333 ohm\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.10, Page Number:1333" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "ratio=1/4.0\n", + "slip=3.0\n", + "ratio2=4.0\n", + "\n", + "#calculations\n", + "K=math.sqrt(ratio/((ratio2**2)*0.01*slip))\n", + "\n", + "#result\n", + "print \"Percentage Tapping=\",K*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Percentage Tapping= 72.1687836487 %\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.11, Page Number:1333" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "load=14.92#kW\n", + "v1=400#V\n", + "n=950#rpm\n", + "f=50.0#Hz\n", + "v2=400#V\n", + "ratio=1.8\n", + "i=30#A\n", + "\n", + "#calculations\n", + "v=v1/math.sqrt(ratio)\n", + "If=6*v*i/v1\n", + "K=v/v1\n", + "kisc=K**2*6*i\n", + "ts_tf=(1/6.0)*6**2*(f/1000.0)\n", + "\n", + "#result\n", + "print \"a)voltage=\",v,\"V\"\n", + "print \"b)current=\",If,\"A\"\n", + "print \"c)line current=\",kisc,\"A\"\n", + "print \"d)percentage=\",ts_tf*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a)voltage= 298.142397 V\n", + "b)current= 134.16407865 A\n", + "c)line current= 100.0 A\n", + "d)percentage= 30.0 %\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.12, Page Number:1334" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "ratio=5.0\n", + "per=5\n", + "\n", + "#calculations\n", + "k=math.sqrt(ratio/3)\n", + "tst_tf=(3.0/5)*5**2*0.01*per*100\n", + "\n", + "#result\n", + "print \"auto-transformation ratio=\",tst_tf,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "auto-transformation ratio= 75.0 %\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.13, Page Number:1334" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=400.0#V\n", + "per=3.5\n", + "v2=92.0#V\n", + "\n", + "#calculations\n", + "k=math.sqrt(2/(v/v2))\n", + "ts_tf=k**2*(v/v2)**2*0.01*per\n", + "\n", + "#result\n", + "print \"auto-transformation ratio=\",ts_tf*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "auto-transformation ratio= 30.4347826087 %\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.14, Page Number:1336" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "load=12.0#kW\n", + "v=440.0#V\n", + "efficiency=0.85\n", + "pf=0.8\n", + "i=45.0#A\n", + "v2=220.0#V\n", + "\n", + "#calculations\n", + "isc=i*v/v2\n", + "if_=load*1000/(efficiency*math.sqrt(3)*pf*v)\n", + "ist=isc/math.sqrt(3)\n", + "ratio=ist/if_\n", + "\n", + "#result\n", + "print \"ratio=\",ratio" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "ratio= 2.244\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.15, Page Number:1336" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "i=60.0#A\n", + "n1=940.0#rpm\n", + "t=150.0#N-m\n", + "i2=300.0#A\n", + "\n", + "#calculations\n", + "sf=(1000-n1)/1000\n", + "tst=t*(i2/i)**2*sf\n", + "s_i=i2/3\n", + "sd_tst=tst/3\n", + "\n", + "#result\n", + "print \"Starting torque=\",tst,\"N-m\"\n", + "print\"when star/delta is used:\"\n", + "print \"starting current=\",s_i,\"A\"\n", + "print \"starting torque=\",sd_tst,\"N-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Starting torque= 225.0 N-m\n", + "when star/delta is used:\n", + "starting current= 100.0 A\n", + "starting torque= 75.0 N-m\n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.16, Page Number:1336" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "tapping=70.7\n", + "ratio=6.0\n", + "slip=4.0\n", + "\n", + "#calculation\n", + "tst_tf=(1.0/3.0)*ratio**2.0*slip*0.01\n", + "tst_tf2=(1.0/2)*ratio**2.0*slip*0.01\n", + "\n", + "#result\n", + "print \"star-delta switch:starting torque=\",tst_tf*100,\"%\"\n", + "print \"auto-transformer switch:starting torque=\",tst_tf2*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "star-delta switch:starting torque= 48.0 %\n", + "auto-transformer switch:starting torque= 72.0 %\n" + ] + } + ], + "prompt_number": 48 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.17, Page Number:1337" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "load=11.2#W\n", + "f=50.0#Hz\n", + "v=400.0#V\n", + "n=960.0#rpm\n", + "i=86.4#A\n", + "efficiency=0.88\n", + "pf=0.85\n", + "\n", + "#calculations\n", + "isc=i/math.sqrt(3)\n", + "ist=isc/math.sqrt(3)\n", + "il=load*1000/(efficiency*pf*math.sqrt(3)*v)\n", + "iph=il/math.sqrt(3)\n", + "tst_tf=(ist*math.sqrt(3)/il)**2*0.05\n", + "\n", + "#result\n", + "print \"starting torque=\",tst_tf*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "starting torque= 26.6369577796 %\n" + ] + } + ], + "prompt_number": 49 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.18, Page Number:1337" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "output=10.0#kW\n", + "v=400.0#V\n", + "pf=0.85\n", + "efficiency=0.88\n", + "v2=200.0#V\n", + "i=40.0#A\n", + "\n", + "#calculations\n", + "il=load*1000/(efficiency*math.sqrt(3)*v*pf)\n", + "isc=i*v/v2\n", + "iscp=isc/math.sqrt(3)\n", + "ist=iscp/math.sqrt(3)\n", + "ratio=ist/il\n", + "\n", + "#result\n", + "print \"ratio=\",ratio" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "ratio= 1.23388000387\n" + ] + } + ], + "prompt_number": 53 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.19, Page Number:1337" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=3.73*1000#W\n", + "v=400.0#V\n", + "f=50.0#Hz\n", + "slip=4.5\n", + "t=250.0\n", + "i=650.0\n", + "tap=60.0\n", + "\n", + "#calculation\n", + "il=i/3\n", + "im=i/3\n", + "tst=t/3\n", + "ilm=(tap/100)**2*i\n", + "imk=(tap/100)*i\n", + "tstk=(tap/100)**2*t\n", + "\n", + "#result\n", + "print \"star/delta:\"\n", + "print \"line current=\",il,\"%\"\n", + "print \"motor current=\",im,\"%\"\n", + "print \"starting torque=\",tst,\"%\"\n", + "print \"60% taps:\"\n", + "print \"line current=\",ilm,\"%\"\n", + "print \"motor current=\",imk,\"%\"\n", + "print \"starting torque=\",tstk,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " star/delta:\n", + "line current= 216.666666667 %\n", + "motor current= 216.666666667 %\n", + "starting torque= 83.3333333333 %\n", + "60% taps:\n", + "line current= 234.0 %\n", + "motor current= 390.0 %\n", + "starting torque= 90.0 %\n" + ] + } + ], + "prompt_number": 55 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.20, Page Number:1338" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=180.0\n", + "flt=35.0\n", + "tap=75.0\n", + "\n", + "#calculations\n", + "isc=load*3.0/100\n", + "isck=tap**2*isc/100\n", + "sf=flt*3\n", + "tst_tf=tap**2*sf/100\n", + "#result\n", + "print \"starting current=\",isck,\"%\"\n", + "print \"starting torque=\",tst_tf/100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "starting current= 303.75 %\n", + "starting torque= 59.0625 %\n" + ] + } + ], + "prompt_number": 68 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.21, Page Number:1338" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#variable declaration\n", + "w=7.46#kW\n", + "ic=1.7\n", + "t=35.0\n", + "ratio=60.0\n", + "\n", + "#calculations\n", + "sf=t*3/100\n", + "il1=ic*3\n", + "tst=(ratio/1000)**2*sf*10000\n", + "il2=(ratio/100)*3*ic\n", + "\n", + "#results\n", + "print \"auto-starter:\"\n", + "print \"line-current=\",il1,\"%\"\n", + "print \"torque=\",tst,\"%\"\n", + "print \"voltage decreased to 60%\"\n", + "print \"line-current\",il2,\"%\"\n", + "print \"torque=\",tst,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "auto-starter:\n", + "line-current= 5.1 %\n", + "torque= 37.8 %\n", + "voltage decreased to 60%\n", + "line-current 3.06 %\n", + "torque= 37.8 %\n" + ] + } + ], + "prompt_number": 71 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.22, Page Number:1342" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#variable declaration\n", + "slip=2.0\n", + "r=0.02#ohm\n", + "n=6.0\n", + "#calculations\n", + "smax=r2=slip/100.0\n", + "R1=r2/smax\n", + "K=math.pow(smax,1.0/5)\n", + "R2=K*R1\n", + "R3=K*R2\n", + "R4=K*R3\n", + "R5=K*R4\n", + "p1=R1-R2\n", + "p2=R2-R3\n", + "p3=R3-R4\n", + "p4=R4-R5\n", + "p5=R5-r2\n", + "\n", + "#result\n", + "print \"resistances of various starter sections:\"\n", + "print \"p1=\",p1,\"ohm\"\n", + "print \"p2=\",p2,\"ohm\"\n", + "print \"p3=\",p3,\"ohm\"\n", + "print \"p4=\",p4,\"ohm\"\n", + "print \"p5=\",p5,\"ohm\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resistances of various starter sections:\n", + "p1= 0.542694948073 ohm\n", + "p2= 0.248177141409 ohm\n", + "p3= 0.113492660539 ohm\n", + "p4= 0.0519007670213 ohm\n", + "p5= 0.0237344829577 ohm\n" + ] + } + ], + "prompt_number": 107 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.23, Page Number:1345" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "primary=complex(1,3)\n", + "outer=complex(3,1)\n", + "inner=complex(0.6,5)\n", + "s=4\n", + "outer2=complex(3/(s*0.01),1)\n", + "inner2=complex(0.6/(s*0.01),5)\n", + "v=440#V\n", + "\n", + "\n", + "#calculations\n", + "#s=1\n", + "z01=primary+1/((1/outer)+(1/inner))\n", + "current_per_phase=v/abs(z01)\n", + "torque=3*current_per_phase**2*(z01.real-1)\n", + "\n", + "print \"s=1: torque=\",torque,\"synch watt\"\n", + "\n", + "#s=4\n", + "z01=primary+1/((1/outer2)+(1/inner2))\n", + "current_per_phase=v/abs(z01)\n", + "torque=3*current_per_phase**2*(z01.real-1)\n", + "\n", + "print \"s=4: torque=\",torque,\"synch watt\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "s=1: torque= 35065.3642462 synch watt\n", + "s=4: torque= 32129.9449695 synch watt\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.24, Page Number:1346" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "inner=complex(0.4,2)\n", + "outer=complex(2,0.4)\n", + "s=5\n", + "inner2=complex(0.4/(s*0.01),2)\n", + "outer2=complex(2/(s*0.01),0.4)\n", + "print \n", + "#calculations\n", + "#s=1\n", + "zi=abs(inner)\n", + "zo=abs(outer)\n", + "r_ratio=inner.imag/outer.imag\n", + "to_ti=r_ratio*(zo/zi)**2\n", + "print \"Ratio of torques when s=1:\",to_ti\n", + "\n", + "#s=5\n", + "zi=abs(inner2)\n", + "zo=abs(outer2)\n", + "print zi\n", + "r_ratio=inner2.imag/outer2.imag\n", + "to_ti=r_ratio*(zi/zo)**2\n", + "\n", + "print \"Ratio of torques when s=5:\",to_ti" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "Ratio of torques when s=1: 5.0\n", + "8.24621125124\n", + "Ratio of torques when s=5: 0.212478752125\n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.25, Page Number:1346" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "s=5\n", + "zi=complex(0.05,0.4)\n", + "zo=complex(0.5,0.1)\n", + "v=100#V\n", + "\n", + "#calculations\n", + "#s=1\n", + "z=zo*zi/(zo+zi)\n", + "r2=z.real\n", + "z=abs(z)\n", + "i2=v/z\n", + "t=i2**2*r2\n", + "print \"s=1:torque=\",t,\"synch watts\"\n", + "\n", + "#s=0.01\n", + "zi=complex(0.05/(s*0.01),0.4)\n", + "zo=complex(0.5/(s*0.01),0.1)\n", + "z=zo*zi/(zo+zi)\n", + "r2=z.real\n", + "z=abs(z)\n", + "i2=v/z\n", + "t=i2**2*r2\n", + "print \"s=5:torque=\",t,\"synch watts\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "s=1:torque= 22307.6923077 synch watts\n", + "s=5:torque= 9620.58966517 synch watts\n" + ] + } + ], + "prompt_number": 43 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.27, Page Number:1347" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "zo=complex(1,0)\n", + "zi=complex(0.15,3)\n", + "v=250#V\n", + "n=1000#rpm\n", + "\n", + "#calculations\n", + "z2=zo*zi/(zo+zi)\n", + "stator=complex(0.25,3.5)\n", + "z01=z2+stator\n", + "i=complex(v,0)/z01\n", + "i=abs(i)\n", + "cu_loss=i**2*z01.real\n", + "T=cu_loss*3/(2*math.pi*(n/60))\n", + "#result\n", + "print \"torque=\",T,\"N-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "torque= 135.560320318 N-m\n" + ] + } + ], + "prompt_number": 49 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.28, Page Number:1348" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "z1=complex(1,2.8)\n", + "zo=complex(3,1)\n", + "zi=complex(0.5,5)\n", + "v=440#V\n", + "s=0.04\n", + "\n", + "#calculations\n", + "#s=1\n", + "z2=zo*zi/(zo+zi)\n", + "z01=z1+z2\n", + "i2=v/z01\n", + "r2=z2.real\n", + "t=abs(i2)**2*r2\n", + "\n", + "print \"s=1:torque=\",t,\"synch. watt\"\n", + "\n", + "#s=0.04\n", + "zo=complex(3.0/s,1.0)\n", + "zi=complex(0.5/s,5.0)\n", + "z2=zo*zi/(zo+zi)\n", + "z01=z1+z2\n", + "i2=v/z01\n", + "r2=z2.real\n", + "t=abs(i2)**2*r2\n", + "print \"s=4:torque=\",t,\"synch. watt\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "s=1:torque= 12388.3258184 synch. watt\n", + "s=4:torque= 11489.1141244 synch. watt\n" + ] + } + ], + "prompt_number": 58 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.29, Page Number:1351" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "f=50.0#Hz\n", + "r=0.30#ohm\n", + "n1=1440.0#rpm\n", + "n2=1320.0#rpm\n", + "ns=120.0*f/4.0\n", + "#calculations\n", + "s1=(ns-n1)/ns\n", + "s2=(ns-n2)/ns\n", + "r=s2*r/s1-r\n", + "\n", + "#result\n", + "print \"external resistance=\",r,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "external resistance= 0.6 ohm\n" + ] + } + ], + "prompt_number": 60 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.30, Page Number:1348" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "f=50.0#Hz\n", + "s=0.03\n", + "ratio=10.0\n", + "r=0.2\n", + "\n", + "#calculations\n", + "ns=120*f/6\n", + "s1=s\n", + "n1=ns*(1-s1)\n", + "n2=n1-10*n1/100\n", + "s2=(ns-n2)/ns\n", + "r=s2*r/s1-r\n", + "\n", + "#result\n", + "print \"external resistance=\",r,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "external resistance= 0.646666666667 ohm\n" + ] + } + ], + "prompt_number": 61 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.31, Page Number:1354" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "f=50#Hz\n", + "s=0.02\n", + "\n", + "#calculations\n", + "nsc=120*f/10\n", + "n=(1-s)*nsc\n", + "nsa=120*f/6\n", + "sa=(nsa-n)/nsa\n", + "f_=sa*f\n", + "n_=(120*f_)/4\n", + "sb=(n_-n)/n_\n", + "f__=sb*f_\n", + "\n", + "#resu;t\n", + "print \"f_=\",f_,\"Hz\"\n", + "print \"f_ _=\",f__,\"Hz\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "f_= 20.6 Hz\n", + "f_ _= 1.0 Hz\n" + ] + } + ], + "prompt_number": 69 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.32, Page Number:1354" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "f=50.0#Hz\n", + "f2=1.0#Hz\n", + "\n", + "#calculations\n", + "nsc=120*f/10\n", + "s=f2/f\n", + "n=nsc-s*nsc\n", + "nsa=120*f/4\n", + "sa=(nsa-n)/nsa\n", + "f1=sa*f\n", + "n2=120*f1/6\n", + "sb=(n2-n)/n2\n", + "\n", + "#result\n", + "print \"sa=\",sa*100,\"%\"\n", + "print \"sb=\",sb*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "sa= 60.8 %\n", + "sb= 3.28947368421 %\n" + ] + } + ], + "prompt_number": 75 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.33, Page Number:1354" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "f=50#Hz\n", + "load=74.6#kW\n", + "\n", + "#calculations\n", + "nsc=120*f/10\n", + "output=load*4/10\n", + "\n", + "#result\n", + "print \"speed of set=\",nsc,\"rpm\"\n", + "print \"electric power transferred=\",output,\"kW\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed of set= 600 rpm\n", + "electric power transferred= 29.84 kW\n" + ] + } + ], + "prompt_number": 79 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 35.34, Page Number:1355" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "f=50#Hz\n", + "load=25#kW\n", + "\n", + "#calculations\n", + "nsc=120*f/10\n", + "output=load*4/10\n", + "\n", + "#result\n", + "print \"speed of set=\",nsc,\"rpm\"\n", + "print \"electric power transferred=\",output,\"kW\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed of set= 600 rpm\n", + "electric power transferred= 10 kW\n" + ] + } + ], + "prompt_number": 78 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter36_2.ipynb b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter36_2.ipynb new file mode 100644 index 00000000..a28f10ba --- /dev/null +++ b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter36_2.ipynb @@ -0,0 +1,393 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a362cd0373fe77cde513a2a109a4d7c05a5dbd87d086b1227fbc532438b6bbb6" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 36: Single-Phase Motors" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 36.1, Page Number:1374" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#variable declaration\n", + "R1=1.86\n", + "X1=2.56\n", + "R2=3.56\n", + "X2=2.56\n", + "Xm=53.5\n", + "r1=R1/2\n", + "x1=X1/2\n", + "r2=R2/2\n", + "x2=X2/2\n", + "xm=Xm/2\n", + "v=110\n", + "f=60\n", + "s=0.05\n", + "\n", + "#calculations\n", + "xo=xm+x2\n", + "\n", + "zf=(((r2/s)*xm)/(((r2/s)*(r2/s))+(xo*xo)))*xm\n", + "jf=(((r2/s)*(r2/s)+(x2*xo))/(((r2/s)*(r2/s))+(xo*xo)))*xm\n", + "Jf=math.degrees(math.atan(jf/zf))\n", + "\n", + "zb=(((r2/(2-s))*xm)/(((r2/s)*(r2/(2-s)))+(xo*xo)))*xm\n", + "jb=(((r2/(2-s))*(r2/(2-s))+(x2*xo))/(((r2/(2-s))*(r2/(2-s)))+(xo*xo)))*xm\n", + "Jb=math.degrees(math.atan(jb/zb))\n", + "\n", + "Z1=R1\n", + "J1=X1\n", + "z01=Z1+zf+zb\n", + "j01=jf+jb+J1\n", + "J01=math.degrees(math.atan(j01/z01))\n", + "\n", + "i1=v/z01\n", + "vf=i1*zf\n", + "vb=i1*zb\n", + "z3=math.sqrt(((r2/s)*(r2/s))+(x2*x2))\n", + "z5=math.sqrt(((r2/(2-s))*(r2/(2-s)))+(x2*x2))\n", + "\n", + "i3=vf/z3\n", + "i5=vb/z5\n", + "tf=(i3*i3*r2)/s\n", + "tb=t5=(i5*i5*r2)/(2-s)\n", + "t=tf-tb\n", + "output=t*(1-s)\n", + "\n", + "#result\n", + "print \"output = \",output" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "output = 206.798750547\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Example Number 36.2, Page Number:1375" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#variable declaration\n", + "p=185\n", + "v=110\n", + "f=50\n", + "s=0.05\n", + "R1=1.86\n", + "X1=2.56\n", + "Xo=53.5\n", + "R2=3.56\n", + "X2=2.56\n", + "Xm=53.5\n", + "cl=3.5#core loss\n", + "fl=13.5#friction loss\n", + "vf=(82.5/100)*v\n", + "ic=(cl*100)/vf\n", + "r1=R1/2\n", + "x1=X1/2\n", + "r2=R2/2\n", + "x2=X2/2\n", + "xm=Xm/2\n", + "rc=vf/ic\n", + "\n", + "#calculations\n", + "\n", + "#motor 1\n", + "c=1/rc #conductance of corebranch\n", + "s=-(1/xm)#susceptance\n", + "a1=(r2/s)/(((r2/s)*r2/s)+(x2*x2))#admittance\n", + "a1j=-x2/(((r2/s)*r2/s)+(x2*x2))#admittance j\n", + "yf=c+a1\n", + "yfj=s+a1j\n", + "zf=(yf*yf)+(yfj*yfj)\n", + "zfr=yf/zf\n", + "zfj=yfj/zf\n", + "\n", + "#motor 2\n", + "a2=(r2/2-s)/(((r2/(2-s))*(r2/(2-s)))+(x2*x2))\n", + "a2j=-x2/(((r2/(2-s))*(r2/(2-s)))+(x2*x2))\n", + "Z1=R1\n", + "J1=X1\n", + "yb=yf+a2\n", + "ybj=yfj+a2j\n", + "zb1=(yb*yb)+(ybj*ybj)\n", + "zbr=yb/zb1\n", + "zbj=ybj/zb1\n", + "z01=Z1+zf+zbr\n", + "z01j=J1+zfj+zbj\n", + "\n", + "i1=v/z01\n", + "vf=i1*zf\n", + "vb=i1*zbr\n", + "z3=math.sqrt(((r2/s)*(r2/s))+(x2*x2))\n", + "z5=math.sqrt(((r2/(2-s))*(r2/(2-s)))+(x2*x2))\n", + "\n", + "i3=vf/z3\n", + "i5=vb/z5\n", + "tf=(i3*i3*r2)/s\n", + "tb=t5=(i5*i5*r2)/(2-s)\n", + "t=tf-tb\n", + "watt=t*(1-s)\n", + "net_output=watt-fl\n", + "\n", + "#result\n", + "print \"Net output = \",net_output" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Net output = -446.423232085\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 36.3, Page Number:1376" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#variable declaration\n", + "w=250\n", + "v=230\n", + "f=50\n", + "zm=4.5\n", + "zmj=3.7\n", + "za=9.5\n", + "zaj=3.5\n", + "\n", + "#calculations\n", + "zma=math.degrees(math.atan(zmj/zm))\n", + "ialeadv=90-zma\n", + "x=za*(math.tan(math.radians(ialeadv)))\n", + "xc=x+zaj\n", + "c=1000000/(xc*2*50*3.14)\n", + "\n", + "#result\n", + "print \"C= \",c,\" uf\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "C= 211.551875951 uf\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 36.4, Page Number:1393" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#variable declaration\n", + "\n", + "p=250\n", + "f=50\n", + "v=220\n", + "ndc=2000\n", + "ia=1\n", + "ra=20\n", + "la=0.4\n", + "\n", + "#calculations\n", + "ebdc=v-(ia*ra)\n", + "#ac\n", + "xa=2*3.14*f*la\n", + "ebac=-(ia*ra)+math.sqrt((v*v)-((ia*xa)*(ia*xa)))\n", + "nac=(ebac*ndc)/ebdc\n", + "cos_phi=(ebac+(ia*ra))/v\n", + "pmech=ebac*ia\n", + "T=(pmech*9.55)/nac\n", + "\n", + "#result\n", + "print \"Speed= \",nac,\" rpm\"\n", + "print \"Torque= \",T,\" N-m\"\n", + "print \"Power Factor= \",cos_phi,\" lag\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Speed= 1606.22922133 rpm\n", + "Torque= 0.955 N-m\n", + "Power Factor= 0.821013282424 lag\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Example Number 36.5, Page Number:1394" + ], + "language": "python", + "metadata": {}, + "outputs": [] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#variable declaration\n", + "r=30\n", + "l=0.5\n", + "v=250\n", + "idc=0.8\n", + "ndc=2000\n", + "f=50\n", + "ia=0.8\n", + "\n", + "#calculations\n", + "\n", + "xa=2*3.14*f*l\n", + "ra=r\n", + "ebac=-(ia*ra)+math.sqrt((v*v)-((ia*xa)*(ia*xa)))\n", + "ebdc=v-(r*idc)\n", + "nac=(ndc*ebac)/ebdc\n", + "cos_phi=(ebac+(ia*ra))/v\n", + "\n", + "#result\n", + "print \"Speed= \",nac,\" rpm\"\n", + "print \"Power Factor= \",cos_phi,\" lag\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Speed= 1700.52062383 rpm\n", + "Power Factor= 0.864635321971 lag\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 36.6, Page Number:1396" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#variable declaration\n", + "f=50\n", + "a=30\n", + "w=8\n", + "v=220\n", + "v2=205\n", + "pole=4\n", + "\n", + "#calculations\n", + "\n", + "ns=(120*f)/pole\n", + "tsh=(9.55*w*1000)/ns\n", + "alpha=0.5*(math.degrees(math.asin((v*v*math.sin(math.radians(2*a)))/(v2*v2))))\n", + "\n", + "#result\n", + "print \"Torque angle if voltage drops to 205 V = \",alpha,\" degrees\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Torque angle if voltage drops to 205 V = 42.9327261097 degrees\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter37_2.ipynb b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter37_2.ipynb new file mode 100644 index 00000000..7e0be0a9 --- /dev/null +++ b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter37_2.ipynb @@ -0,0 +1,2781 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:3f52bfdb4973d016ec59d44992f6a2ce15bb8cca394c854d00d33c6af91049f3" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 37: Alternators" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.1, Page Number:1412" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "s1=36.0\n", + "p1=4.0\n", + "span1=8.0\n", + "s2=72.0\n", + "p2=6.0\n", + "span2=10.0\n", + "s3=96.0\n", + "p3=6.0\n", + "span3=12.0\n", + "\n", + "#calculations\n", + "alpha1=2*p1*180/s1\n", + "alpha2=3*p2*180/s2\n", + "alpha3=5*p3*180/s3\n", + "kc1=math.cos(math.radians(alpha1/2))\n", + "kc2=math.cos(math.radians(alpha2/2))\n", + "kc3=math.cos(math.radians(alpha3/2))\n", + "\n", + "#result\n", + "print \"a)kc=\",kc1\n", + "print \"b)kc=\",kc2\n", + "print \"c)kc=\",kc3" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a)kc= 0.939692620786\n", + "b)kc= 0.923879532511\n", + "c)kc= 0.881921264348\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.2, Page Number:1414" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "s=36.0\n", + "p=4.0\n", + "\n", + "#calculations\n", + "n=s/p\n", + "beta=180/n\n", + "m=s/(p*3)\n", + "kd=math.sin(m*math.radians(beta/2))/(m*math.sin(math.radians(beta/2)))\n", + "\n", + "#result\n", + "print \"distribution factor=\",kd" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "distribution factor= 0.959795080524\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.3, Page Number:1414" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=10.0#V\n", + "beta=30.0#degrees\n", + "m=6.0\n", + "\n", + "#calculations\n", + "kd=math.sin(m*math.radians(beta/2))/(m*math.sin(math.radians(beta/2)))\n", + "arith_sum=6*v\n", + "vector_sum=kd*arith_sum\n", + "\n", + "#calculation\n", + "print \"emf of six coils in series=\",vector_sum,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "emf of six coils in series= 38.6370330516 V\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.4, Page Number:1414" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "beta=180/9\n", + "ratio=2.0/3.0\n", + "m1=9\n", + "m2=6\n", + "m3=3\n", + "\n", + "#calculation\n", + "kd1=math.sin(m1*math.radians(beta/2))/(m1*math.sin(math.radians(beta/2)))\n", + "kd2=math.sin(m2*math.radians(beta/2))/(m2*math.sin(math.radians(beta/2)))\n", + "kd3=math.sin(m3*math.radians(beta/2))/(m3*math.sin(math.radians(beta/2)))\n", + "\n", + "#result\n", + "print \"i) kd=\",kd1\n", + "print \"ii)kd=\",kd2\n", + "print \"iii)kd=\",kd3" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i) kd= 0.639863387016\n", + "ii)kd= 0.831206922161\n", + "iii)kd= 0.959795080524\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.5, Page Number:1416" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "slot=18.0\n", + "s=16.0\n", + "m1=3.0\n", + "m2=5.0\n", + "m3=7.0\n", + "\n", + "#calculations\n", + "span=(s-1)\n", + "alpha=180*3/slot\n", + "kc1=math.cos(math.radians(alpha/2))\n", + "kc3=math.cos(math.radians(m1*alpha/2))\n", + "kc5=math.cos(math.radians(m2*alpha/2))\n", + "kc7=math.cos(math.radians(m3*alpha/2))\n", + "\n", + "#result\n", + "print \"kc1=\",kc1\n", + "print \"kc3=\",kc3\n", + "print \"kc5=\",kc5\n", + "print \"kc7=\",kc7" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "kc1= 0.965925826289\n", + "kc3= 0.707106781187\n", + "kc5= 0.258819045103\n", + "kc7= -0.258819045103\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.6, Page Number:1416" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=16.0\n", + "s=144.0\n", + "z=10.0\n", + "phi=0.03#Wb\n", + "n=375.0#rpm\n", + "\n", + "#calculation\n", + "f=p*n/120\n", + "n=s/p\n", + "beta=180/9\n", + "m=s/(p*3)\n", + "kd=math.sin(m*math.radians(beta/2))/(m*math.sin(math.radians(beta/2)))\n", + "t=s*z/(3*2)\n", + "eph=4.44*1*0.96*f*phi*t\n", + "el=3**0.5*eph\n", + "#result\n", + "print \"frequency=\",f,\"Hz\"\n", + "print \"phase emf=\",eph,\"V\"\n", + "print \"line emf=\",el,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "frequency= 50.0 Hz\n", + "phase emf= 1534.464 V\n", + "line emf= 2657.76961039 V\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.7, Page Number:1416" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=6\n", + "s=54\n", + "phi=0.1#Wb\n", + "n=1200#rpm\n", + "t=8\n", + "#calculations\n", + "beta=180/9\n", + "kc=math.cos(beta/2)\n", + "f=p*n/120\n", + "n=s/p\n", + "m=s/(p*3)\n", + "kd=math.sin(m*math.radians(beta/2))/(m*math.sin(math.radians(beta/2)))\n", + "z=s*8/3\n", + "t=z/2\n", + "eph=4.44*0.98*0.96*f*phi*t\n", + "el=3**0.*eph\n", + "\n", + "#result\n", + "print \"eph=\",eph,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "eph= 1804.529664 V\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.8, Page Number:1416" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=16.0\n", + "slots=144.0\n", + "z=4.0\n", + "n=375.0\n", + "airgap=5*0.01\n", + "theta=150.0\n", + "\n", + "#calculation\n", + "kf=1.11\n", + "alpha=(180-theta)\n", + "kc=math.cos(math.radians(alpha/2))\n", + "beta=180/9\n", + "m=slots/(p*3)\n", + "kd=math.sin(m*math.radians(beta/2))/(m*math.sin(math.radians(beta/2)))\n", + "f=p*n/120\n", + "s=slots/3\n", + "eph=4*kf*kc*kd*f*airgap*s*4/2\n", + "\n", + "#result\n", + "print \"emf per phase=\",eph,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "emf per phase= 987.908016392 V\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.9, Page Number:1417" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=10\n", + "f=50#Hz\n", + "n=600#rpm\n", + "slots=180\n", + "s=15\n", + "d=1.2#m\n", + "l=0.4#m\n", + "m=6\n", + "beta=180/18\n", + "#calculations\n", + "area=(1.2*3.14/p)*l\n", + "phi1=area*0.637\n", + "vr=1.1*2*f*phi1\n", + "vp=2**0.5*vr\n", + "v3=0.4*vp\n", + "v5=0.2*vp\n", + "vf=6*vp*0.966\n", + "vf3=6*v3*0.707\n", + "vf5=6*v5*0.259\n", + "kd1=math.sin(m*math.radians(beta/2))/(m*math.sin(math.radians(beta/2)))\n", + "kd2=math.sin(math.radians(3*m*beta/2))/(6*math.sin(3*math.radians(beta/2)))\n", + "kd3=math.sin(math.radians(5*m*beta/2))/(6*math.sin(5*math.radians(beta/2)))\n", + "vph=vf*2**0.5*60*kd1\n", + "vph3=vf3*2**0.5*60*kd2\n", + "vph5=vf5*2**0.5*60*kd3\n", + "rmsv=(vph**2+vph3**2+vph5**2)**0.5\n", + "rmsvl=3**0.5*(vph**2+vph5**2)**0.5\n", + "\n", + "#result\n", + "print \"i)e=\",vp,\"sin theta+\",v3,\"sin 3theta+\",v5,\"sin 5theta\"\n", + "print \"ii)e=\",vf,\"sin theta+\",vf3,\"sin 3theta+\",vf5,\"sin 5theta\"\n", + "print \"iii)rms value of phase voltage=\",rmsv,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)e= 14.9354392872 sin theta+ 5.97417571489 sin 3theta+ 2.98708785745 sin 5theta\n", + "ii)e= 86.5658061088 sin theta+ 25.3424533826 sin 3theta+ 4.64193453047 sin 5theta\n", + "iii)rms value of phase voltage= 7158.83679423 V\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.10, Page Number:1418" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "p=4\n", + "f=50.0#Hz\n", + "slot=60.0\n", + "z=4.0\n", + "s=3.0\n", + "theta=60.0\n", + "phi=0.943#Wb\n", + "\n", + "#calculation\n", + "m=slot/(p*s)\n", + "beta=slot/5\n", + "kd=math.sin(m*math.radians(beta/2))/(m*math.sin(math.radians(beta/2)))\n", + "alpha=(s/15)*180\n", + "kc=math.cos(math.radians(alpha/2))\n", + "z=slot*z/s\n", + "t=z/2\n", + "kf=1.11\n", + "eph=z*kf*kc*kd*f*phi*t/2\n", + "el=3**0.5*eph*0.1\n", + "\n", + "#result\n", + "print \"line voltage=\",el,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "line voltage= 13196.4478482 V\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.11, Page Number:1418" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=4.0\n", + "f=50.0#Hz\n", + "slot=15.0\n", + "z=10.0\n", + "kd=0.95\n", + "e=1825#v\n", + "kc=1\n", + "kf=1.11\n", + "#calculations\n", + "slots=p*slot\n", + "slotsp=slots/3\n", + "turnp=20*z/2\n", + "phi=e/(3**0.5*p*kc*kf*kd*f*turnp)\n", + "z=slots*z\n", + "n=120*f/p\n", + "eg=(phi*0.001*z*n)/slots\n", + "\n", + "#result\n", + "print \"emf=\",eg*1000,\"V\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "emf= 749.405577006 V\n" + ] + } + ], + "prompt_number": 47 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.12, Page Number:1419" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=360#V\n", + "f=60.0#Hz\n", + "i=3.6#A\n", + "f2=40#Hz\n", + "i2=2.4#A\n", + "\n", + "#calculations\n", + "e2=v*i2*f2/(f*i)\n", + "\n", + "#result\n", + "print \"e2=\",e2,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "e2= 160.0 V\n" + ] + } + ], + "prompt_number": 49 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.13, Page Number:1418" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=0\n", + "f=50.0#Hz\n", + "slot=2\n", + "z=4\n", + "theta=150#degrees\n", + "phi=0.12#Wb\n", + "per=20#%\n", + "\n", + "#calculations\n", + "alpha=180-theta\n", + "slotp=6\n", + "m=2\n", + "beta=180/slotp\n", + "kd1=math.sin(m*math.radians(beta/2))/(m*math.sin(math.radians(beta/2)))\n", + "z=10*slot*z\n", + "t=z/2\n", + "e1=4.44*kd1*kd1*f*0.12*t\n", + "kc3=math.cos(3*math.radians(alpha/2))\n", + "f2=f*3\n", + "phi3=(1.0/3)*per*0.12\n", + "e3=4.44*kd3*kd3*theta*0.008*40\n", + "e=(e1**2+e3**2)**0.5\n", + "\n", + "#result\n", + "print \"e=\",e,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "e= 994.25286629 V\n" + ] + } + ], + "prompt_number": 50 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.14, Page Number:1419" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=230.0#V\n", + "per=10.0#%\n", + "per2=6.0#%\n", + "f=50.0#Hz\n", + "r=10.0#ohm\n", + "\n", + "#calculation\n", + "#star connection\n", + "e5=per*v/100\n", + "e=(v**2+e5**2)**0.5\n", + "eph=3**0.5*e\n", + "\n", + "#delta\n", + "e3=10*v/100\n", + "f3=10*3\n", + "i=e3/f3\n", + "\n", + "#result\n", + "print \"line voltage for star=\",eph,\"V\"\n", + "print \"line voltage for delta=\",e3,\"V\"\n", + "print \"current=\",i,\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "line voltage for star= 400.358589267 V\n", + "line voltage for delta= 23.0 V\n", + "current= 0.766666666667 A\n" + ] + } + ], + "prompt_number": 55 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.15(a), Page Number:1420" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=10.0\n", + "p1=24.0\n", + "f=25#Hz\n", + "p3=6.0\n", + "s=0.05\n", + "\n", + "#calculation\n", + "n=120*f/p\n", + "f1=p1*n/120\n", + "n2=120*f1/6\n", + "n3=(1-s)*n2\n", + "f2=s*f1p\n", + "\n", + "\n", + "#result\n", + "print \"frequency=\",f1,\"Hz\"\n", + "print \"speed=\",n3,\"rpm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "frequency= 60.0 Hz\n", + "speed= 1140.0 rpm\n" + ] + } + ], + "prompt_number": 56 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.15(b), Page Number:1420" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "p=4\n", + "phi=0.12#Wb\n", + "slotsp=4\n", + "cp=4\n", + "theta=150#degrees\n", + "\n", + "#calculation\n", + "slots=slotsp*3*p\n", + "c=cp*slots\n", + "turns=32\n", + "kb=math.sin(math.radians(60/2))/(p*math.sin(math.radians(7.5)))\n", + "kp=math.cos(math.radians(15))\n", + "eph=4.44*50*0.12*kb*0.966*turns\n", + "el=eph*3**0.5\n", + "\n", + "#result\n", + "print \"line voltage\",el,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "line voltage 1365.94840977 V\n" + ] + } + ], + "prompt_number": 62 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.16, Page Number:1426" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=10#MW\n", + "pf=0.85\n", + "v=11#kV\n", + "r=0.1#ohm\n", + "x=0.66#ohm\n", + "\n", + "#calculation\n", + "i=load*10**6/(3**0.5*v*1000*pf)\n", + "iradrop=i*r\n", + "ixsdrop=i*x\n", + "vp=v*1000/3**0.5\n", + "phi=math.acos(pf)\n", + "sinphi=math.sin(phi)\n", + "e0=((vp*pf+i*r)**2+(vp*sinphi+i*x)**2)**0.5\n", + "el=3**0.5*e0\n", + "\n", + "#result\n", + "print \"linevalue of emf=\",el,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "linevalue of emf= 11475.6408913 V\n" + ] + } + ], + "prompt_number": 69 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.17(a), Page Number:1428" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=2200.0#V\n", + "f=50.0#Hz\n", + "load=440.0#KVA\n", + "r=0.5#ohm\n", + "i=40.0#A\n", + "il=200.0#A\n", + "vf=1160.0#V\n", + "\n", + "#calculations\n", + "zs=vf/200\n", + "xs=(zs**2-r**2)**0.5\n", + "\n", + "#result\n", + "print \"synchronous impedence=\",zs,\"ohm\"\n", + "print \"synchronous reactance=\",xs,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "synchronous impedence= 5.8 ohm\n", + "synchronous reactance= 5.77840808528 ohm\n" + ] + } + ], + "prompt_number": 71 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.17(b), Page Number:1428" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=60.0#kVA\n", + "v=220.0#V\n", + "f=50.0#Hz\n", + "r=0.016#ohm\n", + "x=0.07#ohm\n", + "pf=0.7\n", + "\n", + "#calculations\n", + "i=load*1000/v\n", + "ira=i*r\n", + "ixl=i*x\n", + "#unity pf\n", + "e=((v+ira)**2+(ixl)**2)**0.5\n", + "#pf of 0.7 lag\n", + "e2=((v*pf+ira)**2+(v*pf+ixl)**2)**0.5\n", + "#pf of 0.7 lead\n", + "e3=((v*pf+ira)**2+(v*pf-ixl)**2)**0.5\n", + "\n", + "#result\n", + "print \"voltage with pf=1\",e,\"V\"\n", + "print \"voltage with pf=0.7 lag\",e2,\"V\"\n", + "print \"voltage with pf=0.7 lead\",e3,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "voltage with pf=1 225.174386048 V\n", + "voltage with pf=0.7 lag 234.604995966 V\n", + "voltage with pf=0.7 lead 208.03726621 V\n" + ] + } + ], + "prompt_number": 75 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.18(a), Page Number:1429" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=50.0#KVA\n", + "v1=440.0#V\n", + "f=50.0#Hz\n", + "r=0.25#ohm\n", + "x=3.2#ohm\n", + "xl=0.5#ohm\n", + "\n", + "#calculation\n", + "v=v1/3**0.5\n", + "i=load*1000/(3**0.5*v1)\n", + "rd=i*r\n", + "ixl=i*xl\n", + "ea=((v+rd)**2+(ixl)**2)**0.5\n", + "el=3**0.5*ea\n", + "e0=((v+rd)**2+(i*x)**2)**0.5\n", + "e0l=e0*3**0.5\n", + "per=(e0-v)/v\n", + "xa=x-xl\n", + "#result\n", + "print \"internal emf Ea=\",el,\"V\"\n", + "print \"no load emf=\",e0l,\"V\"\n", + "print \"percentage regulation=\",per*100,\"%\"\n", + "print \"valueof synchronous reactance=\",xa,\"ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "internal emf Ea= 471.842539659 V\n", + "no load emf= 592.991130967 V\n", + "percentage regulation= 34.7707115833 %\n", + "valueof synchronous reactance= 2.7 ohm\n" + ] + } + ], + "prompt_number": 87 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.19, Page Number:1432" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "i=200.0#A\n", + "v=50.0#V\n", + "r=0.1#ohm\n", + "il=100.0#A\n", + "pf=0.8\n", + "vt=200.0#V\n", + "\n", + "#calculation\n", + "zs=v/vt\n", + "xs=(zs**2-r**2)**0.5\n", + "ira=il*r\n", + "ixs=il*xs\n", + "sinphi=math.sin(math.acos(pf))\n", + "e0=((vt*pf+ira)**2+(vt*sinphi+ixs)**2)**0.5\n", + "\n", + "#result\n", + "print \"induced voltage=\",e0,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "induced voltage= 222.090276316 V\n" + ] + } + ], + "prompt_number": 90 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.20, Page Number:1433" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=2000.0#V\n", + "i=100.0#A\n", + "pf=0.8\n", + "pf2=0.71\n", + "i2=2.5#A\n", + "v2=500.0#V\n", + "r=0.8#ohm\n", + "\n", + "#calculations\n", + "sinphi1=math.sin(math.acos(pf))\n", + "sinphi2=math.sin(math.acos(pf2))\n", + "zs=v2/i\n", + "xs=(zs**2-r**2)**.5\n", + "#unity pf\n", + "e01=((v+r*i)**2+(i*xs)**2)**0.5\n", + "reg1=(e01-v)*100/v\n", + "#at pf=0.8\n", + "e02=((v*pf+r*i)**2+(v*sinphi1-i*xs)**2)**0.5\n", + "reg2=(e02-v)*100/v\n", + "#at pf=0.71\n", + "e03=((v*pf2+r*i)**2+(v*sinphi2+i*xs)**2)**0.5\n", + "reg3=(e03-v)*100/v\n", + "\n", + "#result\n", + "print \"voltage regulation unity pf=\",reg1,\"%\"\n", + "print \"voltage regulation 0.8 lag pf=\",reg2,\"%\"\n", + "print \"voltage regulation 0.71 lead pf=\",reg3,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "0.6\n", + "voltage regulation unity pf= 6.88779163216 %\n", + "voltage regulation 0.8 lag pf= -8.875640156 %\n", + "voltage regulation 0.71 lead pf= 21.1141910671 %\n" + ] + } + ], + "prompt_number": 100 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.21, Page Number:1433" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=3000.0#V\n", + "load=100.0#kVA\n", + "f=50.0#Hz\n", + "r=0.2\n", + "i1=40.0#A\n", + "i2=200.0#A\n", + "v2=1040.0#V\n", + "pf=0.8\n", + "v1=v/3**0.5\n", + "#calculations\n", + "sinphi1=math.sin(math.acos(pf))\n", + "zs=v2/(3**0.5*i2)\n", + "xs=(zs**2-r**2)**.5\n", + "i=load*1000/(3**0.5*v)\n", + "\n", + "\n", + "#at pf=0.8 lag\n", + "e01=((v1*pf+r*i)**2+(v1*sinphi1+i*xs)**2)**0.5\n", + "reg1=(e01-v1)*100/v1\n", + "#at pf=0.8 lead\n", + "e02=((v1*pf+r*i)**2+(v1*sinphi1-i*xs)**2)**0.5\n", + "reg2=(e02-v1)*100/v1\n", + "\n", + "#result\n", + "print \"voltage regulation 0.8 lag pf=\",reg1,\"%\"\n", + "print \"voltage regulation 0.8 lag pf=\",reg2,\"%\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "voltage regulation 0.8 lag pf= 2.20611574348 %\n", + "voltage regulation 0.8 lag pf= -1.77945143824 %\n" + ] + } + ], + "prompt_number": 112 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.22, Page Number:1434" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=1600.0#kVA\n", + "v=13500.0#V\n", + "r=1.5#ohm\n", + "x=30.0#ohm\n", + "load1=1280.0#kW\n", + "pf=0.8\n", + "\n", + "#calculation\n", + "sinphi1=math.sin(math.acos(pf))\n", + "i=load1*1000/(3**0.5*v*pf)\n", + "ira=i*r\n", + "ixs=i*x\n", + "vp=v/3**0.5\n", + "e0=((vp*pf+ira)**2+(vp*sinphi1-ixs)**2)**0.5\n", + "regn=(e0-vp)*100/vp\n", + "\n", + "#result\n", + "print \"percentage regulation=\",regn,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "percentage regulation= -11.9909032489 %\n" + ] + } + ], + "prompt_number": 122 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.23, Page Number:1435" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=10.0#kVA\n", + "v=400.0#V\n", + "f=50.0#Hz\n", + "pf=0.8\n", + "r=0.5#ohm\n", + "x=10.0#ohm\n", + "\n", + "#calculations\n", + "i=load*1000/(3**0.5*v)\n", + "ira=i*r\n", + "ixs=i*x\n", + "vp=v/3**0.5\n", + "sinphi=math.sin(math.acos(pf))\n", + "e0=((vp*pf+ira)**2+(vp*sinphi+ixs)**2)**0.5\n", + "regn=(e0-vp)/vp\n", + "thetadel=math.atan((vp*sinphi+ixs)/(vp*pf+ira))\n", + "delta=math.degrees(thetadel)-math.degrees(math.acos(pf))\n", + "\n", + "#result\n", + "print \"voltage regulation=\",regn*100,\"%\"\n", + "print \"power angle=\",delta,\"degrees\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "voltage regulation= 48.0405877623 %\n", + "power angle= 18.9704078085 degrees\n" + ] + } + ], + "prompt_number": 127 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.24, Page Number:1435" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=6000.0#KVA\n", + "v=6600.0#V\n", + "p=2.0\n", + "f=50.0#Hz\n", + "i2=125.0#A\n", + "v1=8000.0#V\n", + "i3=800.0#A\n", + "d=0.03\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "sinphi=math.sin(math.acos(pf))\n", + "zs=v1/(3**0.5*i3)\n", + "vp=v/3**0.5\n", + "rd=d*vp\n", + "il=load*1000/(3**0.5*v)\n", + "ira=rd\n", + "ra=ira/il\n", + "xs=(zs**2-ra**2)**0.5\n", + "e0=((vp*pf+ira)**2+(vp*sinphi+il*xs)**2)**0.5\n", + "reg=(e0-vp)/vp\n", + "\n", + "#result\n", + "print \"percentage regulation=\",reg*100,\"%\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "percentage regulation= 62.2972136768 %\n" + ] + } + ], + "prompt_number": 133 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.25, Page Number:1435" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "f=50.0#Hz\n", + "load=2000#KVA\n", + "v=2300#V\n", + "i=600#A\n", + "v2=900#V\n", + "r=0.12#ohm\n", + "pf=0.8\n", + "\n", + "#calculation\n", + "sinphi=math.sin(math.acos(pf))\n", + "zs=v2/(3**0.5*i)\n", + "rp=r/2\n", + "re=rp*1.5\n", + "xs=(zs**2-re**2)**0.5\n", + "il=load*1000/(3**0.5*v)\n", + "ira=il*rp\n", + "ixs=il*xs\n", + "vp=v/3**0.5\n", + "e0=((vp+ira)**2+(ixs)**2)**0.5\n", + "reg1=(e0-vp)/vp\n", + "e0=((vp*pf+ira)**2+(vp*sinphi+ixs)**2)**0.5\n", + "reg2=(e0-vp)/vp\n", + "#result\n", + "print \"regulation at pf=1\",reg1*100,\"%\"\n", + "print \"regulation at pf=0.8\",reg2*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "regulation at pf=1 7.32796146323 %\n", + "regulation at pf=0.8 23.8398862235 %\n" + ] + } + ], + "prompt_number": 134 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.26, Page Number:1436" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "v=Symbol('v')\n", + "load=2000#KVA\n", + "load1=11#KV\n", + "r=0.3#ohm\n", + "x=5#ohm\n", + "pf=0.8\n", + "\n", + "#calculation\n", + "sinphi=math.sin(math.acos(pf))\n", + "i=load*1000/(3**0.5*load1*1000)\n", + "vt=load1*1000/3**0.5\n", + "ira=i*r\n", + "ixs=i*x\n", + "e0=((vt*pf+ira)**2+(vt*sinphi+ixs)**2)**0.5\n", + "v=solve(((pf*v+ira)**2+(sinphi*v-ixs)**2)**0.5-e0,v)\n", + "\n", + "#result\n", + "print \"terminal voltage=\",v[1],\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "terminal voltage= 6978.31767618569 V\n" + ] + } + ], + "prompt_number": 150 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.27, Page Number:1436" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=1200#KVA\n", + "load1=3.3#KV\n", + "f=50#Hz\n", + "r=0.25#ohm\n", + "i=35#A\n", + "i2=200#A\n", + "v=1.1#kV\n", + "pf=0.8\n", + "\n", + "#calculation\n", + "zs=v*1000/(3**0.5*i2)\n", + "xs=(zs**2-r**2)**0.5\n", + "v=load1*1000/3**0.5\n", + "theta=math.atan(xs/r)\n", + "ia=load*1000/(3**0.5*load1*1000)\n", + "e=v+ia*zs\n", + "change=(e-v)/v\n", + "\n", + "#result\n", + "print \"per unit change=\",change" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "per unit change= 0.349909254054\n" + ] + } + ], + "prompt_number": 151 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.28, Page Number:1437" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "f=50#Hz\n", + "v1=11#kV\n", + "load=3#MVA\n", + "i=100#A\n", + "v2=12370#V\n", + "vt=11000#V\n", + "pf=0.8\n", + "r=0.4#ohm\n", + "\n", + "#calculation\n", + "E0=v1*1000/3**0.5\n", + "v=v2/3**0.5\n", + "pf=0\n", + "sinphi=1\n", + "xs=(v-(E0**2-(i*r)**2)**0.5)/i\n", + "il=load*10**6/(3**0.5*v1*1000)\n", + "ira=il*r\n", + "ixs=il*xs\n", + "e0=((E0*pf+ira)**2+(E0*sinphi+ixs)**2)**0.5\n", + "regn=(e0-E0)*100/E0\n", + "#result\n", + "print \"regulation=\",regn,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "regulation= 19.6180576177 %\n" + ] + } + ], + "prompt_number": 175 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.29, Page Number:1437" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "pf=0.8\n", + "vt=3500#v\n", + "load=2280#KW\n", + "v1=3300#V\n", + "r=8#ohm\n", + "x=6#ohm\n", + "\n", + "#calculation\n", + "vl=vt/3**0.5\n", + "vp=v1/3**0.5\n", + "il=load*1000/(3**0.5*v1*pf)\n", + "drop=vl-vp\n", + "z=(r**2+x**2)**0.5\n", + "x=vl/(z+drop/il)\n", + "vtp=vl-x*drop/il\n", + "vtpl=vtp*3**0.5\n", + "\n", + "#result\n", + "print \"terminal voltage=\",vtpl,\"V\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "terminal voltage= 3420.781893 V\n" + ] + } + ], + "prompt_number": 176 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.30, Page Number:1441" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=3.5#MVA\n", + "v=4160#V\n", + "f=50#Hz\n", + "i=200#A\n", + "pf=0.8\n", + "\n", + "#calculation\n", + "il=load*10**6/(3**0.5*v)\n", + "zs=4750/(3**0.5*il)\n", + "ra=0\n", + "ixs=il*zs\n", + "vp=v/3**0.5\n", + "sinphi=math.sin(math.acos(pf))\n", + "e0=((vp*pf)**2+(vp*sinphi+ixs)**2)**0.5\n", + "regn=(e0-vp)/vp\n", + "#result\n", + "print \"regulation=\",regn,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "regulation= 0.91675794767 %\n" + ] + } + ], + "prompt_number": 184 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.39, Page Number:1455" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "xd=0.7\n", + "xq=0.4\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "v=1\n", + "sinphi=math.sin(math.acos(pf))\n", + "ia=1\n", + "tandelta=ia*xq*pf/(v+xq*sinphi)\n", + "delta=math.atan(tandelta)\n", + "i_d=ia*math.sin(math.radians(36.9)+delta)\n", + "e0=v*math.cos(delta)+i_d*xd\n", + "\n", + "#result\n", + "print \"load angle=\",math.degrees(delta),\"degrees\"\n", + "print \"no load voltage=\",e0,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "load angle= 14.4702941001 degrees\n", + "no load voltage= 1.51511515874 V\n" + ] + } + ], + "prompt_number": 185 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.40, Page Number:1455" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "f=50.0#Hz\n", + "xd=0.6\n", + "xq=0.45\n", + "ra=0.015\n", + "pf=0.8\n", + "ia=1\n", + "v=1\n", + "sinphi=math.sin(math.acos(pf))\n", + "#calculation\n", + "tanpsi=(v*sinphi+ia*xq)/(v*pf+ia*ra)\n", + "psi=math.atan(tanpsi)\n", + "delta=psi-math.acos(pf)\n", + "i_d=ia*math.sin(psi)\n", + "iq=ia*math.cos(psi)\n", + "e0=v*math.cos(delta)+iq*ra+i_d*xd\n", + "regn=(e0-v)*100/v\n", + "\n", + "#result\n", + "print \"open circuit voltage=\",e0,\"V\"\n", + "print \"regulation=\",regn,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "open circuit voltage= 1.44767600311 V\n", + "regulation= 44.7676003107 %\n" + ] + } + ], + "prompt_number": 187 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.41, Page Number:1455" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "ia=10#A\n", + "phi=math.radians(20)\n", + "v=400#V\n", + "xd=10#ohm\n", + "xq=6.5#ohm\n", + "\n", + "#calculations\n", + "pf=math.cos(phi)\n", + "sinphi=math.sin(phi)\n", + "tandelta=ia*xq*pf/(v+ia*xq*sinphi)\n", + "delta=math.atan(tandelta)\n", + "i_d=ia*math.sin(phi+delta)\n", + "iq=ia*math.cos(phi+delta)\n", + "e0=v*math.cos(delta)+i_d*xd\n", + "regn=(e0-v)/v\n", + "\n", + "#result\n", + "print \"load angle=\",math.degrees(delta),\"degrees\"\n", + "print \"id=\",i_d,\"A\"\n", + "print \"iq=\",iq,\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "load angle= 8.23131209115 degrees\n", + "id= 4.7303232581 A\n", + "iq= 8.81045071911 A\n" + ] + } + ], + "prompt_number": 189 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.42, Page Number:1459" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "e1=220#V\n", + "f1=60#Hz\n", + "e2=222#V\n", + "f2=59#Hz\n", + "\n", + "#calculation\n", + "emax=(e1+e2)/2\n", + "emin=(e2-e1)/2\n", + "f=(f1-f2)\n", + "epeak=emax/0.707\n", + "pulse=(f1-f2)*60\n", + "\n", + "#result\n", + "print \"max voltage=\",emax,\"V\"\n", + "print \"min voltage=\",emin,\"V\"\n", + "print \"frequency=\",f,\"Hz\"\n", + "print \"peak value of voltage=\",epeak,\"V\"\n", + "print \"number of maximum light pulsations/minute=\",pulse" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "max voltage= 221 V\n", + "min voltage= 1 V\n", + "frequency= 1 Hz\n", + "peak value of voltage= 312.588401697 V\n", + "number of maximum light pulsations/minute= 60\n" + ] + } + ], + "prompt_number": 190 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.43, Page Number:1462" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "power=1500#kVA\n", + "v=6.6#kV\n", + "r=0.4#ohm\n", + "x=6#ohm\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "i=power*1000/(3**0.5*v*1000)\n", + "ira=i*r\n", + "ixs=i*x\n", + "vp=v*1000/3**0.5\n", + "phi=math.acos(pf)\n", + "tanphialpha=(vp*math.sin(phi)+ixs)/(vp*pf+ira)\n", + "phialpha=math.atan(tanphialpha)\n", + "alpha=phialpha-phi\n", + "\n", + "#result\n", + "print \"power angle=\",math.degrees(alpha)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "power angle= 7.87684146241\n" + ] + } + ], + "prompt_number": 198 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.44, Page Number:1464" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=3000#KVA\n", + "p=6\n", + "n=1000#rpm\n", + "v=3300#v\n", + "x=0.25\n", + "\n", + "#calculation\n", + "vp=v/3**0.5\n", + "i=load*1000/(3**0.5*v)\n", + "ixs=x*vp\n", + "xs=x*vp/i\n", + "alpha=1*p/2\n", + "psy=3*3.14*vp**2/(60*xs*n)\n", + "tsy=9.55*psy/n\n", + "\n", + "#result\n", + "print \"synchronizing power=\",psy,\"kW\"\n", + "print \"torque=\",tsy*1000,\"N-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "synchronizing power= 628.0 kW\n", + "torque= 5997.4 N-m\n" + ] + } + ], + "prompt_number": 202 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.45, Page Number:1465" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=3#MVA\n", + "n=1000#rpm\n", + "v1=3.3#kV\n", + "r=0.25\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "vp=v1*1000/3**0.5\n", + "i=load*1000000/(3**0.5*v1*1000)\n", + "ixs=complex(0,r*vp)\n", + "xs=ixs/i\n", + "v=vp*complex(pf,math.sin(math.acos(pf)))\n", + "e0=v+ixs\n", + "alpha=math.atan(e0.imag/e0.real)-math.acos(pf)\n", + "p=6/2\n", + "psy=abs(e0)*vp*math.cos(alpha)*math.sin(math.radians(3))/xs\n", + "tsy=9.55*3*psy*100/n\n", + "\n", + "#result\n", + "print \"synchronous power=\",-psy*3/1000,\"kW\"\n", + "print \"toque=\",-tsy/100,\"N-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "synchronous power= 722.236196153j kW\n", + "toque= 6897.35567326j N-m\n" + ] + } + ], + "prompt_number": 221 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.46, Page Number:1465" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=750#KVA\n", + "v=11#kV\n", + "p=4\n", + "r=1#%\n", + "x=15#%\n", + "pf=0.8\n", + "#calculation\n", + "i=load*1000/(3**0.5*v*1000)\n", + "vph=v*1000/3**0.5\n", + "ira=r*vph/1000\n", + "ra=ira/i\n", + "xs=x*vph/(100*i)\n", + "zs=(ra**2+xs**2)**0.5\n", + "#no load\n", + "alpha=p/2\n", + "psy=math.radians(alpha)*vph**2/xs\n", + "#fl 0.8 pf\n", + "e=((vph*pf+i*ra)**2+(vph*math.sin(math.acos(pf)+i*xs))**2)**0.5\n", + "psy2=math.radians(alpha)*e*vph/xs\n", + "\n", + "#result\n", + "print \"Synchronous power at:\"\n", + "print \"no load=\",psy,\"W\"\n", + "print \"at pf of 0.8=\",psy2,\"w\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Synchronous power at:\n", + "no load= 58177.6417331 W\n", + "at pf of 0.8= 73621.2350169 w\n" + ] + } + ], + "prompt_number": 225 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.47, Page Number:1466" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=2000#KVA\n", + "p=8\n", + "n=750#rpm\n", + "v1=6000#V\n", + "pf=0.8\n", + "r=6#ohm\n", + "\n", + "#calculations\n", + "alpha=math.radians(4)\n", + "v=v1/3**0.5\n", + "i=load*1000/(3**0.5*v1)\n", + "e0=((v*pf)**2+(v*math.sin(math.acos(pf))+i*r)**2)**0.5\n", + "psy=alpha*e0*v*3/r\n", + "tsy=9.55*psy/n\n", + "\n", + "#result\n", + "print \"synchronous power=\",psy,\"W\"\n", + "print \"synchronous torque=\",tsy,\"N-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "synchronous power= 514916.500204 W\n", + "synchronous torque= 6556.60343593 N-m\n" + ] + } + ], + "prompt_number": 226 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.48, Page Number:1467" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=5000#KVA\n", + "v=10000#V\n", + "n=1500#rpm\n", + "f=50#Hz\n", + "r=20#%\n", + "pf=0.8\n", + "phi=0.5\n", + "\n", + "#calculations\n", + "vp=v/3**0.5\n", + "i=load*1000/(3**0.5*v)\n", + "xs=r*vp/(1000*i)\n", + "p=120*f/n\n", + "alpha=math.radians(2)\n", + "#no load\n", + "psy=3*alpha*vp**2/(p*1000)\n", + "tsy=9.55*psy*1000/(n*2)\n", + "#pf=0.8\n", + "v2=vp*complex(pf,math.sin(math.acos(pf)))\n", + "ixs=complex(0,i*4)\n", + "e0=v+ixs\n", + "psy2=abs(e0)*vp*math.cos(math.radians(8.1))*math.sin(math.radians(2))*3/4\n", + "tsy2=9.55*psy2/(n*20)\n", + "\n", + "#result\n", + "print \"synchronous power:\"\n", + "print \"atno load=\",psy,\"w\"\n", + "print \"at 0.8 pf=\",psy2,\"w\"\n", + "print \"torque:\"\n", + "print \"at no load=\",tsy,\"N-m\"\n", + "print \"at pf=0.8=\",tsy2,\"N-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "synchronous power:\n", + "atno load= 872.664625997 w\n", + "at 0.8 pf= 1506057.44405 w\n", + "torque:\n", + "at no load= 2777.98239276 N-m\n", + "at pf=0.8= 479.428286357 N-m\n" + ] + } + ], + "prompt_number": 229 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.49, Page Number:1468" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=6.6#kW\n", + "load1=3000#kW\n", + "pf=0.8\n", + "xa=complex(0.5,10)\n", + "xb=complex(0.4,12)\n", + "i0=150#A\n", + "\n", + "#calculation\n", + "v=complex(load*1000/3**0.5,0)\n", + "cosphi1=1500*1000/(load*1000*i0*3**0.5)\n", + "phi1=math.acos(cosphi1)\n", + "sinphi1=math.sin(phi1)\n", + "i=328*complex(pf,-math.sin(math.acos(pf)))\n", + "i1=i0*complex(cosphi1,-sinphi1)\n", + "i2=i-i1\n", + "coshi2=i2.real/181\n", + "ea=v+i1*xa\n", + "eal=3**0.5*abs(ea)\n", + "eb=v+i2*xb\n", + "ebl=3**0.5*abs(eb)\n", + "alpha1=(ea.imag/ea.real)\n", + "alpha2=(eb.imag/eb.real)\n", + "#result\n", + "print \"Ea=\",ea,\"V\"\n", + "print \"Eb=\",eb,\"V\"\n", + "print \"alpha1=\",math.degrees(alpha1),\"degrees\"\n", + "print \"alpha2=\",math.degrees(alpha2),\"degrees\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ea= (4602.91884998+1275.81974829j) V\n", + "Eb= (5352.42648271+1524.56032028j) V\n", + "alpha1= 15.8810288383 degrees\n", + "alpha2= 16.3198639435 degrees\n" + ] + } + ], + "prompt_number": 245 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.50, Page Number:1468" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declration\n", + "e1=complex(230,0)\n", + "e2=230*complex(0.985,0.174)\n", + "z1=complex(0,2)\n", + "z2=complex(0,3)\n", + "z=6\n", + "i1=((e1-e2)*z+e1*z2)/(z*(z1+z2)+z1*z2)\n", + "i2=((e2-e1)*z+e2*z1)/(z*(z1+z2)+z1*z2)\n", + "i=i1+i2\n", + "v=i*z\n", + "p1=abs(v)*abs(i1)*math.cos(math.atan(i1.imag/i1.real))\n", + "p2=abs(v)*abs(i2)*math.cos(math.atan(i2.imag/i2.real))\n", + "\n", + "#result\n", + "print \"terminal voltage=\",v,\"V\"\n", + "print \"current\",i,\"A\"\n", + "print \"power 1=\",p1,\"W\"\n", + "print \"power 2=\",p2,\"W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "terminal voltage= (222.905384615-28.5730769231j) V\n", + "current (37.1508974359-4.76217948718j) A\n", + "power 1= 3210.60292765 W\n", + "power 2= 5138.29001053 W\n" + ] + } + ], + "prompt_number": 249 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.51, Page Number:1471" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=1500#kW\n", + "v=11#KV\n", + "pf=0.867\n", + "x=50#ohm\n", + "r=4#ohm\n", + "i=50#A\n", + "\n", + "#calculations\n", + "il=load*1000/(3**0.5*v*1000*pf)\n", + "phi=math.acos(pf)\n", + "sinphi=math.sin(phi)\n", + "iwatt=il*pf\n", + "iwattless=il*sinphi\n", + "i1=il/2\n", + "i2=iwatt/2\n", + "iw1=(i**2-i1**2)**0.5\n", + "iw2=i2-iw1\n", + "ia=(i2**2+iw2**2)**0.5\n", + "vt=v*1000/3**0.5\n", + "ir=i*r\n", + "ix=x*i\n", + "cosphi=i2/i\n", + "sinphi=math.sin(math.acos(cosphi))\n", + "e=((vt*cosphi+ir)**2+(vt*sinphi+ix)**2)**0.5\n", + "el=3**0.5*e\n", + "\n", + "#result\n", + "print \"armature current=\",ia,\"A\"\n", + "print \"line voltage=\",el,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "armature current= 43.4628778514 A\n", + "line voltage= 14304.0798593 V\n" + ] + } + ], + "prompt_number": 251 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.52, Page Number:1472" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=10#MW\n", + "pf=0.8\n", + "output=6000#kW\n", + "pfa=0.92\n", + "\n", + "#calculations\n", + "phi=math.acos(pf)\n", + "phia=math.acos(pfa)\n", + "tanphi=math.tan(phi)\n", + "tanphia=math.tan(phia)\n", + "loadkvar=load*1000*tanphi\n", + "akvar=output*tanphia\n", + "kwb=(load*1000-output)\n", + "kvarb=loadkvar-akvar\n", + "kvab=complex(kwb,kvarb)\n", + "pfb=math.cos(math.atan(kvab.imag/kvab.real))\n", + "kvarb=kwb*pfb\n", + "kvara=-loadkvar-kvarb\n", + "kvaa=complex(output,kvara)\n", + "pfa=math.cos(math.atan(kvaa.imag/kvaa.real))\n", + "\n", + "#result\n", + "print \"new pfb=\",pfb\n", + "print \"new pfa=\",pfa" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "new pfb= 0.628980253433\n", + "new pfa= 0.513894032194\n" + ] + } + ], + "prompt_number": 253 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.54, Page Number:1473" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=6600#V\n", + "load=1000#KVA\n", + "x=20#%\n", + "pf=0.8\n", + "\n", + "#calculation\n", + "i=87.5\n", + "x=8.7\n", + "vp=3810\n", + "e0=4311\n", + "ir=70\n", + "ix=52.5\n", + "IX=762\n", + "vb1=(e0**2-vp**2)**0.5\n", + "i1x=vb1\n", + "i1=i1x/x\n", + "output=3**0.5*v*i1/1000\n", + "b2v=(vp**2+e0**2)**0.5\n", + "i2z=b2v\n", + "i2=b2v/x\n", + "i2rx=e0\n", + "i2r=i2rx/x\n", + "i2x=vp/x\n", + "tanphi2=i2x/i2r\n", + "phi2=math.atan(tanphi2)\n", + "cosphi2=math.cos(phi2)\n", + "output1=3**0.5*v*i2*cosphi2/1000\n", + "\n", + "#result\n", + "print \"power output at unity pf=\",output,\"kW\"\n", + "print \"max power output=\",output1,\"kW\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " power output at unity pf= 2650.38477722 kW\n", + "max power output= 5664.52285143 kW\n" + ] + } + ], + "prompt_number": 255 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.55, Page Number:1474" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "x=10.0#ohm\n", + "i=220.0#A\n", + "load=11.0#kV\n", + "per=25.0#%\n", + "\n", + "#calculations\n", + "oa1=load*1000/3**0.5\n", + "a1c1=i*x\n", + "e0=(oa1**2+a1c1**2)**0.5\n", + "emf=(1+per/100)*e0\n", + "a1a2=(emf**2-a1c1**2)**0.5-oa1\n", + "ix=a1a2/x\n", + "i1=(i**2+ix**2)**0.5\n", + "pf=i/i1\n", + "bv=(oa1**2+emf**2)**0.5\n", + "imax=bv/x\n", + "ir=emf/x\n", + "ix=oa1/x\n", + "pfmax=ir/imax\n", + "output=3**0.5*load*1000*imax*pfmax*0.001\n", + "#result\n", + "print \"new current=\",i1,\"A\"\n", + "print \"new power factor=\",pf\n", + "print \"max power output=\",output,\"kW\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "new current= 281.573453399 A\n", + "new power factor= 0.781323655849\n", + "max power output= 16006.7954319 kW\n" + ] + } + ], + "prompt_number": 258 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.56, Page Number:1475" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=20.0#MVA\n", + "load1=35.0#MVA\n", + "pf=0.8\n", + "output=25.0#MVA\n", + "cosphi1=0.9\n", + "\n", + "#calculations\n", + "loadmw=load1*pf\n", + "loadmvar=load1*0.6\n", + "sinphi=math.sin(math.acos(cosphi))\n", + "mva1=25\n", + "mw1=mva1*cosphi1\n", + "mvar1=25*sinphi1\n", + "mw2=loadmw-mw1\n", + "mvar2=loadmvar-mvar1\n", + "mva2=(mw2**2+mvar2**2)**0.5\n", + "cosphi2=mw2/mva2\n", + "\n", + "#result\n", + "print \"output=\",mva2\n", + "print \"pf=\",cosphi2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "output= 10.4509862952\n", + "pf= 0.52626611926\n" + ] + } + ], + "prompt_number": 260 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.57, Page Number:1475" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declarations\n", + "load=600#KW\n", + "loadm=707#kW\n", + "pf=0.707\n", + "output=900#kW\n", + "pf1=0.9\n", + "\n", + "#calculation\n", + "kva=1000\n", + "kvar=kva*(1-pf1**2)**0.5\n", + "active_p=1307-output\n", + "reactive_p=loadm-kvar\n", + "\n", + "#result\n", + "print \"active power shared by second machine=\",active_p,\"kW\"\n", + "print \"reactive power shared by second machine=\",reactive_p,\"kVAR\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "active power shared by second machine= 407 kW\n", + "reactive power shared by second machine= 271.110105646 kVAR\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.58, Page Number:1476" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "l1=500#kW\n", + "l2=1000#kW\n", + "pf1=0.9\n", + "l3=800#kW\n", + "pf2=0.8\n", + "l4=500#kW\n", + "pf3=0.9\n", + "output=1500#kW\n", + "pf=0.95\n", + "\n", + "#calculation\n", + "kw1=l1\n", + "kw2=l2\n", + "kw3=l3\n", + "kw4=500\n", + "kvar2=kw2*0.436/pf1\n", + "kvar3=kw3*0.6/pf2\n", + "kvar4=kw4*0.436/pf3\n", + "kvar=output/pf\n", + "kw=kw1+kw2+kw3+kw4-output\n", + "kvar=kvar2+kvar3+kvar4-kvar\n", + "cosphi=math.cos(math.atan(kvar/kw))\n", + "\n", + "#result\n", + "print \"kW output=\",kw\n", + "print \"pf=\",cosphi" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "kW output= 1300\n", + "pf= 0.981685651341\n" + ] + } + ], + "prompt_number": 264 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.59, Page Number:1476" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "z=complex(0.2,2)\n", + "ze=complex(3,4)\n", + "emf1=complex(2000,0)\n", + "emf2=complex(22000,100)\n", + "\n", + "#calculations\n", + "i1=complex(68.2,-102.5)\n", + "i2=complex(127,-196.4)\n", + "i=i1+i2\n", + "v=i*ze\n", + "pva1=v*i1\n", + "kw1=pva1.real*3\n", + "a11=math.atan(-i1.imag/i1.real)\n", + "a12=math.atan(-v.imag/v.real)\n", + "pf1=math.cos(a11-a12)\n", + "pva2=v*i2\n", + "kw2=pva2.real*3\n", + "a21=math.atan(-i2.imag/i2.real)\n", + "a22=math.atan(-v.imag/v.real)\n", + "pf2=math.cos(a21-a22)\n", + "\n", + "#result\n", + "print \"kw output 1=\",kw1/1000\n", + "print \"pf 1=\",pf1\n", + "print \"kw output 2=\",kw2/1000\n", + "print \"pf 2=\",pf2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "kw output 1= 328.79427\n", + "pf 1= 0.606839673468\n", + "kw output 2= 610.34892\n", + "pf 2= 0.596381892841\n" + ] + } + ], + "prompt_number": 273 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.63, Page Number:1481" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=5000#KVA\n", + "v=10000#V\n", + "f=50#Hz\n", + "ns=1500#rpm\n", + "j=1.5*10**4#khm2\n", + "ratio=5\n", + "\n", + "#calculation\n", + "t=0.0083*ns*(j/(load*ratio*f))**0.5\n", + "\n", + "#result\n", + "print \"natural time period of oscillation=\",round(t,3),\"s\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "natural time period of oscillation= 1.364 s\n" + ] + } + ], + "prompt_number": 275 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.64, Page Number:1481" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=10000#KVA\n", + "p=4\n", + "v=6600#V\n", + "f=50#Hz\n", + "xs=25#%\n", + "pf=1.5\n", + "\n", + "#calculations\n", + "ratio=100/xs\n", + "ns=120*f/p\n", + "j=(pf/(0.0083*ns))**2*load*ratio*f\n", + "\n", + "#result\n", + "print \"moment of inertia=\",j/1000,\"x10^4 kg-m2\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "moment of inertia= 29.0317898098 x10^4 kg-m2\n" + ] + } + ], + "prompt_number": 277 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.65, Page Number:1481" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=10.0#MVA\n", + "v=10.0#kV\n", + "f=50.0#Hz\n", + "ns=1500.0#rpm\n", + "j=2.0*10**5#kgm2\n", + "x=40.0\n", + "\n", + "#calculation\n", + "ratio=100.0/x\n", + "t=0.0083*ns*(j/(load*1000*ratio*f))**0.5\n", + "\n", + "#result\n", + "print \"frequency of oscillation of the rotor=\",round(1/t,1),\"Hz\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "frequency of oscillation of the rotor= 0.2 Hz\n" + ] + } + ], + "prompt_number": 283 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.66, Page Number:1483" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "v=11#kV\n", + "z=complex(1,10)\n", + "emf=14#kV\n", + "\n", + "#calculations\n", + "e=emf*1000/3**0.5\n", + "v=v*1000/3**0.5\n", + "costheta=z.real/abs(z)\n", + "pmax=e*v*3/(z.imag*1000)\n", + "pmax_per_phase=(v/abs(z))*(e-(v/abs(z)))*3\n", + "\n", + "#result\n", + "print \"max output =\",pmax_per_phase/1000,\"kW\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "max output = 14125.5529273 kW\n" + ] + } + ], + "prompt_number": 285 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 37.67, Page Number:1484" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "load=11#kVA\n", + "load1=10#MW\n", + "z=complex(0.8,8.0)\n", + "v=14#kV\n", + "\n", + "#calculations\n", + "pmax=(load*1000/3**0.5)*(v*1000/3**0.5)*3/z.imag\n", + "imax=((v*1000/3**0.5)**2+(load*1000/3**0.5)**2)**0.5/z.imag\n", + "pf=(v/3**0.5)*1000/((v*1000/3**0.5)**2+(load*1000/3**0.5)**2)**0.5\n", + "\n", + "#result\n", + "print \"maximum output=\",pmax/1000000,\"MW\"\n", + "print \"current=\",imax,\"A\"\n", + "print \"pf=\",pf" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "maximum output= 19.25 MW\n", + "current= 1284.92866209 A\n", + "pf= 0.786318338822\n" + ] + } + ], + "prompt_number": 289 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter38_2.ipynb b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter38_2.ipynb new file mode 100644 index 00000000..eb91f537 --- /dev/null +++ b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter38_2.ipynb @@ -0,0 +1,1682 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a6bbecd88376ba06b11df7bbad39447a579ab954844d7c4715263117b7255967" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 38: Synchronous Motor" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.1, Page Number:1495" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "p=75#kW\n", + "f=50#Hz\n", + "v=440#V\n", + "pf=0.8\n", + "loss=0.95\n", + "xs=2.5#ohm\n", + "\n", + "#calculations\n", + "ns=120*f/4\n", + "pm=p*1000/loss\n", + "ia=pm/(math.sqrt(3)*v*pf)\n", + "vol_phase=v/math.sqrt(3)\n", + "\n", + "#calculations\n", + "print \"mechanical power=\",pm,\"W\"\n", + "print \"armature current=\",ia,\"A\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "mechanical power= 78947.3684211 W\n", + "armature current= 129.489444346 A\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.2, Page Number:1498" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import cmath\n", + "#variable declaration\n", + "p=20\n", + "vl=693#V\n", + "r=10#ohm\n", + "lag=0.5#degrees\n", + "\n", + "#calculations\n", + "#lag=0.5\n", + "alpha=p*lag/2\n", + "eb=vp=vl/math.sqrt(3)\n", + "er=complex(vp-eb*math.cos(math.radians(alpha)),eb*math.sin(math.radians(alpha)))\n", + "zs=complex(0,10)\n", + "ia=er/zs\n", + "power_input=3*vp*abs(ia)*math.cos(math.radians(cmath.phase(ia)))\n", + "print \"displacement:0.5%\"\n", + "print \"alpha=\",alpha,\"degrees\"\n", + "print \"armature emf/phase=\",eb,\"V\"\n", + "print \"armature current/phase=\",ia,\"A\"\n", + "print \"power drawn=\",power_input,\"W\"\n", + "print \"\"\n", + "\n", + "#lag=5\n", + "lag=5\n", + "alpha=p*lag/2\n", + "eb=vp=vl/math.sqrt(3)\n", + "er=complex(vp-eb*math.cos(math.radians(alpha)),eb*math.sin(math.radians(alpha)))\n", + "zs=complex(0,10)\n", + "ia=er/zs\n", + "power_input=3*vp*abs(ia)*math.cos(math.radians(cmath.phase(ia)))\n", + "\n", + "print \"displacement:5%\"\n", + "print \"alpha=\",alpha,\"degrees\"\n", + "print \"armature emf/phase=\",eb,\"V\"\n", + "print \"armature current/phase=\",ia,\"A\"\n", + "print \"power drawn=\",power_input,\"W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "displacement:0.5%\n", + "alpha= 5.0 degrees\n", + "armature emf/phase= 400.103736548 V\n", + "armature current/phase= (3.4871338335-0.152251551219j) A\n", + "power drawn= 4189.63221768 W\n", + "\n", + "displacement:5%\n", + "alpha= 50 degrees\n", + "armature emf/phase= 400.103736548 V\n", + "armature current/phase= (30.6497244054-14.2922012106j) A\n", + "power drawn= 40591.222447 W\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.3, Page Number:1499" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=400.0#V/ph\n", + "i=32.0#A/ph\n", + "xs=10.0#ohm\n", + "\n", + "#calculations\n", + "e=math.sqrt(v**2+(i*xs)**2)\n", + "delta=math.atan((i*xs)/v)\n", + "power=3*v*i\n", + "power_other=3*(v*e/10)*math.sin(delta)*0.001\n", + "\n", + "#result\n", + "print \"E=\",e,\"V\"\n", + "print \"delta=\",math.degrees(delta),\"degrees\"\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "E= 512.249938995 V\n", + "delta= 38.6598082541 degrees\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.4, Page Number:1506" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "w=150#kW\n", + "f=50#Hz\n", + "v=2300#V\n", + "n=1000#rpm\n", + "xd=32#ohm\n", + "xq=20#ohm\n", + "alpha=16#degrees\n", + "\n", + "#calculations\n", + "vp=v/math.sqrt(3)\n", + "eb=2*vp\n", + "ex_power=eb*vp*math.sin(math.radians(alpha))/xd\n", + "rel_power=(vp**2*(xd-xq)*math.sin(math.radians(2*alpha)))/(2*xd*xq)\n", + "pm=3*(ex_power+rel_power)\n", + "tg=9.55*pm/1000\n", + "\n", + "#result\n", + "print \"torque=\",tg,\"N-m\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "torque= 1121.29686485 N-m\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.6, Page Number:1506" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=11000#V\n", + "ia=60#A\n", + "r=1#ohm\n", + "x=30#ohm\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "p2=math.sqrt(3)*v*ia*pf\n", + "cu_loss=ia**2*3\n", + "pm=p2-cu_loss\n", + "vp=v/math.sqrt(3)\n", + "phi=math.acos(pf)\n", + "theta=math.atan(x/r)\n", + "zs=x\n", + "z_drop=ia*zs\n", + "eb=math.sqrt((vp**2+z_drop**2-(2*vp*z_drop*math.cos(theta+phi))))*math.sqrt(3)\n", + "\n", + "#result\n", + "print \"power supplied=\",p2/1000,\"kW\"\n", + "print \"mechanical power=\",pm/1000,\"KW\"\n", + "print \"induced emf=\",eb,\"V\"\n", + "\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "power supplied= 914.522826396 kW\n", + "mechanical power= 903.722826396 KW\n", + "induced emf= 13039.2734763 V\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.7, Page Number:1507" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=400#V\n", + "i=32#A\n", + "pf=1\n", + "xd=10#ohm\n", + "xq=6.5#ohm\n", + "\n", + "#calculations\n", + "e=math.sqrt(v**2+(i*xq)**2)+((xd-xq)*14.8)\n", + "delta=math.atan((i*xq)/v)\n", + "power=3*v*i\n", + "power_other=3*(v*e/10)*math.sin(delta)*0.001\n", + "\n", + "#result\n", + "print \"E=\",e,\"V\"\n", + "print \"delta=\",math.degrees(delta),\"degrees\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "E= 502.648089715 V\n", + "delta= 27.4744316263 degrees\n" + ] + } + ], + "prompt_number": 60 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.8, Page Number:1508" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=500#V\n", + "output=7.46#kW\n", + "pf=0.9\n", + "r=0.8#ohm\n", + "loss=500#W\n", + "ex_loss=800#W\n", + "\n", + "#calculations\n", + "pm=output*1000+loss+ex_loss\n", + "ia=(v*pf-math.sqrt(v**2*pf**2-4*r*pm))/(2*r)\n", + "m_input=loss*ia*pf\n", + "efficiency=output*1000/m_input\n", + "\n", + "#result\n", + "print \"commercial efficiency=\",efficiency*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "commercial efficiency= 82.1029269497 %\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.9, Page Number:1509" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=2300#V\n", + "r=0.2#ohm\n", + "x=2.2#ohm\n", + "pf=0.5\n", + "il=200#A\n", + "\n", + "#calculations\n", + "phi=math.acos(pf)\n", + "theta=math.atan(x//r)\n", + "v=v/math.sqrt(3)\n", + "zs=math.sqrt(r**2+x**2)\n", + "eb=math.sqrt(v**2+(il*zs)**2-(2*v*il*zs*math.cos(phi+theta)))\n", + "\n", + "#result\n", + "print \"Eb=\",eb,\"volt/phase\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Eb= 1708.04482042 volt/phase\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.10, Page Number:1509" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "vl=6600#V\n", + "f=50#Hz\n", + "il=50#A\n", + "r=1#ohm\n", + "x=20#ohm\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "#0.8 lagging\n", + "power_i=math.sqrt(3)*v*f*pf\n", + "v=vl/math.sqrt(3)\n", + "phi=math.acos(pf)\n", + "theta=math.atan(x/r)\n", + "zs=math.sqrt(x**2+r**2)\n", + "eb=math.sqrt(v**2+(il*zs)**2-(2*v*il*zs*math.cos(phi-theta)))*math.sqrt(3)\n", + "\n", + "print \"0.8 lag: Eb=\",eb\n", + "\n", + "#0.8 leading\n", + "power_i=math.sqrt(3)*v*f*pf\n", + "v=vl/math.sqrt(3)\n", + "phi=math.acos(pf)\n", + "theta=math.atan(x/r)\n", + "zs=math.sqrt(x**2+r**2)\n", + "eb=math.sqrt(v**2+(il*zs)**2-(2*v*il*zs*math.cos(phi+theta)))*math.sqrt(3)\n", + "\n", + "print \"0.8 leading:Eb=\",eb" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "0.8 lag: Eb= 5651.1180113\n", + "0.8 leading:Eb= 7705.24623679\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.11, Page Number:1510" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "x=0.4\n", + "pf=0.8\n", + "v=100#V\n", + "phi=math.acos(pf)\n", + "#calculations\n", + "#pf=1\n", + "eb=math.sqrt(v**2+(x*v)**2)\n", + "#pf=0.8 lag\n", + "eb2=math.sqrt(v**2+(x*v)**2-(2*v*x*v*math.cos(math.radians(90)-phi)))\n", + "#pf=0.8 lead\n", + "eb3=math.sqrt(v**2+(x*v)**2-(2*v*x*v*math.cos(math.radians(90)+phi)))\n", + "#result\n", + "print \"pf=1: Eb=\",eb,\"V\"\n", + "print \"pf=0.8 lag:Eb=\",eb2,\"V\"\n", + "print \"pf=0.8 lead:Eb=\",eb3,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "pf=1: Eb= 107.703296143 V\n", + "pf=0.8 lag:Eb= 82.4621125124 V\n", + "pf=0.8 lead:Eb= 128.062484749 V\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.12, Page Number:1510" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaraion\n", + "load=1000#kVA\n", + "v=11000#V\n", + "r=3.5#ohm\n", + "x=40#ohm\n", + "pf=0.8\n", + "\n", + "#calculations\n", + "ia=load*1000/(math.sqrt(3)*v)\n", + "vp=v/math.sqrt(3)\n", + "phi=math.acos(pf)\n", + "ra=ia*r\n", + "xa=ia*x\n", + "za=math.sqrt(ra**2+xa**2)\n", + "theta=math.atan(x/r)\n", + "\n", + "#pf=1\n", + "eb1=math.sqrt(vp**2+za**2-(2*vp*za*math.cos(theta)))\n", + "alpha1=math.asin(xa*math.sin(theta)/eb1)\n", + "\n", + "#pf=0.8 lag\n", + "eb2=math.sqrt(vp**2+xa**2-(2*vp*xa*math.cos(theta-phi)))*math.sqrt(3)\n", + "alpha2=math.asin(xa*math.sin(theta-phi)/eb2)\n", + "#pf=1\n", + "eb3=math.sqrt(vp**2+xa**2-(2*vp*xa*math.cos(theta+phi)))*math.sqrt(3)\n", + "alpha3=math.asin(xa*math.sin(theta+phi)/eb3)\n", + "\n", + "#result\n", + "print \"at pf=1\"\n", + "print \"Eb=\",eb1*math.sqrt(3),\"V\"\n", + "print \"alpha=\",math.degrees(alpha1),\"degrees\"\n", + "print \"at pf=0.8 lagging\"\n", + "print \"Eb=\",eb2,\"V\"\n", + "print \"alpha=\",math.degrees(alpha2),\"degrees\"\n", + "print \"at pf=0.8 leading\"\n", + "print \"Eb=\",eb3,\"V\"\n", + "print \"alpha=\",math.degrees(alpha3),\"degrees\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "at pf=1\n", + "Eb= 11283.8105339 V\n", + "alpha= 18.7256601694 degrees\n", + "at pf=0.8 lagging\n", + "Eb= 8990.39249633 V\n", + "alpha= 10.0142654731 degrees\n", + "at pf=0.8 leading\n", + "Eb= 13283.8907748 V\n", + "alpha= 7.71356041367 degrees\n" + ] + } + ], + "prompt_number": 56 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.14, Page Number:1513" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "z=complex(0.5,0.866)\n", + "v=200#V\n", + "output=6000#W\n", + "loss=500#W\n", + "i=50#A\n", + "\n", + "#calculations\n", + "cu_loss=i**2*z.real\n", + "motor_intake=output+loss+cu_loss\n", + "phi=math.acos(motor_intake/(v*i))\n", + "theta=math.atan(z.imag/z.real)\n", + "zs=abs(z)*i\n", + "eb1=math.sqrt(v**2+zs**2-(2*v*zs*math.cos(math.radians(60)-phi)))\n", + "eb2=math.sqrt(v**2+zs**2-(2*v*zs*math.cos(math.radians(60)+phi)))\n", + "#result\n", + "print \"lag:eb=\",eb1,\"V\"\n", + "print \"lag:eb=\",eb2,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "lag:eb= 154.286783862 V\n", + "lag:eb= 213.765547573 V\n" + ] + } + ], + "prompt_number": 65 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.15, Page Number:1513" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=2200#V\n", + "f=50#Hz\n", + "z=complex(0.4,6)\n", + "lag=3#degrees\n", + "\n", + "#calculations\n", + "eb=v/math.sqrt(3)\n", + "alpha=lag*8/2\n", + "er=math.sqrt(eb**2+eb**2-(2*eb*eb*(math.cos(math.radians(alpha)))))\n", + "zs=abs(z)\n", + "ia=er/zs\n", + "theta=math.atan(z.imag/z.real)\n", + "phi=theta-(math.asin(eb*math.sin(math.radians(alpha))/er))\n", + "pf=math.cos(phi)\n", + "total_input=3*eb*ia*pf\n", + "cu_loss=3*ia**2*z.real\n", + "pm=total_input-cu_loss\n", + "pm_max=(eb*eb/zs)-(eb**2*z.real/(zs**2))\n", + "#result\n", + "print \"armature current=\",ia,\"A\"\n", + "print \"power factor=\",pf\n", + "print \"power of the motor=\",pm/1000,\"kW\"\n", + "print \"max power of motor=\",pm_max/1000,\"kW\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "armature current= 44.1583059199 A\n", + "power factor= 0.99927231631\n", + "power of the motor= 165.803353329 kW\n", + "max power of motor= 250.446734776 kW\n" + ] + } + ], + "prompt_number": 72 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.16, Page Number:1514" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "eb=250#V\n", + "lead=150#degrees\n", + "v=200#V\n", + "x=2.5#times resistance\n", + "alpha=lead/3\n", + "#calculations\n", + "er=math.sqrt(v**2+eb**2-(2*v*eb*math.cos(math.radians(alpha))))\n", + "theta=math.atan(x)\n", + "phi=math.radians(90)-theta\n", + "pf=math.cos(phi)\n", + "\n", + "#results\n", + "print \"pf at which the motor is operating=\",pf" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "pf at which the motor is operating= 0.928476690885\n" + ] + } + ], + "prompt_number": 73 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.17, Page Number:1514" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=6600#V\n", + "r=10#ohm\n", + "inpt=900#kW\n", + "e=8900#V\n", + "\n", + "#calculations\n", + "vp=v/math.sqrt(3)\n", + "eb=e/math.sqrt(3)\n", + "icos=inpt*1000/(math.sqrt(3)*v)\n", + "bc=r*icos\n", + "ac=math.sqrt(eb**2-bc**2)\n", + "oc=ac-vp\n", + "phi=math.atan(oc/bc)\n", + "i=icos/math.cos(phi)\n", + "\n", + "#result\n", + "print \"Line current=\",i,\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Line current= 149.188331836 A\n" + ] + } + ], + "prompt_number": 82 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.18, Page Number:1515" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=6600#V\n", + "x=20#ohm\n", + "inpt=1000#kW\n", + "pf=0.8\n", + "inpt2=1500#kW\n", + "\n", + "#variable declaration\n", + "va=v/math.sqrt(3)\n", + "ia1=inpt*1000/(math.sqrt(3)*v*pf)\n", + "zs=x\n", + "phi=math.acos(pf)\n", + "ia1zs=ia1*zs\n", + "eb=math.sqrt(va**2+ia1zs**2-(2*va*ia1zs*math.cos(math.radians(90)+phi)))\n", + "ia2cosphi2=inpt2*1000/(math.sqrt(3)*v)\n", + "cosphi2=x*ia2cosphi2\n", + "ac=math.sqrt(eb**2-cosphi2*2)\n", + "phi2=math.atan(ac/cosphi2)\n", + "pf=math.cos(phi2)\n", + "alpha2=math.atan(cosphi2/ac)\n", + "\n", + "#results\n", + "print \"new power angle=\",math.degrees(alpha2),\"degrees\"\n", + "print \"new power factor=\",pf" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "new power angle= 25.8661450552 degrees\n", + "new power factor= 0.436270181217\n" + ] + } + ], + "prompt_number": 97 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.19, Page Number:1515" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=400#V\n", + "inpt=5472#W\n", + "x=10#ohm\n", + "\n", + "#calculations\n", + "va=v/math.sqrt(3)\n", + "iacosphi=inpt/(math.sqrt(3)*v)\n", + "zs=x\n", + "iazs=iacosphi*zs\n", + "ac=math.sqrt(va**2-iazs**2)\n", + "oc=va-ac\n", + "bc=iazs\n", + "phi=math.atan(oc/iazs)\n", + "pf=math.cos(phi)\n", + "ia=iacosphi/pf\n", + "alpha=math.atan(bc/ac)\n", + "#result\n", + "print \"load angle=\",math.degrees(alpha),\"degrees\"\n", + "print \"power factor=\",pf\n", + "print \"armature current=\",ia,\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "load angle= 19.9987718079 degrees\n", + "power factor= 0.984809614116\n", + "armature current= 8.01997824686 A\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.20, Page Number:1515" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "import scipy\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "i2=Symbol('i2')\n", + "v=2000.0#V\n", + "r=0.2#ohm\n", + "xs=2.2#ohm\n", + "inpt=800.0#kW\n", + "e=2500.0#V\n", + "\n", + "#calculations\n", + "i1=inpt*1000/(math.sqrt(3)*v)\n", + "vp=v/math.sqrt(3)\n", + "ep=e/math.sqrt(3)\n", + "theta=math.atan(xs/r)\n", + "i2=solve(((i1*xs+r*i2)**2+(vp+i1*r-xs*i2)**2)-ep**2,i2)\n", + "i=math.sqrt(i1**2+i2[0]**2)\n", + "pf=i1/i\n", + "\n", + "#result\n", + "print \"line currrent=\",i,\"A\"\n", + "print \"power factor=\",pf" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "line currrent= 241.492937915 A\n", + "power factor= 0.956301702525\n" + ] + } + ], + "prompt_number": 152 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.21, Page Number:1516" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=440#V\n", + "f=50#Hz\n", + "inpt=7.46#kW\n", + "r=0.5#ohm\n", + "pf=0.75\n", + "loss=500#W\n", + "ex_loss=650#W\n", + "\n", + "#calculations\n", + "ia=inpt*1000/(math.sqrt(3)*v*pf)\n", + "cu_loss=3*ia**2*r\n", + "power=inpt*1000+ex_loss\n", + "output=inpt*1000-cu_loss-loss\n", + "efficiency=output/power\n", + "\n", + "#result\n", + "print \"armature current=\",ia,\"A\"\n", + "print \"power=\",power,\"W\"\n", + "print \"efficiency=\",efficiency*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "armature current= 13.0516151762 A\n", + "power= 8110.0 W\n", + "efficiency= 82.6693343026 %\n" + ] + } + ], + "prompt_number": 156 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.22, Page Number:1517" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "v=3300#V\n", + "x=18#ohm\n", + "pf=0.707\n", + "inpt=800#kW\n", + "\n", + "#calculations\n", + "ia=inpt*1000/(math.sqrt(3)*v*pf)\n", + "ip=ia/math.sqrt(3)\n", + "zs=x\n", + "iazs=ip*zs\n", + "phi=math.acos(pf)\n", + "theta=math.radians(90)\n", + "eb=math.sqrt(v**2+iazs**2-(2*v*iazs*(-1)*pf))\n", + "alpha=math.asin(iazs*math.sin(theta+phi)/eb)\n", + "\n", + "#result\n", + "print \"excitation emf=\",eb,\"V\"\n", + "print \"rotor angle=\",math.degrees(alpha),\"degrees\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "excitation emf= 4972.19098879 V\n", + "rotor angle= 17.0098509277 degrees\n" + ] + } + ], + "prompt_number": 157 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.23, Page Number:1517" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "inpt=75#kW\n", + "v=400#V\n", + "r=0.04#ohm\n", + "x=0.4#ohm\n", + "pf=0.8\n", + "efficiency=0.925\n", + "\n", + "#calculations\n", + "input_m=inpt*1000/efficiency\n", + "ia=input_m/(math.sqrt(3)*v)\n", + "zs=math.sqrt(r**2+x**2)\n", + "iazs=ia*zs\n", + "phi=math.atan(x/r)\n", + "theta=math.radians(90)-phi\n", + "vp=v/math.sqrt(3)\n", + "eb=math.sqrt(vp**2+iazs**2-(2*vp*iazs*math.cos(theta+phi)))\n", + "cu_loss=3*ia**2*r\n", + "ns=120*50/40\n", + "pm=input_m-cu_loss\n", + "tg=9.55*pm/ns\n", + "\n", + "#result\n", + "print \"emf=\",eb,\"eb\"\n", + "print \"mechanical power=\",pm,\"W\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "emf= 235.683320812 eb\n", + "mechanical power= 79437.5456538 W\n" + ] + } + ], + "prompt_number": 158 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.24, Page Number:1517" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=400#V\n", + "f=50#Hz\n", + "r=0.5#ohm\n", + "zs=x=4#ohm\n", + "i=15#A\n", + "i2=60#A\n", + "\n", + "#calculations\n", + "vp=v/math.sqrt(3)\n", + "iazs=i*zs\n", + "xs=math.sqrt(x**2-r**2)\n", + "theta=math.atan(xs/r)\n", + "eb=math.sqrt(vp**2+iazs**2-(2*vp*iazs*math.cos(theta)))\n", + "iazs2=i2*zs\n", + "phi=theta-math.acos(vp**2-vp**2+iazs2**2/(2*vp*iazs2))\n", + "pf=math.cos(phi)\n", + "input_m=math.sqrt(3)*v*i2*pf\n", + "cu_loss=3*i2**2*r\n", + "pm=input_m-cu_loss\n", + "ns=120*50/6\n", + "tg=9.55*pm/ns\n", + "\n", + "#result\n", + "print \"gross torque developed=\",tg,\"N-m\"\n", + "print \"new power factor=\",pf" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "gross torque developed= 310.739709828 N-m\n", + "new power factor= 0.912650996943\n" + ] + } + ], + "prompt_number": 161 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.25, Page Number:1518" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=400#V\n", + "inpt=7.46#kW\n", + "xs=10#W/phase\n", + "efficiency=0.85\n", + "\n", + "#calculations\n", + "input_m=inpt*1000/efficiency\n", + "il=input_m/(math.sqrt(3)*v)\n", + "zs=il*xs\n", + "vp=v/math.sqrt(3)\n", + "eb=math.sqrt(vp**2+zs**2)\n", + "\n", + "#result\n", + "print \"minimum current=\",il,\"A\"\n", + "print \"inducedemf=\",eb,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "minimum current= 12.6677441416 A\n", + "inducedemf= 263.401798584 V\n" + ] + } + ], + "prompt_number": 164 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.26, Page Number:1518" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=400#V\n", + "f=50#Hz\n", + "inpt=37.5#kW\n", + "efficiency=0.88\n", + "zs=complex(0.2,1.6)\n", + "pf=0.9\n", + "\n", + "#calculations\n", + "input_m=inpt/efficiency\n", + "ia=input_m*1000/(math.sqrt(3)*v*pf)\n", + "vp=v/math.sqrt(3)\n", + "er=ia*abs(zs)\n", + "phi=math.acos(pf)\n", + "theta=math.atan(zs.imag/zs.real)\n", + "eb=math.sqrt(vp**2+er**2-(2*vp*er*math.cos(theta+phi)))\n", + "alpha=math.asin(math.sin(theta+phi)*er/eb)\n", + "pm=3*eb*vp*math.sin(alpha)/abs(zs)\n", + "#result\n", + "print \"excitation emf=\",eb*math.sqrt(3),\"V\"\n", + "print \"total mechanical power developed=\",pm,\"W\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "excitation emf= 495.407915636 V\n", + "total mechanical power developed= 44844.4875189 W\n" + ] + } + ], + "prompt_number": 206 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.27, Page Number:1519" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "import scipy\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "#variable declaration\n", + "v=6600.0#V\n", + "xs=20.0#ohm\n", + "inpt=1000.0#kW\n", + "pf=0.8\n", + "inpt2=1500.0#kW\n", + "phi2=Symbol('phi2')\n", + "#calculations\n", + "vp=v/math.sqrt(3)\n", + "ia=inpt*1000/(math.sqrt(3)*v*pf)\n", + "theta=math.radians(90)\n", + "er=ia*xs\n", + "zs=xs\n", + "phi=math.acos(pf)\n", + "eb=math.sqrt(vp**2+er**2-(2*vp*er*math.cos(theta+phi)))\n", + "alpha=math.asin(inpt2*1000*zs/(3*eb*vp))\n", + "#vp/eb=cos(alpha+phi2)/cos(phi2)\n", + "#solving we get\n", + "phi2=math.radians(19.39)\n", + "pf=math.cos(phi2)\n", + "#result\n", + "print \"new power factor=\",pf\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "new power factor= 0.943280616635\n" + ] + } + ], + "prompt_number": 228 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.28, Page Number:1519" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=400#V\n", + "x=4#ohms/phase\n", + "r=0.5#ohms/phase\n", + "ia=60#A\n", + "pf=0.866\n", + "loss=2#kW\n", + "\n", + "#calculations\n", + "vp=v/math.sqrt(3)\n", + "zs=abs(complex(r,x))\n", + "phi=math.acos(pf)\n", + "iazs=ia*zs\n", + "theta=math.atan(x/r)\n", + "eb=math.sqrt(vp**2+iazs**2-(2*vp*iazs*math.cos(theta+phi)))\n", + "pm_max=(eb*vp/zs)-(eb**2*r/zs**2)\n", + "pm=3*pm_max\n", + "output=pm-loss*1000\n", + "\n", + "#result\n", + "print \"maximum power output=\",output/1000,\"kW\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "maximum power output= 51.3898913442 kW\n" + ] + } + ], + "prompt_number": 229 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.29, Page Number:1519" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "z=10#ohm\n", + "x=0.5#ohm\n", + "v=2000#V\n", + "f=25#Hz\n", + "eb=1600#V\n", + "\n", + "#calculations\n", + "pf=x/z\n", + "pm_max=(eb*v/z)-(eb**2*pf/zs)\n", + "ns=120*f/6\n", + "tg_max=9.55*pm_max/ns\n", + "\n", + "#result\n", + "print \"maximum total torque=\",tg_max,\"N-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "maximum total torque= 5505.51976175 N-m\n" + ] + } + ], + "prompt_number": 231 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.30, Page Number:1520" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variabke declaration\n", + "v=2000#V\n", + "n=1500#rpm\n", + "x=3#ohm/phase\n", + "ia=200#A\n", + "\n", + "#calculations\n", + "eb=vp=v/math.sqrt(3)\n", + "zs=ia*x\n", + "sinphi=(eb**2-vp**2-zs**2)/(2*zs*vp)\n", + "phi=math.asin(sinphi)\n", + "pf=math.cos(phi)\n", + "pi=math.sqrt(3)*v*ia*pf/1000\n", + "tg=9.55*pi*1000/n\n", + "\n", + "#result\n", + "print \"power input=\",pi,\"kW\"\n", + "print \"power factor=\",pf\n", + "print \"torque=\",tg,\"N-m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "power input= 669.029147347 kW\n", + "power factor= 0.965660395791\n", + "torque= 4259.48557144 N-m\n" + ] + } + ], + "prompt_number": 234 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.31, Page Number:1520" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=3300#V\n", + "r=2#ohm\n", + "x=18#ohm\n", + "e=3800#V\n", + "\n", + "#calculations\n", + "theta=math.atan(x/r)\n", + "vp=v/math.sqrt(3)\n", + "eb=e/math.sqrt(3)\n", + "alpha=theta\n", + "er=math.sqrt(vp**2+eb**2-(2*vp*eb*math.cos(theta)))\n", + "zs=math.sqrt(r**2+x**2)\n", + "ia=er/zs\n", + "pm_max=((eb*vp/zs)-(eb**2*r/zs**2))*3\n", + "cu_loss=3*ia**2*r\n", + "input_m=pm_max+cu_loss\n", + "pf=input_m/(math.sqrt(3)*v*ia)\n", + "\n", + "#result\n", + "print \"maximum total mechanical power=\",pm_max,\"W\"\n", + "print \"current=\",ia,\"A\"\n", + "print \"pf=\",pf\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "maximum total mechanical power= 604356.888001 W\n", + "current= 151.417346198 A\n", + "pf= 0.857248980398\n" + ] + } + ], + "prompt_number": 235 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.32, Page Number:1521" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=415#V\n", + "e=520#V\n", + "z=complex(0.5,4)\n", + "loss=1000#W\n", + "\n", + "#calculations\n", + "theta=math.atan(z.imag/z.real)\n", + "er=math.sqrt(v**2+e**2-(2*v*e*math.cos(theta)))\n", + "zs=abs(z)\n", + "i=er/zs\n", + "il=math.sqrt(3)*i\n", + "pm_max=((e*v/zs)-(e**2*z.real/zs**2))*3\n", + "output=pm_max-loss\n", + "cu_loss=3*i**2*z.real\n", + "input_m=pm_max+cu_loss\n", + "pf=input_m/(math.sqrt(3)*il*v)\n", + "efficiency=output/input_m\n", + "\n", + "#result\n", + "print \"power output=\",output/1000,\"kW\"\n", + "print \"line current=\",il,\"A\"\n", + "print \"power factor=\",pf\n", + "print \"efficiency=\",efficiency*100,\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "power output= 134.640174346 kW\n", + "line current= 268.015478962 A\n", + "power factor= 0.890508620247\n", + "efficiency= 78.4816159071 %\n" + ] + } + ], + "prompt_number": 240 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.33, Page Number:1524" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "v=400#V\n", + "inpt=37.3#kW\n", + "efficiency=0.88\n", + "z=complex(0.2,1.6)\n", + "pf=0.9\n", + "\n", + "#calculations\n", + "vp=v/math.sqrt(3)\n", + "zs=abs(z)\n", + "il=inpt*1000/(math.sqrt(3)*v*efficiency*pf)\n", + "izs=zs*il\n", + "theta=math.atan(z.imag/z.real)\n", + "phi=math.acos(pf)\n", + "eb=math.sqrt(vp**2+izs**2-(2*vp*izs*math.cos(theta+phi)))\n", + "input_m=inpt*1000/efficiency\n", + "cu_loss=3*il**2*z.real\n", + "pm=input_m-cu_loss\n", + "\n", + "#result\n", + "print \"induced emf=\",eb*math.sqrt(3),\"V\"\n", + "print \"total mechanical power=\",pm/1000,\"kW\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "induced emf= 494.75258624 V\n", + "total mechanical power= 39.6138268735 kW\n" + ] + } + ], + "prompt_number": 243 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.34, Page Number:1525" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "inpt=48#kW\n", + "v=693#V\n", + "pf=0.8\n", + "ratio=0.3\n", + "x=2#W/phase\n", + "\n", + "#calculations\n", + "il=inpt*1000/(math.sqrt(3)*v*pf)\n", + "vp=v/math.sqrt(3)\n", + "zs=x\n", + "izs=zs*il\n", + "theta=math.atan(float(\"inf\"))\n", + "phi=math.acos(pf)\n", + "eb=math.sqrt(vp**2+izs**2-(2*vp*izs*math.cos(theta-phi)))\n", + "i_cosphi=pf*il\n", + "bc=i_cosphi*x\n", + "eb=eb+(ratio*eb)\n", + "ac=math.sqrt(eb**2-bc**2)\n", + "oc=ac-vp\n", + "phi2=math.atan(oc/bc)\n", + "pf=math.cos(phi2)\n", + "i2=i_cosphi/pf\n", + "\n", + "#result\n", + "print \"current=\",i2,\"A\"\n", + "print \"pf=\",pf" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "current= 46.3871111945 A\n", + "pf= 0.862084919821\n" + ] + } + ], + "prompt_number": 251 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 38.35, Page Number:1526" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#variable declaration\n", + "load=60.0#kW\n", + "inpt=240.0#kW\n", + "pf=0.8\n", + "pf2=0.9\n", + "\n", + "#calculations\n", + "total_load=inpt+load\n", + "phi=math.acos(pf2)\n", + "kVAR=total_load*math.tan(phi)\n", + "#factory load\n", + "phil=math.acos(pf)\n", + "kVAR=inpt*math.tan(phil)\n", + "kVA=inpt/pf\n", + "kVAR1=total_load*math.sin(phil)\n", + "lead_kVAR=kVAR1-kVAR\n", + "#synchronous motor\n", + "phim=math.atan(lead_kVAR/load)\n", + "motorpf=math.cos(phim)\n", + "motorkVA=math.sqrt(load**2+lead_kVAR**2)\n", + "\n", + "#result\n", + "print \"leading kVAR supplied by the motor=\",motorkVA\n", + "print \"pf=\",pf" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "leading kVAR supplied by the motor= 60.0\n", + "pf= 0.8\n" + ] + } + ], + "prompt_number": 253 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter39_2.ipynb b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter39_2.ipynb new file mode 100644 index 00000000..e889465f --- /dev/null +++ b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/chapter39_2.ipynb @@ -0,0 +1,256 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:c262c33cbbcf1d1756b9358f8cf1d8ed92f53825858905e2598fd8e15870c7ca" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 39: Special Machines" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 39.1, Page Number:1537" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable description\n", + "p=8.0 #number of poles\n", + "tp=5.0 #number of teeth for each pole\n", + "nr=50.0 #number of rotor teeth\n", + "\n", + "#calculation\n", + "ns=p*tp #number of stator teeth\n", + "B=((nr-ns)*360)/(nr*ns) #stepping angle\n", + "\n", + "#result\n", + "print \"stepping angle is \",B,\"degrees\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "stepping angle is 1.8 degrees\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 39.2, Page Number:1537" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "B=2.5\n", + "rn=25\n", + "f=3600\n", + "\n", + "#calculation\n", + "r=360/B\n", + "s=r*rn\n", + "n=(B*f)/360\n", + "\n", + "#result\n", + "print \"Resolution =\",int(r),\"steps/revolution\"\n", + "print \" Number of steps required for the shaft to make 25 revolutions =\",int(s)\n", + "print \" Shaft speed\", int(n),\"rps\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Resolution = 144 steps/revolution\n", + "Number of steps required for the shaft to make 25 revolutions = 3600\n", + "Shaft speed 25 rps\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 39.3, Page Number:1544" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "B=15 #stepping angle\n", + "pn=3 #number of phases\n", + "nr=360/(pn*B) #number of rotor teeth\n", + "\n", + "#number of stator teeth\n", + "ns1=((360*nr)/(360-(nr*B))) #ns>nr\n", + "ns2=((360*nr)/(360+(nr*B))) #nr>ns\n", + "\n", + "#result\n", + "print \"When ns>nr: ns= \",ns1\n", + "print \"When nr>ns: ns= \",ns2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "When ns>nr: ns= 12\n", + "When nr>ns: ns= 6\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 39.4, Page Number:1545" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "B=1.8\n", + "pn=4\n", + "\n", + "#calculation\n", + "nr=360/(pn*B) #number of rotor teeth\n", + "ns=nr\n", + "\n", + "#result\n", + "print \"Number of rotor teeth = \",int(nr)\n", + "print \"Number of statot teeth = \",int(ns)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of rotor teeth = 50.0\n", + "Number of statot teeth = 50.0\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example Number 39.5, Page Number:1555" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#variable declaration\n", + "er=20\n", + "\n", + "#calculation\n", + "a=40\n", + "e2=er*math.cos(math.radians(a))\n", + "e1=er*math.cos(math.radians(a-120))\n", + "e3=er*math.cos(math.radians(a+120))\n", + "\n", + "#result\n", + "print \"a) For a=40 degrees\"\n", + "print \" e2s=\" ,e2,\"V\"\n", + "print \" e1s=\" ,e1,\"V\"\n", + "print \" e3s=\" ,e3,\"V\"\n", + "\n", + "#calculation\n", + "a=(-40)\n", + "e2=er*math.cos(math.radians(a))\n", + "e1=er*math.cos(math.radians(a-120))\n", + "e3=er*math.cos(math.radians(a+120))\n", + "\n", + "#result\n", + "print \"b) For a=-40 degrees\"\n", + "print \" e2s=\" ,e2,\"V\"\n", + "print \" e1s=\" ,e1,\"V\"\n", + "print \" e3s=\" ,e3,\"V\"\n", + "\n", + "#calculation\n", + "a=30\n", + "e12=math.sqrt(3)*er*math.cos(math.radians(a-150))\n", + "e23=math.sqrt(3)*er*math.cos(math.radians(a-30))\n", + "e31=math.sqrt(3)*er*math.cos(math.radians(a+90))\n", + "\n", + "#result\n", + "print \"c) For a=30 degrees\"\n", + "print \" e12=\" ,e12,\"V\"\n", + "print \" e23=\" ,e23,\"V\"\n", + "print \" e31=\" ,e31,\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a) For a=40 degrees\n", + " e2s= 15.3208888624 V\n", + " e1s= 3.47296355334 V\n", + " e3s= -18.7938524157 V\n", + "b) For a=-40 degrees\n", + " e2s= 15.3208888624 V\n", + " e1s= -18.7938524157 V\n", + " e3s= 3.47296355334 V\n", + "c) For a=30 degrees\n", + " e12= -17.3205080757 V\n", + " e23= 34.6410161514 V\n", + " e31= -17.3205080757 V\n" + ] + } + ], + "prompt_number": 41 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/screenshots/chapter29example32_2.png b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/screenshots/chapter29example32_2.png Binary files differnew file mode 100644 index 00000000..0c99fa16 --- /dev/null +++ b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/screenshots/chapter29example32_2.png diff --git a/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/screenshots/chapter29example33_2.png b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/screenshots/chapter29example33_2.png Binary files differnew file mode 100644 index 00000000..3db6b46d --- /dev/null +++ b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/screenshots/chapter29example33_2.png diff --git a/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/screenshots/chapter32example30_2.png b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/screenshots/chapter32example30_2.png Binary files differnew file mode 100644 index 00000000..1e7a1724 --- /dev/null +++ b/A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A._K._Theraja_B.L_Thereja/screenshots/chapter32example30_2.png diff --git a/Advance_Semiconductor_Devices_by_K._C._Nandi/chapter1_1.ipynb b/Advance_Semiconductor_Devices_by_K._C._Nandi/chapter1_1.ipynb new file mode 100644 index 00000000..4e65fcec --- /dev/null +++ b/Advance_Semiconductor_Devices_by_K._C._Nandi/chapter1_1.ipynb @@ -0,0 +1,2149 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter-1 Review of fundamentals of semiconductor" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.5.1 Pg 1-7" + ] + }, + { + "cell_type": "code", + "execution_count": 92, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Forbidden gap for Si : \n", + "at 35 degree C = 1.099 eV\n", + "at 60 degree C = 1.090 eV\n", + "\n", + "Forbidden gap for Ge : \n", + "at 35 degree C = 0.7163 eV\n", + "at 60 degree C = 0.7107 eV\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Given : \n", + "t1=35##degreeC\n", + "t2=60##degreeC\n", + "T1=t1+273##K\n", + "T2=t2+273##K\n", + "print \"Forbidden gap for Si : \"\n", + "EG1_Si=1.21-3.6*10**-4*T1##eV\n", + "print \"at 35 degree C = %0.3f eV\"%EG1_Si\n", + "EG2_Si=1.21-3.6*10**-4*T2##eV\n", + "print \"at 60 degree C = %0.3f eV\"%EG2_Si\n", + "print \"\\nForbidden gap for Ge : \"\n", + "EG1_Ge=0.785-2.23*10**-4*T1##eV\n", + "print \"at 35 degree C = %0.4f eV\"%EG1_Ge\n", + "EG2_Ge=0.785-2.23*10**-4*T2##eV\n", + "print \"at 60 degree C = %0.4f eV\"%EG2_Ge" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.9.1 Pg 1-22" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Concentration of electron = 2.88e+21 m**3 \n", + "(ii) Drift velocity = 2.167 m/s \n" + ] + } + ], + "source": [ + "#Given : \n", + "l=6*10**-2##m\n", + "V=1##Volt\n", + "A=10*10**-6##m**2\n", + "I=10*10**-3##A\n", + "q=1.602*10**-19##Coulomb\n", + "mu_n=1300*10**-4##m**2/V-s\n", + "E=V/l##V/m\n", + "v=mu_n*E##m/s\n", + "J=I/A##A/m**2\n", + "n=J/(q*mu_n*E)##per m**3\n", + "print \"(i) Concentration of electron = %0.2e m**3 \"%n\n", + "print \"(ii) Drift velocity = %0.3f m/s \"%v" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.9.2 Pg 1-23" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Electron mobility = 0.365 m**2/V-s\n" + ] + } + ], + "source": [ + "#Given : \n", + "l=6*10**-2##m\n", + "V=12##Volt\n", + "v=73##m/s\n", + "E=V/l##V/m\n", + "mu=v/E##m**2/V-s\n", + "print \"Electron mobility = %0.3f m**2/V-s\"%mu" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.10.1 Pg 1-25" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Concentration of electron = 9.615e+20 per m**3 \n", + "Electron velocity = 3.250 m/s \n" + ] + } + ], + "source": [ + "#Given : \n", + "l=4*10**-2##m\n", + "A=10*10**-6##m**2\n", + "V=1##Volt\n", + "I=5*10**-3##A\n", + "q=1.6*10**-19##Coulomb\n", + "mu=1300##cm**2/V-s\n", + "J=I/A##A/m**2\n", + "E=V/l##V/m\n", + "n=J/(q*mu*10**-4*E)#\n", + "v=mu*10**-4*E##m/s\n", + "print \"Concentration of electron = %0.3e per m**3 \"%n\n", + "print \"Electron velocity = %0.3f m/s \"%v" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.11.1 Pg 1-28" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Conductivity = 5.520e-04 (ohm-m)**-1 \n", + "Resistivity = 1811.6 ohm-m\n" + ] + } + ], + "source": [ + "#Given : \n", + "ni=1.5*10**10/10**-6##per m**3\n", + "mu_n=1800*10**-4##m**2/V-s\n", + "mu_p=500*10**-4##m**2/V-s\n", + "q=1.6*10**-19##Coulomb\n", + "sigma_i=ni*(mu_n+mu_p)*q##(ohm-m)**-1\n", + "print \"Conductivity = %0.3e (ohm-m)**-1 \"%sigma_i\n", + "rho_i=1/sigma_i##ohm-m\n", + "print \"Resistivity = %0.1f ohm-m\"%rho_i" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.11.2 Pg 1-28" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Intrinsic carrier concentration = 1.5e+10 per cm**3\n" + ] + } + ], + "source": [ + "from math import sqrt, exp\n", + "#Given : \n", + "T=300##K\n", + "Ao=2.735*10**31##constant for Si\n", + "k=86*10**-6##boltzman constant\n", + "EGO=1.1##volt(Bandgap energy)\n", + "ni=sqrt(Ao*T**3*exp(-EGO/k/T))##per cm**3\n", + "print \"Intrinsic carrier concentration = %0.1e per cm**3\"%ni" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.11.3 Pg 1-29" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Current = 3.00 A\n", + "(ii) Conductivity = 2.78e+07 (ohm-m)**-1\n", + "(iii) velocity of free electrons = 2.08e-02 m/s\n", + "(iv) Mobility = 0.193 m**2/V-s \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Given : \n", + "A=1*10**-6##m**2\n", + "R=3.6*10**-4/10**-2##ohm/m\n", + "n=9*10**26##electrons/m**3\n", + "J=3*10**6##A/m**2\n", + "q=1.6*10**-19##Coulomb\n", + "I=J*A##A\n", + "print \"(i) Current = %0.2f A\"%I\n", + "rho=R*A##ohm-m\n", + "sigma=1/rho##(ohm-m)**-1\n", + "print \"(ii) Conductivity = %0.2e (ohm-m)**-1\"%sigma\n", + "v=J/n/q##m/s\n", + "print \"(iii) velocity of free electrons = %0.2e m/s\"%v\n", + "mu=sigma/n/q##m**2/V-s\n", + "print \"(iv) Mobility = %0.3f m**2/V-s \"%mu" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.11.4 Pg 1-31" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Intrinsic concentration at 30 degree C = 1.16e+16 per m**3) \n", + "Intrinsic concentration at 100 degree C = 1.221e+18 (per m**3)\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt,exp\n", + "#Given : \n", + "rho=3*10**5*10**-2##ohm-m\n", + "T1=30+273##K\n", + "mu_n=0.13##m**2/V-s\n", + "mu_p=0.05##m**2/V-s\n", + "q=1.6*10**-19##Coulomb\n", + "T2=100+273##K\n", + "sigma_i=1/rho##(ohm-m)**-1\n", + "ni1=sigma_i/q/(mu_n+mu_p)##electrons/m**3\n", + "print \"Intrinsic concentration at 30 degree C = %0.2e per m**3) \"%ni1\n", + "k=8.62*10**-5##eV/K(Boltzman constant)\n", + "EGO=1.21##V(Energy band gap)\n", + "Ao=ni1**2/(T1**3*exp(-EGO/k/T1))##constant\n", + "ni2=sqrt(Ao*T2**3*exp(-EGO/k/T2))##per cm**3\n", + "print \"Intrinsic concentration at 100 degree C = %0.3e (per m**3)\"%ni2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.11.5 Pg 1-32" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Majority Carrier density = 3.720e+20 per m**3\n" + ] + } + ], + "source": [ + "#Given : \n", + "l=0.1*10**-2##m\n", + "R=1.5*10**3##ohm\n", + "mu_n=0.14##m**2/V-s\n", + "mu_p=0.05##m**2/V-s\n", + "A=8*10**-8##m**2\n", + "ni=1.5*10**10*10**6## per m**3\n", + "q=1.6*10**-19##Coulomb\n", + "rho_n=R*A/l##ohm-m\n", + "sigma_n=1/rho_n##(ohm-m)**-1\n", + "ND=sigma_n/mu_n/q##\n", + "print \"Majority Carrier density = %0.3e per m**3\"%ND" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.11.6 Pg 1-32" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Length of the bar = 0.855 mm\n" + ] + } + ], + "source": [ + "#Given : \n", + "A=2.5*10**-4##m**2\n", + "n=1.5*10**16##per m**3\n", + "q=1.6*10**-19##Coulomb\n", + "mu_n=0.14##m**2/V-s\n", + "mu_p=0.05##m**2/V-s\n", + "I=1.2*10**-3##A\n", + "V=9##Volts\n", + "ni=n## per m**3\n", + "sigma_i=ni*q*(mu_n+mu_p)##(ohm-m)**-1\n", + "rho_i=1/sigma_i##ohm-m\n", + "R=V/I##ohm\n", + "l=R*A/rho_i##m\n", + "print \"Length of the bar = %0.3f mm\"%(l*1000)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.11.7 Pg 1-34" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistivity = 2.3148e+05 ohm-cm\n", + "Ratio of donor impurity atom to Si atom : 1e-08\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Given : \n", + "n=5*10**22##per cm**3\n", + "mu_n=1300##cm**2/V-s\n", + "mu_p=500##cm**2/V-s\n", + "ni=1.5*10**10##per cm**3\n", + "T=300##K\n", + "rho_n=9.5##ohm-cm\n", + "q=1.6*10**-19##Coulomb\n", + "sigma_i=ni*q*(mu_n+mu_p)##(ohm-cm)**-1\n", + "rho_i=1/sigma_i##ohm-cm\n", + "print \"Resistivity = %0.4e ohm-cm\"%rho_i\n", + "sigma_n=1/rho_n##(ohm-cm)**-1\n", + "ND=sigma_n/mu_n/q##per m**3\n", + "Ratio=ND/n#\n", + "print \"Ratio of donor impurity atom to Si atom : %0.e\"%(Ratio)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.11.8 Pg 1-35" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistivity of intrinsic Si = 2246.91 ohm-m \n", + "Resistivity of doped Si = 9.259e-02 ohm-m\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Given : \n", + "n=5*10**22##per cm**3\n", + "ni=1.52*10**10*10**6##per m**3\n", + "q=1.6*10**-19##Coulomb\n", + "mu_n=0.135##m**2/V-s\n", + "mu_p=0.048##m**2/V-s\n", + "impurity=1/10**8##atoms\n", + "sigma_i=ni*q*(mu_n+mu_p)##(ohm-cm)**-1\n", + "rho_i=1/sigma_i##ohm-cm\n", + "print \"Resistivity of intrinsic Si = %0.2f ohm-m \"%rho_i\n", + "ND=n*impurity*10**6##per m**3\n", + "sigma_n=ND*mu_n*q##(ohm-m)**-1\n", + "rho_n=1/sigma_n##ohm-m\n", + "print \"Resistivity of doped Si = %0.3e ohm-m\"%rho_n\n", + "#Answer in the book is not accurate." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.11.9 Pg 1-36" + ] + }, + { + "cell_type": "code", + "execution_count": 104, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ratio of doner atom to Si atom per unit volume : 1e-08\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Given : \n", + "rho=9.6*10**-2##ohm-m\n", + "mu_n=1300*10**-4##m**2/V-s\n", + "sigma_n=1/rho##(ohm-cm)**-1\n", + "TotalAtoms=5*10**28##per m**3\n", + "q=1.6*10**-19##Coulomb\n", + "ND=sigma_n/mu_n/q##per m**3\n", + "ratio=ND/TotalAtoms#\n", + "print \"Ratio of doner atom to Si atom per unit volume : %0.e\"%ratio" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.11.10 Pg 1-37" + ] + }, + { + "cell_type": "code", + "execution_count": 105, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistivity of Ge = 45.0 ohm-cm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Given : \n", + "ni=2.5*10**13##per cm**3\n", + "mu_p=1800##cm**2/V-s\n", + "mu_n=3800##cm**2/V-s\n", + "q=1.6*10**-19##Coulomb\n", + "sigma_i=ni*q*(mu_n+mu_p)##(ohm-cm)**-1\n", + "rho_i=1/sigma_i##ohm-cm\n", + "print \"Resistivity of Ge =\",round(rho_i),\"ohm-cm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.11.11 Pg 1-37" + ] + }, + { + "cell_type": "code", + "execution_count": 106, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Electron concentration = 1.440e+10 per m**3 \n", + "Conductivity of Si = 80 (ohm-m)**-1\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Given : \n", + "ni=1.2*10**16##per m**3\n", + "p=10**22##per m**3\n", + "mu_p=500*10**-4##cm**2/V-s\n", + "mu_n=1350*10**-4##cm**2/V-s\n", + "q=1.6*10**-19##Coulomb\n", + "n=ni**2/p##per m**3\n", + "print \"Electron concentration = %0.3e per m**3 \"%n\n", + "sigma=q*(n*mu_n+p*mu_p)##(ohm-m)**-1\n", + "print \"Conductivity of Si = %0.f (ohm-m)**-1\"%sigma" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.12.1 Pg 1-39" + ] + }, + { + "cell_type": "code", + "execution_count": 107, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Electron concentration = 1e+17 per cm**3 \n", + "Holes = 2.25e+03 per cm**3 : \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Given : \n", + "T=27+273##K\n", + "ND=10**17##per cm**3\n", + "ni=1.5*10**10##per cm**3\n", + "n=ND##per m**3#ND>>n\n", + "print \"Electron concentration = %0.e per cm**3 \"%n\n", + "p=ni**2/n##per m**3\n", + "print \"Holes = %0.2e per cm**3 : \"%p" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.12.2 Pg 1-39" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total free electrons = 2.20e+17 per m**3\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Given : \n", + "Vol=4*50*1.5##mm**3\n", + "ni=2.4*10**19##per m**3\n", + "p=7.85*10**14##per m**3\n", + "n=ni**2/p##per m**3\n", + "Vol=Vol*10**-9##m**3\n", + "TotalElectron=n*Vol##no. of electrons\n", + "print \"Total free electrons = %0.2e per m**3\"%TotalElectron" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.13.1 Pg 1-41" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total current density = 524.21 A/m**2 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from sympy import symbols, solve\n", + "#Given : \n", + "ND=10**14##per cm**3\n", + "NA=7*10**13##per cm**3\n", + "rho_i=60##ohm-cm\n", + "E=2##V/cm\n", + "q=1.6*10**-19##Coulomb\n", + "mu_p=1800##cm**2/V-s\n", + "mu_n=3800##cm**2/V-s\n", + "sigma_i=1/rho_i##(ohm-cm)**-1\n", + "ni=sigma_i/q/(mu_n+mu_p)##per cm**3\n", + "p = symbols('p')\n", + "n=p+(ND-NA)##per cm**3\n", + "#n*p=ni**2 \n", + "expr = n*p-ni**2 \n", + "#m=[1 (ND-NA) -ni**2]##polynomial\n", + "p=solve(expr,p)[1]##per m**3 #taking only +ve value\n", + "n=ni**2/p##per m**3\n", + "J=(n*mu_n+p*mu_p)*q*E/10**-4##A/m**2\n", + "print \"Total current density = %0.2f A/m**2 \"%J\n", + "#Answer in the textbook is not accurate." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.13.2 Pg 1-43" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Value of electrical field, E = 0.8150 V/cm \n" + ] + } + ], + "source": [ + "from sympy import symbols, solve\n", + "from __future__ import division\n", + "#Given : \n", + "ND=10**14##per cm**3\n", + "NA=7*10**3##per cm**3\n", + "rho_i=60##ohm-cm\n", + "J=52##mA/cm**2\n", + "q=1.6*10**-19##Coulomb\n", + "mu_p=1800##cm**2/V-s\n", + "mu_n=3800##cm**2/V-s\n", + "sigma_i=1/rho_i##(ohm-cm)**-1\n", + "ni=sigma_i/q/(mu_n+mu_p)##per cm**3\n", + "p = symbols('p')\n", + "n=p+(ND-NA)##per cm**3\n", + "#n*p=ni**2 \n", + "expr = n*p-ni**2 \n", + "#m=[1 (ND-NA) -ni**2]##polynomial\n", + "p=solve(expr,p)[1]##per m**3 #taking only +ve value\n", + "n=ni**2/p##per m**3\n", + "E=J*10**-3/q/(n*mu_n+p*mu_p)##V/m\n", + "print \"Value of electrical field, E = %0.4f V/cm \"%E" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.13.3 Pg 1-45" + ] + }, + { + "cell_type": "code", + "execution_count": 111, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total current density = 382.75 A/m**2\n" + ] + } + ], + "source": [ + "from sympy import symbols, solve\n", + "from __future__ import division\n", + "#Given : \n", + "ND=10**14##per cm**3\n", + "NA=7*10**13##per cm**3\n", + "rho_i=60##ohm-cm\n", + "E=2##V/cm\n", + "q=1.6*10**-19##Coulomb\n", + "mu_p=500##cm**2/V-s\n", + "mu_n=1300##cm**2/V-s\n", + "sigma_i=1/rho_i##(ohm-cm)**-1\n", + "ni=sigma_i/q/(mu_n+mu_p)##per cm**3\n", + "p = symbols('p')\n", + "n=p+(ND-NA)##per cm**3\n", + "#n*p=ni**2 \n", + "expr = n*p-ni**2 \n", + "#m=[1 (ND-NA) -ni**2]##polynomial\n", + "p=solve(expr,p)[1]##per m**3 #taking only +ve value\n", + "n=ni**2/p##per m**3\n", + "J=(n*mu_n+p*mu_p)*q*E/10**-4##A/m**2\n", + "print \"Total current density = %0.2f A/m**2\"%J" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.13.4 Pg 1-46" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Electron mobility = 0.365 m**2/V-s \n" + ] + } + ], + "source": [ + "#Given : \n", + "l=6*10**-2##m\n", + "V=12##volts\n", + "v=73##m/s\n", + "E=V/l##V/m\n", + "mu=v/E##m**2/V-s\n", + "print \"Electron mobility = %0.3f m**2/V-s \"%mu" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.15.1 Pg 1-54" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Magnitude of hall voltage = 65 mV \n" + ] + } + ], + "source": [ + "#Given : \n", + "ND=10**13##per cm**3\n", + "Bz=0.2##Wb/m**2\n", + "d=5##mm\n", + "E=5##V/cm\n", + "q=1.6*10**-19##Coulomb\n", + "mu_n=1300##cm**2/V-s\n", + "rho=ND*q##Coulomb/cm**3\n", + "J=rho*mu_n*E##A/cm**2\n", + "VH=Bz*10**-4*J*d*10**-1/rho##V\n", + "print \"Magnitude of hall voltage = %0.f mV \"%(VH*10**3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.15.2 Pg 1-55" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Mobility = 0.050909 m**2/V-s\n" + ] + } + ], + "source": [ + "#Given : \n", + "rho=220*10**3*10**-2##ohm/m\n", + "d=2.2*10**-3##m\n", + "w=2*10**-3##m\n", + "B=0.1##Wb/m**2\n", + "I=5*10**-6##A\n", + "VH=28*10**-3##V\n", + "sigma=1/rho##(ohm-m)**-1\n", + "RH=VH*w/(B*I)##ohm\n", + "mu=sigma*RH##m**2/V-s\n", + "print \"Mobility = %0.6f m**2/V-s\"%mu" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.16.1 Pg 1-59" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Concentration of electron = 9.615e+20 per m**3\n", + "Electron velocity = 3.25 m/s\n" + ] + } + ], + "source": [ + "#Given : \n", + "l=4*10**-2##m\n", + "A=10*10**-6##m**2\n", + "V=1##Volt\n", + "I=5*10**-3##A\n", + "q=1.6*10**-19##Coulomb\n", + "mu=1300##cm**2/V-s\n", + "J=I/A##A/m**2\n", + "E=V/l##V/m\n", + "n=J/(q*mu*10**-4*E)\n", + "v=mu*10**-4*E##m/s\n", + "print \"Concentration of electron = %0.3e per m**3\"%n\n", + "print \"Electron velocity = %0.2f m/s\"%v" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.16.2 Pg 1-59" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistivity of doped Ge = 3.738 ohm-cm\n" + ] + } + ], + "source": [ + "#Given : \n", + "mu_n=3800##cm**2/V-s\n", + "mu_p=1300##cm**2/V-s\n", + "ni=2.5*10**13##per cm**3\n", + "q=1.6*10**-19##Coulomb\n", + "ND=4.4*10**22/10**8##per cm**3\n", + "sigma_n=ND*q*mu_n##(ohm-m)**-1\n", + "rho_n=1/sigma_n##ohm-cm\n", + "print \"Resistivity of doped Ge = %0.3f ohm-cm\"%rho_n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.16.3 Pg 1-60" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Minor carrier density = 4.5e+11 per m**3\n" + ] + } + ], + "source": [ + "#Given : \n", + "ni=1.5*10**16##per m**3\n", + "n=5*10**20##per m**3\n", + "p=ni**2/n##per m**3\n", + "print \"Minor carrier density = %0.1e per m**3\"%p" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.16.4 Pg 1-60" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Electron concentration, n = 8.00e+15 per cm**3\n", + "Hole concentration, p = 2.8125e+04 per cm**3 \n", + "Total electron concentration, nT = 8e+15 per cm**3\n", + "Total hole concentration, pT = 1.00e+14 per cm**3\n", + "Current, I = 36.45 mA\n" + ] + } + ], + "source": [ + "#Given : \n", + "ni=1.5*10**10##per cm**3\n", + "mu_n=1400##cm**2/V-s\n", + "mu_p=500##cm**2/V-s\n", + "l=1##cm\n", + "a=1##mm**2\n", + "q=1.6*10**-19##Coulomb\n", + "del_n=10**14##per cm**3\n", + "del_p=10**14##per cm**3\n", + "Nd=8*10**15##per cm**3\n", + "n=Nd##per cm**3(Nd>>ni)\n", + "print \"Electron concentration, n = %0.2e per cm**3\"%n\n", + "p=ni**2/n##per m**3\n", + "print \"Hole concentration, p = %0.4e per cm**3 \"%p\n", + "nT=Nd+del_n##per cm**3\n", + "print \"Total electron concentration, nT = %0.e per cm**3\"%nT\n", + "pT=p+del_p##per cm**3\n", + "print \"Total hole concentration, pT = %0.2e per cm**3\"%pT\n", + "sigma=(nT*mu_n+pT*mu_p)*q##(ohm-cm)**-1\n", + "rho=1/sigma##ohm-cm\n", + "R=rho*l/(a*10**-2)##ohm\n", + "V=2##volt\n", + "I=V/R##A\n", + "print \"Current, I = %0.2f mA\"%(I*1000)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.16.5 Pg 1-61" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current, I = 0.944 mA\n" + ] + } + ], + "source": [ + "#Given : \n", + "A=2.3*10**-4##m**2\n", + "n=1.5*10**16##per m**3\n", + "l=1##mm\n", + "mu_n=1400##cm**2/V-s\n", + "mu_p=500##cm**2/V-s\n", + "p=n##per m**3\n", + "ni=n##per m**3\n", + "q=1.6*10**-19##Coulomb\n", + "sigma_i=ni*(mu_n*10**-4+mu_p*10**-4)*q##(ohm-m)**-1\n", + "rho_i=1/sigma_i##ohm-m\n", + "R=rho_i*l*10**-3/A##ohm\n", + "V=9##volt\n", + "I=V/R##A\n", + "print \"Current, I = %0.3f mA\"%(I*1000)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.16.6 Pg 1-62" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Concentration gradient, dn/dx = 1.644e+19\n" + ] + } + ], + "source": [ + "#Given : \n", + "ND=10**14##per m**3\n", + "Jn=10##mA/cm**2\n", + "E=3##V/cm\n", + "T=27+273##K\n", + "q=1.6*10**-19##Coulomb\n", + "mu_n=1500##cm**2/V-s\n", + "Dn=mu_n/39##Diffusion constant\n", + "n=ND##per m**3\n", + "dnBYdx=((Jn*10**-3/10**-4)-n*q*mu_n*E)/q/Dn##concentration gradient\n", + "print \"Concentration gradient, dn/dx = %0.3e\"%dnBYdx" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.16.7 Pg 1-63" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total current density = 672.0 A/m**2\n" + ] + } + ], + "source": [ + "#Given : \n", + "ND=10**13##per m**3\n", + "NA=10**14##per m**3\n", + "rho_i=44##ohm-cm\n", + "E=3##V/cm\n", + "q=1.6*10**-19##Coulomb\n", + "mu_n=0.38##m**2/V-s\n", + "mu_p=0.18##m**2/V-s\n", + "ni=2.5*10**19##per m**3\n", + "from sympy import symbols, solve\n", + "p = symbols('p')\n", + "n=p+(ND-NA)##per cm**3\n", + "#n*p=ni**2 \n", + "expr = n*p-ni**2 \n", + "#m=[1 (ND-NA) -ni**2]##polynomial\n", + "p=solve(expr,p)[1]##per m**3 #taking only +ve value\n", + "n=ni**2/p##per m**3\n", + "J=(n*mu_n+p*mu_p)*q*(E/10**-2)##A/m**2\n", + "print \"Total current density = %0.1f A/m**2\"%J\n", + "#Ans in the textbook is not accurate." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.16.8 Pg 1-64" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Conductivity of intrinsic Ge = 2.24 (ohm-m)**-1\n", + "Conductivity after adding donor impurity = 267.52 (ohm-m)**-1\n", + "Conductivity after adding acceptor impurity = 126.72 (ohm-m)**-1 \n" + ] + } + ], + "source": [ + "#Given : \n", + "T=300##K\n", + "ni=2.5*10**13##per cm**3\n", + "mu_n=3800##cm**2/V-s\n", + "mu_p=1800##cm**2/V-s\n", + "q=1.6*10**-19##Coulomb\n", + "sigma_i=ni*(mu_n+mu_p)*q/10**-2##(ohm-m)**-1\n", + "print \"Conductivity of intrinsic Ge = %0.2f (ohm-m)**-1\"%sigma_i\n", + "ND=4.4*10**22/10**7##per cm**3\n", + "n=ND##per cm**3\n", + "sigma_n=n*mu_n*q/10**-2##(ohm-m)**-1\n", + "print \"Conductivity after adding donor impurity = %0.2f (ohm-m)**-1\"%sigma_n\n", + "NA=4.4*10**22/10**7##per cm**3\n", + "p=NA##per cm**3\n", + "sigma_p=p*mu_p*q/10**-2##(ohm-m)**-1\n", + "print \"Conductivity after adding acceptor impurity = %0.2f (ohm-m)**-1 \"%sigma_p" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.40.1 Pg 1-102" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Equilibrium hole concentration = 2.25e+03 per cm**3\n" + ] + } + ], + "source": [ + "#Given : \n", + "ND=10**17##per cm**3\n", + "ni=1.5*10**10##per cm**3\n", + "no=ND##per cm**3#/Nd>>ni\n", + "po=ni**2/no##per cm**3\n", + "print \"Equilibrium hole concentration = %0.2e per cm**3\"%po" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.40.3 Pg 1-103" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Fermi level, Ef = Ei + 0.407 eV\n" + ] + } + ], + "source": [ + "from math import log\n", + "#Given : \n", + "ni=1.5*10**10##per cm**3\n", + "ND=10**17##per cm**3\n", + "no=ND##per cm**3#/Nd>>ni\n", + "po=ni**2/no##per cm**3\n", + "KT=0.0259##constant\n", + "delEf=KT*log(no/ni)##eV\n", + "print \"Fermi level, Ef = Ei +\",round(delEf,3),\"eV\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.40.4 Pg 1-104" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Diffusion coffiecients of electron = 4.40e-03 m**2/s\n", + "Diffusion coffiecients of holes = 6.47e-04 m**2/s\n" + ] + } + ], + "source": [ + "#Given : \n", + "K=1.38*10**-23##J/K\n", + "T=27+273##K\n", + "e=1.6*10**-19##constant\n", + "mu_n=0.17##m**2/V-s\n", + "mu_p=0.025##m**2/V-s\n", + "Dn=K*T/e*mu_n##m**2/s\n", + "print \"Diffusion coffiecients of electron = %0.2e m**2/s\"%Dn\n", + "Dp=K*T/e*mu_p##m**2/s\n", + "print \"Diffusion coffiecients of holes = %0.2e m**2/s\"%Dp" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.40.5 Pg 1-105" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Diffusion current density = 3.152e+03 A/m**2\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "#Given : \n", + "K=1.38*10**-23##J/K\n", + "T=27+273##K\n", + "e=1.6*10**-19##constant\n", + "del_no=10**20##per.m**3\n", + "tau_n=10**-7##s\n", + "mu_n=0.15##m**2/V-s\n", + "Dn=K*T/e*mu_n##m**2/s\n", + "Ln=sqrt(Dn*tau_n)##m\n", + "Jn=e*Dn*del_no/Ln##A/m**2\n", + "print \"Diffusion current density = %0.3e A/m**2\"%Jn" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.40.6 Pg 1-105" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Concentration of holes = 4.680e+11 per m**3\n" + ] + } + ], + "source": [ + "#Given : \n", + "sigma_n=0.1##(ohm-cm)**-1\n", + "mu_n=1300##m**2/V-s\n", + "ni=1.5*10**10##per cm**3\n", + "q=1.6*10**-19##Coulomb\n", + "n_n=sigma_n/q/mu_n##per cm**3\n", + "p_n=ni**2/n_n##per cm**3\n", + "p_n=p_n*10**6##per m**3\n", + "print \"Concentration of holes = %0.3e per m**3\"%p_n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.40.7 Pg 1-106" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Electron transit time = 6.41e-09 s\n", + "Photoconductor gain : 216.0\n" + ] + } + ], + "source": [ + "#Given : \n", + "L=100*10**-6##m\n", + "A=10**-7*10**-6##m**2\n", + "mu_e=0.13##m**2/V-s\n", + "mu_h=0.05##m**2/V-s\n", + "tau_h=10**-6##sec\n", + "V=12##volt\n", + "E=V/L##v/m\n", + "tn=L**2/(mu_e*V)##sec\n", + "print \"Electron transit time = %0.2e s\"%tn\n", + "Gain=tau_h/tn*(1+mu_h/mu_e)##\n", + "print \"Photoconductor gain :\",Gain" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.40.8 Pg 1-106" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistivity of intrinsic Ge at 300K = 45.00 ohm-cm\n", + "Resistivity of doped Ge = 3.74 ohm-cm \n" + ] + } + ], + "source": [ + "#Given : \n", + "T=300##K\n", + "rho_i=45##ohm-cm\n", + "#part (i)\n", + "mu_n=3800##cm**2/V-s\n", + "mu_p=1800##cm**2/V-s\n", + "ni=2.5*10**13##per cm**3\n", + "q=1.6*10**-19##Coulomb\n", + "sigma=ni*q*(mu_n+mu_p)##(ohm-cm)**-1\n", + "rho=1/sigma##ohm-cm\n", + "print \"Resistivity of intrinsic Ge at 300K = %0.2f ohm-cm\"%round(rho)\n", + "#part (ii)\n", + "ND=4.4*10**22/10**8##per cm**3\n", + "sigma=ND*q*mu_n##(ohm-cm)**-1\n", + "rho=1/sigma##ohm-cm\n", + "print \"Resistivity of doped Ge = %0.2f ohm-cm \"%rho" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.40.9 Pg 1-107" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Electron concentration = 1e+22 per m**3\n", + "Electron concentration = 1e+10 per m**3\n" + ] + } + ], + "source": [ + "#Given : \n", + "ni=10**16##per m**3\n", + "ND=10**22##per m**3\n", + "n=ND##per m**3#ND>>ni\n", + "print \"Electron concentration = %0.e per m**3\"%n\n", + "p=ni**2/n##per m**3\n", + "print \"Electron concentration = %0.e per m**3\"%p" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.40.10 Pg 1-107" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ratio of donor atom to Si atom : 1e-08\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Given : \n", + "rho=9.6*10**-2##ohm-m\n", + "mu_n=1300##cm**2/V-s\n", + "q=1.6*10**-19##Coulomb\n", + "sigma_n=1/rho##(ohm-m)**-1\n", + "ND=sigma_n/q/(mu_n*10**-4)##per m**3\n", + "ni=5*10**22*10**6##per m**3\n", + "print \"Ratio of donor atom to Si atom : %0.e\"%(ND/ni)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.40.11 Pg 1-108" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Equillibrium electron density = 2.25e+15 per cm**3\n", + "Equillibrium hole density = 1e+05 per cm**3 \n" + ] + } + ], + "source": [ + "#Given : \n", + "ni=1.5*10**10##per cm**3\n", + "n_n=2.25*10**15##per cm**3\n", + "print \"Equillibrium electron density = %0.2e per cm**3\"%n_n\n", + "p_n=ni**2/n_n##per cm**3\n", + "print \"Equillibrium hole density = %0.e per cm**3 \"%p_n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.40.12 Pg 1-108" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Material is p-type & Carrier concentration = 1e+16 holes per cm**3\n" + ] + } + ], + "source": [ + "#Given : \n", + "NA=2*10**16##per cm**3\n", + "ND=10**16##per cm**3\n", + "p=NA-ND##per cm**3\n", + "print \"Material is p-type & Carrier concentration = %0.e holes per cm**3\"%p" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.40.13 Pg 1-108" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of generation of minority carrier = 1e+20 electron hole pair/sec/cm**3\n" + ] + } + ], + "source": [ + "#Given : \n", + "del_n=10**15##per cm**3\n", + "tau_p=10*10**-6##sec\n", + "rate=del_n/tau_p##rate of generation minority carrier\n", + "print \"Rate of generation of minority carrier = %0.e electron hole pair/sec/cm**3\"%rate" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.40.14 Pg 1-109" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Mobility = 5000 cm**2/V-s\n" + ] + } + ], + "source": [ + "#Given : \n", + "E=10##V/cm\n", + "v=1/(20*10**-6)##m/s\n", + "mu=v/E##cm**2/V-s\n", + "print \"Mobility = %02.f cm**2/V-s\"%mu" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.40.15 Pg 1-109" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Hole diffucion current, Jp = -707.17 A/cm**2)\n", + "Electron diffucion current, Jp = 2552.08 A/cm**2\n" + ] + } + ], + "source": [ + "from math import sqrt,exp\n", + "#Given : \n", + "ND=4.5*10**15 #per cm**3\n", + "A=1*10**-2 #cm**2\n", + "l=10 #cm\n", + "tau_p=1*10**-6 #sec\n", + "tau_n=1*10**-6 #sec\n", + "Dp=12 #cm**2/sec\n", + "Dn=30 #cm**2/sec\n", + "q=1.6*10**-19 #coulamb\n", + "del_p=10**21 #electron hole pair/cm**3/sec\n", + "x=34.6*10**-4 #cm\n", + "Kdash=26 #mV(Kdash is taken as K*T/q)\n", + "ni=1.5*10**10 #per cm**3\n", + "no=ND #per cm**3#ND<<ni\n", + "po=ni**2/no #per cm**3\n", + "ln=sqrt(Dn*tau_n) #cm\n", + "lp=sqrt(Dp*tau_p) #cm\n", + "dpBYdx=del_p*exp(-x/lp) #per cm**4\n", + "dnBYdx=del_p*exp(-x/ln) #per cm**4\n", + "Jp=-q*Dp*dpBYdx #A/cm**2\n", + "print \"Hole diffucion current, Jp = %0.2f A/cm**2)\"%Jp\n", + "Jn=q*Dn*dnBYdx #A/cm**2\n", + "print \"Electron diffucion current, Jp = %0.2f A/cm**2\"%Jn\n", + "#Answer is wrong in the book. Wrong calculation for dpBYdx and dnBYdx." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.40.16 Pg 1-111" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Energy band gap = 2.26 eV\n" + ] + } + ], + "source": [ + "#Given : \n", + "h=6.626*10**-34##J-s\n", + "lamda=5490##Angstrum\n", + "c=3*10**8##m/s(speed of light)\n", + "f=c/(lamda*10**-10)##Hz\n", + "E=(h/1.6/10**-19)*f##eV\n", + "print \"Energy band gap = %0.2f eV\"%E" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.40.17 Pg 1-111" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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XSprW6jp2k1qfZzqZ8vl0QuViSRe1o57dQNI1kn4jaVmVMv5uZlDrs/T3sj6S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\ng2FmZpm4wTAzs0zcYJiZWSb/H7crH3T5904sAAAAAElFTkSuQmCC\n", + "text/plain": [ + "<matplotlib.figure.Figure at 0x7f40ec3e2310>" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current density = 1120.00 A/cm**2\n" + ] + } + ], + "source": [ + "%matplotlib inline\n", + "from matplotlib.pyplot import plot, title, show, xlabel, ylabel\n", + "\n", + "#Given : \n", + "q=1.6*10**-19##Coulomb\n", + "Dn=35##cm**2/s\n", + "x=[0, 2]##micro meter\n", + "n=[10**17 ,6*10**16]##per cm**3\n", + "plot(x,n)#\n", + "title('n Vs x')#\n", + "xlabel('x(micro meter)')#\n", + "ylabel('n(electrons per cm**3)')#\n", + "show()\n", + "dnBYdx=(n[1]-n[0])/(x[0]-x[1])/10**-4##gradient\n", + "Jn=q*Dn*dnBYdx##A/cm**2\n", + "print \"Current density = %0.2f A/cm**2\"%Jn" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.40.18 Pg 1-112" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance of the bar = 1 Mohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Given : \n", + "q=1.6*10**-19##Coulomb\n", + "l=0.1##cm\n", + "A=100*10**-8##cm**2\n", + "n_n=5*10**20*10**-6##per cm**3\n", + "mu_n=0.13*10**4##cm**2/V-s\n", + "sigma_n=q*n_n*mu_n##(ohm-cm)**-1\n", + "rho=1/sigma_n##ohm-cm\n", + "R=rho*l/A##ohm\n", + "print \"Resistance of the bar = %0.f Mohm\"%round(R/10**6)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.40.19 Pg 1-113" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Answer is (B). Depletion width on p-side = 0.33 micro meter\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Given : \n", + "NA=9*10**16##per cm**3\n", + "ND=1*10**16##per cm**3\n", + "w_total=3##micro meter\n", + "w_p=w_total*ND/NA##micro meter\n", + "print \"Answer is (B). Depletion width on p-side = %0.2f micro meter\"%w_p" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.40.20 Pg 1-113" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Majority carrier density = 4.50e+11 per m**3\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Given : \n", + "ni=1.5*10**16##per m**3\n", + "n_n=5*10**20##per m**3\n", + "p_n=ni**2/n_n##per m**3\n", + "print \"Majority carrier density = %0.2e per m**3\"%p_n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.40.21 Pg 1-113" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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EKEp6iEFU6aokSfOBpwLzJ/bZ/nWNdTW/fxJDDLykh+i2Oq9KejvwW+B7wDcaHhExC0kP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\n", + "text/plain": [ + "<matplotlib.figure.Figure at 0x7fd3b998cb90>" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current density = 8.00 A/cm**2\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "%matplotlib inline\n", + "from matplotlib.pyplot import plot, title, show, xlabel, ylabel\n", + "#Given : \n", + "q=1.6*10**-19##Coulomb\n", + "Dn=25##cm**2/s\n", + "x=[0 ,0.5]##micro meter(base width)\n", + "n=[10**14 ,0]##per cm**3\n", + "plot(x,n)#\n", + "title('n Vs x')#\n", + "xlabel('x(micro meter)')#\n", + "ylabel('n(electrons per cm**3)')#\n", + "show()\n", + "dnBYdx=(n[1]-n[0])/(x[0]-x[1])/10**-4##gradient\n", + "Jn=q*Dn*dnBYdx##A/cm**2\n", + "print \"Current density = %0.2f A/cm**2\"%Jn" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.40.22 Pg 1-114" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Band gap = 1.431 eV\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Given : \n", + "h=6.64*10**-34##planks constant\n", + "c=3*10**8##m/s(speed of light)\n", + "lamda=0.87*10**-6##m\n", + "Eg=h*c/lamda/(1.6*10**-19)##eV\n", + "print \"Band gap = %0.3f eV\"%Eg" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.40.23 Pg 1-114" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) Absorbed power = 9.00 mW\n", + "(b) Rate of excess thermal energy = 2.564e-03 J/s\n", + "(c) No. of photons per sec : 2.81e+16\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import exp\n", + "#Given : \n", + "t=0.46*10**-4##cm\n", + "E=2##eV\n", + "alfa=5*10**4##cm**-1\n", + "Io=10##mW\n", + "q=1.6*10**-19##Coulomb\n", + "It=Io*exp(-alfa*t)##mW\n", + "Pabs=Io-It##mW\n", + "print \"(a) Absorbed power = %0.2f mW\"%round(Pabs)\n", + "Eg=1.43##eV(Band gap)\n", + "heat_fraction=(E-Eg)/E#\n", + "E_heat=heat_fraction*Pabs*10**-3##J/s(energy converted to heat)\n", + "print \"(b) Rate of excess thermal energy = %0.3e J/s\"%E_heat\n", + "photons=Pabs*10**-3/q/E##no. of photons per sec\n", + "print \"(c) No. of photons per sec : %0.2e\"%photons" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.40.24 Pg 1-115" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Steady state separation between Fp & Ec = 0.70 eV\n", + "Hole current = 1900.55 A\n", + "Excess stored hole charge = 1.44e-07 Coulomb\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log,exp,sqrt,pi\n", + "#Given : \n", + "Kdash=0.0259##constant(taken as K*T/q)\n", + "A=0.5##cm**2\n", + "Na=10**17##per cm**3\n", + "ni=1.5*10**10##per cm**3\n", + "delta_p=5*10**16##per cm**3\n", + "x=1000##Angstrum\n", + "mu_p=500##cm**2/V-s\n", + "tau_p=10**-10##sec\n", + "q=1.6*10**-19##Coulomb\n", + "\n", + "Dp=Kdash*mu_p##cm/s\n", + "Lp=sqrt(Dp*tau_p)##cm\n", + "p0=Na##per cm**3\n", + "p=p0+delta_p*exp(x*10**-8/Lp)##per cm**3\n", + "delE1=log(p/ni)*Kdash##eV(taken as Ei-Fp)\n", + "Eg=1.12##eV(Band gap)\n", + "delE2=Eg-delE1##eV(taken as Ec-Fp)\n", + "print \"Steady state separation between Fp & Ec = %0.2f eV\"%delE2\n", + "Ip=q*A*Dp/Lp*delta_p*exp(x*10**-8/Lp)##A\n", + "print \"Hole current = %0.2f A\"%Ip\n", + "Qp=q*A*delta_p*Lp##C\n", + "print \"Excess stored hole charge = %0.2e Coulomb\"%Qp\n", + "#Answer in the book is wrong beacause of calculation mistake in the value of p & Ip." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 1.40.25 Pg 1-116" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Steady state separation between Fp & Ec = 0.70 eV\n", + "Hole current = 1900.55 A \n", + "Excess stored hole charge = 1.44e-07 Coulomb\n" + ] + } + ], + "source": [ + "from math import sqrt,log,exp,pi\n", + "from __future__ import division\n", + "#Given : \n", + "Kdash=0.0259##constant(taken as K*T/q)\n", + "A=0.5##cm**2\n", + "Na=10**17##per cm**3\n", + "ni=1.5*10**10##per cm**3\n", + "delta_p=5*10**16##per cm**3\n", + "x=1000##Angstrum\n", + "mu_p=500##cm**2/V-s\n", + "tau_p=10**-10##sec\n", + "q=1.6*10**-19##Coulomb\n", + "\n", + "Dp=Kdash*mu_p##cm/s\n", + "Lp=sqrt(Dp*tau_p)##cm\n", + "p0=Na##per cm**3\n", + "p=p0+delta_p*exp(x*10**-8/Lp)##per cm**3\n", + "delE1=log(p/ni)*Kdash##eV(taken as Ei-Fp)\n", + "Eg=1.12##eV(Band gap)\n", + "delE2=Eg-delE1##eV(taken as Ec-Fp)\n", + "print \"Steady state separation between Fp & Ec = %0.2f eV\"%delE2\n", + "Ip=q*A*Dp/Lp*delta_p*exp(x*10**-8/Lp)##A\n", + "print \"Hole current = %0.2f A \"%Ip\n", + "Qp=q*A*delta_p*Lp##C\n", + "print \"Excess stored hole charge = %0.2e Coulomb\"%Qp\n", + "#Answer in the book is wrong beacause of calculation mistake in the value of p & Ip." + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Advance_Semiconductor_Devices_by_K._C._Nandi/chapter2_1.ipynb b/Advance_Semiconductor_Devices_by_K._C._Nandi/chapter2_1.ipynb new file mode 100644 index 00000000..6d06b941 --- /dev/null +++ b/Advance_Semiconductor_Devices_by_K._C._Nandi/chapter2_1.ipynb @@ -0,0 +1,1304 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter-2 Junctions and interfaces" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.6.1 Pg 2-21" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Junction potential = 328.7 mV\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log\n", + "#Given : \n", + "Ge=4.4*10**22##atoms/cm**3\n", + "NA=Ge/10**8##per cm**3\n", + "NA=NA*10**6##per m**3\n", + "ND=NA*10**3##per m**3\n", + "ni=2.5*10**13##per cm**3\n", + "ni=ni*10**6##per m**3\n", + "VT=26##mV\n", + "Vj=VT*log(NA*ND/ni**2)##mV\n", + "print \"Junction potential = %0.1f mV\"%Vj" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.6.2 Pg 2-22" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Contact potential = 0.0888 V\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log\n", + "\n", + "#Given : \n", + "ni=2.5*10**15##per cm**3\n", + "Ge=4.4*10**22##atoms/cm**3\n", + "NA=Ge/10**8##per cm**3\n", + "NA=NA*10**6##per m**3\n", + "ND=NA*10**3##per m**3\n", + "ni=ni*10**6##per m**3\n", + "T=27+273##K\n", + "VT=T/11600##V\n", + "Vo=VT*log(NA*ND/ni**2)##V\n", + "print \"Contact potential = %0.4f V\"%Vo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.6.3 Pg 2-23" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Height of potential barrier = 0.564 V\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log\n", + "\n", + "#Given : \n", + "mu_n=1500*10**-4##m**2/V-s\n", + "mu_p=475*10**-4##m**2/V-s\n", + "ni=1.45*10**10*10**6##per m**3\n", + "q=1.6*10**-19##Coulomb\n", + "rho_p=10##ohm-cm\n", + "rho_p=rho_p*10**-2##ohm-m\n", + "rho_n=3.5##ohm-cm\n", + "rho_n=rho_n*10**-2##ohm-m\n", + "sigma_p=1/rho_p##(ohm-m)**-1\n", + "NA=sigma_p/q/mu_p##m**3\n", + "sigma_n=1/rho_n##(ohm-m)**-1\n", + "ND=sigma_p/q/mu_n##m**3\n", + "VT=26*10**-3##V\n", + "Vj=VT*log(NA*ND/ni**2)##V\n", + "print \"Height of potential barrier = %0.3f V\"%Vj\n", + "#Answer in the book is wrong." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.6.4 Pg 2-24" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Height of potential barrier = 0.9347 V\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log\n", + "\n", + "#Given : \n", + "rho_p=2##ohm-cm\n", + "rho_p=rho_p*10**-2##ohm-m\n", + "rho_n=1##ohm-cm\n", + "rho_n=rho_n*10**-2##ohm-m\n", + "mu_n=1500*10**-4##m**2/V-s\n", + "mu_p=2100*10**-4##m**2/V-s\n", + "ni=2.5*10**13##per m**3\n", + "q=1.6*10**-19##Coulomb\n", + "sigma_p=1/rho_p##(ohm-m)**-1\n", + "NA=sigma_p/q/mu_p##m**3\n", + "sigma_n=1/rho_n##(ohm-m)**-1\n", + "ND=sigma_p/q/mu_n##m**3\n", + "T=27+273##K\n", + "VT=T/11600##V\n", + "Vj=VT*log(NA*ND/ni**2)##V\n", + "print \"Height of potential barrier = %0.4f V\"%Vj\n", + "#Anser in the book is wrong." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.7.1 Pg 2-27" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Diode current = 4.3 mA\n", + "Diode voltage = 0.65 Volts\n" + ] + } + ], + "source": [ + "#Given : \n", + "Vgamma=0.6##Volt\n", + "rf=12##ohm\n", + "V=5##Volts\n", + "R=1##kohm\n", + "IF=(V-Vgamma)/(R*1000+rf)##A\n", + "print \"Diode current = %0.1f mA\"%(IF*1000)\n", + "VF=Vgamma+IF*rf##volts\n", + "print \"Diode voltage = %0.2f Volts\"%VF" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.7.2 Pg 2-35" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Alternating component of voltage across RL, Vo(ac) = 0.1989 *sin(omega*t)\n", + "Total load voltage = 9.3 + 0.1989 *sin(omega*t)\n" + ] + } + ], + "source": [ + "#Given : \n", + "Vgamma=0.6##Volt\n", + "Rf=10##ohm\n", + "Eta=2#\n", + "Vm=0.2##Volts\n", + "Vdc=10##Volts\n", + "RL=1##kohm\n", + "IDQ=(Vdc-Vgamma)/(RL*1000+Rf)##A\n", + "VT=25*10**-3##Volts\n", + "rd=Eta*VT/IDQ##ohm\n", + "print \"Alternating component of voltage across RL, Vo(ac) = \",round((RL*1000/(RL*1000+rd)*Vm),4),\"*sin(omega*t)\"\n", + "Vo_DC=IDQ*RL*1000##Volts\n", + "print \"Total load voltage = \",round(Vo_DC,1),\"+\",round((RL*1000/(RL*1000+rd)*Vm),4),\"*sin(omega*t)\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.7.3 Pg 2-37" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Saturation value of voltage = 33.20 mV\n", + "Ratio of forward to reverse current = -15782.65\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import exp,log\n", + "#Given : \n", + "Eta=2##for Si diode\n", + "T=300##K\n", + "VT=T/11600##V\n", + "IbyIo=90/100#\n", + "#I=Io*(exp(V/Eta/VT)-1)\n", + "V=log(IbyIo+1)*Eta*VT##V\n", + "print \"Saturation value of voltage = %0.2f mV\"%(V*1000)\n", + "VF=0.5##Volts\n", + "VR=-0.5##Volts\n", + "IFbyIR=(exp(VF/Eta/VT)-1)/(exp(VR/Eta/VT)-1)##ratio\n", + "print \"Ratio of forward to reverse current = %0.2f\"%IFbyIR\n", + "#Answer in the book is wrong." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.7.4 Pg 2-37" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Saturation value of voltage = 33.20 mV\n", + "Ratio of forward to reverse current : -47.78 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log,exp\n", + "\n", + "#Given : \n", + "Eta=2##for Si diode\n", + "T=300##K\n", + "VT=T/11600##V\n", + "IbyIo=90/100#\n", + "#I=Io*(exp(V/Eta/VT)-1)\n", + "V=log(IbyIo+1)*Eta*VT##V\n", + "print \"Saturation value of voltage = %0.2f mV\"%(V*1000)\n", + "VF=0.2##Volts\n", + "VR=-0.2##Volts\n", + "IFbyIR=(exp(VF/Eta/VT)-1)/(exp(VR/Eta/VT)-1)##ratio\n", + "print \"Ratio of forward to reverse current : %0.2f \"%IFbyIR\n", + "#Answer in the book is wrong." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.9.1 Pg 2-61" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Reverse saturation current = 5.043 nA\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log,exp\n", + "\n", + "#Given : \n", + "IF=10##mA\n", + "VF=0.75##volts\n", + "T=27+273##K\n", + "Eta=2##for Si diode\n", + "VT=T/11600##V\n", + "Io=IF/(exp(VF/Eta/VT)-1)##mA\n", + "print \"Reverse saturation current = %0.3f nA\"%(Io*10**6)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.9.2 Pg -61" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Reverse saturation current = 91.66 nA\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log,exp\n", + "\n", + "#Given : \n", + "IF=10##mA\n", + "VF=0.3##Volts\n", + "T=27+273##K\n", + "Eta=1##for Ge diode\n", + "VT=T/11600##V\n", + "Io=IF/(exp(VF/Eta/VT)-1)##mA\n", + "print \"Reverse saturation current = %0.2f nA\"%(Io*10**6)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.9.3 Pg 2-61" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Forwad current = 0.1091 mA\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log,exp\n", + "\n", + "#Given : \n", + "Io=1*10**-9##A\n", + "T=27+273##K\n", + "VT=T/11600##V\n", + "VF=0.3##Volts\n", + "Eta=1##for Ge diode\n", + "IF=Io*(exp(VF/Eta/VT)-1)##mA\n", + "print \"Forwad current = %0.4f mA\"%(IF*10**3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.9.4 Pg 2-62" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Value of Eta : 1.12\n", + "Current, Io = 9.54 nA\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log,exp\n", + "\n", + "#Given : \n", + "T=27+273##K\n", + "V1=0.4##V\n", + "V2=0.42##V\n", + "I1=10##mA\n", + "I2=20##mA\n", + "VT=T/11600##V\n", + "Eta=1/log(I1/I2)*(V1-V2)/VT\n", + "print \"Value of Eta : %0.2f\"%Eta\n", + "Io=I1/(exp(V1/Eta/VT)-1)*10**-3##A\n", + "print \"Current, Io = %0.2f nA\"%(Io*10**9)\n", + "#Ans in the book is not accurate." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.9.5 Pg 2-63" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage across the diodes = 0.4628 V\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log,exp\n", + "\n", + "#Given : \n", + "Io1=10**-12##A\n", + "Io2=10**-10##A\n", + "I=2##mA\n", + "Eta=1##constant\n", + "T=27+273##K\n", + "VT=26/1000##V\n", + "#I=I1+I2\n", + "V=(log(I*10**-3/(Io1+Io2))+1)*Eta*VT##V\n", + "print \"Voltage across the diodes = %0.4f V\"%V\n", + "#Ans in the book is not accurate." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.9.6 Pg 2-64" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Source current = 23.68 micro Ampere\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log,exp\n", + "\n", + "#Given : \n", + "Io1=10*10**-9##A\n", + "Io2=10*10**-9##A\n", + "Eta=1.1##constant\n", + "T=25+273##K\n", + "V=0.2##V(assumed)\n", + "VT=T/11600##V\n", + "I1=Io1*(exp(V/Eta/VT)-1)##A\n", + "I2=Io2*(exp(V/Eta/VT)-1)##A\n", + "I=I1+I2##A\n", + "print \"Source current = %0.2f micro Ampere\"%(I*10**6)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.9.7 Pg 2-65" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage Vin = 1.823 V\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log,exp\n", + "\n", + "#Given : \n", + "Io=10**-13##A\n", + "T=27+273##K\n", + "Eta=1##constant\n", + "V=0.6##V\n", + "VT=26/1000##V\n", + "I3=Io*(exp(V/Eta/VT)-1)##A\n", + "R=1*1000##ohm\n", + "Ir=V/R##A\n", + "Itotal=I3+Ir##A\n", + "VD1=log(Itotal/Io)*Eta*VT##V\n", + "VD2=VD1##V\n", + "Vin=VD1+VD2+V##V\n", + "print \"Voltage Vin = %0.3f V\"%Vin" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.9.8 Pg 2-66" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Case(i) : Vb=9.8V\n", + "VD1 = 0.200 V\n", + "VD2 = 9.800 V\n", + "Case(ii) : Vb=10.2V\n", + "VD1 = 0.018 V\n", + "VD2 = 9.982 V\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log,exp\n", + "\n", + "#Given : \n", + "Vs=10##V\n", + "print \"Case(i) : Vb=9.8V\"\n", + "Vb=9.8##V\n", + "#D1 forward & D2 reverse biased: Breakdown D2\n", + "VD2=Vb##V\n", + "VD1=Vs-Vb##V\n", + "print \"VD1 = %0.3f V\"%VD1\n", + "print \"VD2 = %0.3f V\"%VD2\n", + "print \"Case(ii) : Vb=10.2V\"\n", + "Vb=10.2##V\n", + "#D1 forward & D2 reverse biased: none will be breakdown\n", + "VD2=Vb##V\n", + "#I=I0 so exp(V1/Eta/VT)-1=1\n", + "Eta=1##constant\n", + "VT=26/1000##V\n", + "VD1=log(1+1)*Eta*VT##V\n", + "VD2=Vs-VD1##V\n", + "print \"VD1 = %0.3f V\"%VD1\n", + "print \"VD2 = %0.3f V\"%VD2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.9.9 Pg 2-67" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage across diode D1 = 0.018 V\n", + "Voltage across diode D2 = 4.982 V\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log,exp\n", + "\n", + "#Given : \n", + "Vs=5##Volt\n", + "Eta=1##constant\n", + "VT=26/1000##V\n", + "#I=I0 so exp(V1/Eta/VT)-1=1\n", + "V1=log(1+1)*Eta*VT##Volt\n", + "V2=Vs-V1##Volt\n", + "print \"Voltage across diode D1 = %0.3f V\"%V1\n", + "print \"Voltage across diode D2 = %0.3f V\"%V2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.10.2 Pg 2-70" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature of junction = 287.28 degree K\n", + "Temperature of junction = 14.28 degree C\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log,exp\n", + "\n", + "#Given : \n", + "rho_n=10##ohm-cm\n", + "rho_p=3.5##ohm-cm\n", + "ni=1.5*10**10##per cm**3\n", + "Vj=0.56##volt\n", + "q=1.6*10**-19##Coulomb\n", + "mu_n=1500##cm**2/V-s\n", + "mu_p=500##cm**2/V-s\n", + "sigma_p=1/rho_p##(ohm-cm)**-1\n", + "NA=sigma_p/q/mu_p##per cm**3\n", + "sigma_n=1/rho_n##(ohm-cm)**-1\n", + "ND=sigma_n/q/mu_n##per cm**3\n", + "VT=Vj/log(NA*ND/ni**2)##V\n", + "T=11600*VT##K\n", + "print \"Temperature of junction = %0.2f degree K\"%T\n", + "t=T-273##degree C\n", + "print \"Temperature of junction = %0.2f degree C\"%t" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.11.1 Pg 2-75" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Reverse saturation current at 87 degree C = 648.69 nA\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log,exp\n", + "\n", + "#Given : \n", + "Io=10##nA\n", + "T1=27+273##K\n", + "T2=87+273##K\n", + "VT=T1/11600##V\n", + "Eta=2##for Si\n", + "m=1.5##for Si\n", + "VGO=-1.21##volt\n", + "K=Io*10**-9/T1**m/exp(VGO/Eta/VT)##constant\n", + "VT=T2/11600##V\n", + "Io2=K*T2**m*exp(VGO/Eta/VT)##A\n", + "print \"Reverse saturation current at 87 degree C = %0.2f nA\"%(Io2*10**9)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.11.2 Pg 2-76" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Factor by which current increases : 56.94 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log,exp\n", + "\n", + "#Given : \n", + "V=0.45##volt\n", + "Eta=2##for Si\n", + "T1=27+273##K\n", + "T2=125+273##K\n", + "VT1=T1/11600##V\n", + "VT2=T2/11600##V\n", + "I1BYIo1=exp(V/Eta/VT1)#\n", + "I2BYIo2=exp(V/Eta/VT2)#\n", + "m=1.5##for Si\n", + "VGO=1.21##volt\n", + "Io1BYIo2=(T1/T2)**m*exp(-VGO/Eta/VT1+VGO/Eta/VT2)##constant\n", + "I2BYI1=I2BYIo2/I1BYIo1/Io1BYIo2#\n", + "print \"Factor by which current increases : %0.2f \"%I2BYI1\n", + "#Answer is wrong in the textbook." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.11.3 Pg 2-78" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "% change = 2332.39 diode current\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log,exp\n", + "\n", + "#Given : \n", + "Io1=2##nA\n", + "T1=10+273##K\n", + "V=0.4##volt\n", + "VT1=T1/11600##V\n", + "m=1.5##for Si\n", + "Eta=2##for Si\n", + "VGO=-1.21##volt\n", + "K=Io1*10**-9/T1**m/exp(VGO/Eta/VT1)##constant\n", + "I1=Io1*10**-9*(exp(V/Eta/VT1)-1)##nA\n", + "T2=70+273##K\n", + "VT2=T2/11600##V\n", + "Io2=K*T2**m*(exp(VGO/Eta/VT2))##A\n", + "I2=Io2*(exp(V/Eta/VT2)-1)##nA\n", + "change=(I2-I1)/I1*100##%\n", + "print \"%% change = %0.2f diode current\"%change\n", + "#Answer is wrong in the textbook." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.11.4 Pg 2-79" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Part(i)\n", + "d(log(Io))/dt for Ge = 0.11 per degree C\n", + "d(log(Io))/dt for Si = 0.08 per degree C \n", + "Part(ii)\n", + "dV/dt for Si = -2.08 mV per degree C \n", + "dV/dt for Si = -2.29 mV per degree C \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log,exp\n", + "\n", + "#Given : \n", + "T=300##K\n", + "m_Si=1.5##for Si\n", + "m_Ge=1.5##for Ge\n", + "EGO_Si=1.21##Volt\n", + "EGO_Ge=0.785##Volt\n", + "Eta_Si=2#\n", + "Eta_Ge=1#\n", + "VT=26/1000##V\n", + "print \"Part(i)\"\n", + "d_logIoBYdt_Ge=m_Ge/T+EGO_Ge/(Eta_Ge*T*VT)##per degree C\n", + "print \"d(log(Io))/dt for Ge = %0.2f per degree C\"%d_logIoBYdt_Ge\n", + "d_logIoBYdt_Si=m_Si/T+EGO_Si/(Eta_Si*T*VT)##per degree C\n", + "print \"d(log(Io))/dt for Si = %0.2f per degree C \"%d_logIoBYdt_Si\n", + "print \"Part(ii)\"\n", + "V=0.2##Volt\n", + "dVBYdt_Ge=V/T-Eta_Ge*VT*d_logIoBYdt_Ge#\n", + "print \"dV/dt for Si = %0.2f mV per degree C \"%(dVBYdt_Ge*1000)\n", + "V=0.6##Volt\n", + "dVBYdt_Si=V/T-Eta_Si*VT*d_logIoBYdt_Si\n", + "print \"dV/dt for Si = %0.2f mV per degree C \"%(dVBYdt_Si*1000)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.12.1 Pg 2-85" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Width of depletion region = 2.30 micro meter \n", + "New value of Width of depletion region = 4.80 micro meter \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt\n", + "#Given : \n", + "NA=4*10**20##per m**3\n", + "Vj=0.2##Volt\n", + "V1=-1##Volts\n", + "V2=-5##Volts\n", + "epsilon_r=16##for Ge\n", + "epsilon_o=8.85*10**-12##permitivity\n", + "q=1.6*10**-19##Coulomb\n", + "W1=sqrt(2*epsilon_r*epsilon_o*(Vj-V1)/q/NA)##m\n", + "print \"Width of depletion region = %0.2f micro meter \"%(W1*10**6)\n", + "W2=sqrt(2*epsilon_r*epsilon_o*(Vj-V2)/q/NA)##m\n", + "print \"New value of Width of depletion region = %0.2f micro meter \"%(W2*10**6)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.12.2 Pg 2-86" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Transition capacitance = 49.16 pF \n", + "New value of Transition capacitance = 23.62 pF \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt\n", + "\n", + "#Given : \n", + "NA=4*10**20##per m**3\n", + "Vj=0.2##Volt\n", + "V1=-1##Volts\n", + "V2=-5##Volts\n", + "A=0.8*10**-6##m**2\n", + "epsilon_r=16##for Ge\n", + "epsilon_o=8.85*10**-12##permitivity\n", + "q=1.6*10**-19##Coulomb\n", + "W1=sqrt(2*epsilon_r*epsilon_o*(Vj-V1)/q/NA)##m\n", + "CT1=epsilon_r*epsilon_o*A/W1##\n", + "print \"Transition capacitance = %0.2f pF \"%(CT1*10**12)\n", + "W2=sqrt(2*epsilon_r*epsilon_o*(Vj-V2)/q/NA)##m\n", + "CT2=epsilon_r*epsilon_o*A/W2##\n", + "print \"New value of Transition capacitance = %0.2f pF \"%( CT2*10**12)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.12.3 Pg 2-87" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Width of depletion region = 7.76 micro meter\n", + "Transition capacitance = 18.26 pF\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt\n", + "\n", + "#Given : \n", + "NA=3*10**20##per m**3\n", + "Vj=0.2##Volt\n", + "V=-10##Volts\n", + "A=1*10**-6##m**2\n", + "epsilon_r=16##for Ge\n", + "epsilon_o=8.854*10**-12##permitivity\n", + "q=1.6*10**-19##Coulomb\n", + "W=sqrt(2*epsilon_r*epsilon_o*(Vj-V)/q/NA)##m\n", + "print \"Width of depletion region = %0.2f micro meter\"%(W*10**6)\n", + "CT=epsilon_r*epsilon_o*A/W##\n", + "print \"Transition capacitance = %0.2f pF\"%(CT*10**12)\n", + "#Answer is wrong in the textbook." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.12.4 Pg 2-88" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Barrier capacitance = 70.83 pF \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt\n", + "\n", + "#Given : \n", + "W=2*10**-4*10**-2##m\n", + "A=1*10**-6##m**2\n", + "epsilon_r=16##for Ge\n", + "epsilon_o=8.854*10**-12##permitivity\n", + "q=1.6*10**-19##Coulomb\n", + "CT=epsilon_r*epsilon_o*A/W##\n", + "print \"Barrier capacitance = %0.2f pF \"%(CT*10**12)\n", + "#Answer is wrong in the textbook." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.12.5 Pg 2-88" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Diameter = 1397.53 micro meter\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt,pi\n", + "#Given : \n", + "Vj=0.5##Volt\n", + "V=-4.5##Volt\n", + "rho_p=5*10**-2##ohm-m\n", + "epsilon_r=12##for Si\n", + "epsilon_o=8.854*10**-12##permitivity\n", + "q=1.6*10**-19##Coulomb\n", + "CT=100*10**-12##F\n", + "mu_p=500*10**-4##m**2/V-s\n", + "sigma_p=1/rho_p##(ohm-m)**-1\n", + "NA=sigma_p/q/mu_p##per m**3\n", + "W=sqrt(2*epsilon_r*epsilon_o*(Vj-V)/q/NA)##m\n", + "A=CT*W/(epsilon_r*epsilon_o)##\n", + "r=sqrt(A/pi)##m\n", + "D=2*r##m\n", + "print \"Diameter = %0.2f micro meter\"%(D*10**6)\n", + "#Answer is wrong = %0.2f the textbook. Sqrt is not taken while calculatng W value and also other mistakes." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.12.6 Pg 2-90" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Value of reverse voltage = 33.38 mV\n", + "Ratio of forward bias current to reverse saturation current = -46.81 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt,log,exp\n", + "#Given : \n", + "Eta=2##for Si\n", + "T=300##K\n", + "VT=26/1000##V\n", + "IbyIo=0.9#\n", + "#part (i)\n", + "V=log(IbyIo+1)*Eta*VT##Volt\n", + "print \"Value of reverse voltage = %0.2f mV\"%(V*1000)\n", + "#part (ii)\n", + "VF=0.2##Volt\n", + "VR=-0.2##Volt\n", + "IFbyIR=(exp(VF/Eta/VT)-1)/(exp(VR/Eta/VT)-1)#\n", + "print \"Ratio of forward bias current to reverse saturation current = %0.2f \"%IFbyIR\n", + "#Answer is wrong in the textbook." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.12.7 Pg 2-91" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "For R=10 kohm\n", + "I1 = 9.96 mA\n", + "I2 = 0.00 mA\n", + "For R=1 kohm\n", + "I1 = -2883.74 mA\n", + "I2 = 56.54 mA\n" + ] + } + ], + "source": [ + "from numpy import mat, linalg\n", + "#Given : \n", + "Vs=100##V\n", + "Rf1=20##ohm\n", + "Vgamma1=0.2##Volts\n", + "Rf2=15##ohm\n", + "Vgamma2=0.6##Volts\n", + "Vb_Ge=0.2##Volts\n", + "Vb_Si=0.6##Volts\n", + "R1=10*10**3##ohm\n", + "R2=1*10**3##ohm\n", + "#Case(i)\n", + "Imax=Vs/R1##A\n", + "#D1 ON & D2 off\n", + "V=Vb_Ge+Rf1*Imax##Volt\n", + "#D2 off as V<Vb_Si\n", + "I2=0##A\n", + "I1=(Vs-V)/(R1+Rf1)##A\n", + "print \"For R=10 kohm\"\n", + "print \"I1 = %0.2f mA\"%(I1*1000)\n", + "print \"I2 = %0.2f mA\"%I2\n", + "#Case(ii)\n", + "R=R2##ohm#D1 & D2 ON \n", + "#V=Vb_Ge+Rf1*I1#V=Vb_Si+Rf2*I2\n", + "#V=Vs-I*R#V=Vs-(I1+I2)*R\n", + "#20*I1-15*I2=Vb_Si-Vb_Ge\n", + "#1020*I1+1000*I2=99.8\n", + "A=mat([[20, 1020],[-Rf2, R]])#\n", + "B=mat([[Vb_Ge-Vb_Ge],[Vs-Vb_Ge]])#\n", + "X = linalg.solve(A,B)\n", + "I1=X[0]*1000##mA\n", + "I2=X[1]*1000##mA\n", + "print \"For R=1 kohm\"\n", + "print \"I1 = %0.2f mA\"%I1\n", + "print \"I2 = %0.2f mA\"%I2\n", + "#Answer for 2nd part is not accurate in the book." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 2.12.8 Pg 2-93" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "For R=1 kohm\n", + "DC source = 2499.50 Volts\n" + ] + } + ], + "source": [ + "import numpy as np\n", + "\n", + "#Given : \n", + "Rf=10##ohm\n", + "Vgamma=0.5##Volt\n", + "RL=20##ohm\n", + "V=3##Volt\n", + "#Loop 1: 75*I1-50*I=V-Vgamma\n", + "#Loop 2: -50*I1+80*I=-Vgamma\n", + "\n", + "A=np.mat([[75 ,-50],[-50, 80]])#\n", + "B=np.mat([[V-Vgamma], [-Vgamma]])#\n", + "X = np.linalg.solve(A,B)\n", + "I1=X[0]*1000##mA\n", + "I2=X[1]*1000##mA\n", + "print \"For R=1 kohm\"\n", + "Vx=-Vgamma+50*I1##Volt\n", + "print \"DC source = %0.2f Volts\"%Vx[0,0]\n", + "#Answer is wrong in the textbook." + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Advance_Semiconductor_Devices_by_K._C._Nandi/chapter5_1.ipynb b/Advance_Semiconductor_Devices_by_K._C._Nandi/chapter5_1.ipynb new file mode 100644 index 00000000..ca979975 --- /dev/null +++ b/Advance_Semiconductor_Devices_by_K._C._Nandi/chapter5_1.ipynb @@ -0,0 +1,72 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter-5 Metal semiconductor field effect transistors" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 5.6.1 Pg 5-22" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Kn = 0.302 mA/V**2\n", + "Current = 0.15 mA\n" + ] + } + ], + "source": [ + "#Given : \n", + "VTN=0.7##V\n", + "W=45##micro m\n", + "L=4##micro m\n", + "mu_n=700##cm**2/V-s\n", + "t_ox=450##Angstrum\n", + "epsilon_ox=3.9*8.85*10**-14##F/cm\n", + "VGS=2*VTN##V\n", + "Kn=(W*10**-4)*mu_n*epsilon_ox/(2*(L*10**-4)*(t_ox*10**-8))##A/V**2\n", + "Kn=Kn*10**3##mA/V**2\n", + "print \"Kn = %0.3f mA/V**2\"%Kn\n", + "ID=Kn*(VGS-VTN)**2##A\n", + "print \"Current = %0.2f mA\"%ID\n", + "#Answer is wrong in the book. Calculation mistake whle calculating value for Kn." + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Advance_Semiconductor_Devices_by_K._C._Nandi/screenshots/KC_econ1_chapter1_1.png b/Advance_Semiconductor_Devices_by_K._C._Nandi/screenshots/KC_econ1_chapter1_1.png Binary files differnew file mode 100644 index 00000000..63a8c57c --- /dev/null +++ b/Advance_Semiconductor_Devices_by_K._C._Nandi/screenshots/KC_econ1_chapter1_1.png diff --git a/Advance_Semiconductor_Devices_by_K._C._Nandi/screenshots/KC_econ_ch1_1.png b/Advance_Semiconductor_Devices_by_K._C._Nandi/screenshots/KC_econ_ch1_1.png Binary files differnew file mode 100644 index 00000000..5cd8e98d --- /dev/null +++ b/Advance_Semiconductor_Devices_by_K._C._Nandi/screenshots/KC_econ_ch1_1.png diff --git a/Advance_Semiconductor_Devices_by_K._C._Nandi/screenshots/KC_percentChangeinDiodeCurrent_chapter2_1.png b/Advance_Semiconductor_Devices_by_K._C._Nandi/screenshots/KC_percentChangeinDiodeCurrent_chapter2_1.png Binary files differnew file mode 100644 index 00000000..a378a6f1 --- /dev/null +++ b/Advance_Semiconductor_Devices_by_K._C._Nandi/screenshots/KC_percentChangeinDiodeCurrent_chapter2_1.png diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch15_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch15_1.ipynb deleted file mode 100644 index 1c77fdc3..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch15_1.ipynb +++ /dev/null @@ -1,179 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 15 : Electric forces and electric fields" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 15.1 Page No : 502" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The attractive force = 8.19e-08 N\n", - "The gravitational force = 3.61e-47 N\n" - ] - } - ], - "source": [ - "k_e=8.99*10**9\n", - "e=1.6*10**-19\n", - "r=5.3*10**-11\n", - "F_e= (k_e*e*e)/(r*r)\n", - "print \"The attractive force = %0.2e N\"%F_e\n", - "G=6.67*10**-11\n", - "m_e=9.11*10**-31\n", - "m_p=1.67*10**-27\n", - "F_g=(G*m_e*m_p)/(r*r)\n", - "print \"The gravitational force = %0.2e N\"%F_g" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 15.2 Page No : 503" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The force = 5.62e-09 N\n", - "The force = 1.08e-08 N\n" - ] - } - ], - "source": [ - "k_e=8.99*10**9 #N.m**2/c**2\n", - "q2=2*10**-9# = %0.2f c\n", - "q3=5*10**-9# = %0.2f c\n", - "r1=4#in m\n", - "F_23=(q2*q3*k_e)/(r1*r1)\n", - "print \"The force = %0.2e N\"%F_23\n", - "q1=6*10**-9\n", - "r2=5#in m\n", - "F_13=(q1*q3*k_e)/(r2*r2)\n", - "print \"The force = %0.2e N\"%F_13" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 15.4 Page No: 507" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The magnitude of force = 3.20e-15 N\n" - ] - } - ], - "source": [ - "q=1.6*10**-19#in c\n", - "E=2*10**4# = %0.2f N/C\n", - "F=q*E\n", - "print \"The magnitude of force = %0.2e N\"%F" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 15.5 Page No: 509" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Magnitude of E1 = 3.93e+05 N/C\n", - "Magnitude of E2 = 1.80e+05 N/C\n", - "Magnitude in x direction = 1.80e+05 N/C\n", - "Magnitude in y direction = 2.49e+05 N/C\n", - "Angle = 54.17 degree\n" - ] - } - ], - "source": [ - "from math import degrees, atan\n", - "k_e=8.99*10**9 #N.m**2/c**2\n", - "q1=7*10**-6# = %0.2f C\n", - "q2=5*10**-6#in C\n", - "r1=0.4\n", - "r2=0.5\n", - "E1=(k_e*q1)/(r1**2)\n", - "E2=(k_e*q2)/(r2**2)\n", - "Ex=(k_e*q2)/(r2**2)\n", - "print \"Magnitude of E1 = %0.2e N/C\"%E1\n", - "print \"Magnitude of E2 = %0.2e N/C\"%E2\n", - "print \"Magnitude in x direction = %0.2e N/C\"%Ex\n", - "Ey=(3.93*10**5)+(-1.44*10**5)\n", - "print \"Magnitude in y direction = %0.2e N/C\"%Ey\n", - "phi=degrees(atan(Ey/Ex))\n", - "print \"Angle = %0.2f degree\"%phi\n", - "#Answer given in the book is wrong" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch15_2.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch15_2.ipynb deleted file mode 100644 index 1c77fdc3..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch15_2.ipynb +++ /dev/null @@ -1,179 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 15 : Electric forces and electric fields" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 15.1 Page No : 502" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The attractive force = 8.19e-08 N\n", - "The gravitational force = 3.61e-47 N\n" - ] - } - ], - "source": [ - "k_e=8.99*10**9\n", - "e=1.6*10**-19\n", - "r=5.3*10**-11\n", - "F_e= (k_e*e*e)/(r*r)\n", - "print \"The attractive force = %0.2e N\"%F_e\n", - "G=6.67*10**-11\n", - "m_e=9.11*10**-31\n", - "m_p=1.67*10**-27\n", - "F_g=(G*m_e*m_p)/(r*r)\n", - "print \"The gravitational force = %0.2e N\"%F_g" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 15.2 Page No : 503" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The force = 5.62e-09 N\n", - "The force = 1.08e-08 N\n" - ] - } - ], - "source": [ - "k_e=8.99*10**9 #N.m**2/c**2\n", - "q2=2*10**-9# = %0.2f c\n", - "q3=5*10**-9# = %0.2f c\n", - "r1=4#in m\n", - "F_23=(q2*q3*k_e)/(r1*r1)\n", - "print \"The force = %0.2e N\"%F_23\n", - "q1=6*10**-9\n", - "r2=5#in m\n", - "F_13=(q1*q3*k_e)/(r2*r2)\n", - "print \"The force = %0.2e N\"%F_13" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 15.4 Page No: 507" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The magnitude of force = 3.20e-15 N\n" - ] - } - ], - "source": [ - "q=1.6*10**-19#in c\n", - "E=2*10**4# = %0.2f N/C\n", - "F=q*E\n", - "print \"The magnitude of force = %0.2e N\"%F" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 15.5 Page No: 509" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Magnitude of E1 = 3.93e+05 N/C\n", - "Magnitude of E2 = 1.80e+05 N/C\n", - "Magnitude in x direction = 1.80e+05 N/C\n", - "Magnitude in y direction = 2.49e+05 N/C\n", - "Angle = 54.17 degree\n" - ] - } - ], - "source": [ - "from math import degrees, atan\n", - "k_e=8.99*10**9 #N.m**2/c**2\n", - "q1=7*10**-6# = %0.2f C\n", - "q2=5*10**-6#in C\n", - "r1=0.4\n", - "r2=0.5\n", - "E1=(k_e*q1)/(r1**2)\n", - "E2=(k_e*q2)/(r2**2)\n", - "Ex=(k_e*q2)/(r2**2)\n", - "print \"Magnitude of E1 = %0.2e N/C\"%E1\n", - "print \"Magnitude of E2 = %0.2e N/C\"%E2\n", - "print \"Magnitude in x direction = %0.2e N/C\"%Ex\n", - "Ey=(3.93*10**5)+(-1.44*10**5)\n", - "print \"Magnitude in y direction = %0.2e N/C\"%Ey\n", - "phi=degrees(atan(Ey/Ex))\n", - "print \"Angle = %0.2f degree\"%phi\n", - "#Answer given in the book is wrong" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch15_3.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch15_3.ipynb deleted file mode 100644 index 1c77fdc3..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch15_3.ipynb +++ /dev/null @@ -1,179 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 15 : Electric forces and electric fields" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 15.1 Page No : 502" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The attractive force = 8.19e-08 N\n", - "The gravitational force = 3.61e-47 N\n" - ] - } - ], - "source": [ - "k_e=8.99*10**9\n", - "e=1.6*10**-19\n", - "r=5.3*10**-11\n", - "F_e= (k_e*e*e)/(r*r)\n", - "print \"The attractive force = %0.2e N\"%F_e\n", - "G=6.67*10**-11\n", - "m_e=9.11*10**-31\n", - "m_p=1.67*10**-27\n", - "F_g=(G*m_e*m_p)/(r*r)\n", - "print \"The gravitational force = %0.2e N\"%F_g" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 15.2 Page No : 503" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The force = 5.62e-09 N\n", - "The force = 1.08e-08 N\n" - ] - } - ], - "source": [ - "k_e=8.99*10**9 #N.m**2/c**2\n", - "q2=2*10**-9# = %0.2f c\n", - "q3=5*10**-9# = %0.2f c\n", - "r1=4#in m\n", - "F_23=(q2*q3*k_e)/(r1*r1)\n", - "print \"The force = %0.2e N\"%F_23\n", - "q1=6*10**-9\n", - "r2=5#in m\n", - "F_13=(q1*q3*k_e)/(r2*r2)\n", - "print \"The force = %0.2e N\"%F_13" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 15.4 Page No: 507" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The magnitude of force = 3.20e-15 N\n" - ] - } - ], - "source": [ - "q=1.6*10**-19#in c\n", - "E=2*10**4# = %0.2f N/C\n", - "F=q*E\n", - "print \"The magnitude of force = %0.2e N\"%F" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 15.5 Page No: 509" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Magnitude of E1 = 3.93e+05 N/C\n", - "Magnitude of E2 = 1.80e+05 N/C\n", - "Magnitude in x direction = 1.80e+05 N/C\n", - "Magnitude in y direction = 2.49e+05 N/C\n", - "Angle = 54.17 degree\n" - ] - } - ], - "source": [ - "from math import degrees, atan\n", - "k_e=8.99*10**9 #N.m**2/c**2\n", - "q1=7*10**-6# = %0.2f C\n", - "q2=5*10**-6#in C\n", - "r1=0.4\n", - "r2=0.5\n", - "E1=(k_e*q1)/(r1**2)\n", - "E2=(k_e*q2)/(r2**2)\n", - "Ex=(k_e*q2)/(r2**2)\n", - "print \"Magnitude of E1 = %0.2e N/C\"%E1\n", - "print \"Magnitude of E2 = %0.2e N/C\"%E2\n", - "print \"Magnitude in x direction = %0.2e N/C\"%Ex\n", - "Ey=(3.93*10**5)+(-1.44*10**5)\n", - "print \"Magnitude in y direction = %0.2e N/C\"%Ey\n", - "phi=degrees(atan(Ey/Ex))\n", - "print \"Angle = %0.2f degree\"%phi\n", - "#Answer given in the book is wrong" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch16_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch16_1.ipynb deleted file mode 100644 index 0e794d3d..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch16_1.ipynb +++ /dev/null @@ -1,365 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 16 : Electrical Energy & Capacitance" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.1 Page No : 533" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The value of E = 4000.00 v/m\n" - ] - } - ], - "source": [ - "v_bminusv_a=-12\n", - "d=0.3*10**-2#in m\n", - "E=-(v_bminusv_a)/d\n", - "print \"The value of E = %0.2f v/m\"%E" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.2 Page No : 533" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "solution a\n", - "Electric potential from A to B = -40000.00 V\n", - "solution b\n", - "Change in electric potential = -0.00 joules\n", - "velocity = 2768514.16 m/s\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "print \"solution a\"\n", - "E=8*10**4#in V/m\n", - "d=0.5#in m\n", - "delta_V=-E*d\n", - "print \"Electric potential from A to B = %0.2f V\"%delta_V\n", - "print \"solution b\"\n", - "q=1.6*10**-19#in C\n", - "delta_PE=q*delta_V\n", - "print \"Change in electric potential = %0.2f joules\"%delta_PE\n", - "m_p=1.67*10**-27#in kg\n", - "vf=sqrt((2*-delta_PE)/m_p)\n", - "print \"velocity = %0.2f m/s\"%vf" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.3 Page No: 534" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Magnitude of V1 = 112375.00 v\n", - "Magnitude of V2 = -35960.00 v\n", - "solution b\n", - "Magnitude of Vp = 76415.00 v\n", - "work done = 0.31 Joule\n" - ] - } - ], - "source": [ - "k_e=8.99*10**9 #N.m**2/c**2\n", - "q1=5*10**-6# in C\n", - "q2=-2*10**-6#in C\n", - "r1=0.4\n", - "r2=0.5\n", - "V1=(k_e*q1)/(r1)\n", - "V2=(k_e*q2)/(r2)\n", - "print \"Solution a\"\n", - "print \"Magnitude of V1 = %0.2f v\"%V1\n", - "print \"Magnitude of V2 = %0.2f v\"%V2\n", - "print \"solution b\"\n", - "vp=V1+V2\n", - "print \"Magnitude of Vp = %0.2f v\"%vp\n", - "q3=4*10**-6#in C\n", - "w=vp*q3\n", - "print \"work done = %0.2f Joule\"%w" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.4 Page No: 535" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Capacitance = 1.77e-12 farad\n" - ] - } - ], - "source": [ - "e0=8.85*10**-12#in c2/N.m2\n", - "A=2*10**-4#in m2\n", - "d=1*10**-3#in m\n", - "c=(e0*A)/d\n", - "print \"Capacitance = %0.2e farad\"%c" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.5 Page No : 535" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "capacitance = 4.50e-05 farad\n", - "voltage between battery = 2.16e-04 c\n" - ] - } - ], - "source": [ - "c1=3*10**-6\n", - "c2=6*10**-6\n", - "c3=12*10**-6\n", - "c4=24*10**-6\n", - "delta_v=18\n", - "c_eq=c1+c2+c3+c4\n", - "print \"capacitance = %0.2e farad\"%c_eq\n", - "q=delta_v*c3\n", - "print \"voltage between battery = %0.2e c\"%q" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.6 Page No : 536" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "solution a\n", - "capacitance = 1.60e-06 farad\n", - "solution b\n", - "voltage between battery = 2.88e-05 c\n" - ] - } - ], - "source": [ - "c1=3*10**-6\n", - "c2=6*10**-6\n", - "c3=12*10**-6\n", - "c4=24*10**-6\n", - "delta_v=18\n", - "print \"solution a\"\n", - "c_eq=1/((1/c1)+(1/c2)+(1/c3)+(1/c4))\n", - "print \"capacitance = %0.2e farad\"%c_eq\n", - "q=delta_v*c_eq\n", - "print \"solution b\"\n", - "print \"voltage between battery = %0.2e c\"%q" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.7 Page No: 536" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "solution a\n", - "capacitance = 2.00e-06 farad\n" - ] - } - ], - "source": [ - "c1=4*10**-6\n", - "c2=4*10**-6\n", - "print \"solution a\"\n", - "c_eq=1/((1/c1)+(1/c2))\n", - "print \"capacitance = %0.2e farad\"%c_eq" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.8 Page No: 537" - ] - }, - { - "cell_type": "code", - "execution_count": 8, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "solution a\n", - "Energy stored = 4671 volt\n", - "solution b\n", - "power = 240000 watt\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "Energy=1.2*10**3#in J\n", - "c=1.1*10**-4#in f\n", - "delta_v=sqrt((2*Energy)/c)\n", - "print \"solution a\"\n", - "print \"Energy stored = %0.f volt\"%delta_v\n", - "print \"solution b\"\n", - "Energy_deliverd=600#in j\n", - "delta_t=2.5*10**-3#in s\n", - "p=(Energy_deliverd)/delta_t\n", - "print \"power = %0.f watt\"%p" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.9 Page No: 538" - ] - }, - { - "cell_type": "code", - "execution_count": 9, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "solution a\n", - "Capacitance = 1.96e-11 farad\n", - "solution b\n", - "Voltage = 16000.0 volt\n", - "Maximum charge = 3.14e-07 columb\n" - ] - } - ], - "source": [ - "k=3.7\n", - "e0=8.85*10**-12#in c2/N.m2\n", - "A=6*10**-4#in m2\n", - "d=1*10**-3#in m\n", - "c=(k*e0*A)/d\n", - "print \"solution a\"\n", - "print \"Capacitance = %0.2e farad\"%c\n", - "print \"solution b\"\n", - "E_max=16*10**6#in v/m\n", - "delta_v_max=E_max*d\n", - "print \"Voltage = %0.1f volt\"%delta_v_max\n", - "Q_max=delta_v_max*c\n", - "print \"Maximum charge = %0.2e columb\"%Q_max" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch16_2.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch16_2.ipynb deleted file mode 100644 index 0e794d3d..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch16_2.ipynb +++ /dev/null @@ -1,365 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 16 : Electrical Energy & Capacitance" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.1 Page No : 533" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The value of E = 4000.00 v/m\n" - ] - } - ], - "source": [ - "v_bminusv_a=-12\n", - "d=0.3*10**-2#in m\n", - "E=-(v_bminusv_a)/d\n", - "print \"The value of E = %0.2f v/m\"%E" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.2 Page No : 533" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "solution a\n", - "Electric potential from A to B = -40000.00 V\n", - "solution b\n", - "Change in electric potential = -0.00 joules\n", - "velocity = 2768514.16 m/s\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "print \"solution a\"\n", - "E=8*10**4#in V/m\n", - "d=0.5#in m\n", - "delta_V=-E*d\n", - "print \"Electric potential from A to B = %0.2f V\"%delta_V\n", - "print \"solution b\"\n", - "q=1.6*10**-19#in C\n", - "delta_PE=q*delta_V\n", - "print \"Change in electric potential = %0.2f joules\"%delta_PE\n", - "m_p=1.67*10**-27#in kg\n", - "vf=sqrt((2*-delta_PE)/m_p)\n", - "print \"velocity = %0.2f m/s\"%vf" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.3 Page No: 534" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Magnitude of V1 = 112375.00 v\n", - "Magnitude of V2 = -35960.00 v\n", - "solution b\n", - "Magnitude of Vp = 76415.00 v\n", - "work done = 0.31 Joule\n" - ] - } - ], - "source": [ - "k_e=8.99*10**9 #N.m**2/c**2\n", - "q1=5*10**-6# in C\n", - "q2=-2*10**-6#in C\n", - "r1=0.4\n", - "r2=0.5\n", - "V1=(k_e*q1)/(r1)\n", - "V2=(k_e*q2)/(r2)\n", - "print \"Solution a\"\n", - "print \"Magnitude of V1 = %0.2f v\"%V1\n", - "print \"Magnitude of V2 = %0.2f v\"%V2\n", - "print \"solution b\"\n", - "vp=V1+V2\n", - "print \"Magnitude of Vp = %0.2f v\"%vp\n", - "q3=4*10**-6#in C\n", - "w=vp*q3\n", - "print \"work done = %0.2f Joule\"%w" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.4 Page No: 535" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Capacitance = 1.77e-12 farad\n" - ] - } - ], - "source": [ - "e0=8.85*10**-12#in c2/N.m2\n", - "A=2*10**-4#in m2\n", - "d=1*10**-3#in m\n", - "c=(e0*A)/d\n", - "print \"Capacitance = %0.2e farad\"%c" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.5 Page No : 535" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "capacitance = 4.50e-05 farad\n", - "voltage between battery = 2.16e-04 c\n" - ] - } - ], - "source": [ - "c1=3*10**-6\n", - "c2=6*10**-6\n", - "c3=12*10**-6\n", - "c4=24*10**-6\n", - "delta_v=18\n", - "c_eq=c1+c2+c3+c4\n", - "print \"capacitance = %0.2e farad\"%c_eq\n", - "q=delta_v*c3\n", - "print \"voltage between battery = %0.2e c\"%q" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.6 Page No : 536" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "solution a\n", - "capacitance = 1.60e-06 farad\n", - "solution b\n", - "voltage between battery = 2.88e-05 c\n" - ] - } - ], - "source": [ - "c1=3*10**-6\n", - "c2=6*10**-6\n", - "c3=12*10**-6\n", - "c4=24*10**-6\n", - "delta_v=18\n", - "print \"solution a\"\n", - "c_eq=1/((1/c1)+(1/c2)+(1/c3)+(1/c4))\n", - "print \"capacitance = %0.2e farad\"%c_eq\n", - "q=delta_v*c_eq\n", - "print \"solution b\"\n", - "print \"voltage between battery = %0.2e c\"%q" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.7 Page No: 536" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "solution a\n", - "capacitance = 2.00e-06 farad\n" - ] - } - ], - "source": [ - "c1=4*10**-6\n", - "c2=4*10**-6\n", - "print \"solution a\"\n", - "c_eq=1/((1/c1)+(1/c2))\n", - "print \"capacitance = %0.2e farad\"%c_eq" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.8 Page No: 537" - ] - }, - { - "cell_type": "code", - "execution_count": 8, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "solution a\n", - "Energy stored = 4671 volt\n", - "solution b\n", - "power = 240000 watt\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "Energy=1.2*10**3#in J\n", - "c=1.1*10**-4#in f\n", - "delta_v=sqrt((2*Energy)/c)\n", - "print \"solution a\"\n", - "print \"Energy stored = %0.f volt\"%delta_v\n", - "print \"solution b\"\n", - "Energy_deliverd=600#in j\n", - "delta_t=2.5*10**-3#in s\n", - "p=(Energy_deliverd)/delta_t\n", - "print \"power = %0.f watt\"%p" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.9 Page No: 538" - ] - }, - { - "cell_type": "code", - "execution_count": 9, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "solution a\n", - "Capacitance = 1.96e-11 farad\n", - "solution b\n", - "Voltage = 16000.0 volt\n", - "Maximum charge = 3.14e-07 columb\n" - ] - } - ], - "source": [ - "k=3.7\n", - "e0=8.85*10**-12#in c2/N.m2\n", - "A=6*10**-4#in m2\n", - "d=1*10**-3#in m\n", - "c=(k*e0*A)/d\n", - "print \"solution a\"\n", - "print \"Capacitance = %0.2e farad\"%c\n", - "print \"solution b\"\n", - "E_max=16*10**6#in v/m\n", - "delta_v_max=E_max*d\n", - "print \"Voltage = %0.1f volt\"%delta_v_max\n", - "Q_max=delta_v_max*c\n", - "print \"Maximum charge = %0.2e columb\"%Q_max" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch16_3.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch16_3.ipynb deleted file mode 100644 index 0e794d3d..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch16_3.ipynb +++ /dev/null @@ -1,365 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 16 : Electrical Energy & Capacitance" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.1 Page No : 533" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The value of E = 4000.00 v/m\n" - ] - } - ], - "source": [ - "v_bminusv_a=-12\n", - "d=0.3*10**-2#in m\n", - "E=-(v_bminusv_a)/d\n", - "print \"The value of E = %0.2f v/m\"%E" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.2 Page No : 533" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "solution a\n", - "Electric potential from A to B = -40000.00 V\n", - "solution b\n", - "Change in electric potential = -0.00 joules\n", - "velocity = 2768514.16 m/s\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "print \"solution a\"\n", - "E=8*10**4#in V/m\n", - "d=0.5#in m\n", - "delta_V=-E*d\n", - "print \"Electric potential from A to B = %0.2f V\"%delta_V\n", - "print \"solution b\"\n", - "q=1.6*10**-19#in C\n", - "delta_PE=q*delta_V\n", - "print \"Change in electric potential = %0.2f joules\"%delta_PE\n", - "m_p=1.67*10**-27#in kg\n", - "vf=sqrt((2*-delta_PE)/m_p)\n", - "print \"velocity = %0.2f m/s\"%vf" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.3 Page No: 534" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Magnitude of V1 = 112375.00 v\n", - "Magnitude of V2 = -35960.00 v\n", - "solution b\n", - "Magnitude of Vp = 76415.00 v\n", - "work done = 0.31 Joule\n" - ] - } - ], - "source": [ - "k_e=8.99*10**9 #N.m**2/c**2\n", - "q1=5*10**-6# in C\n", - "q2=-2*10**-6#in C\n", - "r1=0.4\n", - "r2=0.5\n", - "V1=(k_e*q1)/(r1)\n", - "V2=(k_e*q2)/(r2)\n", - "print \"Solution a\"\n", - "print \"Magnitude of V1 = %0.2f v\"%V1\n", - "print \"Magnitude of V2 = %0.2f v\"%V2\n", - "print \"solution b\"\n", - "vp=V1+V2\n", - "print \"Magnitude of Vp = %0.2f v\"%vp\n", - "q3=4*10**-6#in C\n", - "w=vp*q3\n", - "print \"work done = %0.2f Joule\"%w" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.4 Page No: 535" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Capacitance = 1.77e-12 farad\n" - ] - } - ], - "source": [ - "e0=8.85*10**-12#in c2/N.m2\n", - "A=2*10**-4#in m2\n", - "d=1*10**-3#in m\n", - "c=(e0*A)/d\n", - "print \"Capacitance = %0.2e farad\"%c" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.5 Page No : 535" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "capacitance = 4.50e-05 farad\n", - "voltage between battery = 2.16e-04 c\n" - ] - } - ], - "source": [ - "c1=3*10**-6\n", - "c2=6*10**-6\n", - "c3=12*10**-6\n", - "c4=24*10**-6\n", - "delta_v=18\n", - "c_eq=c1+c2+c3+c4\n", - "print \"capacitance = %0.2e farad\"%c_eq\n", - "q=delta_v*c3\n", - "print \"voltage between battery = %0.2e c\"%q" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.6 Page No : 536" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "solution a\n", - "capacitance = 1.60e-06 farad\n", - "solution b\n", - "voltage between battery = 2.88e-05 c\n" - ] - } - ], - "source": [ - "c1=3*10**-6\n", - "c2=6*10**-6\n", - "c3=12*10**-6\n", - "c4=24*10**-6\n", - "delta_v=18\n", - "print \"solution a\"\n", - "c_eq=1/((1/c1)+(1/c2)+(1/c3)+(1/c4))\n", - "print \"capacitance = %0.2e farad\"%c_eq\n", - "q=delta_v*c_eq\n", - "print \"solution b\"\n", - "print \"voltage between battery = %0.2e c\"%q" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.7 Page No: 536" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "solution a\n", - "capacitance = 2.00e-06 farad\n" - ] - } - ], - "source": [ - "c1=4*10**-6\n", - "c2=4*10**-6\n", - "print \"solution a\"\n", - "c_eq=1/((1/c1)+(1/c2))\n", - "print \"capacitance = %0.2e farad\"%c_eq" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.8 Page No: 537" - ] - }, - { - "cell_type": "code", - "execution_count": 8, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "solution a\n", - "Energy stored = 4671 volt\n", - "solution b\n", - "power = 240000 watt\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "Energy=1.2*10**3#in J\n", - "c=1.1*10**-4#in f\n", - "delta_v=sqrt((2*Energy)/c)\n", - "print \"solution a\"\n", - "print \"Energy stored = %0.f volt\"%delta_v\n", - "print \"solution b\"\n", - "Energy_deliverd=600#in j\n", - "delta_t=2.5*10**-3#in s\n", - "p=(Energy_deliverd)/delta_t\n", - "print \"power = %0.f watt\"%p" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 16.9 Page No: 538" - ] - }, - { - "cell_type": "code", - "execution_count": 9, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "solution a\n", - "Capacitance = 1.96e-11 farad\n", - "solution b\n", - "Voltage = 16000.0 volt\n", - "Maximum charge = 3.14e-07 columb\n" - ] - } - ], - "source": [ - "k=3.7\n", - "e0=8.85*10**-12#in c2/N.m2\n", - "A=6*10**-4#in m2\n", - "d=1*10**-3#in m\n", - "c=(k*e0*A)/d\n", - "print \"solution a\"\n", - "print \"Capacitance = %0.2e farad\"%c\n", - "print \"solution b\"\n", - "E_max=16*10**6#in v/m\n", - "delta_v_max=E_max*d\n", - "print \"Voltage = %0.1f volt\"%delta_v_max\n", - "Q_max=delta_v_max*c\n", - "print \"Maximum charge = %0.2e columb\"%Q_max" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch17_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch17_1.ipynb deleted file mode 100644 index d0eee390..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch17_1.ipynb +++ /dev/null @@ -1,311 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 17 : Current and resistance" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 17.1 Page No: 571" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "solution a : \n", - "Current = 0.83 Amp\n", - "solution b : \n", - "Number of electrons = 0.84 C\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "print \"solution a : \"\n", - "delta_q=1.67 # in c\n", - "delta_t=2 # in s\n", - "I=delta_q/delta_t\n", - "print \"Current = %0.2f Amp\"%I\n", - "print \"solution b : \"\n", - "N=5.22*10**18\n", - "N_q=(1.6*10**-19)*N\n", - "\n", - "print \"Number of electrons = %0.2f C\"%N_q" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 17.2 Page No: 573" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a :\n", - "The drift speed = 2.46e-04 m/s=\n", - "Drift speed of electron = 1.15e+05 m/s\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "M=63.5 # IN G\n", - "rho=8.95\n", - "v=M/rho\n", - "electrons=6.02*10**23\n", - "n=(electrons*10**6)/v\n", - "I=10 # in c/s\n", - "q=1.60*10**-19 # in c\n", - "A=3*10**-6 # in m2\n", - "vd=(I)/(n*q*A)\n", - "print \"Solution a :\"\n", - "print \"The drift speed = %0.2e m/s=\"%vd\n", - "k_b=1.38*10**-23\n", - "T=293\n", - "m=9.11*10**-31\n", - "v_rms=sqrt((3*k_b*T)/m)\n", - "print \"Drift speed of electron = %0.2e m/s\"%v_rms" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 17.3 Page No: 578" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The resistance = 18.75 ohm\n" - ] - } - ], - "source": [ - "delta_v=120\n", - "I=6.4\n", - "R=(delta_v)/I\n", - "print \"The resistance = %0.2f ohm\"%R" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 17.4 Page No: 580" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a : \n", - "Area = 3.24e-07 m**2\n", - "Resistance = 4.63 ohm/m\n", - "solution b : \n", - "The current = 2.16 Amps\n" - ] - } - ], - "source": [ - "from math import pi\n", - "r=0.321*10**-3\n", - "A=pi*(r*r)\n", - "print \"Solution a : \"\n", - "print \"Area = %0.2e m**2\"%A\n", - "rho=1.5*10**-6 # in ohm=m\n", - "l=rho/A\n", - "print\"Resistance = %0.2f ohm/m\"% l\n", - "print \"solution b : \"\n", - "Delta_v=10\n", - "I=(Delta_v)/l\n", - "print \"The current = %0.2f Amps\"%I\n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 17.5 Page No: 582" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Temperature = 156.73 C\n" - ] - } - ], - "source": [ - "R=76.8\n", - "Ro=50\n", - "alpha=3.92*10**-3\n", - "t=(R-Ro)/(alpha*Ro)\n", - "T=t+20\n", - "print \"Temperature = %0.2f C\"%T" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 17.6 Page No: 583" - ] - }, - { - "cell_type": "code", - "execution_count": 16, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The current = 6.00 A\n", - "Power = 288.00 Watt\n" - ] - } - ], - "source": [ - "delta_v=50\n", - "R=8\n", - "I=(delta_v)/R\n", - "print \"The current = %0.2f A\"%I\n", - "P=I*I*R\n", - "print \"Power = %0.2f Watt\"%P" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 17.7 Page No: 585" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Number of bulbs = 32\n" - ] - } - ], - "source": [ - "I=20 # in A\n", - "delta_v=120\n", - "p_bulb=75 # inwatt\n", - "p_total=I*delta_v\n", - "N=p_total/p_bulb\n", - "print \"Number of bulbs = %d\"%N" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 17.8 Page No: 587" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Energy = 2.40 kwh\n", - "Cost = 0.29 dollars\n" - ] - } - ], - "source": [ - "p=0.10 # in w\n", - "t=24 # in h\n", - "Energy=p*t\n", - "print \"Energy = %0.2f kwh\"%Energy\n", - "cost=Energy*0.12\n", - "print \"Cost = %0.2f dollars\"%cost" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch17_2.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch17_2.ipynb deleted file mode 100644 index d0eee390..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch17_2.ipynb +++ /dev/null @@ -1,311 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 17 : Current and resistance" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 17.1 Page No: 571" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "solution a : \n", - "Current = 0.83 Amp\n", - "solution b : \n", - "Number of electrons = 0.84 C\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "print \"solution a : \"\n", - "delta_q=1.67 # in c\n", - "delta_t=2 # in s\n", - "I=delta_q/delta_t\n", - "print \"Current = %0.2f Amp\"%I\n", - "print \"solution b : \"\n", - "N=5.22*10**18\n", - "N_q=(1.6*10**-19)*N\n", - "\n", - "print \"Number of electrons = %0.2f C\"%N_q" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 17.2 Page No: 573" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a :\n", - "The drift speed = 2.46e-04 m/s=\n", - "Drift speed of electron = 1.15e+05 m/s\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "M=63.5 # IN G\n", - "rho=8.95\n", - "v=M/rho\n", - "electrons=6.02*10**23\n", - "n=(electrons*10**6)/v\n", - "I=10 # in c/s\n", - "q=1.60*10**-19 # in c\n", - "A=3*10**-6 # in m2\n", - "vd=(I)/(n*q*A)\n", - "print \"Solution a :\"\n", - "print \"The drift speed = %0.2e m/s=\"%vd\n", - "k_b=1.38*10**-23\n", - "T=293\n", - "m=9.11*10**-31\n", - "v_rms=sqrt((3*k_b*T)/m)\n", - "print \"Drift speed of electron = %0.2e m/s\"%v_rms" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 17.3 Page No: 578" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The resistance = 18.75 ohm\n" - ] - } - ], - "source": [ - "delta_v=120\n", - "I=6.4\n", - "R=(delta_v)/I\n", - "print \"The resistance = %0.2f ohm\"%R" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 17.4 Page No: 580" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a : \n", - "Area = 3.24e-07 m**2\n", - "Resistance = 4.63 ohm/m\n", - "solution b : \n", - "The current = 2.16 Amps\n" - ] - } - ], - "source": [ - "from math import pi\n", - "r=0.321*10**-3\n", - "A=pi*(r*r)\n", - "print \"Solution a : \"\n", - "print \"Area = %0.2e m**2\"%A\n", - "rho=1.5*10**-6 # in ohm=m\n", - "l=rho/A\n", - "print\"Resistance = %0.2f ohm/m\"% l\n", - "print \"solution b : \"\n", - "Delta_v=10\n", - "I=(Delta_v)/l\n", - "print \"The current = %0.2f Amps\"%I\n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 17.5 Page No: 582" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Temperature = 156.73 C\n" - ] - } - ], - "source": [ - "R=76.8\n", - "Ro=50\n", - "alpha=3.92*10**-3\n", - "t=(R-Ro)/(alpha*Ro)\n", - "T=t+20\n", - "print \"Temperature = %0.2f C\"%T" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 17.6 Page No: 583" - ] - }, - { - "cell_type": "code", - "execution_count": 16, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The current = 6.00 A\n", - "Power = 288.00 Watt\n" - ] - } - ], - "source": [ - "delta_v=50\n", - "R=8\n", - "I=(delta_v)/R\n", - "print \"The current = %0.2f A\"%I\n", - "P=I*I*R\n", - "print \"Power = %0.2f Watt\"%P" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 17.7 Page No: 585" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Number of bulbs = 32\n" - ] - } - ], - "source": [ - "I=20 # in A\n", - "delta_v=120\n", - "p_bulb=75 # inwatt\n", - "p_total=I*delta_v\n", - "N=p_total/p_bulb\n", - "print \"Number of bulbs = %d\"%N" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 17.8 Page No: 587" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Energy = 2.40 kwh\n", - "Cost = 0.29 dollars\n" - ] - } - ], - "source": [ - "p=0.10 # in w\n", - "t=24 # in h\n", - "Energy=p*t\n", - "print \"Energy = %0.2f kwh\"%Energy\n", - "cost=Energy*0.12\n", - "print \"Cost = %0.2f dollars\"%cost" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch17_3.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch17_3.ipynb deleted file mode 100644 index d0eee390..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch17_3.ipynb +++ /dev/null @@ -1,311 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 17 : Current and resistance" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 17.1 Page No: 571" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "solution a : \n", - "Current = 0.83 Amp\n", - "solution b : \n", - "Number of electrons = 0.84 C\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "print \"solution a : \"\n", - "delta_q=1.67 # in c\n", - "delta_t=2 # in s\n", - "I=delta_q/delta_t\n", - "print \"Current = %0.2f Amp\"%I\n", - "print \"solution b : \"\n", - "N=5.22*10**18\n", - "N_q=(1.6*10**-19)*N\n", - "\n", - "print \"Number of electrons = %0.2f C\"%N_q" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 17.2 Page No: 573" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a :\n", - "The drift speed = 2.46e-04 m/s=\n", - "Drift speed of electron = 1.15e+05 m/s\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "M=63.5 # IN G\n", - "rho=8.95\n", - "v=M/rho\n", - "electrons=6.02*10**23\n", - "n=(electrons*10**6)/v\n", - "I=10 # in c/s\n", - "q=1.60*10**-19 # in c\n", - "A=3*10**-6 # in m2\n", - "vd=(I)/(n*q*A)\n", - "print \"Solution a :\"\n", - "print \"The drift speed = %0.2e m/s=\"%vd\n", - "k_b=1.38*10**-23\n", - "T=293\n", - "m=9.11*10**-31\n", - "v_rms=sqrt((3*k_b*T)/m)\n", - "print \"Drift speed of electron = %0.2e m/s\"%v_rms" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 17.3 Page No: 578" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The resistance = 18.75 ohm\n" - ] - } - ], - "source": [ - "delta_v=120\n", - "I=6.4\n", - "R=(delta_v)/I\n", - "print \"The resistance = %0.2f ohm\"%R" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 17.4 Page No: 580" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a : \n", - "Area = 3.24e-07 m**2\n", - "Resistance = 4.63 ohm/m\n", - "solution b : \n", - "The current = 2.16 Amps\n" - ] - } - ], - "source": [ - "from math import pi\n", - "r=0.321*10**-3\n", - "A=pi*(r*r)\n", - "print \"Solution a : \"\n", - "print \"Area = %0.2e m**2\"%A\n", - "rho=1.5*10**-6 # in ohm=m\n", - "l=rho/A\n", - "print\"Resistance = %0.2f ohm/m\"% l\n", - "print \"solution b : \"\n", - "Delta_v=10\n", - "I=(Delta_v)/l\n", - "print \"The current = %0.2f Amps\"%I\n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 17.5 Page No: 582" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Temperature = 156.73 C\n" - ] - } - ], - "source": [ - "R=76.8\n", - "Ro=50\n", - "alpha=3.92*10**-3\n", - "t=(R-Ro)/(alpha*Ro)\n", - "T=t+20\n", - "print \"Temperature = %0.2f C\"%T" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 17.6 Page No: 583" - ] - }, - { - "cell_type": "code", - "execution_count": 16, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The current = 6.00 A\n", - "Power = 288.00 Watt\n" - ] - } - ], - "source": [ - "delta_v=50\n", - "R=8\n", - "I=(delta_v)/R\n", - "print \"The current = %0.2f A\"%I\n", - "P=I*I*R\n", - "print \"Power = %0.2f Watt\"%P" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 17.7 Page No: 585" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Number of bulbs = 32\n" - ] - } - ], - "source": [ - "I=20 # in A\n", - "delta_v=120\n", - "p_bulb=75 # inwatt\n", - "p_total=I*delta_v\n", - "N=p_total/p_bulb\n", - "print \"Number of bulbs = %d\"%N" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 17.8 Page No: 587" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Energy = 2.40 kwh\n", - "Cost = 0.29 dollars\n" - ] - } - ], - "source": [ - "p=0.10 # in w\n", - "t=24 # in h\n", - "Energy=p*t\n", - "print \"Energy = %0.2f kwh\"%Energy\n", - "cost=Energy*0.12\n", - "print \"Cost = %0.2f dollars\"%cost" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch18_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch18_1.ipynb deleted file mode 100644 index 941bf10e..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch18_1.ipynb +++ /dev/null @@ -1,284 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 18 : Direct current circuits" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 18.1 Page No: 597" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Equivalent resistance = 18.00 ohm\n", - "Solution b\n", - "Current = 0.33 Amps\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "R1=2\n", - "R2=4\n", - "R3=5\n", - "R4=7\n", - "R_eq=R1+R2+R3+R4\n", - "v=6#in v\n", - "print \"Solution a\"\n", - "print \"Equivalent resistance = %0.2f ohm\"%R_eq\n", - "print \"Solution b\"\n", - "I=v/R_eq\n", - "print \"Current = %0.2f Amps\"%I" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example18.2 Page No: 599" - ] - }, - { - "cell_type": "code", - "execution_count": 8, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "solution a\n", - "Current = 6.00 amps\n", - "Current = 3.00 amps\n", - "Current = 2.00 amps\n", - "solution B\n", - "Power = 108.00 watt\n", - "Power = 54.00 watt\n", - "Power = 36.00 watt\n" - ] - } - ], - "source": [ - "delta_V=18#in volt\n", - "R1=3#in ohm\n", - "R2=6#in ohm\n", - "R3=9#in ohm\n", - "I1=delta_V/R1\n", - "I2=delta_V/R2\n", - "I3=delta_V/R3\n", - "print \"solution a\"\n", - "print \"Current = %0.2f amps\"%I1\n", - "print \"Current = %0.2f amps\"%I2\n", - "print \"Current = %0.2f amps\"%I3\n", - "P1=(I1**2)*R1\n", - "P2=(I2**2)*R2\n", - "P3=(I3**2)*R3\n", - "print \"solution B\"\n", - "print \"Power = %0.2f watt\"%P1\n", - "print \"Power = %0.2f watt\"%P2\n", - "print \"Power = %0.2f watt\"%P3" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 18.3 Page No: 602" - ] - }, - { - "cell_type": "code", - "execution_count": 9, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "solution b\n", - "Current = 3.00 amps\n" - ] - } - ], - "source": [ - "delta_Vac=42#in volt\n", - "R_eq=14#in ohm\n", - "I=delta_Vac/R_eq\n", - "print \"solution b\"\n", - "print \"Current = %0.2f amps\"%I" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 18.4 Page No: 605" - ] - }, - { - "cell_type": "code", - "execution_count": 10, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Current value I1 = -0.83, I2 = -0.53 & I3 = -0.30 amps\n" - ] - } - ], - "source": [ - "from numpy import mat\n", - "#formula used x=inv(a)*b\n", - "I=mat([[1 ,-1, -1],[-4, 0 ,-9],[0, -5, 9]])\n", - "V=mat([[0],[6],[0]])\n", - "X=(I**-1)\n", - "a=X*V\n", - "\n", - "print \"Current value I1 = %0.2f, I2 = %0.2f & I3 = %0.2f amps\"%(a[0],a[1],a[2])" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 18.5 Page No: 606" - ] - }, - { - "cell_type": "code", - "execution_count": 11, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Current value I1 = 2.00, I2 = -3.00 & I3 = -1.00 amps\n" - ] - } - ], - "source": [ - "from numpy import mat\n", - "#prob\n", - "#formula used x=inv(a)*b\n", - "I=mat([[8, 2, 0],[-3, 2, 0],[1, 1, -1]])\n", - "V=mat([[10],[-12],[0]])\n", - "X=I**-1\n", - "a=X*V\n", - "\n", - "print \"Current value I1 = %0.2f, I2 = %0.2f & I3 = %0.2f amps\"%(a[0],a[1],a[2])" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 18.6 Page No: 609" - ] - }, - { - "cell_type": "code", - "execution_count": 12, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Constant of the circuit = 4.00 s\n", - "Charge = 6.00e-05 columb\n", - "Charge = 3.79e-05 columb when capacitance 63.2%\n" - ] - } - ], - "source": [ - "R=8*10**5#in ohms\n", - "C=5*10**-6#in Farad\n", - "t=R*C\n", - "print \"Constant of the circuit = %0.2f s\"%t\n", - "\n", - "Q=C*12\n", - "print \"Charge = %0.2e columb\"%Q\n", - "q=0.632*Q\n", - "print \"Charge = %0.2e columb when capacitance 63.2%%\"%q" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 18.7 Page No: 610" - ] - }, - { - "cell_type": "code", - "execution_count": 13, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "time = 1.39 s \n" - ] - } - ], - "source": [ - "from math import log\n", - "x=log(4)\n", - "print \"time = %0.2f s \"%x" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch18_2.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch18_2.ipynb deleted file mode 100644 index 941bf10e..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch18_2.ipynb +++ /dev/null @@ -1,284 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 18 : Direct current circuits" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 18.1 Page No: 597" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Equivalent resistance = 18.00 ohm\n", - "Solution b\n", - "Current = 0.33 Amps\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "R1=2\n", - "R2=4\n", - "R3=5\n", - "R4=7\n", - "R_eq=R1+R2+R3+R4\n", - "v=6#in v\n", - "print \"Solution a\"\n", - "print \"Equivalent resistance = %0.2f ohm\"%R_eq\n", - "print \"Solution b\"\n", - "I=v/R_eq\n", - "print \"Current = %0.2f Amps\"%I" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example18.2 Page No: 599" - ] - }, - { - "cell_type": "code", - "execution_count": 8, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "solution a\n", - "Current = 6.00 amps\n", - "Current = 3.00 amps\n", - "Current = 2.00 amps\n", - "solution B\n", - "Power = 108.00 watt\n", - "Power = 54.00 watt\n", - "Power = 36.00 watt\n" - ] - } - ], - "source": [ - "delta_V=18#in volt\n", - "R1=3#in ohm\n", - "R2=6#in ohm\n", - "R3=9#in ohm\n", - "I1=delta_V/R1\n", - "I2=delta_V/R2\n", - "I3=delta_V/R3\n", - "print \"solution a\"\n", - "print \"Current = %0.2f amps\"%I1\n", - "print \"Current = %0.2f amps\"%I2\n", - "print \"Current = %0.2f amps\"%I3\n", - "P1=(I1**2)*R1\n", - "P2=(I2**2)*R2\n", - "P3=(I3**2)*R3\n", - "print \"solution B\"\n", - "print \"Power = %0.2f watt\"%P1\n", - "print \"Power = %0.2f watt\"%P2\n", - "print \"Power = %0.2f watt\"%P3" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 18.3 Page No: 602" - ] - }, - { - "cell_type": "code", - "execution_count": 9, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "solution b\n", - "Current = 3.00 amps\n" - ] - } - ], - "source": [ - "delta_Vac=42#in volt\n", - "R_eq=14#in ohm\n", - "I=delta_Vac/R_eq\n", - "print \"solution b\"\n", - "print \"Current = %0.2f amps\"%I" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 18.4 Page No: 605" - ] - }, - { - "cell_type": "code", - "execution_count": 10, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Current value I1 = -0.83, I2 = -0.53 & I3 = -0.30 amps\n" - ] - } - ], - "source": [ - "from numpy import mat\n", - "#formula used x=inv(a)*b\n", - "I=mat([[1 ,-1, -1],[-4, 0 ,-9],[0, -5, 9]])\n", - "V=mat([[0],[6],[0]])\n", - "X=(I**-1)\n", - "a=X*V\n", - "\n", - "print \"Current value I1 = %0.2f, I2 = %0.2f & I3 = %0.2f amps\"%(a[0],a[1],a[2])" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 18.5 Page No: 606" - ] - }, - { - "cell_type": "code", - "execution_count": 11, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Current value I1 = 2.00, I2 = -3.00 & I3 = -1.00 amps\n" - ] - } - ], - "source": [ - "from numpy import mat\n", - "#prob\n", - "#formula used x=inv(a)*b\n", - "I=mat([[8, 2, 0],[-3, 2, 0],[1, 1, -1]])\n", - "V=mat([[10],[-12],[0]])\n", - "X=I**-1\n", - "a=X*V\n", - "\n", - "print \"Current value I1 = %0.2f, I2 = %0.2f & I3 = %0.2f amps\"%(a[0],a[1],a[2])" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 18.6 Page No: 609" - ] - }, - { - "cell_type": "code", - "execution_count": 12, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Constant of the circuit = 4.00 s\n", - "Charge = 6.00e-05 columb\n", - "Charge = 3.79e-05 columb when capacitance 63.2%\n" - ] - } - ], - "source": [ - "R=8*10**5#in ohms\n", - "C=5*10**-6#in Farad\n", - "t=R*C\n", - "print \"Constant of the circuit = %0.2f s\"%t\n", - "\n", - "Q=C*12\n", - "print \"Charge = %0.2e columb\"%Q\n", - "q=0.632*Q\n", - "print \"Charge = %0.2e columb when capacitance 63.2%%\"%q" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 18.7 Page No: 610" - ] - }, - { - "cell_type": "code", - "execution_count": 13, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "time = 1.39 s \n" - ] - } - ], - "source": [ - "from math import log\n", - "x=log(4)\n", - "print \"time = %0.2f s \"%x" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch18_3.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch18_3.ipynb deleted file mode 100644 index 941bf10e..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch18_3.ipynb +++ /dev/null @@ -1,284 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 18 : Direct current circuits" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 18.1 Page No: 597" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Equivalent resistance = 18.00 ohm\n", - "Solution b\n", - "Current = 0.33 Amps\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "R1=2\n", - "R2=4\n", - "R3=5\n", - "R4=7\n", - "R_eq=R1+R2+R3+R4\n", - "v=6#in v\n", - "print \"Solution a\"\n", - "print \"Equivalent resistance = %0.2f ohm\"%R_eq\n", - "print \"Solution b\"\n", - "I=v/R_eq\n", - "print \"Current = %0.2f Amps\"%I" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example18.2 Page No: 599" - ] - }, - { - "cell_type": "code", - "execution_count": 8, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "solution a\n", - "Current = 6.00 amps\n", - "Current = 3.00 amps\n", - "Current = 2.00 amps\n", - "solution B\n", - "Power = 108.00 watt\n", - "Power = 54.00 watt\n", - "Power = 36.00 watt\n" - ] - } - ], - "source": [ - "delta_V=18#in volt\n", - "R1=3#in ohm\n", - "R2=6#in ohm\n", - "R3=9#in ohm\n", - "I1=delta_V/R1\n", - "I2=delta_V/R2\n", - "I3=delta_V/R3\n", - "print \"solution a\"\n", - "print \"Current = %0.2f amps\"%I1\n", - "print \"Current = %0.2f amps\"%I2\n", - "print \"Current = %0.2f amps\"%I3\n", - "P1=(I1**2)*R1\n", - "P2=(I2**2)*R2\n", - "P3=(I3**2)*R3\n", - "print \"solution B\"\n", - "print \"Power = %0.2f watt\"%P1\n", - "print \"Power = %0.2f watt\"%P2\n", - "print \"Power = %0.2f watt\"%P3" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 18.3 Page No: 602" - ] - }, - { - "cell_type": "code", - "execution_count": 9, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "solution b\n", - "Current = 3.00 amps\n" - ] - } - ], - "source": [ - "delta_Vac=42#in volt\n", - "R_eq=14#in ohm\n", - "I=delta_Vac/R_eq\n", - "print \"solution b\"\n", - "print \"Current = %0.2f amps\"%I" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 18.4 Page No: 605" - ] - }, - { - "cell_type": "code", - "execution_count": 10, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Current value I1 = -0.83, I2 = -0.53 & I3 = -0.30 amps\n" - ] - } - ], - "source": [ - "from numpy import mat\n", - "#formula used x=inv(a)*b\n", - "I=mat([[1 ,-1, -1],[-4, 0 ,-9],[0, -5, 9]])\n", - "V=mat([[0],[6],[0]])\n", - "X=(I**-1)\n", - "a=X*V\n", - "\n", - "print \"Current value I1 = %0.2f, I2 = %0.2f & I3 = %0.2f amps\"%(a[0],a[1],a[2])" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 18.5 Page No: 606" - ] - }, - { - "cell_type": "code", - "execution_count": 11, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Current value I1 = 2.00, I2 = -3.00 & I3 = -1.00 amps\n" - ] - } - ], - "source": [ - "from numpy import mat\n", - "#prob\n", - "#formula used x=inv(a)*b\n", - "I=mat([[8, 2, 0],[-3, 2, 0],[1, 1, -1]])\n", - "V=mat([[10],[-12],[0]])\n", - "X=I**-1\n", - "a=X*V\n", - "\n", - "print \"Current value I1 = %0.2f, I2 = %0.2f & I3 = %0.2f amps\"%(a[0],a[1],a[2])" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 18.6 Page No: 609" - ] - }, - { - "cell_type": "code", - "execution_count": 12, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Constant of the circuit = 4.00 s\n", - "Charge = 6.00e-05 columb\n", - "Charge = 3.79e-05 columb when capacitance 63.2%\n" - ] - } - ], - "source": [ - "R=8*10**5#in ohms\n", - "C=5*10**-6#in Farad\n", - "t=R*C\n", - "print \"Constant of the circuit = %0.2f s\"%t\n", - "\n", - "Q=C*12\n", - "print \"Charge = %0.2e columb\"%Q\n", - "q=0.632*Q\n", - "print \"Charge = %0.2e columb when capacitance 63.2%%\"%q" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 18.7 Page No: 610" - ] - }, - { - "cell_type": "code", - "execution_count": 13, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "time = 1.39 s \n" - ] - } - ], - "source": [ - "from math import log\n", - "x=log(4)\n", - "print \"time = %0.2f s \"%x" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch19_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch19_1.ipynb deleted file mode 100644 index 3a586dc0..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch19_1.ipynb +++ /dev/null @@ -1,333 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 19 : Magnetism" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.1 Page No: 631" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The force = 7.62e-19 Newton\n" - ] - } - ], - "source": [ - "q=1.6*10**-19#in columb\n", - "v=1*10**5#in m/s\n", - "B=55*10**-6#in T\n", - "F=q*v*B* 0.8660\n", - "print \"The force = %0.2e Newton\"%F" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.2 Page No: 632" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The force = 2.77e-12 Newton\n", - "Acceleration = 1.66e+15 m/s**2\n" - ] - } - ], - "source": [ - "q=1.6*10**-19#in columb\n", - "v=8*10**6#in m/s\n", - "B=2.5#in T\n", - "F=q*v*B* 0.8660\n", - "print \"The force = %0.2e Newton\"%F\n", - "m=1.67*10**-27\n", - "a=F/m\n", - "print \"Acceleration = %0.2e m/s**2\"%a" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.3 Page No: 635" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The maximaum force = 3.96e-02 Newton\n" - ] - } - ], - "source": [ - "l=36#in m\n", - "I=22#in A\n", - "B=0.50*10**-4#in T\n", - "F=B*I*l\n", - "print \"The maximaum force = %0.2e Newton\"%F" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.4 Page No: 637" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The Torque = 0.39 N-m\n" - ] - } - ], - "source": [ - "from math import pi\n", - "A=pi*(0.5)*0.5#in m\n", - "I=2#in A\n", - "B=0.50#in T\n", - "T=B*I*A*0.5\n", - "print \"The Torque = %0.2f N-m\"%T" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.5 Page No: 640" - ] - }, - { - "cell_type": "code", - "execution_count": 8, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Velocity = 4.69e+06 m/s\n" - ] - } - ], - "source": [ - "q=1.6*10**-19\n", - "B=.35\n", - "r=14*10**-2#in m\n", - "m=1.67*10**-27#kg\n", - "v=(q*B*r)/m\n", - "print \"Velocity = %0.2e m/s\"%v" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.6 Page No: 641" - ] - }, - { - "cell_type": "code", - "execution_count": 9, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Radius of lighter istope = 0.10 m\n", - "Radius of heavier istope = 0.21 m\n", - "Distance of seperation = 0.21 m\n" - ] - } - ], - "source": [ - "q=1.6*10**-19\n", - "B=.10#in T\n", - "v=1*10**6#in m/s\n", - "r=14*10**-2#in m\n", - "m1=1.67*10**-27#in kg\n", - "m2=3.34*10**-27#in kg\n", - "r1=(m1*v)/(q*B)\n", - "r2=(m2*v)/(q*B)\n", - "x=(2*r2)-(2*r1)\n", - "print \"Radius of lighter istope = %0.2f m\"%r1\n", - "print \"Radius of heavier istope = %0.2f m\"%r2\n", - "print \"Distance of seperation = %0.2f m\"%x" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.7 Page No: 644" - ] - }, - { - "cell_type": "code", - "execution_count": 12, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Magnetic field = 2.5e-04 T\n", - "Force = 6e-20 Newton\n" - ] - } - ], - "source": [ - "from math import pi\n", - "Uo=(4*pi*10**-7)\n", - "I=5#in A\n", - "r=4*10**-3\n", - "B=(Uo*I)/(2*pi*r)\n", - "print \"Magnetic field = %0.1e T\"%B\n", - "q=1.6*10**-19\n", - "v=1.5*10**3#in m/s\n", - "F=q*v*B\n", - "print \"Force = %0.e Newton\"%F" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.8 Page No: 646" - ] - }, - { - "cell_type": "code", - "execution_count": 14, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Current = 7.07 A\n" - ] - } - ], - "source": [ - "from math import pi, sqrt\n", - "mo=4*pi*10**-7#Tm/A\n", - "d=0.1#in m\n", - "x=1*10**-4#F/l\n", - "I=sqrt((x*2*pi*d)/mo)\n", - "print \"Current = %0.2f A\"%I" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.9 Page No: 649" - ] - }, - { - "cell_type": "code", - "execution_count": 15, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Magnetic field = 6.28e-04 T\n", - "Force = 1.88e-20 N\n" - ] - } - ], - "source": [ - "from math import pi\n", - "N=100#turns\n", - "l=.1#in m\n", - "n=N/l#in turns/m\n", - "mo=4*pi*10**-7#Tm/A\n", - "I=.5#in A\n", - "B=n*I*mo\n", - "q=1.6*10**-19#in c\n", - "v=375#in m/s\n", - "F=q*v*(B/2)\n", - "\n", - "print \"Magnetic field = %0.2e T\"%B\n", - "print \"Force = %0.2e N\"%F" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch19_2.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch19_2.ipynb deleted file mode 100644 index 3a586dc0..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch19_2.ipynb +++ /dev/null @@ -1,333 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 19 : Magnetism" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.1 Page No: 631" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The force = 7.62e-19 Newton\n" - ] - } - ], - "source": [ - "q=1.6*10**-19#in columb\n", - "v=1*10**5#in m/s\n", - "B=55*10**-6#in T\n", - "F=q*v*B* 0.8660\n", - "print \"The force = %0.2e Newton\"%F" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.2 Page No: 632" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The force = 2.77e-12 Newton\n", - "Acceleration = 1.66e+15 m/s**2\n" - ] - } - ], - "source": [ - "q=1.6*10**-19#in columb\n", - "v=8*10**6#in m/s\n", - "B=2.5#in T\n", - "F=q*v*B* 0.8660\n", - "print \"The force = %0.2e Newton\"%F\n", - "m=1.67*10**-27\n", - "a=F/m\n", - "print \"Acceleration = %0.2e m/s**2\"%a" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.3 Page No: 635" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The maximaum force = 3.96e-02 Newton\n" - ] - } - ], - "source": [ - "l=36#in m\n", - "I=22#in A\n", - "B=0.50*10**-4#in T\n", - "F=B*I*l\n", - "print \"The maximaum force = %0.2e Newton\"%F" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.4 Page No: 637" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The Torque = 0.39 N-m\n" - ] - } - ], - "source": [ - "from math import pi\n", - "A=pi*(0.5)*0.5#in m\n", - "I=2#in A\n", - "B=0.50#in T\n", - "T=B*I*A*0.5\n", - "print \"The Torque = %0.2f N-m\"%T" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.5 Page No: 640" - ] - }, - { - "cell_type": "code", - "execution_count": 8, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Velocity = 4.69e+06 m/s\n" - ] - } - ], - "source": [ - "q=1.6*10**-19\n", - "B=.35\n", - "r=14*10**-2#in m\n", - "m=1.67*10**-27#kg\n", - "v=(q*B*r)/m\n", - "print \"Velocity = %0.2e m/s\"%v" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.6 Page No: 641" - ] - }, - { - "cell_type": "code", - "execution_count": 9, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Radius of lighter istope = 0.10 m\n", - "Radius of heavier istope = 0.21 m\n", - "Distance of seperation = 0.21 m\n" - ] - } - ], - "source": [ - "q=1.6*10**-19\n", - "B=.10#in T\n", - "v=1*10**6#in m/s\n", - "r=14*10**-2#in m\n", - "m1=1.67*10**-27#in kg\n", - "m2=3.34*10**-27#in kg\n", - "r1=(m1*v)/(q*B)\n", - "r2=(m2*v)/(q*B)\n", - "x=(2*r2)-(2*r1)\n", - "print \"Radius of lighter istope = %0.2f m\"%r1\n", - "print \"Radius of heavier istope = %0.2f m\"%r2\n", - "print \"Distance of seperation = %0.2f m\"%x" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.7 Page No: 644" - ] - }, - { - "cell_type": "code", - "execution_count": 12, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Magnetic field = 2.5e-04 T\n", - "Force = 6e-20 Newton\n" - ] - } - ], - "source": [ - "from math import pi\n", - "Uo=(4*pi*10**-7)\n", - "I=5#in A\n", - "r=4*10**-3\n", - "B=(Uo*I)/(2*pi*r)\n", - "print \"Magnetic field = %0.1e T\"%B\n", - "q=1.6*10**-19\n", - "v=1.5*10**3#in m/s\n", - "F=q*v*B\n", - "print \"Force = %0.e Newton\"%F" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.8 Page No: 646" - ] - }, - { - "cell_type": "code", - "execution_count": 14, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Current = 7.07 A\n" - ] - } - ], - "source": [ - "from math import pi, sqrt\n", - "mo=4*pi*10**-7#Tm/A\n", - "d=0.1#in m\n", - "x=1*10**-4#F/l\n", - "I=sqrt((x*2*pi*d)/mo)\n", - "print \"Current = %0.2f A\"%I" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.9 Page No: 649" - ] - }, - { - "cell_type": "code", - "execution_count": 15, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Magnetic field = 6.28e-04 T\n", - "Force = 1.88e-20 N\n" - ] - } - ], - "source": [ - "from math import pi\n", - "N=100#turns\n", - "l=.1#in m\n", - "n=N/l#in turns/m\n", - "mo=4*pi*10**-7#Tm/A\n", - "I=.5#in A\n", - "B=n*I*mo\n", - "q=1.6*10**-19#in c\n", - "v=375#in m/s\n", - "F=q*v*(B/2)\n", - "\n", - "print \"Magnetic field = %0.2e T\"%B\n", - "print \"Force = %0.2e N\"%F" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch19_3.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch19_3.ipynb deleted file mode 100644 index 3a586dc0..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch19_3.ipynb +++ /dev/null @@ -1,333 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 19 : Magnetism" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.1 Page No: 631" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The force = 7.62e-19 Newton\n" - ] - } - ], - "source": [ - "q=1.6*10**-19#in columb\n", - "v=1*10**5#in m/s\n", - "B=55*10**-6#in T\n", - "F=q*v*B* 0.8660\n", - "print \"The force = %0.2e Newton\"%F" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.2 Page No: 632" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The force = 2.77e-12 Newton\n", - "Acceleration = 1.66e+15 m/s**2\n" - ] - } - ], - "source": [ - "q=1.6*10**-19#in columb\n", - "v=8*10**6#in m/s\n", - "B=2.5#in T\n", - "F=q*v*B* 0.8660\n", - "print \"The force = %0.2e Newton\"%F\n", - "m=1.67*10**-27\n", - "a=F/m\n", - "print \"Acceleration = %0.2e m/s**2\"%a" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.3 Page No: 635" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The maximaum force = 3.96e-02 Newton\n" - ] - } - ], - "source": [ - "l=36#in m\n", - "I=22#in A\n", - "B=0.50*10**-4#in T\n", - "F=B*I*l\n", - "print \"The maximaum force = %0.2e Newton\"%F" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.4 Page No: 637" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The Torque = 0.39 N-m\n" - ] - } - ], - "source": [ - "from math import pi\n", - "A=pi*(0.5)*0.5#in m\n", - "I=2#in A\n", - "B=0.50#in T\n", - "T=B*I*A*0.5\n", - "print \"The Torque = %0.2f N-m\"%T" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.5 Page No: 640" - ] - }, - { - "cell_type": "code", - "execution_count": 8, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Velocity = 4.69e+06 m/s\n" - ] - } - ], - "source": [ - "q=1.6*10**-19\n", - "B=.35\n", - "r=14*10**-2#in m\n", - "m=1.67*10**-27#kg\n", - "v=(q*B*r)/m\n", - "print \"Velocity = %0.2e m/s\"%v" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.6 Page No: 641" - ] - }, - { - "cell_type": "code", - "execution_count": 9, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Radius of lighter istope = 0.10 m\n", - "Radius of heavier istope = 0.21 m\n", - "Distance of seperation = 0.21 m\n" - ] - } - ], - "source": [ - "q=1.6*10**-19\n", - "B=.10#in T\n", - "v=1*10**6#in m/s\n", - "r=14*10**-2#in m\n", - "m1=1.67*10**-27#in kg\n", - "m2=3.34*10**-27#in kg\n", - "r1=(m1*v)/(q*B)\n", - "r2=(m2*v)/(q*B)\n", - "x=(2*r2)-(2*r1)\n", - "print \"Radius of lighter istope = %0.2f m\"%r1\n", - "print \"Radius of heavier istope = %0.2f m\"%r2\n", - "print \"Distance of seperation = %0.2f m\"%x" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.7 Page No: 644" - ] - }, - { - "cell_type": "code", - "execution_count": 12, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Magnetic field = 2.5e-04 T\n", - "Force = 6e-20 Newton\n" - ] - } - ], - "source": [ - "from math import pi\n", - "Uo=(4*pi*10**-7)\n", - "I=5#in A\n", - "r=4*10**-3\n", - "B=(Uo*I)/(2*pi*r)\n", - "print \"Magnetic field = %0.1e T\"%B\n", - "q=1.6*10**-19\n", - "v=1.5*10**3#in m/s\n", - "F=q*v*B\n", - "print \"Force = %0.e Newton\"%F" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.8 Page No: 646" - ] - }, - { - "cell_type": "code", - "execution_count": 14, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Current = 7.07 A\n" - ] - } - ], - "source": [ - "from math import pi, sqrt\n", - "mo=4*pi*10**-7#Tm/A\n", - "d=0.1#in m\n", - "x=1*10**-4#F/l\n", - "I=sqrt((x*2*pi*d)/mo)\n", - "print \"Current = %0.2f A\"%I" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 19.9 Page No: 649" - ] - }, - { - "cell_type": "code", - "execution_count": 15, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Magnetic field = 6.28e-04 T\n", - "Force = 1.88e-20 N\n" - ] - } - ], - "source": [ - "from math import pi\n", - "N=100#turns\n", - "l=.1#in m\n", - "n=N/l#in turns/m\n", - "mo=4*pi*10**-7#Tm/A\n", - "I=.5#in A\n", - "B=n*I*mo\n", - "q=1.6*10**-19#in c\n", - "v=375#in m/s\n", - "F=q*v*(B/2)\n", - "\n", - "print \"Magnetic field = %0.2e T\"%B\n", - "print \"Force = %0.2e N\"%F" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch20_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch20_1.ipynb deleted file mode 100644 index 475538f9..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch20_1.ipynb +++ /dev/null @@ -1,316 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 20 : Induced voltages and inductance" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 20.1 Page No: 665" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Magnetic flux = 1.62e-04 T.m**2\n", - "Induced emf = 5.06e-03 volt\n" - ] - } - ], - "source": [ - "B=.5 # in T\n", - "A=3.24*10**-4 # in m**2\n", - "Flux=B*A\n", - "N=25\n", - "delta_t=.8\n", - "print \"Magnetic flux = %0.2e T.m**2\"%Flux\n", - "e=(N*Flux)/(delta_t)\n", - "print \"Induced emf = %0.2e volt\"%e" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 20.2 Page No: 667" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Induced emf = 0.45 volt\n" - ] - } - ], - "source": [ - "B=.6*10**-4 # in T\n", - "l=30\n", - "v=250 # in m/s\n", - "e=B*l*v\n", - "print \"Induced emf = %0.2f volt\"%e" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 20.3 Page No: 672" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Induced emf = 0.25 volt\n", - "Solution b\n", - "Current = 0.50 A\n", - "Solution c\n", - "Power = 0.12 watt\n", - "Energy delivered = 0.12 J\n", - "Solution d\n", - "Force = 0.06 N\n" - ] - } - ], - "source": [ - "B=.25 # in T\n", - "l=.5\n", - "v=2 # in m/s\n", - "e=B*l*v\n", - "print \"Solution a\"\n", - "print \"Induced emf = %0.2f volt\"%e\n", - "R=.5 # in ohm\n", - "I=e/R\n", - "\n", - "print \"Solution b\"\n", - "print \"Current = %0.2f A\"%I\n", - "delta_v=.25\n", - "P=I*delta_v\n", - "print \"Solution c\"\n", - "print \"Power = %0.2f watt\"%P\n", - "t=1 # in s\n", - "w=P*t\n", - "print \"Energy delivered = %0.2f J\"%w\n", - " # Answer give for J in textbook is wrong\n", - "d=v*t\n", - "F=w/d\n", - "print \"Solution d\"\n", - "print \"Force = %0.2f N\"%F" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 20.5 Page No: 678" - ] - }, - { - "cell_type": "code", - "execution_count": 9, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Induced emf = 135.72 volt\n", - "Solution b\n", - "Current = 11.31 A\n", - "Solution c\n", - "Emf in Volt 136*sinwt\n" - ] - } - ], - "source": [ - "from math import pi\n", - "f=60 # in Hz\n", - "w=2*pi*f\n", - "N=8\n", - "A=.09 # in m**2\n", - "B=.5 # in T\n", - "emf=N*A*B*w\n", - "print \"Solution a\"\n", - "print \"Induced emf = %0.2f volt\"%emf\n", - "R=12 # in ohm\n", - "I=emf/R\n", - "print \"Solution b\"\n", - "print \"Current = %0.2f A\"%I\n", - "\n", - "print \"Solution c\"\n", - "print \"Emf in Volt 136*sinwt\"" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 20.6 Page No: 680" - ] - }, - { - "cell_type": "code", - "execution_count": 10, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Maximum Current = 12.00 A\n", - "Solution b\n", - "Current = 5.00 A\n" - ] - } - ], - "source": [ - "emf=120 # in Volt\n", - "R=10 # in Ohm\n", - "e_back=70\n", - "I=emf/R\n", - "print \"Solution a\"\n", - "print \"Maximum Current = %0.2f A\"%I\n", - "print \"Solution b\"\n", - "I=(emf-e_back)/R\n", - "print \"Current = %0.2f A\"%I" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 20.8 Page No: 684" - ] - }, - { - "cell_type": "code", - "execution_count": 12, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Inductance = 1.81e-04 H\n", - "Solution b\n", - "Emf = 9.05e-04 Volt\n" - ] - } - ], - "source": [ - "from math import pi\n", - "uo=4*pi*10**-7 # in m/A\n", - "N=300\n", - "A=4*10**-4 # in m**2\n", - "l=25*10**-2\n", - "L=(uo*N*N*A)/l\n", - "print \"Solution a\"\n", - "print \"Inductance = %0.2e H\"%L\n", - "delta_I=-5\n", - "delta_t=1\n", - "e=(-L*delta_I)/(delta_t)\n", - "print \"Solution b\"\n", - "print \"Emf = %0.2e Volt\"%e" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 20.9 Page No: 685" - ] - }, - { - "cell_type": "code", - "execution_count": 14, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Time constant = 5.00e-03 s\n", - "Solution b\n", - "Current = 1.26 Amps\n" - ] - } - ], - "source": [ - "L=30*10**-3 # in Henry\n", - "R=6 # in Ohm\n", - "tou=L/R\n", - "print \"Solution a\"\n", - "print \"Time constant = %0.2e s\"%tou\n", - "\n", - "e=12\n", - "I=(0.632*e)/R\n", - "\n", - "\n", - "print \"Solution b\"\n", - "print \"Current = %0.2f Amps\"%I\n" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch20_2.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch20_2.ipynb deleted file mode 100644 index 475538f9..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch20_2.ipynb +++ /dev/null @@ -1,316 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 20 : Induced voltages and inductance" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 20.1 Page No: 665" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Magnetic flux = 1.62e-04 T.m**2\n", - "Induced emf = 5.06e-03 volt\n" - ] - } - ], - "source": [ - "B=.5 # in T\n", - "A=3.24*10**-4 # in m**2\n", - "Flux=B*A\n", - "N=25\n", - "delta_t=.8\n", - "print \"Magnetic flux = %0.2e T.m**2\"%Flux\n", - "e=(N*Flux)/(delta_t)\n", - "print \"Induced emf = %0.2e volt\"%e" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 20.2 Page No: 667" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Induced emf = 0.45 volt\n" - ] - } - ], - "source": [ - "B=.6*10**-4 # in T\n", - "l=30\n", - "v=250 # in m/s\n", - "e=B*l*v\n", - "print \"Induced emf = %0.2f volt\"%e" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 20.3 Page No: 672" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Induced emf = 0.25 volt\n", - "Solution b\n", - "Current = 0.50 A\n", - "Solution c\n", - "Power = 0.12 watt\n", - "Energy delivered = 0.12 J\n", - "Solution d\n", - "Force = 0.06 N\n" - ] - } - ], - "source": [ - "B=.25 # in T\n", - "l=.5\n", - "v=2 # in m/s\n", - "e=B*l*v\n", - "print \"Solution a\"\n", - "print \"Induced emf = %0.2f volt\"%e\n", - "R=.5 # in ohm\n", - "I=e/R\n", - "\n", - "print \"Solution b\"\n", - "print \"Current = %0.2f A\"%I\n", - "delta_v=.25\n", - "P=I*delta_v\n", - "print \"Solution c\"\n", - "print \"Power = %0.2f watt\"%P\n", - "t=1 # in s\n", - "w=P*t\n", - "print \"Energy delivered = %0.2f J\"%w\n", - " # Answer give for J in textbook is wrong\n", - "d=v*t\n", - "F=w/d\n", - "print \"Solution d\"\n", - "print \"Force = %0.2f N\"%F" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 20.5 Page No: 678" - ] - }, - { - "cell_type": "code", - "execution_count": 9, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Induced emf = 135.72 volt\n", - "Solution b\n", - "Current = 11.31 A\n", - "Solution c\n", - "Emf in Volt 136*sinwt\n" - ] - } - ], - "source": [ - "from math import pi\n", - "f=60 # in Hz\n", - "w=2*pi*f\n", - "N=8\n", - "A=.09 # in m**2\n", - "B=.5 # in T\n", - "emf=N*A*B*w\n", - "print \"Solution a\"\n", - "print \"Induced emf = %0.2f volt\"%emf\n", - "R=12 # in ohm\n", - "I=emf/R\n", - "print \"Solution b\"\n", - "print \"Current = %0.2f A\"%I\n", - "\n", - "print \"Solution c\"\n", - "print \"Emf in Volt 136*sinwt\"" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 20.6 Page No: 680" - ] - }, - { - "cell_type": "code", - "execution_count": 10, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Maximum Current = 12.00 A\n", - "Solution b\n", - "Current = 5.00 A\n" - ] - } - ], - "source": [ - "emf=120 # in Volt\n", - "R=10 # in Ohm\n", - "e_back=70\n", - "I=emf/R\n", - "print \"Solution a\"\n", - "print \"Maximum Current = %0.2f A\"%I\n", - "print \"Solution b\"\n", - "I=(emf-e_back)/R\n", - "print \"Current = %0.2f A\"%I" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 20.8 Page No: 684" - ] - }, - { - "cell_type": "code", - "execution_count": 12, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Inductance = 1.81e-04 H\n", - "Solution b\n", - "Emf = 9.05e-04 Volt\n" - ] - } - ], - "source": [ - "from math import pi\n", - "uo=4*pi*10**-7 # in m/A\n", - "N=300\n", - "A=4*10**-4 # in m**2\n", - "l=25*10**-2\n", - "L=(uo*N*N*A)/l\n", - "print \"Solution a\"\n", - "print \"Inductance = %0.2e H\"%L\n", - "delta_I=-5\n", - "delta_t=1\n", - "e=(-L*delta_I)/(delta_t)\n", - "print \"Solution b\"\n", - "print \"Emf = %0.2e Volt\"%e" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 20.9 Page No: 685" - ] - }, - { - "cell_type": "code", - "execution_count": 14, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Time constant = 5.00e-03 s\n", - "Solution b\n", - "Current = 1.26 Amps\n" - ] - } - ], - "source": [ - "L=30*10**-3 # in Henry\n", - "R=6 # in Ohm\n", - "tou=L/R\n", - "print \"Solution a\"\n", - "print \"Time constant = %0.2e s\"%tou\n", - "\n", - "e=12\n", - "I=(0.632*e)/R\n", - "\n", - "\n", - "print \"Solution b\"\n", - "print \"Current = %0.2f Amps\"%I\n" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch20_3.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch20_3.ipynb deleted file mode 100644 index 475538f9..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch20_3.ipynb +++ /dev/null @@ -1,316 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 20 : Induced voltages and inductance" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 20.1 Page No: 665" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Magnetic flux = 1.62e-04 T.m**2\n", - "Induced emf = 5.06e-03 volt\n" - ] - } - ], - "source": [ - "B=.5 # in T\n", - "A=3.24*10**-4 # in m**2\n", - "Flux=B*A\n", - "N=25\n", - "delta_t=.8\n", - "print \"Magnetic flux = %0.2e T.m**2\"%Flux\n", - "e=(N*Flux)/(delta_t)\n", - "print \"Induced emf = %0.2e volt\"%e" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 20.2 Page No: 667" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Induced emf = 0.45 volt\n" - ] - } - ], - "source": [ - "B=.6*10**-4 # in T\n", - "l=30\n", - "v=250 # in m/s\n", - "e=B*l*v\n", - "print \"Induced emf = %0.2f volt\"%e" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 20.3 Page No: 672" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Induced emf = 0.25 volt\n", - "Solution b\n", - "Current = 0.50 A\n", - "Solution c\n", - "Power = 0.12 watt\n", - "Energy delivered = 0.12 J\n", - "Solution d\n", - "Force = 0.06 N\n" - ] - } - ], - "source": [ - "B=.25 # in T\n", - "l=.5\n", - "v=2 # in m/s\n", - "e=B*l*v\n", - "print \"Solution a\"\n", - "print \"Induced emf = %0.2f volt\"%e\n", - "R=.5 # in ohm\n", - "I=e/R\n", - "\n", - "print \"Solution b\"\n", - "print \"Current = %0.2f A\"%I\n", - "delta_v=.25\n", - "P=I*delta_v\n", - "print \"Solution c\"\n", - "print \"Power = %0.2f watt\"%P\n", - "t=1 # in s\n", - "w=P*t\n", - "print \"Energy delivered = %0.2f J\"%w\n", - " # Answer give for J in textbook is wrong\n", - "d=v*t\n", - "F=w/d\n", - "print \"Solution d\"\n", - "print \"Force = %0.2f N\"%F" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 20.5 Page No: 678" - ] - }, - { - "cell_type": "code", - "execution_count": 9, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Induced emf = 135.72 volt\n", - "Solution b\n", - "Current = 11.31 A\n", - "Solution c\n", - "Emf in Volt 136*sinwt\n" - ] - } - ], - "source": [ - "from math import pi\n", - "f=60 # in Hz\n", - "w=2*pi*f\n", - "N=8\n", - "A=.09 # in m**2\n", - "B=.5 # in T\n", - "emf=N*A*B*w\n", - "print \"Solution a\"\n", - "print \"Induced emf = %0.2f volt\"%emf\n", - "R=12 # in ohm\n", - "I=emf/R\n", - "print \"Solution b\"\n", - "print \"Current = %0.2f A\"%I\n", - "\n", - "print \"Solution c\"\n", - "print \"Emf in Volt 136*sinwt\"" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 20.6 Page No: 680" - ] - }, - { - "cell_type": "code", - "execution_count": 10, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Maximum Current = 12.00 A\n", - "Solution b\n", - "Current = 5.00 A\n" - ] - } - ], - "source": [ - "emf=120 # in Volt\n", - "R=10 # in Ohm\n", - "e_back=70\n", - "I=emf/R\n", - "print \"Solution a\"\n", - "print \"Maximum Current = %0.2f A\"%I\n", - "print \"Solution b\"\n", - "I=(emf-e_back)/R\n", - "print \"Current = %0.2f A\"%I" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 20.8 Page No: 684" - ] - }, - { - "cell_type": "code", - "execution_count": 12, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Inductance = 1.81e-04 H\n", - "Solution b\n", - "Emf = 9.05e-04 Volt\n" - ] - } - ], - "source": [ - "from math import pi\n", - "uo=4*pi*10**-7 # in m/A\n", - "N=300\n", - "A=4*10**-4 # in m**2\n", - "l=25*10**-2\n", - "L=(uo*N*N*A)/l\n", - "print \"Solution a\"\n", - "print \"Inductance = %0.2e H\"%L\n", - "delta_I=-5\n", - "delta_t=1\n", - "e=(-L*delta_I)/(delta_t)\n", - "print \"Solution b\"\n", - "print \"Emf = %0.2e Volt\"%e" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 20.9 Page No: 685" - ] - }, - { - "cell_type": "code", - "execution_count": 14, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Time constant = 5.00e-03 s\n", - "Solution b\n", - "Current = 1.26 Amps\n" - ] - } - ], - "source": [ - "L=30*10**-3 # in Henry\n", - "R=6 # in Ohm\n", - "tou=L/R\n", - "print \"Solution a\"\n", - "print \"Time constant = %0.2e s\"%tou\n", - "\n", - "e=12\n", - "I=(0.632*e)/R\n", - "\n", - "\n", - "print \"Solution b\"\n", - "print \"Current = %0.2f Amps\"%I\n" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch21_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch21_1.ipynb deleted file mode 100644 index 6c83ccae..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch21_1.ipynb +++ /dev/null @@ -1,284 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 21 : Alternating current circuits and electromagnetic waves" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 21.1 Page No: 698" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Voltage = 141.42 V\n", - "Current = 1.41 Amps\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "from math import sqrt\n", - "V_max=200#in V\n", - "V_rms=(V_max)/sqrt(2)\n", - "R=100#in ohm\n", - "I_rms=V_rms/R\n", - "print \"Voltage = %0.2f V\"%V_rms\n", - "print \"Current = %0.2f Amps\"%I_rms" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 21.2 Page No: 700" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Resistance = 331.56 ohm\n", - "Current = 0.45 Amps\n" - ] - } - ], - "source": [ - "C=8*10**-6\n", - "X_c=1/(377*C)\n", - "print \"Resistance = %0.2f ohm\"%X_c\n", - "I_rms=150/X_c\n", - "print \"Current = %0.2f Amps\"%I_rms" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 21.3 Page No: 702" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Resistance = 9.43 ohm\n", - "Current = 15.92 Amps\n" - ] - } - ], - "source": [ - "L=25*10**-3#In H\n", - "w=377\n", - "X_L=w*L#In ohm\n", - "print \"Resistance = %0.2f ohm\"%X_L\n", - "I_rms=150/X_L#In A\n", - "print \"Current = %0.2f Amps\"%I_rms" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 21.4 Page No: 706" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Impedence = 587.81 ohm\n", - "Current = 0.26 Amps\n", - "Angle = -64.83 degree\n", - "Voltage at Resistance = 63.80 Volt\n", - "Voltage at Inductance = 57.67 Volt\n", - "Voltage at Capacitance = 193.43 Volt\n" - ] - } - ], - "source": [ - "from math import atan, degrees, sqrt\n", - "R=250#in ohm\n", - "Xc=758#in ohm\n", - "Xl=226#in Ohm\n", - "X=Xl-Xc\n", - "V_max=150#in Volt\n", - "Z=sqrt(R**2+X**2)\n", - "I=V_max/Z\n", - "q=degrees(atan(X/R))\n", - "print \"Impedence = %0.2f ohm\"%Z\n", - "print \"Current = %0.2f Amps\"%I\n", - "print \"Angle = %0.2f degree\"%q\n", - "V_R=I*R\n", - "V_C=I*Xc\n", - "V_L=I*Xl\n", - "print \"Voltage at Resistance = %0.2f Volt\"%V_R\n", - "print \"Voltage at Inductance = %0.2f Volt\"%V_L\n", - "print \"Voltage at Capacitance = %0.2f Volt\"%V_C" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 21.5 Page No: 708" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Voltage = 106.07 V\n", - "Current = 0.18 Amps\n", - "Power = 8.15 watt\n" - ] - } - ], - "source": [ - "from math import sqrt,cos\n", - "V_max=150#in V\n", - "V_rms=(V_max)/sqrt(2)\n", - "I_max=.255#in ohm\n", - "I_rms=I_max/sqrt(2)\n", - "cos=.426\n", - "P=V_rms*I_rms*cos\n", - "print \"Voltage = %0.2f V\"%V_rms\n", - "print \"Current = %0.2f Amps\"%I_rms\n", - "print \"Power = %0.2f watt\"%P" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 21.6 Page No: 709" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Capacitance = 2e-06 Farad\n" - ] - } - ], - "source": [ - "L=20*10**-3#in H\n", - "C=1/(25*10**6*L)\n", - "print \"Capacitance = %0.e Farad\"%C" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 21.7 Page No: 711" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Percentage of power lost = 0.02\n", - "Solution B\n", - "Percentage of power lost = 75.00\n" - ] - } - ], - "source": [ - "I1=100\n", - "v1=4*10**3\n", - "v2=2.40*10**5\n", - "I2=(I1*v1)/v2\n", - "R=30#in ohm\n", - "p_lost=I2*I2*R\n", - "P_output=I1*v1\n", - "p_per=(p_lost*100/P_output)\n", - "print \"Solution a\"\n", - "print \"Percentage of power lost = %0.2f\"%p_per\n", - "P_lost=I1*I1*R\n", - "per=(P_lost*100)/(4*10**5)\n", - "print \"Solution B\"\n", - "print \"Percentage of power lost = %0.2f\"%per" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch21_2.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch21_2.ipynb deleted file mode 100644 index 6c83ccae..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch21_2.ipynb +++ /dev/null @@ -1,284 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 21 : Alternating current circuits and electromagnetic waves" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 21.1 Page No: 698" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Voltage = 141.42 V\n", - "Current = 1.41 Amps\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "from math import sqrt\n", - "V_max=200#in V\n", - "V_rms=(V_max)/sqrt(2)\n", - "R=100#in ohm\n", - "I_rms=V_rms/R\n", - "print \"Voltage = %0.2f V\"%V_rms\n", - "print \"Current = %0.2f Amps\"%I_rms" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 21.2 Page No: 700" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Resistance = 331.56 ohm\n", - "Current = 0.45 Amps\n" - ] - } - ], - "source": [ - "C=8*10**-6\n", - "X_c=1/(377*C)\n", - "print \"Resistance = %0.2f ohm\"%X_c\n", - "I_rms=150/X_c\n", - "print \"Current = %0.2f Amps\"%I_rms" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 21.3 Page No: 702" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Resistance = 9.43 ohm\n", - "Current = 15.92 Amps\n" - ] - } - ], - "source": [ - "L=25*10**-3#In H\n", - "w=377\n", - "X_L=w*L#In ohm\n", - "print \"Resistance = %0.2f ohm\"%X_L\n", - "I_rms=150/X_L#In A\n", - "print \"Current = %0.2f Amps\"%I_rms" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 21.4 Page No: 706" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Impedence = 587.81 ohm\n", - "Current = 0.26 Amps\n", - "Angle = -64.83 degree\n", - "Voltage at Resistance = 63.80 Volt\n", - "Voltage at Inductance = 57.67 Volt\n", - "Voltage at Capacitance = 193.43 Volt\n" - ] - } - ], - "source": [ - "from math import atan, degrees, sqrt\n", - "R=250#in ohm\n", - "Xc=758#in ohm\n", - "Xl=226#in Ohm\n", - "X=Xl-Xc\n", - "V_max=150#in Volt\n", - "Z=sqrt(R**2+X**2)\n", - "I=V_max/Z\n", - "q=degrees(atan(X/R))\n", - "print \"Impedence = %0.2f ohm\"%Z\n", - "print \"Current = %0.2f Amps\"%I\n", - "print \"Angle = %0.2f degree\"%q\n", - "V_R=I*R\n", - "V_C=I*Xc\n", - "V_L=I*Xl\n", - "print \"Voltage at Resistance = %0.2f Volt\"%V_R\n", - "print \"Voltage at Inductance = %0.2f Volt\"%V_L\n", - "print \"Voltage at Capacitance = %0.2f Volt\"%V_C" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 21.5 Page No: 708" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Voltage = 106.07 V\n", - "Current = 0.18 Amps\n", - "Power = 8.15 watt\n" - ] - } - ], - "source": [ - "from math import sqrt,cos\n", - "V_max=150#in V\n", - "V_rms=(V_max)/sqrt(2)\n", - "I_max=.255#in ohm\n", - "I_rms=I_max/sqrt(2)\n", - "cos=.426\n", - "P=V_rms*I_rms*cos\n", - "print \"Voltage = %0.2f V\"%V_rms\n", - "print \"Current = %0.2f Amps\"%I_rms\n", - "print \"Power = %0.2f watt\"%P" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 21.6 Page No: 709" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Capacitance = 2e-06 Farad\n" - ] - } - ], - "source": [ - "L=20*10**-3#in H\n", - "C=1/(25*10**6*L)\n", - "print \"Capacitance = %0.e Farad\"%C" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 21.7 Page No: 711" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Percentage of power lost = 0.02\n", - "Solution B\n", - "Percentage of power lost = 75.00\n" - ] - } - ], - "source": [ - "I1=100\n", - "v1=4*10**3\n", - "v2=2.40*10**5\n", - "I2=(I1*v1)/v2\n", - "R=30#in ohm\n", - "p_lost=I2*I2*R\n", - "P_output=I1*v1\n", - "p_per=(p_lost*100/P_output)\n", - "print \"Solution a\"\n", - "print \"Percentage of power lost = %0.2f\"%p_per\n", - "P_lost=I1*I1*R\n", - "per=(P_lost*100)/(4*10**5)\n", - "print \"Solution B\"\n", - "print \"Percentage of power lost = %0.2f\"%per" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch21_3.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch21_3.ipynb deleted file mode 100644 index 6c83ccae..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch21_3.ipynb +++ /dev/null @@ -1,284 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 21 : Alternating current circuits and electromagnetic waves" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 21.1 Page No: 698" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Voltage = 141.42 V\n", - "Current = 1.41 Amps\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "from math import sqrt\n", - "V_max=200#in V\n", - "V_rms=(V_max)/sqrt(2)\n", - "R=100#in ohm\n", - "I_rms=V_rms/R\n", - "print \"Voltage = %0.2f V\"%V_rms\n", - "print \"Current = %0.2f Amps\"%I_rms" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 21.2 Page No: 700" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Resistance = 331.56 ohm\n", - "Current = 0.45 Amps\n" - ] - } - ], - "source": [ - "C=8*10**-6\n", - "X_c=1/(377*C)\n", - "print \"Resistance = %0.2f ohm\"%X_c\n", - "I_rms=150/X_c\n", - "print \"Current = %0.2f Amps\"%I_rms" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 21.3 Page No: 702" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Resistance = 9.43 ohm\n", - "Current = 15.92 Amps\n" - ] - } - ], - "source": [ - "L=25*10**-3#In H\n", - "w=377\n", - "X_L=w*L#In ohm\n", - "print \"Resistance = %0.2f ohm\"%X_L\n", - "I_rms=150/X_L#In A\n", - "print \"Current = %0.2f Amps\"%I_rms" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 21.4 Page No: 706" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Impedence = 587.81 ohm\n", - "Current = 0.26 Amps\n", - "Angle = -64.83 degree\n", - "Voltage at Resistance = 63.80 Volt\n", - "Voltage at Inductance = 57.67 Volt\n", - "Voltage at Capacitance = 193.43 Volt\n" - ] - } - ], - "source": [ - "from math import atan, degrees, sqrt\n", - "R=250#in ohm\n", - "Xc=758#in ohm\n", - "Xl=226#in Ohm\n", - "X=Xl-Xc\n", - "V_max=150#in Volt\n", - "Z=sqrt(R**2+X**2)\n", - "I=V_max/Z\n", - "q=degrees(atan(X/R))\n", - "print \"Impedence = %0.2f ohm\"%Z\n", - "print \"Current = %0.2f Amps\"%I\n", - "print \"Angle = %0.2f degree\"%q\n", - "V_R=I*R\n", - "V_C=I*Xc\n", - "V_L=I*Xl\n", - "print \"Voltage at Resistance = %0.2f Volt\"%V_R\n", - "print \"Voltage at Inductance = %0.2f Volt\"%V_L\n", - "print \"Voltage at Capacitance = %0.2f Volt\"%V_C" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 21.5 Page No: 708" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Voltage = 106.07 V\n", - "Current = 0.18 Amps\n", - "Power = 8.15 watt\n" - ] - } - ], - "source": [ - "from math import sqrt,cos\n", - "V_max=150#in V\n", - "V_rms=(V_max)/sqrt(2)\n", - "I_max=.255#in ohm\n", - "I_rms=I_max/sqrt(2)\n", - "cos=.426\n", - "P=V_rms*I_rms*cos\n", - "print \"Voltage = %0.2f V\"%V_rms\n", - "print \"Current = %0.2f Amps\"%I_rms\n", - "print \"Power = %0.2f watt\"%P" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 21.6 Page No: 709" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Capacitance = 2e-06 Farad\n" - ] - } - ], - "source": [ - "L=20*10**-3#in H\n", - "C=1/(25*10**6*L)\n", - "print \"Capacitance = %0.e Farad\"%C" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 21.7 Page No: 711" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Percentage of power lost = 0.02\n", - "Solution B\n", - "Percentage of power lost = 75.00\n" - ] - } - ], - "source": [ - "I1=100\n", - "v1=4*10**3\n", - "v2=2.40*10**5\n", - "I2=(I1*v1)/v2\n", - "R=30#in ohm\n", - "p_lost=I2*I2*R\n", - "P_output=I1*v1\n", - "p_per=(p_lost*100/P_output)\n", - "print \"Solution a\"\n", - "print \"Percentage of power lost = %0.2f\"%p_per\n", - "P_lost=I1*I1*R\n", - "per=(P_lost*100)/(4*10**5)\n", - "print \"Solution B\"\n", - "print \"Percentage of power lost = %0.2f\"%per" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch22_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch22_1.ipynb deleted file mode 100644 index e4ba19da..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch22_1.ipynb +++ /dev/null @@ -1,169 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 22 : Reflection and refraction of light" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 22.2 Page No: 739" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Angle = 19.20 degree\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "from math import sin, pi, degrees, asin\n", - "n1=1\n", - "n2=1.52\n", - "x=sin(pi/180*30)\n", - "theta_2=degrees(asin((n1*x)/n2))\n", - "print \"Angle = %0.2f degree\"%theta_2" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 22.3 Page No: 739" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Velocity = 2.06e+08 m/s\n", - "Solution b\n", - "Wavelength in Fused quartz = 403.98 nm\n" - ] - } - ], - "source": [ - "print \"Solution a\"\n", - "c=3*10**8# Constant in m/s\n", - "n=1.458\n", - "v=c/n\n", - "print \"Velocity = %0.2e m/s\"%v\n", - "print \"Solution b\"\n", - "lambda_o=589#in nm\n", - "lambda_n=lambda_o/n\n", - "print \"Wavelength in Fused quartz = %0.2f nm\"%lambda_n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 22.5 Page No: 741" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Angle = 16.24 degree\n", - "Angle = 25.69 degree\n" - ] - } - ], - "source": [ - "from math import atan, degrees, asin\n", - "x=699#in micrometer(w-a)\n", - "t=1200 #in micrometer\n", - "b=x/2\n", - "theta_2=degrees(atan(b/t))\n", - "print \"Angle = %0.2f degree\"%theta_2\n", - "y=sin(pi/180*theta_2)\n", - "n1=1\n", - "n2=1.55\n", - "theta_1=degrees(asin((n2*y)/n1))\n", - "print \"Angle = %0.2f degree\"%theta_1" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 22.6 Page No: 744" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Angle(theta_c) = 48.75 degree\n" - ] - } - ], - "source": [ - "from math import asin, degrees\n", - "n1=1.33\n", - "n2=1\n", - "x=degrees(asin(n2/n1))\n", - "\n", - "print \"Angle(theta_c) = %0.2f degree\"%x" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch22_2.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch22_2.ipynb deleted file mode 100644 index e4ba19da..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch22_2.ipynb +++ /dev/null @@ -1,169 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 22 : Reflection and refraction of light" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 22.2 Page No: 739" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Angle = 19.20 degree\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "from math import sin, pi, degrees, asin\n", - "n1=1\n", - "n2=1.52\n", - "x=sin(pi/180*30)\n", - "theta_2=degrees(asin((n1*x)/n2))\n", - "print \"Angle = %0.2f degree\"%theta_2" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 22.3 Page No: 739" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Velocity = 2.06e+08 m/s\n", - "Solution b\n", - "Wavelength in Fused quartz = 403.98 nm\n" - ] - } - ], - "source": [ - "print \"Solution a\"\n", - "c=3*10**8# Constant in m/s\n", - "n=1.458\n", - "v=c/n\n", - "print \"Velocity = %0.2e m/s\"%v\n", - "print \"Solution b\"\n", - "lambda_o=589#in nm\n", - "lambda_n=lambda_o/n\n", - "print \"Wavelength in Fused quartz = %0.2f nm\"%lambda_n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 22.5 Page No: 741" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Angle = 16.24 degree\n", - "Angle = 25.69 degree\n" - ] - } - ], - "source": [ - "from math import atan, degrees, asin\n", - "x=699#in micrometer(w-a)\n", - "t=1200 #in micrometer\n", - "b=x/2\n", - "theta_2=degrees(atan(b/t))\n", - "print \"Angle = %0.2f degree\"%theta_2\n", - "y=sin(pi/180*theta_2)\n", - "n1=1\n", - "n2=1.55\n", - "theta_1=degrees(asin((n2*y)/n1))\n", - "print \"Angle = %0.2f degree\"%theta_1" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 22.6 Page No: 744" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Angle(theta_c) = 48.75 degree\n" - ] - } - ], - "source": [ - "from math import asin, degrees\n", - "n1=1.33\n", - "n2=1\n", - "x=degrees(asin(n2/n1))\n", - "\n", - "print \"Angle(theta_c) = %0.2f degree\"%x" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch22_3.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch22_3.ipynb deleted file mode 100644 index e4ba19da..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch22_3.ipynb +++ /dev/null @@ -1,169 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 22 : Reflection and refraction of light" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 22.2 Page No: 739" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Angle = 19.20 degree\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "from math import sin, pi, degrees, asin\n", - "n1=1\n", - "n2=1.52\n", - "x=sin(pi/180*30)\n", - "theta_2=degrees(asin((n1*x)/n2))\n", - "print \"Angle = %0.2f degree\"%theta_2" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 22.3 Page No: 739" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Solution a\n", - "Velocity = 2.06e+08 m/s\n", - "Solution b\n", - "Wavelength in Fused quartz = 403.98 nm\n" - ] - } - ], - "source": [ - "print \"Solution a\"\n", - "c=3*10**8# Constant in m/s\n", - "n=1.458\n", - "v=c/n\n", - "print \"Velocity = %0.2e m/s\"%v\n", - "print \"Solution b\"\n", - "lambda_o=589#in nm\n", - "lambda_n=lambda_o/n\n", - "print \"Wavelength in Fused quartz = %0.2f nm\"%lambda_n" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 22.5 Page No: 741" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Angle = 16.24 degree\n", - "Angle = 25.69 degree\n" - ] - } - ], - "source": [ - "from math import atan, degrees, asin\n", - "x=699#in micrometer(w-a)\n", - "t=1200 #in micrometer\n", - "b=x/2\n", - "theta_2=degrees(atan(b/t))\n", - "print \"Angle = %0.2f degree\"%theta_2\n", - "y=sin(pi/180*theta_2)\n", - "n1=1\n", - "n2=1.55\n", - "theta_1=degrees(asin((n2*y)/n1))\n", - "print \"Angle = %0.2f degree\"%theta_1" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 22.6 Page No: 744" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Angle(theta_c) = 48.75 degree\n" - ] - } - ], - "source": [ - "from math import asin, degrees\n", - "n1=1.33\n", - "n2=1\n", - "x=degrees(asin(n2/n1))\n", - "\n", - "print \"Angle(theta_c) = %0.2f degree\"%x" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch23_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch23_1.ipynb deleted file mode 100644 index 3f646e85..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch23_1.ipynb +++ /dev/null @@ -1,367 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 23 : Mirrors and lenses" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 23.1 Page No: 760" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The hight = 0.90 m\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "AC= 1.8-.1#in m\n", - "AD=.5*AC\n", - "CF=.10#/in m\n", - "X=.5*CF#in m\n", - "FA=1.8#in m\n", - "d=FA-AD-X\n", - "print \"The hight = %0.2f m\"%d" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 23.2 Page No : 767" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "part a\n", - "The magnification when object is at 25cm : -0.67\n", - "part c\n", - "The magnification when object is at 5cm : 2.00\n" - ] - } - ], - "source": [ - "p=25#in cm\n", - "f=10#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "p=25\n", - "M=-(q/p)\n", - "print \"part a\"\n", - "print \"The magnification when object is at 25cm : %0.2f\"%M\n", - "p=5#in cm\n", - "f=10#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "p=5\n", - "M=-(q/p)\n", - "print \"part c\"\n", - "print \"The magnification when object is at 5cm : %0.2f\"%M" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 23.3 Page No: 768" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "part a\n", - "The position of final image = -5.71 cm\n", - "part b\n", - "The magnification : 0.23\n" - ] - } - ], - "source": [ - "p=20#in cm\n", - "f=-8#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "p=25\n", - "M=-(q/p)\n", - "print \"part a\"\n", - "print \"The position of final image = %0.2f cm\"%q\n", - "print \"part b\"\n", - "print \"The magnification : %0.2f\"%M" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 23.4 Page No: 769" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The focal length = 26.67 cm\n" - ] - } - ], - "source": [ - "p=40#in cm\n", - "q=-(2*p)\n", - "\n", - "x=(1/p)-(1/q)\n", - "f=1/x\n", - "print \"The focal length = %0.2f cm\"%f\n", - "#Answer given in book is wrong" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 23.5 Page No: 770" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The position of final image = -17.14 cm\n", - "The magnification when object = -1.29 cm\n", - "The Position of image = 2.57 cm\n" - ] - } - ], - "source": [ - "p=20#in cm\n", - "n1=1.5#in cm\n", - "n2=1#in cm\n", - "R=-30#in cm\n", - "x=(n2-n1)/R\n", - "y=n1/p\n", - "s=x-y\n", - "q=1/s\n", - "print \"The position of final image = %0.2f cm\"%q\n", - "M=(n1*q)/(n2*p)\n", - "print \"The magnification when object = %0.2f cm\"%M\n", - "h=2#in cm\n", - "h1=-M*h\n", - "print \"The Position of image = %0.2f cm\"%h1" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 23.7 Page No: 777" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "part a\n", - "The position of final image = 15.00 cm\n", - "The magnification : -0.50\n", - "part b\n", - "The position of final image = -10.00 cm\n", - "The magnification : 2.00\n" - ] - } - ], - "source": [ - "p=30#in cm\n", - "f=10#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "\n", - "M=-(q/p)\n", - "print \"part a\"\n", - "print \"The position of final image = %0.2f cm\"%q\n", - "print \"The magnification : %0.2f\"%M\n", - "p=5#in cm\n", - "f=10#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "M=-(q/p)\n", - "print \"part b\"\n", - "print \"The position of final image = %0.2f cm\"%q\n", - "print \"The magnification : %0.2f\"%M" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 23.8 Page No: 778" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "part a\n", - "The position of final image = -7.50 cm\n", - "The magnification : 0.25\n", - "part b\n", - "The position of final image = -5.00 cm\n", - "The magnification : 0.50\n", - "part c\n", - "The position of final image = -3.33 cm\n", - "The magnification : 0.67\n" - ] - } - ], - "source": [ - "p=30#in cm\n", - "f=-10#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "\n", - "M=-(q/p)\n", - "print \"part a\"\n", - "print \"The position of final image = %0.2f cm\"%q\n", - "print \"The magnification : %0.2f\"%M\n", - "p=10#in cm\n", - "f=-10#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "M=-(q/p)\n", - "print \"part b\"\n", - "print \"The position of final image = %0.2f cm\"%q\n", - "print \"The magnification : %0.2f\"%M\n", - "p=5#in cm\n", - "f=-10#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "M=-(q/p)\n", - "print \"part c\"\n", - "print \"The position of final image = %0.2f cm\"%q\n", - "print \"The magnification : %0.2f\"%M" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 23.9 Page No: 779" - ] - }, - { - "cell_type": "code", - "execution_count": 8, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The magnification : -0.67\n" - ] - } - ], - "source": [ - "p=30#in cm\n", - "f=10#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "\n", - "M1=-(q/p)\n", - "\n", - "p=5#in cm\n", - "f=20#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "\n", - "M2=-(q/p)\n", - "\n", - "\n", - "M=M1*M2\n", - "print \"The magnification : %0.2f\"%M" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch23_2.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch23_2.ipynb deleted file mode 100644 index 3f646e85..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch23_2.ipynb +++ /dev/null @@ -1,367 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 23 : Mirrors and lenses" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 23.1 Page No: 760" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The hight = 0.90 m\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "AC= 1.8-.1#in m\n", - "AD=.5*AC\n", - "CF=.10#/in m\n", - "X=.5*CF#in m\n", - "FA=1.8#in m\n", - "d=FA-AD-X\n", - "print \"The hight = %0.2f m\"%d" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 23.2 Page No : 767" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "part a\n", - "The magnification when object is at 25cm : -0.67\n", - "part c\n", - "The magnification when object is at 5cm : 2.00\n" - ] - } - ], - "source": [ - "p=25#in cm\n", - "f=10#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "p=25\n", - "M=-(q/p)\n", - "print \"part a\"\n", - "print \"The magnification when object is at 25cm : %0.2f\"%M\n", - "p=5#in cm\n", - "f=10#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "p=5\n", - "M=-(q/p)\n", - "print \"part c\"\n", - "print \"The magnification when object is at 5cm : %0.2f\"%M" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 23.3 Page No: 768" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "part a\n", - "The position of final image = -5.71 cm\n", - "part b\n", - "The magnification : 0.23\n" - ] - } - ], - "source": [ - "p=20#in cm\n", - "f=-8#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "p=25\n", - "M=-(q/p)\n", - "print \"part a\"\n", - "print \"The position of final image = %0.2f cm\"%q\n", - "print \"part b\"\n", - "print \"The magnification : %0.2f\"%M" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 23.4 Page No: 769" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The focal length = 26.67 cm\n" - ] - } - ], - "source": [ - "p=40#in cm\n", - "q=-(2*p)\n", - "\n", - "x=(1/p)-(1/q)\n", - "f=1/x\n", - "print \"The focal length = %0.2f cm\"%f\n", - "#Answer given in book is wrong" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 23.5 Page No: 770" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The position of final image = -17.14 cm\n", - "The magnification when object = -1.29 cm\n", - "The Position of image = 2.57 cm\n" - ] - } - ], - "source": [ - "p=20#in cm\n", - "n1=1.5#in cm\n", - "n2=1#in cm\n", - "R=-30#in cm\n", - "x=(n2-n1)/R\n", - "y=n1/p\n", - "s=x-y\n", - "q=1/s\n", - "print \"The position of final image = %0.2f cm\"%q\n", - "M=(n1*q)/(n2*p)\n", - "print \"The magnification when object = %0.2f cm\"%M\n", - "h=2#in cm\n", - "h1=-M*h\n", - "print \"The Position of image = %0.2f cm\"%h1" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 23.7 Page No: 777" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "part a\n", - "The position of final image = 15.00 cm\n", - "The magnification : -0.50\n", - "part b\n", - "The position of final image = -10.00 cm\n", - "The magnification : 2.00\n" - ] - } - ], - "source": [ - "p=30#in cm\n", - "f=10#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "\n", - "M=-(q/p)\n", - "print \"part a\"\n", - "print \"The position of final image = %0.2f cm\"%q\n", - "print \"The magnification : %0.2f\"%M\n", - "p=5#in cm\n", - "f=10#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "M=-(q/p)\n", - "print \"part b\"\n", - "print \"The position of final image = %0.2f cm\"%q\n", - "print \"The magnification : %0.2f\"%M" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 23.8 Page No: 778" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "part a\n", - "The position of final image = -7.50 cm\n", - "The magnification : 0.25\n", - "part b\n", - "The position of final image = -5.00 cm\n", - "The magnification : 0.50\n", - "part c\n", - "The position of final image = -3.33 cm\n", - "The magnification : 0.67\n" - ] - } - ], - "source": [ - "p=30#in cm\n", - "f=-10#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "\n", - "M=-(q/p)\n", - "print \"part a\"\n", - "print \"The position of final image = %0.2f cm\"%q\n", - "print \"The magnification : %0.2f\"%M\n", - "p=10#in cm\n", - "f=-10#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "M=-(q/p)\n", - "print \"part b\"\n", - "print \"The position of final image = %0.2f cm\"%q\n", - "print \"The magnification : %0.2f\"%M\n", - "p=5#in cm\n", - "f=-10#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "M=-(q/p)\n", - "print \"part c\"\n", - "print \"The position of final image = %0.2f cm\"%q\n", - "print \"The magnification : %0.2f\"%M" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 23.9 Page No: 779" - ] - }, - { - "cell_type": "code", - "execution_count": 8, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The magnification : -0.67\n" - ] - } - ], - "source": [ - "p=30#in cm\n", - "f=10#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "\n", - "M1=-(q/p)\n", - "\n", - "p=5#in cm\n", - "f=20#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "\n", - "M2=-(q/p)\n", - "\n", - "\n", - "M=M1*M2\n", - "print \"The magnification : %0.2f\"%M" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch23_3.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch23_3.ipynb deleted file mode 100644 index 3f646e85..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch23_3.ipynb +++ /dev/null @@ -1,367 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 23 : Mirrors and lenses" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 23.1 Page No: 760" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The hight = 0.90 m\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "AC= 1.8-.1#in m\n", - "AD=.5*AC\n", - "CF=.10#/in m\n", - "X=.5*CF#in m\n", - "FA=1.8#in m\n", - "d=FA-AD-X\n", - "print \"The hight = %0.2f m\"%d" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 23.2 Page No : 767" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "part a\n", - "The magnification when object is at 25cm : -0.67\n", - "part c\n", - "The magnification when object is at 5cm : 2.00\n" - ] - } - ], - "source": [ - "p=25#in cm\n", - "f=10#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "p=25\n", - "M=-(q/p)\n", - "print \"part a\"\n", - "print \"The magnification when object is at 25cm : %0.2f\"%M\n", - "p=5#in cm\n", - "f=10#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "p=5\n", - "M=-(q/p)\n", - "print \"part c\"\n", - "print \"The magnification when object is at 5cm : %0.2f\"%M" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 23.3 Page No: 768" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "part a\n", - "The position of final image = -5.71 cm\n", - "part b\n", - "The magnification : 0.23\n" - ] - } - ], - "source": [ - "p=20#in cm\n", - "f=-8#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "p=25\n", - "M=-(q/p)\n", - "print \"part a\"\n", - "print \"The position of final image = %0.2f cm\"%q\n", - "print \"part b\"\n", - "print \"The magnification : %0.2f\"%M" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 23.4 Page No: 769" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The focal length = 26.67 cm\n" - ] - } - ], - "source": [ - "p=40#in cm\n", - "q=-(2*p)\n", - "\n", - "x=(1/p)-(1/q)\n", - "f=1/x\n", - "print \"The focal length = %0.2f cm\"%f\n", - "#Answer given in book is wrong" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 23.5 Page No: 770" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The position of final image = -17.14 cm\n", - "The magnification when object = -1.29 cm\n", - "The Position of image = 2.57 cm\n" - ] - } - ], - "source": [ - "p=20#in cm\n", - "n1=1.5#in cm\n", - "n2=1#in cm\n", - "R=-30#in cm\n", - "x=(n2-n1)/R\n", - "y=n1/p\n", - "s=x-y\n", - "q=1/s\n", - "print \"The position of final image = %0.2f cm\"%q\n", - "M=(n1*q)/(n2*p)\n", - "print \"The magnification when object = %0.2f cm\"%M\n", - "h=2#in cm\n", - "h1=-M*h\n", - "print \"The Position of image = %0.2f cm\"%h1" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 23.7 Page No: 777" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "part a\n", - "The position of final image = 15.00 cm\n", - "The magnification : -0.50\n", - "part b\n", - "The position of final image = -10.00 cm\n", - "The magnification : 2.00\n" - ] - } - ], - "source": [ - "p=30#in cm\n", - "f=10#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "\n", - "M=-(q/p)\n", - "print \"part a\"\n", - "print \"The position of final image = %0.2f cm\"%q\n", - "print \"The magnification : %0.2f\"%M\n", - "p=5#in cm\n", - "f=10#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "M=-(q/p)\n", - "print \"part b\"\n", - "print \"The position of final image = %0.2f cm\"%q\n", - "print \"The magnification : %0.2f\"%M" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 23.8 Page No: 778" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "part a\n", - "The position of final image = -7.50 cm\n", - "The magnification : 0.25\n", - "part b\n", - "The position of final image = -5.00 cm\n", - "The magnification : 0.50\n", - "part c\n", - "The position of final image = -3.33 cm\n", - "The magnification : 0.67\n" - ] - } - ], - "source": [ - "p=30#in cm\n", - "f=-10#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "\n", - "M=-(q/p)\n", - "print \"part a\"\n", - "print \"The position of final image = %0.2f cm\"%q\n", - "print \"The magnification : %0.2f\"%M\n", - "p=10#in cm\n", - "f=-10#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "M=-(q/p)\n", - "print \"part b\"\n", - "print \"The position of final image = %0.2f cm\"%q\n", - "print \"The magnification : %0.2f\"%M\n", - "p=5#in cm\n", - "f=-10#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "M=-(q/p)\n", - "print \"part c\"\n", - "print \"The position of final image = %0.2f cm\"%q\n", - "print \"The magnification : %0.2f\"%M" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 23.9 Page No: 779" - ] - }, - { - "cell_type": "code", - "execution_count": 8, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The magnification : -0.67\n" - ] - } - ], - "source": [ - "p=30#in cm\n", - "f=10#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "\n", - "M1=-(q/p)\n", - "\n", - "p=5#in cm\n", - "f=20#in cm\n", - "x=(1/f)-(1/p)\n", - "q=1/x\n", - "\n", - "M2=-(q/p)\n", - "\n", - "\n", - "M=M1*M2\n", - "print \"The magnification : %0.2f\"%M" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch24_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch24_1.ipynb deleted file mode 100644 index 82f29c9f..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch24_1.ipynb +++ /dev/null @@ -1,233 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 24 : Wave optics" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 1 Page No: 794" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "(A) wavelength of light = 5.63e-07 meters\n", - "(B) Distance between adjacent fringes = 0.022 meters\n" - ] - } - ], - "source": [ - "L=1.2 # Seperation between screen and double-slit in meter\n", - "d=3*10**-5 #distance between the two slits\n", - "m=2 #second order bright fringe\n", - "Y=4.5*10**-2 #distance of second order bright fringe from centerline\n", - "#wavelength of light\n", - "lamda=(Y*d)/(m*L)\n", - "print \"(A) wavelength of light = %0.2e meters\"%lamda\n", - "#distance between adjacent bright fringes\n", - "#delta_Y=Y(m+1)-Ym\n", - "delta_Y=lamda*L/d\n", - "print \"(B) Distance between adjacent fringes = %0.3f meters\"%delta_Y" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 2 Page No: 798" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Minimum thickness of soap bubble film = 113.16 nm is\n" - ] - } - ], - "source": [ - "n=1.33 #refractive index of soap bubble\n", - "lamda=602 #wavelength of light in nm\n", - "#for constructive interference we have 2nt=lamda/2\n", - "t=lamda/(4*n)\n", - "print \"Minimum thickness of soap bubble film = %0.2f nm is\"%t" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 3 Page No: 799" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Minimum thickness of film = 95.17 nm is\n" - ] - } - ], - "source": [ - "n=1.45 #refractive index of silicon monoxide\n", - "lamda=552 #wavelength of light in nm\n", - "#for destructive interference we have condition for minimn thickness 2t=lamda/2n\n", - "t=lamda/(4*n)\n", - "print \"Minimum thickness of film = %0.2f nm is\"%t" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 5 Page No: 801" - ] - }, - { - "cell_type": "code", - "execution_count": 8, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Pit depth in a CD = 121.88 nm\n" - ] - } - ], - "source": [ - "n=1.6 #refractive index of plastic transparent layer\n", - "lamda=780 #wavelength of laser light in nm\n", - "#for destructive interference we have condition for minimn thickness 2t=lamda/2n\n", - "t=lamda/(4*n)\n", - "print \"Pit depth in a CD = %0.2f nm\"%t" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 6 Page No: 804" - ] - }, - { - "cell_type": "code", - "execution_count": 10, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - " Position of first dark fringe = 3.87e-03 meters\n" - ] - } - ], - "source": [ - "lamda=580*10**-9 #wavelength of incident light in meter\n", - "a=0.30*10**-3 #slit width in meter\n", - "L=2 #distance of screen from slit in meters\n", - "#The first dark fringe corresponds to m=+1 or -1\n", - "m=1\n", - "sin_theta=m*lamda/a\n", - "#From fig 24.16 tan_theta=y/L and since theta is very small we have sin_theta=tan_theta hence sin_theta=y/L\n", - "y=L*sin_theta \n", - "print \" Position of first dark fringe = %0.2e meters\"%y" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 7 Page No: 808" - ] - }, - { - "cell_type": "code", - "execution_count": 11, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Angle in degrees at which first order maxima is observed : 22.32\n", - "Angle in degrees at which second order maxima is observed : 49.43\n", - "for higher order number of diffraction the the solutions are non realistic\n" - ] - } - ], - "source": [ - "from math import asin, degrees\n", - "lamda=632.8 #wavelength of monochromatic light from helium-neon laser in meter\n", - "a=6000 #lines in diffraction grating per cm\n", - "d=10**7/a#slit seperation in nm\n", - "#for the first order maximum we have m=1\n", - "sin_theta1=lamda/d\n", - "theta1=degrees(asin(sin_theta1))\n", - "print \"Angle in degrees at which first order maxima is observed : %0.2f\"%theta1\n", - "#for the second order maximum we have m=2\n", - "sin_theta2=2*lamda/d\n", - "theta2=degrees(asin(sin_theta2))\n", - "print \"Angle in degrees at which second order maxima is observed : %0.2f\"%theta2\n", - "print \"for higher order number of diffraction the the solutions are non realistic\"" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch24_2.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch24_2.ipynb deleted file mode 100644 index 82f29c9f..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch24_2.ipynb +++ /dev/null @@ -1,233 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 24 : Wave optics" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 1 Page No: 794" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "(A) wavelength of light = 5.63e-07 meters\n", - "(B) Distance between adjacent fringes = 0.022 meters\n" - ] - } - ], - "source": [ - "L=1.2 # Seperation between screen and double-slit in meter\n", - "d=3*10**-5 #distance between the two slits\n", - "m=2 #second order bright fringe\n", - "Y=4.5*10**-2 #distance of second order bright fringe from centerline\n", - "#wavelength of light\n", - "lamda=(Y*d)/(m*L)\n", - "print \"(A) wavelength of light = %0.2e meters\"%lamda\n", - "#distance between adjacent bright fringes\n", - "#delta_Y=Y(m+1)-Ym\n", - "delta_Y=lamda*L/d\n", - "print \"(B) Distance between adjacent fringes = %0.3f meters\"%delta_Y" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 2 Page No: 798" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Minimum thickness of soap bubble film = 113.16 nm is\n" - ] - } - ], - "source": [ - "n=1.33 #refractive index of soap bubble\n", - "lamda=602 #wavelength of light in nm\n", - "#for constructive interference we have 2nt=lamda/2\n", - "t=lamda/(4*n)\n", - "print \"Minimum thickness of soap bubble film = %0.2f nm is\"%t" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 3 Page No: 799" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Minimum thickness of film = 95.17 nm is\n" - ] - } - ], - "source": [ - "n=1.45 #refractive index of silicon monoxide\n", - "lamda=552 #wavelength of light in nm\n", - "#for destructive interference we have condition for minimn thickness 2t=lamda/2n\n", - "t=lamda/(4*n)\n", - "print \"Minimum thickness of film = %0.2f nm is\"%t" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 5 Page No: 801" - ] - }, - { - "cell_type": "code", - "execution_count": 8, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Pit depth in a CD = 121.88 nm\n" - ] - } - ], - "source": [ - "n=1.6 #refractive index of plastic transparent layer\n", - "lamda=780 #wavelength of laser light in nm\n", - "#for destructive interference we have condition for minimn thickness 2t=lamda/2n\n", - "t=lamda/(4*n)\n", - "print \"Pit depth in a CD = %0.2f nm\"%t" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 6 Page No: 804" - ] - }, - { - "cell_type": "code", - "execution_count": 10, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - " Position of first dark fringe = 3.87e-03 meters\n" - ] - } - ], - "source": [ - "lamda=580*10**-9 #wavelength of incident light in meter\n", - "a=0.30*10**-3 #slit width in meter\n", - "L=2 #distance of screen from slit in meters\n", - "#The first dark fringe corresponds to m=+1 or -1\n", - "m=1\n", - "sin_theta=m*lamda/a\n", - "#From fig 24.16 tan_theta=y/L and since theta is very small we have sin_theta=tan_theta hence sin_theta=y/L\n", - "y=L*sin_theta \n", - "print \" Position of first dark fringe = %0.2e meters\"%y" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 7 Page No: 808" - ] - }, - { - "cell_type": "code", - "execution_count": 11, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Angle in degrees at which first order maxima is observed : 22.32\n", - "Angle in degrees at which second order maxima is observed : 49.43\n", - "for higher order number of diffraction the the solutions are non realistic\n" - ] - } - ], - "source": [ - "from math import asin, degrees\n", - "lamda=632.8 #wavelength of monochromatic light from helium-neon laser in meter\n", - "a=6000 #lines in diffraction grating per cm\n", - "d=10**7/a#slit seperation in nm\n", - "#for the first order maximum we have m=1\n", - "sin_theta1=lamda/d\n", - "theta1=degrees(asin(sin_theta1))\n", - "print \"Angle in degrees at which first order maxima is observed : %0.2f\"%theta1\n", - "#for the second order maximum we have m=2\n", - "sin_theta2=2*lamda/d\n", - "theta2=degrees(asin(sin_theta2))\n", - "print \"Angle in degrees at which second order maxima is observed : %0.2f\"%theta2\n", - "print \"for higher order number of diffraction the the solutions are non realistic\"" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch24_3.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch24_3.ipynb deleted file mode 100644 index 82f29c9f..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch24_3.ipynb +++ /dev/null @@ -1,233 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 24 : Wave optics" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 1 Page No: 794" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "(A) wavelength of light = 5.63e-07 meters\n", - "(B) Distance between adjacent fringes = 0.022 meters\n" - ] - } - ], - "source": [ - "L=1.2 # Seperation between screen and double-slit in meter\n", - "d=3*10**-5 #distance between the two slits\n", - "m=2 #second order bright fringe\n", - "Y=4.5*10**-2 #distance of second order bright fringe from centerline\n", - "#wavelength of light\n", - "lamda=(Y*d)/(m*L)\n", - "print \"(A) wavelength of light = %0.2e meters\"%lamda\n", - "#distance between adjacent bright fringes\n", - "#delta_Y=Y(m+1)-Ym\n", - "delta_Y=lamda*L/d\n", - "print \"(B) Distance between adjacent fringes = %0.3f meters\"%delta_Y" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 2 Page No: 798" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Minimum thickness of soap bubble film = 113.16 nm is\n" - ] - } - ], - "source": [ - "n=1.33 #refractive index of soap bubble\n", - "lamda=602 #wavelength of light in nm\n", - "#for constructive interference we have 2nt=lamda/2\n", - "t=lamda/(4*n)\n", - "print \"Minimum thickness of soap bubble film = %0.2f nm is\"%t" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 3 Page No: 799" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Minimum thickness of film = 95.17 nm is\n" - ] - } - ], - "source": [ - "n=1.45 #refractive index of silicon monoxide\n", - "lamda=552 #wavelength of light in nm\n", - "#for destructive interference we have condition for minimn thickness 2t=lamda/2n\n", - "t=lamda/(4*n)\n", - "print \"Minimum thickness of film = %0.2f nm is\"%t" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 5 Page No: 801" - ] - }, - { - "cell_type": "code", - "execution_count": 8, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Pit depth in a CD = 121.88 nm\n" - ] - } - ], - "source": [ - "n=1.6 #refractive index of plastic transparent layer\n", - "lamda=780 #wavelength of laser light in nm\n", - "#for destructive interference we have condition for minimn thickness 2t=lamda/2n\n", - "t=lamda/(4*n)\n", - "print \"Pit depth in a CD = %0.2f nm\"%t" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 6 Page No: 804" - ] - }, - { - "cell_type": "code", - "execution_count": 10, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - " Position of first dark fringe = 3.87e-03 meters\n" - ] - } - ], - "source": [ - "lamda=580*10**-9 #wavelength of incident light in meter\n", - "a=0.30*10**-3 #slit width in meter\n", - "L=2 #distance of screen from slit in meters\n", - "#The first dark fringe corresponds to m=+1 or -1\n", - "m=1\n", - "sin_theta=m*lamda/a\n", - "#From fig 24.16 tan_theta=y/L and since theta is very small we have sin_theta=tan_theta hence sin_theta=y/L\n", - "y=L*sin_theta \n", - "print \" Position of first dark fringe = %0.2e meters\"%y" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 7 Page No: 808" - ] - }, - { - "cell_type": "code", - "execution_count": 11, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Angle in degrees at which first order maxima is observed : 22.32\n", - "Angle in degrees at which second order maxima is observed : 49.43\n", - "for higher order number of diffraction the the solutions are non realistic\n" - ] - } - ], - "source": [ - "from math import asin, degrees\n", - "lamda=632.8 #wavelength of monochromatic light from helium-neon laser in meter\n", - "a=6000 #lines in diffraction grating per cm\n", - "d=10**7/a#slit seperation in nm\n", - "#for the first order maximum we have m=1\n", - "sin_theta1=lamda/d\n", - "theta1=degrees(asin(sin_theta1))\n", - "print \"Angle in degrees at which first order maxima is observed : %0.2f\"%theta1\n", - "#for the second order maximum we have m=2\n", - "sin_theta2=2*lamda/d\n", - "theta2=degrees(asin(sin_theta2))\n", - "print \"Angle in degrees at which second order maxima is observed : %0.2f\"%theta2\n", - "print \"for higher order number of diffraction the the solutions are non realistic\"" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch25_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch25_1.ipynb deleted file mode 100644 index 918e8b63..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch25_1.ipynb +++ /dev/null @@ -1,291 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 25 : Optical Instruments" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 1 Page No: 827" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) focal length f = 50.00 cm\n", - "b) Power of the lens = 2.00 diopters\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "q=-50 # Near point of an eye in cm\n", - "p=25 #object location in cm\n", - "#a) focal length calculation\n", - "#Using Thin Lens equation 1/f=((1/p)+(1/q))\n", - "f=p*q/(p+q)\n", - "print \"a) focal length f = %0.2f cm\"%f\n", - "#b) power of the lens\n", - "f1=50*10**-2# focal length in meters\n", - "P=1/f1\n", - "print \"b) Power of the lens = %0.2f diopters\"%P" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 3 Page No: 830" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) Maximum angular magnification of the lens : 3.50\n", - "Angular Magnification of lens when eye is relaxed : 2.50\n" - ] - } - ], - "source": [ - "f=10 # focal length in cm\n", - "#a)Maximum angular magnification\n", - "M_max=1+(25/f)\n", - "print \"a) Maximum angular magnification of the lens : %0.2f\"%M_max\n", - "m=25/f\n", - "print \"Angular Magnification of lens when eye is relaxed : %0.2f\"%m" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 4 Page No: 832" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Magnification of microscope with two long focal lengths : -45.00\n", - "Magnification of microscope with a combination of 20 mm objective and 2.5 cm eyepiece : -90.00\n", - "Magnification of microscope with a combination of 20 mm objective and 2.5 cm eyepiece : -450.00 \n", - "Possible magnification of microscope with two short focal lengths : -900.00\n" - ] - } - ], - "source": [ - "#interchangeable objectives\n", - "f1=2 # focal length in cm\n", - "f2=0.2 #focal length in cm\n", - "#data of two eye pieces\n", - "f3=5 #focal length in cm\n", - "f4=2.5 #focal length in cm\n", - "L=18 # length of microscope\n", - "#Calculation of magnification for four combinations of lens\n", - "#magnification of compound microscope m =-(L/fo)*(25cm/fe) where fo is shortest focal length compared to fe\n", - "#combination of two long focal lengths\n", - "m1=-(L/f1)*(25/f3)\n", - "print \"Magnification of microscope with two long focal lengths : %0.2f\"%m1\n", - "#combination of 20 mm objective and 2.5 cm eyepiece\n", - "m2=-(L/f1)*(25/f4)\n", - "print \"Magnification of microscope with a combination of 20 mm objective and 2.5 cm eyepiece : %0.2f\"%m2\n", - "#combination of 2 mm objective and 5 cm eyepiece\n", - "m3=-(L/f2)*(25/f3)\n", - "print \"Magnification of microscope with a combination of 20 mm objective and 2.5 cm eyepiece : %0.2f \"%m3\n", - "#combination of two short focal lengths\n", - "m4=-(L/f2)*(25/f4)\n", - "print \"Possible magnification of microscope with two short focal lengths : %0.2f\"%m4" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 5 Page No: 834" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Angular magnification of the telescope : 83.33\n" - ] - } - ], - "source": [ - "d=8 #diameter of objective mirror of reflecting telescope in inches\n", - "fo=1500 #focal length of objective mirror of reflecting telescope in mm\n", - "fe=18 #focal length of eyepiece\n", - "m=fo/fe\n", - "print \"Angular magnification of the telescope : %0.2f\"%m" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 6 Page No: 837" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) Limiting angle of resolution in radians : 7.98e-07\n", - "b) Maximum limit of resolution for the microscope in radians : 5.42e-07\n", - "c) Limiting angle of resolution for the microscope when water filled the space b/w the object and objective in radians : 6.00e-07\n" - ] - } - ], - "source": [ - "l=589*10**-9 #Wavelength of sodium light m\n", - "d=90*10**-2 #diameter of the aperture in m \n", - "L=400*10**-9 #Wavelength of desirable Visble light\n", - "n=1.33 #refractive index of water\n", - "#a) Calculation of limiting angle of resolution\n", - "#Limiting angle of resolution of the circular aperture is Theta_min=1.22*(l/d)\n", - "Theta_min1=1.22*(l/d)\n", - "print \"a) Limiting angle of resolution in radians : %0.2e\"%Theta_min1\n", - "#b) Calculation of maximum limit of resolution for the microscope\n", - "Theta_min2=1.22*(L/d)\n", - "print \"b) Maximum limit of resolution for the microscope in radians : %0.2e\"%Theta_min2\n", - "#c)Effect of water b/w the object and objective on resolving power of microscope\n", - "lw=l/n\n", - "Theta_min3=1.22*(lw/d)\n", - "print \"c) Limiting angle of resolution for the microscope when water filled the space b/w the object and objective in radians : %0.2e\"%Theta_min3" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 7 Page No: 838" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Magnification of telescope A is : 166.67\n", - "Magnification of telescope B is : 50.00\n" - ] - } - ], - "source": [ - "f1=1000# focal length of objective of telescope A in mm\n", - "f2=1250# focal length of objective of telescope B in mm\n", - "f3=6# focal length of eyepiece of telescope A in mm\n", - "f4=25# focal length of eyepiece of telescope Bin mm\n", - "#C) Calculation of magnification of the telescope\n", - "m_A=f1/f3\n", - "m_B=f2/f4\n", - "print \"Magnification of telescope A is : %0.2f\"%m_A\n", - "print \"Magnification of telescope B is : %0.2f\"%m_B" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 8 Page No: 839" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) Resolving poer of grating inorder to distinguish the wavelengths = 998.31\n", - "b) No.of lines of the grating illuminated to resolve the lines in the second order spectrum are 499\n" - ] - } - ], - "source": [ - "L1=589 # wavelength of first bright line in sodium spectrum in nm\n", - "L2=589.59 # wavelength of second bright line in sodium spectrum in nm\n", - "m=2 # order of the spectrum\n", - "delta_L=L2-L1\n", - "R=L1/delta_L\n", - "print \"a) Resolving poer of grating inorder to distinguish the wavelengths = %0.2f\"% R\n", - "N=R/m\n", - "print \"b) No.of lines of the grating illuminated to resolve the lines in the second order spectrum are %d\"%N" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch25_2.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch25_2.ipynb deleted file mode 100644 index 918e8b63..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch25_2.ipynb +++ /dev/null @@ -1,291 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 25 : Optical Instruments" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 1 Page No: 827" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) focal length f = 50.00 cm\n", - "b) Power of the lens = 2.00 diopters\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "q=-50 # Near point of an eye in cm\n", - "p=25 #object location in cm\n", - "#a) focal length calculation\n", - "#Using Thin Lens equation 1/f=((1/p)+(1/q))\n", - "f=p*q/(p+q)\n", - "print \"a) focal length f = %0.2f cm\"%f\n", - "#b) power of the lens\n", - "f1=50*10**-2# focal length in meters\n", - "P=1/f1\n", - "print \"b) Power of the lens = %0.2f diopters\"%P" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 3 Page No: 830" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) Maximum angular magnification of the lens : 3.50\n", - "Angular Magnification of lens when eye is relaxed : 2.50\n" - ] - } - ], - "source": [ - "f=10 # focal length in cm\n", - "#a)Maximum angular magnification\n", - "M_max=1+(25/f)\n", - "print \"a) Maximum angular magnification of the lens : %0.2f\"%M_max\n", - "m=25/f\n", - "print \"Angular Magnification of lens when eye is relaxed : %0.2f\"%m" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 4 Page No: 832" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Magnification of microscope with two long focal lengths : -45.00\n", - "Magnification of microscope with a combination of 20 mm objective and 2.5 cm eyepiece : -90.00\n", - "Magnification of microscope with a combination of 20 mm objective and 2.5 cm eyepiece : -450.00 \n", - "Possible magnification of microscope with two short focal lengths : -900.00\n" - ] - } - ], - "source": [ - "#interchangeable objectives\n", - "f1=2 # focal length in cm\n", - "f2=0.2 #focal length in cm\n", - "#data of two eye pieces\n", - "f3=5 #focal length in cm\n", - "f4=2.5 #focal length in cm\n", - "L=18 # length of microscope\n", - "#Calculation of magnification for four combinations of lens\n", - "#magnification of compound microscope m =-(L/fo)*(25cm/fe) where fo is shortest focal length compared to fe\n", - "#combination of two long focal lengths\n", - "m1=-(L/f1)*(25/f3)\n", - "print \"Magnification of microscope with two long focal lengths : %0.2f\"%m1\n", - "#combination of 20 mm objective and 2.5 cm eyepiece\n", - "m2=-(L/f1)*(25/f4)\n", - "print \"Magnification of microscope with a combination of 20 mm objective and 2.5 cm eyepiece : %0.2f\"%m2\n", - "#combination of 2 mm objective and 5 cm eyepiece\n", - "m3=-(L/f2)*(25/f3)\n", - "print \"Magnification of microscope with a combination of 20 mm objective and 2.5 cm eyepiece : %0.2f \"%m3\n", - "#combination of two short focal lengths\n", - "m4=-(L/f2)*(25/f4)\n", - "print \"Possible magnification of microscope with two short focal lengths : %0.2f\"%m4" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 5 Page No: 834" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Angular magnification of the telescope : 83.33\n" - ] - } - ], - "source": [ - "d=8 #diameter of objective mirror of reflecting telescope in inches\n", - "fo=1500 #focal length of objective mirror of reflecting telescope in mm\n", - "fe=18 #focal length of eyepiece\n", - "m=fo/fe\n", - "print \"Angular magnification of the telescope : %0.2f\"%m" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 6 Page No: 837" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) Limiting angle of resolution in radians : 7.98e-07\n", - "b) Maximum limit of resolution for the microscope in radians : 5.42e-07\n", - "c) Limiting angle of resolution for the microscope when water filled the space b/w the object and objective in radians : 6.00e-07\n" - ] - } - ], - "source": [ - "l=589*10**-9 #Wavelength of sodium light m\n", - "d=90*10**-2 #diameter of the aperture in m \n", - "L=400*10**-9 #Wavelength of desirable Visble light\n", - "n=1.33 #refractive index of water\n", - "#a) Calculation of limiting angle of resolution\n", - "#Limiting angle of resolution of the circular aperture is Theta_min=1.22*(l/d)\n", - "Theta_min1=1.22*(l/d)\n", - "print \"a) Limiting angle of resolution in radians : %0.2e\"%Theta_min1\n", - "#b) Calculation of maximum limit of resolution for the microscope\n", - "Theta_min2=1.22*(L/d)\n", - "print \"b) Maximum limit of resolution for the microscope in radians : %0.2e\"%Theta_min2\n", - "#c)Effect of water b/w the object and objective on resolving power of microscope\n", - "lw=l/n\n", - "Theta_min3=1.22*(lw/d)\n", - "print \"c) Limiting angle of resolution for the microscope when water filled the space b/w the object and objective in radians : %0.2e\"%Theta_min3" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 7 Page No: 838" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Magnification of telescope A is : 166.67\n", - "Magnification of telescope B is : 50.00\n" - ] - } - ], - "source": [ - "f1=1000# focal length of objective of telescope A in mm\n", - "f2=1250# focal length of objective of telescope B in mm\n", - "f3=6# focal length of eyepiece of telescope A in mm\n", - "f4=25# focal length of eyepiece of telescope Bin mm\n", - "#C) Calculation of magnification of the telescope\n", - "m_A=f1/f3\n", - "m_B=f2/f4\n", - "print \"Magnification of telescope A is : %0.2f\"%m_A\n", - "print \"Magnification of telescope B is : %0.2f\"%m_B" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 8 Page No: 839" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) Resolving poer of grating inorder to distinguish the wavelengths = 998.31\n", - "b) No.of lines of the grating illuminated to resolve the lines in the second order spectrum are 499\n" - ] - } - ], - "source": [ - "L1=589 # wavelength of first bright line in sodium spectrum in nm\n", - "L2=589.59 # wavelength of second bright line in sodium spectrum in nm\n", - "m=2 # order of the spectrum\n", - "delta_L=L2-L1\n", - "R=L1/delta_L\n", - "print \"a) Resolving poer of grating inorder to distinguish the wavelengths = %0.2f\"% R\n", - "N=R/m\n", - "print \"b) No.of lines of the grating illuminated to resolve the lines in the second order spectrum are %d\"%N" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch25_3.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch25_3.ipynb deleted file mode 100644 index 918e8b63..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch25_3.ipynb +++ /dev/null @@ -1,291 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 25 : Optical Instruments" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 1 Page No: 827" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) focal length f = 50.00 cm\n", - "b) Power of the lens = 2.00 diopters\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "q=-50 # Near point of an eye in cm\n", - "p=25 #object location in cm\n", - "#a) focal length calculation\n", - "#Using Thin Lens equation 1/f=((1/p)+(1/q))\n", - "f=p*q/(p+q)\n", - "print \"a) focal length f = %0.2f cm\"%f\n", - "#b) power of the lens\n", - "f1=50*10**-2# focal length in meters\n", - "P=1/f1\n", - "print \"b) Power of the lens = %0.2f diopters\"%P" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 3 Page No: 830" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) Maximum angular magnification of the lens : 3.50\n", - "Angular Magnification of lens when eye is relaxed : 2.50\n" - ] - } - ], - "source": [ - "f=10 # focal length in cm\n", - "#a)Maximum angular magnification\n", - "M_max=1+(25/f)\n", - "print \"a) Maximum angular magnification of the lens : %0.2f\"%M_max\n", - "m=25/f\n", - "print \"Angular Magnification of lens when eye is relaxed : %0.2f\"%m" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 4 Page No: 832" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Magnification of microscope with two long focal lengths : -45.00\n", - "Magnification of microscope with a combination of 20 mm objective and 2.5 cm eyepiece : -90.00\n", - "Magnification of microscope with a combination of 20 mm objective and 2.5 cm eyepiece : -450.00 \n", - "Possible magnification of microscope with two short focal lengths : -900.00\n" - ] - } - ], - "source": [ - "#interchangeable objectives\n", - "f1=2 # focal length in cm\n", - "f2=0.2 #focal length in cm\n", - "#data of two eye pieces\n", - "f3=5 #focal length in cm\n", - "f4=2.5 #focal length in cm\n", - "L=18 # length of microscope\n", - "#Calculation of magnification for four combinations of lens\n", - "#magnification of compound microscope m =-(L/fo)*(25cm/fe) where fo is shortest focal length compared to fe\n", - "#combination of two long focal lengths\n", - "m1=-(L/f1)*(25/f3)\n", - "print \"Magnification of microscope with two long focal lengths : %0.2f\"%m1\n", - "#combination of 20 mm objective and 2.5 cm eyepiece\n", - "m2=-(L/f1)*(25/f4)\n", - "print \"Magnification of microscope with a combination of 20 mm objective and 2.5 cm eyepiece : %0.2f\"%m2\n", - "#combination of 2 mm objective and 5 cm eyepiece\n", - "m3=-(L/f2)*(25/f3)\n", - "print \"Magnification of microscope with a combination of 20 mm objective and 2.5 cm eyepiece : %0.2f \"%m3\n", - "#combination of two short focal lengths\n", - "m4=-(L/f2)*(25/f4)\n", - "print \"Possible magnification of microscope with two short focal lengths : %0.2f\"%m4" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 5 Page No: 834" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Angular magnification of the telescope : 83.33\n" - ] - } - ], - "source": [ - "d=8 #diameter of objective mirror of reflecting telescope in inches\n", - "fo=1500 #focal length of objective mirror of reflecting telescope in mm\n", - "fe=18 #focal length of eyepiece\n", - "m=fo/fe\n", - "print \"Angular magnification of the telescope : %0.2f\"%m" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 6 Page No: 837" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) Limiting angle of resolution in radians : 7.98e-07\n", - "b) Maximum limit of resolution for the microscope in radians : 5.42e-07\n", - "c) Limiting angle of resolution for the microscope when water filled the space b/w the object and objective in radians : 6.00e-07\n" - ] - } - ], - "source": [ - "l=589*10**-9 #Wavelength of sodium light m\n", - "d=90*10**-2 #diameter of the aperture in m \n", - "L=400*10**-9 #Wavelength of desirable Visble light\n", - "n=1.33 #refractive index of water\n", - "#a) Calculation of limiting angle of resolution\n", - "#Limiting angle of resolution of the circular aperture is Theta_min=1.22*(l/d)\n", - "Theta_min1=1.22*(l/d)\n", - "print \"a) Limiting angle of resolution in radians : %0.2e\"%Theta_min1\n", - "#b) Calculation of maximum limit of resolution for the microscope\n", - "Theta_min2=1.22*(L/d)\n", - "print \"b) Maximum limit of resolution for the microscope in radians : %0.2e\"%Theta_min2\n", - "#c)Effect of water b/w the object and objective on resolving power of microscope\n", - "lw=l/n\n", - "Theta_min3=1.22*(lw/d)\n", - "print \"c) Limiting angle of resolution for the microscope when water filled the space b/w the object and objective in radians : %0.2e\"%Theta_min3" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 7 Page No: 838" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Magnification of telescope A is : 166.67\n", - "Magnification of telescope B is : 50.00\n" - ] - } - ], - "source": [ - "f1=1000# focal length of objective of telescope A in mm\n", - "f2=1250# focal length of objective of telescope B in mm\n", - "f3=6# focal length of eyepiece of telescope A in mm\n", - "f4=25# focal length of eyepiece of telescope Bin mm\n", - "#C) Calculation of magnification of the telescope\n", - "m_A=f1/f3\n", - "m_B=f2/f4\n", - "print \"Magnification of telescope A is : %0.2f\"%m_A\n", - "print \"Magnification of telescope B is : %0.2f\"%m_B" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 8 Page No: 839" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) Resolving poer of grating inorder to distinguish the wavelengths = 998.31\n", - "b) No.of lines of the grating illuminated to resolve the lines in the second order spectrum are 499\n" - ] - } - ], - "source": [ - "L1=589 # wavelength of first bright line in sodium spectrum in nm\n", - "L2=589.59 # wavelength of second bright line in sodium spectrum in nm\n", - "m=2 # order of the spectrum\n", - "delta_L=L2-L1\n", - "R=L1/delta_L\n", - "print \"a) Resolving poer of grating inorder to distinguish the wavelengths = %0.2f\"% R\n", - "N=R/m\n", - "print \"b) No.of lines of the grating illuminated to resolve the lines in the second order spectrum are %d\"%N" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch26_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch26_1.ipynb deleted file mode 100644 index 461f6400..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch26_1.ipynb +++ /dev/null @@ -1,350 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 26 : Relativity" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example1 Page No: 855" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Period of the pendulum w.r.t to observer = 9.61 \n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "from math import sqrt\n", - "Tp=3 #proper time in sec\n", - "c=3*10**8 #velocity of light in m/sec\n", - "v=0.95*c\n", - "gamma=1/sqrt(1-(v**2/c**2))\n", - "T=gamma*Tp\n", - "print \"Period of the pendulum w.r.t to observer = %.2f \"%T" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example2 Page No: 857" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Length of spaceship measured by moving observer = 16.93 meters\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "Lp=120 # length of space ship in meters\n", - "c=3*10**8 #velocity of light in m/sec\n", - "v=0.99*c\n", - "gamma=1/sqrt(1-(v**2/c**2))\n", - "L=Lp/gamma\n", - "print \"Length of spaceship measured by moving observer = %0.2f meters\"%L" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example3 Page No: 859" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Distance from spaceship to the groung measured by an observer in spaceship = 105.75 meters\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "Lp=435 # length of space ship in meters\n", - "c=3*10**8 #velocity of light in m/sec\n", - "v=0.970*c\n", - "gamma=1/sqrt(1-(v**2/c**2))\n", - "L=Lp/gamma\n", - "print \"Distance from spaceship to the groung measured by an observer in spaceship = %0.2f meters\"%L" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example4 Page No: 861" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The observer sees the horizontal dimension of the spaceship gets contracted to a length of 16.24 meters\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "c=3*10**8 #velocity of light in m/sec\n", - "#when the spaceship is at rest\n", - "x=52 # diatance in x direction in meters\n", - "y=25 #measurement in y direction\n", - "v=0.95*c\n", - "#when the spaceship moves to an observer at rest only x dimension looks contracted\n", - "gamma=1/sqrt(1-(v**2/c**2))\n", - "L=x/gamma\n", - "print \"The observer sees the horizontal dimension of the spaceship gets contracted to a length of %0.2f meters\"%L" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example5 Page No: 862" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "relativistic momentum = 3.10e-22 kg.m/s\n", - "classical momentum = 2.05e-22 kg.m/s\n", - "the relativistic result is 51 percent greater than classical result\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "c=3*10**8 #velocity of light in m/sec\n", - "m=9.11*10**-31 #mass of electron in kg\n", - "v=0.75*c\n", - "gamma=1/sqrt(1-(v**2/c**2))\n", - "#relativistic momentum\n", - "p=m*v*gamma\n", - "print \"relativistic momentum = %0.2e kg.m/s\"%p\n", - "#classical approach\n", - "P=m*v\n", - "print \"classical momentum = %0.2e kg.m/s\"%P\n", - "Z=(p-P)*100/P\n", - "print \"the relativistic result is %d percent greater than classical result\"%Z" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example6 Page No: 864" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "velocity of light w.r.t stationary observer = 3.00e+08 m/sec\n" - ] - } - ], - "source": [ - "c=3*10**8 #velocity of light in m/sec\n", - "Vmo=0.80*c # velocity of motocycle w.r.t stationary observer \n", - "Vlm=c # velocity of motocycle w.r.t motorcycle\n", - "#velocity of light w.r.t stationary observer \n", - "Vlo=(Vlm+Vmo)/(1+(Vlm*Vmo)/c**2)\n", - "print \"velocity of light w.r.t stationary observer = %0.2e m/sec\"%Vlo" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example7 Page No: 865" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The energy equivalent of baseball = 4.50e+16 joules\n" - ] - } - ], - "source": [ - "c=3*10**8 #velocity of light in m/sec\n", - "m=0.50 #mass of baseball in kg\n", - "E=m*c**2\n", - "print \"The energy equivalent of baseball = %0.2e joules\"%E" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 8 Page No: 866" - ] - }, - { - "cell_type": "code", - "execution_count": 8, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "total energy of an electron = 0.97 Mev\n", - "Kinetic energy of electron = 0.46 Mev\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "c=3*10**8 #velocity of light in m/sec\n", - "m=0.511 #rest energy of electron in Mev\n", - "v=0.85*c\n", - "gamma=1/sqrt(1-(v**2/c**2))\n", - "E=(m)*gamma\n", - "print \"total energy of an electron = %0.2f Mev\"%E\n", - "K=E-m\n", - "print \"Kinetic energy of electron = %0.2f Mev\"%K" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 9 Page No: 867" - ] - }, - { - "cell_type": "code", - "execution_count": 9, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) Kinetic energy released in fission = 200.62 Mev\n", - " Speed of Barium fragment = 1.66e+07 Mev\n", - " Speed of krypton fragment = 2.05e+07 Mev\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "m_n=1.008665 #mass of neutron in amu\n", - "m_U=235.043924 #atomic mass of uranium in amu\n", - "m_Ba=140.903496 #atomic mass of barium in amu\n", - "m_Kr=91.907720 #atomic mass of krypton in amu\n", - "c=3*10**8 # velocity of light in m/s\n", - "#a) Kinetic energy released in fission of uranium\n", - "KE_final_=((m_n+m_U)-(m_Ba+m_Kr+(3*m_n)))*c**2\n", - "#1 amu = 931.494 Mev/c**2\n", - "KE_final=KE_final_*931.494/c**2\n", - "print \"a) Kinetic energy released in fission = %0.2f Mev\"%KE_final\n", - "#b) velocities of barium and krypton\n", - "#E=mc2/sqrt(1-v2/c2)\n", - "KE_Ba=KE_final\n", - "m_Ba_=m_Ba*931.494/c**2 # mass of barium in Mev\n", - "E_Ba=KE_Ba+m_Ba_*c**2\n", - "V_Ba=(sqrt(1-(((m_Ba_*c**2)**2)/E_Ba**2)))*c\n", - "print \" Speed of Barium fragment = %0.2e Mev\"%V_Ba\n", - "KE_Kr=KE_final\n", - "m_Kr_=m_Kr*931.494/c**2 # mass of krypton in Mev\n", - "E_Kr=KE_Kr+m_Kr_*c**2\n", - "V_Kr=(sqrt(1-((m_Kr_*c**2)**2)/E_Kr**2))*c\n", - "print \" Speed of krypton fragment = %0.2e Mev\"%V_Kr\n", - "#The difference in answer is because of round off" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch26_2.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch26_2.ipynb deleted file mode 100644 index 461f6400..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch26_2.ipynb +++ /dev/null @@ -1,350 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 26 : Relativity" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example1 Page No: 855" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Period of the pendulum w.r.t to observer = 9.61 \n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "from math import sqrt\n", - "Tp=3 #proper time in sec\n", - "c=3*10**8 #velocity of light in m/sec\n", - "v=0.95*c\n", - "gamma=1/sqrt(1-(v**2/c**2))\n", - "T=gamma*Tp\n", - "print \"Period of the pendulum w.r.t to observer = %.2f \"%T" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example2 Page No: 857" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Length of spaceship measured by moving observer = 16.93 meters\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "Lp=120 # length of space ship in meters\n", - "c=3*10**8 #velocity of light in m/sec\n", - "v=0.99*c\n", - "gamma=1/sqrt(1-(v**2/c**2))\n", - "L=Lp/gamma\n", - "print \"Length of spaceship measured by moving observer = %0.2f meters\"%L" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example3 Page No: 859" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Distance from spaceship to the groung measured by an observer in spaceship = 105.75 meters\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "Lp=435 # length of space ship in meters\n", - "c=3*10**8 #velocity of light in m/sec\n", - "v=0.970*c\n", - "gamma=1/sqrt(1-(v**2/c**2))\n", - "L=Lp/gamma\n", - "print \"Distance from spaceship to the groung measured by an observer in spaceship = %0.2f meters\"%L" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example4 Page No: 861" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The observer sees the horizontal dimension of the spaceship gets contracted to a length of 16.24 meters\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "c=3*10**8 #velocity of light in m/sec\n", - "#when the spaceship is at rest\n", - "x=52 # diatance in x direction in meters\n", - "y=25 #measurement in y direction\n", - "v=0.95*c\n", - "#when the spaceship moves to an observer at rest only x dimension looks contracted\n", - "gamma=1/sqrt(1-(v**2/c**2))\n", - "L=x/gamma\n", - "print \"The observer sees the horizontal dimension of the spaceship gets contracted to a length of %0.2f meters\"%L" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example5 Page No: 862" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "relativistic momentum = 3.10e-22 kg.m/s\n", - "classical momentum = 2.05e-22 kg.m/s\n", - "the relativistic result is 51 percent greater than classical result\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "c=3*10**8 #velocity of light in m/sec\n", - "m=9.11*10**-31 #mass of electron in kg\n", - "v=0.75*c\n", - "gamma=1/sqrt(1-(v**2/c**2))\n", - "#relativistic momentum\n", - "p=m*v*gamma\n", - "print \"relativistic momentum = %0.2e kg.m/s\"%p\n", - "#classical approach\n", - "P=m*v\n", - "print \"classical momentum = %0.2e kg.m/s\"%P\n", - "Z=(p-P)*100/P\n", - "print \"the relativistic result is %d percent greater than classical result\"%Z" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example6 Page No: 864" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "velocity of light w.r.t stationary observer = 3.00e+08 m/sec\n" - ] - } - ], - "source": [ - "c=3*10**8 #velocity of light in m/sec\n", - "Vmo=0.80*c # velocity of motocycle w.r.t stationary observer \n", - "Vlm=c # velocity of motocycle w.r.t motorcycle\n", - "#velocity of light w.r.t stationary observer \n", - "Vlo=(Vlm+Vmo)/(1+(Vlm*Vmo)/c**2)\n", - "print \"velocity of light w.r.t stationary observer = %0.2e m/sec\"%Vlo" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example7 Page No: 865" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The energy equivalent of baseball = 4.50e+16 joules\n" - ] - } - ], - "source": [ - "c=3*10**8 #velocity of light in m/sec\n", - "m=0.50 #mass of baseball in kg\n", - "E=m*c**2\n", - "print \"The energy equivalent of baseball = %0.2e joules\"%E" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 8 Page No: 866" - ] - }, - { - "cell_type": "code", - "execution_count": 8, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "total energy of an electron = 0.97 Mev\n", - "Kinetic energy of electron = 0.46 Mev\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "c=3*10**8 #velocity of light in m/sec\n", - "m=0.511 #rest energy of electron in Mev\n", - "v=0.85*c\n", - "gamma=1/sqrt(1-(v**2/c**2))\n", - "E=(m)*gamma\n", - "print \"total energy of an electron = %0.2f Mev\"%E\n", - "K=E-m\n", - "print \"Kinetic energy of electron = %0.2f Mev\"%K" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 9 Page No: 867" - ] - }, - { - "cell_type": "code", - "execution_count": 9, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) Kinetic energy released in fission = 200.62 Mev\n", - " Speed of Barium fragment = 1.66e+07 Mev\n", - " Speed of krypton fragment = 2.05e+07 Mev\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "m_n=1.008665 #mass of neutron in amu\n", - "m_U=235.043924 #atomic mass of uranium in amu\n", - "m_Ba=140.903496 #atomic mass of barium in amu\n", - "m_Kr=91.907720 #atomic mass of krypton in amu\n", - "c=3*10**8 # velocity of light in m/s\n", - "#a) Kinetic energy released in fission of uranium\n", - "KE_final_=((m_n+m_U)-(m_Ba+m_Kr+(3*m_n)))*c**2\n", - "#1 amu = 931.494 Mev/c**2\n", - "KE_final=KE_final_*931.494/c**2\n", - "print \"a) Kinetic energy released in fission = %0.2f Mev\"%KE_final\n", - "#b) velocities of barium and krypton\n", - "#E=mc2/sqrt(1-v2/c2)\n", - "KE_Ba=KE_final\n", - "m_Ba_=m_Ba*931.494/c**2 # mass of barium in Mev\n", - "E_Ba=KE_Ba+m_Ba_*c**2\n", - "V_Ba=(sqrt(1-(((m_Ba_*c**2)**2)/E_Ba**2)))*c\n", - "print \" Speed of Barium fragment = %0.2e Mev\"%V_Ba\n", - "KE_Kr=KE_final\n", - "m_Kr_=m_Kr*931.494/c**2 # mass of krypton in Mev\n", - "E_Kr=KE_Kr+m_Kr_*c**2\n", - "V_Kr=(sqrt(1-((m_Kr_*c**2)**2)/E_Kr**2))*c\n", - "print \" Speed of krypton fragment = %0.2e Mev\"%V_Kr\n", - "#The difference in answer is because of round off" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch26_3.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch26_3.ipynb deleted file mode 100644 index 461f6400..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch26_3.ipynb +++ /dev/null @@ -1,350 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 26 : Relativity" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example1 Page No: 855" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Period of the pendulum w.r.t to observer = 9.61 \n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "from math import sqrt\n", - "Tp=3 #proper time in sec\n", - "c=3*10**8 #velocity of light in m/sec\n", - "v=0.95*c\n", - "gamma=1/sqrt(1-(v**2/c**2))\n", - "T=gamma*Tp\n", - "print \"Period of the pendulum w.r.t to observer = %.2f \"%T" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example2 Page No: 857" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Length of spaceship measured by moving observer = 16.93 meters\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "Lp=120 # length of space ship in meters\n", - "c=3*10**8 #velocity of light in m/sec\n", - "v=0.99*c\n", - "gamma=1/sqrt(1-(v**2/c**2))\n", - "L=Lp/gamma\n", - "print \"Length of spaceship measured by moving observer = %0.2f meters\"%L" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example3 Page No: 859" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Distance from spaceship to the groung measured by an observer in spaceship = 105.75 meters\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "Lp=435 # length of space ship in meters\n", - "c=3*10**8 #velocity of light in m/sec\n", - "v=0.970*c\n", - "gamma=1/sqrt(1-(v**2/c**2))\n", - "L=Lp/gamma\n", - "print \"Distance from spaceship to the groung measured by an observer in spaceship = %0.2f meters\"%L" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example4 Page No: 861" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The observer sees the horizontal dimension of the spaceship gets contracted to a length of 16.24 meters\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "c=3*10**8 #velocity of light in m/sec\n", - "#when the spaceship is at rest\n", - "x=52 # diatance in x direction in meters\n", - "y=25 #measurement in y direction\n", - "v=0.95*c\n", - "#when the spaceship moves to an observer at rest only x dimension looks contracted\n", - "gamma=1/sqrt(1-(v**2/c**2))\n", - "L=x/gamma\n", - "print \"The observer sees the horizontal dimension of the spaceship gets contracted to a length of %0.2f meters\"%L" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example5 Page No: 862" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "relativistic momentum = 3.10e-22 kg.m/s\n", - "classical momentum = 2.05e-22 kg.m/s\n", - "the relativistic result is 51 percent greater than classical result\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "c=3*10**8 #velocity of light in m/sec\n", - "m=9.11*10**-31 #mass of electron in kg\n", - "v=0.75*c\n", - "gamma=1/sqrt(1-(v**2/c**2))\n", - "#relativistic momentum\n", - "p=m*v*gamma\n", - "print \"relativistic momentum = %0.2e kg.m/s\"%p\n", - "#classical approach\n", - "P=m*v\n", - "print \"classical momentum = %0.2e kg.m/s\"%P\n", - "Z=(p-P)*100/P\n", - "print \"the relativistic result is %d percent greater than classical result\"%Z" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example6 Page No: 864" - ] - }, - { - "cell_type": "code", - "execution_count": 6, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "velocity of light w.r.t stationary observer = 3.00e+08 m/sec\n" - ] - } - ], - "source": [ - "c=3*10**8 #velocity of light in m/sec\n", - "Vmo=0.80*c # velocity of motocycle w.r.t stationary observer \n", - "Vlm=c # velocity of motocycle w.r.t motorcycle\n", - "#velocity of light w.r.t stationary observer \n", - "Vlo=(Vlm+Vmo)/(1+(Vlm*Vmo)/c**2)\n", - "print \"velocity of light w.r.t stationary observer = %0.2e m/sec\"%Vlo" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example7 Page No: 865" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The energy equivalent of baseball = 4.50e+16 joules\n" - ] - } - ], - "source": [ - "c=3*10**8 #velocity of light in m/sec\n", - "m=0.50 #mass of baseball in kg\n", - "E=m*c**2\n", - "print \"The energy equivalent of baseball = %0.2e joules\"%E" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 8 Page No: 866" - ] - }, - { - "cell_type": "code", - "execution_count": 8, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "total energy of an electron = 0.97 Mev\n", - "Kinetic energy of electron = 0.46 Mev\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "c=3*10**8 #velocity of light in m/sec\n", - "m=0.511 #rest energy of electron in Mev\n", - "v=0.85*c\n", - "gamma=1/sqrt(1-(v**2/c**2))\n", - "E=(m)*gamma\n", - "print \"total energy of an electron = %0.2f Mev\"%E\n", - "K=E-m\n", - "print \"Kinetic energy of electron = %0.2f Mev\"%K" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 9 Page No: 867" - ] - }, - { - "cell_type": "code", - "execution_count": 9, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) Kinetic energy released in fission = 200.62 Mev\n", - " Speed of Barium fragment = 1.66e+07 Mev\n", - " Speed of krypton fragment = 2.05e+07 Mev\n" - ] - } - ], - "source": [ - "from math import sqrt\n", - "m_n=1.008665 #mass of neutron in amu\n", - "m_U=235.043924 #atomic mass of uranium in amu\n", - "m_Ba=140.903496 #atomic mass of barium in amu\n", - "m_Kr=91.907720 #atomic mass of krypton in amu\n", - "c=3*10**8 # velocity of light in m/s\n", - "#a) Kinetic energy released in fission of uranium\n", - "KE_final_=((m_n+m_U)-(m_Ba+m_Kr+(3*m_n)))*c**2\n", - "#1 amu = 931.494 Mev/c**2\n", - "KE_final=KE_final_*931.494/c**2\n", - "print \"a) Kinetic energy released in fission = %0.2f Mev\"%KE_final\n", - "#b) velocities of barium and krypton\n", - "#E=mc2/sqrt(1-v2/c2)\n", - "KE_Ba=KE_final\n", - "m_Ba_=m_Ba*931.494/c**2 # mass of barium in Mev\n", - "E_Ba=KE_Ba+m_Ba_*c**2\n", - "V_Ba=(sqrt(1-(((m_Ba_*c**2)**2)/E_Ba**2)))*c\n", - "print \" Speed of Barium fragment = %0.2e Mev\"%V_Ba\n", - "KE_Kr=KE_final\n", - "m_Kr_=m_Kr*931.494/c**2 # mass of krypton in Mev\n", - "E_Kr=KE_Kr+m_Kr_*c**2\n", - "V_Kr=(sqrt(1-((m_Kr_*c**2)**2)/E_Kr**2))*c\n", - "print \" Speed of krypton fragment = %0.2e Mev\"%V_Kr\n", - "#The difference in answer is because of round off" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch27_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch27_1.ipynb deleted file mode 100644 index cc3bdd59..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch27_1.ipynb +++ /dev/null @@ -1,401 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 27 : Quantum physics" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 1 Page No: 874" - ] - }, - { - "cell_type": "code", - "execution_count": 9, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Wavelength at which radiation emitted from the skin reaches its peak = 9.41e-06 meters\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "T=35 #Temperature of the skin in celsius\n", - "T1=T+273 #Temperature in kelvin\n", - "#From Wien's displacement law \n", - "Lambda_max=(0.2898*10**-2)/T1\n", - "print \"Wavelength at which radiation emitted from the skin reaches its peak = %0.2e meters\"%Lambda_max" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 2 Page No: 878" - ] - }, - { - "cell_type": "code", - "execution_count": 12, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) Total energy of Simple harmonic oscillator with given amplitude = 2.00 Joules\n", - " Frequency of oscillation = 0.56 Hertz\n", - "b) Quantum number for the given macroscopic system : 5.36e+33\n", - "c) Energy carried away by a one-quantum charge = 3.73e-34 joules\n" - ] - } - ], - "source": [ - "from math import sqrt, pi\n", - "m=2 # mass of the object in Kg\n", - "k=25 #force constant of spring in N/m\n", - "A=0.4 #Amplitude of Simple harmonic oscillation by spring in meters\n", - "h=6.63*10**-34#js\n", - "#a) Total energy and frequency of SHO calculation\n", - "E=(1/2)*k*A**2\n", - "f=(1/(2*pi))*sqrt(k/m)\n", - "print \"a) Total energy of Simple harmonic oscillator with given amplitude = %0.2f Joules\"%E\n", - "print \" Frequency of oscillation = %0.2f Hertz\"%f\n", - "#b) Calculation of quantum number for the system\n", - "n=E/(h*f)\n", - "print \"b) Quantum number for the given macroscopic system : %0.2e\"%n\n", - "#c) Calculation of energy carried away in a quantum charge\n", - "delta_E=h*f\n", - "print \"c) Energy carried away by a one-quantum charge = %0.2e joules\"%delta_E" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 3 Page No: 879" - ] - }, - { - "cell_type": "code", - "execution_count": 14, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Energy carried by a photon with the given frequency = 3.98e-19 Joules\n" - ] - } - ], - "source": [ - "f=6*10**14 #frequency of yellow light in hertz\n", - "h=6.63*10**-34 #plancks constant J.s\n", - "E=h*f\n", - "print \"Energy carried by a photon with the given frequency = %0.2e Joules\"%E" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 4 Page No: 882" - ] - }, - { - "cell_type": "code", - "execution_count": 16, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) Maximum Kinetic energy of th eejected photoelectrons = 1.68 ev is\n", - "b) Cut off wavelength for sodium = 5.05e-07 meters\n" - ] - } - ], - "source": [ - "l=0.3*10**-6 #wavelength of light in meters\n", - "W=2.46 #work function for sodium in ev\n", - "c=3*10**8 #velocity of light in m/s\n", - "h=6.63*10**-34#js\n", - "#a) Maximum KE of the ejected photoelectrons\n", - "E=(h*c/l)/(1.6*10**-19) #energy of each photon of th eilluminating light beam in ev\n", - "KE_max=E-W\n", - "print \"a) Maximum Kinetic energy of th eejected photoelectrons = %0.2f ev is\"%KE_max\n", - "#b) Cut off wavelength for sodium \n", - "W1=W*1.6*10**-19\n", - "lc=h*c/W1\n", - "print \"b) Cut off wavelength for sodium = %0.2e meters\"%lc" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 5 Page No: 885" - ] - }, - { - "cell_type": "code", - "execution_count": 18, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Minimum wavelength produced = 1.24e-11 meters\n" - ] - } - ], - "source": [ - "V=10**5 #potential difference in Volts\n", - "h=6.63*10**-34 # plancks constant in J.s\n", - "c=3*10**8# velocity of light in m/s\n", - "e=1.6*10**-19# elelctronic charge in coulombs\n", - "L_min=(h*c)/(e*V)\n", - "print \"Minimum wavelength produced = %0.2e meters\"%L_min" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 6 Page No: 886" - ] - }, - { - "cell_type": "code", - "execution_count": 20, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Grazing angle at first order of interference = 6.40 degree\n", - "Grazing angle at third order of interference = 19.54 degree\n" - ] - } - ], - "source": [ - "from math import asin, degrees\n", - "d=0.314 #spacing between certain planes in a crystal of calcite in nm\n", - "l=0.070 #wavelength of X-rays in nm\n", - "m=1# first order of interference\n", - "theta1=degrees(asin((m*l)/(2*d)))\n", - "print \"Grazing angle at first order of interference = %0.2f degree\"%theta1\n", - "m=3 #third order of interference\n", - "theta2=degrees(asin((m*l)/(2*d)))\n", - "print \"Grazing angle at third order of interference = %0.2f degree\"%theta2" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 7 Page No: 887" - ] - }, - { - "cell_type": "code", - "execution_count": 23, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Wavelength of the scattered X-rays at the given angle in 0.20 nm\n" - ] - } - ], - "source": [ - "from math import pi, cos\n", - "Lo=0.200000 #wavelength of X-rays in nm\n", - "h=6.63*10**-34 #in J.s\n", - "m_e=9.11*10**-31 # in Kg\n", - "c=3*10**8 #in m/s\n", - "theta=45 #in degrees\n", - "#wavelength is represented by d\n", - "delta_L=(h/(m_e*c))*(1-cos(pi/180*theta))\n", - "L=delta_L+Lo\n", - "print \"Wavelength of the scattered X-rays at the given angle in %.2f nm\"%L\n", - "\n", - "#Answer given in textbook is wrong" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 8 Page No: 887" - ] - }, - { - "cell_type": "code", - "execution_count": 25, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "de Broglie wavelength for an electron = 7.28e-11 meters\n" - ] - } - ], - "source": [ - "h=6.63*10**-34 #in J.s\n", - "m_e=9.11*10**-31 # in Kg\n", - "v=1*10**7 #in m/s\n", - "lamda=h/(m_e*v)\n", - "print \"de Broglie wavelength for an electron = %0.2e meters\"%lamda" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 9 Page No: 888" - ] - }, - { - "cell_type": "code", - "execution_count": 27, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "de Broglie wavelength of the ball = 1.14e-34 meters\n" - ] - } - ], - "source": [ - "h=6.63*10**-34 #in J.s\n", - "m=0.145 # in Kg\n", - "v=40 #in m/s\n", - "lamda=h/(m*v)\n", - "print \"de Broglie wavelength of the ball = %0.2e meters\"%lamda" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 10 Page No: 889" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Uncertainity in position of electron = 3.86e-06 Meters\n" - ] - } - ], - "source": [ - "from math import pi\n", - "h=6.63*10**-34#js\n", - "v=5*10**3 #speed of the electron in m/s\n", - "m_e=9.11*10**-31 # mass of electron in Kg\n", - "p=m_e*v\n", - "delta_p=0.00300*p\n", - "#Uncertainity principle states delta_x*delta_p >=h/(4*pi)\n", - "delta_x=h/(4*pi*delta_p)\n", - "print \"Uncertainity in position of electron = %0.2e Meters\"%delta_x" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 11 Page No: 889" - ] - }, - { - "cell_type": "code", - "execution_count": 8, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - " Minimum uncertainity in energy of the excited states = 5.28e-27 Joules\n" - ] - } - ], - "source": [ - "from math import pi\n", - "h=6.63*10**-34 # plancks constant in J.s\n", - "delta_t=1.00*10**-8 # Average time that an ellectron exists in the excited states in sec\n", - "delta_E=h/(4*pi*delta_t)\n", - "print \" Minimum uncertainity in energy of the excited states = %0.2e Joules\"%delta_E" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch27_2.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch27_2.ipynb deleted file mode 100644 index cc3bdd59..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch27_2.ipynb +++ /dev/null @@ -1,401 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 27 : Quantum physics" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 1 Page No: 874" - ] - }, - { - "cell_type": "code", - "execution_count": 9, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Wavelength at which radiation emitted from the skin reaches its peak = 9.41e-06 meters\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "T=35 #Temperature of the skin in celsius\n", - "T1=T+273 #Temperature in kelvin\n", - "#From Wien's displacement law \n", - "Lambda_max=(0.2898*10**-2)/T1\n", - "print \"Wavelength at which radiation emitted from the skin reaches its peak = %0.2e meters\"%Lambda_max" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 2 Page No: 878" - ] - }, - { - "cell_type": "code", - "execution_count": 12, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) Total energy of Simple harmonic oscillator with given amplitude = 2.00 Joules\n", - " Frequency of oscillation = 0.56 Hertz\n", - "b) Quantum number for the given macroscopic system : 5.36e+33\n", - "c) Energy carried away by a one-quantum charge = 3.73e-34 joules\n" - ] - } - ], - "source": [ - "from math import sqrt, pi\n", - "m=2 # mass of the object in Kg\n", - "k=25 #force constant of spring in N/m\n", - "A=0.4 #Amplitude of Simple harmonic oscillation by spring in meters\n", - "h=6.63*10**-34#js\n", - "#a) Total energy and frequency of SHO calculation\n", - "E=(1/2)*k*A**2\n", - "f=(1/(2*pi))*sqrt(k/m)\n", - "print \"a) Total energy of Simple harmonic oscillator with given amplitude = %0.2f Joules\"%E\n", - "print \" Frequency of oscillation = %0.2f Hertz\"%f\n", - "#b) Calculation of quantum number for the system\n", - "n=E/(h*f)\n", - "print \"b) Quantum number for the given macroscopic system : %0.2e\"%n\n", - "#c) Calculation of energy carried away in a quantum charge\n", - "delta_E=h*f\n", - "print \"c) Energy carried away by a one-quantum charge = %0.2e joules\"%delta_E" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 3 Page No: 879" - ] - }, - { - "cell_type": "code", - "execution_count": 14, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Energy carried by a photon with the given frequency = 3.98e-19 Joules\n" - ] - } - ], - "source": [ - "f=6*10**14 #frequency of yellow light in hertz\n", - "h=6.63*10**-34 #plancks constant J.s\n", - "E=h*f\n", - "print \"Energy carried by a photon with the given frequency = %0.2e Joules\"%E" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 4 Page No: 882" - ] - }, - { - "cell_type": "code", - "execution_count": 16, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) Maximum Kinetic energy of th eejected photoelectrons = 1.68 ev is\n", - "b) Cut off wavelength for sodium = 5.05e-07 meters\n" - ] - } - ], - "source": [ - "l=0.3*10**-6 #wavelength of light in meters\n", - "W=2.46 #work function for sodium in ev\n", - "c=3*10**8 #velocity of light in m/s\n", - "h=6.63*10**-34#js\n", - "#a) Maximum KE of the ejected photoelectrons\n", - "E=(h*c/l)/(1.6*10**-19) #energy of each photon of th eilluminating light beam in ev\n", - "KE_max=E-W\n", - "print \"a) Maximum Kinetic energy of th eejected photoelectrons = %0.2f ev is\"%KE_max\n", - "#b) Cut off wavelength for sodium \n", - "W1=W*1.6*10**-19\n", - "lc=h*c/W1\n", - "print \"b) Cut off wavelength for sodium = %0.2e meters\"%lc" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 5 Page No: 885" - ] - }, - { - "cell_type": "code", - "execution_count": 18, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Minimum wavelength produced = 1.24e-11 meters\n" - ] - } - ], - "source": [ - "V=10**5 #potential difference in Volts\n", - "h=6.63*10**-34 # plancks constant in J.s\n", - "c=3*10**8# velocity of light in m/s\n", - "e=1.6*10**-19# elelctronic charge in coulombs\n", - "L_min=(h*c)/(e*V)\n", - "print \"Minimum wavelength produced = %0.2e meters\"%L_min" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 6 Page No: 886" - ] - }, - { - "cell_type": "code", - "execution_count": 20, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Grazing angle at first order of interference = 6.40 degree\n", - "Grazing angle at third order of interference = 19.54 degree\n" - ] - } - ], - "source": [ - "from math import asin, degrees\n", - "d=0.314 #spacing between certain planes in a crystal of calcite in nm\n", - "l=0.070 #wavelength of X-rays in nm\n", - "m=1# first order of interference\n", - "theta1=degrees(asin((m*l)/(2*d)))\n", - "print \"Grazing angle at first order of interference = %0.2f degree\"%theta1\n", - "m=3 #third order of interference\n", - "theta2=degrees(asin((m*l)/(2*d)))\n", - "print \"Grazing angle at third order of interference = %0.2f degree\"%theta2" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 7 Page No: 887" - ] - }, - { - "cell_type": "code", - "execution_count": 23, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Wavelength of the scattered X-rays at the given angle in 0.20 nm\n" - ] - } - ], - "source": [ - "from math import pi, cos\n", - "Lo=0.200000 #wavelength of X-rays in nm\n", - "h=6.63*10**-34 #in J.s\n", - "m_e=9.11*10**-31 # in Kg\n", - "c=3*10**8 #in m/s\n", - "theta=45 #in degrees\n", - "#wavelength is represented by d\n", - "delta_L=(h/(m_e*c))*(1-cos(pi/180*theta))\n", - "L=delta_L+Lo\n", - "print \"Wavelength of the scattered X-rays at the given angle in %.2f nm\"%L\n", - "\n", - "#Answer given in textbook is wrong" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 8 Page No: 887" - ] - }, - { - "cell_type": "code", - "execution_count": 25, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "de Broglie wavelength for an electron = 7.28e-11 meters\n" - ] - } - ], - "source": [ - "h=6.63*10**-34 #in J.s\n", - "m_e=9.11*10**-31 # in Kg\n", - "v=1*10**7 #in m/s\n", - "lamda=h/(m_e*v)\n", - "print \"de Broglie wavelength for an electron = %0.2e meters\"%lamda" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 9 Page No: 888" - ] - }, - { - "cell_type": "code", - "execution_count": 27, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "de Broglie wavelength of the ball = 1.14e-34 meters\n" - ] - } - ], - "source": [ - "h=6.63*10**-34 #in J.s\n", - "m=0.145 # in Kg\n", - "v=40 #in m/s\n", - "lamda=h/(m*v)\n", - "print \"de Broglie wavelength of the ball = %0.2e meters\"%lamda" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 10 Page No: 889" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Uncertainity in position of electron = 3.86e-06 Meters\n" - ] - } - ], - "source": [ - "from math import pi\n", - "h=6.63*10**-34#js\n", - "v=5*10**3 #speed of the electron in m/s\n", - "m_e=9.11*10**-31 # mass of electron in Kg\n", - "p=m_e*v\n", - "delta_p=0.00300*p\n", - "#Uncertainity principle states delta_x*delta_p >=h/(4*pi)\n", - "delta_x=h/(4*pi*delta_p)\n", - "print \"Uncertainity in position of electron = %0.2e Meters\"%delta_x" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 11 Page No: 889" - ] - }, - { - "cell_type": "code", - "execution_count": 8, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - " Minimum uncertainity in energy of the excited states = 5.28e-27 Joules\n" - ] - } - ], - "source": [ - "from math import pi\n", - "h=6.63*10**-34 # plancks constant in J.s\n", - "delta_t=1.00*10**-8 # Average time that an ellectron exists in the excited states in sec\n", - "delta_E=h/(4*pi*delta_t)\n", - "print \" Minimum uncertainity in energy of the excited states = %0.2e Joules\"%delta_E" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch27_3.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch27_3.ipynb deleted file mode 100644 index cc3bdd59..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch27_3.ipynb +++ /dev/null @@ -1,401 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 27 : Quantum physics" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 1 Page No: 874" - ] - }, - { - "cell_type": "code", - "execution_count": 9, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Wavelength at which radiation emitted from the skin reaches its peak = 9.41e-06 meters\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "T=35 #Temperature of the skin in celsius\n", - "T1=T+273 #Temperature in kelvin\n", - "#From Wien's displacement law \n", - "Lambda_max=(0.2898*10**-2)/T1\n", - "print \"Wavelength at which radiation emitted from the skin reaches its peak = %0.2e meters\"%Lambda_max" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 2 Page No: 878" - ] - }, - { - "cell_type": "code", - "execution_count": 12, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) Total energy of Simple harmonic oscillator with given amplitude = 2.00 Joules\n", - " Frequency of oscillation = 0.56 Hertz\n", - "b) Quantum number for the given macroscopic system : 5.36e+33\n", - "c) Energy carried away by a one-quantum charge = 3.73e-34 joules\n" - ] - } - ], - "source": [ - "from math import sqrt, pi\n", - "m=2 # mass of the object in Kg\n", - "k=25 #force constant of spring in N/m\n", - "A=0.4 #Amplitude of Simple harmonic oscillation by spring in meters\n", - "h=6.63*10**-34#js\n", - "#a) Total energy and frequency of SHO calculation\n", - "E=(1/2)*k*A**2\n", - "f=(1/(2*pi))*sqrt(k/m)\n", - "print \"a) Total energy of Simple harmonic oscillator with given amplitude = %0.2f Joules\"%E\n", - "print \" Frequency of oscillation = %0.2f Hertz\"%f\n", - "#b) Calculation of quantum number for the system\n", - "n=E/(h*f)\n", - "print \"b) Quantum number for the given macroscopic system : %0.2e\"%n\n", - "#c) Calculation of energy carried away in a quantum charge\n", - "delta_E=h*f\n", - "print \"c) Energy carried away by a one-quantum charge = %0.2e joules\"%delta_E" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 3 Page No: 879" - ] - }, - { - "cell_type": "code", - "execution_count": 14, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Energy carried by a photon with the given frequency = 3.98e-19 Joules\n" - ] - } - ], - "source": [ - "f=6*10**14 #frequency of yellow light in hertz\n", - "h=6.63*10**-34 #plancks constant J.s\n", - "E=h*f\n", - "print \"Energy carried by a photon with the given frequency = %0.2e Joules\"%E" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 4 Page No: 882" - ] - }, - { - "cell_type": "code", - "execution_count": 16, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) Maximum Kinetic energy of th eejected photoelectrons = 1.68 ev is\n", - "b) Cut off wavelength for sodium = 5.05e-07 meters\n" - ] - } - ], - "source": [ - "l=0.3*10**-6 #wavelength of light in meters\n", - "W=2.46 #work function for sodium in ev\n", - "c=3*10**8 #velocity of light in m/s\n", - "h=6.63*10**-34#js\n", - "#a) Maximum KE of the ejected photoelectrons\n", - "E=(h*c/l)/(1.6*10**-19) #energy of each photon of th eilluminating light beam in ev\n", - "KE_max=E-W\n", - "print \"a) Maximum Kinetic energy of th eejected photoelectrons = %0.2f ev is\"%KE_max\n", - "#b) Cut off wavelength for sodium \n", - "W1=W*1.6*10**-19\n", - "lc=h*c/W1\n", - "print \"b) Cut off wavelength for sodium = %0.2e meters\"%lc" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 5 Page No: 885" - ] - }, - { - "cell_type": "code", - "execution_count": 18, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Minimum wavelength produced = 1.24e-11 meters\n" - ] - } - ], - "source": [ - "V=10**5 #potential difference in Volts\n", - "h=6.63*10**-34 # plancks constant in J.s\n", - "c=3*10**8# velocity of light in m/s\n", - "e=1.6*10**-19# elelctronic charge in coulombs\n", - "L_min=(h*c)/(e*V)\n", - "print \"Minimum wavelength produced = %0.2e meters\"%L_min" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 6 Page No: 886" - ] - }, - { - "cell_type": "code", - "execution_count": 20, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Grazing angle at first order of interference = 6.40 degree\n", - "Grazing angle at third order of interference = 19.54 degree\n" - ] - } - ], - "source": [ - "from math import asin, degrees\n", - "d=0.314 #spacing between certain planes in a crystal of calcite in nm\n", - "l=0.070 #wavelength of X-rays in nm\n", - "m=1# first order of interference\n", - "theta1=degrees(asin((m*l)/(2*d)))\n", - "print \"Grazing angle at first order of interference = %0.2f degree\"%theta1\n", - "m=3 #third order of interference\n", - "theta2=degrees(asin((m*l)/(2*d)))\n", - "print \"Grazing angle at third order of interference = %0.2f degree\"%theta2" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 7 Page No: 887" - ] - }, - { - "cell_type": "code", - "execution_count": 23, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Wavelength of the scattered X-rays at the given angle in 0.20 nm\n" - ] - } - ], - "source": [ - "from math import pi, cos\n", - "Lo=0.200000 #wavelength of X-rays in nm\n", - "h=6.63*10**-34 #in J.s\n", - "m_e=9.11*10**-31 # in Kg\n", - "c=3*10**8 #in m/s\n", - "theta=45 #in degrees\n", - "#wavelength is represented by d\n", - "delta_L=(h/(m_e*c))*(1-cos(pi/180*theta))\n", - "L=delta_L+Lo\n", - "print \"Wavelength of the scattered X-rays at the given angle in %.2f nm\"%L\n", - "\n", - "#Answer given in textbook is wrong" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 8 Page No: 887" - ] - }, - { - "cell_type": "code", - "execution_count": 25, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "de Broglie wavelength for an electron = 7.28e-11 meters\n" - ] - } - ], - "source": [ - "h=6.63*10**-34 #in J.s\n", - "m_e=9.11*10**-31 # in Kg\n", - "v=1*10**7 #in m/s\n", - "lamda=h/(m_e*v)\n", - "print \"de Broglie wavelength for an electron = %0.2e meters\"%lamda" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 9 Page No: 888" - ] - }, - { - "cell_type": "code", - "execution_count": 27, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "de Broglie wavelength of the ball = 1.14e-34 meters\n" - ] - } - ], - "source": [ - "h=6.63*10**-34 #in J.s\n", - "m=0.145 # in Kg\n", - "v=40 #in m/s\n", - "lamda=h/(m*v)\n", - "print \"de Broglie wavelength of the ball = %0.2e meters\"%lamda" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 10 Page No: 889" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Uncertainity in position of electron = 3.86e-06 Meters\n" - ] - } - ], - "source": [ - "from math import pi\n", - "h=6.63*10**-34#js\n", - "v=5*10**3 #speed of the electron in m/s\n", - "m_e=9.11*10**-31 # mass of electron in Kg\n", - "p=m_e*v\n", - "delta_p=0.00300*p\n", - "#Uncertainity principle states delta_x*delta_p >=h/(4*pi)\n", - "delta_x=h/(4*pi*delta_p)\n", - "print \"Uncertainity in position of electron = %0.2e Meters\"%delta_x" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 11 Page No: 889" - ] - }, - { - "cell_type": "code", - "execution_count": 8, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - " Minimum uncertainity in energy of the excited states = 5.28e-27 Joules\n" - ] - } - ], - "source": [ - "from math import pi\n", - "h=6.63*10**-34 # plancks constant in J.s\n", - "delta_t=1.00*10**-8 # Average time that an ellectron exists in the excited states in sec\n", - "delta_E=h/(4*pi*delta_t)\n", - "print \" Minimum uncertainity in energy of the excited states = %0.2e Joules\"%delta_E" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch28_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch28_1.ipynb deleted file mode 100644 index 941b57e6..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch28_1.ipynb +++ /dev/null @@ -1,197 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 28 : Atomic Physics" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 1 Page No: 897" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Wavelength of the emitted photon = 1.22e-07 meters\n", - "frequency of the emitted photon = 2.47e+15 meters\n" - ] - } - ], - "source": [ - "RH=1.097*10**7 #Rydberg constant in per meter\n", - "lamda=4/(3*RH)\n", - "c=3*10**8 # m/sec\n", - "f=c/lamda\n", - "print \"Wavelength of the emitted photon = %0.2e meters\"%lamda\n", - "print \"frequency of the emitted photon = %0.2e meters\"%f" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 2 Page No: 898" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "longest wavelength that photon emmited = 6.56e-07 meters\n", - "Energy emmited by the photon = 3.03e-19 Joules\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "RH=1.097*10**7 #Rydberg constant in per meter\n", - "h=6.626*10**-34 #plancks constant in j.s\n", - "c=3*10**8 # velocity of light in m/s\n", - "nf=2 #quantum number\n", - "ni=3# quantum number\n", - "#assuming k=1/lamda\n", - "k=RH*((1/nf**2-1/ni**2))\n", - "lamda=1/k\n", - "print \"longest wavelength that photon emmited = %0.2e meters\"%lamda\n", - "E_photon=h*c/lamda\n", - "print \"Energy emmited by the photon = %0.2e Joules\"%E_photon" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 3 Page No: 901" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) Energy of the atom in ground state = -54.40 eV\n", - "b) Radius of the ground state orbit = 0.000 nm\n" - ] - } - ], - "source": [ - "Z=2 #atomic number of helium\n", - "n=1 #principal quantum number\n", - "E=-Z**2*13.6/n**2\n", - "print \"a) Energy of the atom in ground state = %0.2f eV\"%E\n", - "r=(n**2/Z)*0.0529#in nm\n", - "print \"b) Radius of the ground state orbit = %0.3f nm\"%r" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 4 Page No: 906" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Energy of the states with quantum number 2 = -3.40 ev\n" - ] - } - ], - "source": [ - "n=2# principal quantum number \n", - "E=-13.6/n**2\n", - "print \"Energy of the states with quantum number 2 = %0.2f ev\"%E" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 6 Page No: 906" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Energy of the characteristic emiited from tungsten target when electron drops from M shell to K shell = 6.61e+04 ev\n" - ] - } - ], - "source": [ - "Z=74 #atomic number of tungsten\n", - "Eo=13.6 #ground state enenrgy in ev\n", - "E_K=-(Z-1)**2*(13.6) #Energy of the electron in K shell\n", - "n=3\n", - "Z_eff=Z-n**2\n", - "E3=Eo/n**2\n", - "E_M=-Z_eff**2*E3\n", - "E=E_M-E_K\n", - "print \"Energy of the characteristic emiited from tungsten target when electron drops from M shell to K shell = %0.2e ev\"%E\n", - "#Difference in answer is because of roundoff" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch28_2.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch28_2.ipynb deleted file mode 100644 index 941b57e6..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch28_2.ipynb +++ /dev/null @@ -1,197 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 28 : Atomic Physics" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 1 Page No: 897" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Wavelength of the emitted photon = 1.22e-07 meters\n", - "frequency of the emitted photon = 2.47e+15 meters\n" - ] - } - ], - "source": [ - "RH=1.097*10**7 #Rydberg constant in per meter\n", - "lamda=4/(3*RH)\n", - "c=3*10**8 # m/sec\n", - "f=c/lamda\n", - "print \"Wavelength of the emitted photon = %0.2e meters\"%lamda\n", - "print \"frequency of the emitted photon = %0.2e meters\"%f" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 2 Page No: 898" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "longest wavelength that photon emmited = 6.56e-07 meters\n", - "Energy emmited by the photon = 3.03e-19 Joules\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "RH=1.097*10**7 #Rydberg constant in per meter\n", - "h=6.626*10**-34 #plancks constant in j.s\n", - "c=3*10**8 # velocity of light in m/s\n", - "nf=2 #quantum number\n", - "ni=3# quantum number\n", - "#assuming k=1/lamda\n", - "k=RH*((1/nf**2-1/ni**2))\n", - "lamda=1/k\n", - "print \"longest wavelength that photon emmited = %0.2e meters\"%lamda\n", - "E_photon=h*c/lamda\n", - "print \"Energy emmited by the photon = %0.2e Joules\"%E_photon" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 3 Page No: 901" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) Energy of the atom in ground state = -54.40 eV\n", - "b) Radius of the ground state orbit = 0.000 nm\n" - ] - } - ], - "source": [ - "Z=2 #atomic number of helium\n", - "n=1 #principal quantum number\n", - "E=-Z**2*13.6/n**2\n", - "print \"a) Energy of the atom in ground state = %0.2f eV\"%E\n", - "r=(n**2/Z)*0.0529#in nm\n", - "print \"b) Radius of the ground state orbit = %0.3f nm\"%r" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 4 Page No: 906" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Energy of the states with quantum number 2 = -3.40 ev\n" - ] - } - ], - "source": [ - "n=2# principal quantum number \n", - "E=-13.6/n**2\n", - "print \"Energy of the states with quantum number 2 = %0.2f ev\"%E" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 6 Page No: 906" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Energy of the characteristic emiited from tungsten target when electron drops from M shell to K shell = 6.61e+04 ev\n" - ] - } - ], - "source": [ - "Z=74 #atomic number of tungsten\n", - "Eo=13.6 #ground state enenrgy in ev\n", - "E_K=-(Z-1)**2*(13.6) #Energy of the electron in K shell\n", - "n=3\n", - "Z_eff=Z-n**2\n", - "E3=Eo/n**2\n", - "E_M=-Z_eff**2*E3\n", - "E=E_M-E_K\n", - "print \"Energy of the characteristic emiited from tungsten target when electron drops from M shell to K shell = %0.2e ev\"%E\n", - "#Difference in answer is because of roundoff" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch28_3.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch28_3.ipynb deleted file mode 100644 index 941b57e6..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch28_3.ipynb +++ /dev/null @@ -1,197 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 28 : Atomic Physics" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 1 Page No: 897" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Wavelength of the emitted photon = 1.22e-07 meters\n", - "frequency of the emitted photon = 2.47e+15 meters\n" - ] - } - ], - "source": [ - "RH=1.097*10**7 #Rydberg constant in per meter\n", - "lamda=4/(3*RH)\n", - "c=3*10**8 # m/sec\n", - "f=c/lamda\n", - "print \"Wavelength of the emitted photon = %0.2e meters\"%lamda\n", - "print \"frequency of the emitted photon = %0.2e meters\"%f" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 2 Page No: 898" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "longest wavelength that photon emmited = 6.56e-07 meters\n", - "Energy emmited by the photon = 3.03e-19 Joules\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "RH=1.097*10**7 #Rydberg constant in per meter\n", - "h=6.626*10**-34 #plancks constant in j.s\n", - "c=3*10**8 # velocity of light in m/s\n", - "nf=2 #quantum number\n", - "ni=3# quantum number\n", - "#assuming k=1/lamda\n", - "k=RH*((1/nf**2-1/ni**2))\n", - "lamda=1/k\n", - "print \"longest wavelength that photon emmited = %0.2e meters\"%lamda\n", - "E_photon=h*c/lamda\n", - "print \"Energy emmited by the photon = %0.2e Joules\"%E_photon" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 3 Page No: 901" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) Energy of the atom in ground state = -54.40 eV\n", - "b) Radius of the ground state orbit = 0.000 nm\n" - ] - } - ], - "source": [ - "Z=2 #atomic number of helium\n", - "n=1 #principal quantum number\n", - "E=-Z**2*13.6/n**2\n", - "print \"a) Energy of the atom in ground state = %0.2f eV\"%E\n", - "r=(n**2/Z)*0.0529#in nm\n", - "print \"b) Radius of the ground state orbit = %0.3f nm\"%r" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 4 Page No: 906" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Energy of the states with quantum number 2 = -3.40 ev\n" - ] - } - ], - "source": [ - "n=2# principal quantum number \n", - "E=-13.6/n**2\n", - "print \"Energy of the states with quantum number 2 = %0.2f ev\"%E" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 6 Page No: 906" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Energy of the characteristic emiited from tungsten target when electron drops from M shell to K shell = 6.61e+04 ev\n" - ] - } - ], - "source": [ - "Z=74 #atomic number of tungsten\n", - "Eo=13.6 #ground state enenrgy in ev\n", - "E_K=-(Z-1)**2*(13.6) #Energy of the electron in K shell\n", - "n=3\n", - "Z_eff=Z-n**2\n", - "E3=Eo/n**2\n", - "E_M=-Z_eff**2*E3\n", - "E=E_M-E_K\n", - "print \"Energy of the characteristic emiited from tungsten target when electron drops from M shell to K shell = %0.2e ev\"%E\n", - "#Difference in answer is because of roundoff" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch29_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch29_1.ipynb deleted file mode 100644 index e3458b6c..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch29_1.ipynb +++ /dev/null @@ -1,263 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 29 : Nuclear Physics" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 1 Page No: 916" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Nuclear density = 2.31e+17 kg/m3\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "from math import pi\n", - "m=1.67*10**-27 #mass of nucleus in kg\n", - "ro=1.2*10**-15 #in meter\n", - "p=(3*m)/(4*pi*(ro)**3)\n", - "print \"Nuclear density = %0.2e kg/m3\"%p" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 2 Page No: 920" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Binding energy of Deuteron = 2.22 Mev\n" - ] - } - ], - "source": [ - "mp=1.007825 #in u\n", - "mn=1.008665 #in u\n", - "md=2.014102 #in u\n", - "u=931.494 #Mev\n", - "M=mp+mn\n", - "delta_m=(M-md) #in u\n", - "E=delta_m*u\n", - "print \"Binding energy of Deuteron = %0.2f Mev\"%E" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 3 Page No: 922" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Activity or decay rate at t=0 = 1.11e-05 Ci\n" - ] - } - ], - "source": [ - "No=3*10**16 #no.of radioactive nuclei present at t=0\n", - "t_half=1.6*10**3 #years\n", - "T_half=t_half*3.16*10**7 #in sec\n", - "d=0.693/T_half\n", - "R_o=d*No # decays/s\n", - "Ci=3.7*10**10\n", - "Ro=R_o/Ci\n", - "print \"Activity or decay rate at t=0 = %0.2e Ci\"%Ro" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 4 Page No: 923" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) No.of atoms remaining after 12 days : 45612654\n", - "Initial activity of the radon sample = 837.68 decay/sec\n" - ] - } - ], - "source": [ - "from math import exp\n", - "T_half=3.83 #half life time of Radon in days\n", - "No=4*10**8 #Initial No .of Radon atoms \n", - "lamda=0.693/T_half # in days\n", - "t=12 \n", - "N=No*exp(-(lamda*t))\n", - "print \"a) No.of atoms remaining after 12 days : %0.f\"%N\n", - "lamda_=lamda/(8.64*10**4)\n", - "R=lamda_*No\n", - "print \"Initial activity of the radon sample = %0.2f decay/sec\"%R" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 5 Page No: 925" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Energy liberated = 4.87 Mev\n" - ] - } - ], - "source": [ - "m_d=222.017571 #mass of daughter nuclei in atomic units\n", - "m_alpha=4.002602 #mass of alpha particle in atomic units\n", - "M_p=226.025402 #mass of parent nuclei in atomic units\n", - "m=m_d+m_alpha\n", - "delta_m=(M_p-m)\n", - "E=delta_m*931.494\n", - "print \"Energy liberated = %0.2f Mev\"%E" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 6 Page No: 927" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Energy released in beta decay = 0.156 Mev\n" - ] - } - ], - "source": [ - "M_C=14.003242 #mass of carbon in atomic mass units\n", - "M_N=14.003074 #mass of nitogen in atomic mass units\n", - "delta_M=M_C-M_N\n", - "E=delta_M*(931.494)\n", - "print \"Energy released in beta decay = %0.3f Mev\"%E" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 7 Page No: 928" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Age of the skeleton = 10915.43 years\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "from math import log\n", - "T_half=3.01*10**9 #half life time in min\n", - "lamda=0.693/T_half\n", - "R=200 # in decay/min\n", - "R0_=15 #decay rate in decay/min.g\n", - "m=50 #weight of carbon\n", - "R0=R0_*m #in decay/min\n", - "t1=-(log(R/R0)/lamda) #im min\n", - "t=t1/525949\n", - "print \"Age of the skeleton = %0.2f years\"%t" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch29_2.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch29_2.ipynb deleted file mode 100644 index e3458b6c..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch29_2.ipynb +++ /dev/null @@ -1,263 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 29 : Nuclear Physics" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 1 Page No: 916" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Nuclear density = 2.31e+17 kg/m3\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "from math import pi\n", - "m=1.67*10**-27 #mass of nucleus in kg\n", - "ro=1.2*10**-15 #in meter\n", - "p=(3*m)/(4*pi*(ro)**3)\n", - "print \"Nuclear density = %0.2e kg/m3\"%p" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 2 Page No: 920" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Binding energy of Deuteron = 2.22 Mev\n" - ] - } - ], - "source": [ - "mp=1.007825 #in u\n", - "mn=1.008665 #in u\n", - "md=2.014102 #in u\n", - "u=931.494 #Mev\n", - "M=mp+mn\n", - "delta_m=(M-md) #in u\n", - "E=delta_m*u\n", - "print \"Binding energy of Deuteron = %0.2f Mev\"%E" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 3 Page No: 922" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Activity or decay rate at t=0 = 1.11e-05 Ci\n" - ] - } - ], - "source": [ - "No=3*10**16 #no.of radioactive nuclei present at t=0\n", - "t_half=1.6*10**3 #years\n", - "T_half=t_half*3.16*10**7 #in sec\n", - "d=0.693/T_half\n", - "R_o=d*No # decays/s\n", - "Ci=3.7*10**10\n", - "Ro=R_o/Ci\n", - "print \"Activity or decay rate at t=0 = %0.2e Ci\"%Ro" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 4 Page No: 923" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) No.of atoms remaining after 12 days : 45612654\n", - "Initial activity of the radon sample = 837.68 decay/sec\n" - ] - } - ], - "source": [ - "from math import exp\n", - "T_half=3.83 #half life time of Radon in days\n", - "No=4*10**8 #Initial No .of Radon atoms \n", - "lamda=0.693/T_half # in days\n", - "t=12 \n", - "N=No*exp(-(lamda*t))\n", - "print \"a) No.of atoms remaining after 12 days : %0.f\"%N\n", - "lamda_=lamda/(8.64*10**4)\n", - "R=lamda_*No\n", - "print \"Initial activity of the radon sample = %0.2f decay/sec\"%R" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 5 Page No: 925" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Energy liberated = 4.87 Mev\n" - ] - } - ], - "source": [ - "m_d=222.017571 #mass of daughter nuclei in atomic units\n", - "m_alpha=4.002602 #mass of alpha particle in atomic units\n", - "M_p=226.025402 #mass of parent nuclei in atomic units\n", - "m=m_d+m_alpha\n", - "delta_m=(M_p-m)\n", - "E=delta_m*931.494\n", - "print \"Energy liberated = %0.2f Mev\"%E" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 6 Page No: 927" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Energy released in beta decay = 0.156 Mev\n" - ] - } - ], - "source": [ - "M_C=14.003242 #mass of carbon in atomic mass units\n", - "M_N=14.003074 #mass of nitogen in atomic mass units\n", - "delta_M=M_C-M_N\n", - "E=delta_M*(931.494)\n", - "print \"Energy released in beta decay = %0.3f Mev\"%E" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 7 Page No: 928" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Age of the skeleton = 10915.43 years\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "from math import log\n", - "T_half=3.01*10**9 #half life time in min\n", - "lamda=0.693/T_half\n", - "R=200 # in decay/min\n", - "R0_=15 #decay rate in decay/min.g\n", - "m=50 #weight of carbon\n", - "R0=R0_*m #in decay/min\n", - "t1=-(log(R/R0)/lamda) #im min\n", - "t=t1/525949\n", - "print \"Age of the skeleton = %0.2f years\"%t" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch29_3.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch29_3.ipynb deleted file mode 100644 index e3458b6c..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch29_3.ipynb +++ /dev/null @@ -1,263 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 29 : Nuclear Physics" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 1 Page No: 916" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Nuclear density = 2.31e+17 kg/m3\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "from math import pi\n", - "m=1.67*10**-27 #mass of nucleus in kg\n", - "ro=1.2*10**-15 #in meter\n", - "p=(3*m)/(4*pi*(ro)**3)\n", - "print \"Nuclear density = %0.2e kg/m3\"%p" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 2 Page No: 920" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Binding energy of Deuteron = 2.22 Mev\n" - ] - } - ], - "source": [ - "mp=1.007825 #in u\n", - "mn=1.008665 #in u\n", - "md=2.014102 #in u\n", - "u=931.494 #Mev\n", - "M=mp+mn\n", - "delta_m=(M-md) #in u\n", - "E=delta_m*u\n", - "print \"Binding energy of Deuteron = %0.2f Mev\"%E" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 3 Page No: 922" - ] - }, - { - "cell_type": "code", - "execution_count": 2, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Activity or decay rate at t=0 = 1.11e-05 Ci\n" - ] - } - ], - "source": [ - "No=3*10**16 #no.of radioactive nuclei present at t=0\n", - "t_half=1.6*10**3 #years\n", - "T_half=t_half*3.16*10**7 #in sec\n", - "d=0.693/T_half\n", - "R_o=d*No # decays/s\n", - "Ci=3.7*10**10\n", - "Ro=R_o/Ci\n", - "print \"Activity or decay rate at t=0 = %0.2e Ci\"%Ro" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 4 Page No: 923" - ] - }, - { - "cell_type": "code", - "execution_count": 3, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "a) No.of atoms remaining after 12 days : 45612654\n", - "Initial activity of the radon sample = 837.68 decay/sec\n" - ] - } - ], - "source": [ - "from math import exp\n", - "T_half=3.83 #half life time of Radon in days\n", - "No=4*10**8 #Initial No .of Radon atoms \n", - "lamda=0.693/T_half # in days\n", - "t=12 \n", - "N=No*exp(-(lamda*t))\n", - "print \"a) No.of atoms remaining after 12 days : %0.f\"%N\n", - "lamda_=lamda/(8.64*10**4)\n", - "R=lamda_*No\n", - "print \"Initial activity of the radon sample = %0.2f decay/sec\"%R" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 5 Page No: 925" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Energy liberated = 4.87 Mev\n" - ] - } - ], - "source": [ - "m_d=222.017571 #mass of daughter nuclei in atomic units\n", - "m_alpha=4.002602 #mass of alpha particle in atomic units\n", - "M_p=226.025402 #mass of parent nuclei in atomic units\n", - "m=m_d+m_alpha\n", - "delta_m=(M_p-m)\n", - "E=delta_m*931.494\n", - "print \"Energy liberated = %0.2f Mev\"%E" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 6 Page No: 927" - ] - }, - { - "cell_type": "code", - "execution_count": 5, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Energy released in beta decay = 0.156 Mev\n" - ] - } - ], - "source": [ - "M_C=14.003242 #mass of carbon in atomic mass units\n", - "M_N=14.003074 #mass of nitogen in atomic mass units\n", - "delta_M=M_C-M_N\n", - "E=delta_M*(931.494)\n", - "print \"Energy released in beta decay = %0.3f Mev\"%E" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 7 Page No: 928" - ] - }, - { - "cell_type": "code", - "execution_count": 7, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Age of the skeleton = 10915.43 years\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "from math import log\n", - "T_half=3.01*10**9 #half life time in min\n", - "lamda=0.693/T_half\n", - "R=200 # in decay/min\n", - "R0_=15 #decay rate in decay/min.g\n", - "m=50 #weight of carbon\n", - "R0=R0_*m #in decay/min\n", - "t1=-(log(R/R0)/lamda) #im min\n", - "t=t1/525949\n", - "print \"Age of the skeleton = %0.2f years\"%t" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch30_1.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch30_1.ipynb deleted file mode 100644 index 3ab7db90..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch30_1.ipynb +++ /dev/null @@ -1,107 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 30 : Nuclear energy and elementary particles" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 2 Page No: 943" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Disintegration energy = 5.33e+26 Mev is\n", - "or = 2.37e+07 KWh\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "Q=208 #disintegration energy per event in Mev\n", - "m=1*10**3 #mass of uranium\n", - "A=235 #mass number or uranium in g/mol\n", - "a=6.02*10**23 #avagadro number nuclei/mol\n", - "N=(a/A)*m #nuclei\n", - "E=N*Q\n", - "P=E*4.45*10**-20\n", - "print \"Disintegration energy = %0.2e Mev is\"%E\n", - "print \"or = %0.2e KWh\"%P" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 3 Page No: 947" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - " Energy release in deuterium-deuterium reaction = 4.03 Mev\n" - ] - } - ], - "source": [ - "m1=2.014102 # mass of deuterium in atomic mass unit\n", - "m2=3.016049 #mass of tritium in atomic mass unit\n", - "m3=1.007825 # mass of hydrogen in atomic mass unit\n", - "#referring to the deuterium-deuterium reaction\n", - "#mass before reaction\n", - "M1=2*m1\n", - "#mass after reaction\n", - "M2=m2+m3\n", - "#excessive mass\n", - "m=M1-M2\n", - "#converting mass into energy\n", - "#1 u = 931.494 Mev\n", - "E=m*931.494\n", - "print \" Energy release in deuterium-deuterium reaction = %0.2f Mev\"%E" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch30_2.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch30_2.ipynb deleted file mode 100644 index 3ab7db90..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch30_2.ipynb +++ /dev/null @@ -1,107 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 30 : Nuclear energy and elementary particles" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 2 Page No: 943" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Disintegration energy = 5.33e+26 Mev is\n", - "or = 2.37e+07 KWh\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "Q=208 #disintegration energy per event in Mev\n", - "m=1*10**3 #mass of uranium\n", - "A=235 #mass number or uranium in g/mol\n", - "a=6.02*10**23 #avagadro number nuclei/mol\n", - "N=(a/A)*m #nuclei\n", - "E=N*Q\n", - "P=E*4.45*10**-20\n", - "print \"Disintegration energy = %0.2e Mev is\"%E\n", - "print \"or = %0.2e KWh\"%P" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 3 Page No: 947" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - " Energy release in deuterium-deuterium reaction = 4.03 Mev\n" - ] - } - ], - "source": [ - "m1=2.014102 # mass of deuterium in atomic mass unit\n", - "m2=3.016049 #mass of tritium in atomic mass unit\n", - "m3=1.007825 # mass of hydrogen in atomic mass unit\n", - "#referring to the deuterium-deuterium reaction\n", - "#mass before reaction\n", - "M1=2*m1\n", - "#mass after reaction\n", - "M2=m2+m3\n", - "#excessive mass\n", - "m=M1-M2\n", - "#converting mass into energy\n", - "#1 u = 931.494 Mev\n", - "E=m*931.494\n", - "print \" Energy release in deuterium-deuterium reaction = %0.2f Mev\"%E" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch30_3.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch30_3.ipynb deleted file mode 100644 index 3ab7db90..00000000 --- a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch30_3.ipynb +++ /dev/null @@ -1,107 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "# Chapter 30 : Nuclear energy and elementary particles" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 2 Page No: 943" - ] - }, - { - "cell_type": "code", - "execution_count": 4, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "Disintegration energy = 5.33e+26 Mev is\n", - "or = 2.37e+07 KWh\n" - ] - } - ], - "source": [ - "from __future__ import division\n", - "Q=208 #disintegration energy per event in Mev\n", - "m=1*10**3 #mass of uranium\n", - "A=235 #mass number or uranium in g/mol\n", - "a=6.02*10**23 #avagadro number nuclei/mol\n", - "N=(a/A)*m #nuclei\n", - "E=N*Q\n", - "P=E*4.45*10**-20\n", - "print \"Disintegration energy = %0.2e Mev is\"%E\n", - "print \"or = %0.2e KWh\"%P" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Example 3 Page No: 947" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "collapsed": false - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - " Energy release in deuterium-deuterium reaction = 4.03 Mev\n" - ] - } - ], - "source": [ - "m1=2.014102 # mass of deuterium in atomic mass unit\n", - "m2=3.016049 #mass of tritium in atomic mass unit\n", - "m3=1.007825 # mass of hydrogen in atomic mass unit\n", - "#referring to the deuterium-deuterium reaction\n", - "#mass before reaction\n", - "M1=2*m1\n", - "#mass after reaction\n", - "M2=m2+m3\n", - "#excessive mass\n", - "m=M1-M2\n", - "#converting mass into energy\n", - "#1 u = 931.494 Mev\n", - "E=m*931.494\n", - "print \" Energy release in deuterium-deuterium reaction = %0.2f Mev\"%E" - ] - } - ], - "metadata": { - "kernelspec": { - "display_name": "Python 2", - "language": "python", - "name": "python2" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 2 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython2", - "version": "2.7.9" - } - }, - "nbformat": 4, - "nbformat_minor": 0 -} diff --git a/Electronic_Devices_and_Circuits_by_J._Paul/Ch10_1.ipynb b/Electronic_Devices_and_Circuits_by_J._Paul/Ch10_1.ipynb new file mode 100644 index 00000000..a6339334 --- /dev/null +++ b/Electronic_Devices_and_Circuits_by_J._Paul/Ch10_1.ipynb @@ -0,0 +1,226 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 10 - Negative Feedback Amplifiers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 467 example 1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "reverse transmission = 0.10\n", + "gain with feedback = 10.00\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "av=1000#\n", + "chvoga=0.001##change in voltage gain\n", + "beta1=1/((chvoga)/(100/av))-1#\n", + "beta1=beta1/av#\n", + "fegain=(av)/(1+(av*(beta1)))#\n", + "print \"reverse transmission = %0.2f\"%((beta1))\n", + "print \"gain with feedback = %0.2f\"%((fegain))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 467 example 2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "output voltage = 2.19\n", + "input voltage = 0.25 volt\n" + ] + } + ], + "source": [ + "voltag=36##volt\n", + "w=0.07##harmonic distortion\n", + "inpvol=0.028##volt\n", + "beta1=0.012#\n", + "a=voltag/inpvol#\n", + "fegain=a/(1+beta1*a)##correction in book\n", + "volta1=fegain*inpvol#\n", + "print \"output voltage = %0.2f\"%((volta1))\n", + "#decrease of gain 9\n", + "inpvol=9*inpvol#\n", + "print \"input voltage = %0.2f\"%((inpvol)),\"volt\"#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 468 example 3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "required input = 1.00 volt\n", + "harmonic distortion = 0.10\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "volgain=2000##voltage gain\n", + "outpower=20##watts\n", + "inpsig=10*10**-3##volts\n", + "fedbac=40##decibel\n", + "fedgai=volgain/100#\n", + "outvol=volgain*inpsig##output voltage\n", + "inpvol=outvol/fedgai##required input\n", + "#10 second harmonic distortion\n", + "distor=(10/100)#\n", + "print \"required input = %0.2f\"%((inpvol)),\"volt\"#\n", + "print \"harmonic distortion = %0.2f\"%((distor))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 469 example 5" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "feedback factor = 0.019\n", + "over gain = 0.005\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "fedgai=60##decibel\n", + "outimp=10*10**3##ohm\n", + "outim1=500##ohm modified impedance\n", + "fedgai=1000#\n", + "fedbac=((outimp/outim1)-(1))/fedgai#\n", + "#10 change in gain\n", + "overga=1/((1+(fedgai*fedbac))/0.1)##over gain\n", + "print \"feedback factor = %0.3f\"%((fedbac))\n", + "print \"over gain = %0.3f\"%((overga))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 470 example 6" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current gain = -11.60\n", + "voltage gain = -46.40\n", + "transconductance = -0.01 ampere per volt\n", + "transresistance = -46403.71 ohm\n", + "input resistance = 1042.65 ohm\n", + "output resistance = 3636.36 ohm\n" + ] + } + ], + "source": [ + "colres=4*10**3##ohm\n", + "r=4*10**3##ohm\n", + "basres=20*10**3##ohm\n", + "r1=1*10**3##ohm\n", + "hie=1.1*10**3#\n", + "hfe=50#\n", + "hoe=(40*10**3)#\n", + "ri=basres*hie/(basres+hie)#\n", + "curgai=((r1/(r1+ri)))*((basres/(basres+hie)))*((-hfe*colres)/(colres+r))#\n", + "volgai=curgai*r/r1#\n", + "tranco=volgai/r#\n", + "tranre=r1*volgai#\n", + "outres=hoe*colres/(hoe+colres)#\n", + "print \"current gain = %0.2f\"%((curgai))\n", + "print \"voltage gain = %0.2f\"%((volgai))\n", + "print \"transconductance = %0.2f\"%((tranco)),\"ampere per volt\"#\n", + "print \"transresistance = %0.2f\"%((tranre)),\"ohm\"#\n", + "print \"input resistance = %0.2f\"%((ri)),\"ohm\"#\n", + "print \"output resistance = %0.2f\"%((outres)),\"ohm\"#" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronic_Devices_and_Circuits_by_J._Paul/Ch11_1.ipynb b/Electronic_Devices_and_Circuits_by_J._Paul/Ch11_1.ipynb new file mode 100644 index 00000000..fc78f77f --- /dev/null +++ b/Electronic_Devices_and_Circuits_by_J._Paul/Ch11_1.ipynb @@ -0,0 +1,322 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 11 - Sinusoidal Oscillators" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 514 example 2" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "min frequency = 1769.29hertz\n", + "max frequency = 17692.85hertz\n", + "resistance r3 = 20000.00ohm\n" + ] + } + ], + "source": [ + "macapa=900*10**-12##farad\n", + "micapa=90*10**-12##farad\n", + "r=100*10**3##ohm\n", + "#(a) frequency range\n", + "fremin=1/(2*3.14*r*macapa)\n", + "print \"min frequency = %0.2f\"%((fremin))+\"hertz\"\n", + "fremax=1/(2*3.14*r*micapa)\n", + "print \"max frequency = %0.2f\"%((fremax))+\"hertz\"\n", + "#(b) r3\n", + "r=10*10**3##ohm\n", + "r3=2*r\n", + "print \"resistance r3 = %0.2f\"%((r3))+\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 516 example 3" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "min voltage >= 7.50volt\n", + "frequency = 42379.83hertz\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "c1=0.004*10**-6##farad\n", + "c2=0.03*10**-6##farad\n", + "induct=4*10**-3##henry\n", + "#min voltage\n", + "mivolt=c2/c1\n", + "print \"min voltage >= %0.2f\"%((mivolt))+\"volt\"\n", + "#frequency\n", + "freque=(((1/(2*3.14)))*sqrt((c1+c2)/(induct*c1*c2)))\n", + "print \"frequency = %0.2f\"%((freque))+\"hertz\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 517 example 5" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "frequency = 166467.63hertz\n", + "ratio1 greater than 1 so oscillations possible\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "induct=500*10**-6##henry\n", + "induc1=5000*10**-6##henry\n", + "mutuin=300*10**-6##henry\n", + "c1=150*10**-12##farad\n", + "#(a) frequency\n", + "indcto=induct+induc1+2*mutuin\n", + "freque=1/((2)*3.14*sqrt(indcto*c1))\n", + "#(b) condition\n", + "r=10*10**3##ohm\n", + "conduc=8*10**-3##ampere per volt\n", + "r1=50*10**3##ohm\n", + "r_=r*r1/(r+r1)\n", + "volgai=conduc*r_\n", + "print \"frequency = %0.2f\"%((freque))+\"hertz\"\n", + "ratio1=(induc1+mutuin)/(induct+mutuin)\n", + "ratio1=ratio1*volgai\n", + "print \"ratio1 greater than 1 so oscillations possible\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 518 example 6" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resonanting capacitance = 5.00e-14farad\n", + "resonant frequency = 1.42e+06hertz\n", + "parallel resonant frequency = 1.03e+06hertz\n", + "series resonant frequency = 1.01e+06hertz\n", + "quality factor = 3162.28\n", + "loop gain = 100.00\n", + "bias = 6.00e-05second\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "cgs=5*10**-12##farad\n", + "cds=1*10**-12##farad\n", + "conduct=10*10**-3##ampere per volt\n", + "rd=50*10**3##ohm\n", + "r=10*10**6##ohm\n", + "induct=0.5##henry\n", + "c1=0.05*10**-12##farad\n", + "rse=1*10**3##ohm\n", + "c=1*10**-12##farad\n", + "#(1) c11\n", + "c11=((((cds*cgs)/(cds+cgs))+1)*c1)/(((cds*cgs)/(cds+cgs))+1+c1)\n", + "print \"resonanting capacitance = %0.2e\"%((c11))+\"farad\"\n", + "#(2) frequency\n", + "freque=((sqrt(2))/(2*3.14*sqrt(induct*c11)))\n", + "print \"resonant frequency = %0.2e\"%((freque))+\"hertz\"\n", + "#(3) frequency parallel\n", + "\n", + "freque=1/(2*3.14*sqrt(((induct*c*c1))/(c+c1)))\n", + "print \"parallel resonant frequency = %0.2e\"%((freque))+\"hertz\"\n", + "#frequency series\n", + "freque=1/((2*3.14*sqrt(induct*c1)))\n", + "print \"series resonant frequency = %0.2e\"%((freque))+\"hertz\"\n", + "qualit=((induct/c1)**(0.5))/rse\n", + "print \"quality factor = %0.2f\"%((qualit))\n", + "#correction required in book\n", + "#(4) loop gain\n", + "abeta1=conduct*rd*cds/cgs\n", + "print \"loop gain = %0.2f\"%((abeta1))\n", + "#(5)\n", + "w=r*(cds+cgs)\n", + "print \"bias = %0.2e\"%((w))+\"second\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 519 example 7" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "frequency = 1.23e+06hertz\n", + "gain = 5.00\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "c=200*10**-12##farad\n", + "c1=1000*10**-12##farad\n", + "induct=100*10**-6##henry\n", + "#(1) frequency\n", + "ceq=(c*c1)/(c+c1)\n", + "freque=1/(2*3.14*(sqrt(induct*ceq)))\n", + "print \"frequency = %0.2e\"%((freque))+\"hertz\"##correction in the book\n", + "gaimin=c1/c\n", + "print \"gain = %0.2f\"%((gaimin))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 520 example 8" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "inductance = 4.02e-05henry\n" + ] + } + ], + "source": [ + "induc1=0.4*10**-3##henry\n", + "c=0.004*10**-6##farad\n", + "freque=120*10**3##hertz\n", + "induct=((1/(4*3.14**2*freque**2*c)))-induc1\n", + "print \"inductance = %0.2e\"%((induct))+\"henry\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 520 example 9" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "frequency = 1087243.22hertz\n", + "ratio parallel series = 1.03\n", + "quality factor = 409.67\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "induct=0.33##henry\n", + "c=0.065*10**-12##farad\n", + "c1=1*10**-12##farad\n", + "r=5.5*10**3##ohm\n", + "#(1) series resonant frequency\n", + "freque=(1/(2*(3.14)))*sqrt(1/((induct)*c))\n", + "print \"frequency = %0.2f\"%((freque))+\"hertz\"\n", + "#(2)exceed of frequency\n", + "ratio1=sqrt((1+(c/c1)))\n", + "print \"ratio parallel series = %0.2f\"%((ratio1))\n", + "#correction required in the book\n", + "#(3) quality factor\n", + "qualit=(1/r)*sqrt(induct/c)\n", + "print \"quality factor = %0.2f\"%((qualit))" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronic_Devices_and_Circuits_by_J._Paul/Ch12_1.ipynb b/Electronic_Devices_and_Circuits_by_J._Paul/Ch12_1.ipynb new file mode 100644 index 00000000..58504fac --- /dev/null +++ b/Electronic_Devices_and_Circuits_by_J._Paul/Ch12_1.ipynb @@ -0,0 +1,351 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 12 - Power Electronic devices " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 553 example 1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "source resistance = 111.87 ohm\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "slope1=130\n", + "trivol=15##volt\n", + "d=0.5##watts\n", + "ig=sqrt(d/slope1)\n", + "vg=slope1*ig\n", + "r=(trivol-vg)/ig\n", + "print \"source resistance = %0.2f\"%((r)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 553 example 2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current = 0.010 ampere\n", + "input current less than required current\n" + ] + } + ], + "source": [ + "from math import exp\n", + "latcur=50*10**-3##ampere\n", + "durpul=50*10**-6##second\n", + "induct=0.5##henry\n", + "r=20##ohm\n", + "voltag=100##volt\n", + "w=induct/r\n", + "inpcur=-(voltag/r)*((1)-exp(-durpul/w))\n", + "print \"current = %0.3f\"%(abs(inpcur)),\"ampere\"\n", + "print \"input current less than required current\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 554 example 3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "min duration = 4.00e-06 second\n" + ] + } + ], + "source": [ + "latcur=4*10**-3##ampere\n", + "induct=0.1##henry\n", + "voltag=100##volt\n", + "durmin=induct*latcur/voltag\n", + "print \"min duration = %0.2e\"%((durmin)),\"second\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 554 example 4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "source resistance = 2000.00 ohm\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "slope1=3*10**3\n", + "egs=10##volt\n", + "d=0.012##watts\n", + "ig=sqrt(d/slope1)\n", + "vg=slope1*ig\n", + "r=(egs-vg)/ig\n", + "\n", + "print \"source resistance = %0.2f\"%((r)),\"ohm\"##it is not given in the book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 554 example 5" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistance = 14.00 ohm\n", + "frequency = 18750.00 hertz\n" + ] + } + ], + "source": [ + "slope1=16\n", + "durmax=4*10**-6##second\n", + "curmin=500*10**-3##ampere\n", + "voltag=15##volt\n", + "#(1) resistance\n", + "vg=slope1*curmin\n", + "r=(voltag-vg)/curmin\n", + "#(2)\n", + "d=vg*curmin\n", + "freque=0.3/(d*durmax)\n", + "print \"resistance = %0.2f\"%((r)),\"ohm\"\n", + "print \"frequency = %0.2f\"%((freque)),\"hertz\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 555 example 6" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "change of voltage = 800.00 volt per microsecond\n" + ] + } + ], + "source": [ + "c1=20*10**-12##farad\n", + "limcur=16*10**-3##ampere\n", + "w=(limcur/c1)*10**-6##convert second to microsecond\n", + "print \"change of voltage = %0.2f\"%((w)),\"volt per microsecond\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 555 example 7" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current = 2121.32 ampere\n", + "current = 45000.00 ampere square second\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "ratcur=3000##ampere\n", + "freque=50##hertz\n", + "i=sqrt(ratcur**2/2)\n", + "print \"current = %0.2f\"%((i)),\"ampere\"\n", + "i=((ratcur)/sqrt(2))**2/(2*freque)\n", + "print \"current = %0.2f\"%((i)),\"ampere square second\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 556 example 9" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "max limit resistance = 1410000.00 ohm\n", + "min limit resistance = 2650.00 ohm\n", + "resistance = 4.67e+04 ohm\n", + "rb1 = 100.00 ohm\n", + "rb2 = 653.59 ohm\n", + "peak voltage = 15.90 volt\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log\n", + "voltag=30##volt\n", + "w=0.51\n", + "i1=10*10**-6##ampere\n", + "v1=3.5##volt\n", + "curen1=10*10**-3##ampere\n", + "freque=60##hertz\n", + "tridun=50*10**-6##second\n", + "pinvol=w*voltag+0.6\n", + "r=(voltag-pinvol)/i1\n", + "print \"max limit resistance = %0.2f\"%((r)),\"ohm\"\n", + "r=(voltag-v1)/(curen1)\n", + "print \"min limit resistance = %0.2f\"%((r)),\"ohm\"\n", + "capac1=0.5*10**-6##farad\n", + "r=(1/freque)*(1/(capac1*log(1/(1-w))))\n", + "print \"resistance = %0.2e\"%((r)),\"ohm\"\n", + "rb2=10**4/(w*voltag)\n", + "rb1=tridun/capac1\n", + "print \"rb1 = %0.2f\"%((rb1)),\"ohm\"\n", + "print \"rb2 = %0.2f\"%((rb2)),\"ohm\"\n", + "print \"peak voltage = %0.2f\"%((pinvol)),\"volt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 557 example 10" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage = 7.00 volt\n", + "this voltage makes to off\n" + ] + } + ], + "source": [ + "re=1*10**3##ohm\n", + "i1=5*10**-3##ampere\n", + "\n", + "voltag=re*i1+2\n", + "print \"voltage = %0.2f\"%((voltag)),\"volt\"\n", + "\n", + "\n", + "print \"this voltage makes to off\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronic_Devices_and_Circuits_by_J._Paul/Ch13_1.ipynb b/Electronic_Devices_and_Circuits_by_J._Paul/Ch13_1.ipynb new file mode 100644 index 00000000..b0a888e4 --- /dev/null +++ b/Electronic_Devices_and_Circuits_by_J._Paul/Ch13_1.ipynb @@ -0,0 +1,357 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter13 - Cathode Ray Oscilloscope" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber : 578 example 1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power to electrons = 8.0 watts\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "quanti=3*10**17#\n", + "voltag=10*10**3##volt\n", + "distan=40*10**-3##metre per minute\n", + "w=quanti*1.6*10**-19*voltag\n", + "w=w/60##per second\n", + "\n", + "print \"power to electrons = \",round((w),2),\"watts\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber : 578 example 2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "input voltage = 9.55 volt\n", + "frequency = 4761.9 hertz\n", + "vm1coswt vm2sinwt squaring and adding gives ellipse\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "sensit=5## per centimetre\n", + "q=50*10**-6##second per centimetre\n", + "petope=5.4##centimetre\n", + "horiax=8.4##centimetre\n", + "voltag=petope*sensit#\n", + "voltag=voltag/((2)*sqrt(2))#\n", + "#one cycle\n", + "horiax=(horiax/2)*q#\n", + "freque=1/horiax#\n", + "print \"input voltage = \",round((voltag),2),\"volt\"\n", + "print \"frequency = \",round((freque),2),\"hertz\"\n", + "\n", + "\n", + "print \"vm1coswt vm2sinwt squaring and adding gives ellipse\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber : 579 example 3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "velocity x = 1.874e+07 metre per second\n", + "velocity x = 3.10e+05 metre per second\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "voltag=1000##volt\n", + "#(1) velocity\n", + "vx=sqrt(2*1.6*10**-19*(voltag)/(9.11*10**-31))#\n", + "print \"velocity x = %0.3e\"%vx,\"metre per second\"\n", + "vox=1*10**5##metre per second intial velocity\n", + "vx=sqrt((vox)+((2*1.6*10**-19*voltag)/(2.01*1.66*10**-27)))#\n", + "\n", + "print \"velocity x = %0.2e\"%vx,\"metre per second\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber : 580 example 4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "transverse magnetic field = 3.87e-04 weber per metre square\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "voltag=2000##volt\n", + "d=15##centimetre\n", + "d1=3##centimetre\n", + "r1=((d**2+d1**2)/(6))*10**-2##centimetre to metre\n", + "vox=sqrt(2*1.6*10**-19*(voltag)/(9.11*10**-31))#\n", + "b=vox/((1.6*10**-19*r1)/(9.11*10**-31))#\n", + "\n", + "print \"transverse magnetic field = %0.2e\"%b,\"weber per metre square\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber : 581 example 5" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "max frequency = 6.63e+08 hertz\n", + "duration electron between the plates = 4.53e-08 second\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "voltag=2000##volt\n", + "d=2*10**-2##metre\n", + "#(1) frequency\n", + "vx=sqrt(2*1.6*10**-19*(voltag)/(9.11*10**-31))#\n", + "durati=d/vx#\n", + "freque=1/(2*durati)#\n", + "print \"max frequency = %0.2e\"%freque,\"hertz\"\n", + "#(2)\n", + "durati=60*durati#\n", + "print \"duration electron between the plates = %0.2e\"%durati,\"second\"#correction in book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber : 582 example 7" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "max velocity = 1.68e+07 metre per second\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "voltag=800##volt\n", + "\n", + "\n", + "q=1.6*10**-19##coulomb\n", + "m=9.11*10**-31##kilogram\n", + "vox=sqrt(2*q*voltag/m)#\n", + "\n", + "print \"max velocity = %0.2e\"%vox,\"metre per second\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber : 582 example 8" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "velocity = 2.65e+07 metre per second\n", + "sensitivity = 3.75e-04 metre per volt\n", + "deflection factor = 2666.67 volt per metre\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "voltag=2000##volt\n", + "d=1.5*10**-2##centimetre\n", + "d1=5*10**-3##metre\n", + "distan=50*10**-2##metre\n", + "#(1) velocity\n", + "vox=sqrt(2*1.6*10**-19*(voltag)/(9.11*10**-31))#\n", + "#(2) sensitivity\n", + "defsen=distan*d/(2*d1*voltag)#\n", + "#deflection factor\n", + "g=1/defsen#\n", + "print \"velocity = %0.2e\"%vox,\"metre per second\"\n", + "print \"sensitivity = %0.2e\"%defsen,\"metre per volt\"\n", + "\n", + "print \"deflection factor = \",round((g),2),\"volt per metre\"#correction in the book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber : 582 example 9" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "velocity = 2.65e+07 metre per second\n", + "fc = 1.33e+08 hertz\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "voltag=2000##volt\n", + "d=50*10**-3##metre\n", + "#(1) velocity\n", + "vox=sqrt(2*1.6*10**-19*(voltag)/(9.11*10**-31))#\n", + "print \"velocity = %0.2e\"%vox,\"metre per second\"\n", + "#(2) fc\n", + "fc=vox/(4*d)#\n", + "\n", + "print \"fc = %0.2e\"%fc,\"hertz\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber : 582 example 10" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "phase angle = 30.0 degre\n" + ] + } + ], + "source": [ + "from math import asin, degrees\n", + "y=2.5##divisions\n", + "y1=1.25##divisions\n", + "y=y1/y#\n", + "w=degrees(asin(y))\n", + "\n", + "print \"phase angle = \",round((w),2),\"degre\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronic_Devices_and_Circuits_by_J._Paul/Ch1_1.ipynb b/Electronic_Devices_and_Circuits_by_J._Paul/Ch1_1.ipynb new file mode 100644 index 00000000..fd90a687 --- /dev/null +++ b/Electronic_Devices_and_Circuits_by_J._Paul/Ch1_1.ipynb @@ -0,0 +1,992 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1 - Semiconductor Physics" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 24 example 1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "minority concentration = 2.25e+12 per metre square\n", + "shift in fermi = 0.23 volt\n", + "minority concentration when n doubled = 9.00e+12 per cubic metre\n" + ] + } + ], + "source": [ + "from math import log\n", + "incaco=1.5*10**16##cubic metre\n", + "resist=2*10**3##ohm metre\n", + "dopcon=10**20##metre\n", + "q=26*10**-3##electron volt\n", + "#(1)\n", + "w=2.25*10**32/dopcon#\n", + "#(3)\n", + "shifer=q*log(dopcon/incaco)##shift in fermi level\n", + "ni=9*10**32#\n", + "#(3)\n", + "w1=ni/dopcon#\n", + "print \"minority concentration = %0.2e\"%((w)),\"per metre square\"#\n", + "print \"shift in fermi = %0.2f\"%((shifer)),\"volt\"#\n", + "print \"minority concentration when n doubled = %0.2e\"%((w1)),\"per cubic metre\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 25 example 2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "conductivity = 7.12e+24 second per metre\n", + "drift velocity = 10.44 metre per second\n", + "density = 2.14e+28 ampere per cubic metre\n" + ] + } + ], + "source": [ + "numfre=7.87*10**28##per cubic metre\n", + "molity=34.8##square centimetre/velocity second\n", + "e=30##volt per centimetre\n", + "#(1)\n", + "molity=molity*10**-4#q=1.6*10**-19#\n", + "conduc=numfre*q*molity#\n", + "#(2)\n", + "e=e*10**2#\n", + "veloci=(molity*e)#\n", + "curden=conduc*e#\n", + "print \"conductivity = %0.2e\"%((conduc)),\"second per metre\"#\n", + "print \"drift velocity = %0.2f\"%((veloci)),\"metre per second\"#\n", + "print \"density = %0.2e\"%((curden)),\"ampere per cubic metre\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 26 example 3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "conductivity = 0.0224 second per centimetre\n", + "conductivity at extent of 1 impurity = 0.30 second per centimetre\n", + "conductivity acceptor to extent of 1 impurity = 1.30 second per centimetre\n" + ] + } + ], + "source": [ + "ni=2.5*10**13##per square centimetre\n", + "moe=3800#square centimetre/velocity second\n", + "mo1=1800##square centimetre/velocity second\n", + "num=4.51*10**22##number of atoms\n", + "q=1.6*10**-19#\n", + "conduc=ni*q*(moe+mo1)#\n", + "num=num/10**7#\n", + "impura=(ni**2)/num#\n", + "ni=5*10**14#\n", + "condu1=ni*q*moe#\n", + "print \"conductivity = %0.4f\"%((conduc)),\"second per centimetre\"#\n", + "print \"conductivity at extent of 1 impurity = %0.2f\"%((condu1)),\"second per centimetre\"##there is mistake in book as 3.04s/cm\n", + "conduc=num*q*mo1#\n", + "print \"conductivity acceptor to extent of 1 impurity = %0.2f\"%((conduc)),\"second per centimetre\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 27 example 4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "conductivity intrinisc at 300kelvin = 4.32e-06 second per centimetre\n", + "conductivity when donor atom added to extent of 1 impurity = 0.104 second per centimetre\n", + "conductivity when acceptor added to extent of 1 impurity = 0.040 second per centimetre\n" + ] + } + ], + "source": [ + "ni=1.5*10**10##per cubic centimetre\n", + "moe=1300##square centimetre/velocity second\n", + "mo1=500##square centimetre/velocity second\n", + "w=5*10**22##atoms per cubic centimetre\n", + "q=1.6*10**-19#\n", + "#(a) conductivity intrinisc at 300kelvin\n", + "conduc=ni*q*(moe+mo1)##conductivity\n", + "u=((ni)/(5*10**14))#\n", + "ni=5*10**14#\n", + "#(b)conductivity when donor atom added to extent of 1 impurity\n", + "condu1=ni*q*moe#\n", + "print \"conductivity intrinisc at 300kelvin = %0.2e\"%((conduc)),\"second per centimetre\"#\n", + "print \"conductivity when donor atom added to extent of 1 impurity = %0.3f\"%((condu1)),\"second per centimetre\"#\n", + "#conductivity when acceptor added to extent of 1 impurity\n", + "conduc=ni*q*mo1#\n", + "print \"conductivity when acceptor added to extent of 1 impurity = %0.3f\"%((conduc)),\"second per centimetre\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 28 example 5" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "conductivity intrinisc at 300kelvin = 0.022 second per centimetre\n", + "conductivity with donor impurity 1 = 27.36 second per centimetre\n", + "conductivity with acceptor impurity 1 = 2.88e-09 second per centimetre\n", + "conductivity on both = 24.62 second per centimetre\n" + ] + } + ], + "source": [ + "ni=2.5*10**13##per cubic centimetre\n", + "moe=3800##square centimetre/velocity second\n", + "mo1=1800##square centimetre/velocity second\n", + "w=4.5*10**22##atoms per cubic centimetre\n", + "q=1.6*10**-19#\n", + "#(1) conductivity intrinisc at 300kelvin\n", + "conduc=ni*q*(moe+mo1)#\n", + "u=10**6#\n", + "u=((w)/(u))#\n", + "#(2) conductivity with donor impurity 1\n", + "condu1=u*q*moe#\n", + "print \"conductivity intrinisc at 300kelvin = %0.3f\"%((conduc)),\"second per centimetre\"#\n", + "print \"conductivity with donor impurity 1 = %0.2f\"%((condu1)),\"second per centimetre\"#\n", + "u=10**7#u=w/u#\n", + "#(3) conductivity with acceptor impurity 1\n", + "conduc=u*q*mo1#\n", + "print \"conductivity with acceptor impurity 1 = %0.2e\"%((conduc)),\"second per centimetre\"#\n", + "u=0.9*(w/10**6)#\n", + "#(4) conductivity on both\n", + "conduc=u*q*moe#\n", + "print \"conductivity on both = %0.2f\"%((conduc)),\"second per centimetre\"#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 29 example 6" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "fermi = 0.33 electron volt\n", + "fermi below the conduction band\n" + ] + } + ], + "source": [ + "ferlev=0.3##electron volt\n", + "u=300##kelvin\n", + "u1=330##kelvin\n", + "ferlev=ferlev*u1/u#\n", + "print \"fermi = %0.2f\"%((ferlev)),\"electron volt\"#\n", + "print \"fermi below the conduction band\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 29 example 7" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "fermi = 0.17 electron volt\n" + ] + } + ], + "source": [ + "from math import log\n", + "ferlev=0.02##electron volt\n", + "q=4##donor impurity added\n", + "w=0.025##electron volt\n", + "ferlev=-((log(q)-8))/40#\n", + "print \"fermi = %0.2f\"%((ferlev)),\"electron volt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 30 example 8" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistance = 1570.39 ohm\n" + ] + } + ], + "source": [ + "from sympy import symbols, solve\n", + "area=1.5*10**-2##centimetre square\n", + "w=1.6##centimetre\n", + "resist=20##ohm centimetre\n", + "durati=60*10**-6##second in book given as mili\n", + "quanti=8*10**15##photons per second\n", + "\n", + "\n", + "#(1) resistance at each photon gives a electron hole pair\n", + "up=1800##centimetre square per velocity second\n", + "un=3800##centimetre square per velocity second\n", + "q=1.6*10**-19##coulomb\n", + "ni=2.5*10**13##per cubic centimetre\n", + "sigma1=1/resist#\n", + "z1=3800#\n", + "z=-sigma1/q#\n", + "u=ni**2/up#\n", + "#n=poly([(z1) z u],'n')#\n", + "n=symbols('n')\n", + "expr=z1*n**2+z*n+u\n", + "n=solve(expr,n)[1]\n", + "n=7.847*10**13##n>ni taken so it is admissible\n", + "p1=ni**2/n#\n", + "volume=w*area#\n", + "nchang=quanti*durati/volume#\n", + "pchang=nchang#\n", + "sigm11=q*((n+nchang)*un+(pchang+p1)*up)#\n", + "resis1=1/sigm11#\n", + "r1=resis1*w/area#\n", + "print \"resistance = %0.2f\"%((r1)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 31 example 9" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "concentration of electron = 8660254037.84 per cubic centimetre\n", + "concentration of holes = 25980762113.53 per cubic centimetre\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt\n", + "moe=1350##square centimetre/velocity second\n", + "mo1=450##square centimetre/velocity second\n", + "ni=1.5*10**10##per cubic centimetre\n", + "concn1=ni*((sqrt(mo1/moe)))##concentration\n", + "concne=((ni**2)/(concn1))\n", + "\n", + "print \"concentration of electron = %0.2f\"%((concn1)),\"per cubic centimetre\"#\n", + "print \"concentration of holes = %0.2f\"%((concne)),\"per cubic centimetre\"#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 32 example 10" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "concentration of hole = 1.09e+21 per cubic centimetre\n", + "concentration of electron = 2.07e+11 per cubic centimetre\n" + ] + } + ], + "source": [ + "resist=0.12##ohm metre\n", + "q=1.6*10**-19#\n", + "concn1=((1/resist)/(0.048*q))##concentration of hole\n", + "concne=((1.5*10**16)**(2))/concn1##concentration of electron\n", + "print \"concentration of hole = %0.2e\"%((concn1)),\"per cubic centimetre\"#\n", + "print \"concentration of electron = %0.2e\"%((concne)),\"per cubic centimetre\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 32 example 11" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "concentration of acceptor atoms = 6.25e+19 per cubic metre\n" + ] + } + ], + "source": [ + "resist=1*10**3##ohm\n", + "w=20*10**-6##wide metre\n", + "w1=400*10**-6##long metre\n", + "mo1=500##square centimetre/velocity second\n", + "q=1.6*10**-19#\n", + "conduc=(resist*w*4*10**-6)/w1#\n", + "concentration=((1)/(conduc*mo1*q))#\n", + "print \"concentration of acceptor atoms = %0.2e\"%((concentration)),\"per cubic metre\"##correction in the book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 32 example 12" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "dn constants = 98.80 square metre per second\n", + "dp constants = 33.80 square metre per second\n" + ] + } + ], + "source": [ + "w=0.026#\n", + "moe=3800##square centimetre/velocitysecond\n", + "mo1=1300##square centimetre/velocitysecond\n", + "u=(moe*w)#\n", + "u1=(mo1*w)#\n", + "print \"dn constants = %0.2f\"%((u)),\"square metre per second\"##correction in the book\n", + "print \"dp constants = %0.2f\"%((u1)),\"square metre per second\"##correction in the book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 33 example 13" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "distance of fermi level from center = 0.021 electron volt\n" + ] + } + ], + "source": [ + "from math import log\n", + "w=0.026*(3/2)*log(3)/2#\n", + "print \"distance of fermi level from center = %0.3f\"%((w)),\" electron volt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 33 example 14" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistivity = 44.64 ohm centimetre\n", + "resistivity equal to 45\n", + "resistivity = 32.42 ohm centimetre\n", + "resistivity equal to 32.4\n" + ] + } + ], + "source": [ + "up=1800##centimetre square per velocity second\n", + "un=3800##centimetre square per velocity second\n", + "\n", + "#(1) resistivity is 45 ohm\n", + "q=1.6*10**-19##coulomb\n", + "ni=2.5*10**13#\n", + "sigma1=(un+up)*q*ni#\n", + "resist=1/sigma1#\n", + "print \"resistivity = %0.2f\"%((resist)),\" ohm centimetre\"#\n", + "print \"resistivity equal to 45\"#\n", + "#(2) impurity added to extent of 1 atom per 10**9\n", + "n=4.4*10**22/10**9\n", + "p1=ni**2/n#\n", + "sigma1=(n*un+p1*up)*q#\n", + "resist=1/sigma1\n", + "print \"resistivity = %0.2f\"%((resist)),\" ohm centimetre\"#\n", + "print \"resistivity equal to 32.4\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 34 example 15" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "concentration of the a free electrons = 1.05e+04\n", + "concentration of the a free holes = 1.00e+14\n", + "sample p\n", + "n = 1.00e+15 electrons per cubic centimetre\n", + "p = 1.10e+15 holes per cubic centimetre\n", + "essentially intrinsic\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "from sympy import symbols, solve, exp\n", + "nd=4*10**14##atoms per cubic centimetre\n", + "na=5*10**14##atoms per cubic centimetre\n", + "#(1) concentration\n", + "ni=2.5*10**13#\n", + "np=ni**2#\n", + "#p1=n+10**14\n", + "z=1#\n", + "z1=10**14#\n", + "u=-ni**2#\n", + "#n=poly([z z1 u],'q')#\n", + "n=symbols('n')\n", + "expr = z*n**2+z1*n+u\n", + "n = solve(expr,n)[1]\n", + "n=1.05*10**4#\n", + "print \"concentration of the a free electrons = %0.2e\"%((n))\n", + "p1=n+10**14#\n", + "print \"concentration of the a free holes = %0.2e\"%((p1))\n", + "#(2)\n", + "print \"sample p\"#\n", + "a=ni**2/(300**3*exp(-(0.785/0.026)))#\n", + "w=400##kelvin\n", + "ni=sqrt(a*w**3*exp(-0.786/(8.62*10**-5*w)))#\n", + "ni=((n)*(n+10**14))/10**3#\n", + "n=ni-0.05*10**15#\n", + "print \"n = %0.2e\"%((n)),\"electrons per cubic centimetre\"\n", + "p1=n+10**14#\n", + "print \"p = %0.2e\"%((p1)),\"holes per cubic centimetre\"\n", + "\n", + "print \"essentially intrinsic\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 35 example 16" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "concentration of n = 6.00e+08 electrons per cubic centimetre\n", + "concentration of holes = 1.04e+18 holes per cubic centimetre\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "w=300##kelvin\n", + "conduc=300##ohm centimetre inverse\n", + "u=1800#\n", + "p=conduc/(u*1.6*10**-19)##concentration holes\n", + "n=(2.5*10**13)**2/(p)#\n", + "print \"concentration of n = %0.2e\"%((n)),\"electrons per cubic centimetre\"\n", + "print \"concentration of holes = %0.2e\"%((p)),\"holes per cubic centimetre\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 35 example 17" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current density = 0.17 ampere per square centimetre\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from sympy import symbols, solve\n", + "nd=10**14##atoms per cubic centimetre\n", + "na=5*10**13##atoms per cubic centimetre\n", + "un=3800#\n", + "up=1800#\n", + "q=1.6*10**-19##coulomb\n", + "resist=80##ohm metre\n", + "e1=5##volt per metre\n", + "w=nd-na#\n", + "ni=(un+up)*q*resist#\n", + "n=symbols('n')\n", + "#p1=oly([1 w -ni**2],'q')#\n", + "expr = n**2+w*n-ni**2\n", + "##p1=taken as 3.65*19**12\n", + "p1=solve(expr, p1)\n", + "p1=3.65*10**12#\n", + "n=p1+w#\n", + "j=(n*un+p1*up)*q*e1#\n", + "print \"current density = %0.2f\"%((j)),\"ampere per square centimetre\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 36 example 18" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistivity = 1.25 ohm centimetre\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "na=1*10**16##per cubic centimetre correction in the book\n", + "ni=1.48*10**10##per cubic centimetre\n", + "un=0.13*10**4##centimetre square per velocity second\n", + "u=0.05*10**4##centimetre square per velocity second\n", + "n=ni**2/na#\n", + "q=1/(1.6*10**-19*(un*n+(u*na)))#\n", + "print \"resistivity = %0.2f\"%((q)),\"ohm centimetre\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 37 example 19" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage across sample = 9.38 volt\n", + "drift velocity = 37.50 metre per second\n", + "transverse force per coulomb = 1.88 newton per coulomb\n", + "transverse electric field = 1.88 volt per metre\n", + "hall voltage = 0.02 volt\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "e1=750##volt per metre\n", + "b=0.05##metre square per velocity second\n", + "un=0.05##metre square per velocity second\n", + "up=0.14##metre square per velocity second\n", + "#(1) voltage\n", + "w=1.25*10**-2##metre\n", + "v1=e1*w#\n", + "print \"voltage across sample = %0.2f\"%((v1)),\"volt\"#\n", + "#(2) drift velocity\n", + "vd=un*e1#\n", + "print \"drift velocity = %0.2f\"%((vd)),\"metre per second\"#\n", + "#transverse force per coulomb\n", + "f1=vd*b#\n", + "print \"transverse force per coulomb = %0.2f\"%((f1)),\"newton per coulomb\"#\n", + "#(4) transverse electric field\n", + "e1=vd*b#\n", + "print \"transverse electric field = %0.2f\"%((e1)),\"volt per metre\"#\n", + "#(5) hall voltage\n", + "q=0.9*10**-2#\n", + "vh=e1*q\n", + "print \"hall voltage = %0.2f\"%((vh)),\"volt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 37 example 20" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistivity at 300kelvin = 2.31e+05 ohm centimetre\n", + "resistivity at impurity of 1 atom included per 10**5 atoms = 0.010 ohm centimetre\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "un=1300##centimetre square per velocity second\n", + "#at 300kelvin\n", + "ni=1.5*10**10#\n", + "u=500##centimetre square per velocity second\n", + "conduc=1.6*10**-19*1.5*10**10*(un+u)#\n", + "q=1/conduc#\n", + "#impurity of 1 atom included per 10**5 atoms\n", + "print \"resistivity at 300kelvin = %0.2e\"%((q)),\"ohm centimetre\"#\n", + "n=5*10**22/10**5#\n", + "p=ni**2/n#\n", + "q=1/(1.6*10**-19*(un*n+(u*p)))\n", + "\n", + "print \"resistivity at impurity of 1 atom included per 10**5 atoms = %0.3f\"%((q)),\"ohm centimetre\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 38 example 21" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ec-ef = -0.20\n", + "ec-ef = 0.04 electron volt ef above ec\n", + "impurities included per germanium atoms = 0.0002\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt, log, log10\n", + "n=4.4*10**22#\n", + "nd=n/10**7#\n", + "w=300##kelvin\n", + "nc=4.82*10**15*w**(3/2)/1/sqrt(8)#\n", + "ec_ef1=-0.026*log((nc/(nd)))#\n", + "print \"ec-ef = %0.2f\"%((ec_ef1))\n", + "#(2) impurities included inratio 1 to 10**3\n", + "n=4.4*10**22#\n", + "nd=n/(10**3)#\n", + "ec_ef1=-0.026*log(nc/nd)#\n", + "print \"ec-ef = %0.2f\"%((ec_ef1)),\"electron volt ef above ec\"#\n", + "q=log10(nd/nc)/log10(10)#\n", + "print \"impurities included per germanium atoms = 0.0002\"#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 39 example 22" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ef-ec = 0.15 electron volt\n", + "ef-ec = 0.03 electron volt\n", + "temperature = 240.33 kelvin\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log\n", + "n=5*10**22##atoms per cubic centimetre\n", + "#(1) 1 atom per 10**6\n", + "m=0.8##metre\n", + "na=n/10**6#\n", + "w=300##kelvin\n", + "nv=4.82*10**15*(m)**(3/2)*w**(3/2)#\n", + "ef_ec=0.026*log(nv/na)#\n", + "print \"ef-ec = %0.2f\"%((ef_ec)),\"electron volt\"#\n", + "#(2) impurity included 10*10**3 per atom\n", + "na=n/(10*10**3)#\n", + "ef_ec=0.026*log(nv/na)#\n", + "print \"ef-ec = %0.2f\"%((ef_ec)),\"electron volt\"#\n", + "#(3) condition to concide ec=ef\n", + "na=4.81*10**15#\n", + "w=(nv/na)**(2/3)#\n", + "print \"temperature = %0.2f\"%((w)),\"kelvin\"##correction in the book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 40 example 23 " + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "hall voltage = 0.17 volt\n", + "remains the same but there change in polarity\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#figure is not given in the book\n", + "nd=10**7##per cubic centimetre\n", + "na=10**17##per cubic centimetre\n", + "voltag=0.1*3800*10**-4*1500*3*10**-3#\n", + "print \"hall voltage = %0.2f\"%((voltag)),\"volt\"#\n", + "print \"remains the same but there change in polarity\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 40 example 24" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "mobilty = 0.12 metre square per velocity second\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "vh=60*10**-3##volt\n", + "w=6*10**-3##metre\n", + "bz=0.1##weber per metre square\n", + "i1=10*10**-6##ampere\n", + "resist=300000*10**-2##ohm metre\n", + "#(1)\n", + "#mobility\n", + "rh=vh*w/(bz*i1)#\n", + "u1=rh/resist#\n", + "print \"mobilty = %0.2f\"%((u1)),\"metre square per velocity second\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronic_Devices_and_Circuits_by_J._Paul/Ch2_1.ipynb b/Electronic_Devices_and_Circuits_by_J._Paul/Ch2_1.ipynb new file mode 100644 index 00000000..c4421823 --- /dev/null +++ b/Electronic_Devices_and_Circuits_by_J._Paul/Ch2_1.ipynb @@ -0,0 +1,851 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2 - Semiconductor Diodes" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 99 example 1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ratio of reverse saturation current = 4963.36\n" + ] + } + ], + "source": [ + "q=0.01##centimetre\n", + "sigma1=1##ohm centimetre inverse\n", + "q1=0.01##centimetre\n", + "sigm11=0.01##ohm centimetre inverse\n", + "iratio=(0.0224**2*2.11*20)*3.6**2/((3.11*(4.3**2*10**-6)**2*2.6*20*10**3))#\n", + "for q in range(0,2):\n", + " if q==1:\n", + " un=3800#\n", + " up=1500#\n", + " q=1.6*10**-19#\n", + " ni=2.5*10#\n", + " else:\n", + " q=1.6*10**-19#\n", + " up=500\n", + " un=1300#\n", + " ni=1.5*10\n", + "\n", + " \n", + " b=un/up#\n", + " sigmai=(un+up)*q*ni#\n", + "\n", + "print \"ratio of reverse saturation current = %0.2f\"%((iratio))\n", + "##correction required in the book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 100 example 2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "reverse current ratio = 7.79e-09\n" + ] + } + ], + "source": [ + "sigma1=0.01##ohm centimetre inverse\n", + "area11=4*10**-3##metre square\n", + "q=0.01*10**-2##metre\n", + "un=1300.0#\n", + "up=500.0#\n", + "ni=1.5*10**15##per cubic centimetre\n", + "sigma1=(un+up)*1.6*10**-19*ni#\n", + "iratio=(4*10**-10*0.026*sigma1**2*2.6*2/10**-4)/3.6**2#\n", + "print \"reverse current ratio = %0.2e\"%((iratio))\n", + "##correction required in the book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 100 example 3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "reverse saturation current = 3.48e-06 ampere\n" + ] + } + ], + "source": [ + "a=4*10**-4##metre square\n", + "sigmap=1#\n", + "sigman=0.1#\n", + "de=0.15#\n", + "vtem=26*10**-3#\n", + "i=(a*vtem*((2.11)*(0.224))/((3.22)**(2)))*((1/de*sigman)+(1/de*sigmap))#\n", + "print \"reverse saturation current = %0.2e\"%(i),\"ampere\"#correction in the book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 101 example 4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage at which the reverse saturation current at saturate = -0.06 volt\n", + "reverse saturation current = -6.84 ampere\n", + "reverse saturation current 0.10 = 0.000 ampere\n", + "reverse saturation current 0.20 = 0.022 ampere\n", + "reverse saturation current 0.30 = 1.026 ampere\n" + ] + } + ], + "source": [ + "from math import log, exp\n", + "w=0.9#\n", + "voltaf=0.05##volt\n", + "revcur=10*10**-6##ampere\n", + "#(1) voltage\n", + "volrev=0.026*(log((-w+1)))##voltage at which the reverse saturation current at saturate\n", + "resacu=((exp(voltaf/0.026)-1)/((exp(-voltaf/0.026)-1)))##reverse saturation current\n", + "print \"voltage at which the reverse saturation current at saturate = %0.2f\"%((volrev)),\"volt\"\n", + "print \"reverse saturation current = %0.2f\"%((resacu)),\"ampere\"\n", + "u=0.1#\n", + "for q in range(0,3):\n", + " reverc=revcur*(exp((u/0.026))-1)\n", + " print \"reverse saturation current %0.2f\"%((u)),\" = %0.3f\"%((reverc)),\"ampere\"\n", + " u=u+0.1#\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 103 example 6" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "capacitance = 7.08e-11 farad\n" + ] + } + ], + "source": [ + "a=1*10**-6##metre square\n", + "w=2*10**-6##thick centimetre\n", + "re=16#\n", + "eo=8.854*10**-12#\n", + "c=(eo*re*a)/w#\n", + "print \"capacitance = %0.2e\"%(c),\"farad\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 105 example 7" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "width of depletion layer at -10.00 = 7.73e-06 metre\n", + "width of depletion layer at -0.10 = 1.33e-06 metre\n", + "width of depletion layer at 0.10 = 7.65e-07 metre\n", + "capacitance at -10.00 = 1.57e-11 farad\n", + "capacitance at -0.10 = 9.13e-11 farad\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "volbar=0.2##barrier voltage for germanium volt\n", + "na=3*10**20##atoms per metre\n", + "#(1) width of depletion layer at 10 and 0.1 volt\n", + "\n", + "for q in [-10, -0.1, 0.1]:\n", + " w=2.42*10**-6*sqrt((0.2-(q)))#\n", + " print \"width of depletion layer at %0.2f\"%((q)),\" = %0.2e\"%((w)),\"metre\"#for -0.1volt correction in the book\n", + "\n", + "#(d) capacitance\n", + "for q in [-10, -0.1]:\n", + " capaci=0.05*10**-9/sqrt(0.2-q)#\n", + " print \"capacitance at %0.2f\"%((q)),\" = %0.2e\"%((capaci)),\"farad\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 104 example 8" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "maximum forward current = 2.22 ampere\n", + "forward diode resistance = 0.40 ohm\n" + ] + } + ], + "source": [ + "p=2##watts\n", + "voltaf=900*10**-3##volt\n", + "i1=p/voltaf#\n", + "r1=voltaf/i1#\n", + "print \"maximum forward current = %0.2f\"%(i1),\"ampere\"\n", + "\n", + "\n", + "print \"forward diode resistance = %0.2f\"%(r1),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 108 example 11" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "alpha = 104.86 degree\n" + ] + } + ], + "source": [ + "from math import atan, degrees\n", + "r=250##ohm\n", + "c=40*10**-6##farad\n", + "alpha1=180-degrees(atan(377*r*c))\n", + "print \"alpha = %0.2f\"%(alpha1),\"degree\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 109 example 12" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "inductance = 3022899.27 henry\n", + "output voltage = 31.03 volt\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "i1=0.1##current in ampere\n", + "vms=40##rms voltage in volts\n", + "c=40*10**-6##capacitance in farad\n", + "r1=50##resistance in ohms\n", + "ripple=0.0001#\n", + "induct=((1.76/c)*sqrt(0.472/ripple))##inductance\n", + "outv=(2*sqrt(2)*vms)/3.14-i1*r1##output voltage\n", + "print \"inductance = %0.2f\"%(induct),\"henry\"#correction in the book\n", + "print \"output voltage = %0.2f\"%(outv),\"volt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 109 example 14" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ripple voltage = 0.093 volt\n", + "ripple voltage including filters = 118.49 volt\n", + "ripple voltage = 0.0040 volt\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "voltag=40##volt\n", + "i1=0.2##ampere\n", + "c1=40*10**-6##farad\n", + "c2=c1#\n", + "induct=2##henry\n", + "#(1) ripple\n", + "vdc=2*sqrt(2)*voltag/3.14#\n", + "r1=vdc/i1#\n", + "induc1=r1/1130#\n", + "v1=voltag/(3*3.14**3*120**2*4*induct*c1)#\n", + "print \"ripple voltage = %0.3f\"%((v1)),\"volt\"\n", + "#(2) with two filter\n", + "v1=4*voltag/((3*3.14**5)*(16*120**2*induct**2*c1**2))#\n", + "print \"ripple voltage including filters = %0.2f\"%((v1)),\"volt\"#correction in the book\n", + "#(3)ripple voltage\n", + "v1=4*voltag/(5*3.14*1.414*2*3.14*240*240*3.14*induct*c1)#\n", + "v1=v1/20#\n", + "print \"ripple voltage = %0.4f\"%((v1)),\"volt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 111 example 15" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage and ripple with load\n", + "vdc = 250.21 volt\n", + "ripple = 3.13e-02\n", + "capacitance connected across load\n", + "vdc = 497.91 volt\n", + "ripple = 3.76e-02\n", + "filter containing two inductors and capacitors in parallel\n", + "vdc = 250.00 volt\n", + "ripple = 6.48e-04\n", + "two filter\n", + "vdc = 250.00 volt\n", + "ripple = 4.76e-06\n", + "vdc = 358.26 volt\n", + "ripple = 1.61e-04\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt\n", + "voltag=375##volt\n", + "r1=2000##ohm\n", + "induct=20##henry\n", + "c1=16*10**-6##farad\n", + "r11=100##ohm\n", + "r=200##ohm\n", + "#(1) voltage and ripple with load\n", + "print \"voltage and ripple with load\"\n", + "r=r+r11+400#\n", + "vdc=((2*sqrt(2)*voltag/3.14))/1.35#\n", + "ripple=r1/(3*sqrt(2)*(377)*induct*2)#\n", + "print \"vdc = %0.2f\"%((vdc)),\"volt\"\n", + "print \"ripple = %0.2e\"%((ripple))\n", + "#(2) capacitance connected across load\n", + "print \"capacitance connected across load\"\n", + "vdc=sqrt(2)*voltag/(1+1/(4*(60)*r1*2*c1))#\n", + "ripple=1/(4*sqrt(3)*(60)*r1*2*c1)#\n", + "print \"vdc = %0.2f\"%((vdc)),\"volt\"\n", + "print \"ripple = %0.2e\"%((ripple))\n", + "#(3) filter containing two inductors and capacitors in parallel\n", + "print \"filter containing two inductors and capacitors in parallel\"\n", + "vdc=250##volt\n", + "ripple=0.83*10**-6/(2*induct*2*c1)##correction in the book\n", + "print \"vdc = %0.2f\"%((vdc)),\"volt\"\n", + "print \"ripple = %0.2e\"%((ripple))\n", + "#(4) two filter\n", + "print \"two filter\"\n", + "vdc=250#\n", + "ripple=sqrt(2)/(3*16*3.14**2*60**2*induct*c1)**2##correction in the book\n", + "print \"vdc = %0.2f\"%((vdc)),\"volt\"\n", + "print \"ripple = %0.2e\"%((ripple))\n", + "vdc=sqrt(2)*voltag/(1+(4170/(r1*16))+(r/r1))#\n", + "ripple=3300/(16**2*2*20*r1)#\n", + "print \"vdc = %0.2f\"%((vdc)),\"volt\"\n", + "print \"ripple = %0.2e\"%((ripple))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 112 example 16" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "output voltage = 362.14 volt\n", + "ripple voltage = 1.46e-03\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "capaci=4##farad\n", + "induct=20##henry\n", + "i1=50*10**-3##ampere\n", + "resist=200##ohm\n", + "maxvol=300*sqrt(2)#\n", + "vdc=maxvol-((4170/capaci)*(i1))-(i1*resist)#\n", + "ripple=(3300*i1)/((capaci**2)*(induct)*353)#\n", + "print \"output voltage = %0.2f\"%((vdc)),\"volt\"\n", + "print \"ripple voltage = %0.2e\"%((ripple))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 113 example 17" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "inductance of filter = 4.98 henry\n", + "resistance of filter = 250.00 ohm\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "voltag=25##volt\n", + "c1=10*10**-6##farad\n", + "i1=100*10**-3##ampere\n", + "ripple=0.001#\n", + "w=754##radians\n", + "#(1) inductance and resistance\n", + "\n", + "\n", + "r1=voltag/i1#\n", + "induct=40/(sqrt(2)*w**2*(c1))#\n", + "print \"inductance of filter = %0.2f\"%((induct)),\"henry\"#correction in the book\n", + "print \"resistance of filter = %0.2f\"%((r1)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 113 example 18" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current = 2.83e-04 ampere\n", + "current at 100celsius rise\n", + "current = 6.81e-04 ampere\n" + ] + } + ], + "source": [ + "from math import exp\n", + "resacu=0.1*10**-12##ampere\n", + "u=20+273##kelvin\n", + "voltaf=0.55##volt\n", + "w=1.38*10**-23#\n", + "q=1.6*10**-19#\n", + "for z in range(1,3):\n", + " if z==2 :\n", + " u=100+273#\n", + " print \"current at 100celsius rise\"\n", + " \n", + " voltag=w*u/q#\n", + " i1=(10**-13)*(exp((voltaf/voltag))-1)#\n", + " if z==2:\n", + " i1=(256*10**-13)*((exp(voltaf/voltag)-1))#\n", + " \n", + " print \"current = %0.2e\"%((i1)),\"ampere\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 114 example 19" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "thermal voltage = 0.026 volt\n", + "barrier voltage = 0.535 volt\n" + ] + } + ], + "source": [ + "from math import log\n", + "na=10*22##atoms per cubic metre\n", + "nd=1.2*10**21##donor per cubic metre\n", + "voltag=1.38*10**-23*(273+298)/(1.6*10**-19)##correction in the book\n", + "voltag=0.026#\n", + "ni=1.5*10**16#\n", + "ni=ni**2#\n", + "v1=voltag*log((na*nd)/(ni))#\n", + "print \"thermal voltage = %0.3f\"%((voltag)),\"volt\"\n", + "print \"barrier voltage = %0.3f\"%(abs(v1)),\"volt\"#correction in the book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 114 example 20" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current = 9.16e-06 ampere\n" + ] + } + ], + "source": [ + "from math import exp\n", + "i1=2*10**-7##ampere\n", + "voltag=0.026##volt\n", + "i=i1*((exp(0.1/voltag)-1))#\n", + "print \"current = %0.2e\"%((i)),\"ampere\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 115 example 21" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistance at 150mvolt = 80.74 ohm\n" + ] + } + ], + "source": [ + "from math import exp\n", + "resacu=1*10**-6##ampere\n", + "voltaf=150*10**-3##volt\n", + "w=8.62*10**-5#\n", + "voltag=0.026##volt\n", + "u=300##kelvin\n", + "uw=u*w#\n", + "resist=(uw)/((resacu)*exp(voltaf/voltag))#\n", + "print \"resistance at 150mvolt = %0.2f\"%((resist)),\"ohm\"#correction in the book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 115 example 22" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "change in barrier = 0.18 volt\n" + ] + } + ], + "source": [ + "from math import log\n", + "dopfac=1000#\n", + "w=300##kelvin\n", + "q=0.026*log(dopfac)#\n", + "print \"change in barrier = %0.2f\"%((q)),\"volt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 116 example 23" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "depletion capacitance = 1.09e-11 farad\n", + "capacitance = 3.85e-07 farad\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "area12=1*10**-8##metre square\n", + "volre1=-1##reverse voltage\n", + "capac1=5*10**-12##farad\n", + "volbu1=0.9##volt\n", + "voltag=0.5##volt\n", + "i1=10*10**-3##ampere\n", + "durmin=1*10**-6##ssecond\n", + "#(1) capacitance\n", + "capac1=capac1*sqrt((volre1-volbu1)/(voltag-volbu1))#\n", + "print \"depletion capacitance = %0.2e\"%((capac1)),\"farad\"\n", + "#(2) capacitance\n", + "capac1=i1*durmin/(0.026)#\n", + "\n", + "print \"capacitance = %0.2e\"%((capac1)),\"farad\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 116 example 24" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "potential germanium = 0.34 volt\n", + "potential silicon = 0.74 volt\n" + ] + } + ], + "source": [ + "from math import log\n", + "quantg=4*10**22##atoms per cubic centimetre\n", + "quants=5*10**22##atoms per cubic centimetre\n", + "w=2.5*10**13##per cubic centimetre\n", + "w1=1.5*10**10##per cubic centimetre\n", + "for q in [quantg, quants]:\n", + " na=2*q/(10**8)\n", + " nd=500*na#\n", + " if q==quantg :\n", + " w=w#\n", + " voltag=0.026*log(na*nd/w**2)#\n", + " print \"potential germanium = %0.2f\"%((voltag)),\"volt\"\n", + " \n", + " if q==quants:\n", + " w=w1#\n", + " voltag=0.026*log(na*nd/w**2)#\n", + " print \"potential silicon = %0.2f\"%((voltag)),\"volt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 117 example 25" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "electrons density = 9.62e+20 per cubic metre\n", + "holes density = 1.25e+23 per cubic metre\n" + ] + } + ], + "source": [ + "u=0.05##metre square per velocity second correction in the book\n", + "un=0.13##metre square per velocity second\n", + "condun=20##second per metre conductivity of n region\n", + "condup=1000##second per metre conductivity of p region\n", + "p=condup/(1.6*10**-19*u)#\n", + "no=condun/(1.6*10**-19*un)#\n", + "print \"electrons density = %0.2e\"%((no)),\"per cubic metre\"\n", + "print \"holes density = %0.2e\"%((p)),\"per cubic metre\"#others to find is not in the book" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronic_Devices_and_Circuits_by_J._Paul/Ch3_1.ipynb b/Electronic_Devices_and_Circuits_by_J._Paul/Ch3_1.ipynb new file mode 100644 index 00000000..3939306a --- /dev/null +++ b/Electronic_Devices_and_Circuits_by_J._Paul/Ch3_1.ipynb @@ -0,0 +1,261 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3 - Special Semiconductor Diodes " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 138 example 1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistance range from 196.00 to 217.78 ohms\n", + "resistance range at 50 from 146.00 to 167.78 ohms\n" + ] + } + ], + "source": [ + "#zener diode\n", + "voltag=5.2##volts\n", + "w=260*10**-3##watts\n", + "appv=15##voltsw1=50##watts\n", + "imax=w/voltag*0.1#\n", + "#to maitain a constant voltage\n", + "imax1=(w/voltag)-imax#\n", + "resmin=(appv-voltag)/(w/voltag)#\n", + "resmax=(appv-voltag)/imax1#\n", + "#load 50\n", + "resmax1=((9.8)/(45*10**-3))-50#\n", + "resmin1=((9.8)/(50*10**-3))-50#\n", + "res50=resmax1-resmin1#\n", + "print \"resistance range from %0.2f\"%(resmin),\" to %0.2f\"%(resmax),\"ohms\"\n", + "print \"resistance range at 50 from %0.2f\"%(resmin1),\" to %0.2f\"%(resmax1),\"ohms\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 139 example 2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage drop = 5.67volts\n" + ] + } + ], + "source": [ + "i1=20*10**-3##ampere\n", + "i=30*10**-3##ampere\n", + "v1=5.6##volts\n", + "v=5.65##volts\n", + "#condition\n", + "u=35*10**-3##ampere\n", + "voltag=5*u+5.5#\n", + "print \"voltage drop = %0.2f\"%(voltag)+\"volts\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 139 example 3 " + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "fi0 = 4.96\n" + ] + } + ], + "source": [ + "from math import log\n", + "v=4.3##volt\n", + "q=4##volt\n", + "dop=10**17##per cubic centimetre\n", + "fi0=0.254*log(dop/(5.1*10**10))#\n", + "fi01=0.407+q+0.55#\n", + "print 'fi0 = %0.2f'%(fi01)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 140example 4 " + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current = 9.95e-05 ampere\n", + "resistance = 1.61e+05 ohm\n", + "r1 = 1.60e+05 ohm\n", + "r2 = 4.02e+04 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "v1=20##volt\n", + "i1=((v1)/(200+1))*10**-3#\n", + "print 'current = %0.2e'%(i1),'ampere'\n", + "#greater than 20\n", + "vone=16#\n", + "r=vone/i1#\n", + "r1=r-1*10**3#\n", + "r11=200*10**3-r1#\n", + "print 'resistance = %0.2e'%(r),'ohm'\n", + "print \"r1 = %0.2e\"%((r1)),\"ohm\"\n", + "print \"r2 = %0.2e\"%((r11)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 142 example 6 " + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "maximum current = 0.035 ampere\n", + "v1 minimum = 262.50 volt\n", + "v1 maximum = 393.75 volt\n" + ] + } + ], + "source": [ + "v1=150##volt\n", + "vone=300#volt\n", + "idmax=40*10**-3##ampere\n", + "idmin=5*10**-3##ampere\n", + "r=(vone-v1)/idmax#\n", + "imax=idmax-idmin#\n", + "print 'maximum current = %0.3f'%(imax),'ampere'\n", + "#minimum\n", + "zq=1#\n", + "while (zq<=2):\n", + " if zq==1 :\n", + " ione=25*10**-3#\n", + " i1=ione+idmin#\n", + " vmin=(i1*r)+v1#\n", + " print 'v1 minimum = %0.2f'%(vmin),'volt'\n", + " else:\n", + " ione=25*10**-3#\n", + " i1=ione+idmax#\n", + " vmin=(i1*r)+v1#\n", + " print 'v1 maximum = %0.2f'%(vmin),'volt'\n", + " \n", + " \n", + " zq=zq+1#\n", + " " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 142 example 7" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "width = 2.22e-08 metre\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "q=4.5*10**22##atoms per cubic metre\n", + "na=q/(10**4)#\n", + "eo=0.026*24.16#\n", + "e=1.6*10**-19#\n", + "W=sqrt((4*16*0.628)/(36*3.14*10**9*na*10**6*e))#\n", + "print 'width = %0.2e'%(W),'metre'" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronic_Devices_and_Circuits_by_J._Paul/Ch4_1.ipynb b/Electronic_Devices_and_Circuits_by_J._Paul/Ch4_1.ipynb new file mode 100644 index 00000000..de047efc --- /dev/null +++ b/Electronic_Devices_and_Circuits_by_J._Paul/Ch4_1.ipynb @@ -0,0 +1,1732 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4 - Bipolar Junction Transistor" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 201 example 1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "r1 = 19990.91 ohm\n" + ] + } + ], + "source": [ + "alpha=0.98#\n", + "vbe=0.7##base emitter voltage volt\n", + "ie=-4*10**-3##emitter current\n", + "vc=12##colector voltage volt\n", + "colr=3.3*10**3##ohms\n", + "colCurrent=ie*(-alpha)#\n", + "baseCurrent=0.02*ie#\n", + "vbn=vbe+(-4*10**-3*100)#\n", + "i2=-vbn/(10*10**3)#\n", + "i1=-(baseCurrent+i2)#\n", + "vcn=(vc-((colCurrent+i1)*colr))#\n", + "v1=vcn-0.9#\n", + "r1=v1/i1#\n", + "print \"r1 = %0.2f\"%(abs(r1)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 202 example 2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rb = 21500.00 ohm\n", + "rc = 700.00 ohm\n", + "rb at emitter resistance 100ohm = 18950.00 ohm\n" + ] + } + ], + "source": [ + "colvoltag=12##volts\n", + "vbe=5##volts\n", + "colcur=10*10**-3##ampere\n", + "vce=5##volts\n", + "beta1=50#\n", + "ib=colcur/beta1#\n", + "rb=(vbe-0.7)/ib#\n", + "rc=(12-vbe)/colcur#\n", + "#when 100ohm included\n", + "print \"rb = %0.2f\"%(rb),\"ohm\"\n", + "print \"rc = %0.2f\"%(rc),\"ohm\"\n", + "rb=(vce-0.7-(colcur+ib)*beta1)/ib#\n", + "\n", + "print \"rb at emitter resistance 100ohm = %0.2f\"%(rb),\"ohm\"#correction in the book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 205 example 5" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "max resistance = 76562.50 ohm\n", + "baseresistance = 100000.00 ohm\n", + "temperature = 46.70 celsius\n" + ] + } + ], + "source": [ + "from math import log\n", + "#given\n", + "reveri=2*10**-6##ampere at 25\n", + "icb=2*10**-6*2**5##ampere at 75\n", + "basevoltag=5##volt\n", + "#(1)\n", + "rb=(-0.1+basevoltag)/(icb)#\n", + "print \"max resistance = %0.2f\"%((rb)),\"ohm\"#correction in the book\n", + "#(2)\n", + "basevoltag=1#\n", + "rb=100*10**3#\n", + "reveri=(-0.1+basevoltag)/rb#\n", + "q=reveri/(2*10**-6)#\n", + "w=q**10#\n", + "u=log(w)\n", + "t=25+(u/log((2)))#\n", + "print \"baseresistance = %0.2f\"%((rb)),\"ohm\"\n", + "print \"temperature = %0.2f\"%((t)),\"celsius\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 205 example 6" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "min resistance = 3769.23 ohm\n" + ] + } + ], + "source": [ + "#given\n", + "vbe=0.8##volt\n", + "beta1=100#\n", + "vce=0.2##volt\n", + "rb=200*10**3##ohm\n", + "bascur=(6-vbe)/rb#\n", + "colres=(10-vce)/(beta1*bascur)#\n", + "print \"min resistance = %0.2f\"%((colres)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 206 example 7" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "in saturation mode\n", + "vo = 3.51 volt\n", + "emitter resistance < 688.58 ohm\n" + ] + } + ], + "source": [ + "beta1=100#\n", + "colres=3*10**3##collector resistance #ohm\n", + "rb=8*10**3##ohm\n", + "r1=500##ohm\n", + "voltag=5##volt\n", + "#(1)\n", + "ib=(-voltag+0.7)/((1+beta1)*r1+(rb))#\n", + "ic=beta1*ib#\n", + "vce=(-10-ic*(colres)+r1*(ib+ic))#\n", + "vcb=vce+0.7#\n", + "#(2)\n", + "volmin=-0.2+abs(ib+ic)*r1#\n", + "re=-(0.7+rb*ib+voltag)/((1+(beta1))*ib)#\n", + "print \"in saturation mode\"\n", + "print \"vo = %0.2f\"%((volmin)),\"volt\"#correction in the book\n", + "print \"emitter resistance < %0.2f\"%((re)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 207 example 9" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the operating point at vbb = 1.00 volt ic = 2.50e-03 ampere vce = 7.00 volt\n", + "the operating point at vbb = 12.00 volt ic = 5.95e-03 ampere vce = -176.33 volt\n", + "beta at saturation = 6.32\n" + ] + } + ], + "source": [ + "vcc=12##volt\n", + "rb=12*10**3##ohm\n", + "colres=2*10**3##ohm\n", + "beta1=100#\n", + "vb=0.7##volt\n", + "vce=0.1##volt\n", + "\n", + "for q in range(1,3):\n", + " if q==1:\n", + " vbb=1\n", + " else:\n", + " vbb=12\n", + " \n", + " ib=(vbb-vb)/rb\n", + " ic=beta1*ib\n", + " ie=ic+ib\n", + " vce=vcc-ic*colres\n", + " if q==2 :\n", + " ic=(vcc-0.1)/colres\n", + " \n", + "\n", + " print \"the operating point at vbb = %0.2f\"%((vbb)),\"volt ic = %0.2e\"%((ic)),\"ampere vce = %0.2f\"%((vce)),\" volt\"\n", + "\n", + "beta1=ic/ib#\n", + "\n", + "print \"beta at saturation = %0.2f\"%((beta1))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 208 example 11" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current = 1.00e-03 ampere\n" + ] + } + ], + "source": [ + "vbe=0.65##volt\n", + "colres=2*10**3##ohm\n", + "voltag=10##volt\n", + "i1=voltag/10#\n", + "q=(1.65-vbe)/(1*10**3)#\n", + "\n", + "\n", + "print \"current = %0.2e\"%((q)),\"ampere\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 208 example 12" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "vce = 5.70 volt\n", + "collector current = 1.05e-03 ampere\n" + ] + } + ], + "source": [ + "vcc=12##volt\n", + "r1=10*10**3##ohm\n", + "colres=1*10**3##ohm\n", + "re=5*10**3##ohm\n", + "rb=5*10**3##ohm\n", + "beta1=100#\n", + "vbe=0.7##volt\n", + "basvol=vcc*10/20#\n", + "ib=((basvol-vbe)/(rb+beta1*rb))#\n", + "ic=beta1*ib#\n", + "vce=vcc-ic*(colres+re)#\n", + "print \"vce = %0.2f\"%((vce)),\"volt\"\n", + "print \"collector current = %0.2e\"%((ic)),\"ampere\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 209 example 13" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "r1 = 4.22 times r2\n", + "if r2 is 1200ohm\n", + "r1 = 5061.77 ohm\n", + "r2 = 1200.00 ohm\n" + ] + } + ], + "source": [ + "colres=330##ohm\n", + "re=0.1*10**3##ohm\n", + "vcc=12##volt\n", + "vce=0.2##volt\n", + "revcur=18*10**-3#ampere\n", + "ib=0.3*10**-3##ampere\n", + "stability=10#\n", + "beta1=100#\n", + "colres=0.330##ohm\n", + "re=0.1*10**3##ohm\n", + "vbe=0.2#\n", + "rb=(((1+beta1)*re)/10-((1+beta1)*re))/(1-10.1)#\n", + "vb=2+ib*rb#\n", + "w=vcc/vb#\n", + "q=w-1#\n", + "r1=1.2*10**3#\n", + "r=q*1.2*10**3#\n", + "print \"r1 = %0.2f\"%((q)),\"times r2\"\n", + "print \"if r2 is 1200ohm\"\n", + "print \"r1 = %0.2f\"%((r)),\"ohm\"\n", + "\n", + "print \"r2 = %0.2f\"%((r1)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 210 example 14" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "collector current = 2.49e-03 ampere\n", + "emitter current = 2.52e-03 ampere\n", + "collector current with ib = 2.47e-03 ampere\n", + "emitter current = 2.50e-03 ampere\n", + "error = 7.94e-03\n" + ] + } + ], + "source": [ + "alpha1=0.99#\n", + "ib=25*10**-6##ampere\n", + "icb=200*10**-9##ampere\n", + "beta1=alpha1/(1-alpha1)#\n", + "ic=beta1*ib+(beta1+1)*icb#\n", + "print \"collector current = %0.2e\"%((ic)),\"ampere\"\n", + "ie1=(ic-icb)/alpha1#\n", + "print \"emitter current = %0.2e\"%((ie1)),\"ampere\"\n", + "ic=beta1*ib#\n", + "print \"collector current with ib = %0.2e\"%((ic)),\"ampere\"\n", + "ie=ic/alpha1#\n", + "print \"emitter current = %0.2e\"%((ie)),\"ampere\"\n", + "w=(ie1-ie)/ie1#\n", + "print \"error = %0.2e\"%((w))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 211 example 15" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistance = 381879.22 ohm\n", + "stability = 22.62\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "vcc=26##volt\n", + "colres=20*10**3##ohm\n", + "re=470##ohm\n", + "beta1=45#\n", + "vce=8##volt\n", + "ib=(vcc-vce)/((1+beta1)*(colres+re))#\n", + "ic=beta1*ib#\n", + "r1=((vcc-colres*(ib+ic)-re*(ib+ic)-(0.7)))/ib#\n", + "print \"resistance = %0.2f\"%((r1)),\"ohm\"\n", + "stability=(1+beta1)/(1+(beta1*re)/(re+colres))#\n", + "print \"stability = %0.2f\"%((stability))\n", + "#correction required in the book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 211 example 16" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current = 6.69e-04 ampere\n", + "vce = 2.69 volt\n", + "resistance = 6.90e+02 ohm\n", + "current = 6.36e-04 ampere\n", + "vce = 2.63 volt\n" + ] + } + ], + "source": [ + "vcc=1.5#volt in book should be changed as 1.5\n", + "colres=1.5*10**3##ohm\n", + "emresi=0.27*10**3##ohm\n", + "r1=2.7*10**3##ohm\n", + "r=2.7*10**3##ohm\n", + "beta1=45#\n", + "basre1=690##ohm\n", + "voltag=r*vcc/(r*r1)#\n", + "basres=(r*r1)/(r+r1)#\n", + "vbe=0.2#\n", + "for q in range (1,3):\n", + " if q==2 :\n", + " print \"resistance = %0.2e\"%((basre1)),\"ohm\"\n", + " basres=basres+basre1\n", + " \n", + " bascur=(((voltag+vbe)))/(basres+(45*(emresi)))\n", + " colcur=beta1*bascur\n", + " vce=(vcc+colcur*colres+(bascur+colcur)*emresi)\n", + " print \"current = %0.2e\"%((colcur)),\"ampere\"\n", + " print \"vce = %0.2f\"%((vce)),\"volt\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 212 example 17" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "base resistance = 62500.00 ohm\n", + "stability = 26.00\n" + ] + } + ], + "source": [ + "beta1=25#\n", + "colres=2.5*10**3##ohm\n", + "vcc=10##volt\n", + "vce=-5##volt\n", + "ic=-(vcc+vce)/colres#\n", + "ib=ic/beta1#\n", + "rb=vce/ib#\n", + "stability=(1+beta1)/((1+beta1)*((colres)/(colres+rb)))#\n", + "print \"base resistance = %0.2f\"%((rb)),\"ohm\"#correction in book\n", + "print \"stability = %0.2f\"%((stability))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 212 example 18" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "junction temperature = 51.00 celsius\n" + ] + } + ], + "source": [ + "therre=8##celsius per watts\n", + "tepera=27##celsius ambient temperature\n", + "potran=3##watt\n", + "tejunc=tepera+(therre*potran)#\n", + "print \"junction temperature = %0.2f\"%((tejunc)),\"celsius\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 213 example 19" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "dissipation = 9.75 watt\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "ambtep=40##celsius\n", + "juntep=160##celsius\n", + "hs_a=8#\n", + "j_c=5#\n", + "c_a=85#\n", + "j_a=(j_c)+(c_a*hs_a)/(c_a+hs_a)#\n", + "podiss=(juntep-ambtep)/j_a#\n", + "print \"dissipation = %0.2f\"%((podiss)),\"watt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 213 example 21" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "alpha = 0.99\n", + "beta = 199.00\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "emicur=1*10**-3##ampere\n", + "colcur=0.995*10**-3##ampere\n", + "alpha1=colcur/emicur#\n", + "beta1=alpha1/(1-alpha1)#\n", + "print \"alpha = %0.2f\"%((alpha1))\n", + "print \"beta = %0.2f\"%((beta1))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 213 example 22" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "alpha = 0.99\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "beta1=100#\n", + "alpha1=beta1/(beta1+1)#\n", + "\n", + "print \"alpha = %0.2f\"%((alpha1))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 213 example.23" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ic = 0.0067 ampere\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "rb=200*10**3##ohm\n", + "rc=2*10**3##ohm\n", + "vcc=20##volt\n", + "ib=(vcc)/(rb+200*rc)#\n", + "ic=200*ib#\n", + "print \"ic = %0.4f\"%((ic)),\"ampere\"\n", + "#correction required in book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 214 example 24" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "collector current = 9.81e-04 ampere\n", + "base current = 1.90e-05 ampere\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "alpha1=0.98#\n", + "revcur=1*10**-6##ampere\n", + "emicur=1*10**-3##ampere\n", + "colcur=alpha1*emicur+revcur#\n", + "bascur=emicur-colcur#\n", + "print \"collector current = %0.2e\"%((colcur)),\"ampere\"\n", + "print \"base current = %0.2e\"%((bascur)),\"ampere\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 214 example 25" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "vce = 8.00 volt\n", + "emitter resistance = 50.00 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "colcur=100*10**-3##ampere\n", + "ouresi=20##ohm\n", + "r=200##ohm\n", + "r1=100##ohm\n", + "vcc=15##volt\n", + "basvol=((r1)/(r+r1))*vcc#\n", + "em1res=basvol/colcur#\n", + "vce=vcc-(ouresi+em1res)*colcur#\n", + "print \"vce = %0.2f\"%((vce)),\"volt\"\n", + "print \"emitter resistance = %0.2f\"%((em1res)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 214 example 26" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "emitter current = 0.019 ampere\n", + "collector to emitter = 3.112 volt\n", + "collector to emitter = -15.18 volt\n", + "collector current = 0.005 ampere\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "colres=1*10**3##ohm\n", + "beta1=50#\n", + "vbe=0.3##volt\n", + "vcc=6##volt\n", + "rb=10*10**3##ohm\n", + "re=100##ohm\n", + "em1cur=((vcc-vbe)*(beta1+1))/((rb+((beta1+1)*re)))#\n", + "for q in range(1,3):\n", + " if q==2 :\n", + " colres=1*10**3#\n", + " vce=vcc-(colres+re)*em1cur#\n", + " ic=vcc/(colres+re)#\n", + " print \"collector to emitter = %0.2f\"%((vce)),\"volt\"\n", + " print \"collector current = %0.3f\"%((ic)),\"ampere\"\n", + " \n", + " if q==1 :\n", + " colres=50#\n", + " rb=100#\n", + " vce=vcc-(colres+rb)*em1cur#\n", + " print \"emitter current = %0.3f\"%((em1cur)),\"ampere\"\n", + " print \"collector to emitter = %0.3f\"%((vce)),\"volt\"\n", + " " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 216 example 27" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistance = 3928.57 ohm\n", + "resistance r1 = 51281.87 ohm\n", + "resistance r2 = 34645.75 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "beta1=99#\n", + "stability=5#\n", + "vbe=0.2##volt\n", + "colres=2.5*10**3##ohm\n", + "vce=6##volt\n", + "ven=5.5##volt\n", + "vcc=15##volt\n", + "vcn=vce+ven#\n", + "colvol=vcc-vcn##voltage across collector resistance\n", + "ic=colvol/colres#\n", + "ib=ic/beta1#\n", + "colre1=ven/ic#\n", + "rb=stability*colre1/(1-(stability/(1+beta1)))##correction in the book taken collector resistance as 3.13*10**3ohm but it is 3.93*10**3ohm\n", + "v1=(ib*rb)+(vbe)+((ib+ic)*colre1)#\n", + "r=rb*vcc/v1#\n", + "r1=r*v1/(vcc-v1)#\n", + "print \"resistance = %0.2f\"%((colre1)),\"ohm\"\n", + "print \"resistance r1 = %0.2f\"%((r)),\"ohm\"\n", + "print \"resistance r2 = %0.2f\"%((r1)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 216 example 28" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "base current = -4.22e-05 ampere\n", + "collector current = -2.11e-03 ampere\n", + "emitter current = -2.15e-03 ampere\n", + "vcb = 5.99 volt\n", + "the collector base junction is reverse biased the transistor in active region\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "beta1=50#\n", + "vbb=5##volt\n", + "rb=10*10**3##ohm\n", + "colres=800##ohm\n", + "re=1.8*10**3##ohm\n", + "vcc=5##volt\n", + "ib=(0.7-vbb)/((rb)+(beta1+1)*re)##correction in book\n", + "re=beta1*ib#\n", + "ie=(ib+re)#\n", + "vce=vcc-colres*re-re*ie#\n", + "vcb=(vce-0.7)#\n", + "print \"base current = %0.2e\"%((ib)),\"ampere\"\n", + "print \"collector current = %0.2e\"%((re)),\"ampere\"\n", + "print \"emitter current = %0.2e\"%((ie)),\"ampere\"\n", + "print \"vcb = %0.2f\"%((vcb)),\"volt\"#correction in book\n", + "print \"the collector base junction is reverse biased the transistor in active region\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 217 example 29" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "collector current = 9.49e-04 ampere\n", + "collector emitter voltage = 6.31 volt\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "r=40*10**3##ohm\n", + "r1=5*10**3##ohm\n", + "colres=r1#\n", + "beta1=50#\n", + "em1res=1*10**3##ohm\n", + "vcc=12##volt\n", + "rth=r*r1/(r+r1)#\n", + "v1=r1*vcc/(r1+r)#\n", + "bascur=(v1-0.3)/(rth+(beta1*em1res))#\n", + "colcur=beta1*bascur#\n", + "vce=vcc-(colres+em1res)*colcur#\n", + "print \"collector current = %0.2e\"%((colcur)),\"ampere\"\n", + "print \"collector emitter voltage = %0.2f\"%((vce)),\"volt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 217 example 30" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " base resistance = 49500 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "colcur=8*10**-3##ampere\n", + "re=500##ohm\n", + "vce=3##volt\n", + "beta1=80#\n", + "vcc=9##volt\n", + "ib=colcur/beta1#\n", + "rb=(vcc-(1+beta1)*(ib*re))/ib#\n", + "print \" base resistance = %0.f\"%((rb)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 217 example 31" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "base current = 9.08e-06 ampere\n", + "collector current = 9.08e-04 ampere\n", + "emitter current = 9.17e-04 ampere\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "vcc=10##volt\n", + "basres=1*10**6##ohm\n", + "colres=2*10**3##ohm\n", + "em1res=1*10**3##ohm\n", + "beta1=100#\n", + "bascur=vcc/(basres+(beta1+1)*(em1res))#\n", + "colcur=beta1*bascur#\n", + "em1cur=colcur+bascur#\n", + "print \"base current = %0.2e\"%((bascur)),\"ampere\"\n", + "print \"collector current = %0.2e\"%((colcur)),\"ampere\"#correction in book\n", + "print \"emitter current = %0.2e\"%((em1cur)),\"ampere\"#correction in book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 218 example 32" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "collector current = 1.98e-03 ampere\n", + "emitter current = -2.00e-03 ampere\n", + "collector emitter voltage = 5.34 volt\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "alpha1=0.99#\n", + "rebacu=1*10**-11##ampere\n", + "colres=2*10**3##ohm\n", + "vcc=10##volt\n", + "bascur=20*10**-6##ampere\n", + "beta1=alpha1/(1-alpha1)#\n", + "i1=(1+beta1)*rebacu#\n", + "colcur=beta1*bascur+i1#\n", + "em1cur=-(bascur+colcur)#\n", + "vcb=vcc-colcur*colres#\n", + "vce=vcb-0.7#\n", + "print \"collector current = %0.2e\"%((colcur)),\"ampere\"\n", + "print \"emitter current = %0.2e\"%((em1cur)),\"ampere\"\n", + "print \"collector emitter voltage = %0.2f\"%((vce)),\"volt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 220 example 33" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "base current = 2.15e-05 ampere\n", + "collector current = 2.15e-03 ampere\n", + "emitter current = 2.17e-03 ampere\n", + "base current = 2.56e-05 ampere\n", + "collector current = 2.56e-03 ampere\n", + "emitter current = 2.59e-03 ampere\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "beta1=100#\n", + "revcur=20*10**-9##ampere\n", + "colres=3*10**3##ohm\n", + "rb=200*10**3##ohm\n", + "vbb=5##volt\n", + "vcc=11##volt\n", + "em1res=2*10**3##ohm\n", + "ib=(vbb-0.7)/rb#\n", + "ic=beta1*ib#\n", + "ie=ib+ic#\n", + "print \"base current = %0.2e\"%((ib)),\"ampere\"\n", + "print \"collector current = %0.2e\"%((ic)),\"ampere\"\n", + "print \"emitter current = %0.2e\"%((ie)),\"ampere\"#question asked only currents\n", + "#2*10**3 ohm added to emitter\n", + "ib=-(0.7-vcc)/(rb+((1+beta1)*em1res))#\n", + "ic=beta1*ib#\n", + "ie=ib+ic#\n", + "print \"base current = %0.2e\"%((ib)),\"ampere\"#correction in book\n", + "print \"collector current = %0.2e\"%((ic)),\"ampere\"\n", + "print \"emitter current = %0.2e\"%((ie)),\"ampere\"#question asked only currents" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 221 example 34" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "emitter current = 2.00e-03 ampere\n", + "collector current = 2.00e-03 ampere\n", + "voltage = 10.00 volt\n", + "vcb = 2.00 volt\n", + "emitter resistance = 6000.00 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "em1cur=2*10**-3##ampere\n", + "v1=12##volt\n", + "vcc=12##volt\n", + "format(12)#\n", + "colres=5*10**3##ohm\n", + "em1res=v1/em1cur#\n", + "colcur=em1cur#\n", + "voltag=colcur*colres##ic*r\n", + "v1=vcc-(colres*colcur)#\n", + "print \"emitter current = %0.2e\"%((em1cur)),\"ampere\"\n", + "print \"collector current = %0.2e\"%((colcur)),\"ampere\"\n", + "print \"voltage = %0.2f\"%((voltag)),\"volt\"\n", + "print \"vcb = %0.2f\"%(abs(v1)),\"volt\"\n", + "print \"emitter resistance = %0.2f\"%((em1res)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 221 example 35" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistance at 0.0 volt 80000.00 ohm\n", + "resistance at 0.70 volt 66000.00 ohm\n", + "vbb at 12volt\n", + "resistance at 0.00 volt 240000.00 ohm\n", + "resistance at 0.70 volt 226000.00 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "vbb=4##volt\n", + "ib=50*10**-6##ampere\n", + "for q in [0, 0.7, 4, 12]:\n", + " if q==0 :\n", + " rb=(vbb-q)/ib#\n", + " print \"resistance at %0.1f\"%((q)),\"volt %0.2f\"%((rb)),\"ohm\"\n", + " elif q==0.7:\n", + " rb=(vbb-q)/ib\n", + " print \"resistance at %0.2f\"%((q)),\"volt %0.2f\"%((rb)),\"ohm\"\n", + " elif q==4:\n", + " print \"vbb at 12volt\"\n", + " q=0\n", + " vbb=12\n", + " rb=(vbb-q)/ib\n", + " print \"resistance at %0.2f\"%((q)),\"volt %0.2f\"%((rb)),\"ohm\"\n", + " else:\n", + " q=0.7#\n", + " vbb=12#\n", + " rb=(vbb-q)/ib#\n", + " \n", + " \n", + " print \"resistance at %0.2f\"%((q)),\"volt %0.2f\"%((rb)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 222 example 36" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "beta = 99.96\n", + "ie = 0.005 ampere\n", + "alpha = 1.00\n", + "ib = 9.80e-05 ampere\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "ic=5.2*10**-3##ampere\n", + "ib=50*10**-6##ampere\n", + "icb=2*10**-6##ampere\n", + "beta1=(ic-icb)/(ib+icb)#\n", + "print \"beta = %0.2f\"%((beta1))\n", + "ie=ib+ic#\n", + "\n", + "print \"ie = %0.3f\"%((ie)),\"ampere\"\n", + "alpha1=(ic-icb)/ic#\n", + "print \"alpha = %0.2f\"%((alpha1))\n", + "\n", + "\n", + "\n", + "ic=10*10**-3##ampere\n", + "ib=(ic-(beta1+1)*(icb))/beta1#\n", + "\n", + "\n", + "print \"ib = %0.2e\"%((ib)),\"ampere\"\n", + "#correction required in the book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 222 example 37" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "collector current at beta 160.00 = 1.07e-03 ampere\n", + "vce at beta 160.00 = -5.69 volt\n", + "collector current at beta 80.00 = 1.07e-03 ampere\n", + "vce at beta 80.00 = -3.03 volt\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "beta1=160\n", + "vb=-0.8##volt\n", + "re=2.5*10**3##ohm\n", + "vcc=10##volt\n", + "for q in [160, 80]:\n", + " ib=(vcc-vb)*10**2/((re)*(1+q)*400)#\n", + " ic=q*ib#\n", + " colres=1.5*10**3##ohm\n", + " print \"collector current at beta %0.2f\"%((q)),\" = %0.2e\"%((ic)),\"ampere\"\n", + " #correction required in the book\n", + " ie=(1+beta1)*ib#\n", + " vce=-(vcc-colres*ic-re*ie)#\n", + " print \"vce at beta %0.2f\"%((q)),\" = %0.2f\"%((vce)),\"volt\"\n", + " #correction required in the book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 222 example 38" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rb = 630000.00 ohm\n", + "stability = 56.51\n", + "new point\n", + "ic = 6.42e-04 ampere\n", + "vce = 8.79 volt\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "vb=0.7##volt\n", + "vce=7##volt\n", + "ic=1*10**-3##ampere\n", + "vcc=12##volt\n", + "beta1=100#\n", + "colres=(vcc-vce)/ic#\n", + "ib=ic/beta1#\n", + "#rb\n", + "rb=(vcc-vb-ic*colres)/ib#\n", + "print \"rb = %0.2f\"%((rb)),\" ohm\"\n", + "#stability\n", + "stability=(1+beta1)/(1+beta1*(colres/(colres+rb)))#\n", + "print \"stability = %0.2f\"%((stability))\n", + "#beta=50\n", + "beta1=50#\n", + "print \"new point\"\n", + "ib=(vcc-vb)/(beta1*colres+rb)#\n", + "ic=beta1*ib#\n", + "print \"ic = %0.2e\"%((ic)),\" ampere\"\n", + "vce=vcc-(ic*colres)#\n", + "print \"vce = %0.2f\"%((vce)),\" volt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 223 example 39" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "new point\n", + "vce = 4.21 volt\n", + "ic = 1.93e-03 ampere\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "vcc=16##volt\n", + "colres=3*10**3##ohm\n", + "re=2*10**3##ohm\n", + "r1=56*10**3##ohm\n", + "r2=20*10**3##ohm\n", + "alpha1=0.985#\n", + "vb=0.3##volt\n", + "#coordinates\n", + "beta1=alpha1/(1-alpha1)#\n", + "v1=vcc*r2/(r1+r2)#\n", + "rb=r2/(r1+r2)#\n", + "ic=(v1-vb)/((rb/beta1)+(re/beta1)+re)#\n", + "print \"new point\"\n", + "print \"vce = %0.2f\"%((v1)),\" volt\"\n", + "print \"ic = %0.2e\"%((ic)),\" ampere\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 224 example 40" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "re = 1300.00 ohm\n", + "r1 = 26233.18 ohm\n", + "r2 = 3725.66 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "vce=12##volt\n", + "ic=2*10**-3##ampere\n", + "vcc=24##volt\n", + "vb=0.7##volt\n", + "beta1=50#\n", + "colres=4.7*10**3##ohm\n", + "#re\n", + "re=((vcc-vce)/(ic))-colres#\n", + "print \"re = %0.2f\"%((re)),\" ohm\"\n", + "#r1\n", + "ib=ic/beta1#\n", + "v1=ib*3.25*10**3+vb+(ib+1.5*10**3)#\n", + "r1=3.25*18*10**3/2.23#\n", + "print \"r1 = %0.2f\"%((r1)),\" ohm\"\n", + "#r2\n", + "r2=26.23*2.23*10**3/(18-2.3)#\n", + "print \"r2 = %0.2f\"%((r2)),\" ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 225 example 41" + ] + }, + { + "cell_type": "code", + "execution_count": 36, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ib = 2.87e-05 ampere\n", + "ic = 3.58e-03 ampere\n", + "ie = 3.61e-03 ampere\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "colres=3*10**3##ohm\n", + "rb=150*10**3##ohm\n", + "beta1=125#\n", + "vcc=10##volt\n", + "v1=5##volt\n", + "vb=0.7##volt\n", + "ib=(v1-vb)/rb#\n", + "print \"ib = %0.2e\"%((ib)),\" ampere\"\n", + "ic=beta1*ib#\n", + "ie=ic+ib#\n", + "print \"ic = %0.2e\"%((ic)),\" ampere\"\n", + "print \"ie = %0.2e\"%((ie)),\" ampere\"#correction in the book in question to find only currents" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 226 example 42" + ] + }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "re = 1033.33 ohm\n", + "rb = 4485.11 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "beta1=50#\n", + "vb=0.6##volt\n", + "vcc=18##volt\n", + "colres=4.3*10**3##ohm\n", + "ic=1.5*10**-3##ampere\n", + "vce=10##volt\n", + "stability=4#\n", + "r1=(vcc-vce)/ic#\n", + "re=r1-colres#\n", + "w=(beta1+1)*(stability)*re/(1+beta1-stability)#\n", + "print \"re = %0.2f\"%((re)),\"ohm\"\n", + "print \"rb = %0.2f\"%((w)),\"ohm\"#correction in the book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 226 example 43" + ] + }, + { + "cell_type": "code", + "execution_count": 38, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "r1 = 3.8*r2\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "re=100##ohm\n", + "beta1=100#\n", + "rb=1*10**3##ohm\n", + "stability=(1+beta1)/(1+beta1*(re/(re+rb)))#\n", + "r1=3.8#r2\n", + "print \"r1 = 3.8*r2\"#correction in the book not given in question" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 228 example 45" + ] + }, + { + "cell_type": "code", + "execution_count": 39, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rb at 75 celsius = 14062.50 ohm\n", + "icb = 1.80e-05 ampere\n", + "temperature at which current till max = 58.40 celsius\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log10\n", + "icb=2*10**-6##ampere\n", + "vbb=1##volt\n", + "r1=50*10**3##ohm\n", + "#current increases every 10celsius rb at 75celsius\n", + "vb=-0.1##volt\n", + "icb=2**6*10**-6##at 75celsius\n", + "rb=(vb+vbb)/icb#\n", + "print \"rb at 75 celsius = %0.2f\"%((rb)),\"ohm\"\n", + "icb=(vb+vbb)/r1#\n", + "print \"icb = %0.2e\"%((icb)),\"ampere\"\n", + "w=(log10(icb*10**6)*20/log10(2))-25#\n", + "print \"temperature at which current till max = %0.2f\"%((w)),\"celsius\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 228 example 46" + ] + }, + { + "cell_type": "code", + "execution_count": 40, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "min collector resistance = 4558.14 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "vb=0.8##volt\n", + "beta1=100#\n", + "vce=0.2##volt\n", + "vcc=10##volt\n", + "rb=200*10**3##ohm\n", + "#collector resistance\n", + "ib=(5-0.7)/rb#\n", + "colres=(vcc-vce)/(beta1*ib)#\n", + "print \"min collector resistance = %0.2f\"%((colres)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 229 example 47" + ] + }, + { + "cell_type": "code", + "execution_count": 41, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ib2 = 4.00e-03 ampere\n", + "ie1 = -4.00e-03 ampere\n", + "ic2 = 9.60e-02 ampere\n", + "ib1 = 8.00e-05 ampere\n", + "ic1 = 3.92e-03 ampere\n", + "ic = 9.99e-02 ampere\n", + "ic/ib = 1249.00\n", + "ic/ie = -1.00\n", + "vce = 12.01 volt\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "alpha1=0.98#\n", + "alph11=0.96#\n", + "vcc=24##volt\n", + "colres=120##ohm\n", + "ie=100*10**-3##ampere\n", + "beta1=alpha1/(1-alpha1)#\n", + "bet11=alph11/(1-alph11)#\n", + "ib2=ie/(1+bet11)#\n", + "ie1=-ib2#\n", + "print \"ib2 = %0.2e\"%((ib2)),\"ampere\"\n", + "print \"ie1 = %0.2e\"%((ie1)),\"ampere\"\n", + "\n", + "\n", + "ic2=bet11*ib2#\n", + "ib1=ib2/(1+beta1)#\n", + "ic1=beta1*ib1#\n", + "print \"ic2 = %0.2e\"%((ic2)),\"ampere\"\n", + "print \"ib1 = %0.2e\"%((ib1)),\"ampere\"\n", + "print \"ic1 = %0.2e\"%((ic1)),\"ampere\"\n", + "ic=ic1+ic2#\n", + "vce=vcc-ic*colres#\n", + "ib=ib1#\n", + "w=ic/ib#\n", + "q=-ic/ie#\n", + "print \"ic = %0.2e\"%((ic)),\"ampere\"\n", + "print \"ic/ib = %0.2f\"%((w))\n", + "print \"ic/ie = %0.2f\"%((q))\n", + "#correction required in the book\n", + "print \"vce = %0.2f\"%((vce)),\"volt\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronic_Devices_and_Circuits_by_J._Paul/Ch5_1.ipynb b/Electronic_Devices_and_Circuits_by_J._Paul/Ch5_1.ipynb new file mode 100644 index 00000000..0176b717 --- /dev/null +++ b/Electronic_Devices_and_Circuits_by_J._Paul/Ch5_1.ipynb @@ -0,0 +1,1346 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5 - BJT Amplifier" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 283 example 1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage gain = -50.00\n", + "input resistance = 2600.00 ohm\n", + "ce removed\n", + "voltage gain = -50.00\n", + "input resistance = 12600.00 ohm\n" + ] + } + ], + "source": [ + "ic=1*10**-3##ampere\n", + "vcc=5##volt\n", + "colres=2*10**3##ohm\n", + "r1=1.4*10**3##ohm\n", + "re=100##ohm\n", + "beta1=100\n", + "rb=100##ohm\n", + "v1=0.026\n", + "c1=25*10**-6##farad\n", + "g1=ic/v1\n", + "freque=10*10**3##hertz\n", + "xc=1/(2*freque*3.14*c1)\n", + "volgai=-beta1*colres/(r1+0.1*10**3+2.5*10**3)\n", + "print \"voltage gain = %0.2f\"%((volgai))\n", + "ri=(0.1+2.5)*10**3-((xc.imag)*(1+beta1))\n", + "print \"input resistance = %0.2f\"%((ri)),\"ohm\"\n", + "#ce removed\n", + "volgai=-beta1*colres/((r1+0.1*10**3+2.5*10**3)+(101/1000)*10**3*100)\n", + "print \"ce removed\"\n", + "print \"voltage gain = %0.2f\"%((volgai))\n", + "ri=(0.1+2.5)*10**3+100*101/1000*10**3\n", + "print \"input resistance = %0.2f\"%((ri)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 285 example 2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage gain = 260.00 <180\n", + "voltage gain reduced ce removed\n", + "when cb is short circuited the voltage gain increased\n" + ] + } + ], + "source": [ + "ic=1.3*10**-3##ampere\n", + "colres=2*10**3##ohm\n", + "re=500##ohm\n", + "v1=0.026##volt\n", + "beta1=100\n", + "vcc=15##volt\n", + "c1=10*10**-6##farad\n", + "ib=ic/beta1\n", + "ri=0.01/ib\n", + "volgai=beta1*colres*ib/0.01\n", + "print \"voltage gain = %0.2f\"%((volgai)),\"<180\"\n", + "print \"voltage gain reduced ce removed\"\n", + "print \"when cb is short circuited the voltage gain increased\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 286 example 3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current gain = -11.62\n", + "voltage gain = -46.46\n", + "transconductance = -0.01 ampere per volt\n", + "transresistance = -46464.08 ohm\n", + "input resistance = 1042.65 ohm\n", + "output resistance = 3636.00 ohm\n" + ] + } + ], + "source": [ + "colres=4*10**3##ohm\n", + "r1=4*10**3##ohm\n", + "\n", + "rb=20*10**3##ohm\n", + "r=1*10**3##ohm\n", + "hie=1.1*10**3##ohm\n", + "\n", + "#current gain\n", + "ri=rb*hie/(rb+hie)\n", + "curgai=(1/2.04)*(rb/(rb+(hie)))*(-50*colres/(colres+(r1)))\n", + "print \"current gain = %0.2f\"%((curgai))\n", + "#voltage gain\n", + "volgai=curgai*r1/r\n", + "print \"voltage gain = %0.2f\"%((volgai))\n", + "#transconductance\n", + "conduc=volgai/r1\n", + "print \"transconductance = %0.2f\"%((conduc)),\"ampere per volt\"\n", + "#transresistance\n", + "resist=volgai*r\n", + "print \"transresistance = %0.2f\"%((resist)),\"ohm\"\n", + "#input resistance\n", + "print \"input resistance = %0.2f\"%((ri)),\"ohm\"\n", + "#output resistance\n", + "resist=40*10**3*colres/(40*10**3+colres)\n", + "\n", + "\n", + "\n", + "print \"output resistance = %0.2f\"%((resist)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 287 example 4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rb = 235000.00 ohm\n", + "rb including emitter resistance = 230000.00 ohm\n" + ] + } + ], + "source": [ + "ib=20*10**-6##ampere\n", + "beta1=500\n", + "re=10##ohm correction in the book\n", + "r1=4.7*10**2##ohm correction in the book\n", + "ic=ib*beta1\n", + "voltag=ic*r1##voltage drop at 4.7*10**3ohm\n", + "vc=(10-voltag)\n", + "rb=(vc-0.6)/ib\n", + "print \"rb = %0.2f\"%((rb)),\"ohm\"\n", + "#re included\n", + "voltag=ic*re##voltage drop at re\n", + "vb=(0.6+voltag)\n", + "rb=(vc-vb)/ib\n", + "print \"rb including emitter resistance = %0.2f\"%((rb)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 288 example 5" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage gain with fedback = 73.92 decibel\n", + "beta = 1.20e-04\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log10\n", + "av=12480\n", + "fedbac=8##decibel\n", + "volgai=20*log10(av)##gain without fedback\n", + "volga1=volgai-fedbac\n", + "beta1=((av/5000)-1)/av\n", + "\n", + "print \"voltage gain with fedback = %0.2f\"%((volga1)),\"decibel\"\n", + "print \"beta = %0.2e\"%((beta1))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 288 example 6" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "collector current = 3.00e-03 ampere\n", + "emitter current = 3.00e-03 ampere\n", + "base current = 3.00e-05 ampere\n" + ] + } + ], + "source": [ + "beta1=100\n", + "r1=1.5*10**3##ohm\n", + "vcc=10##volt\n", + "r=100*10**3##ohm\n", + "vb=((vcc)/(r+10*10**3))*10*10**3\n", + "ie=0.3/100\n", + "ib=ie/beta1\n", + "print \"collector current = %0.2e\"%((ie)),\"ampere\"\n", + "print \"emitter current = %0.2e\"%((ie)),\"ampere\"\n", + "print \"base current = %0.2e\"%((ib)),\"ampere\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 268 example 7" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage gain = -125.00\n", + "power gain = 6250.00\n", + "error without hoe = 10.00\n", + "error = 21.00\n" + ] + } + ], + "source": [ + "hie=800##ohm\n", + "he=50*10**-6##mho\n", + "hfe=-55\n", + "z1=2*10**3##ohm\n", + "curgai=hfe/(1+he*z1)\n", + "zi=hie\n", + "volgai=curgai*z1/zi\n", + "powgai=volgai*curgai\n", + "#if hoe neglected\n", + "av=137.5\n", + "hfe=-55\n", + "w=((av-abs(volgai))*100)/abs(volgai)\n", + "ap=hfe*(-av)\n", + "w1=((ap-powgai)*100)/powgai\n", + "print \"voltage gain = %0.2f\"%((volgai))\n", + "\n", + "\n", + "print \"power gain = %0.2f\"%((powgai))\n", + "print \"error without hoe = %0.2f\"%((w))\n", + "print \"error = %0.2f\"%((w1))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 289 example 8" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "emitter current = 1.92e-03 ampere\n", + "vc = 18.12 volt\n", + "collector emitter voltage = 8.53 volt\n" + ] + } + ], + "source": [ + "rb=5*10**3##ohm\n", + "vcc=20##volt\n", + "r=10*10**3##ohm\n", + "colres=5*10**3##ohm\n", + "vb=vcc*r/(r+r)\n", + "beta1=50\n", + "v1=0.6##volt\n", + "ib=(vb-v1)/(1+beta1*colres)\n", + "ic=beta1*ib\n", + "vc=vcc-ic*1*10**3\n", + "vce=vc-rb*(ic+ib)\n", + "print \"emitter current = %0.2e\"%((ic+ib)),\"ampere\"\n", + "print \"vc = %0.2f\"%((vc)),\"volt\"\n", + "print \"collector emitter voltage = %0.2f\"%((vce)),\"volt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 290 example 9" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage gain = 395.49\n" + ] + } + ], + "source": [ + "hib=25##ohm\n", + "hfb=0.999\n", + "hob=10**-6##ohm\n", + "colres=10*10**3##ohm\n", + "#voltage gain\n", + "curgai=hfb/(1+hob*colres)\n", + "zi=hib+hob*colres*curgai\n", + "volgai=curgai*colres/(zi)\n", + "print \"voltage gain = %0.2f\"%((volgai))\n", + "#correction required in the book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 290 example 10" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage gain = 1.00\n", + "input resistance = 101050.00 ohm\n" + ] + } + ], + "source": [ + "re=1*10**3##ohm\n", + "hie=100##ohm\n", + "hfe=100\n", + "#voltage gain\n", + "volgai=1/((1+(hie/(2*(1+hfe)*re))))\n", + "#ri\n", + "ri=(hie/2)+(1+hfe)*re\n", + "print \"voltage gain = %0.2f\"%((volgai))\n", + "print \"input resistance = %0.2f\"%((ri)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 292 example 11" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "emitter current = 4.16e-03 ampere\n", + "vce = 11.68 volt\n" + ] + } + ], + "source": [ + "beta1=90\n", + "re=2*10**3##ohm\n", + "rb=240*10**3##ohm\n", + "vcc=20\n", + "ib=(vcc-0.7)/(rb+(1+beta1)*(re))\n", + "ic=beta1*ib\n", + "vce=vcc-(ib+ic)*re\n", + "print \"emitter current = %0.2e\"%((ib+ic)),\"ampere\"\n", + "print \"vce = %0.2f\"%((vce)),\"volt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 292 example 12" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage gain = -313.90\n", + "current gain = -100.55\n", + "impedance = 86666.67 ohm\n", + "parameters using approxmiate\n", + "voltage gain = -323.12\n", + "current gain = -110.00\n", + "impedance = 86666.67 ohm\n" + ] + } + ], + "source": [ + "hfe=110\n", + "hie=1.6*10**3##ohm\n", + "hoe=20*10**-6##ohm\n", + "colres=4.7*10**3##ohm\n", + "hre=2*10**-4\n", + "r1=470*10**3##ohm\n", + "curgai=-hfe/(1+hoe*colres)\n", + "ri=hie+hre*curgai*colres\n", + "volgai=curgai*colres/ri\n", + "y1=hoe-((hfe*hre)/(hie+1*10**3))\n", + "z1=1/y1\n", + "print \"voltage gain = %0.2f\"%((volgai))\n", + "print \"current gain = %0.2f\"%((curgai))\n", + "print \"impedance = %0.2f\"%((z1)),\"ohm\"\n", + "r0=z1*colres/(z1+colres)\n", + "curgai=-hfe\n", + "ri=hie\n", + "print \"parameters using approxmiate\"\n", + "volgai=curgai*(colres)/ri\n", + "print \"voltage gain = %0.2f\"%((volgai))\n", + "#correction required in the book\n", + "print \"current gain = %0.2f\"%((curgai))\n", + "print \"impedance = %0.2f\"%((z1)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 293 example 13" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "input resistance = 1.00e+07 ohm\n", + "voltage gain = 1.00\n", + "current gain = -10000.00\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "re=1*10**3##ohm\n", + "hie=1000##ohm\n", + "hfe=99\n", + "#inptut resistance\n", + "ri=hie+((1+hfe)*(hie+1+hfe*re))\n", + "\n", + "\n", + "print \"input resistance = %0.2e\"%((ri)),\"ohm\"##correction in the book\n", + "#voltage gain\n", + "volgai=((1+hfe)*(1+hfe)*re)/ri\n", + "print \"voltage gain = %0.2f\"%((volgai))\n", + "\n", + "\n", + "#current gain\n", + "curgai=-((1+hfe)*(1+hfe))\n", + "\n", + "\n", + "print \"current gain = %0.2f\"%((curgai))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 294 example 14" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage gain = 250.00 <180\n", + "input impedance = 2000.00 ohm\n", + "current gain = 100.00\n" + ] + } + ], + "source": [ + "hie=2*10**3##ohm\n", + "beta1=100\n", + "colres=5*10**3##ohm\n", + "volgai=beta1*colres/hie\n", + "print \"voltage gain = %0.2f\"%((volgai)),\"<180\"\n", + "print \"input impedance = %0.2f\"%((hie)),\"ohm\"\n", + "print \"current gain = %0.2f\"%((beta1))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 294 example 15" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage gain = 253.29\n", + "input impedance = 1546.39\n", + "coordinates ic = 1.14e-03 ampere vce = 8.30 volt\n" + ] + } + ], + "source": [ + "colres=4.7*10**3##ohm\n", + "beta1=150\n", + "r1=12*10**3##ohm\n", + "vcc=15##volt\n", + "re=1.2*10**3##ohm\n", + "rac=colres*r1/(colres+r1)\n", + "r=2*10**3##ohm\n", + "#voltage gain\n", + "volgai=beta1*rac/r\n", + "print \"voltage gain = %0.2f\"%((volgai))\n", + "r1=75*10**3##ohm\n", + "r2=7.5*10**3##ohm\n", + "#input impedance\n", + "zin=(r1*r2)/(r1+r2)\n", + "zin=zin*r/(zin+r)\n", + "print \"input impedance = %0.2f\"%((zin))\n", + "#coordinates\n", + "vb=vcc*r2/(r1+r2)\n", + "ie=vb/re\n", + "vce=vcc-((colres+re)*(ie))\n", + "print \"coordinates ic = %0.2e\"%((ie)),\"ampere vce = %0.2f\"%((vce)),\"volt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 296 example 16" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current gain = 57.14\n", + "input impedance = 115485.71 ohm\n", + "voltage gain = 0.99\n", + "output resistance = 32558.14 ohm\n" + ] + } + ], + "source": [ + "r1=2000##ohm\n", + "r=900##ohm\n", + "hie=1200##ohm\n", + "hre=2*10**-4\n", + "hfe=60\n", + "hoe=25*10**-6##ampere per volt\n", + "curgai=(hfe)/(1+hoe*r1)\n", + "print \"current gain = %0.2f\"%((curgai))\n", + "ri=hie+(curgai*r1)\n", + "print \"input impedance = %0.2f\"%((ri)),\"ohm\"\n", + "volgai=curgai*r1/ri\n", + "print \"voltage gain = %0.2f\"%((volgai))\n", + "admita=1/ri\n", + "admita=hoe-(-hfe*hre)/(hie+r)\n", + "r=1/admita\n", + "print \"output resistance = %0.2f\"%((r)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 296 example 17" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage gain = -609.86\n", + "current gain = -60.00\n", + "input impedance = 501.75 ohm\n", + "output impedance = 5100.00 ohm\n" + ] + } + ], + "source": [ + "hfe=60\n", + "hie=500##ohm\n", + "ic=3*10**-3##ampere\n", + "zi=hie\n", + "rb=220*10**3##ohm\n", + "colres=5.1*10**3##ohm\n", + "z=colres\n", + "volgai=-hfe*colres/hie\n", + "curgai=-hfe\n", + "vcc=12##volt\n", + "ib=(vcc-0.6)/rb\n", + "ie=hfe*ib\n", + "re=0.026/ie\n", + "zi=hfe*re\n", + "z=colres\n", + "volgai=-colres/re\n", + "curgai=-hfe\n", + "print \"voltage gain = %0.2f\"%((volgai))\n", + "print \"current gain = %0.2f\"%((curgai))\n", + "print \"input impedance = %0.2f\"%((zi)),\"ohm\"\n", + "print \"output impedance = %0.2f\"%((z)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 297 example 18" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "input impedance = 1817.30 ohm\n", + "output impedance = 4000.00 ohm\n", + "voltage gain = -125.00\n", + "current gain = -56.79\n", + "parameters in re\n", + "input impedance = 1745.41 ohm\n", + "output impedance = 4000.00 ohm\n", + "voltage gain = -134.07\n", + "current gain = -58.50\n" + ] + } + ], + "source": [ + "hie=3.2*10**3##ohm\n", + "hfe=100\n", + "r=40*10**3##ohm\n", + "r1=4.7*10**3##ohm\n", + "colres=4*10**3##ohm\n", + "rb=r*r1/(r+r1)\n", + "zi=hie*rb/(hie+rb)\n", + "z=colres\n", + "re=1.2*10**3##ohm\n", + "volgai=-hfe*colres/hie\n", + "print \"input impedance = %0.2f\"%((zi)),\"ohm\"\n", + "print \"output impedance = %0.2f\"%((z)),\"ohm\"\n", + "print \"voltage gain = %0.2f\"%((volgai))\n", + "curgai=-hfe*rb/(rb+hie)\n", + "print \"current gain = %0.2f\"%((curgai))\n", + "hie=833\n", + "#(1) load open\n", + "vi=1\n", + "ib=vi/hie\n", + "volgai=hfe*ib*1.5*10**3\n", + "#load closed\n", + "hoe=50\n", + "r2=2*10**3##ohm\n", + "ib=vi/(r2+hie)\n", + "vb=1.682\n", + "ib=(vb-0.6)/(rb+(1+hfe)*(re))\n", + "ic=hfe*ib\n", + "ie=ic+ib\n", + "re=0.026/ie\n", + "zi=rb*hfe*re/((rb)+(hfe*re))\n", + "print \"parameters in re\"\n", + "print \"input impedance = %0.2f\"%((zi)),\"ohm\"\n", + "z=colres\n", + "print \"output impedance = %0.2f\"%((z)),\"ohm\"\n", + "volgai=colres/(-re)\n", + "print \"voltage gain = %0.2f\"%((volgai))\n", + "curgai=-hfe*rb/(rb+hfe*re)\n", + "print \"current gain = %0.2f\"%((curgai))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 299 example 19" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage gain = -0.006\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "hfe=120\n", + "hie=0.02##ohm\n", + "r1=5.8*10**3##ohm\n", + "r=27*10**3##ohm\n", + "colres=1.5*10**3##ohm\n", + "re=330*10**3##ohm\n", + "vcc=10##volt\n", + "vb=vcc*r1/(r1+r)\n", + "rb=(r*r1)/(r+r1)\n", + "ib=(vb-0.7)/(rb+((1+hfe)*re))\n", + "volgai=-hfe*ib*2*10**3\n", + "print \"voltage gain = %0.3f\"%((volgai))\n", + "#correction required in the book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 300 example 20" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "upper frequency voltage gain = 7.21e+06 hertz\n", + "upper current gain = 3.61e+06 hertz\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "freque=6*10**6##hertz\n", + "hfe=50\n", + "r1=500##ohm\n", + "g=0.04\n", + "rbb=100##ohm\n", + "\n", + "\n", + "c1=10*10**-12##farad\n", + "r=1000##ohm\n", + "rbe=hfe/g\n", + "ce=g/(2*3.14*freque)\n", + "c1=ce+c1*(1+g*r)\n", + "hie=rbb+rbe\n", + "resist=(r1+rbb)*rbe/(r1+rbb+rbe)\n", + "frequ2=1/(2*3.14*resist*c1)\n", + "curgai=-hfe*r1/(r1+hie)\n", + "volgai=(-hfe*r)/(r1+hie)\n", + "q=volgai*frequ2\n", + "print \"upper frequency voltage gain = %0.2e\"%(abs(q)),\"hertz\"##correction in the book\n", + "q=curgai*frequ2\n", + "print \"upper current gain = %0.2e\"%(abs(q)),\"hertz\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 301 example 21" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current gain = -48.78\n", + "input resistance = 990.24 ohm\n", + "voltage gain = -49.26\n", + "output resistance = 51428.57 ohm\n", + "approximate\n", + "current gain = -50.00\n", + "input resistance = 1000.00 ohm\n", + "voltage gain = -50.00\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "hie=1*10**3##ohm\n", + "hre=2*10**-4\n", + "hoe=25*10**-6##ampere per volt\n", + "hfe=50\n", + "colres=1*10**3##ohm\n", + "curgai=-hfe/(1+hoe*colres)\n", + "print \"current gain = %0.2f\"%((curgai))\n", + "ri=hie-hfe*hre/(hoe+1/colres)\n", + "print \"input resistance = %0.2f\"%((ri)),\"ohm\"\n", + "volgai=curgai*colres/ri\n", + "print \"voltage gain = %0.2f\"%((volgai))\n", + "y1=hoe-((hfe*hre)/(hie+800))\n", + "r1=1/y1\n", + "print \"output resistance = %0.2f\"%((r1)),\"ohm\"\n", + "#approximate\n", + "print \"approximate\"\n", + "curgai=-hfe\n", + "print \"current gain = %0.2f\"%((curgai))\n", + "ri=hie\n", + "print \"input resistance = %0.2f\"%((ri)),\"ohm\"\n", + "volgai=-hfe*colres/hie\n", + "print \"voltage gain = %0.2f\"%((volgai))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 301 example 22" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage gain = 174.11\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "rb1=7.5*10**3##ohm\n", + "rb2=6.8*10**3##ohm\n", + "\n", + "rb3=3.3*10**3##ohm\n", + "re=1.3*10**3##ohm\n", + "colres=2.2*10**3##ohm\n", + "beta1=120\n", + "vcc=18##volt\n", + "vb1=rb3*vcc/(rb3+rb2+rb1)\n", + "ie1=(vb1-0.7)/(re)\n", + "re1=0.026/ie1\n", + "re2=0.026/ie1\n", + "volgai=colres/re2\n", + "print \"voltage gain = %0.2f\"%((volgai))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 302 example 23" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "forced beta = 114.29\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "vcc=5##volt\n", + "colres=250##ohm\n", + "v1=5##volt\n", + "rb=25*10**3##ohm\n", + "beta1=200\n", + "vbs=0.8##volt\n", + "vcon=0.3##volt\n", + "icon=(vcc-vcon)/colres\n", + "ibon=icon/beta1\n", + "ibs=(v1-vbs)/rb\n", + "ic=(vcc-0.2)/colres\n", + "beta1=ic/ibs\n", + "print \"forced beta = %0.2f\"%((beta1))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 303 example 24" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistance r1 = 40847.46 ohm\n", + "resistance r3 = 2197.00 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "vb=0.6##volt\n", + "beta1=100\n", + "ic=1*10**-3##ampere\n", + "vce=2.5##volt\n", + "re=300##ohm\n", + "vcc=5##volt\n", + "ib=ic/beta1\n", + "ie=ic+ib\n", + "ve=ie*re\n", + "vce=vce+ve\n", + "r3=(vcc-vce)/ic\n", + "vb=ve+vb\n", + "r1=(vcc-vb)/(vb/(10*10**3)+(ib))\n", + "print \"resistance r1 = %0.2f\"%((r1)),\"ohm\"\n", + "print \"resistance r3 = %0.2f\"%((r3)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 304 example 25" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "input impedance q1 = 1.75e+04 ohm\n", + "input impedance q2 = 3.50e+06 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "vce2=7.5##volt\n", + "vb=0.7##volt\n", + "beta1=200\n", + "v1=25##volt\n", + "r1=10*10**3##ohm\n", + "vcc=15##volt\n", + "i1=(vcc-vb)/r1\n", + "r=(vcc-vce2)/i1\n", + "z1=beta1*v1/i1\n", + "z=v1/i1\n", + "print \"input impedance q1 = %0.2e\"%((z)),\"ohm\"##correction in the book\n", + "print \"input impedance q2 = %0.2e\"%((z1)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 305 example 26" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "make input = 0\n", + "ground dc\n", + "output resistance = 19.61 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "beta1=99\n", + "r1=1*10**3##ohm\n", + "g=beta1/r1\n", + "r=r1*((r1+r1)/(100))/((r1+((r1+r1)/(100))))\n", + "print \"make input = 0\"\n", + "print \"ground dc\"\n", + "print \"output resistance = %0.2f\"%((r)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 305 example 27" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "output resistance = 1000.00 ohm\n", + "input resistance very low\n", + "voltage gain = 19.23\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "ic=0.5*10**-3##ampere\n", + "rb=100*10**3##ohm\n", + "v1=0.026##volt\n", + "r1=50##ohm\n", + "colres=1*10**3##ohm\n", + "g=ic/v1\n", + "volgai=g*colres\n", + "print \"output resistance = %0.2f\"%((colres)),\"ohm\"\n", + "print \"input resistance very low\"##not given in the book\n", + "print \"voltage gain = %0.2f\"%((volgai))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 306 example 28" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current gain = -12.27\n", + "voltage gain = -49.07\n", + "transconductance = -0.01 ampere per volt\n", + "transresistance = -490686.77 ohm\n", + "input resistance = 990.99 ohm\n", + "output resistance = 4444.44 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "re=4*10**3##ohm\n", + "r1=4*10**3##ohm\n", + "hie=1.1*10**3##ohm\n", + "resist=10*10**3##ohm\n", + "hfe=50\n", + "rb=10*10**3##ohm\n", + "r=1*10**3##ohm\n", + "colres=5*10**3##ohm\n", + "#(1) current gain\n", + "ri=rb*hie/(rb+hie)\n", + "curgai=(1/2.04)*((rb)/(rb+hie))*((-hfe*colres)/(colres+r1))\n", + "print \"current gain = %0.2f\"%((curgai))\n", + "#(2) voltage gain\n", + "volgai=curgai*r1/r\n", + "print \"voltage gain = %0.2f\"%((volgai))\n", + "#(3) tranconductance\n", + "conduc=volgai/r1\n", + "print \"transconductance = %0.2f\"%((conduc)),\"ampere per volt\"\n", + "#transresistance\n", + "resist=resist*volgai\n", + "print \"transresistance = %0.2f\"%((resist)),\"ohm\"\n", + "print \"input resistance = %0.2f\"%((ri)),\"ohm\"\n", + "r=(40*10**3*colres)/(40*10**3+colres)\n", + "print \"output resistance = %0.2f\"%((r)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 307 example 29" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "base resistance = 235000.00 ohm\n", + "base resistance with re\n", + "base resistance = 230000.00 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "beta1=500\n", + "ib=20*10**-6##ampere\n", + "re=100##ohm\n", + "ic=beta1*ib\n", + "vc=ic*0.47*10**3##voltage drop across collector resistance\n", + "v1=(10-vc)\n", + "vb=v1-0.6\n", + "rb=vc/ib\n", + "print \"base resistance = %0.2f\"%((rb)),\"ohm\"\n", + "ve=re*ic\n", + "print \"base resistance with re\"\n", + "b=0.6+0.1\n", + "rb=(v1-b)/ib\n", + "print \"base resistance = %0.2f\"%((rb)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 308 example 30" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "collector current = 3.00e-03 ampere\n", + "base current = 3.00e-05 ampere\n", + "emitter current = 3.00e-03 ampere\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "beta1=100\n", + "re=100##ohm\n", + "vcc=10##volt\n", + "colres=1.5*10**3##ohm\n", + "r=100*10**3##ohm\n", + "r1=10*10**3##ohm\n", + "vb=vcc*r1/(r1+r)\n", + "ie=0.3/re\n", + "ib=ie/beta1\n", + "print \"collector current = %0.2e\"%((ie)),\"ampere\"\n", + "print \"base current = %0.2e\"%((ib)),\"ampere\"\n", + "print \"emitter current = %0.2e\"%((ie)),\"ampere\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronic_Devices_and_Circuits_by_J._Paul/Ch6_1.ipynb b/Electronic_Devices_and_Circuits_by_J._Paul/Ch6_1.ipynb new file mode 100644 index 00000000..4ac77cf6 --- /dev/null +++ b/Electronic_Devices_and_Circuits_by_J._Paul/Ch6_1.ipynb @@ -0,0 +1,246 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6 - BJT at High Frequency" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 337 example 1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "transconductance g = 0.38 ampere/volt\n", + "input conductance gbe = 3.85e-03 ampere/volt\n", + "feedback conductance gbc = 3.85e-07 ampere/volt\n", + "base spread resistance rbb = 240.00 ohm\n", + "output conductance = 1.15e-06 ampere/volt\n", + "transition capacitance cbe = 1.22e-09 farad\n", + "rbc = 2.60e+06 ohm\n", + "rce = 8.67e+05 ohm\n" + ] + } + ], + "source": [ + "colcur=10*10**-3##ampere\n", + "vce=10##volt\n", + "hie=500##ohm\n", + "hoe=4*10**-5\n", + "hfe=100\n", + "hre=1*10**-4\n", + "fqu=50*10**6##hertz\n", + "q=3*10**12##farad\n", + "voltag=26*10**-3##volt\n", + "g=colcur/voltag\n", + "gbe=g/hfe\n", + "gbc=gbe*hre\n", + "rbb=hie-260\n", + "oucond=hoe-(1+hfe)*gbc\n", + "cbe=g/(2*3.14*fqu)\n", + "rbc=1/gbc\n", + "rce=1/oucond\n", + "print \"transconductance g = %0.2f\"%((g)),\"ampere/volt\"\n", + "print \"input conductance gbe = %0.2e\"%((gbe)),\"ampere/volt\"\n", + "print \"feedback conductance gbc = %0.2e\"%((gbc)),\"ampere/volt\"\n", + "print \"base spread resistance rbb = %0.2f\"%((rbb)),\"ohm\"\n", + "print \"output conductance = %0.2e\"%((oucond)),\"ampere/volt\"\n", + "print \"transition capacitance cbe = %0.2e\"%((cbe)),\"farad\"\n", + "print \"rbc = %0.2e\"%((rbc)),\"ohm\"##correction as 2.6mega ohm\n", + "print \"rce = %0.2e\"%((rce)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 337 example 2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "fbeta = 1.00 hertz\n", + "f = 100.00 hertz\n", + "cbe = 3.06e-04 farad\n", + "rbe = 520.00 ohm\n", + "rbb = 80.00 ohm\n" + ] + } + ], + "source": [ + "colcur=5*10**-3##ampere\n", + "vce=10##volt\n", + "hfe=100\n", + "hie=600##ohm\n", + "cugain=10\n", + "fqu=10*10**6##hertz\n", + "\n", + "tracat=3*10**-12##farad\n", + "voltag=26*10**-3##volt\n", + "fbeta1=((((hfe**2)/(cugain**2))-1)/fqu**2)**(1/2)\n", + "fbeta1=1/fbeta1\n", + "fq1=hfe*fbeta1\n", + "cbe=colcur/(2*3.14*fq1*voltag)\n", + "rbe=hfe/(colcur/voltag)\n", + "rbb=hie-rbe\n", + "print \"fbeta = %0.2f\"%((fbeta1)),\"hertz\"\n", + "print \"f = %0.2f\"%((fq1)),\"hertz\"\n", + "print \"cbe = %0.2e\"%((cbe)),\"farad\"\n", + "print \"rbe = %0.2f\"%((rbe)),\"ohm\"\n", + "print \"rbb = %0.2f\"%((rbb)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 338 example 3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "cde = 8.18e-12 farad\n", + "frequency = 1.50e+09 hertz\n" + ] + } + ], + "source": [ + "w=1*10**-4##centimetre\n", + "em1cur=2*10**-3##ampere\n", + "q=47\n", + "voltag=26*10**-3##volt\n", + "cde=(em1cur*w**2)/(voltag*2*q)\n", + "fq1=(em1cur)/(2*3.14*cde*voltag)\n", + "print \"cde = %0.2e\"%((cde)),\"farad\"\n", + "print \"frequency = %0.2e\"%((fq1)),\"hertz\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 339 example 6" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "re = 13.00 ohm\n", + "falpha = 5.99e+07 hertz\n", + "cde = 2.05e-10 farad\n", + "w = 2.66e-09 second\n" + ] + } + ], + "source": [ + "w=5*10**-4##centimetre\n", + "em1cur=2*10**-3##ampere\n", + "q=47\n", + "voltag=26*10**-3##volt\n", + "re=voltag/em1cur\n", + "fq1=2*q/(w**2*2*3.14)\n", + "cde=(em1cur*w**2)/(voltag*2*q)\n", + "w=(w**2)/(2*q)\n", + "print \"re = %0.2f\"%((re)),\"ohm\"\n", + "print \"falpha = %0.2e\"%((fq1)),\"hertz\"\n", + "print \"cde = %0.2e\"%((cde)),\"farad\"\n", + "print \"w = %0.2e\"%((w)),\"second\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber example 8" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "f = 1.50e+13 hertz\n", + "cde = 1.64e-15 farad\n" + ] + } + ], + "source": [ + "w=10**-6##centimetre\n", + "em1cur=4*10**-3##ampere\n", + "voltag=26*10**-3##volt\n", + "q=47\n", + "cde=(em1cur*w**2)/(voltag*2*q)\n", + "fq1=(em1cur)/(2*3.14*cde*voltag)\n", + "print \"f = %0.2e\"%((fq1)),\"hertz\"\n", + "print \"cde = %0.2e\"%((cde)),\"farad\"##correction required in the book." + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronic_Devices_and_Circuits_by_J._Paul/Ch7_1.ipynb b/Electronic_Devices_and_Circuits_by_J._Paul/Ch7_1.ipynb new file mode 100644 index 00000000..51287177 --- /dev/null +++ b/Electronic_Devices_and_Circuits_by_J._Paul/Ch7_1.ipynb @@ -0,0 +1,326 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 7 - Field Effect Transistor" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 370 example 1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "vgs = 0.70 volt\n", + "id = 1.51e-03 ampere\n", + "vds = 11.23 volt\n", + "voltage gain = -21.25\n" + ] + } + ], + "source": [ + "from sympy import symbols, solve\n", + "rd=12*10**3##ohm\n", + "r=1*10**6##ohm\n", + "resour=470##ohm\n", + "vdd=30##volt\n", + "idss=3*10**-3##ampere\n", + "vd=2.4##volt\n", + "v = symbols('v')\n", + "vgs=[0.24, 2.175, 1.41]\n", + "expr = vgs[0]*v**2+vgs[1]*v+vgs[2]\n", + "vgs=-solve(expr,v)[1]\n", + "vgs=0.7\n", + "id=idss*((1-(vgs/vd)))**2\n", + "vds=vdd-id*(rd+resour)\n", + "g=(2*idss/vd)*(1-((vgs/vd)))\n", + "volgai=-g*rd\n", + "print \"vgs = %0.2f\"%((vgs)),\"volt\"\n", + "print \"id = %0.2e\"%((id)),\"ampere\"\n", + "print \"vds = %0.2f\"%((vds)),\"volt\"\n", + "print \"voltage gain = %0.2f\"%((volgai))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 371 example 2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "r1 = 2000.00 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "idss=1*10**-3##ampere\n", + "pinvol=1##volt\n", + "q=10##volt\n", + "rd=56*10**3##ohm\n", + "vdd=24##volt\n", + "dracur=(vdd-q)/rd\n", + "vgs=0.5\n", + "r1=vgs/dracur\n", + "print \"r1 = %0.2f\"%((r1)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 372 example 4" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "id = 2.25e-03 ampere\n", + "vds = 10.50 volt\n" + ] + } + ], + "source": [ + "ids=4*10**-3##ampere\n", + "vp=4##volt\n", + "r=1.3*10**3#ohm\n", + "r1=200*10**3##ohm\n", + "vdd=60##volt\n", + "drares=18*10**3##ohm\n", + "soresi=4*10**3##ohm\n", + "rth=(r*r1)/(r+r1)\n", + "vth=r1*(1-vdd)/(1500*10**3)\n", + "id=-2.25*10**-3\n", + "vds=-vdd-(drares+soresi)*id\n", + "print \"id = %0.2e\"%(abs(id)),\"ampere\"\n", + "print \"vds = %0.2f\"%(abs(vds)),\"volt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 373 example 5" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "source resistance = 156.25 ohm\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "idss=10*10**-3##ampere\n", + "pinvol=-1##volt\n", + "ids=6.4*10**-3##ampere\n", + "vgs=-(sqrt(ids/idss)-(1))*pinvol\n", + "r=pinvol/ids\n", + "print \"source resistance = %0.2f\"%(abs(r)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 374 example 6" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "duration = 1.39e-07 second\n" + ] + } + ], + "source": [ + "from math import log\n", + "v1=2##volt\n", + "vgs=4##volt\n", + "voltag=5##volt\n", + "q=5*10**-3##ampere per volt square\n", + "id=q*(vgs-v1)\n", + "durati=10**-7*log(4)\n", + "\n", + "print \"duration = %0.2e\"%((durati)),\"second\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 7 example 7" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "max transconductance = 4.00e-04 mho\n" + ] + } + ], + "source": [ + "idss=1*10**-3##ampere\n", + "pinvol=-5##volt\n", + "tracon=(2*idss)/abs(pinvol)\n", + "print \"max transconductance = %0.2e\"%((tracon)),\"mho\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 376 example 8" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "vgs = 3.24 volt\n", + "rd = 13527.86 ohm\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "vdd=10##volt\n", + "beta1=10**-4##ampere per square volt\n", + "ids=0.5*10**-3##ampere\n", + "voltag=1##volt\n", + "vgs=(sqrt(ids/beta1)+(1))\n", + "rd=(vdd-vgs)/ids\n", + "\n", + "print \"vgs = %0.2f\"%((vgs)),\"volt\"\n", + "print \"rd = %0.2f\"%((rd)),\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 376 example 9" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "point 10.75 >2.00 volt\n", + "vds greater than 2volt the point in pinch\n" + ] + } + ], + "source": [ + "v1=2##volt\n", + "ids=4*10**-3##ampere\n", + "\n", + "rd=910##ohm\n", + "r1=3*10**3##ohm\n", + "r=12*1**6##ohm\n", + "r11=8.57*10**6##ohm\n", + "vdd=24##volt\n", + "vg=vdd*(r11/(r+(r11)))\n", + "id=3.39*10**-3\n", + "vgsq=vg-id*r1\n", + "vdsq=vdd-id*(rd+r1)\n", + "vdgq=vdsq-vgsq\n", + "print \"point %0.2f\"%(vdsq),\">%0.2f\"%(v1),\"volt\"\n", + "print \"vds greater than 2volt the point in pinch\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronic_Devices_and_Circuits_by_J._Paul/Ch8_1.ipynb b/Electronic_Devices_and_Circuits_by_J._Paul/Ch8_1.ipynb new file mode 100644 index 00000000..b498973e --- /dev/null +++ b/Electronic_Devices_and_Circuits_by_J._Paul/Ch8_1.ipynb @@ -0,0 +1,256 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8 - FET Amplifier" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 399 example 1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage gain = -12.00\n", + "voltage gain = -11.99\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "freque=5*10**3##hertz\n", + "#(1)\n", + "g=2*10**-3##ampere per volt\n", + "rd=10*10**3##ohm\n", + "r1=30*10**3##ohm\n", + "r12=r1*r1/(r1+r1)\n", + "volgai=-(g*r12*rd)/(r12+rd)\n", + "print \"voltage gain = %0.2f\"%((volgai))\n", + "#correction : r12 should be taken as 15*10**3ohm in book\n", + "#(2) capacitance included\n", + "c=0.025*10**-6##farad\n", + "frequ1=1/((2*3.14*(((rd*r1)/(rd+r1))+r1))*c)\n", + "volgai=(volgai/(sqrt((1+(frequ1/freque)**2))))\n", + "\n", + "print \"voltage gain = %0.2f\"%((volgai))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 400 example 2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage gain -18.00\n" + ] + } + ], + "source": [ + "rd=80*10**3##ohm\n", + "r1=8*10**3##ohm\n", + "rd12=5*10**3##ohm\n", + "rd1=rd*r1/(rd+r1)\n", + "u=30\n", + "volgai=-(u*rd1)/(rd1+rd12)\n", + "\n", + "print \"voltage gain %0.2f\"%((volgai))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 401 example 3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage gain = -1.18\n" + ] + } + ], + "source": [ + "r1=60*10**3##ohm\n", + "volgai=-17.7\n", + "rg=80*10**3##ohm\n", + "volgai=((volgai*rg)/(1-volgai))/((rg/(1-volgai))+r1)\n", + "print \"voltage gain = %0.2f\"%((volgai))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 405 example 6" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "r1 = 500.00 ohm\n", + "effective input resistance = 1.25 r3ohm\n", + "r2 = 1500.00 ohm\n", + "voltage gain = 0.98 av`\n" + ] + } + ], + "source": [ + "vds=14##volt\n", + "idq=3*10**-3##ampere\n", + "vdd=20##volt\n", + "g=2*10**-2\n", + "rd=50*10**3##ohm\n", + "vgs=-1.5##volt\n", + "w=(vdd-vds)/idq\n", + "r1=-vgs/idq\n", + "r2=w-r1\n", + "inpres=1/(1-(0.8*((r1)/(r1+r2))))\n", + "volgai=(r1+r2)/(r1+r2+(1/(g)))\n", + "print \"r1 = %0.2f\"%((r1)),\"ohm\"\n", + "print \"effective input resistance = %0.2f\"%((inpres)),\"r3ohm\"\n", + "print \"r2 = %0.2f\"%((r2)),\"ohm\"\n", + "\n", + "\n", + "print \"voltage gain = %0.2f\"%((volgai)),\"av`\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 405 example 7" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "output voltage = -186.17 volt\n" + ] + } + ], + "source": [ + "rg=40*10**3##ohm\n", + "voltag=(1-6*50)*3.3*10**3/(5.3*10**3)\n", + "\n", + "print \"output voltage = %0.2f\"%((voltag)),\"volt\"#\n", + "#correction required in the book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 406 example 9" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage gain = -25.00\n", + "frequency = 1.59e+07 hertz\n", + "output capacitance = 2.00e-12 farad\n", + "req = 5000.00 ohm\n" + ] + } + ], + "source": [ + "u=50\n", + "rd=10*10**3##ohm\n", + "cgs=5*10**-12##farad\n", + "cgd=2*10**-12##farad\n", + "cds=2*10**-12##farad\n", + "freque=3##decibel\n", + "g=u/rd\n", + "volgai=-u*rd/(rd+rd)\n", + "req=rd*rd/(rd+rd)\n", + "frequ1=1/(2*3.14*cgd*req)\n", + "print \"voltage gain = %0.2f\"%((volgai))\n", + "#correction required in book\n", + "print \"frequency = %0.2e\"%((frequ1)),\"hertz\"\n", + "capac1=cgd*(1+g)\n", + "print \"output capacitance = %0.2e\"%((capac1)),\"farad\"\n", + "print \"req = %0.2f\"%((req)),\"ohm\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronic_Devices_and_Circuits_by_J._Paul/Ch9_1.ipynb b/Electronic_Devices_and_Circuits_by_J._Paul/Ch9_1.ipynb new file mode 100644 index 00000000..522ff62b --- /dev/null +++ b/Electronic_Devices_and_Circuits_by_J._Paul/Ch9_1.ipynb @@ -0,0 +1,296 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9 - Multistage Amplifier" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 424 example 1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "frequency1 = 50982.45 hertz\n", + "frequency2 = 196145.92 hertz\n", + "frequency = 269258240.36 hertz\n", + "frequency = 76822.13 hertz\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "from __future__ import division\n", + "#(1) frequency\n", + "freque=100*10**3*sqrt(2**(1/3)-(1))\n", + "frequ2=100*10**3/sqrt(2**(1/3)-(1))\n", + "print \"frequency1 = %0.2f\"%((freque)),\"hertz\"\n", + "print \"frequency2 = %0.2f\"%((frequ2)),\"hertz\"\n", + "#(2)frequency\n", + "freq11=100*10**6##hertz\n", + "freq12=150*10**6##hertz\n", + "freq13=200*10**6##hertz\n", + "freq21=100*10**3##hertz\n", + "freq22=150*10**3##hertz\n", + "freq23=200*10**3##hertz\n", + "frequ1=sqrt(freq11**2+freq12**2+freq13**2)\n", + "print \"frequency = %0.2f\"%((frequ1)),\"hertz\"##correction in the book 269.25mega hertz\n", + "frequ1=1/sqrt((1/(freq21**2))+(1/(freq22**2))+(1/(freq23**2)))\n", + "print \"frequency = %0.2f\"%((frequ1)),\"hertz\"##correction in the book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 424 example 2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "coupling capacitance = 5.48e-06 /r`\n" + ] + } + ], + "source": [ + "freque=60##hertz\n", + "frequ1=freque*0.484\n", + "cb=1/(frequ1*2*3.14*10**3)\n", + "print \"coupling capacitance = %0.2e\"%((cb)),\"/r`\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 425 example 3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "cb = 6.28e-08 farad\n", + "cb = 8.92e-08 farad\n", + "gain of each stage = 1.06e-03\n" + ] + } + ], + "source": [ + "g=10*10**-3##ampere per volt\n", + "rd=5.5*10**3##ohm\n", + "rg=1*10**6##ohm\n", + "#(1) cb frequency 1decibel to 10hertz\n", + "ri=rg\n", + "r1=(rd*8*10**3)/(rd+8*10**3)\n", + "cb=10**-6/(3.14*5.07)\n", + "print \"cb = %0.2e\"%((cb)),\"farad\"\n", + "#(2) cb\n", + "cb=(cb*(5)/(3.52))\n", + "print \"cb = %0.2e\"%((cb)),\"farad\"\n", + "#(3) gain\n", + "a1=g**2*(3.26**2)\n", + "print \"gain of each stage = %0.2e\"%((a1))\n", + "#correction required in the book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 427 example 4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "upper frequency = 78895.46 hertz\n", + "lower frequency = 20408.16 hertz\n" + ] + } + ], + "source": [ + "freque=40*10**3##hertz\n", + "frequ1=freque/0.507\n", + "print \"upper frequency = %0.2f\"%((frequ1)),\"hertz\"\n", + "frequ1=freque/1.96\n", + "print \"lower frequency = %0.2f\"%((frequ1)),\"hertz\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 427 example 5" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "overal voltage gain = 62.01 decibel\n", + "lower frequency of each = 31.70 hertz\n", + "lower frequency overal = 62.13 hertz\n" + ] + } + ], + "source": [ + "from math import log10\n", + "g=2.6*10**-3##ampere per volt\n", + "rd=7.7*10**3##ohm\n", + "rd1=12*10**3##ohm\n", + "cb=0.005*10**-6##farad\n", + "#(1) voltage gain\n", + "volgai=g*((1/rd)+1/rd1+1/(1*10**3))\n", + "volgai=(20*(log10(10.8)))*3\n", + "print \"overal voltage gain = %0.2f\"%((volgai)),\"decibel\"##correction in the book\n", + "#(2) lower frequency\n", + "r=rd*rd1/(rd+rd1)\n", + "freque=1/((2*3.14)*(r+1*10**6)*cb)\n", + "print \"lower frequency of each = %0.2f\"%((freque)),\"hertz\"\n", + "#(3) overal lower frequency\n", + "freque=freque*1.96\n", + "print \"lower frequency overal = %0.2f\"%((freque)),\"hertz\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 429 example 6" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage gain = 2851.60\n", + "cb = 2.61e-06 farad\n", + "cb <= 1.64e-05 farad\n" + ] + } + ], + "source": [ + "hfe=50\n", + "hie=1.1*10**3##ohm\n", + "#(1) gain\n", + "r1=2*10**3##ohm\n", + "volgai=-hfe*r1/(hie)\n", + "r11=25*10**3*hie/(25*10**3+hie)\n", + "r11=r1*r11/(r1+r11)\n", + "volga1=-hfe*r11/hie\n", + "volgai=volgai*volga1\n", + "print \"voltage gain = %0.2f\"%((volgai))\n", + "freque=20##hertz\n", + "ri=25*10**3*hie/(25*10**3+hie)\n", + "cb=1/(2*3.14*(ri+r1)*(freque))\n", + "print \"cb = %0.2e\"%((cb)),\"farad\"\n", + "cb=1/(2*3.14*3.05*10**3*10/3.14)\n", + "print \"cb <= %0.2e\"%((cb)),\"farad\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## PageNumber 432 example 8" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "theta1 = 5.71\n", + "phase constant 10f1<=f<=0.1f11\n" + ] + } + ], + "source": [ + "from math import atan, degrees\n", + "theta1=degrees(atan(0.1))\n", + "print \"theta1 = %0.2f\"%((theta1))\n", + "print \"phase constant 10f1<=f<=0.1f11\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronic_Devices_and_Circuits_by_J._Paul/screenshots/4CollCurr_1.png b/Electronic_Devices_and_Circuits_by_J._Paul/screenshots/4CollCurr_1.png Binary files differnew file mode 100644 index 00000000..e8e0b539 --- /dev/null +++ b/Electronic_Devices_and_Circuits_by_J._Paul/screenshots/4CollCurr_1.png diff --git a/Electronic_Devices_and_Circuits_by_J._Paul/screenshots/4MaxNBasRes_1.png b/Electronic_Devices_and_Circuits_by_J._Paul/screenshots/4MaxNBasRes_1.png Binary files differnew file mode 100644 index 00000000..7ad2da7d --- /dev/null +++ b/Electronic_Devices_and_Circuits_by_J._Paul/screenshots/4MaxNBasRes_1.png diff --git a/Electronic_Devices_and_Circuits_by_J._Paul/screenshots/4saturationMode_1.png b/Electronic_Devices_and_Circuits_by_J._Paul/screenshots/4saturationMode_1.png Binary files differnew file mode 100644 index 00000000..d3931c3f --- /dev/null +++ b/Electronic_Devices_and_Circuits_by_J._Paul/screenshots/4saturationMode_1.png diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch1.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch1.ipynb deleted file mode 100755 index ae565265..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch1.ipynb +++ /dev/null @@ -1,1092 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:b4a0f0559d297d7743b1fffa63bdc97b5eb29843c8d1e8453183351781436bcb" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 1 : De Broglie Matter Waves" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.1 Page No3" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "M = 6e24 # Mass of earth in Kg\n", - "v = 3e4 # Orbital velocity of earth in m/s\n", - "h = 6.625e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(M*v) # calculation of de Broglie wavelength\n", - "\n", - "print \" de Broglie wavelength of earth is %e m.\"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of earth is 3.680556e-63 m.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.2 Page No4" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "M = 1. # Mass of object in Kg\n", - "v = 10. # velocity of object in m/s\n", - "h = 6.625e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(M*v) # calculation of de Broglie wavelength\n", - "\n", - "print \" de Broglie wavelength of body is %e m.\"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of body is 6.625000e-35 m.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.3 Page No7" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# Given that\n", - "m = 1e-30 # Mass of any object in Kg\n", - "v = 1e5 # velocity of object in m/s\n", - "h = 6.625e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(m*v) # calculation of de Broglie wavelength\n", - "\n", - "print \" de Broglie wavelength of body is %e m.\"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of body is 6.625000e-09 m.\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.4 Page No11" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# Given that\n", - "KE = 4.55e-25 # Kinetic energy of an electron in Joule\n", - "m = 9.1e-31 # Mass of any object in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "v = math.sqrt(2*KE/m) # Calculation of velocity of moving electron\n", - "p = m*v #Calculation of momentum of moving electron\n", - "lambda1 = h/p # calculation of de Broglie wavelength\n", - "print \" velocity of electron is %e m/s.\"%(v)\n", - "print \" momentum of electron is %e Kgm/s.\"%(p)\n", - "print \" de Broglie wavelength of electron is %e m.\"%(lambda1)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " velocity of electron is 1.000000e+03 m/s.\n", - " momentum of electron is 9.100000e-28 Kgm/s.\n", - " de Broglie wavelength of electron is 7.274725e-07 m.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.5 Page No14" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "c = 3.e8 # speed of light in m/s\n", - "v = c/20 # Speed of proton in m/s\n", - "m = 1.67e-27 # Mass of proton in Kg\n", - "h = 6.625e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(m*v) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of proton is %e m.\"%(lambda1)\n", - "# Answer in book is 6.645e-14m which is a calculation mistake\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of proton is 2.644711e-14 m.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.6 Page No14" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "e = 12.8 # Energy of neutron in MeV\n", - "c = 3e8 # speed of light in m/s\n", - "m = 1.675e-27 # Mass of neutron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "rest_e = m*c**2/(1e6*1.6e-19)# rest mass energy of neutron in MeV\n", - "if e/rest_e < 0.015 :\n", - " E = e;\n", - "else:\n", - " E = rest_e +e;\n", - "\n", - "lambda1 = h/(math.sqrt(2*m*e*1e6*1.6e-19)) # calculation of de Broglie wavelength\n", - "\n", - "print \" de Broglie wavelength of neutron is %e angstrom.\"%( lambda1*1e10)\n", - "# Answer in book is 8.04e-5 angstrom which is misprinted\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 7.992279e-05 angstrom.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.7 Page No15" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "e = 1.632e-19 # charge on electron in coulomb\n", - "V = 50 # Applied voltage in volts\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of neutron is %f angstrom.\"%( lambda1*1e10)\n", - "# Answer in book is 1.735 angstrom which is misprinted\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 1.717818 angstrom.\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.9 Page No18" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "e = 1.6e-19 # charge on electron in coulomb\n", - "V = 54 # Applied voltage in volts\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of neutron is %f angstrom.\"%( lambda1*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 1.671941 angstrom.\n" - ] - } - ], - "prompt_number": 17 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.10 Page No21" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "E = 10 # Energy of electron in KeV\n", - "m_e = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "v = math.sqrt(2*E*1.6e-16/m_e) # Calculation of velocity of moving electron\n", - "p = m_e*v #Calculation of momentum of moving electron\n", - "lambda1 = h/p # calculation of de Broglie wavelength\n", - "print \" velocity of electron is %0.2e m/s.\"%(v)\n", - "print \" momentum of electron is %.3e Kgm/s.\"%(p)\n", - "print \" de Broglie wavelength of electron is %.2f angstrom.\"%( lambda1*1e10)\n", - " # Answers in book are v = 5.93e6 m/s, p = 5.397e-24 kgm/s, lambda = 1.23 angstrom\n", - "# Which is due to wrong calculation\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " velocity of electron is 5.93e+07 m/s.\n", - " momentum of electron is 5.396e-23 Kgm/s.\n", - " de Broglie wavelength of electron is 0.12 angstrom.\n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.11 Page No22" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 1. # de Broglie wavelength of neutron in angstrom\n", - "m = 1.67e-27 # Mass of electron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "v = h/(m*lambda1*1e-10) # Calculation of velocity of moving neutron\n", - "E = 1./2*m*v**2 # Calculation of kinetic energy of moving neutron\n", - "print \" velocity of neutron is %e m/s.\"%(v)\n", - "print \" Kinetic energy of neutron is %f eV.\"%(E/1.6e-19)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " velocity of neutron is 3.964072e+03 m/s.\n", - " Kinetic energy of neutron is 0.082007 eV.\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.12 Page No26" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#Given that\n", - "E = 2 # Energy of accelerated electron in KeV\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/math.sqrt(2*m*E*1e3*1.6e-19) # Calculation of velocity of moving electron\n", - "print \" Wavelength of electron is %e m.\"%(lambda1)\n", - "# Answer in book is 2.74e-12m\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of electron is 2.743136e-11 m.\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.13 Page No28" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#Given that\n", - "v = 2e8 # speed of moving proton in m/s\n", - "c = 3e8 # speed of light in m/s\n", - "m = 1.67e-27 # Mass of proton in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(m*v/math.sqrt(1-(v/c)**2)) # Calculation of velocity of moving electron\n", - "print \" Wavelength of electron is %e angstrom.\"%( lambda1*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of electron is 1.477322e-05 angstrom.\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.14 Page No30" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1.# wavelength in m/s\n", - "m_e = 9.1e-31 # Mass of electron in Kg\n", - "m_p = 1.67e-27 # Mass of proton in kg\n", - "c = 3e8 # speed of light in m/s\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "p_p = h/(lambda1*1e-10) # Momentum of photon\n", - "p_e = h/(lambda1*1e-10) # Momentum of electron\n", - "E_e = p_e**2/(2*m_e) +m_e*c**2 # Total energy of electron\n", - "E_p = h*c/(lambda1*1e-10) # Total energy of photon\n", - "K_e = p_e**2/(2*m_e) # Kinetic energy of electron \n", - "K_p = h*c/(lambda1*1e-10)# Kinetic energy of photon\n", - "r_K = K_e/K_p # Ratio of kinetic energies\n", - "print \" Momentum of photon is %e Kgm/s while Momentum of electron is %e Kgm/s which are equal.\"%(p_p,p_e)\n", - "print \" Total Energy of photon is %f KeV while Total Energy of electron is %f MeV \"%(E_p/1.6e-19*1e3,(E_e/1.6e-19*1e6))\n", - "print \" Ratio of kinetic energies is %e \"%(r_K)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Momentum of photon is 6.630000e-24 Kgm/s while Momentum of electron is 6.630000e-24 Kgm/s which are equal.\n", - " Total Energy of photon is 12431250.000000 KeV while Total Energy of electron is 512025950892.857117 MeV \n", - " Ratio of kinetic energies is 1.214286e-02 \n" - ] - } - ], - "prompt_number": 32 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.15 Page No31" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "e = 25 # Energy of neutron in eV\n", - "c = 3e8 # speed of light in m/s\n", - "m = 1.67e-27 # Mass of neutron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "rest_e = m*c**2/(1e6*1.6e-19)# rest mass energy of neutron in MeV\n", - "if e/rest_e < 0.015:\n", - " E = e;\n", - "else:\n", - " E = rest_e +e;\n", - "\n", - "lambda1 = h/(math.sqrt(2*m*e*1.6e-19)) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of neutron is %f angstrom.\"%( lambda1*1e10)\n", - "# Answer in book is 8.04e-5 angstrom \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 0.057274 angstrom.\n" - ] - } - ], - "prompt_number": 33 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.16 Page No36" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "e = 2*1.6e-19 # charge on alpha particle in coulomb\n", - "V = 200 # Applied voltage in volts\n", - "m = 4*1.67e-27 # Mass of alpha particle in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of neutron is %f angstrom.\"%( lambda1*1e10)\n", - "# while answer in book is 0.00715 angstrom\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 0.007170 angstrom.\n" - ] - } - ], - "prompt_number": 34 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.17 Page No41" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "M = 20 # Mass of ball in Kg\n", - "V = 5 # velocity of of ball in m/s\n", - "m = 9.1e-31 #Mass of electron in Kg\n", - "v = 1e6 # velocity of of electron in m/s\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "lambda_b = h/(M*V) # calculation of de Broglie wavelength for ball\n", - "lambda_e = h/(m*v) # calculation of de Broglie wavelength electron\n", - "print \" de Broglie wavelength of ball is %e angstrom.\"%(lambda_b*1e10)\n", - "print \" de Broglie wavelength of electron is %f angstrom.\"%(lambda_e*1e10)\n", - "# answer in book is 6.62e-22 angstrom for ball\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of ball is 6.620000e-26 angstrom.\n", - " de Broglie wavelength of electron is 7.274725 angstrom.\n" - ] - } - ], - "prompt_number": 35 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.18 Page No44" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "E = 1 # Energy of neutron in eV\n", - "m = 1.67e-27 # Mass of neutron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "lambda1 = h/math.sqrt(2*m*E*1.6e-19) # Calculation of velocity of moving electron\n", - "print \" Wavelength of electron is %f angstrom.\"%(lambda1*1e10)\n", - "# Answer in book is 6.62e-22 angstrom\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of electron is 0.286368 angstrom.\n" - ] - } - ], - "prompt_number": 36 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.19 Page No44" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 0.5# wavelength of electron in angstrom\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "q = 1.6e-19 # charge on electron in coulomb\n", - "\n", - "V = h**2/(2*m*q*(lambda1*1e-10)**2) # Calculation of velocity of moving electron\n", - "print \" Applied voltage on electron is %f V.\"%(V)\n", - "# Answer in book is 601.6 Volt\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Applied voltage on electron is 601.983516 V.\n" - ] - } - ], - "prompt_number": 37 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.21 Page No46" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "k = 8.6e-5 # Boltzmann consmath.tant\n", - "t = 37 # Temperature in degree Celsius\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "m = 1.67e-27 # Mass of neutron\n", - "\n", - "lambda1 = h/math.sqrt(3*m*(k*1.6e-19)*(t+273))# Calculation of wavelength\n", - "print \" Wavelength of neutron at %d degree Celsius is %f angstrom.\"%(t,lambda1*1e10)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of neutron at 37 degree Celsius is 1.432020 angstrom.\n" - ] - } - ], - "prompt_number": 38 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.22 Page No49" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "k = 8.6e-5 # Boltzmann consmath.tant\n", - "t = 27 # Temperature in degree Celsius\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "m = 6.7e-27 # Mass of helium atom\n", - "\n", - "lambda1 = h/math.sqrt(3*m*(k*1.6e-19)*(t+273))# Calculation of wavelength\n", - "print \" Wavelength of helium at %d degree Celsius is %f angstrom.\"%(t,lambda1*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of helium at 27 degree Celsius is 0.726758 angstrom.\n" - ] - } - ], - "prompt_number": 39 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.23 Page No50" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "E = 200. # energy of electrons in eV\n", - "x = 20. # dismath.tance of screen in cm\n", - "D = 2. # diameter of ring in cm\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # Mass of electron in kg\n", - "\n", - "lambda1 = h/math.sqrt(2*m*E*1.6e-19) # Calculation of wavelength\n", - "theta = (math.atan(D/(2*x)))\n", - "d = lambda1/(2*math.sin(theta))# calculation of interatomic spacing of crystal\n", - "print \" Interatomic spacing of crystal is %f angstrom.\"%(d*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Interatomic spacing of crystal is 8.685393 angstrom.\n" - ] - } - ], - "prompt_number": 43 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.24 Page No52" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "r = 0.5 # Bohr radius of hydrogen in angstrom\n", - "m = 9.1e-31 # Mass of neutron in Kg\n", - "h = 6.6e-34 # Plank consmath.tant\n", - "v = h/(2*math.pi*r*1e-10*m) # velocity of electron in ground state\n", - "print \" Velocity of electron in ground state is %e m/s.\"%(v)\n", - "# Answer in book is 2.31e6 m/s\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Velocity of electron in ground state is 2.308621e+06 m/s.\n" - ] - } - ], - "prompt_number": 45 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.25 Page No55" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 5890 # wavelength of yellow radiation in angstrom\n", - "m = 9.1e-31 # Mass of neutron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "v = h/(lambda1*1e-10*m) # velocity of electron in ground state\n", - "print \" Velocity of electron in ground state is %.2e m/s.\"%(v)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Velocity of electron in ground state is 1.24e+03 m/s.\n" - ] - } - ], - "prompt_number": 49 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.26 Page No56" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 2 # wavelength of neutron in angstrom\n", - "m = 1.67e-27 # Mass of neutron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "v = h/(lambda1*1e-10*m) # velocity of neutron\n", - "k = 0.5*m*v**2 # Kinetic energy of neutron\n", - "print \" Velocity of neutron is %e m/s.\"%(v)\n", - "print \" Kinetic energy of neutron is %.3f eV.\"%(k/1.6e-19)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Velocity of neutron is 1.985030e+03 m/s.\n", - " Kinetic energy of neutron is 0.021 eV.\n" - ] - } - ], - "prompt_number": 53 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.29 Page No61" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "v1 = 50 # Previous applied voltage\n", - "v2 = 65 # final applied voltage\n", - "k = 12.28 \n", - "d = 0.91 # Spacing in a crystal in angstrom\n", - "print \"Example 1.29\"\n", - "\n", - "lambda1 = k/math.sqrt(v1)\n", - "theta= math.asin(lambda1/(2*d))# Angel for initial applied voltage\n", - "lambda1 = k/math.sqrt(v2)# wavelength for final applied voltage\n", - "theta1 = math.asin(lambda1/(2*d))# Angel for final applied voltage\n", - "print \" For first order, (sintheta) is %f For second order sintheta must be %f which is not possible \\\n", - "for any value of angle. So no maxima occur for higher orders \"%(math.sin(theta),2*math.sin(theta))\n", - "print \" Angle of diffraction for first order of beam is %f degree at %d Volts\"%(theta1*180/math.pi,v2)\n", - "# Answer in book is 57.14 degree\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 1.29\n", - " For first order, (sintheta) is 0.954206 For second order sintheta must be 1.908411 which is not possible for any value of angle. So no maxima occur for higher orders \n", - " Angle of diffraction for first order of beam is 56.813542 degree at 65 Volts\n" - ] - } - ], - "prompt_number": 62 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.30 Page No62" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "lambda1 = 680 # Wavelength in m\n", - "g = 9.8 #Acceleration due to gravity\n", - "print \"Example 1.30\"\n", - "v_g = 1/2*math.sqrt(g*lambda1/(2*math.pi)) # Calculation of group velocity\n", - "print \" Group velocity of seawater waves is %f m/s.\"%(v_g)\n", - "# Answer in book is 16.29 m/s\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 1.30\n", - " Group velocity of seawater waves is 0.000000 m/s.\n" - ] - } - ], - "prompt_number": 63 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.32 Page No64" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "lambda1 = 2e-13 # de Broglie wavelength of an electron in m\n", - "c = 3e8 # Speed of light in m/s\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 1.32\"\n", - "E = h*c/(lambda1*1.6e-19) \n", - "E_rest = m*c**2/(1.6e-19) # Calculation of rest mass energy\n", - "E_total = math.sqrt(E**2+E_rest**2) # Total energy in eV\n", - "v_g = c*math.sqrt(1-(E_rest/E_total)**2) # Group velocity\n", - "v_p = c**2/v_g # Phase velocity\n", - "print \" Group velocity of de Broglie waves is %fc and phase velocity is %fc .\"%(v_g/c,v_p/c)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 1.32\n", - " Group velocity of de Broglie waves is 0.996626c and phase velocity is 1.003385c .\n" - ] - } - ], - "prompt_number": 65 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.33 Page No68" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 2e-12 # de Broglie wavelength of an electron in m\n", - "c = 3e8 # Speed of light in m/s\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 1.33\"\n", - "E = h*c/(lambda1*1.6e-19) # Energy due to momentum\n", - "E_rest = m*c**2/(1.6e-19) # Calculation of rest mass energy\n", - "E_total = math.sqrt(E**2+E_rest**2) # Total energy in eV\n", - "KE = E_total - E_rest # Kinetic energy\n", - "v_g = c*math.sqrt(1-(E_rest/E_total)**2) # Group velocity\n", - "v_p = c**2/v_g # Phase velocity\n", - "\n", - "print \" Kinetic energy of electron is %f KeV.\"%(KE/1000)\n", - "print \" Group velocity of de Broglie waves is %fc m/s and phase velocity is %fc m/s.\"%(v_g/c,v_p/c)\n", - "# Answer in book is v_g = 0.6035c & v_p = 1.657c\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 1.33\n", - " Kinetic energy of electron is 293.330537 KeV.\n", - " Group velocity of de Broglie waves is 0.771930c m/s and phase velocity is 1.295454c m/s.\n" - ] - } - ], - "prompt_number": 67 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch1_1.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch1_1.ipynb deleted file mode 100755 index ae565265..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch1_1.ipynb +++ /dev/null @@ -1,1092 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:b4a0f0559d297d7743b1fffa63bdc97b5eb29843c8d1e8453183351781436bcb" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 1 : De Broglie Matter Waves" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.1 Page No3" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "M = 6e24 # Mass of earth in Kg\n", - "v = 3e4 # Orbital velocity of earth in m/s\n", - "h = 6.625e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(M*v) # calculation of de Broglie wavelength\n", - "\n", - "print \" de Broglie wavelength of earth is %e m.\"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of earth is 3.680556e-63 m.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.2 Page No4" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "M = 1. # Mass of object in Kg\n", - "v = 10. # velocity of object in m/s\n", - "h = 6.625e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(M*v) # calculation of de Broglie wavelength\n", - "\n", - "print \" de Broglie wavelength of body is %e m.\"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of body is 6.625000e-35 m.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.3 Page No7" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# Given that\n", - "m = 1e-30 # Mass of any object in Kg\n", - "v = 1e5 # velocity of object in m/s\n", - "h = 6.625e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(m*v) # calculation of de Broglie wavelength\n", - "\n", - "print \" de Broglie wavelength of body is %e m.\"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of body is 6.625000e-09 m.\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.4 Page No11" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# Given that\n", - "KE = 4.55e-25 # Kinetic energy of an electron in Joule\n", - "m = 9.1e-31 # Mass of any object in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "v = math.sqrt(2*KE/m) # Calculation of velocity of moving electron\n", - "p = m*v #Calculation of momentum of moving electron\n", - "lambda1 = h/p # calculation of de Broglie wavelength\n", - "print \" velocity of electron is %e m/s.\"%(v)\n", - "print \" momentum of electron is %e Kgm/s.\"%(p)\n", - "print \" de Broglie wavelength of electron is %e m.\"%(lambda1)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " velocity of electron is 1.000000e+03 m/s.\n", - " momentum of electron is 9.100000e-28 Kgm/s.\n", - " de Broglie wavelength of electron is 7.274725e-07 m.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.5 Page No14" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "c = 3.e8 # speed of light in m/s\n", - "v = c/20 # Speed of proton in m/s\n", - "m = 1.67e-27 # Mass of proton in Kg\n", - "h = 6.625e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(m*v) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of proton is %e m.\"%(lambda1)\n", - "# Answer in book is 6.645e-14m which is a calculation mistake\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of proton is 2.644711e-14 m.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.6 Page No14" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "e = 12.8 # Energy of neutron in MeV\n", - "c = 3e8 # speed of light in m/s\n", - "m = 1.675e-27 # Mass of neutron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "rest_e = m*c**2/(1e6*1.6e-19)# rest mass energy of neutron in MeV\n", - "if e/rest_e < 0.015 :\n", - " E = e;\n", - "else:\n", - " E = rest_e +e;\n", - "\n", - "lambda1 = h/(math.sqrt(2*m*e*1e6*1.6e-19)) # calculation of de Broglie wavelength\n", - "\n", - "print \" de Broglie wavelength of neutron is %e angstrom.\"%( lambda1*1e10)\n", - "# Answer in book is 8.04e-5 angstrom which is misprinted\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 7.992279e-05 angstrom.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.7 Page No15" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "e = 1.632e-19 # charge on electron in coulomb\n", - "V = 50 # Applied voltage in volts\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of neutron is %f angstrom.\"%( lambda1*1e10)\n", - "# Answer in book is 1.735 angstrom which is misprinted\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 1.717818 angstrom.\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.9 Page No18" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "e = 1.6e-19 # charge on electron in coulomb\n", - "V = 54 # Applied voltage in volts\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of neutron is %f angstrom.\"%( lambda1*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 1.671941 angstrom.\n" - ] - } - ], - "prompt_number": 17 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.10 Page No21" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "E = 10 # Energy of electron in KeV\n", - "m_e = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "v = math.sqrt(2*E*1.6e-16/m_e) # Calculation of velocity of moving electron\n", - "p = m_e*v #Calculation of momentum of moving electron\n", - "lambda1 = h/p # calculation of de Broglie wavelength\n", - "print \" velocity of electron is %0.2e m/s.\"%(v)\n", - "print \" momentum of electron is %.3e Kgm/s.\"%(p)\n", - "print \" de Broglie wavelength of electron is %.2f angstrom.\"%( lambda1*1e10)\n", - " # Answers in book are v = 5.93e6 m/s, p = 5.397e-24 kgm/s, lambda = 1.23 angstrom\n", - "# Which is due to wrong calculation\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " velocity of electron is 5.93e+07 m/s.\n", - " momentum of electron is 5.396e-23 Kgm/s.\n", - " de Broglie wavelength of electron is 0.12 angstrom.\n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.11 Page No22" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 1. # de Broglie wavelength of neutron in angstrom\n", - "m = 1.67e-27 # Mass of electron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "v = h/(m*lambda1*1e-10) # Calculation of velocity of moving neutron\n", - "E = 1./2*m*v**2 # Calculation of kinetic energy of moving neutron\n", - "print \" velocity of neutron is %e m/s.\"%(v)\n", - "print \" Kinetic energy of neutron is %f eV.\"%(E/1.6e-19)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " velocity of neutron is 3.964072e+03 m/s.\n", - " Kinetic energy of neutron is 0.082007 eV.\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.12 Page No26" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#Given that\n", - "E = 2 # Energy of accelerated electron in KeV\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/math.sqrt(2*m*E*1e3*1.6e-19) # Calculation of velocity of moving electron\n", - "print \" Wavelength of electron is %e m.\"%(lambda1)\n", - "# Answer in book is 2.74e-12m\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of electron is 2.743136e-11 m.\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.13 Page No28" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#Given that\n", - "v = 2e8 # speed of moving proton in m/s\n", - "c = 3e8 # speed of light in m/s\n", - "m = 1.67e-27 # Mass of proton in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(m*v/math.sqrt(1-(v/c)**2)) # Calculation of velocity of moving electron\n", - "print \" Wavelength of electron is %e angstrom.\"%( lambda1*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of electron is 1.477322e-05 angstrom.\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.14 Page No30" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1.# wavelength in m/s\n", - "m_e = 9.1e-31 # Mass of electron in Kg\n", - "m_p = 1.67e-27 # Mass of proton in kg\n", - "c = 3e8 # speed of light in m/s\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "p_p = h/(lambda1*1e-10) # Momentum of photon\n", - "p_e = h/(lambda1*1e-10) # Momentum of electron\n", - "E_e = p_e**2/(2*m_e) +m_e*c**2 # Total energy of electron\n", - "E_p = h*c/(lambda1*1e-10) # Total energy of photon\n", - "K_e = p_e**2/(2*m_e) # Kinetic energy of electron \n", - "K_p = h*c/(lambda1*1e-10)# Kinetic energy of photon\n", - "r_K = K_e/K_p # Ratio of kinetic energies\n", - "print \" Momentum of photon is %e Kgm/s while Momentum of electron is %e Kgm/s which are equal.\"%(p_p,p_e)\n", - "print \" Total Energy of photon is %f KeV while Total Energy of electron is %f MeV \"%(E_p/1.6e-19*1e3,(E_e/1.6e-19*1e6))\n", - "print \" Ratio of kinetic energies is %e \"%(r_K)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Momentum of photon is 6.630000e-24 Kgm/s while Momentum of electron is 6.630000e-24 Kgm/s which are equal.\n", - " Total Energy of photon is 12431250.000000 KeV while Total Energy of electron is 512025950892.857117 MeV \n", - " Ratio of kinetic energies is 1.214286e-02 \n" - ] - } - ], - "prompt_number": 32 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.15 Page No31" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "e = 25 # Energy of neutron in eV\n", - "c = 3e8 # speed of light in m/s\n", - "m = 1.67e-27 # Mass of neutron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "rest_e = m*c**2/(1e6*1.6e-19)# rest mass energy of neutron in MeV\n", - "if e/rest_e < 0.015:\n", - " E = e;\n", - "else:\n", - " E = rest_e +e;\n", - "\n", - "lambda1 = h/(math.sqrt(2*m*e*1.6e-19)) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of neutron is %f angstrom.\"%( lambda1*1e10)\n", - "# Answer in book is 8.04e-5 angstrom \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 0.057274 angstrom.\n" - ] - } - ], - "prompt_number": 33 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.16 Page No36" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "e = 2*1.6e-19 # charge on alpha particle in coulomb\n", - "V = 200 # Applied voltage in volts\n", - "m = 4*1.67e-27 # Mass of alpha particle in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of neutron is %f angstrom.\"%( lambda1*1e10)\n", - "# while answer in book is 0.00715 angstrom\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 0.007170 angstrom.\n" - ] - } - ], - "prompt_number": 34 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.17 Page No41" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "M = 20 # Mass of ball in Kg\n", - "V = 5 # velocity of of ball in m/s\n", - "m = 9.1e-31 #Mass of electron in Kg\n", - "v = 1e6 # velocity of of electron in m/s\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "lambda_b = h/(M*V) # calculation of de Broglie wavelength for ball\n", - "lambda_e = h/(m*v) # calculation of de Broglie wavelength electron\n", - "print \" de Broglie wavelength of ball is %e angstrom.\"%(lambda_b*1e10)\n", - "print \" de Broglie wavelength of electron is %f angstrom.\"%(lambda_e*1e10)\n", - "# answer in book is 6.62e-22 angstrom for ball\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of ball is 6.620000e-26 angstrom.\n", - " de Broglie wavelength of electron is 7.274725 angstrom.\n" - ] - } - ], - "prompt_number": 35 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.18 Page No44" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "E = 1 # Energy of neutron in eV\n", - "m = 1.67e-27 # Mass of neutron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "lambda1 = h/math.sqrt(2*m*E*1.6e-19) # Calculation of velocity of moving electron\n", - "print \" Wavelength of electron is %f angstrom.\"%(lambda1*1e10)\n", - "# Answer in book is 6.62e-22 angstrom\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of electron is 0.286368 angstrom.\n" - ] - } - ], - "prompt_number": 36 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.19 Page No44" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 0.5# wavelength of electron in angstrom\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "q = 1.6e-19 # charge on electron in coulomb\n", - "\n", - "V = h**2/(2*m*q*(lambda1*1e-10)**2) # Calculation of velocity of moving electron\n", - "print \" Applied voltage on electron is %f V.\"%(V)\n", - "# Answer in book is 601.6 Volt\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Applied voltage on electron is 601.983516 V.\n" - ] - } - ], - "prompt_number": 37 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.21 Page No46" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "k = 8.6e-5 # Boltzmann consmath.tant\n", - "t = 37 # Temperature in degree Celsius\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "m = 1.67e-27 # Mass of neutron\n", - "\n", - "lambda1 = h/math.sqrt(3*m*(k*1.6e-19)*(t+273))# Calculation of wavelength\n", - "print \" Wavelength of neutron at %d degree Celsius is %f angstrom.\"%(t,lambda1*1e10)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of neutron at 37 degree Celsius is 1.432020 angstrom.\n" - ] - } - ], - "prompt_number": 38 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.22 Page No49" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "k = 8.6e-5 # Boltzmann consmath.tant\n", - "t = 27 # Temperature in degree Celsius\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "m = 6.7e-27 # Mass of helium atom\n", - "\n", - "lambda1 = h/math.sqrt(3*m*(k*1.6e-19)*(t+273))# Calculation of wavelength\n", - "print \" Wavelength of helium at %d degree Celsius is %f angstrom.\"%(t,lambda1*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of helium at 27 degree Celsius is 0.726758 angstrom.\n" - ] - } - ], - "prompt_number": 39 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.23 Page No50" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "E = 200. # energy of electrons in eV\n", - "x = 20. # dismath.tance of screen in cm\n", - "D = 2. # diameter of ring in cm\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # Mass of electron in kg\n", - "\n", - "lambda1 = h/math.sqrt(2*m*E*1.6e-19) # Calculation of wavelength\n", - "theta = (math.atan(D/(2*x)))\n", - "d = lambda1/(2*math.sin(theta))# calculation of interatomic spacing of crystal\n", - "print \" Interatomic spacing of crystal is %f angstrom.\"%(d*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Interatomic spacing of crystal is 8.685393 angstrom.\n" - ] - } - ], - "prompt_number": 43 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.24 Page No52" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "r = 0.5 # Bohr radius of hydrogen in angstrom\n", - "m = 9.1e-31 # Mass of neutron in Kg\n", - "h = 6.6e-34 # Plank consmath.tant\n", - "v = h/(2*math.pi*r*1e-10*m) # velocity of electron in ground state\n", - "print \" Velocity of electron in ground state is %e m/s.\"%(v)\n", - "# Answer in book is 2.31e6 m/s\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Velocity of electron in ground state is 2.308621e+06 m/s.\n" - ] - } - ], - "prompt_number": 45 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.25 Page No55" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 5890 # wavelength of yellow radiation in angstrom\n", - "m = 9.1e-31 # Mass of neutron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "v = h/(lambda1*1e-10*m) # velocity of electron in ground state\n", - "print \" Velocity of electron in ground state is %.2e m/s.\"%(v)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Velocity of electron in ground state is 1.24e+03 m/s.\n" - ] - } - ], - "prompt_number": 49 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.26 Page No56" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 2 # wavelength of neutron in angstrom\n", - "m = 1.67e-27 # Mass of neutron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "v = h/(lambda1*1e-10*m) # velocity of neutron\n", - "k = 0.5*m*v**2 # Kinetic energy of neutron\n", - "print \" Velocity of neutron is %e m/s.\"%(v)\n", - "print \" Kinetic energy of neutron is %.3f eV.\"%(k/1.6e-19)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Velocity of neutron is 1.985030e+03 m/s.\n", - " Kinetic energy of neutron is 0.021 eV.\n" - ] - } - ], - "prompt_number": 53 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.29 Page No61" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "v1 = 50 # Previous applied voltage\n", - "v2 = 65 # final applied voltage\n", - "k = 12.28 \n", - "d = 0.91 # Spacing in a crystal in angstrom\n", - "print \"Example 1.29\"\n", - "\n", - "lambda1 = k/math.sqrt(v1)\n", - "theta= math.asin(lambda1/(2*d))# Angel for initial applied voltage\n", - "lambda1 = k/math.sqrt(v2)# wavelength for final applied voltage\n", - "theta1 = math.asin(lambda1/(2*d))# Angel for final applied voltage\n", - "print \" For first order, (sintheta) is %f For second order sintheta must be %f which is not possible \\\n", - "for any value of angle. So no maxima occur for higher orders \"%(math.sin(theta),2*math.sin(theta))\n", - "print \" Angle of diffraction for first order of beam is %f degree at %d Volts\"%(theta1*180/math.pi,v2)\n", - "# Answer in book is 57.14 degree\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 1.29\n", - " For first order, (sintheta) is 0.954206 For second order sintheta must be 1.908411 which is not possible for any value of angle. So no maxima occur for higher orders \n", - " Angle of diffraction for first order of beam is 56.813542 degree at 65 Volts\n" - ] - } - ], - "prompt_number": 62 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.30 Page No62" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "lambda1 = 680 # Wavelength in m\n", - "g = 9.8 #Acceleration due to gravity\n", - "print \"Example 1.30\"\n", - "v_g = 1/2*math.sqrt(g*lambda1/(2*math.pi)) # Calculation of group velocity\n", - "print \" Group velocity of seawater waves is %f m/s.\"%(v_g)\n", - "# Answer in book is 16.29 m/s\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 1.30\n", - " Group velocity of seawater waves is 0.000000 m/s.\n" - ] - } - ], - "prompt_number": 63 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.32 Page No64" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "lambda1 = 2e-13 # de Broglie wavelength of an electron in m\n", - "c = 3e8 # Speed of light in m/s\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 1.32\"\n", - "E = h*c/(lambda1*1.6e-19) \n", - "E_rest = m*c**2/(1.6e-19) # Calculation of rest mass energy\n", - "E_total = math.sqrt(E**2+E_rest**2) # Total energy in eV\n", - "v_g = c*math.sqrt(1-(E_rest/E_total)**2) # Group velocity\n", - "v_p = c**2/v_g # Phase velocity\n", - "print \" Group velocity of de Broglie waves is %fc and phase velocity is %fc .\"%(v_g/c,v_p/c)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 1.32\n", - " Group velocity of de Broglie waves is 0.996626c and phase velocity is 1.003385c .\n" - ] - } - ], - "prompt_number": 65 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.33 Page No68" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 2e-12 # de Broglie wavelength of an electron in m\n", - "c = 3e8 # Speed of light in m/s\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 1.33\"\n", - "E = h*c/(lambda1*1.6e-19) # Energy due to momentum\n", - "E_rest = m*c**2/(1.6e-19) # Calculation of rest mass energy\n", - "E_total = math.sqrt(E**2+E_rest**2) # Total energy in eV\n", - "KE = E_total - E_rest # Kinetic energy\n", - "v_g = c*math.sqrt(1-(E_rest/E_total)**2) # Group velocity\n", - "v_p = c**2/v_g # Phase velocity\n", - "\n", - "print \" Kinetic energy of electron is %f KeV.\"%(KE/1000)\n", - "print \" Group velocity of de Broglie waves is %fc m/s and phase velocity is %fc m/s.\"%(v_g/c,v_p/c)\n", - "# Answer in book is v_g = 0.6035c & v_p = 1.657c\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 1.33\n", - " Kinetic energy of electron is 293.330537 KeV.\n", - " Group velocity of de Broglie waves is 0.771930c m/s and phase velocity is 1.295454c m/s.\n" - ] - } - ], - "prompt_number": 67 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch1_2.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch1_2.ipynb deleted file mode 100644 index ae565265..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch1_2.ipynb +++ /dev/null @@ -1,1092 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:b4a0f0559d297d7743b1fffa63bdc97b5eb29843c8d1e8453183351781436bcb" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 1 : De Broglie Matter Waves" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.1 Page No3" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "M = 6e24 # Mass of earth in Kg\n", - "v = 3e4 # Orbital velocity of earth in m/s\n", - "h = 6.625e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(M*v) # calculation of de Broglie wavelength\n", - "\n", - "print \" de Broglie wavelength of earth is %e m.\"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of earth is 3.680556e-63 m.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.2 Page No4" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "M = 1. # Mass of object in Kg\n", - "v = 10. # velocity of object in m/s\n", - "h = 6.625e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(M*v) # calculation of de Broglie wavelength\n", - "\n", - "print \" de Broglie wavelength of body is %e m.\"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of body is 6.625000e-35 m.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.3 Page No7" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# Given that\n", - "m = 1e-30 # Mass of any object in Kg\n", - "v = 1e5 # velocity of object in m/s\n", - "h = 6.625e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(m*v) # calculation of de Broglie wavelength\n", - "\n", - "print \" de Broglie wavelength of body is %e m.\"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of body is 6.625000e-09 m.\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.4 Page No11" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# Given that\n", - "KE = 4.55e-25 # Kinetic energy of an electron in Joule\n", - "m = 9.1e-31 # Mass of any object in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "v = math.sqrt(2*KE/m) # Calculation of velocity of moving electron\n", - "p = m*v #Calculation of momentum of moving electron\n", - "lambda1 = h/p # calculation of de Broglie wavelength\n", - "print \" velocity of electron is %e m/s.\"%(v)\n", - "print \" momentum of electron is %e Kgm/s.\"%(p)\n", - "print \" de Broglie wavelength of electron is %e m.\"%(lambda1)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " velocity of electron is 1.000000e+03 m/s.\n", - " momentum of electron is 9.100000e-28 Kgm/s.\n", - " de Broglie wavelength of electron is 7.274725e-07 m.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.5 Page No14" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "c = 3.e8 # speed of light in m/s\n", - "v = c/20 # Speed of proton in m/s\n", - "m = 1.67e-27 # Mass of proton in Kg\n", - "h = 6.625e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(m*v) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of proton is %e m.\"%(lambda1)\n", - "# Answer in book is 6.645e-14m which is a calculation mistake\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of proton is 2.644711e-14 m.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.6 Page No14" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "e = 12.8 # Energy of neutron in MeV\n", - "c = 3e8 # speed of light in m/s\n", - "m = 1.675e-27 # Mass of neutron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "rest_e = m*c**2/(1e6*1.6e-19)# rest mass energy of neutron in MeV\n", - "if e/rest_e < 0.015 :\n", - " E = e;\n", - "else:\n", - " E = rest_e +e;\n", - "\n", - "lambda1 = h/(math.sqrt(2*m*e*1e6*1.6e-19)) # calculation of de Broglie wavelength\n", - "\n", - "print \" de Broglie wavelength of neutron is %e angstrom.\"%( lambda1*1e10)\n", - "# Answer in book is 8.04e-5 angstrom which is misprinted\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 7.992279e-05 angstrom.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.7 Page No15" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "e = 1.632e-19 # charge on electron in coulomb\n", - "V = 50 # Applied voltage in volts\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of neutron is %f angstrom.\"%( lambda1*1e10)\n", - "# Answer in book is 1.735 angstrom which is misprinted\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 1.717818 angstrom.\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.9 Page No18" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "e = 1.6e-19 # charge on electron in coulomb\n", - "V = 54 # Applied voltage in volts\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of neutron is %f angstrom.\"%( lambda1*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 1.671941 angstrom.\n" - ] - } - ], - "prompt_number": 17 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.10 Page No21" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "E = 10 # Energy of electron in KeV\n", - "m_e = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "v = math.sqrt(2*E*1.6e-16/m_e) # Calculation of velocity of moving electron\n", - "p = m_e*v #Calculation of momentum of moving electron\n", - "lambda1 = h/p # calculation of de Broglie wavelength\n", - "print \" velocity of electron is %0.2e m/s.\"%(v)\n", - "print \" momentum of electron is %.3e Kgm/s.\"%(p)\n", - "print \" de Broglie wavelength of electron is %.2f angstrom.\"%( lambda1*1e10)\n", - " # Answers in book are v = 5.93e6 m/s, p = 5.397e-24 kgm/s, lambda = 1.23 angstrom\n", - "# Which is due to wrong calculation\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " velocity of electron is 5.93e+07 m/s.\n", - " momentum of electron is 5.396e-23 Kgm/s.\n", - " de Broglie wavelength of electron is 0.12 angstrom.\n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.11 Page No22" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 1. # de Broglie wavelength of neutron in angstrom\n", - "m = 1.67e-27 # Mass of electron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "v = h/(m*lambda1*1e-10) # Calculation of velocity of moving neutron\n", - "E = 1./2*m*v**2 # Calculation of kinetic energy of moving neutron\n", - "print \" velocity of neutron is %e m/s.\"%(v)\n", - "print \" Kinetic energy of neutron is %f eV.\"%(E/1.6e-19)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " velocity of neutron is 3.964072e+03 m/s.\n", - " Kinetic energy of neutron is 0.082007 eV.\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.12 Page No26" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#Given that\n", - "E = 2 # Energy of accelerated electron in KeV\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/math.sqrt(2*m*E*1e3*1.6e-19) # Calculation of velocity of moving electron\n", - "print \" Wavelength of electron is %e m.\"%(lambda1)\n", - "# Answer in book is 2.74e-12m\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of electron is 2.743136e-11 m.\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.13 Page No28" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#Given that\n", - "v = 2e8 # speed of moving proton in m/s\n", - "c = 3e8 # speed of light in m/s\n", - "m = 1.67e-27 # Mass of proton in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(m*v/math.sqrt(1-(v/c)**2)) # Calculation of velocity of moving electron\n", - "print \" Wavelength of electron is %e angstrom.\"%( lambda1*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of electron is 1.477322e-05 angstrom.\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.14 Page No30" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1.# wavelength in m/s\n", - "m_e = 9.1e-31 # Mass of electron in Kg\n", - "m_p = 1.67e-27 # Mass of proton in kg\n", - "c = 3e8 # speed of light in m/s\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "p_p = h/(lambda1*1e-10) # Momentum of photon\n", - "p_e = h/(lambda1*1e-10) # Momentum of electron\n", - "E_e = p_e**2/(2*m_e) +m_e*c**2 # Total energy of electron\n", - "E_p = h*c/(lambda1*1e-10) # Total energy of photon\n", - "K_e = p_e**2/(2*m_e) # Kinetic energy of electron \n", - "K_p = h*c/(lambda1*1e-10)# Kinetic energy of photon\n", - "r_K = K_e/K_p # Ratio of kinetic energies\n", - "print \" Momentum of photon is %e Kgm/s while Momentum of electron is %e Kgm/s which are equal.\"%(p_p,p_e)\n", - "print \" Total Energy of photon is %f KeV while Total Energy of electron is %f MeV \"%(E_p/1.6e-19*1e3,(E_e/1.6e-19*1e6))\n", - "print \" Ratio of kinetic energies is %e \"%(r_K)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Momentum of photon is 6.630000e-24 Kgm/s while Momentum of electron is 6.630000e-24 Kgm/s which are equal.\n", - " Total Energy of photon is 12431250.000000 KeV while Total Energy of electron is 512025950892.857117 MeV \n", - " Ratio of kinetic energies is 1.214286e-02 \n" - ] - } - ], - "prompt_number": 32 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.15 Page No31" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "e = 25 # Energy of neutron in eV\n", - "c = 3e8 # speed of light in m/s\n", - "m = 1.67e-27 # Mass of neutron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "rest_e = m*c**2/(1e6*1.6e-19)# rest mass energy of neutron in MeV\n", - "if e/rest_e < 0.015:\n", - " E = e;\n", - "else:\n", - " E = rest_e +e;\n", - "\n", - "lambda1 = h/(math.sqrt(2*m*e*1.6e-19)) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of neutron is %f angstrom.\"%( lambda1*1e10)\n", - "# Answer in book is 8.04e-5 angstrom \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 0.057274 angstrom.\n" - ] - } - ], - "prompt_number": 33 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.16 Page No36" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "e = 2*1.6e-19 # charge on alpha particle in coulomb\n", - "V = 200 # Applied voltage in volts\n", - "m = 4*1.67e-27 # Mass of alpha particle in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of neutron is %f angstrom.\"%( lambda1*1e10)\n", - "# while answer in book is 0.00715 angstrom\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 0.007170 angstrom.\n" - ] - } - ], - "prompt_number": 34 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.17 Page No41" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "M = 20 # Mass of ball in Kg\n", - "V = 5 # velocity of of ball in m/s\n", - "m = 9.1e-31 #Mass of electron in Kg\n", - "v = 1e6 # velocity of of electron in m/s\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "lambda_b = h/(M*V) # calculation of de Broglie wavelength for ball\n", - "lambda_e = h/(m*v) # calculation of de Broglie wavelength electron\n", - "print \" de Broglie wavelength of ball is %e angstrom.\"%(lambda_b*1e10)\n", - "print \" de Broglie wavelength of electron is %f angstrom.\"%(lambda_e*1e10)\n", - "# answer in book is 6.62e-22 angstrom for ball\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of ball is 6.620000e-26 angstrom.\n", - " de Broglie wavelength of electron is 7.274725 angstrom.\n" - ] - } - ], - "prompt_number": 35 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.18 Page No44" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "E = 1 # Energy of neutron in eV\n", - "m = 1.67e-27 # Mass of neutron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "lambda1 = h/math.sqrt(2*m*E*1.6e-19) # Calculation of velocity of moving electron\n", - "print \" Wavelength of electron is %f angstrom.\"%(lambda1*1e10)\n", - "# Answer in book is 6.62e-22 angstrom\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of electron is 0.286368 angstrom.\n" - ] - } - ], - "prompt_number": 36 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.19 Page No44" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 0.5# wavelength of electron in angstrom\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "q = 1.6e-19 # charge on electron in coulomb\n", - "\n", - "V = h**2/(2*m*q*(lambda1*1e-10)**2) # Calculation of velocity of moving electron\n", - "print \" Applied voltage on electron is %f V.\"%(V)\n", - "# Answer in book is 601.6 Volt\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Applied voltage on electron is 601.983516 V.\n" - ] - } - ], - "prompt_number": 37 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.21 Page No46" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "k = 8.6e-5 # Boltzmann consmath.tant\n", - "t = 37 # Temperature in degree Celsius\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "m = 1.67e-27 # Mass of neutron\n", - "\n", - "lambda1 = h/math.sqrt(3*m*(k*1.6e-19)*(t+273))# Calculation of wavelength\n", - "print \" Wavelength of neutron at %d degree Celsius is %f angstrom.\"%(t,lambda1*1e10)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of neutron at 37 degree Celsius is 1.432020 angstrom.\n" - ] - } - ], - "prompt_number": 38 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.22 Page No49" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "k = 8.6e-5 # Boltzmann consmath.tant\n", - "t = 27 # Temperature in degree Celsius\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "m = 6.7e-27 # Mass of helium atom\n", - "\n", - "lambda1 = h/math.sqrt(3*m*(k*1.6e-19)*(t+273))# Calculation of wavelength\n", - "print \" Wavelength of helium at %d degree Celsius is %f angstrom.\"%(t,lambda1*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of helium at 27 degree Celsius is 0.726758 angstrom.\n" - ] - } - ], - "prompt_number": 39 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.23 Page No50" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "E = 200. # energy of electrons in eV\n", - "x = 20. # dismath.tance of screen in cm\n", - "D = 2. # diameter of ring in cm\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # Mass of electron in kg\n", - "\n", - "lambda1 = h/math.sqrt(2*m*E*1.6e-19) # Calculation of wavelength\n", - "theta = (math.atan(D/(2*x)))\n", - "d = lambda1/(2*math.sin(theta))# calculation of interatomic spacing of crystal\n", - "print \" Interatomic spacing of crystal is %f angstrom.\"%(d*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Interatomic spacing of crystal is 8.685393 angstrom.\n" - ] - } - ], - "prompt_number": 43 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.24 Page No52" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "r = 0.5 # Bohr radius of hydrogen in angstrom\n", - "m = 9.1e-31 # Mass of neutron in Kg\n", - "h = 6.6e-34 # Plank consmath.tant\n", - "v = h/(2*math.pi*r*1e-10*m) # velocity of electron in ground state\n", - "print \" Velocity of electron in ground state is %e m/s.\"%(v)\n", - "# Answer in book is 2.31e6 m/s\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Velocity of electron in ground state is 2.308621e+06 m/s.\n" - ] - } - ], - "prompt_number": 45 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.25 Page No55" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 5890 # wavelength of yellow radiation in angstrom\n", - "m = 9.1e-31 # Mass of neutron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "v = h/(lambda1*1e-10*m) # velocity of electron in ground state\n", - "print \" Velocity of electron in ground state is %.2e m/s.\"%(v)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Velocity of electron in ground state is 1.24e+03 m/s.\n" - ] - } - ], - "prompt_number": 49 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.26 Page No56" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 2 # wavelength of neutron in angstrom\n", - "m = 1.67e-27 # Mass of neutron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "v = h/(lambda1*1e-10*m) # velocity of neutron\n", - "k = 0.5*m*v**2 # Kinetic energy of neutron\n", - "print \" Velocity of neutron is %e m/s.\"%(v)\n", - "print \" Kinetic energy of neutron is %.3f eV.\"%(k/1.6e-19)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Velocity of neutron is 1.985030e+03 m/s.\n", - " Kinetic energy of neutron is 0.021 eV.\n" - ] - } - ], - "prompt_number": 53 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.29 Page No61" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "v1 = 50 # Previous applied voltage\n", - "v2 = 65 # final applied voltage\n", - "k = 12.28 \n", - "d = 0.91 # Spacing in a crystal in angstrom\n", - "print \"Example 1.29\"\n", - "\n", - "lambda1 = k/math.sqrt(v1)\n", - "theta= math.asin(lambda1/(2*d))# Angel for initial applied voltage\n", - "lambda1 = k/math.sqrt(v2)# wavelength for final applied voltage\n", - "theta1 = math.asin(lambda1/(2*d))# Angel for final applied voltage\n", - "print \" For first order, (sintheta) is %f For second order sintheta must be %f which is not possible \\\n", - "for any value of angle. So no maxima occur for higher orders \"%(math.sin(theta),2*math.sin(theta))\n", - "print \" Angle of diffraction for first order of beam is %f degree at %d Volts\"%(theta1*180/math.pi,v2)\n", - "# Answer in book is 57.14 degree\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 1.29\n", - " For first order, (sintheta) is 0.954206 For second order sintheta must be 1.908411 which is not possible for any value of angle. So no maxima occur for higher orders \n", - " Angle of diffraction for first order of beam is 56.813542 degree at 65 Volts\n" - ] - } - ], - "prompt_number": 62 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.30 Page No62" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "lambda1 = 680 # Wavelength in m\n", - "g = 9.8 #Acceleration due to gravity\n", - "print \"Example 1.30\"\n", - "v_g = 1/2*math.sqrt(g*lambda1/(2*math.pi)) # Calculation of group velocity\n", - "print \" Group velocity of seawater waves is %f m/s.\"%(v_g)\n", - "# Answer in book is 16.29 m/s\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 1.30\n", - " Group velocity of seawater waves is 0.000000 m/s.\n" - ] - } - ], - "prompt_number": 63 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.32 Page No64" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "lambda1 = 2e-13 # de Broglie wavelength of an electron in m\n", - "c = 3e8 # Speed of light in m/s\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 1.32\"\n", - "E = h*c/(lambda1*1.6e-19) \n", - "E_rest = m*c**2/(1.6e-19) # Calculation of rest mass energy\n", - "E_total = math.sqrt(E**2+E_rest**2) # Total energy in eV\n", - "v_g = c*math.sqrt(1-(E_rest/E_total)**2) # Group velocity\n", - "v_p = c**2/v_g # Phase velocity\n", - "print \" Group velocity of de Broglie waves is %fc and phase velocity is %fc .\"%(v_g/c,v_p/c)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 1.32\n", - " Group velocity of de Broglie waves is 0.996626c and phase velocity is 1.003385c .\n" - ] - } - ], - "prompt_number": 65 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.33 Page No68" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 2e-12 # de Broglie wavelength of an electron in m\n", - "c = 3e8 # Speed of light in m/s\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 1.33\"\n", - "E = h*c/(lambda1*1.6e-19) # Energy due to momentum\n", - "E_rest = m*c**2/(1.6e-19) # Calculation of rest mass energy\n", - "E_total = math.sqrt(E**2+E_rest**2) # Total energy in eV\n", - "KE = E_total - E_rest # Kinetic energy\n", - "v_g = c*math.sqrt(1-(E_rest/E_total)**2) # Group velocity\n", - "v_p = c**2/v_g # Phase velocity\n", - "\n", - "print \" Kinetic energy of electron is %f KeV.\"%(KE/1000)\n", - "print \" Group velocity of de Broglie waves is %fc m/s and phase velocity is %fc m/s.\"%(v_g/c,v_p/c)\n", - "# Answer in book is v_g = 0.6035c & v_p = 1.657c\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 1.33\n", - " Kinetic energy of electron is 293.330537 KeV.\n", - " Group velocity of de Broglie waves is 0.771930c m/s and phase velocity is 1.295454c m/s.\n" - ] - } - ], - "prompt_number": 67 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch1_3.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch1_3.ipynb deleted file mode 100644 index ae565265..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch1_3.ipynb +++ /dev/null @@ -1,1092 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:b4a0f0559d297d7743b1fffa63bdc97b5eb29843c8d1e8453183351781436bcb" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 1 : De Broglie Matter Waves" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.1 Page No3" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "M = 6e24 # Mass of earth in Kg\n", - "v = 3e4 # Orbital velocity of earth in m/s\n", - "h = 6.625e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(M*v) # calculation of de Broglie wavelength\n", - "\n", - "print \" de Broglie wavelength of earth is %e m.\"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of earth is 3.680556e-63 m.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.2 Page No4" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "M = 1. # Mass of object in Kg\n", - "v = 10. # velocity of object in m/s\n", - "h = 6.625e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(M*v) # calculation of de Broglie wavelength\n", - "\n", - "print \" de Broglie wavelength of body is %e m.\"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of body is 6.625000e-35 m.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.3 Page No7" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# Given that\n", - "m = 1e-30 # Mass of any object in Kg\n", - "v = 1e5 # velocity of object in m/s\n", - "h = 6.625e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(m*v) # calculation of de Broglie wavelength\n", - "\n", - "print \" de Broglie wavelength of body is %e m.\"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of body is 6.625000e-09 m.\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.4 Page No11" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# Given that\n", - "KE = 4.55e-25 # Kinetic energy of an electron in Joule\n", - "m = 9.1e-31 # Mass of any object in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "v = math.sqrt(2*KE/m) # Calculation of velocity of moving electron\n", - "p = m*v #Calculation of momentum of moving electron\n", - "lambda1 = h/p # calculation of de Broglie wavelength\n", - "print \" velocity of electron is %e m/s.\"%(v)\n", - "print \" momentum of electron is %e Kgm/s.\"%(p)\n", - "print \" de Broglie wavelength of electron is %e m.\"%(lambda1)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " velocity of electron is 1.000000e+03 m/s.\n", - " momentum of electron is 9.100000e-28 Kgm/s.\n", - " de Broglie wavelength of electron is 7.274725e-07 m.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.5 Page No14" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "c = 3.e8 # speed of light in m/s\n", - "v = c/20 # Speed of proton in m/s\n", - "m = 1.67e-27 # Mass of proton in Kg\n", - "h = 6.625e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(m*v) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of proton is %e m.\"%(lambda1)\n", - "# Answer in book is 6.645e-14m which is a calculation mistake\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of proton is 2.644711e-14 m.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.6 Page No14" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "e = 12.8 # Energy of neutron in MeV\n", - "c = 3e8 # speed of light in m/s\n", - "m = 1.675e-27 # Mass of neutron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "rest_e = m*c**2/(1e6*1.6e-19)# rest mass energy of neutron in MeV\n", - "if e/rest_e < 0.015 :\n", - " E = e;\n", - "else:\n", - " E = rest_e +e;\n", - "\n", - "lambda1 = h/(math.sqrt(2*m*e*1e6*1.6e-19)) # calculation of de Broglie wavelength\n", - "\n", - "print \" de Broglie wavelength of neutron is %e angstrom.\"%( lambda1*1e10)\n", - "# Answer in book is 8.04e-5 angstrom which is misprinted\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 7.992279e-05 angstrom.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.7 Page No15" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "e = 1.632e-19 # charge on electron in coulomb\n", - "V = 50 # Applied voltage in volts\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of neutron is %f angstrom.\"%( lambda1*1e10)\n", - "# Answer in book is 1.735 angstrom which is misprinted\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 1.717818 angstrom.\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.9 Page No18" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "e = 1.6e-19 # charge on electron in coulomb\n", - "V = 54 # Applied voltage in volts\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of neutron is %f angstrom.\"%( lambda1*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 1.671941 angstrom.\n" - ] - } - ], - "prompt_number": 17 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.10 Page No21" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "E = 10 # Energy of electron in KeV\n", - "m_e = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "v = math.sqrt(2*E*1.6e-16/m_e) # Calculation of velocity of moving electron\n", - "p = m_e*v #Calculation of momentum of moving electron\n", - "lambda1 = h/p # calculation of de Broglie wavelength\n", - "print \" velocity of electron is %0.2e m/s.\"%(v)\n", - "print \" momentum of electron is %.3e Kgm/s.\"%(p)\n", - "print \" de Broglie wavelength of electron is %.2f angstrom.\"%( lambda1*1e10)\n", - " # Answers in book are v = 5.93e6 m/s, p = 5.397e-24 kgm/s, lambda = 1.23 angstrom\n", - "# Which is due to wrong calculation\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " velocity of electron is 5.93e+07 m/s.\n", - " momentum of electron is 5.396e-23 Kgm/s.\n", - " de Broglie wavelength of electron is 0.12 angstrom.\n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.11 Page No22" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 1. # de Broglie wavelength of neutron in angstrom\n", - "m = 1.67e-27 # Mass of electron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "v = h/(m*lambda1*1e-10) # Calculation of velocity of moving neutron\n", - "E = 1./2*m*v**2 # Calculation of kinetic energy of moving neutron\n", - "print \" velocity of neutron is %e m/s.\"%(v)\n", - "print \" Kinetic energy of neutron is %f eV.\"%(E/1.6e-19)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " velocity of neutron is 3.964072e+03 m/s.\n", - " Kinetic energy of neutron is 0.082007 eV.\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.12 Page No26" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#Given that\n", - "E = 2 # Energy of accelerated electron in KeV\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/math.sqrt(2*m*E*1e3*1.6e-19) # Calculation of velocity of moving electron\n", - "print \" Wavelength of electron is %e m.\"%(lambda1)\n", - "# Answer in book is 2.74e-12m\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of electron is 2.743136e-11 m.\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.13 Page No28" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#Given that\n", - "v = 2e8 # speed of moving proton in m/s\n", - "c = 3e8 # speed of light in m/s\n", - "m = 1.67e-27 # Mass of proton in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(m*v/math.sqrt(1-(v/c)**2)) # Calculation of velocity of moving electron\n", - "print \" Wavelength of electron is %e angstrom.\"%( lambda1*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of electron is 1.477322e-05 angstrom.\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.14 Page No30" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1.# wavelength in m/s\n", - "m_e = 9.1e-31 # Mass of electron in Kg\n", - "m_p = 1.67e-27 # Mass of proton in kg\n", - "c = 3e8 # speed of light in m/s\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "p_p = h/(lambda1*1e-10) # Momentum of photon\n", - "p_e = h/(lambda1*1e-10) # Momentum of electron\n", - "E_e = p_e**2/(2*m_e) +m_e*c**2 # Total energy of electron\n", - "E_p = h*c/(lambda1*1e-10) # Total energy of photon\n", - "K_e = p_e**2/(2*m_e) # Kinetic energy of electron \n", - "K_p = h*c/(lambda1*1e-10)# Kinetic energy of photon\n", - "r_K = K_e/K_p # Ratio of kinetic energies\n", - "print \" Momentum of photon is %e Kgm/s while Momentum of electron is %e Kgm/s which are equal.\"%(p_p,p_e)\n", - "print \" Total Energy of photon is %f KeV while Total Energy of electron is %f MeV \"%(E_p/1.6e-19*1e3,(E_e/1.6e-19*1e6))\n", - "print \" Ratio of kinetic energies is %e \"%(r_K)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Momentum of photon is 6.630000e-24 Kgm/s while Momentum of electron is 6.630000e-24 Kgm/s which are equal.\n", - " Total Energy of photon is 12431250.000000 KeV while Total Energy of electron is 512025950892.857117 MeV \n", - " Ratio of kinetic energies is 1.214286e-02 \n" - ] - } - ], - "prompt_number": 32 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.15 Page No31" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "e = 25 # Energy of neutron in eV\n", - "c = 3e8 # speed of light in m/s\n", - "m = 1.67e-27 # Mass of neutron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "rest_e = m*c**2/(1e6*1.6e-19)# rest mass energy of neutron in MeV\n", - "if e/rest_e < 0.015:\n", - " E = e;\n", - "else:\n", - " E = rest_e +e;\n", - "\n", - "lambda1 = h/(math.sqrt(2*m*e*1.6e-19)) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of neutron is %f angstrom.\"%( lambda1*1e10)\n", - "# Answer in book is 8.04e-5 angstrom \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 0.057274 angstrom.\n" - ] - } - ], - "prompt_number": 33 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.16 Page No36" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "e = 2*1.6e-19 # charge on alpha particle in coulomb\n", - "V = 200 # Applied voltage in volts\n", - "m = 4*1.67e-27 # Mass of alpha particle in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of neutron is %f angstrom.\"%( lambda1*1e10)\n", - "# while answer in book is 0.00715 angstrom\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 0.007170 angstrom.\n" - ] - } - ], - "prompt_number": 34 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.17 Page No41" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "M = 20 # Mass of ball in Kg\n", - "V = 5 # velocity of of ball in m/s\n", - "m = 9.1e-31 #Mass of electron in Kg\n", - "v = 1e6 # velocity of of electron in m/s\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "lambda_b = h/(M*V) # calculation of de Broglie wavelength for ball\n", - "lambda_e = h/(m*v) # calculation of de Broglie wavelength electron\n", - "print \" de Broglie wavelength of ball is %e angstrom.\"%(lambda_b*1e10)\n", - "print \" de Broglie wavelength of electron is %f angstrom.\"%(lambda_e*1e10)\n", - "# answer in book is 6.62e-22 angstrom for ball\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of ball is 6.620000e-26 angstrom.\n", - " de Broglie wavelength of electron is 7.274725 angstrom.\n" - ] - } - ], - "prompt_number": 35 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.18 Page No44" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "E = 1 # Energy of neutron in eV\n", - "m = 1.67e-27 # Mass of neutron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "lambda1 = h/math.sqrt(2*m*E*1.6e-19) # Calculation of velocity of moving electron\n", - "print \" Wavelength of electron is %f angstrom.\"%(lambda1*1e10)\n", - "# Answer in book is 6.62e-22 angstrom\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of electron is 0.286368 angstrom.\n" - ] - } - ], - "prompt_number": 36 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.19 Page No44" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 0.5# wavelength of electron in angstrom\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "q = 1.6e-19 # charge on electron in coulomb\n", - "\n", - "V = h**2/(2*m*q*(lambda1*1e-10)**2) # Calculation of velocity of moving electron\n", - "print \" Applied voltage on electron is %f V.\"%(V)\n", - "# Answer in book is 601.6 Volt\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Applied voltage on electron is 601.983516 V.\n" - ] - } - ], - "prompt_number": 37 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.21 Page No46" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "k = 8.6e-5 # Boltzmann consmath.tant\n", - "t = 37 # Temperature in degree Celsius\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "m = 1.67e-27 # Mass of neutron\n", - "\n", - "lambda1 = h/math.sqrt(3*m*(k*1.6e-19)*(t+273))# Calculation of wavelength\n", - "print \" Wavelength of neutron at %d degree Celsius is %f angstrom.\"%(t,lambda1*1e10)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of neutron at 37 degree Celsius is 1.432020 angstrom.\n" - ] - } - ], - "prompt_number": 38 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.22 Page No49" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "k = 8.6e-5 # Boltzmann consmath.tant\n", - "t = 27 # Temperature in degree Celsius\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "m = 6.7e-27 # Mass of helium atom\n", - "\n", - "lambda1 = h/math.sqrt(3*m*(k*1.6e-19)*(t+273))# Calculation of wavelength\n", - "print \" Wavelength of helium at %d degree Celsius is %f angstrom.\"%(t,lambda1*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of helium at 27 degree Celsius is 0.726758 angstrom.\n" - ] - } - ], - "prompt_number": 39 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.23 Page No50" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "E = 200. # energy of electrons in eV\n", - "x = 20. # dismath.tance of screen in cm\n", - "D = 2. # diameter of ring in cm\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # Mass of electron in kg\n", - "\n", - "lambda1 = h/math.sqrt(2*m*E*1.6e-19) # Calculation of wavelength\n", - "theta = (math.atan(D/(2*x)))\n", - "d = lambda1/(2*math.sin(theta))# calculation of interatomic spacing of crystal\n", - "print \" Interatomic spacing of crystal is %f angstrom.\"%(d*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Interatomic spacing of crystal is 8.685393 angstrom.\n" - ] - } - ], - "prompt_number": 43 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.24 Page No52" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "r = 0.5 # Bohr radius of hydrogen in angstrom\n", - "m = 9.1e-31 # Mass of neutron in Kg\n", - "h = 6.6e-34 # Plank consmath.tant\n", - "v = h/(2*math.pi*r*1e-10*m) # velocity of electron in ground state\n", - "print \" Velocity of electron in ground state is %e m/s.\"%(v)\n", - "# Answer in book is 2.31e6 m/s\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Velocity of electron in ground state is 2.308621e+06 m/s.\n" - ] - } - ], - "prompt_number": 45 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.25 Page No55" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 5890 # wavelength of yellow radiation in angstrom\n", - "m = 9.1e-31 # Mass of neutron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "v = h/(lambda1*1e-10*m) # velocity of electron in ground state\n", - "print \" Velocity of electron in ground state is %.2e m/s.\"%(v)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Velocity of electron in ground state is 1.24e+03 m/s.\n" - ] - } - ], - "prompt_number": 49 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.26 Page No56" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 2 # wavelength of neutron in angstrom\n", - "m = 1.67e-27 # Mass of neutron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "v = h/(lambda1*1e-10*m) # velocity of neutron\n", - "k = 0.5*m*v**2 # Kinetic energy of neutron\n", - "print \" Velocity of neutron is %e m/s.\"%(v)\n", - "print \" Kinetic energy of neutron is %.3f eV.\"%(k/1.6e-19)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Velocity of neutron is 1.985030e+03 m/s.\n", - " Kinetic energy of neutron is 0.021 eV.\n" - ] - } - ], - "prompt_number": 53 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.29 Page No61" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "v1 = 50 # Previous applied voltage\n", - "v2 = 65 # final applied voltage\n", - "k = 12.28 \n", - "d = 0.91 # Spacing in a crystal in angstrom\n", - "print \"Example 1.29\"\n", - "\n", - "lambda1 = k/math.sqrt(v1)\n", - "theta= math.asin(lambda1/(2*d))# Angel for initial applied voltage\n", - "lambda1 = k/math.sqrt(v2)# wavelength for final applied voltage\n", - "theta1 = math.asin(lambda1/(2*d))# Angel for final applied voltage\n", - "print \" For first order, (sintheta) is %f For second order sintheta must be %f which is not possible \\\n", - "for any value of angle. So no maxima occur for higher orders \"%(math.sin(theta),2*math.sin(theta))\n", - "print \" Angle of diffraction for first order of beam is %f degree at %d Volts\"%(theta1*180/math.pi,v2)\n", - "# Answer in book is 57.14 degree\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 1.29\n", - " For first order, (sintheta) is 0.954206 For second order sintheta must be 1.908411 which is not possible for any value of angle. So no maxima occur for higher orders \n", - " Angle of diffraction for first order of beam is 56.813542 degree at 65 Volts\n" - ] - } - ], - "prompt_number": 62 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.30 Page No62" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "lambda1 = 680 # Wavelength in m\n", - "g = 9.8 #Acceleration due to gravity\n", - "print \"Example 1.30\"\n", - "v_g = 1/2*math.sqrt(g*lambda1/(2*math.pi)) # Calculation of group velocity\n", - "print \" Group velocity of seawater waves is %f m/s.\"%(v_g)\n", - "# Answer in book is 16.29 m/s\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 1.30\n", - " Group velocity of seawater waves is 0.000000 m/s.\n" - ] - } - ], - "prompt_number": 63 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.32 Page No64" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "lambda1 = 2e-13 # de Broglie wavelength of an electron in m\n", - "c = 3e8 # Speed of light in m/s\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 1.32\"\n", - "E = h*c/(lambda1*1.6e-19) \n", - "E_rest = m*c**2/(1.6e-19) # Calculation of rest mass energy\n", - "E_total = math.sqrt(E**2+E_rest**2) # Total energy in eV\n", - "v_g = c*math.sqrt(1-(E_rest/E_total)**2) # Group velocity\n", - "v_p = c**2/v_g # Phase velocity\n", - "print \" Group velocity of de Broglie waves is %fc and phase velocity is %fc .\"%(v_g/c,v_p/c)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 1.32\n", - " Group velocity of de Broglie waves is 0.996626c and phase velocity is 1.003385c .\n" - ] - } - ], - "prompt_number": 65 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.33 Page No68" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 2e-12 # de Broglie wavelength of an electron in m\n", - "c = 3e8 # Speed of light in m/s\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 1.33\"\n", - "E = h*c/(lambda1*1.6e-19) # Energy due to momentum\n", - "E_rest = m*c**2/(1.6e-19) # Calculation of rest mass energy\n", - "E_total = math.sqrt(E**2+E_rest**2) # Total energy in eV\n", - "KE = E_total - E_rest # Kinetic energy\n", - "v_g = c*math.sqrt(1-(E_rest/E_total)**2) # Group velocity\n", - "v_p = c**2/v_g # Phase velocity\n", - "\n", - "print \" Kinetic energy of electron is %f KeV.\"%(KE/1000)\n", - "print \" Group velocity of de Broglie waves is %fc m/s and phase velocity is %fc m/s.\"%(v_g/c,v_p/c)\n", - "# Answer in book is v_g = 0.6035c & v_p = 1.657c\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 1.33\n", - " Kinetic energy of electron is 293.330537 KeV.\n", - " Group velocity of de Broglie waves is 0.771930c m/s and phase velocity is 1.295454c m/s.\n" - ] - } - ], - "prompt_number": 67 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch1_4.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch1_4.ipynb deleted file mode 100644 index ae565265..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch1_4.ipynb +++ /dev/null @@ -1,1092 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:b4a0f0559d297d7743b1fffa63bdc97b5eb29843c8d1e8453183351781436bcb" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 1 : De Broglie Matter Waves" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.1 Page No3" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "M = 6e24 # Mass of earth in Kg\n", - "v = 3e4 # Orbital velocity of earth in m/s\n", - "h = 6.625e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(M*v) # calculation of de Broglie wavelength\n", - "\n", - "print \" de Broglie wavelength of earth is %e m.\"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of earth is 3.680556e-63 m.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.2 Page No4" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "M = 1. # Mass of object in Kg\n", - "v = 10. # velocity of object in m/s\n", - "h = 6.625e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(M*v) # calculation of de Broglie wavelength\n", - "\n", - "print \" de Broglie wavelength of body is %e m.\"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of body is 6.625000e-35 m.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.3 Page No7" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# Given that\n", - "m = 1e-30 # Mass of any object in Kg\n", - "v = 1e5 # velocity of object in m/s\n", - "h = 6.625e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(m*v) # calculation of de Broglie wavelength\n", - "\n", - "print \" de Broglie wavelength of body is %e m.\"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of body is 6.625000e-09 m.\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.4 Page No11" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# Given that\n", - "KE = 4.55e-25 # Kinetic energy of an electron in Joule\n", - "m = 9.1e-31 # Mass of any object in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "v = math.sqrt(2*KE/m) # Calculation of velocity of moving electron\n", - "p = m*v #Calculation of momentum of moving electron\n", - "lambda1 = h/p # calculation of de Broglie wavelength\n", - "print \" velocity of electron is %e m/s.\"%(v)\n", - "print \" momentum of electron is %e Kgm/s.\"%(p)\n", - "print \" de Broglie wavelength of electron is %e m.\"%(lambda1)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " velocity of electron is 1.000000e+03 m/s.\n", - " momentum of electron is 9.100000e-28 Kgm/s.\n", - " de Broglie wavelength of electron is 7.274725e-07 m.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.5 Page No14" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "c = 3.e8 # speed of light in m/s\n", - "v = c/20 # Speed of proton in m/s\n", - "m = 1.67e-27 # Mass of proton in Kg\n", - "h = 6.625e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(m*v) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of proton is %e m.\"%(lambda1)\n", - "# Answer in book is 6.645e-14m which is a calculation mistake\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of proton is 2.644711e-14 m.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.6 Page No14" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "e = 12.8 # Energy of neutron in MeV\n", - "c = 3e8 # speed of light in m/s\n", - "m = 1.675e-27 # Mass of neutron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "rest_e = m*c**2/(1e6*1.6e-19)# rest mass energy of neutron in MeV\n", - "if e/rest_e < 0.015 :\n", - " E = e;\n", - "else:\n", - " E = rest_e +e;\n", - "\n", - "lambda1 = h/(math.sqrt(2*m*e*1e6*1.6e-19)) # calculation of de Broglie wavelength\n", - "\n", - "print \" de Broglie wavelength of neutron is %e angstrom.\"%( lambda1*1e10)\n", - "# Answer in book is 8.04e-5 angstrom which is misprinted\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 7.992279e-05 angstrom.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.7 Page No15" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "e = 1.632e-19 # charge on electron in coulomb\n", - "V = 50 # Applied voltage in volts\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of neutron is %f angstrom.\"%( lambda1*1e10)\n", - "# Answer in book is 1.735 angstrom which is misprinted\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 1.717818 angstrom.\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.9 Page No18" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "e = 1.6e-19 # charge on electron in coulomb\n", - "V = 54 # Applied voltage in volts\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of neutron is %f angstrom.\"%( lambda1*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 1.671941 angstrom.\n" - ] - } - ], - "prompt_number": 17 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.10 Page No21" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "E = 10 # Energy of electron in KeV\n", - "m_e = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "v = math.sqrt(2*E*1.6e-16/m_e) # Calculation of velocity of moving electron\n", - "p = m_e*v #Calculation of momentum of moving electron\n", - "lambda1 = h/p # calculation of de Broglie wavelength\n", - "print \" velocity of electron is %0.2e m/s.\"%(v)\n", - "print \" momentum of electron is %.3e Kgm/s.\"%(p)\n", - "print \" de Broglie wavelength of electron is %.2f angstrom.\"%( lambda1*1e10)\n", - " # Answers in book are v = 5.93e6 m/s, p = 5.397e-24 kgm/s, lambda = 1.23 angstrom\n", - "# Which is due to wrong calculation\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " velocity of electron is 5.93e+07 m/s.\n", - " momentum of electron is 5.396e-23 Kgm/s.\n", - " de Broglie wavelength of electron is 0.12 angstrom.\n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.11 Page No22" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 1. # de Broglie wavelength of neutron in angstrom\n", - "m = 1.67e-27 # Mass of electron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "v = h/(m*lambda1*1e-10) # Calculation of velocity of moving neutron\n", - "E = 1./2*m*v**2 # Calculation of kinetic energy of moving neutron\n", - "print \" velocity of neutron is %e m/s.\"%(v)\n", - "print \" Kinetic energy of neutron is %f eV.\"%(E/1.6e-19)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " velocity of neutron is 3.964072e+03 m/s.\n", - " Kinetic energy of neutron is 0.082007 eV.\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.12 Page No26" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#Given that\n", - "E = 2 # Energy of accelerated electron in KeV\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/math.sqrt(2*m*E*1e3*1.6e-19) # Calculation of velocity of moving electron\n", - "print \" Wavelength of electron is %e m.\"%(lambda1)\n", - "# Answer in book is 2.74e-12m\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of electron is 2.743136e-11 m.\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.13 Page No28" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#Given that\n", - "v = 2e8 # speed of moving proton in m/s\n", - "c = 3e8 # speed of light in m/s\n", - "m = 1.67e-27 # Mass of proton in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(m*v/math.sqrt(1-(v/c)**2)) # Calculation of velocity of moving electron\n", - "print \" Wavelength of electron is %e angstrom.\"%( lambda1*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of electron is 1.477322e-05 angstrom.\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.14 Page No30" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1.# wavelength in m/s\n", - "m_e = 9.1e-31 # Mass of electron in Kg\n", - "m_p = 1.67e-27 # Mass of proton in kg\n", - "c = 3e8 # speed of light in m/s\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "p_p = h/(lambda1*1e-10) # Momentum of photon\n", - "p_e = h/(lambda1*1e-10) # Momentum of electron\n", - "E_e = p_e**2/(2*m_e) +m_e*c**2 # Total energy of electron\n", - "E_p = h*c/(lambda1*1e-10) # Total energy of photon\n", - "K_e = p_e**2/(2*m_e) # Kinetic energy of electron \n", - "K_p = h*c/(lambda1*1e-10)# Kinetic energy of photon\n", - "r_K = K_e/K_p # Ratio of kinetic energies\n", - "print \" Momentum of photon is %e Kgm/s while Momentum of electron is %e Kgm/s which are equal.\"%(p_p,p_e)\n", - "print \" Total Energy of photon is %f KeV while Total Energy of electron is %f MeV \"%(E_p/1.6e-19*1e3,(E_e/1.6e-19*1e6))\n", - "print \" Ratio of kinetic energies is %e \"%(r_K)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Momentum of photon is 6.630000e-24 Kgm/s while Momentum of electron is 6.630000e-24 Kgm/s which are equal.\n", - " Total Energy of photon is 12431250.000000 KeV while Total Energy of electron is 512025950892.857117 MeV \n", - " Ratio of kinetic energies is 1.214286e-02 \n" - ] - } - ], - "prompt_number": 32 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.15 Page No31" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "e = 25 # Energy of neutron in eV\n", - "c = 3e8 # speed of light in m/s\n", - "m = 1.67e-27 # Mass of neutron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "rest_e = m*c**2/(1e6*1.6e-19)# rest mass energy of neutron in MeV\n", - "if e/rest_e < 0.015:\n", - " E = e;\n", - "else:\n", - " E = rest_e +e;\n", - "\n", - "lambda1 = h/(math.sqrt(2*m*e*1.6e-19)) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of neutron is %f angstrom.\"%( lambda1*1e10)\n", - "# Answer in book is 8.04e-5 angstrom \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 0.057274 angstrom.\n" - ] - } - ], - "prompt_number": 33 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.16 Page No36" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "e = 2*1.6e-19 # charge on alpha particle in coulomb\n", - "V = 200 # Applied voltage in volts\n", - "m = 4*1.67e-27 # Mass of alpha particle in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of neutron is %f angstrom.\"%( lambda1*1e10)\n", - "# while answer in book is 0.00715 angstrom\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 0.007170 angstrom.\n" - ] - } - ], - "prompt_number": 34 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.17 Page No41" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "M = 20 # Mass of ball in Kg\n", - "V = 5 # velocity of of ball in m/s\n", - "m = 9.1e-31 #Mass of electron in Kg\n", - "v = 1e6 # velocity of of electron in m/s\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "lambda_b = h/(M*V) # calculation of de Broglie wavelength for ball\n", - "lambda_e = h/(m*v) # calculation of de Broglie wavelength electron\n", - "print \" de Broglie wavelength of ball is %e angstrom.\"%(lambda_b*1e10)\n", - "print \" de Broglie wavelength of electron is %f angstrom.\"%(lambda_e*1e10)\n", - "# answer in book is 6.62e-22 angstrom for ball\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of ball is 6.620000e-26 angstrom.\n", - " de Broglie wavelength of electron is 7.274725 angstrom.\n" - ] - } - ], - "prompt_number": 35 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.18 Page No44" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "E = 1 # Energy of neutron in eV\n", - "m = 1.67e-27 # Mass of neutron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "lambda1 = h/math.sqrt(2*m*E*1.6e-19) # Calculation of velocity of moving electron\n", - "print \" Wavelength of electron is %f angstrom.\"%(lambda1*1e10)\n", - "# Answer in book is 6.62e-22 angstrom\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of electron is 0.286368 angstrom.\n" - ] - } - ], - "prompt_number": 36 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.19 Page No44" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 0.5# wavelength of electron in angstrom\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "q = 1.6e-19 # charge on electron in coulomb\n", - "\n", - "V = h**2/(2*m*q*(lambda1*1e-10)**2) # Calculation of velocity of moving electron\n", - "print \" Applied voltage on electron is %f V.\"%(V)\n", - "# Answer in book is 601.6 Volt\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Applied voltage on electron is 601.983516 V.\n" - ] - } - ], - "prompt_number": 37 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.21 Page No46" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "k = 8.6e-5 # Boltzmann consmath.tant\n", - "t = 37 # Temperature in degree Celsius\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "m = 1.67e-27 # Mass of neutron\n", - "\n", - "lambda1 = h/math.sqrt(3*m*(k*1.6e-19)*(t+273))# Calculation of wavelength\n", - "print \" Wavelength of neutron at %d degree Celsius is %f angstrom.\"%(t,lambda1*1e10)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of neutron at 37 degree Celsius is 1.432020 angstrom.\n" - ] - } - ], - "prompt_number": 38 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.22 Page No49" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "k = 8.6e-5 # Boltzmann consmath.tant\n", - "t = 27 # Temperature in degree Celsius\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "m = 6.7e-27 # Mass of helium atom\n", - "\n", - "lambda1 = h/math.sqrt(3*m*(k*1.6e-19)*(t+273))# Calculation of wavelength\n", - "print \" Wavelength of helium at %d degree Celsius is %f angstrom.\"%(t,lambda1*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of helium at 27 degree Celsius is 0.726758 angstrom.\n" - ] - } - ], - "prompt_number": 39 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.23 Page No50" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "E = 200. # energy of electrons in eV\n", - "x = 20. # dismath.tance of screen in cm\n", - "D = 2. # diameter of ring in cm\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # Mass of electron in kg\n", - "\n", - "lambda1 = h/math.sqrt(2*m*E*1.6e-19) # Calculation of wavelength\n", - "theta = (math.atan(D/(2*x)))\n", - "d = lambda1/(2*math.sin(theta))# calculation of interatomic spacing of crystal\n", - "print \" Interatomic spacing of crystal is %f angstrom.\"%(d*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Interatomic spacing of crystal is 8.685393 angstrom.\n" - ] - } - ], - "prompt_number": 43 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.24 Page No52" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "r = 0.5 # Bohr radius of hydrogen in angstrom\n", - "m = 9.1e-31 # Mass of neutron in Kg\n", - "h = 6.6e-34 # Plank consmath.tant\n", - "v = h/(2*math.pi*r*1e-10*m) # velocity of electron in ground state\n", - "print \" Velocity of electron in ground state is %e m/s.\"%(v)\n", - "# Answer in book is 2.31e6 m/s\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Velocity of electron in ground state is 2.308621e+06 m/s.\n" - ] - } - ], - "prompt_number": 45 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.25 Page No55" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 5890 # wavelength of yellow radiation in angstrom\n", - "m = 9.1e-31 # Mass of neutron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "v = h/(lambda1*1e-10*m) # velocity of electron in ground state\n", - "print \" Velocity of electron in ground state is %.2e m/s.\"%(v)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Velocity of electron in ground state is 1.24e+03 m/s.\n" - ] - } - ], - "prompt_number": 49 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.26 Page No56" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 2 # wavelength of neutron in angstrom\n", - "m = 1.67e-27 # Mass of neutron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "v = h/(lambda1*1e-10*m) # velocity of neutron\n", - "k = 0.5*m*v**2 # Kinetic energy of neutron\n", - "print \" Velocity of neutron is %e m/s.\"%(v)\n", - "print \" Kinetic energy of neutron is %.3f eV.\"%(k/1.6e-19)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Velocity of neutron is 1.985030e+03 m/s.\n", - " Kinetic energy of neutron is 0.021 eV.\n" - ] - } - ], - "prompt_number": 53 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.29 Page No61" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "v1 = 50 # Previous applied voltage\n", - "v2 = 65 # final applied voltage\n", - "k = 12.28 \n", - "d = 0.91 # Spacing in a crystal in angstrom\n", - "print \"Example 1.29\"\n", - "\n", - "lambda1 = k/math.sqrt(v1)\n", - "theta= math.asin(lambda1/(2*d))# Angel for initial applied voltage\n", - "lambda1 = k/math.sqrt(v2)# wavelength for final applied voltage\n", - "theta1 = math.asin(lambda1/(2*d))# Angel for final applied voltage\n", - "print \" For first order, (sintheta) is %f For second order sintheta must be %f which is not possible \\\n", - "for any value of angle. So no maxima occur for higher orders \"%(math.sin(theta),2*math.sin(theta))\n", - "print \" Angle of diffraction for first order of beam is %f degree at %d Volts\"%(theta1*180/math.pi,v2)\n", - "# Answer in book is 57.14 degree\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 1.29\n", - " For first order, (sintheta) is 0.954206 For second order sintheta must be 1.908411 which is not possible for any value of angle. So no maxima occur for higher orders \n", - " Angle of diffraction for first order of beam is 56.813542 degree at 65 Volts\n" - ] - } - ], - "prompt_number": 62 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.30 Page No62" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "lambda1 = 680 # Wavelength in m\n", - "g = 9.8 #Acceleration due to gravity\n", - "print \"Example 1.30\"\n", - "v_g = 1/2*math.sqrt(g*lambda1/(2*math.pi)) # Calculation of group velocity\n", - "print \" Group velocity of seawater waves is %f m/s.\"%(v_g)\n", - "# Answer in book is 16.29 m/s\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 1.30\n", - " Group velocity of seawater waves is 0.000000 m/s.\n" - ] - } - ], - "prompt_number": 63 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.32 Page No64" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "lambda1 = 2e-13 # de Broglie wavelength of an electron in m\n", - "c = 3e8 # Speed of light in m/s\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 1.32\"\n", - "E = h*c/(lambda1*1.6e-19) \n", - "E_rest = m*c**2/(1.6e-19) # Calculation of rest mass energy\n", - "E_total = math.sqrt(E**2+E_rest**2) # Total energy in eV\n", - "v_g = c*math.sqrt(1-(E_rest/E_total)**2) # Group velocity\n", - "v_p = c**2/v_g # Phase velocity\n", - "print \" Group velocity of de Broglie waves is %fc and phase velocity is %fc .\"%(v_g/c,v_p/c)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 1.32\n", - " Group velocity of de Broglie waves is 0.996626c and phase velocity is 1.003385c .\n" - ] - } - ], - "prompt_number": 65 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.33 Page No68" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 2e-12 # de Broglie wavelength of an electron in m\n", - "c = 3e8 # Speed of light in m/s\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 1.33\"\n", - "E = h*c/(lambda1*1.6e-19) # Energy due to momentum\n", - "E_rest = m*c**2/(1.6e-19) # Calculation of rest mass energy\n", - "E_total = math.sqrt(E**2+E_rest**2) # Total energy in eV\n", - "KE = E_total - E_rest # Kinetic energy\n", - "v_g = c*math.sqrt(1-(E_rest/E_total)**2) # Group velocity\n", - "v_p = c**2/v_g # Phase velocity\n", - "\n", - "print \" Kinetic energy of electron is %f KeV.\"%(KE/1000)\n", - "print \" Group velocity of de Broglie waves is %fc m/s and phase velocity is %fc m/s.\"%(v_g/c,v_p/c)\n", - "# Answer in book is v_g = 0.6035c & v_p = 1.657c\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 1.33\n", - " Kinetic energy of electron is 293.330537 KeV.\n", - " Group velocity of de Broglie waves is 0.771930c m/s and phase velocity is 1.295454c m/s.\n" - ] - } - ], - "prompt_number": 67 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch1_5.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch1_5.ipynb deleted file mode 100644 index ae565265..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch1_5.ipynb +++ /dev/null @@ -1,1092 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:b4a0f0559d297d7743b1fffa63bdc97b5eb29843c8d1e8453183351781436bcb" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 1 : De Broglie Matter Waves" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.1 Page No3" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "M = 6e24 # Mass of earth in Kg\n", - "v = 3e4 # Orbital velocity of earth in m/s\n", - "h = 6.625e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(M*v) # calculation of de Broglie wavelength\n", - "\n", - "print \" de Broglie wavelength of earth is %e m.\"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of earth is 3.680556e-63 m.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.2 Page No4" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "M = 1. # Mass of object in Kg\n", - "v = 10. # velocity of object in m/s\n", - "h = 6.625e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(M*v) # calculation of de Broglie wavelength\n", - "\n", - "print \" de Broglie wavelength of body is %e m.\"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of body is 6.625000e-35 m.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.3 Page No7" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# Given that\n", - "m = 1e-30 # Mass of any object in Kg\n", - "v = 1e5 # velocity of object in m/s\n", - "h = 6.625e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(m*v) # calculation of de Broglie wavelength\n", - "\n", - "print \" de Broglie wavelength of body is %e m.\"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of body is 6.625000e-09 m.\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.4 Page No11" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# Given that\n", - "KE = 4.55e-25 # Kinetic energy of an electron in Joule\n", - "m = 9.1e-31 # Mass of any object in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "v = math.sqrt(2*KE/m) # Calculation of velocity of moving electron\n", - "p = m*v #Calculation of momentum of moving electron\n", - "lambda1 = h/p # calculation of de Broglie wavelength\n", - "print \" velocity of electron is %e m/s.\"%(v)\n", - "print \" momentum of electron is %e Kgm/s.\"%(p)\n", - "print \" de Broglie wavelength of electron is %e m.\"%(lambda1)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " velocity of electron is 1.000000e+03 m/s.\n", - " momentum of electron is 9.100000e-28 Kgm/s.\n", - " de Broglie wavelength of electron is 7.274725e-07 m.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.5 Page No14" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "c = 3.e8 # speed of light in m/s\n", - "v = c/20 # Speed of proton in m/s\n", - "m = 1.67e-27 # Mass of proton in Kg\n", - "h = 6.625e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(m*v) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of proton is %e m.\"%(lambda1)\n", - "# Answer in book is 6.645e-14m which is a calculation mistake\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of proton is 2.644711e-14 m.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.6 Page No14" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "e = 12.8 # Energy of neutron in MeV\n", - "c = 3e8 # speed of light in m/s\n", - "m = 1.675e-27 # Mass of neutron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "rest_e = m*c**2/(1e6*1.6e-19)# rest mass energy of neutron in MeV\n", - "if e/rest_e < 0.015 :\n", - " E = e;\n", - "else:\n", - " E = rest_e +e;\n", - "\n", - "lambda1 = h/(math.sqrt(2*m*e*1e6*1.6e-19)) # calculation of de Broglie wavelength\n", - "\n", - "print \" de Broglie wavelength of neutron is %e angstrom.\"%( lambda1*1e10)\n", - "# Answer in book is 8.04e-5 angstrom which is misprinted\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 7.992279e-05 angstrom.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.7 Page No15" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "e = 1.632e-19 # charge on electron in coulomb\n", - "V = 50 # Applied voltage in volts\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of neutron is %f angstrom.\"%( lambda1*1e10)\n", - "# Answer in book is 1.735 angstrom which is misprinted\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 1.717818 angstrom.\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.9 Page No18" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "e = 1.6e-19 # charge on electron in coulomb\n", - "V = 54 # Applied voltage in volts\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of neutron is %f angstrom.\"%( lambda1*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 1.671941 angstrom.\n" - ] - } - ], - "prompt_number": 17 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.10 Page No21" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "E = 10 # Energy of electron in KeV\n", - "m_e = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "v = math.sqrt(2*E*1.6e-16/m_e) # Calculation of velocity of moving electron\n", - "p = m_e*v #Calculation of momentum of moving electron\n", - "lambda1 = h/p # calculation of de Broglie wavelength\n", - "print \" velocity of electron is %0.2e m/s.\"%(v)\n", - "print \" momentum of electron is %.3e Kgm/s.\"%(p)\n", - "print \" de Broglie wavelength of electron is %.2f angstrom.\"%( lambda1*1e10)\n", - " # Answers in book are v = 5.93e6 m/s, p = 5.397e-24 kgm/s, lambda = 1.23 angstrom\n", - "# Which is due to wrong calculation\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " velocity of electron is 5.93e+07 m/s.\n", - " momentum of electron is 5.396e-23 Kgm/s.\n", - " de Broglie wavelength of electron is 0.12 angstrom.\n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.11 Page No22" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 1. # de Broglie wavelength of neutron in angstrom\n", - "m = 1.67e-27 # Mass of electron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "v = h/(m*lambda1*1e-10) # Calculation of velocity of moving neutron\n", - "E = 1./2*m*v**2 # Calculation of kinetic energy of moving neutron\n", - "print \" velocity of neutron is %e m/s.\"%(v)\n", - "print \" Kinetic energy of neutron is %f eV.\"%(E/1.6e-19)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " velocity of neutron is 3.964072e+03 m/s.\n", - " Kinetic energy of neutron is 0.082007 eV.\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.12 Page No26" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#Given that\n", - "E = 2 # Energy of accelerated electron in KeV\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/math.sqrt(2*m*E*1e3*1.6e-19) # Calculation of velocity of moving electron\n", - "print \" Wavelength of electron is %e m.\"%(lambda1)\n", - "# Answer in book is 2.74e-12m\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of electron is 2.743136e-11 m.\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.13 Page No28" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#Given that\n", - "v = 2e8 # speed of moving proton in m/s\n", - "c = 3e8 # speed of light in m/s\n", - "m = 1.67e-27 # Mass of proton in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(m*v/math.sqrt(1-(v/c)**2)) # Calculation of velocity of moving electron\n", - "print \" Wavelength of electron is %e angstrom.\"%( lambda1*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of electron is 1.477322e-05 angstrom.\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.14 Page No30" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1.# wavelength in m/s\n", - "m_e = 9.1e-31 # Mass of electron in Kg\n", - "m_p = 1.67e-27 # Mass of proton in kg\n", - "c = 3e8 # speed of light in m/s\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "p_p = h/(lambda1*1e-10) # Momentum of photon\n", - "p_e = h/(lambda1*1e-10) # Momentum of electron\n", - "E_e = p_e**2/(2*m_e) +m_e*c**2 # Total energy of electron\n", - "E_p = h*c/(lambda1*1e-10) # Total energy of photon\n", - "K_e = p_e**2/(2*m_e) # Kinetic energy of electron \n", - "K_p = h*c/(lambda1*1e-10)# Kinetic energy of photon\n", - "r_K = K_e/K_p # Ratio of kinetic energies\n", - "print \" Momentum of photon is %e Kgm/s while Momentum of electron is %e Kgm/s which are equal.\"%(p_p,p_e)\n", - "print \" Total Energy of photon is %f KeV while Total Energy of electron is %f MeV \"%(E_p/1.6e-19*1e3,(E_e/1.6e-19*1e6))\n", - "print \" Ratio of kinetic energies is %e \"%(r_K)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Momentum of photon is 6.630000e-24 Kgm/s while Momentum of electron is 6.630000e-24 Kgm/s which are equal.\n", - " Total Energy of photon is 12431250.000000 KeV while Total Energy of electron is 512025950892.857117 MeV \n", - " Ratio of kinetic energies is 1.214286e-02 \n" - ] - } - ], - "prompt_number": 32 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.15 Page No31" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "e = 25 # Energy of neutron in eV\n", - "c = 3e8 # speed of light in m/s\n", - "m = 1.67e-27 # Mass of neutron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "rest_e = m*c**2/(1e6*1.6e-19)# rest mass energy of neutron in MeV\n", - "if e/rest_e < 0.015:\n", - " E = e;\n", - "else:\n", - " E = rest_e +e;\n", - "\n", - "lambda1 = h/(math.sqrt(2*m*e*1.6e-19)) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of neutron is %f angstrom.\"%( lambda1*1e10)\n", - "# Answer in book is 8.04e-5 angstrom \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 0.057274 angstrom.\n" - ] - } - ], - "prompt_number": 33 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.16 Page No36" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "e = 2*1.6e-19 # charge on alpha particle in coulomb\n", - "V = 200 # Applied voltage in volts\n", - "m = 4*1.67e-27 # Mass of alpha particle in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "lambda1 = h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength\n", - "print \" de Broglie wavelength of neutron is %f angstrom.\"%( lambda1*1e10)\n", - "# while answer in book is 0.00715 angstrom\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of neutron is 0.007170 angstrom.\n" - ] - } - ], - "prompt_number": 34 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.17 Page No41" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "M = 20 # Mass of ball in Kg\n", - "V = 5 # velocity of of ball in m/s\n", - "m = 9.1e-31 #Mass of electron in Kg\n", - "v = 1e6 # velocity of of electron in m/s\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "\n", - "lambda_b = h/(M*V) # calculation of de Broglie wavelength for ball\n", - "lambda_e = h/(m*v) # calculation of de Broglie wavelength electron\n", - "print \" de Broglie wavelength of ball is %e angstrom.\"%(lambda_b*1e10)\n", - "print \" de Broglie wavelength of electron is %f angstrom.\"%(lambda_e*1e10)\n", - "# answer in book is 6.62e-22 angstrom for ball\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " de Broglie wavelength of ball is 6.620000e-26 angstrom.\n", - " de Broglie wavelength of electron is 7.274725 angstrom.\n" - ] - } - ], - "prompt_number": 35 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.18 Page No44" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "E = 1 # Energy of neutron in eV\n", - "m = 1.67e-27 # Mass of neutron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "lambda1 = h/math.sqrt(2*m*E*1.6e-19) # Calculation of velocity of moving electron\n", - "print \" Wavelength of electron is %f angstrom.\"%(lambda1*1e10)\n", - "# Answer in book is 6.62e-22 angstrom\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of electron is 0.286368 angstrom.\n" - ] - } - ], - "prompt_number": 36 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.19 Page No44" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 0.5# wavelength of electron in angstrom\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "q = 1.6e-19 # charge on electron in coulomb\n", - "\n", - "V = h**2/(2*m*q*(lambda1*1e-10)**2) # Calculation of velocity of moving electron\n", - "print \" Applied voltage on electron is %f V.\"%(V)\n", - "# Answer in book is 601.6 Volt\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Applied voltage on electron is 601.983516 V.\n" - ] - } - ], - "prompt_number": 37 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.21 Page No46" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "k = 8.6e-5 # Boltzmann consmath.tant\n", - "t = 37 # Temperature in degree Celsius\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "m = 1.67e-27 # Mass of neutron\n", - "\n", - "lambda1 = h/math.sqrt(3*m*(k*1.6e-19)*(t+273))# Calculation of wavelength\n", - "print \" Wavelength of neutron at %d degree Celsius is %f angstrom.\"%(t,lambda1*1e10)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of neutron at 37 degree Celsius is 1.432020 angstrom.\n" - ] - } - ], - "prompt_number": 38 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.22 Page No49" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "k = 8.6e-5 # Boltzmann consmath.tant\n", - "t = 27 # Temperature in degree Celsius\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "m = 6.7e-27 # Mass of helium atom\n", - "\n", - "lambda1 = h/math.sqrt(3*m*(k*1.6e-19)*(t+273))# Calculation of wavelength\n", - "print \" Wavelength of helium at %d degree Celsius is %f angstrom.\"%(t,lambda1*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Wavelength of helium at 27 degree Celsius is 0.726758 angstrom.\n" - ] - } - ], - "prompt_number": 39 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.23 Page No50" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "E = 200. # energy of electrons in eV\n", - "x = 20. # dismath.tance of screen in cm\n", - "D = 2. # diameter of ring in cm\n", - "h = 6.62e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # Mass of electron in kg\n", - "\n", - "lambda1 = h/math.sqrt(2*m*E*1.6e-19) # Calculation of wavelength\n", - "theta = (math.atan(D/(2*x)))\n", - "d = lambda1/(2*math.sin(theta))# calculation of interatomic spacing of crystal\n", - "print \" Interatomic spacing of crystal is %f angstrom.\"%(d*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Interatomic spacing of crystal is 8.685393 angstrom.\n" - ] - } - ], - "prompt_number": 43 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.24 Page No52" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "r = 0.5 # Bohr radius of hydrogen in angstrom\n", - "m = 9.1e-31 # Mass of neutron in Kg\n", - "h = 6.6e-34 # Plank consmath.tant\n", - "v = h/(2*math.pi*r*1e-10*m) # velocity of electron in ground state\n", - "print \" Velocity of electron in ground state is %e m/s.\"%(v)\n", - "# Answer in book is 2.31e6 m/s\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Velocity of electron in ground state is 2.308621e+06 m/s.\n" - ] - } - ], - "prompt_number": 45 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.25 Page No55" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 5890 # wavelength of yellow radiation in angstrom\n", - "m = 9.1e-31 # Mass of neutron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "v = h/(lambda1*1e-10*m) # velocity of electron in ground state\n", - "print \" Velocity of electron in ground state is %.2e m/s.\"%(v)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Velocity of electron in ground state is 1.24e+03 m/s.\n" - ] - } - ], - "prompt_number": 49 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.26 Page No56" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 2 # wavelength of neutron in angstrom\n", - "m = 1.67e-27 # Mass of neutron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "v = h/(lambda1*1e-10*m) # velocity of neutron\n", - "k = 0.5*m*v**2 # Kinetic energy of neutron\n", - "print \" Velocity of neutron is %e m/s.\"%(v)\n", - "print \" Kinetic energy of neutron is %.3f eV.\"%(k/1.6e-19)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " Velocity of neutron is 1.985030e+03 m/s.\n", - " Kinetic energy of neutron is 0.021 eV.\n" - ] - } - ], - "prompt_number": 53 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.29 Page No61" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "v1 = 50 # Previous applied voltage\n", - "v2 = 65 # final applied voltage\n", - "k = 12.28 \n", - "d = 0.91 # Spacing in a crystal in angstrom\n", - "print \"Example 1.29\"\n", - "\n", - "lambda1 = k/math.sqrt(v1)\n", - "theta= math.asin(lambda1/(2*d))# Angel for initial applied voltage\n", - "lambda1 = k/math.sqrt(v2)# wavelength for final applied voltage\n", - "theta1 = math.asin(lambda1/(2*d))# Angel for final applied voltage\n", - "print \" For first order, (sintheta) is %f For second order sintheta must be %f which is not possible \\\n", - "for any value of angle. So no maxima occur for higher orders \"%(math.sin(theta),2*math.sin(theta))\n", - "print \" Angle of diffraction for first order of beam is %f degree at %d Volts\"%(theta1*180/math.pi,v2)\n", - "# Answer in book is 57.14 degree\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 1.29\n", - " For first order, (sintheta) is 0.954206 For second order sintheta must be 1.908411 which is not possible for any value of angle. So no maxima occur for higher orders \n", - " Angle of diffraction for first order of beam is 56.813542 degree at 65 Volts\n" - ] - } - ], - "prompt_number": 62 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.30 Page No62" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "lambda1 = 680 # Wavelength in m\n", - "g = 9.8 #Acceleration due to gravity\n", - "print \"Example 1.30\"\n", - "v_g = 1/2*math.sqrt(g*lambda1/(2*math.pi)) # Calculation of group velocity\n", - "print \" Group velocity of seawater waves is %f m/s.\"%(v_g)\n", - "# Answer in book is 16.29 m/s\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 1.30\n", - " Group velocity of seawater waves is 0.000000 m/s.\n" - ] - } - ], - "prompt_number": 63 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.32 Page No64" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "lambda1 = 2e-13 # de Broglie wavelength of an electron in m\n", - "c = 3e8 # Speed of light in m/s\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 1.32\"\n", - "E = h*c/(lambda1*1.6e-19) \n", - "E_rest = m*c**2/(1.6e-19) # Calculation of rest mass energy\n", - "E_total = math.sqrt(E**2+E_rest**2) # Total energy in eV\n", - "v_g = c*math.sqrt(1-(E_rest/E_total)**2) # Group velocity\n", - "v_p = c**2/v_g # Phase velocity\n", - "print \" Group velocity of de Broglie waves is %fc and phase velocity is %fc .\"%(v_g/c,v_p/c)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 1.32\n", - " Group velocity of de Broglie waves is 0.996626c and phase velocity is 1.003385c .\n" - ] - } - ], - "prompt_number": 65 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 1.33 Page No68" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "lambda1 = 2e-12 # de Broglie wavelength of an electron in m\n", - "c = 3e8 # Speed of light in m/s\n", - "m = 9.1e-31 # Mass of electron in Kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 1.33\"\n", - "E = h*c/(lambda1*1.6e-19) # Energy due to momentum\n", - "E_rest = m*c**2/(1.6e-19) # Calculation of rest mass energy\n", - "E_total = math.sqrt(E**2+E_rest**2) # Total energy in eV\n", - "KE = E_total - E_rest # Kinetic energy\n", - "v_g = c*math.sqrt(1-(E_rest/E_total)**2) # Group velocity\n", - "v_p = c**2/v_g # Phase velocity\n", - "\n", - "print \" Kinetic energy of electron is %f KeV.\"%(KE/1000)\n", - "print \" Group velocity of de Broglie waves is %fc m/s and phase velocity is %fc m/s.\"%(v_g/c,v_p/c)\n", - "# Answer in book is v_g = 0.6035c & v_p = 1.657c\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 1.33\n", - " Kinetic energy of electron is 293.330537 KeV.\n", - " Group velocity of de Broglie waves is 0.771930c m/s and phase velocity is 1.295454c m/s.\n" - ] - } - ], - "prompt_number": 67 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch2.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch2.ipynb deleted file mode 100755 index dc4ec4d6..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch2.ipynb +++ /dev/null @@ -1,1017 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:5ca2861748d43d40b34516562fe47bc041c8b35f6ea545c8e5309115cbc73a04" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 2 : Uncertainty Principle and Schrodinger wave Equation" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.1 Page No71" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_x = 0.2 # Uncertainty in position in angstrom\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "print \"Example 2.1\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = h_bar/(2*del_x*1e-10) # Calculation of uncertainty in momentum\n", - "print \" Uncertainty in momentum of particle is %e kgm/sec \"%(del_p)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.1\n", - " Uncertainty in momentum of particle is 2.637993e-24 kgm/sec \n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.2 Page No75" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#given that\n", - "del_x = 4e-10 # Uncertainty in position in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "print \"Example 2.2\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = h_bar/(2*del_x) # Calculation of uncertainty in momentum\n", - "print \" Uncertainty in momentum of particle is %e kgm/sec.\"%(del_p)\n", - "# Answer in book is given as 1.32e-23 kgm/sec\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.2\n", - " Uncertainty in momentum of particle is 1.318997e-25 kgm/sec.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.3 Page No75" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "v = 3e7 # Velocity of moving electron in m/s\n", - "m = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "c = 3e8 # speed of light in m/s\n", - "print \"Example 2.3\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = m*v/(math.sqrt(1-(v/c)**2)) # calculation of uncertainty in momentum\n", - "del_x = h_bar/(2*del_p) # Calculation of uncertainty in position\n", - "print \" Uncertainty in position of particle is %f angstrom.\"%(del_x*1e10)\n", - "#Answer in book is 0.0194 angstrom which is due to umath.sing approximate values at intermediate steps\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.3\n", - " Uncertainty in position of particle is 0.019229 angstrom.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.5 Page No80" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "v = 1.05e4 # Velocity of moving electron in m/s\n", - "v_error = 0.02 #Percentage error in measurement of velocity\n", - "\n", - "m = 9e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.5\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "p = m*v\n", - "del_p = v_error*p/100 # calculation of uncertainty in momentum\n", - "del_x = h_bar/del_p\n", - "print \" Uncertainty in position of particle is %e m.\"%(del_x)\n", - "# Answer in book is given as 5.58e-3 m\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.5\n", - " Uncertainty in position of particle is 5.583054e-05 m.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.6 Page No82" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "v = 600 # Velocity of moving electron in m/s\n", - "v_error = 0.005 #Percentage error in measurement of velocity\n", - "m = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.6\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "p = m*v\n", - "del_p = v_error*p/100 # calculation of uncertainty in momentum\n", - "del_x = h_bar/(del_p) # Calculation of uncertainty in position\n", - "print \" Uncertainty in position of particle is %e m.\"%(del_x)\n", - "# Answer in book is 0.39e-2 m \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.6\n", - " Uncertainty in position of particle is 3.865191e-03 m.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.7 Page No82" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_x = 1 # let uncertainty in position is unity\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "m_p = 1.67e-27 # mass of proton in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.7\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_v_ratio = m_p/m_e # calculation in uncertainties in the velocity of electron and proton\n", - "print \" Ratio of uncertainties in the velocity of electron to proton is %d.\"%(del_v_ratio)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.7\n", - " Ratio of uncertainties in the velocity of electron to proton is 1835.\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.8 Page No84" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "r = 0.5 # radius of hydrogen atom in angstrom\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.8\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_x = 2*r # calculation of uncertainty in position\n", - "del_p = h_bar/(2*del_x*1e-10) # calculation of uncertainty in momentum\n", - "p = del_p\n", - "E = p**2/(2*m_e*1.6e-19)# Calculation of energy in eV\n", - "print \" Kinetic energy needed by an electron to be confined in electron is %.f eV.\"%((E*100)/100)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.8\n", - " Kinetic energy needed by an electron to be confined in electron is 1 eV.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.9 Page No89" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "v = 5e3 # Velocity of moving electron in m/s\n", - "v_error = 0.003 #Percentage error in measurement of velocity\n", - "\n", - "m = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.9\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "p = m*v\n", - "del_p = v_error*p/100 # calculation of uncertainty in momentum\n", - "del_x = h_bar/(2*del_p) # Calculation of uncertainty in position\n", - "print \" Uncertainty in position of particle is %e m.\"%(del_x)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.9\n", - " Uncertainty in position of particle is 3.865191e-04 m.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.10 Page No90" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "r = 0.53 # radius of hydrogen atom in angstrom\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.10\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_x = 2*r # calculation of uncertainty in position\n", - "del_p = h_bar/(2*del_x*1e-10) # calculation of uncertainty in momentum\n", - "p = del_p\n", - "E = p**2/(2*m_e*1.6e-19)# Calculation of energy in eV\n", - "print \" Kinetic energy needed by an electron to be confined in electron is %f eV.\"%(E)\n", - "# When problem is solved by del_x*del_p = h_bar, then minimum value of kinetic energy will become 13.6eV\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.10\n", - " Kinetic energy needed by an electron to be confined in electron is 0.850754 eV.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.11 Page No92" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "del_t = 2.5e-14 # lifetime in exited state in micro sec\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.11\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_E = h_bar/(1.6e-19*del_t*1e-6) # calculation of uncertainty in momentum\n", - "print \" Minimum error in measurement of energy of this state is %e eV.\"%(del_E)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.11\n", - " Minimum error in measurement of energy of this state is 2.637993e+04 eV.\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.12 Page No92" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "E_eV = 0.5# kinetic energy of electron in KeV\n", - "del_x = 0.4 # Uncertainty in position in nm\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in kg\n", - "print \"Example 2.12\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "E_J = E_eV*1e3*1.6e-19\n", - "p = math.sqrt(2*m*E_J) # Calculation of momentum in kgm/s\n", - "del_p = h_bar/(2*del_x*1e-9) # Calculation of uncertainty in momentum\n", - "per_error = del_p*100 / p # calculation of percentage error in momentum\n", - "print \" Percentage error in momentum is %f percent.\"%(per_error)\n", - "# Answer in book is 1.08 percentage\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.12\n", - " Percentage error in momentum is 1.093108 percent.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.13 Page No95" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "del_x = 2e-9 # Uncertainty in position in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.13\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = h_bar/(2*del_x) # Calculation of uncertainty in momentum\n", - "del_v = del_p/m\n", - "print \" Uncertainty in velocity of particle is %e m/s.\"%(del_v)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.13\n", - " Uncertainty in velocity of particle is 2.898894e+04 m/s.\n" - ] - } - ], - "prompt_number": 17 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.15 Page No97" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "del_x = 5000 # Uncertainty in position in angstrom\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 200 # mass of ball in gram\n", - "v = 6 # velocity of moving ball in m/s\n", - "print \"Example 2.15\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = h_bar/(2*del_x*1e-10) # Calculation of uncertainty in momentum\n", - "p = m*v/1000 # Calculation of momentum\n", - "per_error = del_p*100/p # Calculation of percentage error in calculation of momentum\n", - "print \" Uncertainty in momentum of ball is %e kgm/s.\"%(del_p)\n", - "print \" Percentage error in calculation of momentum is %e.\"%(per_error)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.15\n", - " Uncertainty in momentum of ball is 1.055197e-28 kgm/s.\n", - " Percentage error in calculation of momentum is 1.055197e-26.\n" - ] - } - ], - "prompt_number": 18 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.16 Page No98" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "c = 3e8 # speed of light in m/s\n", - "v = c/10 # Velocity of moving proton in m/s\n", - "v_error = 1 # Percentage error in measurement of velocity \n", - "m = 1.67e-27 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "print \"Example 2.16\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_v = v*v_error/100# calculation of uncertainty in position\n", - "del_x = h_bar/(2*m*del_v) # calculation of uncertainty in momentum\n", - "print \" Uncertainty in position of particle is %e m.\"%(del_x)\n", - "# Answer in book is 1.04e-13 m\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.16\n", - " Uncertainty in position of particle is 1.053091e-13 m.\n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.17 Page No98" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "del_x = 1e-9 # Uncertainty in position in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 200 # mass of ball in gram\n", - "print \"Example 2.17\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_v = h_bar/(2*del_x*m/1000) # Calculation of uncertainty in momentum\n", - "print \" Uncertainty in velocity of ball is %e m/s.\"%(del_v)\n", - "# Answer in book is 2.64e-25 m/s\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.17\n", - " Uncertainty in velocity of ball is 2.637993e-25 m/s.\n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.18 Page No102" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_t = 2e-12 # lifetime of exited state in sec\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.18\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_E = h_bar/(1.6e-19*2*del_t) # calculation of uncertainty in momentum\n", - "print \" Minimum error in measurement of energy of this state is %e eV.\"%(del_E)\n", - "# Answer in book is 1.65e-4 eV\n", - "\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.18\n", - " Minimum error in measurement of energy of this state is 1.648746e-04 eV.\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.19 Page No107" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_t = 1e-8 # lifetime of exited state in sec\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.19\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_nu = h_bar/(2*del_t*h) # calculation of uncertainty in frequency\n", - "print \" Minimum error in measurement of frequency of photon is %e per second.\"%(del_nu)\n", - "# Answer in book is 8e6 per second\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.19\n", - " Minimum error in measurement of frequency of photon is 7.957747e+06 per second.\n" - ] - } - ], - "prompt_number": 22 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.20 Page No108" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_v = 5.5e-20 # Uncertainty in velocity in m/s\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 1 # mass of dust particle in mg\n", - "print \"Example 2.20\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_x = h_bar/(2*del_v*m*1e-6) # Calculation of uncertainty in momentum\n", - "print \" Uncertainty in position of ball is %f angstrom.\"%(del_x*1e10)\n", - "# Answer in book is 9.6 angstrom\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.20\n", - " Uncertainty in position of ball is 9.592702 angstrom.\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.21 Page No110" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#given that\n", - "l = 1 # width of potential well in angstrom\n", - "n = 1 # order corresponding to ground state\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.21\"\n", - "E = n**2*h**2/(8*m*(l*1e-10)**2) # Calculation of energy in Joule\n", - "E_eV = E/1.6e-19 # Calculation of energy in eV\n", - "\n", - "print \" Energy of electron is %f eV.\"%(E_eV)\n", - "# Answer in book is 37.74 eV angstrom\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.21\n", - " Energy of electron is 37.737723 eV.\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.24 Page No113" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "l = 2.5e-10 # width of potential well in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.24\"\n", - "for n in range(1,3):\n", - " E = n**2*h**2/(8*m*l**2) # Calculation of energy in Joule\n", - " E_eV = E/1.6e-19 # Calculation of energy in eV\n", - " print \" Energy of electron for state %d is %f eV.\"%(n,E_eV);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.24\n", - " Energy of electron for state 1 is 6.038036 eV.\n", - " Energy of electron for state 2 is 24.152143 eV.\n" - ] - } - ], - "prompt_number": 33 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.26 Page No117" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# given that\n", - "L = 1# let unit length\n", - "l1 = 0.45*L # initial point\n", - "l2 = 0.55*L # Final point\n", - "\n", - "\n", - "print \"Example 2.26 \"\n", - "p = (1/L)*((l2-(L/(2*math.pi) *math.sin(2*l2*math.pi/L)))- (l1-(L/(2*math.pi) *math.sin(2*l1*math.pi/L)))) # Calculation of probability of finding particle\n", - "p_per = p*100 # probability of finding particle in percentage\n", - "print \" Probability of finding electron between %fL and %fL is %f percent.\"%(l2,l1,p_per)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.26 \n", - " Probability of finding electron between 0.550000L and 0.450000L is 19.836316 percent.\n" - ] - } - ], - "prompt_number": 30 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.27 Page No117" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#given that\n", - "l = 1e-8 # width of potential well in cm\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.27\"\n", - "E_1 = (h)**2/(8*m*(l*1e-2)**2) # Calculation of energy of ground state in Joule\n", - "E_1_eV = E_1/1.6e-19 # Calculation of energy in eV\n", - "E_2 = (2)**2*h**2/(8*m*(l*1e-2)**2) # Calculation of energy of first state in Joule\n", - "E_2_eV = E_2/1.6e-19 # Calculation of energy in eV\n", - "del_E = E_2_eV - E_1_eV # calculation of difference between first state and ground state\n", - "print \" Difference between first state and ground state energies is %f eV.\"%(del_E);\n", - "# Answer in book is 113.04 eV\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.27\n", - " Difference between first state and ground state energies is 113.213170 eV.\n" - ] - } - ], - "prompt_number": 31 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.28 Page No121" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "l = 1 # width of potential well in angstrom\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.28\"\n", - "for n in range(1,4):\n", - " lambda1 = 2.*l/n # Calculation of wavelength\n", - " E = n**2*h**2/(8*m*(l*1e-10)**2) # Calculation of energy in Joule\n", - " E_eV = E/1.6e-19 # Calculation of energy in eV\n", - " print \" For state:%d Energy is %f eV & wavelength is %f angstrom \"%(n,E_eV,lambda1);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.28\n", - " For state:1 Energy is 37.737723 eV & wavelength is 2.000000 angstrom \n", - " For state:2 Energy is 150.950893 eV & wavelength is 1.000000 angstrom \n", - " For state:3 Energy is 339.639509 eV & wavelength is 0.666667 angstrom \n" - ] - } - ], - "prompt_number": 32 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.29 Page No122" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "m = 100 #mass of ball in gram\n", - "l = 1 # length of box in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.29\"\n", - "for n in range(1,4):\n", - " E = (n**2*h**2)/(8*m*1e-3*l**2*1.6e-19)\n", - " print \" Energy state E%d of ball is %e eV\"%(n,E)\n", - "\n", - "print \" As energy difference is very small so we cannot see energy states.\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.29\n", - " Energy state E1 of ball is 3.434133e-48 eV\n", - " Energy state E2 of ball is 1.373653e-47 eV\n", - " Energy state E3 of ball is 3.090720e-47 eV\n", - " As energy difference is very small so we cannot see energy states.\n" - ] - } - ], - "prompt_number": 34 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.30 Page No124" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "l = 30. # width of potential well in angstrom\n", - "x = l/2\n", - "del_x = 2 # interval of length at centre in angstrom\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "n = 1 # ground state\n", - "print \"Example 2.30\"\n", - "phi_x = ((math.sqrt(2/l))*math.sin(n*math.pi*x/l))**2 \n", - "p = phi_x*del_x # Calculation of probability at centre\n", - "print \" Probability of finding particle at centre is %d percent.\"%(p*100)\n", - "# Answer given in book is 16 percent. It is due to wrong calculation \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.30\n", - " Probability of finding particle at centre is 13 percent.\n" - ] - } - ], - "prompt_number": 37 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch2_1.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch2_1.ipynb deleted file mode 100755 index dc4ec4d6..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch2_1.ipynb +++ /dev/null @@ -1,1017 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:5ca2861748d43d40b34516562fe47bc041c8b35f6ea545c8e5309115cbc73a04" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 2 : Uncertainty Principle and Schrodinger wave Equation" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.1 Page No71" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_x = 0.2 # Uncertainty in position in angstrom\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "print \"Example 2.1\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = h_bar/(2*del_x*1e-10) # Calculation of uncertainty in momentum\n", - "print \" Uncertainty in momentum of particle is %e kgm/sec \"%(del_p)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.1\n", - " Uncertainty in momentum of particle is 2.637993e-24 kgm/sec \n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.2 Page No75" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#given that\n", - "del_x = 4e-10 # Uncertainty in position in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "print \"Example 2.2\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = h_bar/(2*del_x) # Calculation of uncertainty in momentum\n", - "print \" Uncertainty in momentum of particle is %e kgm/sec.\"%(del_p)\n", - "# Answer in book is given as 1.32e-23 kgm/sec\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.2\n", - " Uncertainty in momentum of particle is 1.318997e-25 kgm/sec.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.3 Page No75" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "v = 3e7 # Velocity of moving electron in m/s\n", - "m = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "c = 3e8 # speed of light in m/s\n", - "print \"Example 2.3\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = m*v/(math.sqrt(1-(v/c)**2)) # calculation of uncertainty in momentum\n", - "del_x = h_bar/(2*del_p) # Calculation of uncertainty in position\n", - "print \" Uncertainty in position of particle is %f angstrom.\"%(del_x*1e10)\n", - "#Answer in book is 0.0194 angstrom which is due to umath.sing approximate values at intermediate steps\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.3\n", - " Uncertainty in position of particle is 0.019229 angstrom.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.5 Page No80" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "v = 1.05e4 # Velocity of moving electron in m/s\n", - "v_error = 0.02 #Percentage error in measurement of velocity\n", - "\n", - "m = 9e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.5\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "p = m*v\n", - "del_p = v_error*p/100 # calculation of uncertainty in momentum\n", - "del_x = h_bar/del_p\n", - "print \" Uncertainty in position of particle is %e m.\"%(del_x)\n", - "# Answer in book is given as 5.58e-3 m\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.5\n", - " Uncertainty in position of particle is 5.583054e-05 m.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.6 Page No82" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "v = 600 # Velocity of moving electron in m/s\n", - "v_error = 0.005 #Percentage error in measurement of velocity\n", - "m = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.6\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "p = m*v\n", - "del_p = v_error*p/100 # calculation of uncertainty in momentum\n", - "del_x = h_bar/(del_p) # Calculation of uncertainty in position\n", - "print \" Uncertainty in position of particle is %e m.\"%(del_x)\n", - "# Answer in book is 0.39e-2 m \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.6\n", - " Uncertainty in position of particle is 3.865191e-03 m.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.7 Page No82" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_x = 1 # let uncertainty in position is unity\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "m_p = 1.67e-27 # mass of proton in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.7\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_v_ratio = m_p/m_e # calculation in uncertainties in the velocity of electron and proton\n", - "print \" Ratio of uncertainties in the velocity of electron to proton is %d.\"%(del_v_ratio)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.7\n", - " Ratio of uncertainties in the velocity of electron to proton is 1835.\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.8 Page No84" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "r = 0.5 # radius of hydrogen atom in angstrom\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.8\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_x = 2*r # calculation of uncertainty in position\n", - "del_p = h_bar/(2*del_x*1e-10) # calculation of uncertainty in momentum\n", - "p = del_p\n", - "E = p**2/(2*m_e*1.6e-19)# Calculation of energy in eV\n", - "print \" Kinetic energy needed by an electron to be confined in electron is %.f eV.\"%((E*100)/100)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.8\n", - " Kinetic energy needed by an electron to be confined in electron is 1 eV.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.9 Page No89" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "v = 5e3 # Velocity of moving electron in m/s\n", - "v_error = 0.003 #Percentage error in measurement of velocity\n", - "\n", - "m = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.9\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "p = m*v\n", - "del_p = v_error*p/100 # calculation of uncertainty in momentum\n", - "del_x = h_bar/(2*del_p) # Calculation of uncertainty in position\n", - "print \" Uncertainty in position of particle is %e m.\"%(del_x)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.9\n", - " Uncertainty in position of particle is 3.865191e-04 m.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.10 Page No90" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "r = 0.53 # radius of hydrogen atom in angstrom\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.10\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_x = 2*r # calculation of uncertainty in position\n", - "del_p = h_bar/(2*del_x*1e-10) # calculation of uncertainty in momentum\n", - "p = del_p\n", - "E = p**2/(2*m_e*1.6e-19)# Calculation of energy in eV\n", - "print \" Kinetic energy needed by an electron to be confined in electron is %f eV.\"%(E)\n", - "# When problem is solved by del_x*del_p = h_bar, then minimum value of kinetic energy will become 13.6eV\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.10\n", - " Kinetic energy needed by an electron to be confined in electron is 0.850754 eV.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.11 Page No92" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "del_t = 2.5e-14 # lifetime in exited state in micro sec\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.11\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_E = h_bar/(1.6e-19*del_t*1e-6) # calculation of uncertainty in momentum\n", - "print \" Minimum error in measurement of energy of this state is %e eV.\"%(del_E)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.11\n", - " Minimum error in measurement of energy of this state is 2.637993e+04 eV.\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.12 Page No92" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "E_eV = 0.5# kinetic energy of electron in KeV\n", - "del_x = 0.4 # Uncertainty in position in nm\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in kg\n", - "print \"Example 2.12\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "E_J = E_eV*1e3*1.6e-19\n", - "p = math.sqrt(2*m*E_J) # Calculation of momentum in kgm/s\n", - "del_p = h_bar/(2*del_x*1e-9) # Calculation of uncertainty in momentum\n", - "per_error = del_p*100 / p # calculation of percentage error in momentum\n", - "print \" Percentage error in momentum is %f percent.\"%(per_error)\n", - "# Answer in book is 1.08 percentage\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.12\n", - " Percentage error in momentum is 1.093108 percent.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.13 Page No95" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "del_x = 2e-9 # Uncertainty in position in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.13\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = h_bar/(2*del_x) # Calculation of uncertainty in momentum\n", - "del_v = del_p/m\n", - "print \" Uncertainty in velocity of particle is %e m/s.\"%(del_v)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.13\n", - " Uncertainty in velocity of particle is 2.898894e+04 m/s.\n" - ] - } - ], - "prompt_number": 17 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.15 Page No97" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "del_x = 5000 # Uncertainty in position in angstrom\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 200 # mass of ball in gram\n", - "v = 6 # velocity of moving ball in m/s\n", - "print \"Example 2.15\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = h_bar/(2*del_x*1e-10) # Calculation of uncertainty in momentum\n", - "p = m*v/1000 # Calculation of momentum\n", - "per_error = del_p*100/p # Calculation of percentage error in calculation of momentum\n", - "print \" Uncertainty in momentum of ball is %e kgm/s.\"%(del_p)\n", - "print \" Percentage error in calculation of momentum is %e.\"%(per_error)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.15\n", - " Uncertainty in momentum of ball is 1.055197e-28 kgm/s.\n", - " Percentage error in calculation of momentum is 1.055197e-26.\n" - ] - } - ], - "prompt_number": 18 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.16 Page No98" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "c = 3e8 # speed of light in m/s\n", - "v = c/10 # Velocity of moving proton in m/s\n", - "v_error = 1 # Percentage error in measurement of velocity \n", - "m = 1.67e-27 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "print \"Example 2.16\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_v = v*v_error/100# calculation of uncertainty in position\n", - "del_x = h_bar/(2*m*del_v) # calculation of uncertainty in momentum\n", - "print \" Uncertainty in position of particle is %e m.\"%(del_x)\n", - "# Answer in book is 1.04e-13 m\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.16\n", - " Uncertainty in position of particle is 1.053091e-13 m.\n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.17 Page No98" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "del_x = 1e-9 # Uncertainty in position in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 200 # mass of ball in gram\n", - "print \"Example 2.17\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_v = h_bar/(2*del_x*m/1000) # Calculation of uncertainty in momentum\n", - "print \" Uncertainty in velocity of ball is %e m/s.\"%(del_v)\n", - "# Answer in book is 2.64e-25 m/s\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.17\n", - " Uncertainty in velocity of ball is 2.637993e-25 m/s.\n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.18 Page No102" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_t = 2e-12 # lifetime of exited state in sec\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.18\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_E = h_bar/(1.6e-19*2*del_t) # calculation of uncertainty in momentum\n", - "print \" Minimum error in measurement of energy of this state is %e eV.\"%(del_E)\n", - "# Answer in book is 1.65e-4 eV\n", - "\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.18\n", - " Minimum error in measurement of energy of this state is 1.648746e-04 eV.\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.19 Page No107" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_t = 1e-8 # lifetime of exited state in sec\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.19\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_nu = h_bar/(2*del_t*h) # calculation of uncertainty in frequency\n", - "print \" Minimum error in measurement of frequency of photon is %e per second.\"%(del_nu)\n", - "# Answer in book is 8e6 per second\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.19\n", - " Minimum error in measurement of frequency of photon is 7.957747e+06 per second.\n" - ] - } - ], - "prompt_number": 22 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.20 Page No108" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_v = 5.5e-20 # Uncertainty in velocity in m/s\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 1 # mass of dust particle in mg\n", - "print \"Example 2.20\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_x = h_bar/(2*del_v*m*1e-6) # Calculation of uncertainty in momentum\n", - "print \" Uncertainty in position of ball is %f angstrom.\"%(del_x*1e10)\n", - "# Answer in book is 9.6 angstrom\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.20\n", - " Uncertainty in position of ball is 9.592702 angstrom.\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.21 Page No110" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#given that\n", - "l = 1 # width of potential well in angstrom\n", - "n = 1 # order corresponding to ground state\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.21\"\n", - "E = n**2*h**2/(8*m*(l*1e-10)**2) # Calculation of energy in Joule\n", - "E_eV = E/1.6e-19 # Calculation of energy in eV\n", - "\n", - "print \" Energy of electron is %f eV.\"%(E_eV)\n", - "# Answer in book is 37.74 eV angstrom\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.21\n", - " Energy of electron is 37.737723 eV.\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.24 Page No113" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "l = 2.5e-10 # width of potential well in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.24\"\n", - "for n in range(1,3):\n", - " E = n**2*h**2/(8*m*l**2) # Calculation of energy in Joule\n", - " E_eV = E/1.6e-19 # Calculation of energy in eV\n", - " print \" Energy of electron for state %d is %f eV.\"%(n,E_eV);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.24\n", - " Energy of electron for state 1 is 6.038036 eV.\n", - " Energy of electron for state 2 is 24.152143 eV.\n" - ] - } - ], - "prompt_number": 33 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.26 Page No117" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# given that\n", - "L = 1# let unit length\n", - "l1 = 0.45*L # initial point\n", - "l2 = 0.55*L # Final point\n", - "\n", - "\n", - "print \"Example 2.26 \"\n", - "p = (1/L)*((l2-(L/(2*math.pi) *math.sin(2*l2*math.pi/L)))- (l1-(L/(2*math.pi) *math.sin(2*l1*math.pi/L)))) # Calculation of probability of finding particle\n", - "p_per = p*100 # probability of finding particle in percentage\n", - "print \" Probability of finding electron between %fL and %fL is %f percent.\"%(l2,l1,p_per)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.26 \n", - " Probability of finding electron between 0.550000L and 0.450000L is 19.836316 percent.\n" - ] - } - ], - "prompt_number": 30 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.27 Page No117" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#given that\n", - "l = 1e-8 # width of potential well in cm\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.27\"\n", - "E_1 = (h)**2/(8*m*(l*1e-2)**2) # Calculation of energy of ground state in Joule\n", - "E_1_eV = E_1/1.6e-19 # Calculation of energy in eV\n", - "E_2 = (2)**2*h**2/(8*m*(l*1e-2)**2) # Calculation of energy of first state in Joule\n", - "E_2_eV = E_2/1.6e-19 # Calculation of energy in eV\n", - "del_E = E_2_eV - E_1_eV # calculation of difference between first state and ground state\n", - "print \" Difference between first state and ground state energies is %f eV.\"%(del_E);\n", - "# Answer in book is 113.04 eV\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.27\n", - " Difference between first state and ground state energies is 113.213170 eV.\n" - ] - } - ], - "prompt_number": 31 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.28 Page No121" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "l = 1 # width of potential well in angstrom\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.28\"\n", - "for n in range(1,4):\n", - " lambda1 = 2.*l/n # Calculation of wavelength\n", - " E = n**2*h**2/(8*m*(l*1e-10)**2) # Calculation of energy in Joule\n", - " E_eV = E/1.6e-19 # Calculation of energy in eV\n", - " print \" For state:%d Energy is %f eV & wavelength is %f angstrom \"%(n,E_eV,lambda1);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.28\n", - " For state:1 Energy is 37.737723 eV & wavelength is 2.000000 angstrom \n", - " For state:2 Energy is 150.950893 eV & wavelength is 1.000000 angstrom \n", - " For state:3 Energy is 339.639509 eV & wavelength is 0.666667 angstrom \n" - ] - } - ], - "prompt_number": 32 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.29 Page No122" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "m = 100 #mass of ball in gram\n", - "l = 1 # length of box in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.29\"\n", - "for n in range(1,4):\n", - " E = (n**2*h**2)/(8*m*1e-3*l**2*1.6e-19)\n", - " print \" Energy state E%d of ball is %e eV\"%(n,E)\n", - "\n", - "print \" As energy difference is very small so we cannot see energy states.\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.29\n", - " Energy state E1 of ball is 3.434133e-48 eV\n", - " Energy state E2 of ball is 1.373653e-47 eV\n", - " Energy state E3 of ball is 3.090720e-47 eV\n", - " As energy difference is very small so we cannot see energy states.\n" - ] - } - ], - "prompt_number": 34 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.30 Page No124" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "l = 30. # width of potential well in angstrom\n", - "x = l/2\n", - "del_x = 2 # interval of length at centre in angstrom\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "n = 1 # ground state\n", - "print \"Example 2.30\"\n", - "phi_x = ((math.sqrt(2/l))*math.sin(n*math.pi*x/l))**2 \n", - "p = phi_x*del_x # Calculation of probability at centre\n", - "print \" Probability of finding particle at centre is %d percent.\"%(p*100)\n", - "# Answer given in book is 16 percent. It is due to wrong calculation \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.30\n", - " Probability of finding particle at centre is 13 percent.\n" - ] - } - ], - "prompt_number": 37 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch2_2.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch2_2.ipynb deleted file mode 100644 index dc4ec4d6..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch2_2.ipynb +++ /dev/null @@ -1,1017 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:5ca2861748d43d40b34516562fe47bc041c8b35f6ea545c8e5309115cbc73a04" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 2 : Uncertainty Principle and Schrodinger wave Equation" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.1 Page No71" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_x = 0.2 # Uncertainty in position in angstrom\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "print \"Example 2.1\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = h_bar/(2*del_x*1e-10) # Calculation of uncertainty in momentum\n", - "print \" Uncertainty in momentum of particle is %e kgm/sec \"%(del_p)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.1\n", - " Uncertainty in momentum of particle is 2.637993e-24 kgm/sec \n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.2 Page No75" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#given that\n", - "del_x = 4e-10 # Uncertainty in position in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "print \"Example 2.2\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = h_bar/(2*del_x) # Calculation of uncertainty in momentum\n", - "print \" Uncertainty in momentum of particle is %e kgm/sec.\"%(del_p)\n", - "# Answer in book is given as 1.32e-23 kgm/sec\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.2\n", - " Uncertainty in momentum of particle is 1.318997e-25 kgm/sec.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.3 Page No75" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "v = 3e7 # Velocity of moving electron in m/s\n", - "m = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "c = 3e8 # speed of light in m/s\n", - "print \"Example 2.3\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = m*v/(math.sqrt(1-(v/c)**2)) # calculation of uncertainty in momentum\n", - "del_x = h_bar/(2*del_p) # Calculation of uncertainty in position\n", - "print \" Uncertainty in position of particle is %f angstrom.\"%(del_x*1e10)\n", - "#Answer in book is 0.0194 angstrom which is due to umath.sing approximate values at intermediate steps\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.3\n", - " Uncertainty in position of particle is 0.019229 angstrom.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.5 Page No80" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "v = 1.05e4 # Velocity of moving electron in m/s\n", - "v_error = 0.02 #Percentage error in measurement of velocity\n", - "\n", - "m = 9e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.5\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "p = m*v\n", - "del_p = v_error*p/100 # calculation of uncertainty in momentum\n", - "del_x = h_bar/del_p\n", - "print \" Uncertainty in position of particle is %e m.\"%(del_x)\n", - "# Answer in book is given as 5.58e-3 m\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.5\n", - " Uncertainty in position of particle is 5.583054e-05 m.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.6 Page No82" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "v = 600 # Velocity of moving electron in m/s\n", - "v_error = 0.005 #Percentage error in measurement of velocity\n", - "m = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.6\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "p = m*v\n", - "del_p = v_error*p/100 # calculation of uncertainty in momentum\n", - "del_x = h_bar/(del_p) # Calculation of uncertainty in position\n", - "print \" Uncertainty in position of particle is %e m.\"%(del_x)\n", - "# Answer in book is 0.39e-2 m \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.6\n", - " Uncertainty in position of particle is 3.865191e-03 m.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.7 Page No82" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_x = 1 # let uncertainty in position is unity\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "m_p = 1.67e-27 # mass of proton in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.7\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_v_ratio = m_p/m_e # calculation in uncertainties in the velocity of electron and proton\n", - "print \" Ratio of uncertainties in the velocity of electron to proton is %d.\"%(del_v_ratio)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.7\n", - " Ratio of uncertainties in the velocity of electron to proton is 1835.\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.8 Page No84" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "r = 0.5 # radius of hydrogen atom in angstrom\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.8\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_x = 2*r # calculation of uncertainty in position\n", - "del_p = h_bar/(2*del_x*1e-10) # calculation of uncertainty in momentum\n", - "p = del_p\n", - "E = p**2/(2*m_e*1.6e-19)# Calculation of energy in eV\n", - "print \" Kinetic energy needed by an electron to be confined in electron is %.f eV.\"%((E*100)/100)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.8\n", - " Kinetic energy needed by an electron to be confined in electron is 1 eV.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.9 Page No89" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "v = 5e3 # Velocity of moving electron in m/s\n", - "v_error = 0.003 #Percentage error in measurement of velocity\n", - "\n", - "m = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.9\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "p = m*v\n", - "del_p = v_error*p/100 # calculation of uncertainty in momentum\n", - "del_x = h_bar/(2*del_p) # Calculation of uncertainty in position\n", - "print \" Uncertainty in position of particle is %e m.\"%(del_x)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.9\n", - " Uncertainty in position of particle is 3.865191e-04 m.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.10 Page No90" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "r = 0.53 # radius of hydrogen atom in angstrom\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.10\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_x = 2*r # calculation of uncertainty in position\n", - "del_p = h_bar/(2*del_x*1e-10) # calculation of uncertainty in momentum\n", - "p = del_p\n", - "E = p**2/(2*m_e*1.6e-19)# Calculation of energy in eV\n", - "print \" Kinetic energy needed by an electron to be confined in electron is %f eV.\"%(E)\n", - "# When problem is solved by del_x*del_p = h_bar, then minimum value of kinetic energy will become 13.6eV\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.10\n", - " Kinetic energy needed by an electron to be confined in electron is 0.850754 eV.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.11 Page No92" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "del_t = 2.5e-14 # lifetime in exited state in micro sec\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.11\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_E = h_bar/(1.6e-19*del_t*1e-6) # calculation of uncertainty in momentum\n", - "print \" Minimum error in measurement of energy of this state is %e eV.\"%(del_E)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.11\n", - " Minimum error in measurement of energy of this state is 2.637993e+04 eV.\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.12 Page No92" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "E_eV = 0.5# kinetic energy of electron in KeV\n", - "del_x = 0.4 # Uncertainty in position in nm\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in kg\n", - "print \"Example 2.12\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "E_J = E_eV*1e3*1.6e-19\n", - "p = math.sqrt(2*m*E_J) # Calculation of momentum in kgm/s\n", - "del_p = h_bar/(2*del_x*1e-9) # Calculation of uncertainty in momentum\n", - "per_error = del_p*100 / p # calculation of percentage error in momentum\n", - "print \" Percentage error in momentum is %f percent.\"%(per_error)\n", - "# Answer in book is 1.08 percentage\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.12\n", - " Percentage error in momentum is 1.093108 percent.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.13 Page No95" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "del_x = 2e-9 # Uncertainty in position in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.13\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = h_bar/(2*del_x) # Calculation of uncertainty in momentum\n", - "del_v = del_p/m\n", - "print \" Uncertainty in velocity of particle is %e m/s.\"%(del_v)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.13\n", - " Uncertainty in velocity of particle is 2.898894e+04 m/s.\n" - ] - } - ], - "prompt_number": 17 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.15 Page No97" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "del_x = 5000 # Uncertainty in position in angstrom\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 200 # mass of ball in gram\n", - "v = 6 # velocity of moving ball in m/s\n", - "print \"Example 2.15\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = h_bar/(2*del_x*1e-10) # Calculation of uncertainty in momentum\n", - "p = m*v/1000 # Calculation of momentum\n", - "per_error = del_p*100/p # Calculation of percentage error in calculation of momentum\n", - "print \" Uncertainty in momentum of ball is %e kgm/s.\"%(del_p)\n", - "print \" Percentage error in calculation of momentum is %e.\"%(per_error)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.15\n", - " Uncertainty in momentum of ball is 1.055197e-28 kgm/s.\n", - " Percentage error in calculation of momentum is 1.055197e-26.\n" - ] - } - ], - "prompt_number": 18 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.16 Page No98" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "c = 3e8 # speed of light in m/s\n", - "v = c/10 # Velocity of moving proton in m/s\n", - "v_error = 1 # Percentage error in measurement of velocity \n", - "m = 1.67e-27 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "print \"Example 2.16\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_v = v*v_error/100# calculation of uncertainty in position\n", - "del_x = h_bar/(2*m*del_v) # calculation of uncertainty in momentum\n", - "print \" Uncertainty in position of particle is %e m.\"%(del_x)\n", - "# Answer in book is 1.04e-13 m\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.16\n", - " Uncertainty in position of particle is 1.053091e-13 m.\n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.17 Page No98" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "del_x = 1e-9 # Uncertainty in position in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 200 # mass of ball in gram\n", - "print \"Example 2.17\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_v = h_bar/(2*del_x*m/1000) # Calculation of uncertainty in momentum\n", - "print \" Uncertainty in velocity of ball is %e m/s.\"%(del_v)\n", - "# Answer in book is 2.64e-25 m/s\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.17\n", - " Uncertainty in velocity of ball is 2.637993e-25 m/s.\n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.18 Page No102" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_t = 2e-12 # lifetime of exited state in sec\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.18\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_E = h_bar/(1.6e-19*2*del_t) # calculation of uncertainty in momentum\n", - "print \" Minimum error in measurement of energy of this state is %e eV.\"%(del_E)\n", - "# Answer in book is 1.65e-4 eV\n", - "\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.18\n", - " Minimum error in measurement of energy of this state is 1.648746e-04 eV.\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.19 Page No107" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_t = 1e-8 # lifetime of exited state in sec\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.19\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_nu = h_bar/(2*del_t*h) # calculation of uncertainty in frequency\n", - "print \" Minimum error in measurement of frequency of photon is %e per second.\"%(del_nu)\n", - "# Answer in book is 8e6 per second\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.19\n", - " Minimum error in measurement of frequency of photon is 7.957747e+06 per second.\n" - ] - } - ], - "prompt_number": 22 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.20 Page No108" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_v = 5.5e-20 # Uncertainty in velocity in m/s\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 1 # mass of dust particle in mg\n", - "print \"Example 2.20\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_x = h_bar/(2*del_v*m*1e-6) # Calculation of uncertainty in momentum\n", - "print \" Uncertainty in position of ball is %f angstrom.\"%(del_x*1e10)\n", - "# Answer in book is 9.6 angstrom\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.20\n", - " Uncertainty in position of ball is 9.592702 angstrom.\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.21 Page No110" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#given that\n", - "l = 1 # width of potential well in angstrom\n", - "n = 1 # order corresponding to ground state\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.21\"\n", - "E = n**2*h**2/(8*m*(l*1e-10)**2) # Calculation of energy in Joule\n", - "E_eV = E/1.6e-19 # Calculation of energy in eV\n", - "\n", - "print \" Energy of electron is %f eV.\"%(E_eV)\n", - "# Answer in book is 37.74 eV angstrom\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.21\n", - " Energy of electron is 37.737723 eV.\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.24 Page No113" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "l = 2.5e-10 # width of potential well in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.24\"\n", - "for n in range(1,3):\n", - " E = n**2*h**2/(8*m*l**2) # Calculation of energy in Joule\n", - " E_eV = E/1.6e-19 # Calculation of energy in eV\n", - " print \" Energy of electron for state %d is %f eV.\"%(n,E_eV);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.24\n", - " Energy of electron for state 1 is 6.038036 eV.\n", - " Energy of electron for state 2 is 24.152143 eV.\n" - ] - } - ], - "prompt_number": 33 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.26 Page No117" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# given that\n", - "L = 1# let unit length\n", - "l1 = 0.45*L # initial point\n", - "l2 = 0.55*L # Final point\n", - "\n", - "\n", - "print \"Example 2.26 \"\n", - "p = (1/L)*((l2-(L/(2*math.pi) *math.sin(2*l2*math.pi/L)))- (l1-(L/(2*math.pi) *math.sin(2*l1*math.pi/L)))) # Calculation of probability of finding particle\n", - "p_per = p*100 # probability of finding particle in percentage\n", - "print \" Probability of finding electron between %fL and %fL is %f percent.\"%(l2,l1,p_per)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.26 \n", - " Probability of finding electron between 0.550000L and 0.450000L is 19.836316 percent.\n" - ] - } - ], - "prompt_number": 30 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.27 Page No117" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#given that\n", - "l = 1e-8 # width of potential well in cm\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.27\"\n", - "E_1 = (h)**2/(8*m*(l*1e-2)**2) # Calculation of energy of ground state in Joule\n", - "E_1_eV = E_1/1.6e-19 # Calculation of energy in eV\n", - "E_2 = (2)**2*h**2/(8*m*(l*1e-2)**2) # Calculation of energy of first state in Joule\n", - "E_2_eV = E_2/1.6e-19 # Calculation of energy in eV\n", - "del_E = E_2_eV - E_1_eV # calculation of difference between first state and ground state\n", - "print \" Difference between first state and ground state energies is %f eV.\"%(del_E);\n", - "# Answer in book is 113.04 eV\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.27\n", - " Difference between first state and ground state energies is 113.213170 eV.\n" - ] - } - ], - "prompt_number": 31 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.28 Page No121" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "l = 1 # width of potential well in angstrom\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.28\"\n", - "for n in range(1,4):\n", - " lambda1 = 2.*l/n # Calculation of wavelength\n", - " E = n**2*h**2/(8*m*(l*1e-10)**2) # Calculation of energy in Joule\n", - " E_eV = E/1.6e-19 # Calculation of energy in eV\n", - " print \" For state:%d Energy is %f eV & wavelength is %f angstrom \"%(n,E_eV,lambda1);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.28\n", - " For state:1 Energy is 37.737723 eV & wavelength is 2.000000 angstrom \n", - " For state:2 Energy is 150.950893 eV & wavelength is 1.000000 angstrom \n", - " For state:3 Energy is 339.639509 eV & wavelength is 0.666667 angstrom \n" - ] - } - ], - "prompt_number": 32 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.29 Page No122" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "m = 100 #mass of ball in gram\n", - "l = 1 # length of box in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.29\"\n", - "for n in range(1,4):\n", - " E = (n**2*h**2)/(8*m*1e-3*l**2*1.6e-19)\n", - " print \" Energy state E%d of ball is %e eV\"%(n,E)\n", - "\n", - "print \" As energy difference is very small so we cannot see energy states.\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.29\n", - " Energy state E1 of ball is 3.434133e-48 eV\n", - " Energy state E2 of ball is 1.373653e-47 eV\n", - " Energy state E3 of ball is 3.090720e-47 eV\n", - " As energy difference is very small so we cannot see energy states.\n" - ] - } - ], - "prompt_number": 34 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.30 Page No124" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "l = 30. # width of potential well in angstrom\n", - "x = l/2\n", - "del_x = 2 # interval of length at centre in angstrom\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "n = 1 # ground state\n", - "print \"Example 2.30\"\n", - "phi_x = ((math.sqrt(2/l))*math.sin(n*math.pi*x/l))**2 \n", - "p = phi_x*del_x # Calculation of probability at centre\n", - "print \" Probability of finding particle at centre is %d percent.\"%(p*100)\n", - "# Answer given in book is 16 percent. It is due to wrong calculation \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.30\n", - " Probability of finding particle at centre is 13 percent.\n" - ] - } - ], - "prompt_number": 37 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch2_3.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch2_3.ipynb deleted file mode 100644 index dc4ec4d6..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch2_3.ipynb +++ /dev/null @@ -1,1017 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:5ca2861748d43d40b34516562fe47bc041c8b35f6ea545c8e5309115cbc73a04" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 2 : Uncertainty Principle and Schrodinger wave Equation" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.1 Page No71" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_x = 0.2 # Uncertainty in position in angstrom\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "print \"Example 2.1\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = h_bar/(2*del_x*1e-10) # Calculation of uncertainty in momentum\n", - "print \" Uncertainty in momentum of particle is %e kgm/sec \"%(del_p)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.1\n", - " Uncertainty in momentum of particle is 2.637993e-24 kgm/sec \n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.2 Page No75" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#given that\n", - "del_x = 4e-10 # Uncertainty in position in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "print \"Example 2.2\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = h_bar/(2*del_x) # Calculation of uncertainty in momentum\n", - "print \" Uncertainty in momentum of particle is %e kgm/sec.\"%(del_p)\n", - "# Answer in book is given as 1.32e-23 kgm/sec\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.2\n", - " Uncertainty in momentum of particle is 1.318997e-25 kgm/sec.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.3 Page No75" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "v = 3e7 # Velocity of moving electron in m/s\n", - "m = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "c = 3e8 # speed of light in m/s\n", - "print \"Example 2.3\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = m*v/(math.sqrt(1-(v/c)**2)) # calculation of uncertainty in momentum\n", - "del_x = h_bar/(2*del_p) # Calculation of uncertainty in position\n", - "print \" Uncertainty in position of particle is %f angstrom.\"%(del_x*1e10)\n", - "#Answer in book is 0.0194 angstrom which is due to umath.sing approximate values at intermediate steps\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.3\n", - " Uncertainty in position of particle is 0.019229 angstrom.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.5 Page No80" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "v = 1.05e4 # Velocity of moving electron in m/s\n", - "v_error = 0.02 #Percentage error in measurement of velocity\n", - "\n", - "m = 9e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.5\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "p = m*v\n", - "del_p = v_error*p/100 # calculation of uncertainty in momentum\n", - "del_x = h_bar/del_p\n", - "print \" Uncertainty in position of particle is %e m.\"%(del_x)\n", - "# Answer in book is given as 5.58e-3 m\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.5\n", - " Uncertainty in position of particle is 5.583054e-05 m.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.6 Page No82" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "v = 600 # Velocity of moving electron in m/s\n", - "v_error = 0.005 #Percentage error in measurement of velocity\n", - "m = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.6\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "p = m*v\n", - "del_p = v_error*p/100 # calculation of uncertainty in momentum\n", - "del_x = h_bar/(del_p) # Calculation of uncertainty in position\n", - "print \" Uncertainty in position of particle is %e m.\"%(del_x)\n", - "# Answer in book is 0.39e-2 m \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.6\n", - " Uncertainty in position of particle is 3.865191e-03 m.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.7 Page No82" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_x = 1 # let uncertainty in position is unity\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "m_p = 1.67e-27 # mass of proton in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.7\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_v_ratio = m_p/m_e # calculation in uncertainties in the velocity of electron and proton\n", - "print \" Ratio of uncertainties in the velocity of electron to proton is %d.\"%(del_v_ratio)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.7\n", - " Ratio of uncertainties in the velocity of electron to proton is 1835.\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.8 Page No84" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "r = 0.5 # radius of hydrogen atom in angstrom\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.8\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_x = 2*r # calculation of uncertainty in position\n", - "del_p = h_bar/(2*del_x*1e-10) # calculation of uncertainty in momentum\n", - "p = del_p\n", - "E = p**2/(2*m_e*1.6e-19)# Calculation of energy in eV\n", - "print \" Kinetic energy needed by an electron to be confined in electron is %.f eV.\"%((E*100)/100)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.8\n", - " Kinetic energy needed by an electron to be confined in electron is 1 eV.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.9 Page No89" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "v = 5e3 # Velocity of moving electron in m/s\n", - "v_error = 0.003 #Percentage error in measurement of velocity\n", - "\n", - "m = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.9\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "p = m*v\n", - "del_p = v_error*p/100 # calculation of uncertainty in momentum\n", - "del_x = h_bar/(2*del_p) # Calculation of uncertainty in position\n", - "print \" Uncertainty in position of particle is %e m.\"%(del_x)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.9\n", - " Uncertainty in position of particle is 3.865191e-04 m.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.10 Page No90" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "r = 0.53 # radius of hydrogen atom in angstrom\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.10\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_x = 2*r # calculation of uncertainty in position\n", - "del_p = h_bar/(2*del_x*1e-10) # calculation of uncertainty in momentum\n", - "p = del_p\n", - "E = p**2/(2*m_e*1.6e-19)# Calculation of energy in eV\n", - "print \" Kinetic energy needed by an electron to be confined in electron is %f eV.\"%(E)\n", - "# When problem is solved by del_x*del_p = h_bar, then minimum value of kinetic energy will become 13.6eV\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.10\n", - " Kinetic energy needed by an electron to be confined in electron is 0.850754 eV.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.11 Page No92" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "del_t = 2.5e-14 # lifetime in exited state in micro sec\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.11\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_E = h_bar/(1.6e-19*del_t*1e-6) # calculation of uncertainty in momentum\n", - "print \" Minimum error in measurement of energy of this state is %e eV.\"%(del_E)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.11\n", - " Minimum error in measurement of energy of this state is 2.637993e+04 eV.\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.12 Page No92" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "E_eV = 0.5# kinetic energy of electron in KeV\n", - "del_x = 0.4 # Uncertainty in position in nm\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in kg\n", - "print \"Example 2.12\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "E_J = E_eV*1e3*1.6e-19\n", - "p = math.sqrt(2*m*E_J) # Calculation of momentum in kgm/s\n", - "del_p = h_bar/(2*del_x*1e-9) # Calculation of uncertainty in momentum\n", - "per_error = del_p*100 / p # calculation of percentage error in momentum\n", - "print \" Percentage error in momentum is %f percent.\"%(per_error)\n", - "# Answer in book is 1.08 percentage\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.12\n", - " Percentage error in momentum is 1.093108 percent.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.13 Page No95" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "del_x = 2e-9 # Uncertainty in position in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.13\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = h_bar/(2*del_x) # Calculation of uncertainty in momentum\n", - "del_v = del_p/m\n", - "print \" Uncertainty in velocity of particle is %e m/s.\"%(del_v)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.13\n", - " Uncertainty in velocity of particle is 2.898894e+04 m/s.\n" - ] - } - ], - "prompt_number": 17 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.15 Page No97" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "del_x = 5000 # Uncertainty in position in angstrom\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 200 # mass of ball in gram\n", - "v = 6 # velocity of moving ball in m/s\n", - "print \"Example 2.15\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = h_bar/(2*del_x*1e-10) # Calculation of uncertainty in momentum\n", - "p = m*v/1000 # Calculation of momentum\n", - "per_error = del_p*100/p # Calculation of percentage error in calculation of momentum\n", - "print \" Uncertainty in momentum of ball is %e kgm/s.\"%(del_p)\n", - "print \" Percentage error in calculation of momentum is %e.\"%(per_error)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.15\n", - " Uncertainty in momentum of ball is 1.055197e-28 kgm/s.\n", - " Percentage error in calculation of momentum is 1.055197e-26.\n" - ] - } - ], - "prompt_number": 18 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.16 Page No98" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "c = 3e8 # speed of light in m/s\n", - "v = c/10 # Velocity of moving proton in m/s\n", - "v_error = 1 # Percentage error in measurement of velocity \n", - "m = 1.67e-27 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "print \"Example 2.16\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_v = v*v_error/100# calculation of uncertainty in position\n", - "del_x = h_bar/(2*m*del_v) # calculation of uncertainty in momentum\n", - "print \" Uncertainty in position of particle is %e m.\"%(del_x)\n", - "# Answer in book is 1.04e-13 m\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.16\n", - " Uncertainty in position of particle is 1.053091e-13 m.\n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.17 Page No98" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "del_x = 1e-9 # Uncertainty in position in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 200 # mass of ball in gram\n", - "print \"Example 2.17\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_v = h_bar/(2*del_x*m/1000) # Calculation of uncertainty in momentum\n", - "print \" Uncertainty in velocity of ball is %e m/s.\"%(del_v)\n", - "# Answer in book is 2.64e-25 m/s\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.17\n", - " Uncertainty in velocity of ball is 2.637993e-25 m/s.\n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.18 Page No102" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_t = 2e-12 # lifetime of exited state in sec\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.18\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_E = h_bar/(1.6e-19*2*del_t) # calculation of uncertainty in momentum\n", - "print \" Minimum error in measurement of energy of this state is %e eV.\"%(del_E)\n", - "# Answer in book is 1.65e-4 eV\n", - "\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.18\n", - " Minimum error in measurement of energy of this state is 1.648746e-04 eV.\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.19 Page No107" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_t = 1e-8 # lifetime of exited state in sec\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.19\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_nu = h_bar/(2*del_t*h) # calculation of uncertainty in frequency\n", - "print \" Minimum error in measurement of frequency of photon is %e per second.\"%(del_nu)\n", - "# Answer in book is 8e6 per second\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.19\n", - " Minimum error in measurement of frequency of photon is 7.957747e+06 per second.\n" - ] - } - ], - "prompt_number": 22 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.20 Page No108" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_v = 5.5e-20 # Uncertainty in velocity in m/s\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 1 # mass of dust particle in mg\n", - "print \"Example 2.20\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_x = h_bar/(2*del_v*m*1e-6) # Calculation of uncertainty in momentum\n", - "print \" Uncertainty in position of ball is %f angstrom.\"%(del_x*1e10)\n", - "# Answer in book is 9.6 angstrom\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.20\n", - " Uncertainty in position of ball is 9.592702 angstrom.\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.21 Page No110" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#given that\n", - "l = 1 # width of potential well in angstrom\n", - "n = 1 # order corresponding to ground state\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.21\"\n", - "E = n**2*h**2/(8*m*(l*1e-10)**2) # Calculation of energy in Joule\n", - "E_eV = E/1.6e-19 # Calculation of energy in eV\n", - "\n", - "print \" Energy of electron is %f eV.\"%(E_eV)\n", - "# Answer in book is 37.74 eV angstrom\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.21\n", - " Energy of electron is 37.737723 eV.\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.24 Page No113" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "l = 2.5e-10 # width of potential well in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.24\"\n", - "for n in range(1,3):\n", - " E = n**2*h**2/(8*m*l**2) # Calculation of energy in Joule\n", - " E_eV = E/1.6e-19 # Calculation of energy in eV\n", - " print \" Energy of electron for state %d is %f eV.\"%(n,E_eV);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.24\n", - " Energy of electron for state 1 is 6.038036 eV.\n", - " Energy of electron for state 2 is 24.152143 eV.\n" - ] - } - ], - "prompt_number": 33 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.26 Page No117" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# given that\n", - "L = 1# let unit length\n", - "l1 = 0.45*L # initial point\n", - "l2 = 0.55*L # Final point\n", - "\n", - "\n", - "print \"Example 2.26 \"\n", - "p = (1/L)*((l2-(L/(2*math.pi) *math.sin(2*l2*math.pi/L)))- (l1-(L/(2*math.pi) *math.sin(2*l1*math.pi/L)))) # Calculation of probability of finding particle\n", - "p_per = p*100 # probability of finding particle in percentage\n", - "print \" Probability of finding electron between %fL and %fL is %f percent.\"%(l2,l1,p_per)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.26 \n", - " Probability of finding electron between 0.550000L and 0.450000L is 19.836316 percent.\n" - ] - } - ], - "prompt_number": 30 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.27 Page No117" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#given that\n", - "l = 1e-8 # width of potential well in cm\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.27\"\n", - "E_1 = (h)**2/(8*m*(l*1e-2)**2) # Calculation of energy of ground state in Joule\n", - "E_1_eV = E_1/1.6e-19 # Calculation of energy in eV\n", - "E_2 = (2)**2*h**2/(8*m*(l*1e-2)**2) # Calculation of energy of first state in Joule\n", - "E_2_eV = E_2/1.6e-19 # Calculation of energy in eV\n", - "del_E = E_2_eV - E_1_eV # calculation of difference between first state and ground state\n", - "print \" Difference between first state and ground state energies is %f eV.\"%(del_E);\n", - "# Answer in book is 113.04 eV\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.27\n", - " Difference between first state and ground state energies is 113.213170 eV.\n" - ] - } - ], - "prompt_number": 31 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.28 Page No121" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "l = 1 # width of potential well in angstrom\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.28\"\n", - "for n in range(1,4):\n", - " lambda1 = 2.*l/n # Calculation of wavelength\n", - " E = n**2*h**2/(8*m*(l*1e-10)**2) # Calculation of energy in Joule\n", - " E_eV = E/1.6e-19 # Calculation of energy in eV\n", - " print \" For state:%d Energy is %f eV & wavelength is %f angstrom \"%(n,E_eV,lambda1);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.28\n", - " For state:1 Energy is 37.737723 eV & wavelength is 2.000000 angstrom \n", - " For state:2 Energy is 150.950893 eV & wavelength is 1.000000 angstrom \n", - " For state:3 Energy is 339.639509 eV & wavelength is 0.666667 angstrom \n" - ] - } - ], - "prompt_number": 32 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.29 Page No122" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "m = 100 #mass of ball in gram\n", - "l = 1 # length of box in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.29\"\n", - "for n in range(1,4):\n", - " E = (n**2*h**2)/(8*m*1e-3*l**2*1.6e-19)\n", - " print \" Energy state E%d of ball is %e eV\"%(n,E)\n", - "\n", - "print \" As energy difference is very small so we cannot see energy states.\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.29\n", - " Energy state E1 of ball is 3.434133e-48 eV\n", - " Energy state E2 of ball is 1.373653e-47 eV\n", - " Energy state E3 of ball is 3.090720e-47 eV\n", - " As energy difference is very small so we cannot see energy states.\n" - ] - } - ], - "prompt_number": 34 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.30 Page No124" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "l = 30. # width of potential well in angstrom\n", - "x = l/2\n", - "del_x = 2 # interval of length at centre in angstrom\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "n = 1 # ground state\n", - "print \"Example 2.30\"\n", - "phi_x = ((math.sqrt(2/l))*math.sin(n*math.pi*x/l))**2 \n", - "p = phi_x*del_x # Calculation of probability at centre\n", - "print \" Probability of finding particle at centre is %d percent.\"%(p*100)\n", - "# Answer given in book is 16 percent. It is due to wrong calculation \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.30\n", - " Probability of finding particle at centre is 13 percent.\n" - ] - } - ], - "prompt_number": 37 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch2_4.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch2_4.ipynb deleted file mode 100644 index dc4ec4d6..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch2_4.ipynb +++ /dev/null @@ -1,1017 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:5ca2861748d43d40b34516562fe47bc041c8b35f6ea545c8e5309115cbc73a04" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 2 : Uncertainty Principle and Schrodinger wave Equation" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.1 Page No71" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_x = 0.2 # Uncertainty in position in angstrom\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "print \"Example 2.1\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = h_bar/(2*del_x*1e-10) # Calculation of uncertainty in momentum\n", - "print \" Uncertainty in momentum of particle is %e kgm/sec \"%(del_p)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.1\n", - " Uncertainty in momentum of particle is 2.637993e-24 kgm/sec \n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.2 Page No75" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#given that\n", - "del_x = 4e-10 # Uncertainty in position in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "print \"Example 2.2\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = h_bar/(2*del_x) # Calculation of uncertainty in momentum\n", - "print \" Uncertainty in momentum of particle is %e kgm/sec.\"%(del_p)\n", - "# Answer in book is given as 1.32e-23 kgm/sec\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.2\n", - " Uncertainty in momentum of particle is 1.318997e-25 kgm/sec.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.3 Page No75" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "v = 3e7 # Velocity of moving electron in m/s\n", - "m = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "c = 3e8 # speed of light in m/s\n", - "print \"Example 2.3\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = m*v/(math.sqrt(1-(v/c)**2)) # calculation of uncertainty in momentum\n", - "del_x = h_bar/(2*del_p) # Calculation of uncertainty in position\n", - "print \" Uncertainty in position of particle is %f angstrom.\"%(del_x*1e10)\n", - "#Answer in book is 0.0194 angstrom which is due to umath.sing approximate values at intermediate steps\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.3\n", - " Uncertainty in position of particle is 0.019229 angstrom.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.5 Page No80" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "v = 1.05e4 # Velocity of moving electron in m/s\n", - "v_error = 0.02 #Percentage error in measurement of velocity\n", - "\n", - "m = 9e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.5\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "p = m*v\n", - "del_p = v_error*p/100 # calculation of uncertainty in momentum\n", - "del_x = h_bar/del_p\n", - "print \" Uncertainty in position of particle is %e m.\"%(del_x)\n", - "# Answer in book is given as 5.58e-3 m\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.5\n", - " Uncertainty in position of particle is 5.583054e-05 m.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.6 Page No82" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "v = 600 # Velocity of moving electron in m/s\n", - "v_error = 0.005 #Percentage error in measurement of velocity\n", - "m = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.6\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "p = m*v\n", - "del_p = v_error*p/100 # calculation of uncertainty in momentum\n", - "del_x = h_bar/(del_p) # Calculation of uncertainty in position\n", - "print \" Uncertainty in position of particle is %e m.\"%(del_x)\n", - "# Answer in book is 0.39e-2 m \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.6\n", - " Uncertainty in position of particle is 3.865191e-03 m.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.7 Page No82" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_x = 1 # let uncertainty in position is unity\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "m_p = 1.67e-27 # mass of proton in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.7\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_v_ratio = m_p/m_e # calculation in uncertainties in the velocity of electron and proton\n", - "print \" Ratio of uncertainties in the velocity of electron to proton is %d.\"%(del_v_ratio)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.7\n", - " Ratio of uncertainties in the velocity of electron to proton is 1835.\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.8 Page No84" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "r = 0.5 # radius of hydrogen atom in angstrom\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.8\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_x = 2*r # calculation of uncertainty in position\n", - "del_p = h_bar/(2*del_x*1e-10) # calculation of uncertainty in momentum\n", - "p = del_p\n", - "E = p**2/(2*m_e*1.6e-19)# Calculation of energy in eV\n", - "print \" Kinetic energy needed by an electron to be confined in electron is %.f eV.\"%((E*100)/100)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.8\n", - " Kinetic energy needed by an electron to be confined in electron is 1 eV.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.9 Page No89" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "v = 5e3 # Velocity of moving electron in m/s\n", - "v_error = 0.003 #Percentage error in measurement of velocity\n", - "\n", - "m = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.9\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "p = m*v\n", - "del_p = v_error*p/100 # calculation of uncertainty in momentum\n", - "del_x = h_bar/(2*del_p) # Calculation of uncertainty in position\n", - "print \" Uncertainty in position of particle is %e m.\"%(del_x)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.9\n", - " Uncertainty in position of particle is 3.865191e-04 m.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.10 Page No90" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "r = 0.53 # radius of hydrogen atom in angstrom\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.10\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_x = 2*r # calculation of uncertainty in position\n", - "del_p = h_bar/(2*del_x*1e-10) # calculation of uncertainty in momentum\n", - "p = del_p\n", - "E = p**2/(2*m_e*1.6e-19)# Calculation of energy in eV\n", - "print \" Kinetic energy needed by an electron to be confined in electron is %f eV.\"%(E)\n", - "# When problem is solved by del_x*del_p = h_bar, then minimum value of kinetic energy will become 13.6eV\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.10\n", - " Kinetic energy needed by an electron to be confined in electron is 0.850754 eV.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.11 Page No92" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "del_t = 2.5e-14 # lifetime in exited state in micro sec\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.11\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_E = h_bar/(1.6e-19*del_t*1e-6) # calculation of uncertainty in momentum\n", - "print \" Minimum error in measurement of energy of this state is %e eV.\"%(del_E)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.11\n", - " Minimum error in measurement of energy of this state is 2.637993e+04 eV.\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.12 Page No92" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "E_eV = 0.5# kinetic energy of electron in KeV\n", - "del_x = 0.4 # Uncertainty in position in nm\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in kg\n", - "print \"Example 2.12\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "E_J = E_eV*1e3*1.6e-19\n", - "p = math.sqrt(2*m*E_J) # Calculation of momentum in kgm/s\n", - "del_p = h_bar/(2*del_x*1e-9) # Calculation of uncertainty in momentum\n", - "per_error = del_p*100 / p # calculation of percentage error in momentum\n", - "print \" Percentage error in momentum is %f percent.\"%(per_error)\n", - "# Answer in book is 1.08 percentage\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.12\n", - " Percentage error in momentum is 1.093108 percent.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.13 Page No95" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "del_x = 2e-9 # Uncertainty in position in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.13\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = h_bar/(2*del_x) # Calculation of uncertainty in momentum\n", - "del_v = del_p/m\n", - "print \" Uncertainty in velocity of particle is %e m/s.\"%(del_v)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.13\n", - " Uncertainty in velocity of particle is 2.898894e+04 m/s.\n" - ] - } - ], - "prompt_number": 17 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.15 Page No97" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "del_x = 5000 # Uncertainty in position in angstrom\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 200 # mass of ball in gram\n", - "v = 6 # velocity of moving ball in m/s\n", - "print \"Example 2.15\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = h_bar/(2*del_x*1e-10) # Calculation of uncertainty in momentum\n", - "p = m*v/1000 # Calculation of momentum\n", - "per_error = del_p*100/p # Calculation of percentage error in calculation of momentum\n", - "print \" Uncertainty in momentum of ball is %e kgm/s.\"%(del_p)\n", - "print \" Percentage error in calculation of momentum is %e.\"%(per_error)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.15\n", - " Uncertainty in momentum of ball is 1.055197e-28 kgm/s.\n", - " Percentage error in calculation of momentum is 1.055197e-26.\n" - ] - } - ], - "prompt_number": 18 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.16 Page No98" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "c = 3e8 # speed of light in m/s\n", - "v = c/10 # Velocity of moving proton in m/s\n", - "v_error = 1 # Percentage error in measurement of velocity \n", - "m = 1.67e-27 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "print \"Example 2.16\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_v = v*v_error/100# calculation of uncertainty in position\n", - "del_x = h_bar/(2*m*del_v) # calculation of uncertainty in momentum\n", - "print \" Uncertainty in position of particle is %e m.\"%(del_x)\n", - "# Answer in book is 1.04e-13 m\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.16\n", - " Uncertainty in position of particle is 1.053091e-13 m.\n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.17 Page No98" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "del_x = 1e-9 # Uncertainty in position in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 200 # mass of ball in gram\n", - "print \"Example 2.17\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_v = h_bar/(2*del_x*m/1000) # Calculation of uncertainty in momentum\n", - "print \" Uncertainty in velocity of ball is %e m/s.\"%(del_v)\n", - "# Answer in book is 2.64e-25 m/s\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.17\n", - " Uncertainty in velocity of ball is 2.637993e-25 m/s.\n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.18 Page No102" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_t = 2e-12 # lifetime of exited state in sec\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.18\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_E = h_bar/(1.6e-19*2*del_t) # calculation of uncertainty in momentum\n", - "print \" Minimum error in measurement of energy of this state is %e eV.\"%(del_E)\n", - "# Answer in book is 1.65e-4 eV\n", - "\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.18\n", - " Minimum error in measurement of energy of this state is 1.648746e-04 eV.\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.19 Page No107" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_t = 1e-8 # lifetime of exited state in sec\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.19\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_nu = h_bar/(2*del_t*h) # calculation of uncertainty in frequency\n", - "print \" Minimum error in measurement of frequency of photon is %e per second.\"%(del_nu)\n", - "# Answer in book is 8e6 per second\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.19\n", - " Minimum error in measurement of frequency of photon is 7.957747e+06 per second.\n" - ] - } - ], - "prompt_number": 22 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.20 Page No108" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_v = 5.5e-20 # Uncertainty in velocity in m/s\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 1 # mass of dust particle in mg\n", - "print \"Example 2.20\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_x = h_bar/(2*del_v*m*1e-6) # Calculation of uncertainty in momentum\n", - "print \" Uncertainty in position of ball is %f angstrom.\"%(del_x*1e10)\n", - "# Answer in book is 9.6 angstrom\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.20\n", - " Uncertainty in position of ball is 9.592702 angstrom.\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.21 Page No110" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#given that\n", - "l = 1 # width of potential well in angstrom\n", - "n = 1 # order corresponding to ground state\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.21\"\n", - "E = n**2*h**2/(8*m*(l*1e-10)**2) # Calculation of energy in Joule\n", - "E_eV = E/1.6e-19 # Calculation of energy in eV\n", - "\n", - "print \" Energy of electron is %f eV.\"%(E_eV)\n", - "# Answer in book is 37.74 eV angstrom\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.21\n", - " Energy of electron is 37.737723 eV.\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.24 Page No113" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "l = 2.5e-10 # width of potential well in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.24\"\n", - "for n in range(1,3):\n", - " E = n**2*h**2/(8*m*l**2) # Calculation of energy in Joule\n", - " E_eV = E/1.6e-19 # Calculation of energy in eV\n", - " print \" Energy of electron for state %d is %f eV.\"%(n,E_eV);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.24\n", - " Energy of electron for state 1 is 6.038036 eV.\n", - " Energy of electron for state 2 is 24.152143 eV.\n" - ] - } - ], - "prompt_number": 33 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.26 Page No117" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# given that\n", - "L = 1# let unit length\n", - "l1 = 0.45*L # initial point\n", - "l2 = 0.55*L # Final point\n", - "\n", - "\n", - "print \"Example 2.26 \"\n", - "p = (1/L)*((l2-(L/(2*math.pi) *math.sin(2*l2*math.pi/L)))- (l1-(L/(2*math.pi) *math.sin(2*l1*math.pi/L)))) # Calculation of probability of finding particle\n", - "p_per = p*100 # probability of finding particle in percentage\n", - "print \" Probability of finding electron between %fL and %fL is %f percent.\"%(l2,l1,p_per)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.26 \n", - " Probability of finding electron between 0.550000L and 0.450000L is 19.836316 percent.\n" - ] - } - ], - "prompt_number": 30 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.27 Page No117" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#given that\n", - "l = 1e-8 # width of potential well in cm\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.27\"\n", - "E_1 = (h)**2/(8*m*(l*1e-2)**2) # Calculation of energy of ground state in Joule\n", - "E_1_eV = E_1/1.6e-19 # Calculation of energy in eV\n", - "E_2 = (2)**2*h**2/(8*m*(l*1e-2)**2) # Calculation of energy of first state in Joule\n", - "E_2_eV = E_2/1.6e-19 # Calculation of energy in eV\n", - "del_E = E_2_eV - E_1_eV # calculation of difference between first state and ground state\n", - "print \" Difference between first state and ground state energies is %f eV.\"%(del_E);\n", - "# Answer in book is 113.04 eV\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.27\n", - " Difference between first state and ground state energies is 113.213170 eV.\n" - ] - } - ], - "prompt_number": 31 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.28 Page No121" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "l = 1 # width of potential well in angstrom\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.28\"\n", - "for n in range(1,4):\n", - " lambda1 = 2.*l/n # Calculation of wavelength\n", - " E = n**2*h**2/(8*m*(l*1e-10)**2) # Calculation of energy in Joule\n", - " E_eV = E/1.6e-19 # Calculation of energy in eV\n", - " print \" For state:%d Energy is %f eV & wavelength is %f angstrom \"%(n,E_eV,lambda1);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.28\n", - " For state:1 Energy is 37.737723 eV & wavelength is 2.000000 angstrom \n", - " For state:2 Energy is 150.950893 eV & wavelength is 1.000000 angstrom \n", - " For state:3 Energy is 339.639509 eV & wavelength is 0.666667 angstrom \n" - ] - } - ], - "prompt_number": 32 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.29 Page No122" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "m = 100 #mass of ball in gram\n", - "l = 1 # length of box in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.29\"\n", - "for n in range(1,4):\n", - " E = (n**2*h**2)/(8*m*1e-3*l**2*1.6e-19)\n", - " print \" Energy state E%d of ball is %e eV\"%(n,E)\n", - "\n", - "print \" As energy difference is very small so we cannot see energy states.\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.29\n", - " Energy state E1 of ball is 3.434133e-48 eV\n", - " Energy state E2 of ball is 1.373653e-47 eV\n", - " Energy state E3 of ball is 3.090720e-47 eV\n", - " As energy difference is very small so we cannot see energy states.\n" - ] - } - ], - "prompt_number": 34 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.30 Page No124" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "l = 30. # width of potential well in angstrom\n", - "x = l/2\n", - "del_x = 2 # interval of length at centre in angstrom\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "n = 1 # ground state\n", - "print \"Example 2.30\"\n", - "phi_x = ((math.sqrt(2/l))*math.sin(n*math.pi*x/l))**2 \n", - "p = phi_x*del_x # Calculation of probability at centre\n", - "print \" Probability of finding particle at centre is %d percent.\"%(p*100)\n", - "# Answer given in book is 16 percent. It is due to wrong calculation \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.30\n", - " Probability of finding particle at centre is 13 percent.\n" - ] - } - ], - "prompt_number": 37 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch2_5.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch2_5.ipynb deleted file mode 100644 index dc4ec4d6..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch2_5.ipynb +++ /dev/null @@ -1,1017 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:5ca2861748d43d40b34516562fe47bc041c8b35f6ea545c8e5309115cbc73a04" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 2 : Uncertainty Principle and Schrodinger wave Equation" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.1 Page No71" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_x = 0.2 # Uncertainty in position in angstrom\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "print \"Example 2.1\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = h_bar/(2*del_x*1e-10) # Calculation of uncertainty in momentum\n", - "print \" Uncertainty in momentum of particle is %e kgm/sec \"%(del_p)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.1\n", - " Uncertainty in momentum of particle is 2.637993e-24 kgm/sec \n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.2 Page No75" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#given that\n", - "del_x = 4e-10 # Uncertainty in position in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "print \"Example 2.2\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = h_bar/(2*del_x) # Calculation of uncertainty in momentum\n", - "print \" Uncertainty in momentum of particle is %e kgm/sec.\"%(del_p)\n", - "# Answer in book is given as 1.32e-23 kgm/sec\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.2\n", - " Uncertainty in momentum of particle is 1.318997e-25 kgm/sec.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.3 Page No75" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "v = 3e7 # Velocity of moving electron in m/s\n", - "m = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "c = 3e8 # speed of light in m/s\n", - "print \"Example 2.3\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = m*v/(math.sqrt(1-(v/c)**2)) # calculation of uncertainty in momentum\n", - "del_x = h_bar/(2*del_p) # Calculation of uncertainty in position\n", - "print \" Uncertainty in position of particle is %f angstrom.\"%(del_x*1e10)\n", - "#Answer in book is 0.0194 angstrom which is due to umath.sing approximate values at intermediate steps\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.3\n", - " Uncertainty in position of particle is 0.019229 angstrom.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.5 Page No80" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "v = 1.05e4 # Velocity of moving electron in m/s\n", - "v_error = 0.02 #Percentage error in measurement of velocity\n", - "\n", - "m = 9e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.5\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "p = m*v\n", - "del_p = v_error*p/100 # calculation of uncertainty in momentum\n", - "del_x = h_bar/del_p\n", - "print \" Uncertainty in position of particle is %e m.\"%(del_x)\n", - "# Answer in book is given as 5.58e-3 m\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.5\n", - " Uncertainty in position of particle is 5.583054e-05 m.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.6 Page No82" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "v = 600 # Velocity of moving electron in m/s\n", - "v_error = 0.005 #Percentage error in measurement of velocity\n", - "m = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.6\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "p = m*v\n", - "del_p = v_error*p/100 # calculation of uncertainty in momentum\n", - "del_x = h_bar/(del_p) # Calculation of uncertainty in position\n", - "print \" Uncertainty in position of particle is %e m.\"%(del_x)\n", - "# Answer in book is 0.39e-2 m \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.6\n", - " Uncertainty in position of particle is 3.865191e-03 m.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.7 Page No82" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_x = 1 # let uncertainty in position is unity\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "m_p = 1.67e-27 # mass of proton in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.7\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_v_ratio = m_p/m_e # calculation in uncertainties in the velocity of electron and proton\n", - "print \" Ratio of uncertainties in the velocity of electron to proton is %d.\"%(del_v_ratio)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.7\n", - " Ratio of uncertainties in the velocity of electron to proton is 1835.\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.8 Page No84" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "r = 0.5 # radius of hydrogen atom in angstrom\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.8\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_x = 2*r # calculation of uncertainty in position\n", - "del_p = h_bar/(2*del_x*1e-10) # calculation of uncertainty in momentum\n", - "p = del_p\n", - "E = p**2/(2*m_e*1.6e-19)# Calculation of energy in eV\n", - "print \" Kinetic energy needed by an electron to be confined in electron is %.f eV.\"%((E*100)/100)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.8\n", - " Kinetic energy needed by an electron to be confined in electron is 1 eV.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.9 Page No89" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "v = 5e3 # Velocity of moving electron in m/s\n", - "v_error = 0.003 #Percentage error in measurement of velocity\n", - "\n", - "m = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.9\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "p = m*v\n", - "del_p = v_error*p/100 # calculation of uncertainty in momentum\n", - "del_x = h_bar/(2*del_p) # Calculation of uncertainty in position\n", - "print \" Uncertainty in position of particle is %e m.\"%(del_x)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.9\n", - " Uncertainty in position of particle is 3.865191e-04 m.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.10 Page No90" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "r = 0.53 # radius of hydrogen atom in angstrom\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.10\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_x = 2*r # calculation of uncertainty in position\n", - "del_p = h_bar/(2*del_x*1e-10) # calculation of uncertainty in momentum\n", - "p = del_p\n", - "E = p**2/(2*m_e*1.6e-19)# Calculation of energy in eV\n", - "print \" Kinetic energy needed by an electron to be confined in electron is %f eV.\"%(E)\n", - "# When problem is solved by del_x*del_p = h_bar, then minimum value of kinetic energy will become 13.6eV\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.10\n", - " Kinetic energy needed by an electron to be confined in electron is 0.850754 eV.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.11 Page No92" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "del_t = 2.5e-14 # lifetime in exited state in micro sec\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.11\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_E = h_bar/(1.6e-19*del_t*1e-6) # calculation of uncertainty in momentum\n", - "print \" Minimum error in measurement of energy of this state is %e eV.\"%(del_E)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.11\n", - " Minimum error in measurement of energy of this state is 2.637993e+04 eV.\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.12 Page No92" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "E_eV = 0.5# kinetic energy of electron in KeV\n", - "del_x = 0.4 # Uncertainty in position in nm\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in kg\n", - "print \"Example 2.12\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "E_J = E_eV*1e3*1.6e-19\n", - "p = math.sqrt(2*m*E_J) # Calculation of momentum in kgm/s\n", - "del_p = h_bar/(2*del_x*1e-9) # Calculation of uncertainty in momentum\n", - "per_error = del_p*100 / p # calculation of percentage error in momentum\n", - "print \" Percentage error in momentum is %f percent.\"%(per_error)\n", - "# Answer in book is 1.08 percentage\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.12\n", - " Percentage error in momentum is 1.093108 percent.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.13 Page No95" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "del_x = 2e-9 # Uncertainty in position in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.13\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = h_bar/(2*del_x) # Calculation of uncertainty in momentum\n", - "del_v = del_p/m\n", - "print \" Uncertainty in velocity of particle is %e m/s.\"%(del_v)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.13\n", - " Uncertainty in velocity of particle is 2.898894e+04 m/s.\n" - ] - } - ], - "prompt_number": 17 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.15 Page No97" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "del_x = 5000 # Uncertainty in position in angstrom\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 200 # mass of ball in gram\n", - "v = 6 # velocity of moving ball in m/s\n", - "print \"Example 2.15\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_p = h_bar/(2*del_x*1e-10) # Calculation of uncertainty in momentum\n", - "p = m*v/1000 # Calculation of momentum\n", - "per_error = del_p*100/p # Calculation of percentage error in calculation of momentum\n", - "print \" Uncertainty in momentum of ball is %e kgm/s.\"%(del_p)\n", - "print \" Percentage error in calculation of momentum is %e.\"%(per_error)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.15\n", - " Uncertainty in momentum of ball is 1.055197e-28 kgm/s.\n", - " Percentage error in calculation of momentum is 1.055197e-26.\n" - ] - } - ], - "prompt_number": 18 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.16 Page No98" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "c = 3e8 # speed of light in m/s\n", - "v = c/10 # Velocity of moving proton in m/s\n", - "v_error = 1 # Percentage error in measurement of velocity \n", - "m = 1.67e-27 # mass of electron in kg\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "\n", - "print \"Example 2.16\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_v = v*v_error/100# calculation of uncertainty in position\n", - "del_x = h_bar/(2*m*del_v) # calculation of uncertainty in momentum\n", - "print \" Uncertainty in position of particle is %e m.\"%(del_x)\n", - "# Answer in book is 1.04e-13 m\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.16\n", - " Uncertainty in position of particle is 1.053091e-13 m.\n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.17 Page No98" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "del_x = 1e-9 # Uncertainty in position in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 200 # mass of ball in gram\n", - "print \"Example 2.17\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_v = h_bar/(2*del_x*m/1000) # Calculation of uncertainty in momentum\n", - "print \" Uncertainty in velocity of ball is %e m/s.\"%(del_v)\n", - "# Answer in book is 2.64e-25 m/s\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.17\n", - " Uncertainty in velocity of ball is 2.637993e-25 m/s.\n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.18 Page No102" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_t = 2e-12 # lifetime of exited state in sec\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.18\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_E = h_bar/(1.6e-19*2*del_t) # calculation of uncertainty in momentum\n", - "print \" Minimum error in measurement of energy of this state is %e eV.\"%(del_E)\n", - "# Answer in book is 1.65e-4 eV\n", - "\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.18\n", - " Minimum error in measurement of energy of this state is 1.648746e-04 eV.\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.19 Page No107" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_t = 1e-8 # lifetime of exited state in sec\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.19\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_nu = h_bar/(2*del_t*h) # calculation of uncertainty in frequency\n", - "print \" Minimum error in measurement of frequency of photon is %e per second.\"%(del_nu)\n", - "# Answer in book is 8e6 per second\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.19\n", - " Minimum error in measurement of frequency of photon is 7.957747e+06 per second.\n" - ] - } - ], - "prompt_number": 22 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.20 Page No108" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "del_v = 5.5e-20 # Uncertainty in velocity in m/s\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 1 # mass of dust particle in mg\n", - "print \"Example 2.20\"\n", - "h_bar = h / (2*math.pi) # consmath.tant\n", - "del_x = h_bar/(2*del_v*m*1e-6) # Calculation of uncertainty in momentum\n", - "print \" Uncertainty in position of ball is %f angstrom.\"%(del_x*1e10)\n", - "# Answer in book is 9.6 angstrom\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.20\n", - " Uncertainty in position of ball is 9.592702 angstrom.\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.21 Page No110" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#given that\n", - "l = 1 # width of potential well in angstrom\n", - "n = 1 # order corresponding to ground state\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.21\"\n", - "E = n**2*h**2/(8*m*(l*1e-10)**2) # Calculation of energy in Joule\n", - "E_eV = E/1.6e-19 # Calculation of energy in eV\n", - "\n", - "print \" Energy of electron is %f eV.\"%(E_eV)\n", - "# Answer in book is 37.74 eV angstrom\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.21\n", - " Energy of electron is 37.737723 eV.\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.24 Page No113" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "l = 2.5e-10 # width of potential well in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.24\"\n", - "for n in range(1,3):\n", - " E = n**2*h**2/(8*m*l**2) # Calculation of energy in Joule\n", - " E_eV = E/1.6e-19 # Calculation of energy in eV\n", - " print \" Energy of electron for state %d is %f eV.\"%(n,E_eV);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.24\n", - " Energy of electron for state 1 is 6.038036 eV.\n", - " Energy of electron for state 2 is 24.152143 eV.\n" - ] - } - ], - "prompt_number": 33 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.26 Page No117" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# given that\n", - "L = 1# let unit length\n", - "l1 = 0.45*L # initial point\n", - "l2 = 0.55*L # Final point\n", - "\n", - "\n", - "print \"Example 2.26 \"\n", - "p = (1/L)*((l2-(L/(2*math.pi) *math.sin(2*l2*math.pi/L)))- (l1-(L/(2*math.pi) *math.sin(2*l1*math.pi/L)))) # Calculation of probability of finding particle\n", - "p_per = p*100 # probability of finding particle in percentage\n", - "print \" Probability of finding electron between %fL and %fL is %f percent.\"%(l2,l1,p_per)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.26 \n", - " Probability of finding electron between 0.550000L and 0.450000L is 19.836316 percent.\n" - ] - } - ], - "prompt_number": 30 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.27 Page No117" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#given that\n", - "l = 1e-8 # width of potential well in cm\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.27\"\n", - "E_1 = (h)**2/(8*m*(l*1e-2)**2) # Calculation of energy of ground state in Joule\n", - "E_1_eV = E_1/1.6e-19 # Calculation of energy in eV\n", - "E_2 = (2)**2*h**2/(8*m*(l*1e-2)**2) # Calculation of energy of first state in Joule\n", - "E_2_eV = E_2/1.6e-19 # Calculation of energy in eV\n", - "del_E = E_2_eV - E_1_eV # calculation of difference between first state and ground state\n", - "print \" Difference between first state and ground state energies is %f eV.\"%(del_E);\n", - "# Answer in book is 113.04 eV\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.27\n", - " Difference between first state and ground state energies is 113.213170 eV.\n" - ] - } - ], - "prompt_number": 31 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.28 Page No121" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "l = 1 # width of potential well in angstrom\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "m = 9.1e-31 # mass of electron in Kg\n", - "print \"Example 2.28\"\n", - "for n in range(1,4):\n", - " lambda1 = 2.*l/n # Calculation of wavelength\n", - " E = n**2*h**2/(8*m*(l*1e-10)**2) # Calculation of energy in Joule\n", - " E_eV = E/1.6e-19 # Calculation of energy in eV\n", - " print \" For state:%d Energy is %f eV & wavelength is %f angstrom \"%(n,E_eV,lambda1);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.28\n", - " For state:1 Energy is 37.737723 eV & wavelength is 2.000000 angstrom \n", - " For state:2 Energy is 150.950893 eV & wavelength is 1.000000 angstrom \n", - " For state:3 Energy is 339.639509 eV & wavelength is 0.666667 angstrom \n" - ] - } - ], - "prompt_number": 32 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.29 Page No122" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "m = 100 #mass of ball in gram\n", - "l = 1 # length of box in m\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "print \"Example 2.29\"\n", - "for n in range(1,4):\n", - " E = (n**2*h**2)/(8*m*1e-3*l**2*1.6e-19)\n", - " print \" Energy state E%d of ball is %e eV\"%(n,E)\n", - "\n", - "print \" As energy difference is very small so we cannot see energy states.\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.29\n", - " Energy state E1 of ball is 3.434133e-48 eV\n", - " Energy state E2 of ball is 1.373653e-47 eV\n", - " Energy state E3 of ball is 3.090720e-47 eV\n", - " As energy difference is very small so we cannot see energy states.\n" - ] - } - ], - "prompt_number": 34 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.30 Page No124" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "l = 30. # width of potential well in angstrom\n", - "x = l/2\n", - "del_x = 2 # interval of length at centre in angstrom\n", - "h = 6.63e-34 # Plank consmath.tant\n", - "n = 1 # ground state\n", - "print \"Example 2.30\"\n", - "phi_x = ((math.sqrt(2/l))*math.sin(n*math.pi*x/l))**2 \n", - "p = phi_x*del_x # Calculation of probability at centre\n", - "print \" Probability of finding particle at centre is %d percent.\"%(p*100)\n", - "# Answer given in book is 16 percent. It is due to wrong calculation \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 2.30\n", - " Probability of finding particle at centre is 13 percent.\n" - ] - } - ], - "prompt_number": 37 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch3.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch3.ipynb deleted file mode 100755 index 26b2b7d6..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch3.ipynb +++ /dev/null @@ -1,805 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:ce26666b75776c5b07c5f321f02f961249efef8c8458f3dae2110a523ce1ea02" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 3 : X ray and Compton effect" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.1 Page No134" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "d = 2.82 # crystal spacing in angstrom\n", - "n = 2 # order for longest pasmath.sing wavelength\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.1\"\n", - "lambda1 = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n", - "\n", - "print \" Longest wavelength is %f angstrom. \"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.1\n", - " Longest wavelength is 2.820000 angstrom. \n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.2 Page No134" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 0.3 # Wavelength in angstrom\n", - "d = 0.5 # crystal spacing in angstrom\n", - "n = 2 # order \n", - "m = 3 # order\n", - "print \"Example 3.2\"\n", - "theta_n = math.asin(n*lambda1/(2*d))*180/math.pi # Calculation of angle for order n\n", - "theta_m = math.asin(m*lambda1/(2*d))*180/math.pi # Calculation of angle for order m\n", - "\n", - "print \"Angle for %dnd order maxima is %f degree. \"%(n,theta_n)\n", - "print \"Angle for %drd order maxima is %f degree. \"%(m,theta_m)\n", - "# Answers in book are 40.97 degree and 72.29 degree which are due to wrong calculation\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.2\n", - "Angle for 2nd order maxima is 36.869898 degree. \n", - "Angle for 3rd order maxima is 64.158067 degree. \n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.3 Page No139" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#given that\n", - "d = 1.87 # crystal spacing in angstrom\n", - "n = 2 # order for longest pasmath.sing wavelength\n", - "theta = 30 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.3\"\n", - "lambda1 = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n", - "\n", - "print \" Longest wavelength is %f angstrom. \"%(lambda1)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.3\n", - " Longest wavelength is 0.935000 angstrom. \n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.4 Page No143" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 3.6e-9 # Wavelength in cm\n", - "theta = 4.8 # glancing angle in degree\n", - "n = 1 # order \n", - "\n", - "print \"Example 3.4\"\n", - "d = n*lambda1/(2*math.sin(theta*math.pi/180)) # calculation of crystal spacing in angstrom\n", - "\n", - "print \" Crystal spacing in angstrom is %e cm. \"%(d)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.4\n", - " Crystal spacing in angstrom is 2.151107e-08 cm. \n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.5 Page No146" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "d = 2.5 # crystal spacing in angstrom\n", - "n = 1 # order for longest pasmath.sing wavelength\n", - "theta = 20 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.5\"\n", - "lambda1 = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n", - "\n", - "print \"Longest wavelength is %f angstrom. \"%(lambda1)\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.5\n", - "Longest wavelength is 1.710101 angstrom. \n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.6 Page No146" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "d = 2.5 # crystal spacing in angstrom\n", - "n = 1 # order for longest pasmath.sing wavelength\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.6\"\n", - "lambda1 = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n", - "\n", - "print \"Longest wavelength is of %d angstrom. \"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.6\n", - "Longest wavelength is of 5 angstrom. \n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.7 Page No146" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# given that\n", - "theta1_deg = 5 # Absolut degree part of angle for first angle\n", - "theta1_min = 23#remainder minute part of angle for first angle\n", - "theta2_deg = 7 # Absolut degree part of angle for second angle\n", - "theta2_min = 37#remainder minute part of angle for second angle\n", - "theta3_deg = 9 # Absolut degree part of angle for third angle\n", - "theta3_min = 25#remainder minute part of angle for third angle\n", - "\n", - "print \"Example 3.7 \"\n", - "val1 = math.sin((theta1_deg+ theta1_min/60)*math.pi/180)# Sin value for first angle\n", - "val2 = math.sin((theta2_deg+ theta2_min/60)*math.pi/180) #Sin value for second angle\n", - "val3 = math.sin((theta3_deg+ theta3_min/60)*math.pi/180)#Sin value for third angle\n", - "ratio_21 = val2/val1\n", - "ratio_31 = val3/val1\n", - "print \" Interatomic layer separation ratios in crystal are as 1 : %f : %f\"%(ratio_21,ratio_31)\n", - "print \" Above relation shows that crystal is simple cubic crystal structure.\"\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.7 \n", - " Interatomic layer separation ratios in crystal are as 1 : 1.398294 : 1.794884\n", - " Above relation shows that crystal is simple cubic crystal structure.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.8 Page No146" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1.2 # wavelength in angstrom\n", - "theta_deg = 9 # angle fraction in degree\n", - "theta_min = 30 # Angle fraction in minute\n", - "print \"Example 3.8\"\n", - "theta = theta_deg+theta_min/60 # Total angel\n", - "for n in range(1,5):\n", - " d = lambda1/(n*2*math.sin(theta*math.pi/180)) # Inter layer spacing\n", - " print \" If order is %d then spacing is %f angstrom.\"%(n,d)\n", - "\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.8\n", - " If order is 1 then spacing is 3.835472 angstrom.\n", - " If order is 2 then spacing is 1.917736 angstrom.\n", - " If order is 3 then spacing is 1.278491 angstrom.\n", - " If order is 4 then spacing is 0.958868 angstrom.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.9 Page No147" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "h = 6.62e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "e = 1.6e-19 # charge on electron in coulomb\n", - "v = 340 # Applied voltage in volt\n", - "n = 1 # order for longest pasmath.sing wavelength\n", - "theta = 60 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.9\"\n", - "lambda1= h/math.sqrt(2*m_e*e*v) # calculation of wavelength\n", - "d = n*lambda1/(2*math.sin(theta*math.pi/180))# calculation of spacing of crystal\n", - "\n", - "print \"Spacing of crystal is %f angstrom. \"%(d*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.9\n", - "Spacing of crystal is 0.384116 angstrom. \n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.10 Page No149" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "E = 100 # Energy of X ray beam in KeV\n", - "theta = 30 # Scattering angle in degree\n", - "m = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # Speed of light in m/s\n", - "print \"Example 3.10\"\n", - "E_rest = m*c**2/(1.6e-19*1e3) # Rest mass energy in KeV\n", - "k = 1/E + (1-math.cos(theta*math.pi/180))/(E_rest)\n", - "del_e = E - 1/k # Energy of recoiled electron\n", - "print \" Energy of recoiled electron is %f KeV\"%(del_e)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.10\n", - " Energy of recoiled electron is -3720.687014 KeV\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.11 Page No154" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1 #wavelength in angstrom\n", - "h = 6.62e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.11\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))/(m_e*c) # calculation of wavelength shift \n", - "\n", - "print \"Wavelength shift is %f angstrom. \"%(d_lambda*1e10)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.11\n", - "Wavelength shift is 0.024249 angstrom. \n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.12 Page No156" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 0.015 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 60 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.12\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda\n", - "\n", - "print \" Wavelength shift is %f angstrom. \"%(lambda_n)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.12\n", - " Wavelength shift is 0.027143 angstrom. \n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.13 Page No158" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.13\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda # Calculation of recoiled electron wavelength\n", - "d_E = h*c*(lambda_n-lambda1)*1e10/(1.6e-19*lambda_n*lambda1)# Calculation of recoiled electron energy in eV\n", - "print \"Wavelength shift is %f angstrom.\"%(lambda_n)\n", - "print \"Energy of recoiled electron is %.f eV. \"%(( d_E))\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.13\n", - "Wavelength shift is 1.024286 angstrom.\n", - "Energy of recoiled electron is 295 eV. \n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.14 Page No163" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1 #let wavelength in angstrom\n", - "lambda_n = 2*lambda1 # recoiled electron wavelength\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.14\"\n", - "lambda1 = h*1e10/(m_e*c) # calculation of wavelength in angstrom\n", - "E = h*c*1e10/(lambda1*1.6e-19) # calculation of energy of electron\n", - "\n", - "print \"Wavelength shift is %f angstrom. \"%(lambda1)\n", - "print \"Energy of recoiled electron is %f KeV. \"%(E/1e3)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.14\n", - "Wavelength shift is 0.024286 angstrom. \n", - "Energy of recoiled electron is 511.875000 KeV. \n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.15 Page No168" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 2 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 45 # scattering angle \n", - "print \"Example 3.15\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda # Calculation of recoiled electron wavelength\n", - "\n", - "f = d_lambda/lambda1 # Calculation of fraction of energy lost by photon \n", - "\n", - "print \"Fraction of energy lost by photon is %f\"%(f)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.15\n", - "Fraction of energy lost by photon is 0.003557\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.16 Page No171" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#given that\n", - "E_eV = 510 # Energy of gamma ray in keV\n", - "lambda1 = 2 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # scattering angle in degree\n", - "print \"Example 3.16\"\n", - "E_j = E_eV*1e3*1.6e-19 # Energy of gamma ray in Joule\n", - "lambda1 = h*c*1e10/E_j # Calculation of wavelength in angstrom\n", - "\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda # Calculation of recoiled electron wavelength\n", - "print \"Wavelength of scattered radiation is %f Angstrom \"%(lambda_n)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.16\n", - "Wavelength of scattered radiation is 0.048661 Angstrom \n" - ] - } - ], - "prompt_number": 22 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.17 Page No172" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#given that\n", - "lambda1 = 2 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.17\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda # Calculation of recoiled electron wavelength\n", - "d_E = h*c*(lambda_n-lambda1)*1e10/(1.6e-19*lambda_n*lambda1)# Calculation of recoiled electron energy in eV\n", - "print \" Scattered wavelength is %f angstrom.\"%(lambda_n)\n", - "print \" Energy of recoiled electron is %feV. \"%(d_E)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.17\n", - " Scattered wavelength is 2.024286 angstrom.\n", - " Energy of recoiled electron is 74.569954eV. \n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.18 Page No175" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "E_eV = 510 # Energy of gamma ray in keV\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # scattering angle in degree\n", - "print \"Example 3.18\"\n", - "E_j = E_eV*1e3*1.6e-19 # Energy of gamma ray in Joule\n", - "lambda1 = h*c/E_j # Calculation of wavelength in meter\n", - "\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda/1e10 # Calculation of recoiled electron wavelength\n", - "d_E = h*c*(d_lambda/1e10)/(1.6e-19*lambda_n*lambda1)# Calculation of recoiled electron energy in eV\n", - "psi= math.atan(1/(math.tan((theta*math.pi/180)/2)/(1+(h/(lambda1*m_e*c))))) \n", - "phi_deg = 90 - psi*180/math.pi # Calculation of degree part of angle of recoiled electron \n", - "phi_min = 60*(phi_deg - int(phi_deg))# Calculation of minute part of angle of recoiled electron \n", - "print \"Wavelength of scattered radiation is %e m \"%(lambda_n)\n", - "print \"Energy of recoiled electron is %f MeV.\"%(d_E/1e6)\n", - "print \"Recoiled electron angle is %d degree%d minute \"%(phi_deg,phi_min)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.18\n", - "Wavelength of scattered radiation is 4.866071e-12 m \n", - "Energy of recoiled electron is 0.254532 MeV.\n", - "Recoiled electron angle is 26 degree36 minute \n" - ] - } - ], - "prompt_number": 27 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.19 Page No175" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "nu = 2e19 # initial frequency of X ray photon\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # scattering angle in degree\n", - "print \"Example 3.19\"\n", - "d_lambda = h/(m_e*c) # calculation of wavelength shift\n", - "k = 1./nu + d_lambda/c\n", - "nu_1 = 1/k # Frequency after collision\n", - "nu_1 = int(nu_1/1e18)*1e18 # rounding off\n", - "print \"Frequency after collision is %e Hz \"%(nu_1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.19\n", - "Frequency after collision is 1.700000e+19 Hz \n" - ] - } - ], - "prompt_number": 29 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch3_1.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch3_1.ipynb deleted file mode 100755 index 26b2b7d6..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch3_1.ipynb +++ /dev/null @@ -1,805 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:ce26666b75776c5b07c5f321f02f961249efef8c8458f3dae2110a523ce1ea02" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 3 : X ray and Compton effect" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.1 Page No134" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "d = 2.82 # crystal spacing in angstrom\n", - "n = 2 # order for longest pasmath.sing wavelength\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.1\"\n", - "lambda1 = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n", - "\n", - "print \" Longest wavelength is %f angstrom. \"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.1\n", - " Longest wavelength is 2.820000 angstrom. \n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.2 Page No134" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 0.3 # Wavelength in angstrom\n", - "d = 0.5 # crystal spacing in angstrom\n", - "n = 2 # order \n", - "m = 3 # order\n", - "print \"Example 3.2\"\n", - "theta_n = math.asin(n*lambda1/(2*d))*180/math.pi # Calculation of angle for order n\n", - "theta_m = math.asin(m*lambda1/(2*d))*180/math.pi # Calculation of angle for order m\n", - "\n", - "print \"Angle for %dnd order maxima is %f degree. \"%(n,theta_n)\n", - "print \"Angle for %drd order maxima is %f degree. \"%(m,theta_m)\n", - "# Answers in book are 40.97 degree and 72.29 degree which are due to wrong calculation\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.2\n", - "Angle for 2nd order maxima is 36.869898 degree. \n", - "Angle for 3rd order maxima is 64.158067 degree. \n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.3 Page No139" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#given that\n", - "d = 1.87 # crystal spacing in angstrom\n", - "n = 2 # order for longest pasmath.sing wavelength\n", - "theta = 30 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.3\"\n", - "lambda1 = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n", - "\n", - "print \" Longest wavelength is %f angstrom. \"%(lambda1)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.3\n", - " Longest wavelength is 0.935000 angstrom. \n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.4 Page No143" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 3.6e-9 # Wavelength in cm\n", - "theta = 4.8 # glancing angle in degree\n", - "n = 1 # order \n", - "\n", - "print \"Example 3.4\"\n", - "d = n*lambda1/(2*math.sin(theta*math.pi/180)) # calculation of crystal spacing in angstrom\n", - "\n", - "print \" Crystal spacing in angstrom is %e cm. \"%(d)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.4\n", - " Crystal spacing in angstrom is 2.151107e-08 cm. \n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.5 Page No146" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "d = 2.5 # crystal spacing in angstrom\n", - "n = 1 # order for longest pasmath.sing wavelength\n", - "theta = 20 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.5\"\n", - "lambda1 = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n", - "\n", - "print \"Longest wavelength is %f angstrom. \"%(lambda1)\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.5\n", - "Longest wavelength is 1.710101 angstrom. \n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.6 Page No146" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "d = 2.5 # crystal spacing in angstrom\n", - "n = 1 # order for longest pasmath.sing wavelength\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.6\"\n", - "lambda1 = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n", - "\n", - "print \"Longest wavelength is of %d angstrom. \"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.6\n", - "Longest wavelength is of 5 angstrom. \n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.7 Page No146" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# given that\n", - "theta1_deg = 5 # Absolut degree part of angle for first angle\n", - "theta1_min = 23#remainder minute part of angle for first angle\n", - "theta2_deg = 7 # Absolut degree part of angle for second angle\n", - "theta2_min = 37#remainder minute part of angle for second angle\n", - "theta3_deg = 9 # Absolut degree part of angle for third angle\n", - "theta3_min = 25#remainder minute part of angle for third angle\n", - "\n", - "print \"Example 3.7 \"\n", - "val1 = math.sin((theta1_deg+ theta1_min/60)*math.pi/180)# Sin value for first angle\n", - "val2 = math.sin((theta2_deg+ theta2_min/60)*math.pi/180) #Sin value for second angle\n", - "val3 = math.sin((theta3_deg+ theta3_min/60)*math.pi/180)#Sin value for third angle\n", - "ratio_21 = val2/val1\n", - "ratio_31 = val3/val1\n", - "print \" Interatomic layer separation ratios in crystal are as 1 : %f : %f\"%(ratio_21,ratio_31)\n", - "print \" Above relation shows that crystal is simple cubic crystal structure.\"\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.7 \n", - " Interatomic layer separation ratios in crystal are as 1 : 1.398294 : 1.794884\n", - " Above relation shows that crystal is simple cubic crystal structure.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.8 Page No146" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1.2 # wavelength in angstrom\n", - "theta_deg = 9 # angle fraction in degree\n", - "theta_min = 30 # Angle fraction in minute\n", - "print \"Example 3.8\"\n", - "theta = theta_deg+theta_min/60 # Total angel\n", - "for n in range(1,5):\n", - " d = lambda1/(n*2*math.sin(theta*math.pi/180)) # Inter layer spacing\n", - " print \" If order is %d then spacing is %f angstrom.\"%(n,d)\n", - "\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.8\n", - " If order is 1 then spacing is 3.835472 angstrom.\n", - " If order is 2 then spacing is 1.917736 angstrom.\n", - " If order is 3 then spacing is 1.278491 angstrom.\n", - " If order is 4 then spacing is 0.958868 angstrom.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.9 Page No147" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "h = 6.62e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "e = 1.6e-19 # charge on electron in coulomb\n", - "v = 340 # Applied voltage in volt\n", - "n = 1 # order for longest pasmath.sing wavelength\n", - "theta = 60 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.9\"\n", - "lambda1= h/math.sqrt(2*m_e*e*v) # calculation of wavelength\n", - "d = n*lambda1/(2*math.sin(theta*math.pi/180))# calculation of spacing of crystal\n", - "\n", - "print \"Spacing of crystal is %f angstrom. \"%(d*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.9\n", - "Spacing of crystal is 0.384116 angstrom. \n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.10 Page No149" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "E = 100 # Energy of X ray beam in KeV\n", - "theta = 30 # Scattering angle in degree\n", - "m = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # Speed of light in m/s\n", - "print \"Example 3.10\"\n", - "E_rest = m*c**2/(1.6e-19*1e3) # Rest mass energy in KeV\n", - "k = 1/E + (1-math.cos(theta*math.pi/180))/(E_rest)\n", - "del_e = E - 1/k # Energy of recoiled electron\n", - "print \" Energy of recoiled electron is %f KeV\"%(del_e)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.10\n", - " Energy of recoiled electron is -3720.687014 KeV\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.11 Page No154" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1 #wavelength in angstrom\n", - "h = 6.62e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.11\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))/(m_e*c) # calculation of wavelength shift \n", - "\n", - "print \"Wavelength shift is %f angstrom. \"%(d_lambda*1e10)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.11\n", - "Wavelength shift is 0.024249 angstrom. \n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.12 Page No156" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 0.015 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 60 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.12\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda\n", - "\n", - "print \" Wavelength shift is %f angstrom. \"%(lambda_n)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.12\n", - " Wavelength shift is 0.027143 angstrom. \n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.13 Page No158" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.13\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda # Calculation of recoiled electron wavelength\n", - "d_E = h*c*(lambda_n-lambda1)*1e10/(1.6e-19*lambda_n*lambda1)# Calculation of recoiled electron energy in eV\n", - "print \"Wavelength shift is %f angstrom.\"%(lambda_n)\n", - "print \"Energy of recoiled electron is %.f eV. \"%(( d_E))\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.13\n", - "Wavelength shift is 1.024286 angstrom.\n", - "Energy of recoiled electron is 295 eV. \n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.14 Page No163" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1 #let wavelength in angstrom\n", - "lambda_n = 2*lambda1 # recoiled electron wavelength\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.14\"\n", - "lambda1 = h*1e10/(m_e*c) # calculation of wavelength in angstrom\n", - "E = h*c*1e10/(lambda1*1.6e-19) # calculation of energy of electron\n", - "\n", - "print \"Wavelength shift is %f angstrom. \"%(lambda1)\n", - "print \"Energy of recoiled electron is %f KeV. \"%(E/1e3)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.14\n", - "Wavelength shift is 0.024286 angstrom. \n", - "Energy of recoiled electron is 511.875000 KeV. \n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.15 Page No168" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 2 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 45 # scattering angle \n", - "print \"Example 3.15\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda # Calculation of recoiled electron wavelength\n", - "\n", - "f = d_lambda/lambda1 # Calculation of fraction of energy lost by photon \n", - "\n", - "print \"Fraction of energy lost by photon is %f\"%(f)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.15\n", - "Fraction of energy lost by photon is 0.003557\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.16 Page No171" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#given that\n", - "E_eV = 510 # Energy of gamma ray in keV\n", - "lambda1 = 2 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # scattering angle in degree\n", - "print \"Example 3.16\"\n", - "E_j = E_eV*1e3*1.6e-19 # Energy of gamma ray in Joule\n", - "lambda1 = h*c*1e10/E_j # Calculation of wavelength in angstrom\n", - "\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda # Calculation of recoiled electron wavelength\n", - "print \"Wavelength of scattered radiation is %f Angstrom \"%(lambda_n)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.16\n", - "Wavelength of scattered radiation is 0.048661 Angstrom \n" - ] - } - ], - "prompt_number": 22 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.17 Page No172" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#given that\n", - "lambda1 = 2 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.17\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda # Calculation of recoiled electron wavelength\n", - "d_E = h*c*(lambda_n-lambda1)*1e10/(1.6e-19*lambda_n*lambda1)# Calculation of recoiled electron energy in eV\n", - "print \" Scattered wavelength is %f angstrom.\"%(lambda_n)\n", - "print \" Energy of recoiled electron is %feV. \"%(d_E)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.17\n", - " Scattered wavelength is 2.024286 angstrom.\n", - " Energy of recoiled electron is 74.569954eV. \n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.18 Page No175" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "E_eV = 510 # Energy of gamma ray in keV\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # scattering angle in degree\n", - "print \"Example 3.18\"\n", - "E_j = E_eV*1e3*1.6e-19 # Energy of gamma ray in Joule\n", - "lambda1 = h*c/E_j # Calculation of wavelength in meter\n", - "\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda/1e10 # Calculation of recoiled electron wavelength\n", - "d_E = h*c*(d_lambda/1e10)/(1.6e-19*lambda_n*lambda1)# Calculation of recoiled electron energy in eV\n", - "psi= math.atan(1/(math.tan((theta*math.pi/180)/2)/(1+(h/(lambda1*m_e*c))))) \n", - "phi_deg = 90 - psi*180/math.pi # Calculation of degree part of angle of recoiled electron \n", - "phi_min = 60*(phi_deg - int(phi_deg))# Calculation of minute part of angle of recoiled electron \n", - "print \"Wavelength of scattered radiation is %e m \"%(lambda_n)\n", - "print \"Energy of recoiled electron is %f MeV.\"%(d_E/1e6)\n", - "print \"Recoiled electron angle is %d degree%d minute \"%(phi_deg,phi_min)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.18\n", - "Wavelength of scattered radiation is 4.866071e-12 m \n", - "Energy of recoiled electron is 0.254532 MeV.\n", - "Recoiled electron angle is 26 degree36 minute \n" - ] - } - ], - "prompt_number": 27 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.19 Page No175" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "nu = 2e19 # initial frequency of X ray photon\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # scattering angle in degree\n", - "print \"Example 3.19\"\n", - "d_lambda = h/(m_e*c) # calculation of wavelength shift\n", - "k = 1./nu + d_lambda/c\n", - "nu_1 = 1/k # Frequency after collision\n", - "nu_1 = int(nu_1/1e18)*1e18 # rounding off\n", - "print \"Frequency after collision is %e Hz \"%(nu_1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.19\n", - "Frequency after collision is 1.700000e+19 Hz \n" - ] - } - ], - "prompt_number": 29 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch3_2.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch3_2.ipynb deleted file mode 100644 index 26b2b7d6..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch3_2.ipynb +++ /dev/null @@ -1,805 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:ce26666b75776c5b07c5f321f02f961249efef8c8458f3dae2110a523ce1ea02" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 3 : X ray and Compton effect" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.1 Page No134" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "d = 2.82 # crystal spacing in angstrom\n", - "n = 2 # order for longest pasmath.sing wavelength\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.1\"\n", - "lambda1 = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n", - "\n", - "print \" Longest wavelength is %f angstrom. \"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.1\n", - " Longest wavelength is 2.820000 angstrom. \n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.2 Page No134" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 0.3 # Wavelength in angstrom\n", - "d = 0.5 # crystal spacing in angstrom\n", - "n = 2 # order \n", - "m = 3 # order\n", - "print \"Example 3.2\"\n", - "theta_n = math.asin(n*lambda1/(2*d))*180/math.pi # Calculation of angle for order n\n", - "theta_m = math.asin(m*lambda1/(2*d))*180/math.pi # Calculation of angle for order m\n", - "\n", - "print \"Angle for %dnd order maxima is %f degree. \"%(n,theta_n)\n", - "print \"Angle for %drd order maxima is %f degree. \"%(m,theta_m)\n", - "# Answers in book are 40.97 degree and 72.29 degree which are due to wrong calculation\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.2\n", - "Angle for 2nd order maxima is 36.869898 degree. \n", - "Angle for 3rd order maxima is 64.158067 degree. \n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.3 Page No139" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#given that\n", - "d = 1.87 # crystal spacing in angstrom\n", - "n = 2 # order for longest pasmath.sing wavelength\n", - "theta = 30 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.3\"\n", - "lambda1 = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n", - "\n", - "print \" Longest wavelength is %f angstrom. \"%(lambda1)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.3\n", - " Longest wavelength is 0.935000 angstrom. \n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.4 Page No143" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 3.6e-9 # Wavelength in cm\n", - "theta = 4.8 # glancing angle in degree\n", - "n = 1 # order \n", - "\n", - "print \"Example 3.4\"\n", - "d = n*lambda1/(2*math.sin(theta*math.pi/180)) # calculation of crystal spacing in angstrom\n", - "\n", - "print \" Crystal spacing in angstrom is %e cm. \"%(d)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.4\n", - " Crystal spacing in angstrom is 2.151107e-08 cm. \n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.5 Page No146" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "d = 2.5 # crystal spacing in angstrom\n", - "n = 1 # order for longest pasmath.sing wavelength\n", - "theta = 20 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.5\"\n", - "lambda1 = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n", - "\n", - "print \"Longest wavelength is %f angstrom. \"%(lambda1)\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.5\n", - "Longest wavelength is 1.710101 angstrom. \n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.6 Page No146" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "d = 2.5 # crystal spacing in angstrom\n", - "n = 1 # order for longest pasmath.sing wavelength\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.6\"\n", - "lambda1 = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n", - "\n", - "print \"Longest wavelength is of %d angstrom. \"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.6\n", - "Longest wavelength is of 5 angstrom. \n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.7 Page No146" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# given that\n", - "theta1_deg = 5 # Absolut degree part of angle for first angle\n", - "theta1_min = 23#remainder minute part of angle for first angle\n", - "theta2_deg = 7 # Absolut degree part of angle for second angle\n", - "theta2_min = 37#remainder minute part of angle for second angle\n", - "theta3_deg = 9 # Absolut degree part of angle for third angle\n", - "theta3_min = 25#remainder minute part of angle for third angle\n", - "\n", - "print \"Example 3.7 \"\n", - "val1 = math.sin((theta1_deg+ theta1_min/60)*math.pi/180)# Sin value for first angle\n", - "val2 = math.sin((theta2_deg+ theta2_min/60)*math.pi/180) #Sin value for second angle\n", - "val3 = math.sin((theta3_deg+ theta3_min/60)*math.pi/180)#Sin value for third angle\n", - "ratio_21 = val2/val1\n", - "ratio_31 = val3/val1\n", - "print \" Interatomic layer separation ratios in crystal are as 1 : %f : %f\"%(ratio_21,ratio_31)\n", - "print \" Above relation shows that crystal is simple cubic crystal structure.\"\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.7 \n", - " Interatomic layer separation ratios in crystal are as 1 : 1.398294 : 1.794884\n", - " Above relation shows that crystal is simple cubic crystal structure.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.8 Page No146" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1.2 # wavelength in angstrom\n", - "theta_deg = 9 # angle fraction in degree\n", - "theta_min = 30 # Angle fraction in minute\n", - "print \"Example 3.8\"\n", - "theta = theta_deg+theta_min/60 # Total angel\n", - "for n in range(1,5):\n", - " d = lambda1/(n*2*math.sin(theta*math.pi/180)) # Inter layer spacing\n", - " print \" If order is %d then spacing is %f angstrom.\"%(n,d)\n", - "\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.8\n", - " If order is 1 then spacing is 3.835472 angstrom.\n", - " If order is 2 then spacing is 1.917736 angstrom.\n", - " If order is 3 then spacing is 1.278491 angstrom.\n", - " If order is 4 then spacing is 0.958868 angstrom.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.9 Page No147" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "h = 6.62e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "e = 1.6e-19 # charge on electron in coulomb\n", - "v = 340 # Applied voltage in volt\n", - "n = 1 # order for longest pasmath.sing wavelength\n", - "theta = 60 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.9\"\n", - "lambda1= h/math.sqrt(2*m_e*e*v) # calculation of wavelength\n", - "d = n*lambda1/(2*math.sin(theta*math.pi/180))# calculation of spacing of crystal\n", - "\n", - "print \"Spacing of crystal is %f angstrom. \"%(d*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.9\n", - "Spacing of crystal is 0.384116 angstrom. \n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.10 Page No149" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "E = 100 # Energy of X ray beam in KeV\n", - "theta = 30 # Scattering angle in degree\n", - "m = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # Speed of light in m/s\n", - "print \"Example 3.10\"\n", - "E_rest = m*c**2/(1.6e-19*1e3) # Rest mass energy in KeV\n", - "k = 1/E + (1-math.cos(theta*math.pi/180))/(E_rest)\n", - "del_e = E - 1/k # Energy of recoiled electron\n", - "print \" Energy of recoiled electron is %f KeV\"%(del_e)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.10\n", - " Energy of recoiled electron is -3720.687014 KeV\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.11 Page No154" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1 #wavelength in angstrom\n", - "h = 6.62e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.11\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))/(m_e*c) # calculation of wavelength shift \n", - "\n", - "print \"Wavelength shift is %f angstrom. \"%(d_lambda*1e10)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.11\n", - "Wavelength shift is 0.024249 angstrom. \n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.12 Page No156" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 0.015 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 60 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.12\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda\n", - "\n", - "print \" Wavelength shift is %f angstrom. \"%(lambda_n)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.12\n", - " Wavelength shift is 0.027143 angstrom. \n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.13 Page No158" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.13\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda # Calculation of recoiled electron wavelength\n", - "d_E = h*c*(lambda_n-lambda1)*1e10/(1.6e-19*lambda_n*lambda1)# Calculation of recoiled electron energy in eV\n", - "print \"Wavelength shift is %f angstrom.\"%(lambda_n)\n", - "print \"Energy of recoiled electron is %.f eV. \"%(( d_E))\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.13\n", - "Wavelength shift is 1.024286 angstrom.\n", - "Energy of recoiled electron is 295 eV. \n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.14 Page No163" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1 #let wavelength in angstrom\n", - "lambda_n = 2*lambda1 # recoiled electron wavelength\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.14\"\n", - "lambda1 = h*1e10/(m_e*c) # calculation of wavelength in angstrom\n", - "E = h*c*1e10/(lambda1*1.6e-19) # calculation of energy of electron\n", - "\n", - "print \"Wavelength shift is %f angstrom. \"%(lambda1)\n", - "print \"Energy of recoiled electron is %f KeV. \"%(E/1e3)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.14\n", - "Wavelength shift is 0.024286 angstrom. \n", - "Energy of recoiled electron is 511.875000 KeV. \n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.15 Page No168" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 2 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 45 # scattering angle \n", - "print \"Example 3.15\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda # Calculation of recoiled electron wavelength\n", - "\n", - "f = d_lambda/lambda1 # Calculation of fraction of energy lost by photon \n", - "\n", - "print \"Fraction of energy lost by photon is %f\"%(f)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.15\n", - "Fraction of energy lost by photon is 0.003557\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.16 Page No171" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#given that\n", - "E_eV = 510 # Energy of gamma ray in keV\n", - "lambda1 = 2 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # scattering angle in degree\n", - "print \"Example 3.16\"\n", - "E_j = E_eV*1e3*1.6e-19 # Energy of gamma ray in Joule\n", - "lambda1 = h*c*1e10/E_j # Calculation of wavelength in angstrom\n", - "\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda # Calculation of recoiled electron wavelength\n", - "print \"Wavelength of scattered radiation is %f Angstrom \"%(lambda_n)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.16\n", - "Wavelength of scattered radiation is 0.048661 Angstrom \n" - ] - } - ], - "prompt_number": 22 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.17 Page No172" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#given that\n", - "lambda1 = 2 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.17\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda # Calculation of recoiled electron wavelength\n", - "d_E = h*c*(lambda_n-lambda1)*1e10/(1.6e-19*lambda_n*lambda1)# Calculation of recoiled electron energy in eV\n", - "print \" Scattered wavelength is %f angstrom.\"%(lambda_n)\n", - "print \" Energy of recoiled electron is %feV. \"%(d_E)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.17\n", - " Scattered wavelength is 2.024286 angstrom.\n", - " Energy of recoiled electron is 74.569954eV. \n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.18 Page No175" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "E_eV = 510 # Energy of gamma ray in keV\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # scattering angle in degree\n", - "print \"Example 3.18\"\n", - "E_j = E_eV*1e3*1.6e-19 # Energy of gamma ray in Joule\n", - "lambda1 = h*c/E_j # Calculation of wavelength in meter\n", - "\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda/1e10 # Calculation of recoiled electron wavelength\n", - "d_E = h*c*(d_lambda/1e10)/(1.6e-19*lambda_n*lambda1)# Calculation of recoiled electron energy in eV\n", - "psi= math.atan(1/(math.tan((theta*math.pi/180)/2)/(1+(h/(lambda1*m_e*c))))) \n", - "phi_deg = 90 - psi*180/math.pi # Calculation of degree part of angle of recoiled electron \n", - "phi_min = 60*(phi_deg - int(phi_deg))# Calculation of minute part of angle of recoiled electron \n", - "print \"Wavelength of scattered radiation is %e m \"%(lambda_n)\n", - "print \"Energy of recoiled electron is %f MeV.\"%(d_E/1e6)\n", - "print \"Recoiled electron angle is %d degree%d minute \"%(phi_deg,phi_min)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.18\n", - "Wavelength of scattered radiation is 4.866071e-12 m \n", - "Energy of recoiled electron is 0.254532 MeV.\n", - "Recoiled electron angle is 26 degree36 minute \n" - ] - } - ], - "prompt_number": 27 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.19 Page No175" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "nu = 2e19 # initial frequency of X ray photon\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # scattering angle in degree\n", - "print \"Example 3.19\"\n", - "d_lambda = h/(m_e*c) # calculation of wavelength shift\n", - "k = 1./nu + d_lambda/c\n", - "nu_1 = 1/k # Frequency after collision\n", - "nu_1 = int(nu_1/1e18)*1e18 # rounding off\n", - "print \"Frequency after collision is %e Hz \"%(nu_1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.19\n", - "Frequency after collision is 1.700000e+19 Hz \n" - ] - } - ], - "prompt_number": 29 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch3_3.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch3_3.ipynb deleted file mode 100644 index 26b2b7d6..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch3_3.ipynb +++ /dev/null @@ -1,805 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:ce26666b75776c5b07c5f321f02f961249efef8c8458f3dae2110a523ce1ea02" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 3 : X ray and Compton effect" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.1 Page No134" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "d = 2.82 # crystal spacing in angstrom\n", - "n = 2 # order for longest pasmath.sing wavelength\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.1\"\n", - "lambda1 = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n", - "\n", - "print \" Longest wavelength is %f angstrom. \"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.1\n", - " Longest wavelength is 2.820000 angstrom. \n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.2 Page No134" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 0.3 # Wavelength in angstrom\n", - "d = 0.5 # crystal spacing in angstrom\n", - "n = 2 # order \n", - "m = 3 # order\n", - "print \"Example 3.2\"\n", - "theta_n = math.asin(n*lambda1/(2*d))*180/math.pi # Calculation of angle for order n\n", - "theta_m = math.asin(m*lambda1/(2*d))*180/math.pi # Calculation of angle for order m\n", - "\n", - "print \"Angle for %dnd order maxima is %f degree. \"%(n,theta_n)\n", - "print \"Angle for %drd order maxima is %f degree. \"%(m,theta_m)\n", - "# Answers in book are 40.97 degree and 72.29 degree which are due to wrong calculation\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.2\n", - "Angle for 2nd order maxima is 36.869898 degree. \n", - "Angle for 3rd order maxima is 64.158067 degree. \n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.3 Page No139" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#given that\n", - "d = 1.87 # crystal spacing in angstrom\n", - "n = 2 # order for longest pasmath.sing wavelength\n", - "theta = 30 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.3\"\n", - "lambda1 = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n", - "\n", - "print \" Longest wavelength is %f angstrom. \"%(lambda1)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.3\n", - " Longest wavelength is 0.935000 angstrom. \n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.4 Page No143" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 3.6e-9 # Wavelength in cm\n", - "theta = 4.8 # glancing angle in degree\n", - "n = 1 # order \n", - "\n", - "print \"Example 3.4\"\n", - "d = n*lambda1/(2*math.sin(theta*math.pi/180)) # calculation of crystal spacing in angstrom\n", - "\n", - "print \" Crystal spacing in angstrom is %e cm. \"%(d)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.4\n", - " Crystal spacing in angstrom is 2.151107e-08 cm. \n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.5 Page No146" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "d = 2.5 # crystal spacing in angstrom\n", - "n = 1 # order for longest pasmath.sing wavelength\n", - "theta = 20 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.5\"\n", - "lambda1 = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n", - "\n", - "print \"Longest wavelength is %f angstrom. \"%(lambda1)\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.5\n", - "Longest wavelength is 1.710101 angstrom. \n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.6 Page No146" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "d = 2.5 # crystal spacing in angstrom\n", - "n = 1 # order for longest pasmath.sing wavelength\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.6\"\n", - "lambda1 = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n", - "\n", - "print \"Longest wavelength is of %d angstrom. \"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.6\n", - "Longest wavelength is of 5 angstrom. \n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.7 Page No146" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# given that\n", - "theta1_deg = 5 # Absolut degree part of angle for first angle\n", - "theta1_min = 23#remainder minute part of angle for first angle\n", - "theta2_deg = 7 # Absolut degree part of angle for second angle\n", - "theta2_min = 37#remainder minute part of angle for second angle\n", - "theta3_deg = 9 # Absolut degree part of angle for third angle\n", - "theta3_min = 25#remainder minute part of angle for third angle\n", - "\n", - "print \"Example 3.7 \"\n", - "val1 = math.sin((theta1_deg+ theta1_min/60)*math.pi/180)# Sin value for first angle\n", - "val2 = math.sin((theta2_deg+ theta2_min/60)*math.pi/180) #Sin value for second angle\n", - "val3 = math.sin((theta3_deg+ theta3_min/60)*math.pi/180)#Sin value for third angle\n", - "ratio_21 = val2/val1\n", - "ratio_31 = val3/val1\n", - "print \" Interatomic layer separation ratios in crystal are as 1 : %f : %f\"%(ratio_21,ratio_31)\n", - "print \" Above relation shows that crystal is simple cubic crystal structure.\"\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.7 \n", - " Interatomic layer separation ratios in crystal are as 1 : 1.398294 : 1.794884\n", - " Above relation shows that crystal is simple cubic crystal structure.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.8 Page No146" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1.2 # wavelength in angstrom\n", - "theta_deg = 9 # angle fraction in degree\n", - "theta_min = 30 # Angle fraction in minute\n", - "print \"Example 3.8\"\n", - "theta = theta_deg+theta_min/60 # Total angel\n", - "for n in range(1,5):\n", - " d = lambda1/(n*2*math.sin(theta*math.pi/180)) # Inter layer spacing\n", - " print \" If order is %d then spacing is %f angstrom.\"%(n,d)\n", - "\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.8\n", - " If order is 1 then spacing is 3.835472 angstrom.\n", - " If order is 2 then spacing is 1.917736 angstrom.\n", - " If order is 3 then spacing is 1.278491 angstrom.\n", - " If order is 4 then spacing is 0.958868 angstrom.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.9 Page No147" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "h = 6.62e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "e = 1.6e-19 # charge on electron in coulomb\n", - "v = 340 # Applied voltage in volt\n", - "n = 1 # order for longest pasmath.sing wavelength\n", - "theta = 60 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.9\"\n", - "lambda1= h/math.sqrt(2*m_e*e*v) # calculation of wavelength\n", - "d = n*lambda1/(2*math.sin(theta*math.pi/180))# calculation of spacing of crystal\n", - "\n", - "print \"Spacing of crystal is %f angstrom. \"%(d*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.9\n", - "Spacing of crystal is 0.384116 angstrom. \n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.10 Page No149" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "E = 100 # Energy of X ray beam in KeV\n", - "theta = 30 # Scattering angle in degree\n", - "m = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # Speed of light in m/s\n", - "print \"Example 3.10\"\n", - "E_rest = m*c**2/(1.6e-19*1e3) # Rest mass energy in KeV\n", - "k = 1/E + (1-math.cos(theta*math.pi/180))/(E_rest)\n", - "del_e = E - 1/k # Energy of recoiled electron\n", - "print \" Energy of recoiled electron is %f KeV\"%(del_e)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.10\n", - " Energy of recoiled electron is -3720.687014 KeV\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.11 Page No154" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1 #wavelength in angstrom\n", - "h = 6.62e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.11\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))/(m_e*c) # calculation of wavelength shift \n", - "\n", - "print \"Wavelength shift is %f angstrom. \"%(d_lambda*1e10)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.11\n", - "Wavelength shift is 0.024249 angstrom. \n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.12 Page No156" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 0.015 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 60 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.12\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda\n", - "\n", - "print \" Wavelength shift is %f angstrom. \"%(lambda_n)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.12\n", - " Wavelength shift is 0.027143 angstrom. \n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.13 Page No158" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.13\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda # Calculation of recoiled electron wavelength\n", - "d_E = h*c*(lambda_n-lambda1)*1e10/(1.6e-19*lambda_n*lambda1)# Calculation of recoiled electron energy in eV\n", - "print \"Wavelength shift is %f angstrom.\"%(lambda_n)\n", - "print \"Energy of recoiled electron is %.f eV. \"%(( d_E))\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.13\n", - "Wavelength shift is 1.024286 angstrom.\n", - "Energy of recoiled electron is 295 eV. \n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.14 Page No163" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1 #let wavelength in angstrom\n", - "lambda_n = 2*lambda1 # recoiled electron wavelength\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.14\"\n", - "lambda1 = h*1e10/(m_e*c) # calculation of wavelength in angstrom\n", - "E = h*c*1e10/(lambda1*1.6e-19) # calculation of energy of electron\n", - "\n", - "print \"Wavelength shift is %f angstrom. \"%(lambda1)\n", - "print \"Energy of recoiled electron is %f KeV. \"%(E/1e3)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.14\n", - "Wavelength shift is 0.024286 angstrom. \n", - "Energy of recoiled electron is 511.875000 KeV. \n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.15 Page No168" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 2 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 45 # scattering angle \n", - "print \"Example 3.15\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda # Calculation of recoiled electron wavelength\n", - "\n", - "f = d_lambda/lambda1 # Calculation of fraction of energy lost by photon \n", - "\n", - "print \"Fraction of energy lost by photon is %f\"%(f)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.15\n", - "Fraction of energy lost by photon is 0.003557\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.16 Page No171" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#given that\n", - "E_eV = 510 # Energy of gamma ray in keV\n", - "lambda1 = 2 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # scattering angle in degree\n", - "print \"Example 3.16\"\n", - "E_j = E_eV*1e3*1.6e-19 # Energy of gamma ray in Joule\n", - "lambda1 = h*c*1e10/E_j # Calculation of wavelength in angstrom\n", - "\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda # Calculation of recoiled electron wavelength\n", - "print \"Wavelength of scattered radiation is %f Angstrom \"%(lambda_n)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.16\n", - "Wavelength of scattered radiation is 0.048661 Angstrom \n" - ] - } - ], - "prompt_number": 22 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.17 Page No172" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#given that\n", - "lambda1 = 2 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.17\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda # Calculation of recoiled electron wavelength\n", - "d_E = h*c*(lambda_n-lambda1)*1e10/(1.6e-19*lambda_n*lambda1)# Calculation of recoiled electron energy in eV\n", - "print \" Scattered wavelength is %f angstrom.\"%(lambda_n)\n", - "print \" Energy of recoiled electron is %feV. \"%(d_E)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.17\n", - " Scattered wavelength is 2.024286 angstrom.\n", - " Energy of recoiled electron is 74.569954eV. \n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.18 Page No175" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "E_eV = 510 # Energy of gamma ray in keV\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # scattering angle in degree\n", - "print \"Example 3.18\"\n", - "E_j = E_eV*1e3*1.6e-19 # Energy of gamma ray in Joule\n", - "lambda1 = h*c/E_j # Calculation of wavelength in meter\n", - "\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda/1e10 # Calculation of recoiled electron wavelength\n", - "d_E = h*c*(d_lambda/1e10)/(1.6e-19*lambda_n*lambda1)# Calculation of recoiled electron energy in eV\n", - "psi= math.atan(1/(math.tan((theta*math.pi/180)/2)/(1+(h/(lambda1*m_e*c))))) \n", - "phi_deg = 90 - psi*180/math.pi # Calculation of degree part of angle of recoiled electron \n", - "phi_min = 60*(phi_deg - int(phi_deg))# Calculation of minute part of angle of recoiled electron \n", - "print \"Wavelength of scattered radiation is %e m \"%(lambda_n)\n", - "print \"Energy of recoiled electron is %f MeV.\"%(d_E/1e6)\n", - "print \"Recoiled electron angle is %d degree%d minute \"%(phi_deg,phi_min)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.18\n", - "Wavelength of scattered radiation is 4.866071e-12 m \n", - "Energy of recoiled electron is 0.254532 MeV.\n", - "Recoiled electron angle is 26 degree36 minute \n" - ] - } - ], - "prompt_number": 27 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.19 Page No175" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "nu = 2e19 # initial frequency of X ray photon\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # scattering angle in degree\n", - "print \"Example 3.19\"\n", - "d_lambda = h/(m_e*c) # calculation of wavelength shift\n", - "k = 1./nu + d_lambda/c\n", - "nu_1 = 1/k # Frequency after collision\n", - "nu_1 = int(nu_1/1e18)*1e18 # rounding off\n", - "print \"Frequency after collision is %e Hz \"%(nu_1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.19\n", - "Frequency after collision is 1.700000e+19 Hz \n" - ] - } - ], - "prompt_number": 29 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch3_4.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch3_4.ipynb deleted file mode 100644 index 26b2b7d6..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch3_4.ipynb +++ /dev/null @@ -1,805 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:ce26666b75776c5b07c5f321f02f961249efef8c8458f3dae2110a523ce1ea02" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 3 : X ray and Compton effect" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.1 Page No134" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "d = 2.82 # crystal spacing in angstrom\n", - "n = 2 # order for longest pasmath.sing wavelength\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.1\"\n", - "lambda1 = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n", - "\n", - "print \" Longest wavelength is %f angstrom. \"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.1\n", - " Longest wavelength is 2.820000 angstrom. \n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.2 Page No134" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 0.3 # Wavelength in angstrom\n", - "d = 0.5 # crystal spacing in angstrom\n", - "n = 2 # order \n", - "m = 3 # order\n", - "print \"Example 3.2\"\n", - "theta_n = math.asin(n*lambda1/(2*d))*180/math.pi # Calculation of angle for order n\n", - "theta_m = math.asin(m*lambda1/(2*d))*180/math.pi # Calculation of angle for order m\n", - "\n", - "print \"Angle for %dnd order maxima is %f degree. \"%(n,theta_n)\n", - "print \"Angle for %drd order maxima is %f degree. \"%(m,theta_m)\n", - "# Answers in book are 40.97 degree and 72.29 degree which are due to wrong calculation\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.2\n", - "Angle for 2nd order maxima is 36.869898 degree. \n", - "Angle for 3rd order maxima is 64.158067 degree. \n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.3 Page No139" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#given that\n", - "d = 1.87 # crystal spacing in angstrom\n", - "n = 2 # order for longest pasmath.sing wavelength\n", - "theta = 30 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.3\"\n", - "lambda1 = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n", - "\n", - "print \" Longest wavelength is %f angstrom. \"%(lambda1)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.3\n", - " Longest wavelength is 0.935000 angstrom. \n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.4 Page No143" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 3.6e-9 # Wavelength in cm\n", - "theta = 4.8 # glancing angle in degree\n", - "n = 1 # order \n", - "\n", - "print \"Example 3.4\"\n", - "d = n*lambda1/(2*math.sin(theta*math.pi/180)) # calculation of crystal spacing in angstrom\n", - "\n", - "print \" Crystal spacing in angstrom is %e cm. \"%(d)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.4\n", - " Crystal spacing in angstrom is 2.151107e-08 cm. \n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.5 Page No146" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "d = 2.5 # crystal spacing in angstrom\n", - "n = 1 # order for longest pasmath.sing wavelength\n", - "theta = 20 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.5\"\n", - "lambda1 = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n", - "\n", - "print \"Longest wavelength is %f angstrom. \"%(lambda1)\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.5\n", - "Longest wavelength is 1.710101 angstrom. \n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.6 Page No146" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "d = 2.5 # crystal spacing in angstrom\n", - "n = 1 # order for longest pasmath.sing wavelength\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.6\"\n", - "lambda1 = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n", - "\n", - "print \"Longest wavelength is of %d angstrom. \"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.6\n", - "Longest wavelength is of 5 angstrom. \n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.7 Page No146" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# given that\n", - "theta1_deg = 5 # Absolut degree part of angle for first angle\n", - "theta1_min = 23#remainder minute part of angle for first angle\n", - "theta2_deg = 7 # Absolut degree part of angle for second angle\n", - "theta2_min = 37#remainder minute part of angle for second angle\n", - "theta3_deg = 9 # Absolut degree part of angle for third angle\n", - "theta3_min = 25#remainder minute part of angle for third angle\n", - "\n", - "print \"Example 3.7 \"\n", - "val1 = math.sin((theta1_deg+ theta1_min/60)*math.pi/180)# Sin value for first angle\n", - "val2 = math.sin((theta2_deg+ theta2_min/60)*math.pi/180) #Sin value for second angle\n", - "val3 = math.sin((theta3_deg+ theta3_min/60)*math.pi/180)#Sin value for third angle\n", - "ratio_21 = val2/val1\n", - "ratio_31 = val3/val1\n", - "print \" Interatomic layer separation ratios in crystal are as 1 : %f : %f\"%(ratio_21,ratio_31)\n", - "print \" Above relation shows that crystal is simple cubic crystal structure.\"\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.7 \n", - " Interatomic layer separation ratios in crystal are as 1 : 1.398294 : 1.794884\n", - " Above relation shows that crystal is simple cubic crystal structure.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.8 Page No146" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1.2 # wavelength in angstrom\n", - "theta_deg = 9 # angle fraction in degree\n", - "theta_min = 30 # Angle fraction in minute\n", - "print \"Example 3.8\"\n", - "theta = theta_deg+theta_min/60 # Total angel\n", - "for n in range(1,5):\n", - " d = lambda1/(n*2*math.sin(theta*math.pi/180)) # Inter layer spacing\n", - " print \" If order is %d then spacing is %f angstrom.\"%(n,d)\n", - "\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.8\n", - " If order is 1 then spacing is 3.835472 angstrom.\n", - " If order is 2 then spacing is 1.917736 angstrom.\n", - " If order is 3 then spacing is 1.278491 angstrom.\n", - " If order is 4 then spacing is 0.958868 angstrom.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.9 Page No147" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "h = 6.62e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "e = 1.6e-19 # charge on electron in coulomb\n", - "v = 340 # Applied voltage in volt\n", - "n = 1 # order for longest pasmath.sing wavelength\n", - "theta = 60 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.9\"\n", - "lambda1= h/math.sqrt(2*m_e*e*v) # calculation of wavelength\n", - "d = n*lambda1/(2*math.sin(theta*math.pi/180))# calculation of spacing of crystal\n", - "\n", - "print \"Spacing of crystal is %f angstrom. \"%(d*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.9\n", - "Spacing of crystal is 0.384116 angstrom. \n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.10 Page No149" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "E = 100 # Energy of X ray beam in KeV\n", - "theta = 30 # Scattering angle in degree\n", - "m = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # Speed of light in m/s\n", - "print \"Example 3.10\"\n", - "E_rest = m*c**2/(1.6e-19*1e3) # Rest mass energy in KeV\n", - "k = 1/E + (1-math.cos(theta*math.pi/180))/(E_rest)\n", - "del_e = E - 1/k # Energy of recoiled electron\n", - "print \" Energy of recoiled electron is %f KeV\"%(del_e)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.10\n", - " Energy of recoiled electron is -3720.687014 KeV\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.11 Page No154" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1 #wavelength in angstrom\n", - "h = 6.62e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.11\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))/(m_e*c) # calculation of wavelength shift \n", - "\n", - "print \"Wavelength shift is %f angstrom. \"%(d_lambda*1e10)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.11\n", - "Wavelength shift is 0.024249 angstrom. \n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.12 Page No156" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 0.015 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 60 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.12\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda\n", - "\n", - "print \" Wavelength shift is %f angstrom. \"%(lambda_n)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.12\n", - " Wavelength shift is 0.027143 angstrom. \n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.13 Page No158" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.13\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda # Calculation of recoiled electron wavelength\n", - "d_E = h*c*(lambda_n-lambda1)*1e10/(1.6e-19*lambda_n*lambda1)# Calculation of recoiled electron energy in eV\n", - "print \"Wavelength shift is %f angstrom.\"%(lambda_n)\n", - "print \"Energy of recoiled electron is %.f eV. \"%(( d_E))\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.13\n", - "Wavelength shift is 1.024286 angstrom.\n", - "Energy of recoiled electron is 295 eV. \n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.14 Page No163" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1 #let wavelength in angstrom\n", - "lambda_n = 2*lambda1 # recoiled electron wavelength\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.14\"\n", - "lambda1 = h*1e10/(m_e*c) # calculation of wavelength in angstrom\n", - "E = h*c*1e10/(lambda1*1.6e-19) # calculation of energy of electron\n", - "\n", - "print \"Wavelength shift is %f angstrom. \"%(lambda1)\n", - "print \"Energy of recoiled electron is %f KeV. \"%(E/1e3)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.14\n", - "Wavelength shift is 0.024286 angstrom. \n", - "Energy of recoiled electron is 511.875000 KeV. \n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.15 Page No168" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 2 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 45 # scattering angle \n", - "print \"Example 3.15\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda # Calculation of recoiled electron wavelength\n", - "\n", - "f = d_lambda/lambda1 # Calculation of fraction of energy lost by photon \n", - "\n", - "print \"Fraction of energy lost by photon is %f\"%(f)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.15\n", - "Fraction of energy lost by photon is 0.003557\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.16 Page No171" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#given that\n", - "E_eV = 510 # Energy of gamma ray in keV\n", - "lambda1 = 2 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # scattering angle in degree\n", - "print \"Example 3.16\"\n", - "E_j = E_eV*1e3*1.6e-19 # Energy of gamma ray in Joule\n", - "lambda1 = h*c*1e10/E_j # Calculation of wavelength in angstrom\n", - "\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda # Calculation of recoiled electron wavelength\n", - "print \"Wavelength of scattered radiation is %f Angstrom \"%(lambda_n)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.16\n", - "Wavelength of scattered radiation is 0.048661 Angstrom \n" - ] - } - ], - "prompt_number": 22 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.17 Page No172" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#given that\n", - "lambda1 = 2 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.17\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda # Calculation of recoiled electron wavelength\n", - "d_E = h*c*(lambda_n-lambda1)*1e10/(1.6e-19*lambda_n*lambda1)# Calculation of recoiled electron energy in eV\n", - "print \" Scattered wavelength is %f angstrom.\"%(lambda_n)\n", - "print \" Energy of recoiled electron is %feV. \"%(d_E)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.17\n", - " Scattered wavelength is 2.024286 angstrom.\n", - " Energy of recoiled electron is 74.569954eV. \n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.18 Page No175" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "E_eV = 510 # Energy of gamma ray in keV\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # scattering angle in degree\n", - "print \"Example 3.18\"\n", - "E_j = E_eV*1e3*1.6e-19 # Energy of gamma ray in Joule\n", - "lambda1 = h*c/E_j # Calculation of wavelength in meter\n", - "\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda/1e10 # Calculation of recoiled electron wavelength\n", - "d_E = h*c*(d_lambda/1e10)/(1.6e-19*lambda_n*lambda1)# Calculation of recoiled electron energy in eV\n", - "psi= math.atan(1/(math.tan((theta*math.pi/180)/2)/(1+(h/(lambda1*m_e*c))))) \n", - "phi_deg = 90 - psi*180/math.pi # Calculation of degree part of angle of recoiled electron \n", - "phi_min = 60*(phi_deg - int(phi_deg))# Calculation of minute part of angle of recoiled electron \n", - "print \"Wavelength of scattered radiation is %e m \"%(lambda_n)\n", - "print \"Energy of recoiled electron is %f MeV.\"%(d_E/1e6)\n", - "print \"Recoiled electron angle is %d degree%d minute \"%(phi_deg,phi_min)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.18\n", - "Wavelength of scattered radiation is 4.866071e-12 m \n", - "Energy of recoiled electron is 0.254532 MeV.\n", - "Recoiled electron angle is 26 degree36 minute \n" - ] - } - ], - "prompt_number": 27 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.19 Page No175" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "nu = 2e19 # initial frequency of X ray photon\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # scattering angle in degree\n", - "print \"Example 3.19\"\n", - "d_lambda = h/(m_e*c) # calculation of wavelength shift\n", - "k = 1./nu + d_lambda/c\n", - "nu_1 = 1/k # Frequency after collision\n", - "nu_1 = int(nu_1/1e18)*1e18 # rounding off\n", - "print \"Frequency after collision is %e Hz \"%(nu_1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.19\n", - "Frequency after collision is 1.700000e+19 Hz \n" - ] - } - ], - "prompt_number": 29 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch3_5.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch3_5.ipynb deleted file mode 100644 index 26b2b7d6..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch3_5.ipynb +++ /dev/null @@ -1,805 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:ce26666b75776c5b07c5f321f02f961249efef8c8458f3dae2110a523ce1ea02" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 3 : X ray and Compton effect" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.1 Page No134" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "d = 2.82 # crystal spacing in angstrom\n", - "n = 2 # order for longest pasmath.sing wavelength\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.1\"\n", - "lambda1 = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n", - "\n", - "print \" Longest wavelength is %f angstrom. \"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.1\n", - " Longest wavelength is 2.820000 angstrom. \n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.2 Page No134" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 0.3 # Wavelength in angstrom\n", - "d = 0.5 # crystal spacing in angstrom\n", - "n = 2 # order \n", - "m = 3 # order\n", - "print \"Example 3.2\"\n", - "theta_n = math.asin(n*lambda1/(2*d))*180/math.pi # Calculation of angle for order n\n", - "theta_m = math.asin(m*lambda1/(2*d))*180/math.pi # Calculation of angle for order m\n", - "\n", - "print \"Angle for %dnd order maxima is %f degree. \"%(n,theta_n)\n", - "print \"Angle for %drd order maxima is %f degree. \"%(m,theta_m)\n", - "# Answers in book are 40.97 degree and 72.29 degree which are due to wrong calculation\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.2\n", - "Angle for 2nd order maxima is 36.869898 degree. \n", - "Angle for 3rd order maxima is 64.158067 degree. \n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.3 Page No139" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#given that\n", - "d = 1.87 # crystal spacing in angstrom\n", - "n = 2 # order for longest pasmath.sing wavelength\n", - "theta = 30 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.3\"\n", - "lambda1 = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n", - "\n", - "print \" Longest wavelength is %f angstrom. \"%(lambda1)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.3\n", - " Longest wavelength is 0.935000 angstrom. \n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.4 Page No143" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 3.6e-9 # Wavelength in cm\n", - "theta = 4.8 # glancing angle in degree\n", - "n = 1 # order \n", - "\n", - "print \"Example 3.4\"\n", - "d = n*lambda1/(2*math.sin(theta*math.pi/180)) # calculation of crystal spacing in angstrom\n", - "\n", - "print \" Crystal spacing in angstrom is %e cm. \"%(d)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.4\n", - " Crystal spacing in angstrom is 2.151107e-08 cm. \n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.5 Page No146" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "d = 2.5 # crystal spacing in angstrom\n", - "n = 1 # order for longest pasmath.sing wavelength\n", - "theta = 20 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.5\"\n", - "lambda1 = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n", - "\n", - "print \"Longest wavelength is %f angstrom. \"%(lambda1)\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.5\n", - "Longest wavelength is 1.710101 angstrom. \n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.6 Page No146" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "d = 2.5 # crystal spacing in angstrom\n", - "n = 1 # order for longest pasmath.sing wavelength\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.6\"\n", - "lambda1 = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n", - "\n", - "print \"Longest wavelength is of %d angstrom. \"%(lambda1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.6\n", - "Longest wavelength is of 5 angstrom. \n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.7 Page No146" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# given that\n", - "theta1_deg = 5 # Absolut degree part of angle for first angle\n", - "theta1_min = 23#remainder minute part of angle for first angle\n", - "theta2_deg = 7 # Absolut degree part of angle for second angle\n", - "theta2_min = 37#remainder minute part of angle for second angle\n", - "theta3_deg = 9 # Absolut degree part of angle for third angle\n", - "theta3_min = 25#remainder minute part of angle for third angle\n", - "\n", - "print \"Example 3.7 \"\n", - "val1 = math.sin((theta1_deg+ theta1_min/60)*math.pi/180)# Sin value for first angle\n", - "val2 = math.sin((theta2_deg+ theta2_min/60)*math.pi/180) #Sin value for second angle\n", - "val3 = math.sin((theta3_deg+ theta3_min/60)*math.pi/180)#Sin value for third angle\n", - "ratio_21 = val2/val1\n", - "ratio_31 = val3/val1\n", - "print \" Interatomic layer separation ratios in crystal are as 1 : %f : %f\"%(ratio_21,ratio_31)\n", - "print \" Above relation shows that crystal is simple cubic crystal structure.\"\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.7 \n", - " Interatomic layer separation ratios in crystal are as 1 : 1.398294 : 1.794884\n", - " Above relation shows that crystal is simple cubic crystal structure.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.8 Page No146" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1.2 # wavelength in angstrom\n", - "theta_deg = 9 # angle fraction in degree\n", - "theta_min = 30 # Angle fraction in minute\n", - "print \"Example 3.8\"\n", - "theta = theta_deg+theta_min/60 # Total angel\n", - "for n in range(1,5):\n", - " d = lambda1/(n*2*math.sin(theta*math.pi/180)) # Inter layer spacing\n", - " print \" If order is %d then spacing is %f angstrom.\"%(n,d)\n", - "\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.8\n", - " If order is 1 then spacing is 3.835472 angstrom.\n", - " If order is 2 then spacing is 1.917736 angstrom.\n", - " If order is 3 then spacing is 1.278491 angstrom.\n", - " If order is 4 then spacing is 0.958868 angstrom.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.9 Page No147" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "h = 6.62e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "e = 1.6e-19 # charge on electron in coulomb\n", - "v = 340 # Applied voltage in volt\n", - "n = 1 # order for longest pasmath.sing wavelength\n", - "theta = 60 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.9\"\n", - "lambda1= h/math.sqrt(2*m_e*e*v) # calculation of wavelength\n", - "d = n*lambda1/(2*math.sin(theta*math.pi/180))# calculation of spacing of crystal\n", - "\n", - "print \"Spacing of crystal is %f angstrom. \"%(d*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.9\n", - "Spacing of crystal is 0.384116 angstrom. \n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.10 Page No149" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "E = 100 # Energy of X ray beam in KeV\n", - "theta = 30 # Scattering angle in degree\n", - "m = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # Speed of light in m/s\n", - "print \"Example 3.10\"\n", - "E_rest = m*c**2/(1.6e-19*1e3) # Rest mass energy in KeV\n", - "k = 1/E + (1-math.cos(theta*math.pi/180))/(E_rest)\n", - "del_e = E - 1/k # Energy of recoiled electron\n", - "print \" Energy of recoiled electron is %f KeV\"%(del_e)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.10\n", - " Energy of recoiled electron is -3720.687014 KeV\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.11 Page No154" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1 #wavelength in angstrom\n", - "h = 6.62e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.11\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))/(m_e*c) # calculation of wavelength shift \n", - "\n", - "print \"Wavelength shift is %f angstrom. \"%(d_lambda*1e10)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.11\n", - "Wavelength shift is 0.024249 angstrom. \n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.12 Page No156" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 0.015 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 60 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.12\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda\n", - "\n", - "print \" Wavelength shift is %f angstrom. \"%(lambda_n)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.12\n", - " Wavelength shift is 0.027143 angstrom. \n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.13 Page No158" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.13\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda # Calculation of recoiled electron wavelength\n", - "d_E = h*c*(lambda_n-lambda1)*1e10/(1.6e-19*lambda_n*lambda1)# Calculation of recoiled electron energy in eV\n", - "print \"Wavelength shift is %f angstrom.\"%(lambda_n)\n", - "print \"Energy of recoiled electron is %.f eV. \"%(( d_E))\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.13\n", - "Wavelength shift is 1.024286 angstrom.\n", - "Energy of recoiled electron is 295 eV. \n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.14 Page No163" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 1 #let wavelength in angstrom\n", - "lambda_n = 2*lambda1 # recoiled electron wavelength\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.14\"\n", - "lambda1 = h*1e10/(m_e*c) # calculation of wavelength in angstrom\n", - "E = h*c*1e10/(lambda1*1.6e-19) # calculation of energy of electron\n", - "\n", - "print \"Wavelength shift is %f angstrom. \"%(lambda1)\n", - "print \"Energy of recoiled electron is %f KeV. \"%(E/1e3)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.14\n", - "Wavelength shift is 0.024286 angstrom. \n", - "Energy of recoiled electron is 511.875000 KeV. \n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.15 Page No168" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "lambda1 = 2 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 45 # scattering angle \n", - "print \"Example 3.15\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda # Calculation of recoiled electron wavelength\n", - "\n", - "f = d_lambda/lambda1 # Calculation of fraction of energy lost by photon \n", - "\n", - "print \"Fraction of energy lost by photon is %f\"%(f)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.15\n", - "Fraction of energy lost by photon is 0.003557\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.16 Page No171" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#given that\n", - "E_eV = 510 # Energy of gamma ray in keV\n", - "lambda1 = 2 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # scattering angle in degree\n", - "print \"Example 3.16\"\n", - "E_j = E_eV*1e3*1.6e-19 # Energy of gamma ray in Joule\n", - "lambda1 = h*c*1e10/E_j # Calculation of wavelength in angstrom\n", - "\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda # Calculation of recoiled electron wavelength\n", - "print \"Wavelength of scattered radiation is %f Angstrom \"%(lambda_n)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.16\n", - "Wavelength of scattered radiation is 0.048661 Angstrom \n" - ] - } - ], - "prompt_number": 22 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.17 Page No172" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#given that\n", - "lambda1 = 2 #wavelength in angstrom\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # angle for longest pasmath.sing wavelength\n", - "print \"Example 3.17\"\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda # Calculation of recoiled electron wavelength\n", - "d_E = h*c*(lambda_n-lambda1)*1e10/(1.6e-19*lambda_n*lambda1)# Calculation of recoiled electron energy in eV\n", - "print \" Scattered wavelength is %f angstrom.\"%(lambda_n)\n", - "print \" Energy of recoiled electron is %feV. \"%(d_E)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.17\n", - " Scattered wavelength is 2.024286 angstrom.\n", - " Energy of recoiled electron is 74.569954eV. \n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.18 Page No175" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#given that\n", - "E_eV = 510 # Energy of gamma ray in keV\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # scattering angle in degree\n", - "print \"Example 3.18\"\n", - "E_j = E_eV*1e3*1.6e-19 # Energy of gamma ray in Joule\n", - "lambda1 = h*c/E_j # Calculation of wavelength in meter\n", - "\n", - "d_lambda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n", - "lambda_n = lambda1+d_lambda/1e10 # Calculation of recoiled electron wavelength\n", - "d_E = h*c*(d_lambda/1e10)/(1.6e-19*lambda_n*lambda1)# Calculation of recoiled electron energy in eV\n", - "psi= math.atan(1/(math.tan((theta*math.pi/180)/2)/(1+(h/(lambda1*m_e*c))))) \n", - "phi_deg = 90 - psi*180/math.pi # Calculation of degree part of angle of recoiled electron \n", - "phi_min = 60*(phi_deg - int(phi_deg))# Calculation of minute part of angle of recoiled electron \n", - "print \"Wavelength of scattered radiation is %e m \"%(lambda_n)\n", - "print \"Energy of recoiled electron is %f MeV.\"%(d_E/1e6)\n", - "print \"Recoiled electron angle is %d degree%d minute \"%(phi_deg,phi_min)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.18\n", - "Wavelength of scattered radiation is 4.866071e-12 m \n", - "Energy of recoiled electron is 0.254532 MeV.\n", - "Recoiled electron angle is 26 degree36 minute \n" - ] - } - ], - "prompt_number": 27 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.19 Page No175" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#given that\n", - "nu = 2e19 # initial frequency of X ray photon\n", - "h = 6.63e-34 # Planks consmath.tant\n", - "m_e = 9.1e-31 # mass of electron in kg\n", - "c = 3e8 # speed of light in m/sec\n", - "theta = 90 # scattering angle in degree\n", - "print \"Example 3.19\"\n", - "d_lambda = h/(m_e*c) # calculation of wavelength shift\n", - "k = 1./nu + d_lambda/c\n", - "nu_1 = 1/k # Frequency after collision\n", - "nu_1 = int(nu_1/1e18)*1e18 # rounding off\n", - "print \"Frequency after collision is %e Hz \"%(nu_1)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 3.19\n", - "Frequency after collision is 1.700000e+19 Hz \n" - ] - } - ], - "prompt_number": 29 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch4.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch4.ipynb deleted file mode 100755 index 0ea076e1..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch4.ipynb +++ /dev/null @@ -1,557 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:98a33103545b2365c3949736414c5684c66408433d883a4d0f1b451b5a78b395" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 4 : Dielectrics" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.4 Page No177" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "epsilon_r = 1.000074 # Dielectric consmath.tant of He at 0C and 1atm\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "E = 100 # Electric field in V/m\n", - "n = 2.68e27 # Electron density in no,/m**\n", - "N_a = 6e23 # Avogadro number\n", - "V = 22.4 # Volume at STP in litter\n", - "print \"Example 4.4\"\n", - "P = epsilon_0*(epsilon_r-1)*E # Calculation of polarization\n", - "\n", - "N = N_a/(V*1e-3)# Calculation of total number of atoms\n", - "p = P/N # dipole moment per atom\n", - "print \" Dipole moment per atom is %e Coulomb-meter \"%(p)\n", - "# Answer in book is in different form and as 24.45e-40 coulomb-meter\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.4\n", - " Dipole moment per atom is 2.446065e-39 Coulomb-meter \n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.6 Page No181" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "r = 0.055 # Radius of hydrogen atom in nm\n", - "n = 9.8e26 # Number of atoms/cc\n", - "\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.6\"\n", - "alpha_e = 4*math.pi*epsilon_0*(r*1e-9)**3 # Calculation of electronic polarisability\n", - "epsilon_r = 1+n*alpha_e/epsilon_0 # Calculation of relative permeability\n", - "\n", - "print \" Electronic polarisability is %e Fm**2 Relative permeability is %f \"%(alpha_e,epsilon_r)\n", - "\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.6\n", - " Electronic polarisability is 1.851132e-41 Fm**2 Relative permeability is 1.002049 \n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.8 Page No182" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "E = 2000 # Electric field in V/m\n", - "P = 6.4e-8 # Polarization in C/m**2\n", - "print \"Example 4.8\"\n", - "epsilon_r = 1+ P/(epsilon_0*E) # Calculation of relative permittivity\n", - "\n", - "print \" Relative permittivity is %f\"%(epsilon_r)\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.8\n", - " Relative permittivity is 4.614186\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.9 Page No183" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "alpha_e = 2e-40 # Electronic polarisability in Fm**2\n", - "N = 4e28 # density in atoms/m**3\n", - "epsilon_0 = 8.85e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.9\"\n", - "epsilon_r = 1+ N*alpha_e/(epsilon_0) # Calculation of relative permittivity\n", - "print \" Relative permittivity is %f\"%(epsilon_r)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.9\n", - " Relative permittivity is 1.903955\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.10 Page No188" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "epsilon = 2.4e-10 # permitivity of a dielectric material in C**2/N?m**2\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.10\"\n", - "K = epsilon/epsilon_0 # Calculation of dielectric consmath.tant \n", - "zai_e = epsilon_0*(K-1) # Calculation of electrical susceptibility \n", - "\n", - "print \" Relative permittivity is %f\"%(K)\n", - "print \" Electrical susceptibility is %e C**2/Nm**2\"%(zai_e)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.10\n", - " Relative permittivity is 27.106393\n", - " Electrical susceptibility is 2.311460e-10 C**2/Nm**2\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.11 Page No189" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "V = 100 # Applied potential in Volt\n", - "d = 1 # Separation between plates in cm\n", - "k1 = 8 # Dielectric consmath.tant\n", - "k2 = 9 #dielectric consmath.tant\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.11\"\n", - "E_0 = V/(d*1e-2) # Calculation of electric field\n", - "E = E_0/k1*k2 # Calculation of electric field\n", - "D = k1*epsilon_0*E # Calculation of electrical print lacement vector\n", - "P = (k1-1)*epsilon_0*E # Calculation of electrical polarization \n", - "\n", - "print \" Magnitude of Electrical vector is %e Volt/meter\"%(E) # Answer in book is 1.125e3 Volt/meter\n", - "\n", - "print \" Magnitude of Electrical Displacement vector is %e C/m**2\"%(D)# Answer in book is 8.85e-8C/m**2\n", - "\n", - "print \" Magnitude of Electric polarization vector is %e C/m**2\"%(P)# Answer in book is 7.774e-8C/m**2\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.11\n", - " Magnitude of Electrical vector is 1.125000e+04 Volt/meter\n", - " Magnitude of Electrical Displacement vector is 7.968600e-07 C/m**2\n", - " Magnitude of Electric polarization vector is 6.972525e-07 C/m**2\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.12 Page No189" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "alpha_300 = 2.5e-39 # total polarisability in C**2m/N at 300 K\n", - "alpha_600 = 1.75e-39 # total polarisability in C**2m/N at 600 K\n", - "T1 = 300 # Initial temperature in Kelvin\n", - "T2 = 600 # Final Temperature in Kelvin\n", - "print \"Example 4.12\"\n", - "b = (alpha_300-alpha_600)*T2\n", - "al_def_300 = alpha_300 - b/300\n", - "al_oriant_300 = b/300\n", - "al_oriant_600 = b/600\n", - "print \" Deformational Polarizability is %e C**2mN**-1\"%(al_def_300)\n", - "print \" Orientational Polarizability at %d degree Celcius is %e C**2mN**-1\"%(T1,al_oriant_300)\n", - "print \" Orientational Polarizability at %d degree Celcius is %e C**2mN**-1\"%(T2,al_oriant_600)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.12\n", - " Deformational Polarizability is 1.000000e-39 C**2mN**-1\n", - " Orientational Polarizability at 300 degree Celcius is 1.500000e-39 C**2mN**-1\n", - " Orientational Polarizability at 600 degree Celcius is 7.500000e-40 C**2mN**-1\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.13 Page No189" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "alpha_e = 1.5e-40 # Electronic polarizability in Fm**2\n", - "N = 4e28 # density in atoms/m**3\n", - "epsilon_0 = 8.85e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.13\"\n", - "k = N*alpha_e/(3*epsilon_0)\n", - "epsilon_r = (1+ k*2)/(1-k)# Calculation of relative permittivity\n", - "print \" Relative permittivity is %f\"%(epsilon_r)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.13\n", - " Relative permittivity is 1.875912\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.14 Page No191" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "m = 32 # Atomic weight of sulphur\n", - "d = 2.08 # Density in g/cm**3\n", - "alpha_e = 3.5e-40 # Electronic polarizability in Fm**2\n", - "N_a = 6.022e23 # Avogadro Number\n", - "epsilon_0 = 8.85e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.14\"\n", - "N = N_a*d*1e6/m # Calculation of Atoms per unit \n", - "k = N*alpha_e/(3*epsilon_0)\n", - "\n", - "epsilon_r = (1+ k*2)/(1-k)# Calculation of relative permittivity\n", - "print \" Relative permittivity is %f\"%(epsilon_r)\n", - "# Answer in book is 4.17\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.14\n", - " Relative permittivity is 4.198468\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.15 Page No191" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "n = 1.5 # Refractive index\n", - "epsilon = 5.6 # Static dielectric consmath.tant\n", - "print \"Example 4.15\"\n", - "per = (1-((n**2-1)/(n**2+2))*(epsilon+2)/(epsilon-1))*100 # Pecentage ionic polarisability\n", - "print \" Percentage ionic polarizability is %f pecent\"%(per)\n", - "# Answer in book is 5.14 %\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.15\n", - " Percentage ionic polarizability is 51.406650 pecent\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.16 Page No195" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#Given that\n", - "m = 32 # Atomic weight of sulphur\n", - "d = 2050 # Density in Kg/m**3\n", - "N_a = 6.022e23 # Avogadro Number\n", - "epsilon_0 = 8.85e-12 # Permittivity of free space\n", - "epsilon_r = 3.75 # Dielectric consmath.tant of sulphur\n", - "\n", - "print \"Example 4.16\"\n", - "N = N_a*d*1e3/m # Calculation of Atoms per unit \n", - "alpha_e = 3*epsilon_0*((epsilon_r-1)/(epsilon_r+2)) / N \n", - "\n", - "\n", - "print \" Electronic polarizability is %e Fm**2\"%(alpha_e)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.16\n", - " Electronic polarizability is 3.291431e-40 Fm**2\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.17 Page No199" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "n = 1.5 # Refractive index\n", - "epsilon = 4. # Static dielectric consmath.tant\n", - "epsilon_0 = 8.85e-12 # permittivity of free space\n", - "print \"Example 4.17\"\n", - "k1 = (epsilon-1)/(epsilon+2)\n", - "k2 = (n**2-1)/(n**2+2)\n", - "ratio = 1./((k1/k2)-1) \n", - "print \" Ratio of electronic to ionic polarizability is %.2f .\"%(ratio)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.17\n", - " Ratio of electronic to ionic polarizability is 1.43 .\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.18 Page No202" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "t = 1.8e-5 # Relaxation time in second\n", - "epsilon_r = 1 # let\n", - "print \"Example 4.18\"\n", - "f = 1./(2*math.pi*t) # Calculation of frequency\n", - "delta = math.atan(epsilon_r/epsilon_r)\n", - "phi = 90 - delta*180/math.pi # Calculation of phase difference\n", - "print \" Frequency is %f KHz\"%(f/1e3)\n", - "print \" Phase difference between current and voltage is %d degree.\"%(phi)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.18\n", - " Frequency is 8.841941 KHz\n", - " Phase difference between current and voltage is 45 degree.\n" - ] - } - ], - "prompt_number": 18 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch4_1.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch4_1.ipynb deleted file mode 100755 index 0ea076e1..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch4_1.ipynb +++ /dev/null @@ -1,557 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:98a33103545b2365c3949736414c5684c66408433d883a4d0f1b451b5a78b395" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 4 : Dielectrics" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.4 Page No177" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "epsilon_r = 1.000074 # Dielectric consmath.tant of He at 0C and 1atm\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "E = 100 # Electric field in V/m\n", - "n = 2.68e27 # Electron density in no,/m**\n", - "N_a = 6e23 # Avogadro number\n", - "V = 22.4 # Volume at STP in litter\n", - "print \"Example 4.4\"\n", - "P = epsilon_0*(epsilon_r-1)*E # Calculation of polarization\n", - "\n", - "N = N_a/(V*1e-3)# Calculation of total number of atoms\n", - "p = P/N # dipole moment per atom\n", - "print \" Dipole moment per atom is %e Coulomb-meter \"%(p)\n", - "# Answer in book is in different form and as 24.45e-40 coulomb-meter\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.4\n", - " Dipole moment per atom is 2.446065e-39 Coulomb-meter \n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.6 Page No181" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "r = 0.055 # Radius of hydrogen atom in nm\n", - "n = 9.8e26 # Number of atoms/cc\n", - "\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.6\"\n", - "alpha_e = 4*math.pi*epsilon_0*(r*1e-9)**3 # Calculation of electronic polarisability\n", - "epsilon_r = 1+n*alpha_e/epsilon_0 # Calculation of relative permeability\n", - "\n", - "print \" Electronic polarisability is %e Fm**2 Relative permeability is %f \"%(alpha_e,epsilon_r)\n", - "\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.6\n", - " Electronic polarisability is 1.851132e-41 Fm**2 Relative permeability is 1.002049 \n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.8 Page No182" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "E = 2000 # Electric field in V/m\n", - "P = 6.4e-8 # Polarization in C/m**2\n", - "print \"Example 4.8\"\n", - "epsilon_r = 1+ P/(epsilon_0*E) # Calculation of relative permittivity\n", - "\n", - "print \" Relative permittivity is %f\"%(epsilon_r)\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.8\n", - " Relative permittivity is 4.614186\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.9 Page No183" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "alpha_e = 2e-40 # Electronic polarisability in Fm**2\n", - "N = 4e28 # density in atoms/m**3\n", - "epsilon_0 = 8.85e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.9\"\n", - "epsilon_r = 1+ N*alpha_e/(epsilon_0) # Calculation of relative permittivity\n", - "print \" Relative permittivity is %f\"%(epsilon_r)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.9\n", - " Relative permittivity is 1.903955\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.10 Page No188" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "epsilon = 2.4e-10 # permitivity of a dielectric material in C**2/N?m**2\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.10\"\n", - "K = epsilon/epsilon_0 # Calculation of dielectric consmath.tant \n", - "zai_e = epsilon_0*(K-1) # Calculation of electrical susceptibility \n", - "\n", - "print \" Relative permittivity is %f\"%(K)\n", - "print \" Electrical susceptibility is %e C**2/Nm**2\"%(zai_e)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.10\n", - " Relative permittivity is 27.106393\n", - " Electrical susceptibility is 2.311460e-10 C**2/Nm**2\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.11 Page No189" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "V = 100 # Applied potential in Volt\n", - "d = 1 # Separation between plates in cm\n", - "k1 = 8 # Dielectric consmath.tant\n", - "k2 = 9 #dielectric consmath.tant\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.11\"\n", - "E_0 = V/(d*1e-2) # Calculation of electric field\n", - "E = E_0/k1*k2 # Calculation of electric field\n", - "D = k1*epsilon_0*E # Calculation of electrical print lacement vector\n", - "P = (k1-1)*epsilon_0*E # Calculation of electrical polarization \n", - "\n", - "print \" Magnitude of Electrical vector is %e Volt/meter\"%(E) # Answer in book is 1.125e3 Volt/meter\n", - "\n", - "print \" Magnitude of Electrical Displacement vector is %e C/m**2\"%(D)# Answer in book is 8.85e-8C/m**2\n", - "\n", - "print \" Magnitude of Electric polarization vector is %e C/m**2\"%(P)# Answer in book is 7.774e-8C/m**2\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.11\n", - " Magnitude of Electrical vector is 1.125000e+04 Volt/meter\n", - " Magnitude of Electrical Displacement vector is 7.968600e-07 C/m**2\n", - " Magnitude of Electric polarization vector is 6.972525e-07 C/m**2\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.12 Page No189" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "alpha_300 = 2.5e-39 # total polarisability in C**2m/N at 300 K\n", - "alpha_600 = 1.75e-39 # total polarisability in C**2m/N at 600 K\n", - "T1 = 300 # Initial temperature in Kelvin\n", - "T2 = 600 # Final Temperature in Kelvin\n", - "print \"Example 4.12\"\n", - "b = (alpha_300-alpha_600)*T2\n", - "al_def_300 = alpha_300 - b/300\n", - "al_oriant_300 = b/300\n", - "al_oriant_600 = b/600\n", - "print \" Deformational Polarizability is %e C**2mN**-1\"%(al_def_300)\n", - "print \" Orientational Polarizability at %d degree Celcius is %e C**2mN**-1\"%(T1,al_oriant_300)\n", - "print \" Orientational Polarizability at %d degree Celcius is %e C**2mN**-1\"%(T2,al_oriant_600)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.12\n", - " Deformational Polarizability is 1.000000e-39 C**2mN**-1\n", - " Orientational Polarizability at 300 degree Celcius is 1.500000e-39 C**2mN**-1\n", - " Orientational Polarizability at 600 degree Celcius is 7.500000e-40 C**2mN**-1\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.13 Page No189" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "alpha_e = 1.5e-40 # Electronic polarizability in Fm**2\n", - "N = 4e28 # density in atoms/m**3\n", - "epsilon_0 = 8.85e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.13\"\n", - "k = N*alpha_e/(3*epsilon_0)\n", - "epsilon_r = (1+ k*2)/(1-k)# Calculation of relative permittivity\n", - "print \" Relative permittivity is %f\"%(epsilon_r)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.13\n", - " Relative permittivity is 1.875912\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.14 Page No191" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "m = 32 # Atomic weight of sulphur\n", - "d = 2.08 # Density in g/cm**3\n", - "alpha_e = 3.5e-40 # Electronic polarizability in Fm**2\n", - "N_a = 6.022e23 # Avogadro Number\n", - "epsilon_0 = 8.85e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.14\"\n", - "N = N_a*d*1e6/m # Calculation of Atoms per unit \n", - "k = N*alpha_e/(3*epsilon_0)\n", - "\n", - "epsilon_r = (1+ k*2)/(1-k)# Calculation of relative permittivity\n", - "print \" Relative permittivity is %f\"%(epsilon_r)\n", - "# Answer in book is 4.17\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.14\n", - " Relative permittivity is 4.198468\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.15 Page No191" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "n = 1.5 # Refractive index\n", - "epsilon = 5.6 # Static dielectric consmath.tant\n", - "print \"Example 4.15\"\n", - "per = (1-((n**2-1)/(n**2+2))*(epsilon+2)/(epsilon-1))*100 # Pecentage ionic polarisability\n", - "print \" Percentage ionic polarizability is %f pecent\"%(per)\n", - "# Answer in book is 5.14 %\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.15\n", - " Percentage ionic polarizability is 51.406650 pecent\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.16 Page No195" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#Given that\n", - "m = 32 # Atomic weight of sulphur\n", - "d = 2050 # Density in Kg/m**3\n", - "N_a = 6.022e23 # Avogadro Number\n", - "epsilon_0 = 8.85e-12 # Permittivity of free space\n", - "epsilon_r = 3.75 # Dielectric consmath.tant of sulphur\n", - "\n", - "print \"Example 4.16\"\n", - "N = N_a*d*1e3/m # Calculation of Atoms per unit \n", - "alpha_e = 3*epsilon_0*((epsilon_r-1)/(epsilon_r+2)) / N \n", - "\n", - "\n", - "print \" Electronic polarizability is %e Fm**2\"%(alpha_e)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.16\n", - " Electronic polarizability is 3.291431e-40 Fm**2\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.17 Page No199" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "n = 1.5 # Refractive index\n", - "epsilon = 4. # Static dielectric consmath.tant\n", - "epsilon_0 = 8.85e-12 # permittivity of free space\n", - "print \"Example 4.17\"\n", - "k1 = (epsilon-1)/(epsilon+2)\n", - "k2 = (n**2-1)/(n**2+2)\n", - "ratio = 1./((k1/k2)-1) \n", - "print \" Ratio of electronic to ionic polarizability is %.2f .\"%(ratio)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.17\n", - " Ratio of electronic to ionic polarizability is 1.43 .\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.18 Page No202" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "t = 1.8e-5 # Relaxation time in second\n", - "epsilon_r = 1 # let\n", - "print \"Example 4.18\"\n", - "f = 1./(2*math.pi*t) # Calculation of frequency\n", - "delta = math.atan(epsilon_r/epsilon_r)\n", - "phi = 90 - delta*180/math.pi # Calculation of phase difference\n", - "print \" Frequency is %f KHz\"%(f/1e3)\n", - "print \" Phase difference between current and voltage is %d degree.\"%(phi)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.18\n", - " Frequency is 8.841941 KHz\n", - " Phase difference between current and voltage is 45 degree.\n" - ] - } - ], - "prompt_number": 18 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch4_2.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch4_2.ipynb deleted file mode 100644 index 0ea076e1..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch4_2.ipynb +++ /dev/null @@ -1,557 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:98a33103545b2365c3949736414c5684c66408433d883a4d0f1b451b5a78b395" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 4 : Dielectrics" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.4 Page No177" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "epsilon_r = 1.000074 # Dielectric consmath.tant of He at 0C and 1atm\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "E = 100 # Electric field in V/m\n", - "n = 2.68e27 # Electron density in no,/m**\n", - "N_a = 6e23 # Avogadro number\n", - "V = 22.4 # Volume at STP in litter\n", - "print \"Example 4.4\"\n", - "P = epsilon_0*(epsilon_r-1)*E # Calculation of polarization\n", - "\n", - "N = N_a/(V*1e-3)# Calculation of total number of atoms\n", - "p = P/N # dipole moment per atom\n", - "print \" Dipole moment per atom is %e Coulomb-meter \"%(p)\n", - "# Answer in book is in different form and as 24.45e-40 coulomb-meter\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.4\n", - " Dipole moment per atom is 2.446065e-39 Coulomb-meter \n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.6 Page No181" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "r = 0.055 # Radius of hydrogen atom in nm\n", - "n = 9.8e26 # Number of atoms/cc\n", - "\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.6\"\n", - "alpha_e = 4*math.pi*epsilon_0*(r*1e-9)**3 # Calculation of electronic polarisability\n", - "epsilon_r = 1+n*alpha_e/epsilon_0 # Calculation of relative permeability\n", - "\n", - "print \" Electronic polarisability is %e Fm**2 Relative permeability is %f \"%(alpha_e,epsilon_r)\n", - "\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.6\n", - " Electronic polarisability is 1.851132e-41 Fm**2 Relative permeability is 1.002049 \n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.8 Page No182" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "E = 2000 # Electric field in V/m\n", - "P = 6.4e-8 # Polarization in C/m**2\n", - "print \"Example 4.8\"\n", - "epsilon_r = 1+ P/(epsilon_0*E) # Calculation of relative permittivity\n", - "\n", - "print \" Relative permittivity is %f\"%(epsilon_r)\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.8\n", - " Relative permittivity is 4.614186\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.9 Page No183" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "alpha_e = 2e-40 # Electronic polarisability in Fm**2\n", - "N = 4e28 # density in atoms/m**3\n", - "epsilon_0 = 8.85e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.9\"\n", - "epsilon_r = 1+ N*alpha_e/(epsilon_0) # Calculation of relative permittivity\n", - "print \" Relative permittivity is %f\"%(epsilon_r)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.9\n", - " Relative permittivity is 1.903955\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.10 Page No188" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "epsilon = 2.4e-10 # permitivity of a dielectric material in C**2/N?m**2\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.10\"\n", - "K = epsilon/epsilon_0 # Calculation of dielectric consmath.tant \n", - "zai_e = epsilon_0*(K-1) # Calculation of electrical susceptibility \n", - "\n", - "print \" Relative permittivity is %f\"%(K)\n", - "print \" Electrical susceptibility is %e C**2/Nm**2\"%(zai_e)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.10\n", - " Relative permittivity is 27.106393\n", - " Electrical susceptibility is 2.311460e-10 C**2/Nm**2\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.11 Page No189" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "V = 100 # Applied potential in Volt\n", - "d = 1 # Separation between plates in cm\n", - "k1 = 8 # Dielectric consmath.tant\n", - "k2 = 9 #dielectric consmath.tant\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.11\"\n", - "E_0 = V/(d*1e-2) # Calculation of electric field\n", - "E = E_0/k1*k2 # Calculation of electric field\n", - "D = k1*epsilon_0*E # Calculation of electrical print lacement vector\n", - "P = (k1-1)*epsilon_0*E # Calculation of electrical polarization \n", - "\n", - "print \" Magnitude of Electrical vector is %e Volt/meter\"%(E) # Answer in book is 1.125e3 Volt/meter\n", - "\n", - "print \" Magnitude of Electrical Displacement vector is %e C/m**2\"%(D)# Answer in book is 8.85e-8C/m**2\n", - "\n", - "print \" Magnitude of Electric polarization vector is %e C/m**2\"%(P)# Answer in book is 7.774e-8C/m**2\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.11\n", - " Magnitude of Electrical vector is 1.125000e+04 Volt/meter\n", - " Magnitude of Electrical Displacement vector is 7.968600e-07 C/m**2\n", - " Magnitude of Electric polarization vector is 6.972525e-07 C/m**2\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.12 Page No189" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "alpha_300 = 2.5e-39 # total polarisability in C**2m/N at 300 K\n", - "alpha_600 = 1.75e-39 # total polarisability in C**2m/N at 600 K\n", - "T1 = 300 # Initial temperature in Kelvin\n", - "T2 = 600 # Final Temperature in Kelvin\n", - "print \"Example 4.12\"\n", - "b = (alpha_300-alpha_600)*T2\n", - "al_def_300 = alpha_300 - b/300\n", - "al_oriant_300 = b/300\n", - "al_oriant_600 = b/600\n", - "print \" Deformational Polarizability is %e C**2mN**-1\"%(al_def_300)\n", - "print \" Orientational Polarizability at %d degree Celcius is %e C**2mN**-1\"%(T1,al_oriant_300)\n", - "print \" Orientational Polarizability at %d degree Celcius is %e C**2mN**-1\"%(T2,al_oriant_600)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.12\n", - " Deformational Polarizability is 1.000000e-39 C**2mN**-1\n", - " Orientational Polarizability at 300 degree Celcius is 1.500000e-39 C**2mN**-1\n", - " Orientational Polarizability at 600 degree Celcius is 7.500000e-40 C**2mN**-1\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.13 Page No189" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "alpha_e = 1.5e-40 # Electronic polarizability in Fm**2\n", - "N = 4e28 # density in atoms/m**3\n", - "epsilon_0 = 8.85e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.13\"\n", - "k = N*alpha_e/(3*epsilon_0)\n", - "epsilon_r = (1+ k*2)/(1-k)# Calculation of relative permittivity\n", - "print \" Relative permittivity is %f\"%(epsilon_r)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.13\n", - " Relative permittivity is 1.875912\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.14 Page No191" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "m = 32 # Atomic weight of sulphur\n", - "d = 2.08 # Density in g/cm**3\n", - "alpha_e = 3.5e-40 # Electronic polarizability in Fm**2\n", - "N_a = 6.022e23 # Avogadro Number\n", - "epsilon_0 = 8.85e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.14\"\n", - "N = N_a*d*1e6/m # Calculation of Atoms per unit \n", - "k = N*alpha_e/(3*epsilon_0)\n", - "\n", - "epsilon_r = (1+ k*2)/(1-k)# Calculation of relative permittivity\n", - "print \" Relative permittivity is %f\"%(epsilon_r)\n", - "# Answer in book is 4.17\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.14\n", - " Relative permittivity is 4.198468\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.15 Page No191" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "n = 1.5 # Refractive index\n", - "epsilon = 5.6 # Static dielectric consmath.tant\n", - "print \"Example 4.15\"\n", - "per = (1-((n**2-1)/(n**2+2))*(epsilon+2)/(epsilon-1))*100 # Pecentage ionic polarisability\n", - "print \" Percentage ionic polarizability is %f pecent\"%(per)\n", - "# Answer in book is 5.14 %\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.15\n", - " Percentage ionic polarizability is 51.406650 pecent\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.16 Page No195" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#Given that\n", - "m = 32 # Atomic weight of sulphur\n", - "d = 2050 # Density in Kg/m**3\n", - "N_a = 6.022e23 # Avogadro Number\n", - "epsilon_0 = 8.85e-12 # Permittivity of free space\n", - "epsilon_r = 3.75 # Dielectric consmath.tant of sulphur\n", - "\n", - "print \"Example 4.16\"\n", - "N = N_a*d*1e3/m # Calculation of Atoms per unit \n", - "alpha_e = 3*epsilon_0*((epsilon_r-1)/(epsilon_r+2)) / N \n", - "\n", - "\n", - "print \" Electronic polarizability is %e Fm**2\"%(alpha_e)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.16\n", - " Electronic polarizability is 3.291431e-40 Fm**2\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.17 Page No199" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "n = 1.5 # Refractive index\n", - "epsilon = 4. # Static dielectric consmath.tant\n", - "epsilon_0 = 8.85e-12 # permittivity of free space\n", - "print \"Example 4.17\"\n", - "k1 = (epsilon-1)/(epsilon+2)\n", - "k2 = (n**2-1)/(n**2+2)\n", - "ratio = 1./((k1/k2)-1) \n", - "print \" Ratio of electronic to ionic polarizability is %.2f .\"%(ratio)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.17\n", - " Ratio of electronic to ionic polarizability is 1.43 .\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.18 Page No202" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "t = 1.8e-5 # Relaxation time in second\n", - "epsilon_r = 1 # let\n", - "print \"Example 4.18\"\n", - "f = 1./(2*math.pi*t) # Calculation of frequency\n", - "delta = math.atan(epsilon_r/epsilon_r)\n", - "phi = 90 - delta*180/math.pi # Calculation of phase difference\n", - "print \" Frequency is %f KHz\"%(f/1e3)\n", - "print \" Phase difference between current and voltage is %d degree.\"%(phi)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.18\n", - " Frequency is 8.841941 KHz\n", - " Phase difference between current and voltage is 45 degree.\n" - ] - } - ], - "prompt_number": 18 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch4_3.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch4_3.ipynb deleted file mode 100644 index 0ea076e1..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch4_3.ipynb +++ /dev/null @@ -1,557 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:98a33103545b2365c3949736414c5684c66408433d883a4d0f1b451b5a78b395" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 4 : Dielectrics" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.4 Page No177" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "epsilon_r = 1.000074 # Dielectric consmath.tant of He at 0C and 1atm\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "E = 100 # Electric field in V/m\n", - "n = 2.68e27 # Electron density in no,/m**\n", - "N_a = 6e23 # Avogadro number\n", - "V = 22.4 # Volume at STP in litter\n", - "print \"Example 4.4\"\n", - "P = epsilon_0*(epsilon_r-1)*E # Calculation of polarization\n", - "\n", - "N = N_a/(V*1e-3)# Calculation of total number of atoms\n", - "p = P/N # dipole moment per atom\n", - "print \" Dipole moment per atom is %e Coulomb-meter \"%(p)\n", - "# Answer in book is in different form and as 24.45e-40 coulomb-meter\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.4\n", - " Dipole moment per atom is 2.446065e-39 Coulomb-meter \n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.6 Page No181" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "r = 0.055 # Radius of hydrogen atom in nm\n", - "n = 9.8e26 # Number of atoms/cc\n", - "\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.6\"\n", - "alpha_e = 4*math.pi*epsilon_0*(r*1e-9)**3 # Calculation of electronic polarisability\n", - "epsilon_r = 1+n*alpha_e/epsilon_0 # Calculation of relative permeability\n", - "\n", - "print \" Electronic polarisability is %e Fm**2 Relative permeability is %f \"%(alpha_e,epsilon_r)\n", - "\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.6\n", - " Electronic polarisability is 1.851132e-41 Fm**2 Relative permeability is 1.002049 \n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.8 Page No182" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "E = 2000 # Electric field in V/m\n", - "P = 6.4e-8 # Polarization in C/m**2\n", - "print \"Example 4.8\"\n", - "epsilon_r = 1+ P/(epsilon_0*E) # Calculation of relative permittivity\n", - "\n", - "print \" Relative permittivity is %f\"%(epsilon_r)\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.8\n", - " Relative permittivity is 4.614186\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.9 Page No183" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "alpha_e = 2e-40 # Electronic polarisability in Fm**2\n", - "N = 4e28 # density in atoms/m**3\n", - "epsilon_0 = 8.85e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.9\"\n", - "epsilon_r = 1+ N*alpha_e/(epsilon_0) # Calculation of relative permittivity\n", - "print \" Relative permittivity is %f\"%(epsilon_r)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.9\n", - " Relative permittivity is 1.903955\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.10 Page No188" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "epsilon = 2.4e-10 # permitivity of a dielectric material in C**2/N?m**2\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.10\"\n", - "K = epsilon/epsilon_0 # Calculation of dielectric consmath.tant \n", - "zai_e = epsilon_0*(K-1) # Calculation of electrical susceptibility \n", - "\n", - "print \" Relative permittivity is %f\"%(K)\n", - "print \" Electrical susceptibility is %e C**2/Nm**2\"%(zai_e)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.10\n", - " Relative permittivity is 27.106393\n", - " Electrical susceptibility is 2.311460e-10 C**2/Nm**2\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.11 Page No189" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "V = 100 # Applied potential in Volt\n", - "d = 1 # Separation between plates in cm\n", - "k1 = 8 # Dielectric consmath.tant\n", - "k2 = 9 #dielectric consmath.tant\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.11\"\n", - "E_0 = V/(d*1e-2) # Calculation of electric field\n", - "E = E_0/k1*k2 # Calculation of electric field\n", - "D = k1*epsilon_0*E # Calculation of electrical print lacement vector\n", - "P = (k1-1)*epsilon_0*E # Calculation of electrical polarization \n", - "\n", - "print \" Magnitude of Electrical vector is %e Volt/meter\"%(E) # Answer in book is 1.125e3 Volt/meter\n", - "\n", - "print \" Magnitude of Electrical Displacement vector is %e C/m**2\"%(D)# Answer in book is 8.85e-8C/m**2\n", - "\n", - "print \" Magnitude of Electric polarization vector is %e C/m**2\"%(P)# Answer in book is 7.774e-8C/m**2\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.11\n", - " Magnitude of Electrical vector is 1.125000e+04 Volt/meter\n", - " Magnitude of Electrical Displacement vector is 7.968600e-07 C/m**2\n", - " Magnitude of Electric polarization vector is 6.972525e-07 C/m**2\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.12 Page No189" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "alpha_300 = 2.5e-39 # total polarisability in C**2m/N at 300 K\n", - "alpha_600 = 1.75e-39 # total polarisability in C**2m/N at 600 K\n", - "T1 = 300 # Initial temperature in Kelvin\n", - "T2 = 600 # Final Temperature in Kelvin\n", - "print \"Example 4.12\"\n", - "b = (alpha_300-alpha_600)*T2\n", - "al_def_300 = alpha_300 - b/300\n", - "al_oriant_300 = b/300\n", - "al_oriant_600 = b/600\n", - "print \" Deformational Polarizability is %e C**2mN**-1\"%(al_def_300)\n", - "print \" Orientational Polarizability at %d degree Celcius is %e C**2mN**-1\"%(T1,al_oriant_300)\n", - "print \" Orientational Polarizability at %d degree Celcius is %e C**2mN**-1\"%(T2,al_oriant_600)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.12\n", - " Deformational Polarizability is 1.000000e-39 C**2mN**-1\n", - " Orientational Polarizability at 300 degree Celcius is 1.500000e-39 C**2mN**-1\n", - " Orientational Polarizability at 600 degree Celcius is 7.500000e-40 C**2mN**-1\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.13 Page No189" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "alpha_e = 1.5e-40 # Electronic polarizability in Fm**2\n", - "N = 4e28 # density in atoms/m**3\n", - "epsilon_0 = 8.85e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.13\"\n", - "k = N*alpha_e/(3*epsilon_0)\n", - "epsilon_r = (1+ k*2)/(1-k)# Calculation of relative permittivity\n", - "print \" Relative permittivity is %f\"%(epsilon_r)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.13\n", - " Relative permittivity is 1.875912\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.14 Page No191" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "m = 32 # Atomic weight of sulphur\n", - "d = 2.08 # Density in g/cm**3\n", - "alpha_e = 3.5e-40 # Electronic polarizability in Fm**2\n", - "N_a = 6.022e23 # Avogadro Number\n", - "epsilon_0 = 8.85e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.14\"\n", - "N = N_a*d*1e6/m # Calculation of Atoms per unit \n", - "k = N*alpha_e/(3*epsilon_0)\n", - "\n", - "epsilon_r = (1+ k*2)/(1-k)# Calculation of relative permittivity\n", - "print \" Relative permittivity is %f\"%(epsilon_r)\n", - "# Answer in book is 4.17\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.14\n", - " Relative permittivity is 4.198468\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.15 Page No191" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "n = 1.5 # Refractive index\n", - "epsilon = 5.6 # Static dielectric consmath.tant\n", - "print \"Example 4.15\"\n", - "per = (1-((n**2-1)/(n**2+2))*(epsilon+2)/(epsilon-1))*100 # Pecentage ionic polarisability\n", - "print \" Percentage ionic polarizability is %f pecent\"%(per)\n", - "# Answer in book is 5.14 %\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.15\n", - " Percentage ionic polarizability is 51.406650 pecent\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.16 Page No195" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#Given that\n", - "m = 32 # Atomic weight of sulphur\n", - "d = 2050 # Density in Kg/m**3\n", - "N_a = 6.022e23 # Avogadro Number\n", - "epsilon_0 = 8.85e-12 # Permittivity of free space\n", - "epsilon_r = 3.75 # Dielectric consmath.tant of sulphur\n", - "\n", - "print \"Example 4.16\"\n", - "N = N_a*d*1e3/m # Calculation of Atoms per unit \n", - "alpha_e = 3*epsilon_0*((epsilon_r-1)/(epsilon_r+2)) / N \n", - "\n", - "\n", - "print \" Electronic polarizability is %e Fm**2\"%(alpha_e)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.16\n", - " Electronic polarizability is 3.291431e-40 Fm**2\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.17 Page No199" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "n = 1.5 # Refractive index\n", - "epsilon = 4. # Static dielectric consmath.tant\n", - "epsilon_0 = 8.85e-12 # permittivity of free space\n", - "print \"Example 4.17\"\n", - "k1 = (epsilon-1)/(epsilon+2)\n", - "k2 = (n**2-1)/(n**2+2)\n", - "ratio = 1./((k1/k2)-1) \n", - "print \" Ratio of electronic to ionic polarizability is %.2f .\"%(ratio)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.17\n", - " Ratio of electronic to ionic polarizability is 1.43 .\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.18 Page No202" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "t = 1.8e-5 # Relaxation time in second\n", - "epsilon_r = 1 # let\n", - "print \"Example 4.18\"\n", - "f = 1./(2*math.pi*t) # Calculation of frequency\n", - "delta = math.atan(epsilon_r/epsilon_r)\n", - "phi = 90 - delta*180/math.pi # Calculation of phase difference\n", - "print \" Frequency is %f KHz\"%(f/1e3)\n", - "print \" Phase difference between current and voltage is %d degree.\"%(phi)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.18\n", - " Frequency is 8.841941 KHz\n", - " Phase difference between current and voltage is 45 degree.\n" - ] - } - ], - "prompt_number": 18 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch4_4.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch4_4.ipynb deleted file mode 100644 index 0ea076e1..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch4_4.ipynb +++ /dev/null @@ -1,557 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:98a33103545b2365c3949736414c5684c66408433d883a4d0f1b451b5a78b395" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 4 : Dielectrics" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.4 Page No177" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "epsilon_r = 1.000074 # Dielectric consmath.tant of He at 0C and 1atm\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "E = 100 # Electric field in V/m\n", - "n = 2.68e27 # Electron density in no,/m**\n", - "N_a = 6e23 # Avogadro number\n", - "V = 22.4 # Volume at STP in litter\n", - "print \"Example 4.4\"\n", - "P = epsilon_0*(epsilon_r-1)*E # Calculation of polarization\n", - "\n", - "N = N_a/(V*1e-3)# Calculation of total number of atoms\n", - "p = P/N # dipole moment per atom\n", - "print \" Dipole moment per atom is %e Coulomb-meter \"%(p)\n", - "# Answer in book is in different form and as 24.45e-40 coulomb-meter\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.4\n", - " Dipole moment per atom is 2.446065e-39 Coulomb-meter \n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.6 Page No181" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "r = 0.055 # Radius of hydrogen atom in nm\n", - "n = 9.8e26 # Number of atoms/cc\n", - "\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.6\"\n", - "alpha_e = 4*math.pi*epsilon_0*(r*1e-9)**3 # Calculation of electronic polarisability\n", - "epsilon_r = 1+n*alpha_e/epsilon_0 # Calculation of relative permeability\n", - "\n", - "print \" Electronic polarisability is %e Fm**2 Relative permeability is %f \"%(alpha_e,epsilon_r)\n", - "\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.6\n", - " Electronic polarisability is 1.851132e-41 Fm**2 Relative permeability is 1.002049 \n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.8 Page No182" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "E = 2000 # Electric field in V/m\n", - "P = 6.4e-8 # Polarization in C/m**2\n", - "print \"Example 4.8\"\n", - "epsilon_r = 1+ P/(epsilon_0*E) # Calculation of relative permittivity\n", - "\n", - "print \" Relative permittivity is %f\"%(epsilon_r)\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.8\n", - " Relative permittivity is 4.614186\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.9 Page No183" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "alpha_e = 2e-40 # Electronic polarisability in Fm**2\n", - "N = 4e28 # density in atoms/m**3\n", - "epsilon_0 = 8.85e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.9\"\n", - "epsilon_r = 1+ N*alpha_e/(epsilon_0) # Calculation of relative permittivity\n", - "print \" Relative permittivity is %f\"%(epsilon_r)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.9\n", - " Relative permittivity is 1.903955\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.10 Page No188" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "epsilon = 2.4e-10 # permitivity of a dielectric material in C**2/N?m**2\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.10\"\n", - "K = epsilon/epsilon_0 # Calculation of dielectric consmath.tant \n", - "zai_e = epsilon_0*(K-1) # Calculation of electrical susceptibility \n", - "\n", - "print \" Relative permittivity is %f\"%(K)\n", - "print \" Electrical susceptibility is %e C**2/Nm**2\"%(zai_e)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.10\n", - " Relative permittivity is 27.106393\n", - " Electrical susceptibility is 2.311460e-10 C**2/Nm**2\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.11 Page No189" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "V = 100 # Applied potential in Volt\n", - "d = 1 # Separation between plates in cm\n", - "k1 = 8 # Dielectric consmath.tant\n", - "k2 = 9 #dielectric consmath.tant\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.11\"\n", - "E_0 = V/(d*1e-2) # Calculation of electric field\n", - "E = E_0/k1*k2 # Calculation of electric field\n", - "D = k1*epsilon_0*E # Calculation of electrical print lacement vector\n", - "P = (k1-1)*epsilon_0*E # Calculation of electrical polarization \n", - "\n", - "print \" Magnitude of Electrical vector is %e Volt/meter\"%(E) # Answer in book is 1.125e3 Volt/meter\n", - "\n", - "print \" Magnitude of Electrical Displacement vector is %e C/m**2\"%(D)# Answer in book is 8.85e-8C/m**2\n", - "\n", - "print \" Magnitude of Electric polarization vector is %e C/m**2\"%(P)# Answer in book is 7.774e-8C/m**2\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.11\n", - " Magnitude of Electrical vector is 1.125000e+04 Volt/meter\n", - " Magnitude of Electrical Displacement vector is 7.968600e-07 C/m**2\n", - " Magnitude of Electric polarization vector is 6.972525e-07 C/m**2\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.12 Page No189" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "alpha_300 = 2.5e-39 # total polarisability in C**2m/N at 300 K\n", - "alpha_600 = 1.75e-39 # total polarisability in C**2m/N at 600 K\n", - "T1 = 300 # Initial temperature in Kelvin\n", - "T2 = 600 # Final Temperature in Kelvin\n", - "print \"Example 4.12\"\n", - "b = (alpha_300-alpha_600)*T2\n", - "al_def_300 = alpha_300 - b/300\n", - "al_oriant_300 = b/300\n", - "al_oriant_600 = b/600\n", - "print \" Deformational Polarizability is %e C**2mN**-1\"%(al_def_300)\n", - "print \" Orientational Polarizability at %d degree Celcius is %e C**2mN**-1\"%(T1,al_oriant_300)\n", - "print \" Orientational Polarizability at %d degree Celcius is %e C**2mN**-1\"%(T2,al_oriant_600)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.12\n", - " Deformational Polarizability is 1.000000e-39 C**2mN**-1\n", - " Orientational Polarizability at 300 degree Celcius is 1.500000e-39 C**2mN**-1\n", - " Orientational Polarizability at 600 degree Celcius is 7.500000e-40 C**2mN**-1\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.13 Page No189" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "alpha_e = 1.5e-40 # Electronic polarizability in Fm**2\n", - "N = 4e28 # density in atoms/m**3\n", - "epsilon_0 = 8.85e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.13\"\n", - "k = N*alpha_e/(3*epsilon_0)\n", - "epsilon_r = (1+ k*2)/(1-k)# Calculation of relative permittivity\n", - "print \" Relative permittivity is %f\"%(epsilon_r)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.13\n", - " Relative permittivity is 1.875912\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.14 Page No191" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "m = 32 # Atomic weight of sulphur\n", - "d = 2.08 # Density in g/cm**3\n", - "alpha_e = 3.5e-40 # Electronic polarizability in Fm**2\n", - "N_a = 6.022e23 # Avogadro Number\n", - "epsilon_0 = 8.85e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.14\"\n", - "N = N_a*d*1e6/m # Calculation of Atoms per unit \n", - "k = N*alpha_e/(3*epsilon_0)\n", - "\n", - "epsilon_r = (1+ k*2)/(1-k)# Calculation of relative permittivity\n", - "print \" Relative permittivity is %f\"%(epsilon_r)\n", - "# Answer in book is 4.17\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.14\n", - " Relative permittivity is 4.198468\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.15 Page No191" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "n = 1.5 # Refractive index\n", - "epsilon = 5.6 # Static dielectric consmath.tant\n", - "print \"Example 4.15\"\n", - "per = (1-((n**2-1)/(n**2+2))*(epsilon+2)/(epsilon-1))*100 # Pecentage ionic polarisability\n", - "print \" Percentage ionic polarizability is %f pecent\"%(per)\n", - "# Answer in book is 5.14 %\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.15\n", - " Percentage ionic polarizability is 51.406650 pecent\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.16 Page No195" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#Given that\n", - "m = 32 # Atomic weight of sulphur\n", - "d = 2050 # Density in Kg/m**3\n", - "N_a = 6.022e23 # Avogadro Number\n", - "epsilon_0 = 8.85e-12 # Permittivity of free space\n", - "epsilon_r = 3.75 # Dielectric consmath.tant of sulphur\n", - "\n", - "print \"Example 4.16\"\n", - "N = N_a*d*1e3/m # Calculation of Atoms per unit \n", - "alpha_e = 3*epsilon_0*((epsilon_r-1)/(epsilon_r+2)) / N \n", - "\n", - "\n", - "print \" Electronic polarizability is %e Fm**2\"%(alpha_e)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.16\n", - " Electronic polarizability is 3.291431e-40 Fm**2\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.17 Page No199" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "n = 1.5 # Refractive index\n", - "epsilon = 4. # Static dielectric consmath.tant\n", - "epsilon_0 = 8.85e-12 # permittivity of free space\n", - "print \"Example 4.17\"\n", - "k1 = (epsilon-1)/(epsilon+2)\n", - "k2 = (n**2-1)/(n**2+2)\n", - "ratio = 1./((k1/k2)-1) \n", - "print \" Ratio of electronic to ionic polarizability is %.2f .\"%(ratio)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.17\n", - " Ratio of electronic to ionic polarizability is 1.43 .\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.18 Page No202" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "t = 1.8e-5 # Relaxation time in second\n", - "epsilon_r = 1 # let\n", - "print \"Example 4.18\"\n", - "f = 1./(2*math.pi*t) # Calculation of frequency\n", - "delta = math.atan(epsilon_r/epsilon_r)\n", - "phi = 90 - delta*180/math.pi # Calculation of phase difference\n", - "print \" Frequency is %f KHz\"%(f/1e3)\n", - "print \" Phase difference between current and voltage is %d degree.\"%(phi)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.18\n", - " Frequency is 8.841941 KHz\n", - " Phase difference between current and voltage is 45 degree.\n" - ] - } - ], - "prompt_number": 18 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch4_5.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch4_5.ipynb deleted file mode 100644 index 0ea076e1..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch4_5.ipynb +++ /dev/null @@ -1,557 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:98a33103545b2365c3949736414c5684c66408433d883a4d0f1b451b5a78b395" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 4 : Dielectrics" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.4 Page No177" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "epsilon_r = 1.000074 # Dielectric consmath.tant of He at 0C and 1atm\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "E = 100 # Electric field in V/m\n", - "n = 2.68e27 # Electron density in no,/m**\n", - "N_a = 6e23 # Avogadro number\n", - "V = 22.4 # Volume at STP in litter\n", - "print \"Example 4.4\"\n", - "P = epsilon_0*(epsilon_r-1)*E # Calculation of polarization\n", - "\n", - "N = N_a/(V*1e-3)# Calculation of total number of atoms\n", - "p = P/N # dipole moment per atom\n", - "print \" Dipole moment per atom is %e Coulomb-meter \"%(p)\n", - "# Answer in book is in different form and as 24.45e-40 coulomb-meter\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.4\n", - " Dipole moment per atom is 2.446065e-39 Coulomb-meter \n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.6 Page No181" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "r = 0.055 # Radius of hydrogen atom in nm\n", - "n = 9.8e26 # Number of atoms/cc\n", - "\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.6\"\n", - "alpha_e = 4*math.pi*epsilon_0*(r*1e-9)**3 # Calculation of electronic polarisability\n", - "epsilon_r = 1+n*alpha_e/epsilon_0 # Calculation of relative permeability\n", - "\n", - "print \" Electronic polarisability is %e Fm**2 Relative permeability is %f \"%(alpha_e,epsilon_r)\n", - "\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.6\n", - " Electronic polarisability is 1.851132e-41 Fm**2 Relative permeability is 1.002049 \n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.8 Page No182" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "E = 2000 # Electric field in V/m\n", - "P = 6.4e-8 # Polarization in C/m**2\n", - "print \"Example 4.8\"\n", - "epsilon_r = 1+ P/(epsilon_0*E) # Calculation of relative permittivity\n", - "\n", - "print \" Relative permittivity is %f\"%(epsilon_r)\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.8\n", - " Relative permittivity is 4.614186\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.9 Page No183" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "alpha_e = 2e-40 # Electronic polarisability in Fm**2\n", - "N = 4e28 # density in atoms/m**3\n", - "epsilon_0 = 8.85e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.9\"\n", - "epsilon_r = 1+ N*alpha_e/(epsilon_0) # Calculation of relative permittivity\n", - "print \" Relative permittivity is %f\"%(epsilon_r)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.9\n", - " Relative permittivity is 1.903955\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.10 Page No188" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "epsilon = 2.4e-10 # permitivity of a dielectric material in C**2/N?m**2\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.10\"\n", - "K = epsilon/epsilon_0 # Calculation of dielectric consmath.tant \n", - "zai_e = epsilon_0*(K-1) # Calculation of electrical susceptibility \n", - "\n", - "print \" Relative permittivity is %f\"%(K)\n", - "print \" Electrical susceptibility is %e C**2/Nm**2\"%(zai_e)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.10\n", - " Relative permittivity is 27.106393\n", - " Electrical susceptibility is 2.311460e-10 C**2/Nm**2\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.11 Page No189" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "V = 100 # Applied potential in Volt\n", - "d = 1 # Separation between plates in cm\n", - "k1 = 8 # Dielectric consmath.tant\n", - "k2 = 9 #dielectric consmath.tant\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.11\"\n", - "E_0 = V/(d*1e-2) # Calculation of electric field\n", - "E = E_0/k1*k2 # Calculation of electric field\n", - "D = k1*epsilon_0*E # Calculation of electrical print lacement vector\n", - "P = (k1-1)*epsilon_0*E # Calculation of electrical polarization \n", - "\n", - "print \" Magnitude of Electrical vector is %e Volt/meter\"%(E) # Answer in book is 1.125e3 Volt/meter\n", - "\n", - "print \" Magnitude of Electrical Displacement vector is %e C/m**2\"%(D)# Answer in book is 8.85e-8C/m**2\n", - "\n", - "print \" Magnitude of Electric polarization vector is %e C/m**2\"%(P)# Answer in book is 7.774e-8C/m**2\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.11\n", - " Magnitude of Electrical vector is 1.125000e+04 Volt/meter\n", - " Magnitude of Electrical Displacement vector is 7.968600e-07 C/m**2\n", - " Magnitude of Electric polarization vector is 6.972525e-07 C/m**2\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.12 Page No189" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "alpha_300 = 2.5e-39 # total polarisability in C**2m/N at 300 K\n", - "alpha_600 = 1.75e-39 # total polarisability in C**2m/N at 600 K\n", - "T1 = 300 # Initial temperature in Kelvin\n", - "T2 = 600 # Final Temperature in Kelvin\n", - "print \"Example 4.12\"\n", - "b = (alpha_300-alpha_600)*T2\n", - "al_def_300 = alpha_300 - b/300\n", - "al_oriant_300 = b/300\n", - "al_oriant_600 = b/600\n", - "print \" Deformational Polarizability is %e C**2mN**-1\"%(al_def_300)\n", - "print \" Orientational Polarizability at %d degree Celcius is %e C**2mN**-1\"%(T1,al_oriant_300)\n", - "print \" Orientational Polarizability at %d degree Celcius is %e C**2mN**-1\"%(T2,al_oriant_600)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.12\n", - " Deformational Polarizability is 1.000000e-39 C**2mN**-1\n", - " Orientational Polarizability at 300 degree Celcius is 1.500000e-39 C**2mN**-1\n", - " Orientational Polarizability at 600 degree Celcius is 7.500000e-40 C**2mN**-1\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.13 Page No189" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "alpha_e = 1.5e-40 # Electronic polarizability in Fm**2\n", - "N = 4e28 # density in atoms/m**3\n", - "epsilon_0 = 8.85e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.13\"\n", - "k = N*alpha_e/(3*epsilon_0)\n", - "epsilon_r = (1+ k*2)/(1-k)# Calculation of relative permittivity\n", - "print \" Relative permittivity is %f\"%(epsilon_r)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.13\n", - " Relative permittivity is 1.875912\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.14 Page No191" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "m = 32 # Atomic weight of sulphur\n", - "d = 2.08 # Density in g/cm**3\n", - "alpha_e = 3.5e-40 # Electronic polarizability in Fm**2\n", - "N_a = 6.022e23 # Avogadro Number\n", - "epsilon_0 = 8.85e-12 # Permittivity of free space\n", - "\n", - "print \"Example 4.14\"\n", - "N = N_a*d*1e6/m # Calculation of Atoms per unit \n", - "k = N*alpha_e/(3*epsilon_0)\n", - "\n", - "epsilon_r = (1+ k*2)/(1-k)# Calculation of relative permittivity\n", - "print \" Relative permittivity is %f\"%(epsilon_r)\n", - "# Answer in book is 4.17\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.14\n", - " Relative permittivity is 4.198468\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.15 Page No191" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "n = 1.5 # Refractive index\n", - "epsilon = 5.6 # Static dielectric consmath.tant\n", - "print \"Example 4.15\"\n", - "per = (1-((n**2-1)/(n**2+2))*(epsilon+2)/(epsilon-1))*100 # Pecentage ionic polarisability\n", - "print \" Percentage ionic polarizability is %f pecent\"%(per)\n", - "# Answer in book is 5.14 %\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.15\n", - " Percentage ionic polarizability is 51.406650 pecent\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.16 Page No195" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#Given that\n", - "m = 32 # Atomic weight of sulphur\n", - "d = 2050 # Density in Kg/m**3\n", - "N_a = 6.022e23 # Avogadro Number\n", - "epsilon_0 = 8.85e-12 # Permittivity of free space\n", - "epsilon_r = 3.75 # Dielectric consmath.tant of sulphur\n", - "\n", - "print \"Example 4.16\"\n", - "N = N_a*d*1e3/m # Calculation of Atoms per unit \n", - "alpha_e = 3*epsilon_0*((epsilon_r-1)/(epsilon_r+2)) / N \n", - "\n", - "\n", - "print \" Electronic polarizability is %e Fm**2\"%(alpha_e)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.16\n", - " Electronic polarizability is 3.291431e-40 Fm**2\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.17 Page No199" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "n = 1.5 # Refractive index\n", - "epsilon = 4. # Static dielectric consmath.tant\n", - "epsilon_0 = 8.85e-12 # permittivity of free space\n", - "print \"Example 4.17\"\n", - "k1 = (epsilon-1)/(epsilon+2)\n", - "k2 = (n**2-1)/(n**2+2)\n", - "ratio = 1./((k1/k2)-1) \n", - "print \" Ratio of electronic to ionic polarizability is %.2f .\"%(ratio)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.17\n", - " Ratio of electronic to ionic polarizability is 1.43 .\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.18 Page No202" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "t = 1.8e-5 # Relaxation time in second\n", - "epsilon_r = 1 # let\n", - "print \"Example 4.18\"\n", - "f = 1./(2*math.pi*t) # Calculation of frequency\n", - "delta = math.atan(epsilon_r/epsilon_r)\n", - "phi = 90 - delta*180/math.pi # Calculation of phase difference\n", - "print \" Frequency is %f KHz\"%(f/1e3)\n", - "print \" Phase difference between current and voltage is %d degree.\"%(phi)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 4.18\n", - " Frequency is 8.841941 KHz\n", - " Phase difference between current and voltage is 45 degree.\n" - ] - } - ], - "prompt_number": 18 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch6.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch6.ipynb deleted file mode 100755 index 104b616a..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch6.ipynb +++ /dev/null @@ -1,207 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:1843361d9b1f7d4765e522604cbb1bc8ff2515ffb02cdd416946cae448cf39a5" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 6 : Ultrasonic Waves" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.1 Page No207" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "E = 7.9e10 # Young\u2019s modulus in N/m**2\n", - "rho = 2650 # Density in Kg/m**3\n", - "t = 0.003 # Thickness of quartz crystal in m\n", - "print \"Example 6.1\"\n", - "v = math.sqrt(E/rho)# Calculation of velocity \n", - "lambda1 = 2*t # Calculation of fundamental wavelength\n", - "nu = v/lambda1 # Calculation of fundamental frequency\n", - "print \"Fundamental frequency is %e Hz.\"%(nu)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.1\n", - "Fundamental frequency is 9.099957e+05 Hz.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.2 Page No208" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "v = 5760 # Velocity in m/s\n", - "T = 1.6 # Thickness of quartz crystal in mm\n", - "print \"Example 6.2\"\n", - "nu = v/(2*T*1e-3)# Calculation of fundamental frequency\n", - "print \"Fundamental frequency of crystal is %f MHz.\"%(nu/1e6)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.2\n", - "Fundamental frequency of crystal is 1.800000 MHz.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.3 Page No209" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "T =40. # Thickness of steel bar in cm\n", - "t1 = 40. # Time in ms\n", - "t2 = 80. # Time in ms\n", - "print \"Example 6.3\"\n", - "X = T*t1/t2 # Calculation of depth of defect\n", - "print \"Depth of defect is %d cm.\"%(X)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.3\n", - "Depth of defect is 20 cm.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.4 Page No209" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "E = 7.9e10 # Young\u2019s modulus in N/m**2\n", - "rho = 2650 # Density in Kg/m**3\n", - "t = 0.006 # Thickness of quartz crystal in m\n", - "print \"Example 6.4\"\n", - "v = math.sqrt(E/rho)# Calculation of velocity \n", - "lambda1 = 2*t # Calculation of fundamental wavelength\n", - "nu = v/lambda1 # Calculation of fundamental frequency\n", - "print \"Fundamental frequency is %e Hz.\"%(nu)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.4\n", - "Fundamental frequency is 4.549979e+05 Hz.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.5 Page No209" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "L = 1. # Inducmath.tance in Hanery\n", - "nu = 2.e6 # Frequency in Hz\n", - "print \"Example 6.5\"\n", - "C= 1./(4*((math.pi)**2)*nu**2*L) # Calculation of capacimath.tance\n", - "print \"Capacitance is %e microfarad.\"%(C*1e6)\n", - "# Answer in book is 0.00634 micro Farad\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.5\n", - "Capacitance is 6.332574e-09 microfarad.\n" - ] - } - ], - "prompt_number": 8 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch6_1.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch6_1.ipynb deleted file mode 100755 index 104b616a..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch6_1.ipynb +++ /dev/null @@ -1,207 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:1843361d9b1f7d4765e522604cbb1bc8ff2515ffb02cdd416946cae448cf39a5" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 6 : Ultrasonic Waves" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.1 Page No207" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "E = 7.9e10 # Young\u2019s modulus in N/m**2\n", - "rho = 2650 # Density in Kg/m**3\n", - "t = 0.003 # Thickness of quartz crystal in m\n", - "print \"Example 6.1\"\n", - "v = math.sqrt(E/rho)# Calculation of velocity \n", - "lambda1 = 2*t # Calculation of fundamental wavelength\n", - "nu = v/lambda1 # Calculation of fundamental frequency\n", - "print \"Fundamental frequency is %e Hz.\"%(nu)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.1\n", - "Fundamental frequency is 9.099957e+05 Hz.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.2 Page No208" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "v = 5760 # Velocity in m/s\n", - "T = 1.6 # Thickness of quartz crystal in mm\n", - "print \"Example 6.2\"\n", - "nu = v/(2*T*1e-3)# Calculation of fundamental frequency\n", - "print \"Fundamental frequency of crystal is %f MHz.\"%(nu/1e6)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.2\n", - "Fundamental frequency of crystal is 1.800000 MHz.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.3 Page No209" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "T =40. # Thickness of steel bar in cm\n", - "t1 = 40. # Time in ms\n", - "t2 = 80. # Time in ms\n", - "print \"Example 6.3\"\n", - "X = T*t1/t2 # Calculation of depth of defect\n", - "print \"Depth of defect is %d cm.\"%(X)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.3\n", - "Depth of defect is 20 cm.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.4 Page No209" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "E = 7.9e10 # Young\u2019s modulus in N/m**2\n", - "rho = 2650 # Density in Kg/m**3\n", - "t = 0.006 # Thickness of quartz crystal in m\n", - "print \"Example 6.4\"\n", - "v = math.sqrt(E/rho)# Calculation of velocity \n", - "lambda1 = 2*t # Calculation of fundamental wavelength\n", - "nu = v/lambda1 # Calculation of fundamental frequency\n", - "print \"Fundamental frequency is %e Hz.\"%(nu)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.4\n", - "Fundamental frequency is 4.549979e+05 Hz.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.5 Page No209" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "L = 1. # Inducmath.tance in Hanery\n", - "nu = 2.e6 # Frequency in Hz\n", - "print \"Example 6.5\"\n", - "C= 1./(4*((math.pi)**2)*nu**2*L) # Calculation of capacimath.tance\n", - "print \"Capacitance is %e microfarad.\"%(C*1e6)\n", - "# Answer in book is 0.00634 micro Farad\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.5\n", - "Capacitance is 6.332574e-09 microfarad.\n" - ] - } - ], - "prompt_number": 8 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch6_2.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch6_2.ipynb deleted file mode 100644 index 104b616a..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch6_2.ipynb +++ /dev/null @@ -1,207 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:1843361d9b1f7d4765e522604cbb1bc8ff2515ffb02cdd416946cae448cf39a5" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 6 : Ultrasonic Waves" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.1 Page No207" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "E = 7.9e10 # Young\u2019s modulus in N/m**2\n", - "rho = 2650 # Density in Kg/m**3\n", - "t = 0.003 # Thickness of quartz crystal in m\n", - "print \"Example 6.1\"\n", - "v = math.sqrt(E/rho)# Calculation of velocity \n", - "lambda1 = 2*t # Calculation of fundamental wavelength\n", - "nu = v/lambda1 # Calculation of fundamental frequency\n", - "print \"Fundamental frequency is %e Hz.\"%(nu)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.1\n", - "Fundamental frequency is 9.099957e+05 Hz.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.2 Page No208" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "v = 5760 # Velocity in m/s\n", - "T = 1.6 # Thickness of quartz crystal in mm\n", - "print \"Example 6.2\"\n", - "nu = v/(2*T*1e-3)# Calculation of fundamental frequency\n", - "print \"Fundamental frequency of crystal is %f MHz.\"%(nu/1e6)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.2\n", - "Fundamental frequency of crystal is 1.800000 MHz.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.3 Page No209" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "T =40. # Thickness of steel bar in cm\n", - "t1 = 40. # Time in ms\n", - "t2 = 80. # Time in ms\n", - "print \"Example 6.3\"\n", - "X = T*t1/t2 # Calculation of depth of defect\n", - "print \"Depth of defect is %d cm.\"%(X)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.3\n", - "Depth of defect is 20 cm.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.4 Page No209" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "E = 7.9e10 # Young\u2019s modulus in N/m**2\n", - "rho = 2650 # Density in Kg/m**3\n", - "t = 0.006 # Thickness of quartz crystal in m\n", - "print \"Example 6.4\"\n", - "v = math.sqrt(E/rho)# Calculation of velocity \n", - "lambda1 = 2*t # Calculation of fundamental wavelength\n", - "nu = v/lambda1 # Calculation of fundamental frequency\n", - "print \"Fundamental frequency is %e Hz.\"%(nu)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.4\n", - "Fundamental frequency is 4.549979e+05 Hz.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.5 Page No209" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "L = 1. # Inducmath.tance in Hanery\n", - "nu = 2.e6 # Frequency in Hz\n", - "print \"Example 6.5\"\n", - "C= 1./(4*((math.pi)**2)*nu**2*L) # Calculation of capacimath.tance\n", - "print \"Capacitance is %e microfarad.\"%(C*1e6)\n", - "# Answer in book is 0.00634 micro Farad\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.5\n", - "Capacitance is 6.332574e-09 microfarad.\n" - ] - } - ], - "prompt_number": 8 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch6_3.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch6_3.ipynb deleted file mode 100644 index 104b616a..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch6_3.ipynb +++ /dev/null @@ -1,207 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:1843361d9b1f7d4765e522604cbb1bc8ff2515ffb02cdd416946cae448cf39a5" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 6 : Ultrasonic Waves" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.1 Page No207" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "E = 7.9e10 # Young\u2019s modulus in N/m**2\n", - "rho = 2650 # Density in Kg/m**3\n", - "t = 0.003 # Thickness of quartz crystal in m\n", - "print \"Example 6.1\"\n", - "v = math.sqrt(E/rho)# Calculation of velocity \n", - "lambda1 = 2*t # Calculation of fundamental wavelength\n", - "nu = v/lambda1 # Calculation of fundamental frequency\n", - "print \"Fundamental frequency is %e Hz.\"%(nu)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.1\n", - "Fundamental frequency is 9.099957e+05 Hz.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.2 Page No208" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "v = 5760 # Velocity in m/s\n", - "T = 1.6 # Thickness of quartz crystal in mm\n", - "print \"Example 6.2\"\n", - "nu = v/(2*T*1e-3)# Calculation of fundamental frequency\n", - "print \"Fundamental frequency of crystal is %f MHz.\"%(nu/1e6)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.2\n", - "Fundamental frequency of crystal is 1.800000 MHz.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.3 Page No209" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "T =40. # Thickness of steel bar in cm\n", - "t1 = 40. # Time in ms\n", - "t2 = 80. # Time in ms\n", - "print \"Example 6.3\"\n", - "X = T*t1/t2 # Calculation of depth of defect\n", - "print \"Depth of defect is %d cm.\"%(X)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.3\n", - "Depth of defect is 20 cm.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.4 Page No209" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "E = 7.9e10 # Young\u2019s modulus in N/m**2\n", - "rho = 2650 # Density in Kg/m**3\n", - "t = 0.006 # Thickness of quartz crystal in m\n", - "print \"Example 6.4\"\n", - "v = math.sqrt(E/rho)# Calculation of velocity \n", - "lambda1 = 2*t # Calculation of fundamental wavelength\n", - "nu = v/lambda1 # Calculation of fundamental frequency\n", - "print \"Fundamental frequency is %e Hz.\"%(nu)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.4\n", - "Fundamental frequency is 4.549979e+05 Hz.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.5 Page No209" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "L = 1. # Inducmath.tance in Hanery\n", - "nu = 2.e6 # Frequency in Hz\n", - "print \"Example 6.5\"\n", - "C= 1./(4*((math.pi)**2)*nu**2*L) # Calculation of capacimath.tance\n", - "print \"Capacitance is %e microfarad.\"%(C*1e6)\n", - "# Answer in book is 0.00634 micro Farad\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.5\n", - "Capacitance is 6.332574e-09 microfarad.\n" - ] - } - ], - "prompt_number": 8 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch6_4.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch6_4.ipynb deleted file mode 100644 index 104b616a..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch6_4.ipynb +++ /dev/null @@ -1,207 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:1843361d9b1f7d4765e522604cbb1bc8ff2515ffb02cdd416946cae448cf39a5" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 6 : Ultrasonic Waves" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.1 Page No207" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "E = 7.9e10 # Young\u2019s modulus in N/m**2\n", - "rho = 2650 # Density in Kg/m**3\n", - "t = 0.003 # Thickness of quartz crystal in m\n", - "print \"Example 6.1\"\n", - "v = math.sqrt(E/rho)# Calculation of velocity \n", - "lambda1 = 2*t # Calculation of fundamental wavelength\n", - "nu = v/lambda1 # Calculation of fundamental frequency\n", - "print \"Fundamental frequency is %e Hz.\"%(nu)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.1\n", - "Fundamental frequency is 9.099957e+05 Hz.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.2 Page No208" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "v = 5760 # Velocity in m/s\n", - "T = 1.6 # Thickness of quartz crystal in mm\n", - "print \"Example 6.2\"\n", - "nu = v/(2*T*1e-3)# Calculation of fundamental frequency\n", - "print \"Fundamental frequency of crystal is %f MHz.\"%(nu/1e6)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.2\n", - "Fundamental frequency of crystal is 1.800000 MHz.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.3 Page No209" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "T =40. # Thickness of steel bar in cm\n", - "t1 = 40. # Time in ms\n", - "t2 = 80. # Time in ms\n", - "print \"Example 6.3\"\n", - "X = T*t1/t2 # Calculation of depth of defect\n", - "print \"Depth of defect is %d cm.\"%(X)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.3\n", - "Depth of defect is 20 cm.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.4 Page No209" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "E = 7.9e10 # Young\u2019s modulus in N/m**2\n", - "rho = 2650 # Density in Kg/m**3\n", - "t = 0.006 # Thickness of quartz crystal in m\n", - "print \"Example 6.4\"\n", - "v = math.sqrt(E/rho)# Calculation of velocity \n", - "lambda1 = 2*t # Calculation of fundamental wavelength\n", - "nu = v/lambda1 # Calculation of fundamental frequency\n", - "print \"Fundamental frequency is %e Hz.\"%(nu)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.4\n", - "Fundamental frequency is 4.549979e+05 Hz.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.5 Page No209" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "L = 1. # Inducmath.tance in Hanery\n", - "nu = 2.e6 # Frequency in Hz\n", - "print \"Example 6.5\"\n", - "C= 1./(4*((math.pi)**2)*nu**2*L) # Calculation of capacimath.tance\n", - "print \"Capacitance is %e microfarad.\"%(C*1e6)\n", - "# Answer in book is 0.00634 micro Farad\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.5\n", - "Capacitance is 6.332574e-09 microfarad.\n" - ] - } - ], - "prompt_number": 8 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch6_5.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch6_5.ipynb deleted file mode 100644 index 104b616a..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch6_5.ipynb +++ /dev/null @@ -1,207 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:1843361d9b1f7d4765e522604cbb1bc8ff2515ffb02cdd416946cae448cf39a5" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 6 : Ultrasonic Waves" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.1 Page No207" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "E = 7.9e10 # Young\u2019s modulus in N/m**2\n", - "rho = 2650 # Density in Kg/m**3\n", - "t = 0.003 # Thickness of quartz crystal in m\n", - "print \"Example 6.1\"\n", - "v = math.sqrt(E/rho)# Calculation of velocity \n", - "lambda1 = 2*t # Calculation of fundamental wavelength\n", - "nu = v/lambda1 # Calculation of fundamental frequency\n", - "print \"Fundamental frequency is %e Hz.\"%(nu)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.1\n", - "Fundamental frequency is 9.099957e+05 Hz.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.2 Page No208" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "v = 5760 # Velocity in m/s\n", - "T = 1.6 # Thickness of quartz crystal in mm\n", - "print \"Example 6.2\"\n", - "nu = v/(2*T*1e-3)# Calculation of fundamental frequency\n", - "print \"Fundamental frequency of crystal is %f MHz.\"%(nu/1e6)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.2\n", - "Fundamental frequency of crystal is 1.800000 MHz.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.3 Page No209" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "T =40. # Thickness of steel bar in cm\n", - "t1 = 40. # Time in ms\n", - "t2 = 80. # Time in ms\n", - "print \"Example 6.3\"\n", - "X = T*t1/t2 # Calculation of depth of defect\n", - "print \"Depth of defect is %d cm.\"%(X)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.3\n", - "Depth of defect is 20 cm.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.4 Page No209" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "E = 7.9e10 # Young\u2019s modulus in N/m**2\n", - "rho = 2650 # Density in Kg/m**3\n", - "t = 0.006 # Thickness of quartz crystal in m\n", - "print \"Example 6.4\"\n", - "v = math.sqrt(E/rho)# Calculation of velocity \n", - "lambda1 = 2*t # Calculation of fundamental wavelength\n", - "nu = v/lambda1 # Calculation of fundamental frequency\n", - "print \"Fundamental frequency is %e Hz.\"%(nu)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.4\n", - "Fundamental frequency is 4.549979e+05 Hz.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.5 Page No209" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "L = 1. # Inducmath.tance in Hanery\n", - "nu = 2.e6 # Frequency in Hz\n", - "print \"Example 6.5\"\n", - "C= 1./(4*((math.pi)**2)*nu**2*L) # Calculation of capacimath.tance\n", - "print \"Capacitance is %e microfarad.\"%(C*1e6)\n", - "# Answer in book is 0.00634 micro Farad\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 6.5\n", - "Capacitance is 6.332574e-09 microfarad.\n" - ] - } - ], - "prompt_number": 8 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch7.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch7.ipynb deleted file mode 100755 index 1b95e85e..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch7.ipynb +++ /dev/null @@ -1,496 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:5c15c522261c9a1964264eb127a07845e301723248156878641836ea3a61bbf4" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 7 : Maxwells Equations and Electromagnetic Waves" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.1 Page No220" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "p = 1000 # power in watt\n", - "d = 2 # Dismath.tance from lamp in m\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "print \"Example 7.1\"\n", - "s = p/(4*math.pi*d**2)# Calculation of pointing vector\n", - "E_H_ratio = math.sqrt(mu_0/epsilon_0) # Calculation of ratio of Electric field and magnetic field\n", - "E= math.sqrt(E_H_ratio*s) # Calculation of Electric field \n", - "print \" Average value of electric field at distance %d m is %f Volt/m \"%(d,E)\n", - "# Answer in book is 48.87 volt/m which is due to wrong calculation at intermediate steps\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.1\n", - " Average value of electric field at distance 2 m is 86.573038 Volt/m \n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.2 Page No222" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "p = 2 # power in cal/min/cm**2\n", - "\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "mu_0 = 4*math.pi*1e-7 # permeability of free space\n", - "print \"Example 7.2\"\n", - "s = p*4.2e4/60 # Calculation of pointing vector\n", - "E_H_ratio = math.sqrt(mu_0/epsilon_0) # Calculation of ratio of Electric field and magnetic field\n", - "E= math.sqrt(E_H_ratio*s) # Calculation of Electric field \n", - "H = s/E # Calculation of Electric field \n", - "\n", - "print \" Average value of electric field is %f Volt/m \"%(E*math.sqrt(2))\n", - "print \" Average value of magnetic field is %f Amp turn/m \"%(H*math.sqrt(2))\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.2\n", - " Average value of electric field is 1027.061861 Volt/m \n", - " Average value of magnetic field is 2.726223 Amp turn/m \n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.3 Page No225" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # permeability of free space\n", - "mu = mu_0 #permeability of silver\n", - "sigma = 3e7 # conductivity in mhos/m\n", - "f = 1e8 # frequency in Hz\n", - "print \"Example 7.3\"\n", - "omega = 2*math.pi/f # Calculation of time period\n", - "delta = math.sqrt(2/(omega*sigma*mu)) # Calculation of skin depth penetration\n", - "Delta = int(delta/100)*100 # Rounding off\n", - "print \" Skin depth penetration is %e cm. \"%(Delta*1e-6)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.3\n", - " Skin depth penetration is 9.000000e-04 cm. \n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.5 Page No229" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "k = 80 # relative Dielectric consmath.tant of sea water\n", - "epsilon_0 = 1/9e9 # Permittivity of free space\n", - "epsilon = 80*epsilon_0 # Permittivity of free space\n", - "sigma = 4.3 # conductivity in mho/m\n", - "delta = 10 # penetration depth in cm\n", - "mu_0 = 4*math.pi*1e-7 # permeability f free space\n", - "F = 1e8 # Given frequency in Hz\n", - "print \"Example 7.5\"\n", - "f = (1/(math.pi*mu_0*sigma))/(delta*1e-2)**2 # Calculation of frequency\n", - "f1= round(f/1e8)*1e8 # Rounding off\n", - "print \"Frequency required for penetration of depth %d cm is %e Hz\"%(delta,f1)\n", - "omega = 2*math.pi*F\n", - "x = 2*sigma/(epsilon*omega)\n", - "if x>1:\n", - " print \" Sea water is good conductor at frequency lesser than 1e8 Hz \"\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.5\n", - "Frequency required for penetration of depth 10 cm is 0.000000e+00 Hz\n", - " Sea water is good conductor at frequency lesser than 1e8 Hz \n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.7 Page No234" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "k = 12 # relative Dielectric consmath.tant of sea water\n", - "epsilon_0 = 1/9e9 # Permittivity of free space\n", - "sigma = 2 # conductivity in mho/cm\n", - "mu_0 = 4*math.pi*1e-7 # permeability f free space\n", - "f= 1e9 # Given frequency in Hz\n", - "F = 1e6 # Given frequency in Hz\n", - "print \"Example 7.7\"\n", - "delta = math.sqrt(2/(2*math.pi*F*mu_0*sigma*100)) # Calculation of frequency\n", - "print \" For %eHz frequency, Penetration depth is %f cm\"%(F,delta*100)\n", - "omega = 2*math.pi*f\n", - "x = 2*sigma*100/(k*epsilon_0*omega)\n", - "if x>1 :\n", - " print \" Silicon is good conductor at frequency lesser than 1e9 Hz \"\n", - "\n", - "# Answer in book is 3.6 cm\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.7\n", - " For 1.000000e+06Hz frequency, Penetration depth is 3.558813 cm\n", - " Silicon is good conductor at frequency lesser than 1e9 Hz \n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.8 Page No236" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # permeability of free space\n", - "mu = mu_0 #permeability of silver\n", - "sigma = 5.8e7 # conductivity in simens /m\n", - "delta = 0.1 # Skin depth penetration in mm\n", - "\n", - "print \"Example 7.8\"\n", - "f = 2/((delta*1e-3)**2*sigma*mu*2*math.pi) # Calculation of skin depth penetration\n", - "print \" Required frequency is %.2e Hz\"%(f)\n", - "print \" The incident electromagnetic wave is the radio part of spectrum.\"\n", - "# Answer in book is 3.36e5 Hz. Difference is due to approximation at intermediate stages\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.8\n", - " Required frequency is 4.37e+05 Hz\n", - " The incident electromagnetic wave is the radio part of spectrum.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.9 Page No240" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - " \n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "mu = mu_0 #Permeability of silver\n", - "sigma = 3e7 # conductivity in mhos/m\n", - "f = 1e10 # frequency in Hz\n", - "print \"Example 7.9\"\n", - "delta = math.sqrt(1/(math.pi*sigma*f*mu)) # Calculation of skin depth penetration\n", - "print \" Skin depth penetration is %f micrometre. \"%(delta*1e6)\n", - "# Answer in book is 0.93 micrometer\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.9\n", - " Skin depth penetration is 0.918881 micrometre. \n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.10 Page No241" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "p = 500 # power in watt\n", - "d = 1 # Dismath.tance from lamp in m\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "print \"Example 7.10\"\n", - "s = p/(4*math.pi*d**2)# Calculation of pointing vector\n", - "E_H_ratio = math.sqrt(mu_0/epsilon_0) # Calculation of ratio of Electric field and magnetic field\n", - "H = s/E_H_ratio # Calculation of Electric field \n", - "h = (H*100)/100 # rounding off for 2 decimal places\n", - "E= p/(4*math.pi*h) # Calculation of Electric field \n", - "print \" Average value of electric field at distance %d m is %f Volt/m \"%(d,E)\n", - "print \" Average value of magnetic field at distance %d m is %f Amp-turn/m \"%(d,h)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.10\n", - " Average value of electric field at distance 1 m is 376.734309 Volt/m \n", - " Average value of magnetic field at distance 1 m is 0.105615 Amp-turn/m \n" - ] - } - ], - "prompt_number": 18 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.11 Page No243" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "mu = mu_0 #Permeability of silver\n", - "sigma = 3.5e7 # conductivity in simens /m\n", - "delta = 0.03 # Skin depth penetration in mm\n", - "\n", - "print \"Example 7.11\"\n", - "\n", - "f = 2/((delta*1e-3)**2*sigma*mu*2*math.pi) # Calculation of skin depth penetration\n", - "print \" Required frequency is %d MHz.\"%(f/1e6)\n", - "print \" The incident electromagnetic wave is the radio part of spectrum\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.11\n", - " Required frequency is 8 MHz.\n", - " The incident electromagnetic wave is the radio part of spectrum\n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.12 Page No244" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "p = 3.8e26 # power radiated by moon in watt\n", - "d_sun = 1.44e11 # Dismath.tance between sun and earth in meter\n", - "d_moon = 3e8 #Dismath.tance between moon and earth in meter\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "print \"Example 7.12\"\n", - "s = p/(4*math.pi*d_sun**2)# Calculation of solar energy received during solar eclipse in watt /m**2\n", - "S = s*60/(4.2*1e4) # Unit conversion\n", - "\n", - "print \" Solar energy received during solar eclipse is %f Cal per min per m**2 \"%(S)\n", - "# Ansewr in book is 2.1 cal per min per m**2\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.12\n", - " Solar energy received during solar eclipse is 2.083295 Cal per min per m**2 \n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.13 Page No246" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "mu = mu_0 #Permeability of silver\n", - "sigma = 3.5e7 # conductivity in simens /m\n", - "lambda1 = 6328. # Wavelength in angstrom\n", - "c = 3e8# Speed of light in m/sec\n", - "\n", - "print \"Example 7.13\"\n", - "f = c/(lambda1*1e-10)\n", - "omega = 2*math.pi/f # Calculation of time period\n", - "f = c/(lambda1*1e-10) # Calculation of frequency in Hz\n", - "delta = math.sqrt(1/(math.pi*f*sigma*mu)) # Calculation of skin depth penetration\n", - "print \" Skin depth penetration is %f nm. \"%(delta*1e9)\n", - "# Answer in book is 3.9 mm, unit used in book is wrong\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.13\n", - " Skin depth penetration is 3.907138 nm. \n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch7_1.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch7_1.ipynb deleted file mode 100755 index 1b95e85e..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch7_1.ipynb +++ /dev/null @@ -1,496 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:5c15c522261c9a1964264eb127a07845e301723248156878641836ea3a61bbf4" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 7 : Maxwells Equations and Electromagnetic Waves" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.1 Page No220" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "p = 1000 # power in watt\n", - "d = 2 # Dismath.tance from lamp in m\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "print \"Example 7.1\"\n", - "s = p/(4*math.pi*d**2)# Calculation of pointing vector\n", - "E_H_ratio = math.sqrt(mu_0/epsilon_0) # Calculation of ratio of Electric field and magnetic field\n", - "E= math.sqrt(E_H_ratio*s) # Calculation of Electric field \n", - "print \" Average value of electric field at distance %d m is %f Volt/m \"%(d,E)\n", - "# Answer in book is 48.87 volt/m which is due to wrong calculation at intermediate steps\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.1\n", - " Average value of electric field at distance 2 m is 86.573038 Volt/m \n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.2 Page No222" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "p = 2 # power in cal/min/cm**2\n", - "\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "mu_0 = 4*math.pi*1e-7 # permeability of free space\n", - "print \"Example 7.2\"\n", - "s = p*4.2e4/60 # Calculation of pointing vector\n", - "E_H_ratio = math.sqrt(mu_0/epsilon_0) # Calculation of ratio of Electric field and magnetic field\n", - "E= math.sqrt(E_H_ratio*s) # Calculation of Electric field \n", - "H = s/E # Calculation of Electric field \n", - "\n", - "print \" Average value of electric field is %f Volt/m \"%(E*math.sqrt(2))\n", - "print \" Average value of magnetic field is %f Amp turn/m \"%(H*math.sqrt(2))\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.2\n", - " Average value of electric field is 1027.061861 Volt/m \n", - " Average value of magnetic field is 2.726223 Amp turn/m \n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.3 Page No225" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # permeability of free space\n", - "mu = mu_0 #permeability of silver\n", - "sigma = 3e7 # conductivity in mhos/m\n", - "f = 1e8 # frequency in Hz\n", - "print \"Example 7.3\"\n", - "omega = 2*math.pi/f # Calculation of time period\n", - "delta = math.sqrt(2/(omega*sigma*mu)) # Calculation of skin depth penetration\n", - "Delta = int(delta/100)*100 # Rounding off\n", - "print \" Skin depth penetration is %e cm. \"%(Delta*1e-6)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.3\n", - " Skin depth penetration is 9.000000e-04 cm. \n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.5 Page No229" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "k = 80 # relative Dielectric consmath.tant of sea water\n", - "epsilon_0 = 1/9e9 # Permittivity of free space\n", - "epsilon = 80*epsilon_0 # Permittivity of free space\n", - "sigma = 4.3 # conductivity in mho/m\n", - "delta = 10 # penetration depth in cm\n", - "mu_0 = 4*math.pi*1e-7 # permeability f free space\n", - "F = 1e8 # Given frequency in Hz\n", - "print \"Example 7.5\"\n", - "f = (1/(math.pi*mu_0*sigma))/(delta*1e-2)**2 # Calculation of frequency\n", - "f1= round(f/1e8)*1e8 # Rounding off\n", - "print \"Frequency required for penetration of depth %d cm is %e Hz\"%(delta,f1)\n", - "omega = 2*math.pi*F\n", - "x = 2*sigma/(epsilon*omega)\n", - "if x>1:\n", - " print \" Sea water is good conductor at frequency lesser than 1e8 Hz \"\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.5\n", - "Frequency required for penetration of depth 10 cm is 0.000000e+00 Hz\n", - " Sea water is good conductor at frequency lesser than 1e8 Hz \n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.7 Page No234" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "k = 12 # relative Dielectric consmath.tant of sea water\n", - "epsilon_0 = 1/9e9 # Permittivity of free space\n", - "sigma = 2 # conductivity in mho/cm\n", - "mu_0 = 4*math.pi*1e-7 # permeability f free space\n", - "f= 1e9 # Given frequency in Hz\n", - "F = 1e6 # Given frequency in Hz\n", - "print \"Example 7.7\"\n", - "delta = math.sqrt(2/(2*math.pi*F*mu_0*sigma*100)) # Calculation of frequency\n", - "print \" For %eHz frequency, Penetration depth is %f cm\"%(F,delta*100)\n", - "omega = 2*math.pi*f\n", - "x = 2*sigma*100/(k*epsilon_0*omega)\n", - "if x>1 :\n", - " print \" Silicon is good conductor at frequency lesser than 1e9 Hz \"\n", - "\n", - "# Answer in book is 3.6 cm\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.7\n", - " For 1.000000e+06Hz frequency, Penetration depth is 3.558813 cm\n", - " Silicon is good conductor at frequency lesser than 1e9 Hz \n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.8 Page No236" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # permeability of free space\n", - "mu = mu_0 #permeability of silver\n", - "sigma = 5.8e7 # conductivity in simens /m\n", - "delta = 0.1 # Skin depth penetration in mm\n", - "\n", - "print \"Example 7.8\"\n", - "f = 2/((delta*1e-3)**2*sigma*mu*2*math.pi) # Calculation of skin depth penetration\n", - "print \" Required frequency is %.2e Hz\"%(f)\n", - "print \" The incident electromagnetic wave is the radio part of spectrum.\"\n", - "# Answer in book is 3.36e5 Hz. Difference is due to approximation at intermediate stages\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.8\n", - " Required frequency is 4.37e+05 Hz\n", - " The incident electromagnetic wave is the radio part of spectrum.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.9 Page No240" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - " \n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "mu = mu_0 #Permeability of silver\n", - "sigma = 3e7 # conductivity in mhos/m\n", - "f = 1e10 # frequency in Hz\n", - "print \"Example 7.9\"\n", - "delta = math.sqrt(1/(math.pi*sigma*f*mu)) # Calculation of skin depth penetration\n", - "print \" Skin depth penetration is %f micrometre. \"%(delta*1e6)\n", - "# Answer in book is 0.93 micrometer\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.9\n", - " Skin depth penetration is 0.918881 micrometre. \n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.10 Page No241" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "p = 500 # power in watt\n", - "d = 1 # Dismath.tance from lamp in m\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "print \"Example 7.10\"\n", - "s = p/(4*math.pi*d**2)# Calculation of pointing vector\n", - "E_H_ratio = math.sqrt(mu_0/epsilon_0) # Calculation of ratio of Electric field and magnetic field\n", - "H = s/E_H_ratio # Calculation of Electric field \n", - "h = (H*100)/100 # rounding off for 2 decimal places\n", - "E= p/(4*math.pi*h) # Calculation of Electric field \n", - "print \" Average value of electric field at distance %d m is %f Volt/m \"%(d,E)\n", - "print \" Average value of magnetic field at distance %d m is %f Amp-turn/m \"%(d,h)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.10\n", - " Average value of electric field at distance 1 m is 376.734309 Volt/m \n", - " Average value of magnetic field at distance 1 m is 0.105615 Amp-turn/m \n" - ] - } - ], - "prompt_number": 18 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.11 Page No243" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "mu = mu_0 #Permeability of silver\n", - "sigma = 3.5e7 # conductivity in simens /m\n", - "delta = 0.03 # Skin depth penetration in mm\n", - "\n", - "print \"Example 7.11\"\n", - "\n", - "f = 2/((delta*1e-3)**2*sigma*mu*2*math.pi) # Calculation of skin depth penetration\n", - "print \" Required frequency is %d MHz.\"%(f/1e6)\n", - "print \" The incident electromagnetic wave is the radio part of spectrum\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.11\n", - " Required frequency is 8 MHz.\n", - " The incident electromagnetic wave is the radio part of spectrum\n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.12 Page No244" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "p = 3.8e26 # power radiated by moon in watt\n", - "d_sun = 1.44e11 # Dismath.tance between sun and earth in meter\n", - "d_moon = 3e8 #Dismath.tance between moon and earth in meter\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "print \"Example 7.12\"\n", - "s = p/(4*math.pi*d_sun**2)# Calculation of solar energy received during solar eclipse in watt /m**2\n", - "S = s*60/(4.2*1e4) # Unit conversion\n", - "\n", - "print \" Solar energy received during solar eclipse is %f Cal per min per m**2 \"%(S)\n", - "# Ansewr in book is 2.1 cal per min per m**2\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.12\n", - " Solar energy received during solar eclipse is 2.083295 Cal per min per m**2 \n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.13 Page No246" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "mu = mu_0 #Permeability of silver\n", - "sigma = 3.5e7 # conductivity in simens /m\n", - "lambda1 = 6328. # Wavelength in angstrom\n", - "c = 3e8# Speed of light in m/sec\n", - "\n", - "print \"Example 7.13\"\n", - "f = c/(lambda1*1e-10)\n", - "omega = 2*math.pi/f # Calculation of time period\n", - "f = c/(lambda1*1e-10) # Calculation of frequency in Hz\n", - "delta = math.sqrt(1/(math.pi*f*sigma*mu)) # Calculation of skin depth penetration\n", - "print \" Skin depth penetration is %f nm. \"%(delta*1e9)\n", - "# Answer in book is 3.9 mm, unit used in book is wrong\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.13\n", - " Skin depth penetration is 3.907138 nm. \n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch7_2.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch7_2.ipynb deleted file mode 100644 index 1b95e85e..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch7_2.ipynb +++ /dev/null @@ -1,496 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:5c15c522261c9a1964264eb127a07845e301723248156878641836ea3a61bbf4" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 7 : Maxwells Equations and Electromagnetic Waves" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.1 Page No220" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "p = 1000 # power in watt\n", - "d = 2 # Dismath.tance from lamp in m\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "print \"Example 7.1\"\n", - "s = p/(4*math.pi*d**2)# Calculation of pointing vector\n", - "E_H_ratio = math.sqrt(mu_0/epsilon_0) # Calculation of ratio of Electric field and magnetic field\n", - "E= math.sqrt(E_H_ratio*s) # Calculation of Electric field \n", - "print \" Average value of electric field at distance %d m is %f Volt/m \"%(d,E)\n", - "# Answer in book is 48.87 volt/m which is due to wrong calculation at intermediate steps\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.1\n", - " Average value of electric field at distance 2 m is 86.573038 Volt/m \n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.2 Page No222" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "p = 2 # power in cal/min/cm**2\n", - "\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "mu_0 = 4*math.pi*1e-7 # permeability of free space\n", - "print \"Example 7.2\"\n", - "s = p*4.2e4/60 # Calculation of pointing vector\n", - "E_H_ratio = math.sqrt(mu_0/epsilon_0) # Calculation of ratio of Electric field and magnetic field\n", - "E= math.sqrt(E_H_ratio*s) # Calculation of Electric field \n", - "H = s/E # Calculation of Electric field \n", - "\n", - "print \" Average value of electric field is %f Volt/m \"%(E*math.sqrt(2))\n", - "print \" Average value of magnetic field is %f Amp turn/m \"%(H*math.sqrt(2))\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.2\n", - " Average value of electric field is 1027.061861 Volt/m \n", - " Average value of magnetic field is 2.726223 Amp turn/m \n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.3 Page No225" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # permeability of free space\n", - "mu = mu_0 #permeability of silver\n", - "sigma = 3e7 # conductivity in mhos/m\n", - "f = 1e8 # frequency in Hz\n", - "print \"Example 7.3\"\n", - "omega = 2*math.pi/f # Calculation of time period\n", - "delta = math.sqrt(2/(omega*sigma*mu)) # Calculation of skin depth penetration\n", - "Delta = int(delta/100)*100 # Rounding off\n", - "print \" Skin depth penetration is %e cm. \"%(Delta*1e-6)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.3\n", - " Skin depth penetration is 9.000000e-04 cm. \n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.5 Page No229" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "k = 80 # relative Dielectric consmath.tant of sea water\n", - "epsilon_0 = 1/9e9 # Permittivity of free space\n", - "epsilon = 80*epsilon_0 # Permittivity of free space\n", - "sigma = 4.3 # conductivity in mho/m\n", - "delta = 10 # penetration depth in cm\n", - "mu_0 = 4*math.pi*1e-7 # permeability f free space\n", - "F = 1e8 # Given frequency in Hz\n", - "print \"Example 7.5\"\n", - "f = (1/(math.pi*mu_0*sigma))/(delta*1e-2)**2 # Calculation of frequency\n", - "f1= round(f/1e8)*1e8 # Rounding off\n", - "print \"Frequency required for penetration of depth %d cm is %e Hz\"%(delta,f1)\n", - "omega = 2*math.pi*F\n", - "x = 2*sigma/(epsilon*omega)\n", - "if x>1:\n", - " print \" Sea water is good conductor at frequency lesser than 1e8 Hz \"\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.5\n", - "Frequency required for penetration of depth 10 cm is 0.000000e+00 Hz\n", - " Sea water is good conductor at frequency lesser than 1e8 Hz \n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.7 Page No234" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "k = 12 # relative Dielectric consmath.tant of sea water\n", - "epsilon_0 = 1/9e9 # Permittivity of free space\n", - "sigma = 2 # conductivity in mho/cm\n", - "mu_0 = 4*math.pi*1e-7 # permeability f free space\n", - "f= 1e9 # Given frequency in Hz\n", - "F = 1e6 # Given frequency in Hz\n", - "print \"Example 7.7\"\n", - "delta = math.sqrt(2/(2*math.pi*F*mu_0*sigma*100)) # Calculation of frequency\n", - "print \" For %eHz frequency, Penetration depth is %f cm\"%(F,delta*100)\n", - "omega = 2*math.pi*f\n", - "x = 2*sigma*100/(k*epsilon_0*omega)\n", - "if x>1 :\n", - " print \" Silicon is good conductor at frequency lesser than 1e9 Hz \"\n", - "\n", - "# Answer in book is 3.6 cm\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.7\n", - " For 1.000000e+06Hz frequency, Penetration depth is 3.558813 cm\n", - " Silicon is good conductor at frequency lesser than 1e9 Hz \n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.8 Page No236" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # permeability of free space\n", - "mu = mu_0 #permeability of silver\n", - "sigma = 5.8e7 # conductivity in simens /m\n", - "delta = 0.1 # Skin depth penetration in mm\n", - "\n", - "print \"Example 7.8\"\n", - "f = 2/((delta*1e-3)**2*sigma*mu*2*math.pi) # Calculation of skin depth penetration\n", - "print \" Required frequency is %.2e Hz\"%(f)\n", - "print \" The incident electromagnetic wave is the radio part of spectrum.\"\n", - "# Answer in book is 3.36e5 Hz. Difference is due to approximation at intermediate stages\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.8\n", - " Required frequency is 4.37e+05 Hz\n", - " The incident electromagnetic wave is the radio part of spectrum.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.9 Page No240" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - " \n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "mu = mu_0 #Permeability of silver\n", - "sigma = 3e7 # conductivity in mhos/m\n", - "f = 1e10 # frequency in Hz\n", - "print \"Example 7.9\"\n", - "delta = math.sqrt(1/(math.pi*sigma*f*mu)) # Calculation of skin depth penetration\n", - "print \" Skin depth penetration is %f micrometre. \"%(delta*1e6)\n", - "# Answer in book is 0.93 micrometer\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.9\n", - " Skin depth penetration is 0.918881 micrometre. \n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.10 Page No241" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "p = 500 # power in watt\n", - "d = 1 # Dismath.tance from lamp in m\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "print \"Example 7.10\"\n", - "s = p/(4*math.pi*d**2)# Calculation of pointing vector\n", - "E_H_ratio = math.sqrt(mu_0/epsilon_0) # Calculation of ratio of Electric field and magnetic field\n", - "H = s/E_H_ratio # Calculation of Electric field \n", - "h = (H*100)/100 # rounding off for 2 decimal places\n", - "E= p/(4*math.pi*h) # Calculation of Electric field \n", - "print \" Average value of electric field at distance %d m is %f Volt/m \"%(d,E)\n", - "print \" Average value of magnetic field at distance %d m is %f Amp-turn/m \"%(d,h)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.10\n", - " Average value of electric field at distance 1 m is 376.734309 Volt/m \n", - " Average value of magnetic field at distance 1 m is 0.105615 Amp-turn/m \n" - ] - } - ], - "prompt_number": 18 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.11 Page No243" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "mu = mu_0 #Permeability of silver\n", - "sigma = 3.5e7 # conductivity in simens /m\n", - "delta = 0.03 # Skin depth penetration in mm\n", - "\n", - "print \"Example 7.11\"\n", - "\n", - "f = 2/((delta*1e-3)**2*sigma*mu*2*math.pi) # Calculation of skin depth penetration\n", - "print \" Required frequency is %d MHz.\"%(f/1e6)\n", - "print \" The incident electromagnetic wave is the radio part of spectrum\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.11\n", - " Required frequency is 8 MHz.\n", - " The incident electromagnetic wave is the radio part of spectrum\n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.12 Page No244" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "p = 3.8e26 # power radiated by moon in watt\n", - "d_sun = 1.44e11 # Dismath.tance between sun and earth in meter\n", - "d_moon = 3e8 #Dismath.tance between moon and earth in meter\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "print \"Example 7.12\"\n", - "s = p/(4*math.pi*d_sun**2)# Calculation of solar energy received during solar eclipse in watt /m**2\n", - "S = s*60/(4.2*1e4) # Unit conversion\n", - "\n", - "print \" Solar energy received during solar eclipse is %f Cal per min per m**2 \"%(S)\n", - "# Ansewr in book is 2.1 cal per min per m**2\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.12\n", - " Solar energy received during solar eclipse is 2.083295 Cal per min per m**2 \n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.13 Page No246" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "mu = mu_0 #Permeability of silver\n", - "sigma = 3.5e7 # conductivity in simens /m\n", - "lambda1 = 6328. # Wavelength in angstrom\n", - "c = 3e8# Speed of light in m/sec\n", - "\n", - "print \"Example 7.13\"\n", - "f = c/(lambda1*1e-10)\n", - "omega = 2*math.pi/f # Calculation of time period\n", - "f = c/(lambda1*1e-10) # Calculation of frequency in Hz\n", - "delta = math.sqrt(1/(math.pi*f*sigma*mu)) # Calculation of skin depth penetration\n", - "print \" Skin depth penetration is %f nm. \"%(delta*1e9)\n", - "# Answer in book is 3.9 mm, unit used in book is wrong\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.13\n", - " Skin depth penetration is 3.907138 nm. \n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch7_3.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch7_3.ipynb deleted file mode 100644 index 1b95e85e..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch7_3.ipynb +++ /dev/null @@ -1,496 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:5c15c522261c9a1964264eb127a07845e301723248156878641836ea3a61bbf4" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 7 : Maxwells Equations and Electromagnetic Waves" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.1 Page No220" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "p = 1000 # power in watt\n", - "d = 2 # Dismath.tance from lamp in m\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "print \"Example 7.1\"\n", - "s = p/(4*math.pi*d**2)# Calculation of pointing vector\n", - "E_H_ratio = math.sqrt(mu_0/epsilon_0) # Calculation of ratio of Electric field and magnetic field\n", - "E= math.sqrt(E_H_ratio*s) # Calculation of Electric field \n", - "print \" Average value of electric field at distance %d m is %f Volt/m \"%(d,E)\n", - "# Answer in book is 48.87 volt/m which is due to wrong calculation at intermediate steps\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.1\n", - " Average value of electric field at distance 2 m is 86.573038 Volt/m \n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.2 Page No222" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "p = 2 # power in cal/min/cm**2\n", - "\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "mu_0 = 4*math.pi*1e-7 # permeability of free space\n", - "print \"Example 7.2\"\n", - "s = p*4.2e4/60 # Calculation of pointing vector\n", - "E_H_ratio = math.sqrt(mu_0/epsilon_0) # Calculation of ratio of Electric field and magnetic field\n", - "E= math.sqrt(E_H_ratio*s) # Calculation of Electric field \n", - "H = s/E # Calculation of Electric field \n", - "\n", - "print \" Average value of electric field is %f Volt/m \"%(E*math.sqrt(2))\n", - "print \" Average value of magnetic field is %f Amp turn/m \"%(H*math.sqrt(2))\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.2\n", - " Average value of electric field is 1027.061861 Volt/m \n", - " Average value of magnetic field is 2.726223 Amp turn/m \n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.3 Page No225" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # permeability of free space\n", - "mu = mu_0 #permeability of silver\n", - "sigma = 3e7 # conductivity in mhos/m\n", - "f = 1e8 # frequency in Hz\n", - "print \"Example 7.3\"\n", - "omega = 2*math.pi/f # Calculation of time period\n", - "delta = math.sqrt(2/(omega*sigma*mu)) # Calculation of skin depth penetration\n", - "Delta = int(delta/100)*100 # Rounding off\n", - "print \" Skin depth penetration is %e cm. \"%(Delta*1e-6)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.3\n", - " Skin depth penetration is 9.000000e-04 cm. \n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.5 Page No229" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "k = 80 # relative Dielectric consmath.tant of sea water\n", - "epsilon_0 = 1/9e9 # Permittivity of free space\n", - "epsilon = 80*epsilon_0 # Permittivity of free space\n", - "sigma = 4.3 # conductivity in mho/m\n", - "delta = 10 # penetration depth in cm\n", - "mu_0 = 4*math.pi*1e-7 # permeability f free space\n", - "F = 1e8 # Given frequency in Hz\n", - "print \"Example 7.5\"\n", - "f = (1/(math.pi*mu_0*sigma))/(delta*1e-2)**2 # Calculation of frequency\n", - "f1= round(f/1e8)*1e8 # Rounding off\n", - "print \"Frequency required for penetration of depth %d cm is %e Hz\"%(delta,f1)\n", - "omega = 2*math.pi*F\n", - "x = 2*sigma/(epsilon*omega)\n", - "if x>1:\n", - " print \" Sea water is good conductor at frequency lesser than 1e8 Hz \"\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.5\n", - "Frequency required for penetration of depth 10 cm is 0.000000e+00 Hz\n", - " Sea water is good conductor at frequency lesser than 1e8 Hz \n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.7 Page No234" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "k = 12 # relative Dielectric consmath.tant of sea water\n", - "epsilon_0 = 1/9e9 # Permittivity of free space\n", - "sigma = 2 # conductivity in mho/cm\n", - "mu_0 = 4*math.pi*1e-7 # permeability f free space\n", - "f= 1e9 # Given frequency in Hz\n", - "F = 1e6 # Given frequency in Hz\n", - "print \"Example 7.7\"\n", - "delta = math.sqrt(2/(2*math.pi*F*mu_0*sigma*100)) # Calculation of frequency\n", - "print \" For %eHz frequency, Penetration depth is %f cm\"%(F,delta*100)\n", - "omega = 2*math.pi*f\n", - "x = 2*sigma*100/(k*epsilon_0*omega)\n", - "if x>1 :\n", - " print \" Silicon is good conductor at frequency lesser than 1e9 Hz \"\n", - "\n", - "# Answer in book is 3.6 cm\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.7\n", - " For 1.000000e+06Hz frequency, Penetration depth is 3.558813 cm\n", - " Silicon is good conductor at frequency lesser than 1e9 Hz \n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.8 Page No236" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # permeability of free space\n", - "mu = mu_0 #permeability of silver\n", - "sigma = 5.8e7 # conductivity in simens /m\n", - "delta = 0.1 # Skin depth penetration in mm\n", - "\n", - "print \"Example 7.8\"\n", - "f = 2/((delta*1e-3)**2*sigma*mu*2*math.pi) # Calculation of skin depth penetration\n", - "print \" Required frequency is %.2e Hz\"%(f)\n", - "print \" The incident electromagnetic wave is the radio part of spectrum.\"\n", - "# Answer in book is 3.36e5 Hz. Difference is due to approximation at intermediate stages\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.8\n", - " Required frequency is 4.37e+05 Hz\n", - " The incident electromagnetic wave is the radio part of spectrum.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.9 Page No240" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - " \n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "mu = mu_0 #Permeability of silver\n", - "sigma = 3e7 # conductivity in mhos/m\n", - "f = 1e10 # frequency in Hz\n", - "print \"Example 7.9\"\n", - "delta = math.sqrt(1/(math.pi*sigma*f*mu)) # Calculation of skin depth penetration\n", - "print \" Skin depth penetration is %f micrometre. \"%(delta*1e6)\n", - "# Answer in book is 0.93 micrometer\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.9\n", - " Skin depth penetration is 0.918881 micrometre. \n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.10 Page No241" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "p = 500 # power in watt\n", - "d = 1 # Dismath.tance from lamp in m\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "print \"Example 7.10\"\n", - "s = p/(4*math.pi*d**2)# Calculation of pointing vector\n", - "E_H_ratio = math.sqrt(mu_0/epsilon_0) # Calculation of ratio of Electric field and magnetic field\n", - "H = s/E_H_ratio # Calculation of Electric field \n", - "h = (H*100)/100 # rounding off for 2 decimal places\n", - "E= p/(4*math.pi*h) # Calculation of Electric field \n", - "print \" Average value of electric field at distance %d m is %f Volt/m \"%(d,E)\n", - "print \" Average value of magnetic field at distance %d m is %f Amp-turn/m \"%(d,h)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.10\n", - " Average value of electric field at distance 1 m is 376.734309 Volt/m \n", - " Average value of magnetic field at distance 1 m is 0.105615 Amp-turn/m \n" - ] - } - ], - "prompt_number": 18 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.11 Page No243" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "mu = mu_0 #Permeability of silver\n", - "sigma = 3.5e7 # conductivity in simens /m\n", - "delta = 0.03 # Skin depth penetration in mm\n", - "\n", - "print \"Example 7.11\"\n", - "\n", - "f = 2/((delta*1e-3)**2*sigma*mu*2*math.pi) # Calculation of skin depth penetration\n", - "print \" Required frequency is %d MHz.\"%(f/1e6)\n", - "print \" The incident electromagnetic wave is the radio part of spectrum\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.11\n", - " Required frequency is 8 MHz.\n", - " The incident electromagnetic wave is the radio part of spectrum\n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.12 Page No244" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "p = 3.8e26 # power radiated by moon in watt\n", - "d_sun = 1.44e11 # Dismath.tance between sun and earth in meter\n", - "d_moon = 3e8 #Dismath.tance between moon and earth in meter\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "print \"Example 7.12\"\n", - "s = p/(4*math.pi*d_sun**2)# Calculation of solar energy received during solar eclipse in watt /m**2\n", - "S = s*60/(4.2*1e4) # Unit conversion\n", - "\n", - "print \" Solar energy received during solar eclipse is %f Cal per min per m**2 \"%(S)\n", - "# Ansewr in book is 2.1 cal per min per m**2\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.12\n", - " Solar energy received during solar eclipse is 2.083295 Cal per min per m**2 \n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.13 Page No246" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "mu = mu_0 #Permeability of silver\n", - "sigma = 3.5e7 # conductivity in simens /m\n", - "lambda1 = 6328. # Wavelength in angstrom\n", - "c = 3e8# Speed of light in m/sec\n", - "\n", - "print \"Example 7.13\"\n", - "f = c/(lambda1*1e-10)\n", - "omega = 2*math.pi/f # Calculation of time period\n", - "f = c/(lambda1*1e-10) # Calculation of frequency in Hz\n", - "delta = math.sqrt(1/(math.pi*f*sigma*mu)) # Calculation of skin depth penetration\n", - "print \" Skin depth penetration is %f nm. \"%(delta*1e9)\n", - "# Answer in book is 3.9 mm, unit used in book is wrong\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.13\n", - " Skin depth penetration is 3.907138 nm. \n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch7_4.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch7_4.ipynb deleted file mode 100644 index 1b95e85e..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch7_4.ipynb +++ /dev/null @@ -1,496 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:5c15c522261c9a1964264eb127a07845e301723248156878641836ea3a61bbf4" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 7 : Maxwells Equations and Electromagnetic Waves" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.1 Page No220" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "p = 1000 # power in watt\n", - "d = 2 # Dismath.tance from lamp in m\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "print \"Example 7.1\"\n", - "s = p/(4*math.pi*d**2)# Calculation of pointing vector\n", - "E_H_ratio = math.sqrt(mu_0/epsilon_0) # Calculation of ratio of Electric field and magnetic field\n", - "E= math.sqrt(E_H_ratio*s) # Calculation of Electric field \n", - "print \" Average value of electric field at distance %d m is %f Volt/m \"%(d,E)\n", - "# Answer in book is 48.87 volt/m which is due to wrong calculation at intermediate steps\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.1\n", - " Average value of electric field at distance 2 m is 86.573038 Volt/m \n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.2 Page No222" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "p = 2 # power in cal/min/cm**2\n", - "\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "mu_0 = 4*math.pi*1e-7 # permeability of free space\n", - "print \"Example 7.2\"\n", - "s = p*4.2e4/60 # Calculation of pointing vector\n", - "E_H_ratio = math.sqrt(mu_0/epsilon_0) # Calculation of ratio of Electric field and magnetic field\n", - "E= math.sqrt(E_H_ratio*s) # Calculation of Electric field \n", - "H = s/E # Calculation of Electric field \n", - "\n", - "print \" Average value of electric field is %f Volt/m \"%(E*math.sqrt(2))\n", - "print \" Average value of magnetic field is %f Amp turn/m \"%(H*math.sqrt(2))\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.2\n", - " Average value of electric field is 1027.061861 Volt/m \n", - " Average value of magnetic field is 2.726223 Amp turn/m \n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.3 Page No225" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # permeability of free space\n", - "mu = mu_0 #permeability of silver\n", - "sigma = 3e7 # conductivity in mhos/m\n", - "f = 1e8 # frequency in Hz\n", - "print \"Example 7.3\"\n", - "omega = 2*math.pi/f # Calculation of time period\n", - "delta = math.sqrt(2/(omega*sigma*mu)) # Calculation of skin depth penetration\n", - "Delta = int(delta/100)*100 # Rounding off\n", - "print \" Skin depth penetration is %e cm. \"%(Delta*1e-6)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.3\n", - " Skin depth penetration is 9.000000e-04 cm. \n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.5 Page No229" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "k = 80 # relative Dielectric consmath.tant of sea water\n", - "epsilon_0 = 1/9e9 # Permittivity of free space\n", - "epsilon = 80*epsilon_0 # Permittivity of free space\n", - "sigma = 4.3 # conductivity in mho/m\n", - "delta = 10 # penetration depth in cm\n", - "mu_0 = 4*math.pi*1e-7 # permeability f free space\n", - "F = 1e8 # Given frequency in Hz\n", - "print \"Example 7.5\"\n", - "f = (1/(math.pi*mu_0*sigma))/(delta*1e-2)**2 # Calculation of frequency\n", - "f1= round(f/1e8)*1e8 # Rounding off\n", - "print \"Frequency required for penetration of depth %d cm is %e Hz\"%(delta,f1)\n", - "omega = 2*math.pi*F\n", - "x = 2*sigma/(epsilon*omega)\n", - "if x>1:\n", - " print \" Sea water is good conductor at frequency lesser than 1e8 Hz \"\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.5\n", - "Frequency required for penetration of depth 10 cm is 0.000000e+00 Hz\n", - " Sea water is good conductor at frequency lesser than 1e8 Hz \n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.7 Page No234" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "k = 12 # relative Dielectric consmath.tant of sea water\n", - "epsilon_0 = 1/9e9 # Permittivity of free space\n", - "sigma = 2 # conductivity in mho/cm\n", - "mu_0 = 4*math.pi*1e-7 # permeability f free space\n", - "f= 1e9 # Given frequency in Hz\n", - "F = 1e6 # Given frequency in Hz\n", - "print \"Example 7.7\"\n", - "delta = math.sqrt(2/(2*math.pi*F*mu_0*sigma*100)) # Calculation of frequency\n", - "print \" For %eHz frequency, Penetration depth is %f cm\"%(F,delta*100)\n", - "omega = 2*math.pi*f\n", - "x = 2*sigma*100/(k*epsilon_0*omega)\n", - "if x>1 :\n", - " print \" Silicon is good conductor at frequency lesser than 1e9 Hz \"\n", - "\n", - "# Answer in book is 3.6 cm\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.7\n", - " For 1.000000e+06Hz frequency, Penetration depth is 3.558813 cm\n", - " Silicon is good conductor at frequency lesser than 1e9 Hz \n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.8 Page No236" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # permeability of free space\n", - "mu = mu_0 #permeability of silver\n", - "sigma = 5.8e7 # conductivity in simens /m\n", - "delta = 0.1 # Skin depth penetration in mm\n", - "\n", - "print \"Example 7.8\"\n", - "f = 2/((delta*1e-3)**2*sigma*mu*2*math.pi) # Calculation of skin depth penetration\n", - "print \" Required frequency is %.2e Hz\"%(f)\n", - "print \" The incident electromagnetic wave is the radio part of spectrum.\"\n", - "# Answer in book is 3.36e5 Hz. Difference is due to approximation at intermediate stages\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.8\n", - " Required frequency is 4.37e+05 Hz\n", - " The incident electromagnetic wave is the radio part of spectrum.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.9 Page No240" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - " \n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "mu = mu_0 #Permeability of silver\n", - "sigma = 3e7 # conductivity in mhos/m\n", - "f = 1e10 # frequency in Hz\n", - "print \"Example 7.9\"\n", - "delta = math.sqrt(1/(math.pi*sigma*f*mu)) # Calculation of skin depth penetration\n", - "print \" Skin depth penetration is %f micrometre. \"%(delta*1e6)\n", - "# Answer in book is 0.93 micrometer\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.9\n", - " Skin depth penetration is 0.918881 micrometre. \n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.10 Page No241" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "p = 500 # power in watt\n", - "d = 1 # Dismath.tance from lamp in m\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "print \"Example 7.10\"\n", - "s = p/(4*math.pi*d**2)# Calculation of pointing vector\n", - "E_H_ratio = math.sqrt(mu_0/epsilon_0) # Calculation of ratio of Electric field and magnetic field\n", - "H = s/E_H_ratio # Calculation of Electric field \n", - "h = (H*100)/100 # rounding off for 2 decimal places\n", - "E= p/(4*math.pi*h) # Calculation of Electric field \n", - "print \" Average value of electric field at distance %d m is %f Volt/m \"%(d,E)\n", - "print \" Average value of magnetic field at distance %d m is %f Amp-turn/m \"%(d,h)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.10\n", - " Average value of electric field at distance 1 m is 376.734309 Volt/m \n", - " Average value of magnetic field at distance 1 m is 0.105615 Amp-turn/m \n" - ] - } - ], - "prompt_number": 18 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.11 Page No243" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "mu = mu_0 #Permeability of silver\n", - "sigma = 3.5e7 # conductivity in simens /m\n", - "delta = 0.03 # Skin depth penetration in mm\n", - "\n", - "print \"Example 7.11\"\n", - "\n", - "f = 2/((delta*1e-3)**2*sigma*mu*2*math.pi) # Calculation of skin depth penetration\n", - "print \" Required frequency is %d MHz.\"%(f/1e6)\n", - "print \" The incident electromagnetic wave is the radio part of spectrum\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.11\n", - " Required frequency is 8 MHz.\n", - " The incident electromagnetic wave is the radio part of spectrum\n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.12 Page No244" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "p = 3.8e26 # power radiated by moon in watt\n", - "d_sun = 1.44e11 # Dismath.tance between sun and earth in meter\n", - "d_moon = 3e8 #Dismath.tance between moon and earth in meter\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "print \"Example 7.12\"\n", - "s = p/(4*math.pi*d_sun**2)# Calculation of solar energy received during solar eclipse in watt /m**2\n", - "S = s*60/(4.2*1e4) # Unit conversion\n", - "\n", - "print \" Solar energy received during solar eclipse is %f Cal per min per m**2 \"%(S)\n", - "# Ansewr in book is 2.1 cal per min per m**2\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.12\n", - " Solar energy received during solar eclipse is 2.083295 Cal per min per m**2 \n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.13 Page No246" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "mu = mu_0 #Permeability of silver\n", - "sigma = 3.5e7 # conductivity in simens /m\n", - "lambda1 = 6328. # Wavelength in angstrom\n", - "c = 3e8# Speed of light in m/sec\n", - "\n", - "print \"Example 7.13\"\n", - "f = c/(lambda1*1e-10)\n", - "omega = 2*math.pi/f # Calculation of time period\n", - "f = c/(lambda1*1e-10) # Calculation of frequency in Hz\n", - "delta = math.sqrt(1/(math.pi*f*sigma*mu)) # Calculation of skin depth penetration\n", - "print \" Skin depth penetration is %f nm. \"%(delta*1e9)\n", - "# Answer in book is 3.9 mm, unit used in book is wrong\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.13\n", - " Skin depth penetration is 3.907138 nm. \n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch7_5.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch7_5.ipynb deleted file mode 100644 index 1b95e85e..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch7_5.ipynb +++ /dev/null @@ -1,496 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:5c15c522261c9a1964264eb127a07845e301723248156878641836ea3a61bbf4" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 7 : Maxwells Equations and Electromagnetic Waves" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.1 Page No220" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "p = 1000 # power in watt\n", - "d = 2 # Dismath.tance from lamp in m\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "print \"Example 7.1\"\n", - "s = p/(4*math.pi*d**2)# Calculation of pointing vector\n", - "E_H_ratio = math.sqrt(mu_0/epsilon_0) # Calculation of ratio of Electric field and magnetic field\n", - "E= math.sqrt(E_H_ratio*s) # Calculation of Electric field \n", - "print \" Average value of electric field at distance %d m is %f Volt/m \"%(d,E)\n", - "# Answer in book is 48.87 volt/m which is due to wrong calculation at intermediate steps\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.1\n", - " Average value of electric field at distance 2 m is 86.573038 Volt/m \n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.2 Page No222" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "p = 2 # power in cal/min/cm**2\n", - "\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "mu_0 = 4*math.pi*1e-7 # permeability of free space\n", - "print \"Example 7.2\"\n", - "s = p*4.2e4/60 # Calculation of pointing vector\n", - "E_H_ratio = math.sqrt(mu_0/epsilon_0) # Calculation of ratio of Electric field and magnetic field\n", - "E= math.sqrt(E_H_ratio*s) # Calculation of Electric field \n", - "H = s/E # Calculation of Electric field \n", - "\n", - "print \" Average value of electric field is %f Volt/m \"%(E*math.sqrt(2))\n", - "print \" Average value of magnetic field is %f Amp turn/m \"%(H*math.sqrt(2))\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.2\n", - " Average value of electric field is 1027.061861 Volt/m \n", - " Average value of magnetic field is 2.726223 Amp turn/m \n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.3 Page No225" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # permeability of free space\n", - "mu = mu_0 #permeability of silver\n", - "sigma = 3e7 # conductivity in mhos/m\n", - "f = 1e8 # frequency in Hz\n", - "print \"Example 7.3\"\n", - "omega = 2*math.pi/f # Calculation of time period\n", - "delta = math.sqrt(2/(omega*sigma*mu)) # Calculation of skin depth penetration\n", - "Delta = int(delta/100)*100 # Rounding off\n", - "print \" Skin depth penetration is %e cm. \"%(Delta*1e-6)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.3\n", - " Skin depth penetration is 9.000000e-04 cm. \n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.5 Page No229" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "k = 80 # relative Dielectric consmath.tant of sea water\n", - "epsilon_0 = 1/9e9 # Permittivity of free space\n", - "epsilon = 80*epsilon_0 # Permittivity of free space\n", - "sigma = 4.3 # conductivity in mho/m\n", - "delta = 10 # penetration depth in cm\n", - "mu_0 = 4*math.pi*1e-7 # permeability f free space\n", - "F = 1e8 # Given frequency in Hz\n", - "print \"Example 7.5\"\n", - "f = (1/(math.pi*mu_0*sigma))/(delta*1e-2)**2 # Calculation of frequency\n", - "f1= round(f/1e8)*1e8 # Rounding off\n", - "print \"Frequency required for penetration of depth %d cm is %e Hz\"%(delta,f1)\n", - "omega = 2*math.pi*F\n", - "x = 2*sigma/(epsilon*omega)\n", - "if x>1:\n", - " print \" Sea water is good conductor at frequency lesser than 1e8 Hz \"\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.5\n", - "Frequency required for penetration of depth 10 cm is 0.000000e+00 Hz\n", - " Sea water is good conductor at frequency lesser than 1e8 Hz \n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.7 Page No234" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "k = 12 # relative Dielectric consmath.tant of sea water\n", - "epsilon_0 = 1/9e9 # Permittivity of free space\n", - "sigma = 2 # conductivity in mho/cm\n", - "mu_0 = 4*math.pi*1e-7 # permeability f free space\n", - "f= 1e9 # Given frequency in Hz\n", - "F = 1e6 # Given frequency in Hz\n", - "print \"Example 7.7\"\n", - "delta = math.sqrt(2/(2*math.pi*F*mu_0*sigma*100)) # Calculation of frequency\n", - "print \" For %eHz frequency, Penetration depth is %f cm\"%(F,delta*100)\n", - "omega = 2*math.pi*f\n", - "x = 2*sigma*100/(k*epsilon_0*omega)\n", - "if x>1 :\n", - " print \" Silicon is good conductor at frequency lesser than 1e9 Hz \"\n", - "\n", - "# Answer in book is 3.6 cm\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.7\n", - " For 1.000000e+06Hz frequency, Penetration depth is 3.558813 cm\n", - " Silicon is good conductor at frequency lesser than 1e9 Hz \n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.8 Page No236" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # permeability of free space\n", - "mu = mu_0 #permeability of silver\n", - "sigma = 5.8e7 # conductivity in simens /m\n", - "delta = 0.1 # Skin depth penetration in mm\n", - "\n", - "print \"Example 7.8\"\n", - "f = 2/((delta*1e-3)**2*sigma*mu*2*math.pi) # Calculation of skin depth penetration\n", - "print \" Required frequency is %.2e Hz\"%(f)\n", - "print \" The incident electromagnetic wave is the radio part of spectrum.\"\n", - "# Answer in book is 3.36e5 Hz. Difference is due to approximation at intermediate stages\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.8\n", - " Required frequency is 4.37e+05 Hz\n", - " The incident electromagnetic wave is the radio part of spectrum.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.9 Page No240" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - " \n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "mu = mu_0 #Permeability of silver\n", - "sigma = 3e7 # conductivity in mhos/m\n", - "f = 1e10 # frequency in Hz\n", - "print \"Example 7.9\"\n", - "delta = math.sqrt(1/(math.pi*sigma*f*mu)) # Calculation of skin depth penetration\n", - "print \" Skin depth penetration is %f micrometre. \"%(delta*1e6)\n", - "# Answer in book is 0.93 micrometer\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.9\n", - " Skin depth penetration is 0.918881 micrometre. \n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.10 Page No241" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "p = 500 # power in watt\n", - "d = 1 # Dismath.tance from lamp in m\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "print \"Example 7.10\"\n", - "s = p/(4*math.pi*d**2)# Calculation of pointing vector\n", - "E_H_ratio = math.sqrt(mu_0/epsilon_0) # Calculation of ratio of Electric field and magnetic field\n", - "H = s/E_H_ratio # Calculation of Electric field \n", - "h = (H*100)/100 # rounding off for 2 decimal places\n", - "E= p/(4*math.pi*h) # Calculation of Electric field \n", - "print \" Average value of electric field at distance %d m is %f Volt/m \"%(d,E)\n", - "print \" Average value of magnetic field at distance %d m is %f Amp-turn/m \"%(d,h)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.10\n", - " Average value of electric field at distance 1 m is 376.734309 Volt/m \n", - " Average value of magnetic field at distance 1 m is 0.105615 Amp-turn/m \n" - ] - } - ], - "prompt_number": 18 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.11 Page No243" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "mu = mu_0 #Permeability of silver\n", - "sigma = 3.5e7 # conductivity in simens /m\n", - "delta = 0.03 # Skin depth penetration in mm\n", - "\n", - "print \"Example 7.11\"\n", - "\n", - "f = 2/((delta*1e-3)**2*sigma*mu*2*math.pi) # Calculation of skin depth penetration\n", - "print \" Required frequency is %d MHz.\"%(f/1e6)\n", - "print \" The incident electromagnetic wave is the radio part of spectrum\"\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.11\n", - " Required frequency is 8 MHz.\n", - " The incident electromagnetic wave is the radio part of spectrum\n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.12 Page No244" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - " \n", - "#Given that\n", - "p = 3.8e26 # power radiated by moon in watt\n", - "d_sun = 1.44e11 # Dismath.tance between sun and earth in meter\n", - "d_moon = 3e8 #Dismath.tance between moon and earth in meter\n", - "epsilon_0 = 8.854e-12 # Permittivity of free space\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "print \"Example 7.12\"\n", - "s = p/(4*math.pi*d_sun**2)# Calculation of solar energy received during solar eclipse in watt /m**2\n", - "S = s*60/(4.2*1e4) # Unit conversion\n", - "\n", - "print \" Solar energy received during solar eclipse is %f Cal per min per m**2 \"%(S)\n", - "# Ansewr in book is 2.1 cal per min per m**2\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.12\n", - " Solar energy received during solar eclipse is 2.083295 Cal per min per m**2 \n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.13 Page No246" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - " \n", - "#Given that\n", - "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n", - "mu = mu_0 #Permeability of silver\n", - "sigma = 3.5e7 # conductivity in simens /m\n", - "lambda1 = 6328. # Wavelength in angstrom\n", - "c = 3e8# Speed of light in m/sec\n", - "\n", - "print \"Example 7.13\"\n", - "f = c/(lambda1*1e-10)\n", - "omega = 2*math.pi/f # Calculation of time period\n", - "f = c/(lambda1*1e-10) # Calculation of frequency in Hz\n", - "delta = math.sqrt(1/(math.pi*f*sigma*mu)) # Calculation of skin depth penetration\n", - "print \" Skin depth penetration is %f nm. \"%(delta*1e9)\n", - "# Answer in book is 3.9 mm, unit used in book is wrong\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 7.13\n", - " Skin depth penetration is 3.907138 nm. \n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch8.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch8.ipynb deleted file mode 100755 index 31c79ed3..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch8.ipynb +++ /dev/null @@ -1,448 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:497ece365171c1444da23921d16f63e0c1b4e77fbb7335ff607952ae9c43252d" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 8 : Superconductivity" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.1 Page No251" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# Given that\n", - "H_c_0= 0.0306# Critical Field in tesla\n", - "T_c = 3.7 # Critical temperature in kelvin\n", - "T = 2 # Temperature in kelvin\n", - "print \"Example 8.1\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "H_c = H_c_0*(1-(T/T_c)**2) # Calculation of critical field\n", - "\n", - "print \"Magnetic Field at %d K is %f tesla.\"%(T,H_c)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.1\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Magnetic Field at 2 K is 0.021659 tesla.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.2 Page No255" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "H_c= 3.3e4 # # Magnetic field in A/m\n", - "T_c = 7.2 # Critical temperature in kelvin\n", - "T = 5 # Temperature in kelvin\n", - "print \"Example 8.2\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "H_c_0 = H_c*(1-(T/T_c)**2)**(-1) # Calculation of critical field\n", - "print \"Magnetic Field at %d K is %e A/m\"%(T,H_c_0)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.2\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Magnetic Field at 5 K is 6.373770e+04 A/m\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.3 Page No257" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "H_c_0= 1 # Let \n", - "H_c= 0.1 * H_c_0 # Magnetic field in A/m\n", - "T_c = 7.2 # Critical temperature in kelvin\n", - "\n", - "print \"Example 8.3\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T = T_c*math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n", - "\n", - "print \"Required temperature is %f K.\"%(T)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.3\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Required temperature is 6.830520 K.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.4 Page No257" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "# Given that\n", - "H_c_0= 0.0803# Critical Field in tesla\n", - "T_c = 7.2 # Critical temperature in kelvin\n", - "T = 4.2 # Temperature in kelvin\n", - "print \"Example 8.4\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "H_c = H_c_0*(1-(T/T_c)**2) # Calculation of critical field\n", - "\n", - "print \"Magnetic Field at %d K is %f tesla.\"%(T,H_c)\n", - "# Answer in book is 0.0548 tesla\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.4\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Magnetic Field at 4 K is 0.052976 tesla.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.5 Page No261" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "# Given that\n", - "H_c_0= 1.5e5# Critical field in A/m \n", - "H_c= 1.05e5 # Magnetic field in A/m\n", - "T_c = 9.2 # Critical temperature in kelvin\n", - "\n", - "print \"Example 8.5\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T = T_c*math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n", - "\n", - "print \"Required temperature is %f K.\"%(T)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.5\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Required temperature is 5.039048 K.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.6 Page No264" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "# Given that\n", - "H_c_0= 2e5# Critical field in A/m \n", - "H_c= 1e5 # Magnetic field in A/m\n", - "T_c = 8 # Critical temperature in kelvin\n", - "\n", - "print \"Example 8.6\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T = T_c/math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n", - "\n", - "print \"Required temperature is %f K.\"%(T)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.6\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Required temperature is 11.313708 K.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.7 Page No265" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "H_c_0= 8e5# Critical field in A/m \n", - "H_c= 4e4 # Magnetic field in A/m\n", - "T_c = 7.26 # Critical temperature in kelvin\n", - "\n", - "print \"Example 8.7\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T = T_c*math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n", - "\n", - "print \"Required temperature is %f K.\"%(T)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.7\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Required temperature is 7.076173 K.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.8 Page No268" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "T1 = 14 # Temp in K\n", - "T2 = 13 # Temp in K\n", - "T = 4.2 # Temp in K\n", - "Hc_T1 = 0.176 # Critical field at Temp T1\n", - "Hc_T2 = 0.528 # Critical field at Temp T2\n", - "\n", - "print \"Example 8.8\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T_c = math.sqrt((T1**2*(Hc_T2/Hc_T1)- T2**2) /(Hc_T2/Hc_T1 - 1)) # Calculation of transition temperature\n", - "t_c = (T_c*10)/10 # Rounding off two two decimal places\n", - "Hc_0 = Hc_T1/(1-(T1/t_c)**2) # Calculation of critical field\n", - "Hc_T = Hc_0*(1-(T/t_c)**2) # Calculation of critical field \n", - "\n", - "print \" Transition temperature is %f K.\"%(t_c)\n", - "print \"Critical field at %f K is %fT.\"%(T,Hc_0)\n", - "print \"Critical field at 0 K is %fT.\"%(Hc_T)\n", - "# Answer in book is 2.588 T for 0 K and 2.37 for 4.2 K\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.8\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - " Transition temperature is 14.474115 K.\n", - "Critical field at 4.200000 K is 2.731259T.\n", - "Critical field at 0 K is 2.501286T.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.9 Page No273" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "m_0 = 9.1e-31 # Mass of electron in kg\n", - "mu_0 = 1.256e-6# SI\n", - "e = 1.6e-19 # Charge on electron in coulomb\n", - "eta_s = 1e28 # superelectron density in no. per cube\n", - "T_1 = 0 # First temp in kelvin\n", - "T_2 = 1 # Second temp in kelvin\n", - "T_c = 3 # Critical temp in kelvin\n", - "\n", - "print \"Example 8.9\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tlambda_0 = math.sqrtm_0/mu_0*eta_s*e**2\"\n", - "lambda_0 = math.sqrt(m_0/(mu_0*eta_s*e**2))# Calculation of penetration depth at 0K\n", - "lambda_t = lambda_0/math.sqrt(1-(T_2/T_c)**4) # Calculation of penetration depth at 2K\n", - "\n", - "print \"Penetration depth at %d K is %d angestrom.\"%(T_1,lambda_0*1e10)\n", - "print \"Penetration depth at %d K is %f angestrom.\"%(T_2,lambda_t*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.9\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tlambda_0 = math.sqrtm_0/mu_0*eta_s*e**2\n", - "Penetration depth at 0 K is 531 angestrom.\n", - "Penetration depth at 1 K is 531.992971 angestrom.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.10 Page No277" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "T_1 = 3.5 # Temperature in kelvin\n", - "T_c = 4.153 # Critical temp in kelvin\n", - "lambda_t = 750 # Penetration depth at T_1 in angstrom\n", - "print \"Example 8.10\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used lambda_0 = lambda_t*math.sqrt1-T_1/T_c**4 \"\n", - "\n", - "lambda_0 = lambda_t*math.sqrt(1-(T_1/T_c)**4) # Calculation of penetration depth at 3.5K\n", - "print \" Penetration depth at 0 K is %f angstrom.\"%(lambda_0)\n", - "\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.10\n", - "Smath.radians(numpy.arcmath.tan(ard formula used lambda_0 = lambda_t*math.sqrt1-T_1/T_c**4 \n", - " Penetration depth at 0 K is 527.960928 angstrom.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch8_1.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch8_1.ipynb deleted file mode 100755 index 31c79ed3..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch8_1.ipynb +++ /dev/null @@ -1,448 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:497ece365171c1444da23921d16f63e0c1b4e77fbb7335ff607952ae9c43252d" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 8 : Superconductivity" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.1 Page No251" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# Given that\n", - "H_c_0= 0.0306# Critical Field in tesla\n", - "T_c = 3.7 # Critical temperature in kelvin\n", - "T = 2 # Temperature in kelvin\n", - "print \"Example 8.1\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "H_c = H_c_0*(1-(T/T_c)**2) # Calculation of critical field\n", - "\n", - "print \"Magnetic Field at %d K is %f tesla.\"%(T,H_c)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.1\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Magnetic Field at 2 K is 0.021659 tesla.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.2 Page No255" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "H_c= 3.3e4 # # Magnetic field in A/m\n", - "T_c = 7.2 # Critical temperature in kelvin\n", - "T = 5 # Temperature in kelvin\n", - "print \"Example 8.2\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "H_c_0 = H_c*(1-(T/T_c)**2)**(-1) # Calculation of critical field\n", - "print \"Magnetic Field at %d K is %e A/m\"%(T,H_c_0)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.2\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Magnetic Field at 5 K is 6.373770e+04 A/m\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.3 Page No257" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "H_c_0= 1 # Let \n", - "H_c= 0.1 * H_c_0 # Magnetic field in A/m\n", - "T_c = 7.2 # Critical temperature in kelvin\n", - "\n", - "print \"Example 8.3\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T = T_c*math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n", - "\n", - "print \"Required temperature is %f K.\"%(T)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.3\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Required temperature is 6.830520 K.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.4 Page No257" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "# Given that\n", - "H_c_0= 0.0803# Critical Field in tesla\n", - "T_c = 7.2 # Critical temperature in kelvin\n", - "T = 4.2 # Temperature in kelvin\n", - "print \"Example 8.4\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "H_c = H_c_0*(1-(T/T_c)**2) # Calculation of critical field\n", - "\n", - "print \"Magnetic Field at %d K is %f tesla.\"%(T,H_c)\n", - "# Answer in book is 0.0548 tesla\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.4\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Magnetic Field at 4 K is 0.052976 tesla.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.5 Page No261" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "# Given that\n", - "H_c_0= 1.5e5# Critical field in A/m \n", - "H_c= 1.05e5 # Magnetic field in A/m\n", - "T_c = 9.2 # Critical temperature in kelvin\n", - "\n", - "print \"Example 8.5\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T = T_c*math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n", - "\n", - "print \"Required temperature is %f K.\"%(T)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.5\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Required temperature is 5.039048 K.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.6 Page No264" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "# Given that\n", - "H_c_0= 2e5# Critical field in A/m \n", - "H_c= 1e5 # Magnetic field in A/m\n", - "T_c = 8 # Critical temperature in kelvin\n", - "\n", - "print \"Example 8.6\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T = T_c/math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n", - "\n", - "print \"Required temperature is %f K.\"%(T)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.6\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Required temperature is 11.313708 K.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.7 Page No265" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "H_c_0= 8e5# Critical field in A/m \n", - "H_c= 4e4 # Magnetic field in A/m\n", - "T_c = 7.26 # Critical temperature in kelvin\n", - "\n", - "print \"Example 8.7\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T = T_c*math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n", - "\n", - "print \"Required temperature is %f K.\"%(T)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.7\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Required temperature is 7.076173 K.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.8 Page No268" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "T1 = 14 # Temp in K\n", - "T2 = 13 # Temp in K\n", - "T = 4.2 # Temp in K\n", - "Hc_T1 = 0.176 # Critical field at Temp T1\n", - "Hc_T2 = 0.528 # Critical field at Temp T2\n", - "\n", - "print \"Example 8.8\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T_c = math.sqrt((T1**2*(Hc_T2/Hc_T1)- T2**2) /(Hc_T2/Hc_T1 - 1)) # Calculation of transition temperature\n", - "t_c = (T_c*10)/10 # Rounding off two two decimal places\n", - "Hc_0 = Hc_T1/(1-(T1/t_c)**2) # Calculation of critical field\n", - "Hc_T = Hc_0*(1-(T/t_c)**2) # Calculation of critical field \n", - "\n", - "print \" Transition temperature is %f K.\"%(t_c)\n", - "print \"Critical field at %f K is %fT.\"%(T,Hc_0)\n", - "print \"Critical field at 0 K is %fT.\"%(Hc_T)\n", - "# Answer in book is 2.588 T for 0 K and 2.37 for 4.2 K\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.8\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - " Transition temperature is 14.474115 K.\n", - "Critical field at 4.200000 K is 2.731259T.\n", - "Critical field at 0 K is 2.501286T.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.9 Page No273" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "m_0 = 9.1e-31 # Mass of electron in kg\n", - "mu_0 = 1.256e-6# SI\n", - "e = 1.6e-19 # Charge on electron in coulomb\n", - "eta_s = 1e28 # superelectron density in no. per cube\n", - "T_1 = 0 # First temp in kelvin\n", - "T_2 = 1 # Second temp in kelvin\n", - "T_c = 3 # Critical temp in kelvin\n", - "\n", - "print \"Example 8.9\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tlambda_0 = math.sqrtm_0/mu_0*eta_s*e**2\"\n", - "lambda_0 = math.sqrt(m_0/(mu_0*eta_s*e**2))# Calculation of penetration depth at 0K\n", - "lambda_t = lambda_0/math.sqrt(1-(T_2/T_c)**4) # Calculation of penetration depth at 2K\n", - "\n", - "print \"Penetration depth at %d K is %d angestrom.\"%(T_1,lambda_0*1e10)\n", - "print \"Penetration depth at %d K is %f angestrom.\"%(T_2,lambda_t*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.9\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tlambda_0 = math.sqrtm_0/mu_0*eta_s*e**2\n", - "Penetration depth at 0 K is 531 angestrom.\n", - "Penetration depth at 1 K is 531.992971 angestrom.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.10 Page No277" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "T_1 = 3.5 # Temperature in kelvin\n", - "T_c = 4.153 # Critical temp in kelvin\n", - "lambda_t = 750 # Penetration depth at T_1 in angstrom\n", - "print \"Example 8.10\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used lambda_0 = lambda_t*math.sqrt1-T_1/T_c**4 \"\n", - "\n", - "lambda_0 = lambda_t*math.sqrt(1-(T_1/T_c)**4) # Calculation of penetration depth at 3.5K\n", - "print \" Penetration depth at 0 K is %f angstrom.\"%(lambda_0)\n", - "\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.10\n", - "Smath.radians(numpy.arcmath.tan(ard formula used lambda_0 = lambda_t*math.sqrt1-T_1/T_c**4 \n", - " Penetration depth at 0 K is 527.960928 angstrom.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch8_2.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch8_2.ipynb deleted file mode 100644 index 31c79ed3..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch8_2.ipynb +++ /dev/null @@ -1,448 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:497ece365171c1444da23921d16f63e0c1b4e77fbb7335ff607952ae9c43252d" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 8 : Superconductivity" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.1 Page No251" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# Given that\n", - "H_c_0= 0.0306# Critical Field in tesla\n", - "T_c = 3.7 # Critical temperature in kelvin\n", - "T = 2 # Temperature in kelvin\n", - "print \"Example 8.1\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "H_c = H_c_0*(1-(T/T_c)**2) # Calculation of critical field\n", - "\n", - "print \"Magnetic Field at %d K is %f tesla.\"%(T,H_c)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.1\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Magnetic Field at 2 K is 0.021659 tesla.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.2 Page No255" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "H_c= 3.3e4 # # Magnetic field in A/m\n", - "T_c = 7.2 # Critical temperature in kelvin\n", - "T = 5 # Temperature in kelvin\n", - "print \"Example 8.2\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "H_c_0 = H_c*(1-(T/T_c)**2)**(-1) # Calculation of critical field\n", - "print \"Magnetic Field at %d K is %e A/m\"%(T,H_c_0)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.2\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Magnetic Field at 5 K is 6.373770e+04 A/m\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.3 Page No257" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "H_c_0= 1 # Let \n", - "H_c= 0.1 * H_c_0 # Magnetic field in A/m\n", - "T_c = 7.2 # Critical temperature in kelvin\n", - "\n", - "print \"Example 8.3\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T = T_c*math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n", - "\n", - "print \"Required temperature is %f K.\"%(T)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.3\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Required temperature is 6.830520 K.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.4 Page No257" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "# Given that\n", - "H_c_0= 0.0803# Critical Field in tesla\n", - "T_c = 7.2 # Critical temperature in kelvin\n", - "T = 4.2 # Temperature in kelvin\n", - "print \"Example 8.4\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "H_c = H_c_0*(1-(T/T_c)**2) # Calculation of critical field\n", - "\n", - "print \"Magnetic Field at %d K is %f tesla.\"%(T,H_c)\n", - "# Answer in book is 0.0548 tesla\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.4\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Magnetic Field at 4 K is 0.052976 tesla.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.5 Page No261" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "# Given that\n", - "H_c_0= 1.5e5# Critical field in A/m \n", - "H_c= 1.05e5 # Magnetic field in A/m\n", - "T_c = 9.2 # Critical temperature in kelvin\n", - "\n", - "print \"Example 8.5\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T = T_c*math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n", - "\n", - "print \"Required temperature is %f K.\"%(T)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.5\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Required temperature is 5.039048 K.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.6 Page No264" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "# Given that\n", - "H_c_0= 2e5# Critical field in A/m \n", - "H_c= 1e5 # Magnetic field in A/m\n", - "T_c = 8 # Critical temperature in kelvin\n", - "\n", - "print \"Example 8.6\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T = T_c/math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n", - "\n", - "print \"Required temperature is %f K.\"%(T)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.6\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Required temperature is 11.313708 K.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.7 Page No265" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "H_c_0= 8e5# Critical field in A/m \n", - "H_c= 4e4 # Magnetic field in A/m\n", - "T_c = 7.26 # Critical temperature in kelvin\n", - "\n", - "print \"Example 8.7\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T = T_c*math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n", - "\n", - "print \"Required temperature is %f K.\"%(T)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.7\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Required temperature is 7.076173 K.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.8 Page No268" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "T1 = 14 # Temp in K\n", - "T2 = 13 # Temp in K\n", - "T = 4.2 # Temp in K\n", - "Hc_T1 = 0.176 # Critical field at Temp T1\n", - "Hc_T2 = 0.528 # Critical field at Temp T2\n", - "\n", - "print \"Example 8.8\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T_c = math.sqrt((T1**2*(Hc_T2/Hc_T1)- T2**2) /(Hc_T2/Hc_T1 - 1)) # Calculation of transition temperature\n", - "t_c = (T_c*10)/10 # Rounding off two two decimal places\n", - "Hc_0 = Hc_T1/(1-(T1/t_c)**2) # Calculation of critical field\n", - "Hc_T = Hc_0*(1-(T/t_c)**2) # Calculation of critical field \n", - "\n", - "print \" Transition temperature is %f K.\"%(t_c)\n", - "print \"Critical field at %f K is %fT.\"%(T,Hc_0)\n", - "print \"Critical field at 0 K is %fT.\"%(Hc_T)\n", - "# Answer in book is 2.588 T for 0 K and 2.37 for 4.2 K\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.8\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - " Transition temperature is 14.474115 K.\n", - "Critical field at 4.200000 K is 2.731259T.\n", - "Critical field at 0 K is 2.501286T.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.9 Page No273" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "m_0 = 9.1e-31 # Mass of electron in kg\n", - "mu_0 = 1.256e-6# SI\n", - "e = 1.6e-19 # Charge on electron in coulomb\n", - "eta_s = 1e28 # superelectron density in no. per cube\n", - "T_1 = 0 # First temp in kelvin\n", - "T_2 = 1 # Second temp in kelvin\n", - "T_c = 3 # Critical temp in kelvin\n", - "\n", - "print \"Example 8.9\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tlambda_0 = math.sqrtm_0/mu_0*eta_s*e**2\"\n", - "lambda_0 = math.sqrt(m_0/(mu_0*eta_s*e**2))# Calculation of penetration depth at 0K\n", - "lambda_t = lambda_0/math.sqrt(1-(T_2/T_c)**4) # Calculation of penetration depth at 2K\n", - "\n", - "print \"Penetration depth at %d K is %d angestrom.\"%(T_1,lambda_0*1e10)\n", - "print \"Penetration depth at %d K is %f angestrom.\"%(T_2,lambda_t*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.9\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tlambda_0 = math.sqrtm_0/mu_0*eta_s*e**2\n", - "Penetration depth at 0 K is 531 angestrom.\n", - "Penetration depth at 1 K is 531.992971 angestrom.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.10 Page No277" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "T_1 = 3.5 # Temperature in kelvin\n", - "T_c = 4.153 # Critical temp in kelvin\n", - "lambda_t = 750 # Penetration depth at T_1 in angstrom\n", - "print \"Example 8.10\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used lambda_0 = lambda_t*math.sqrt1-T_1/T_c**4 \"\n", - "\n", - "lambda_0 = lambda_t*math.sqrt(1-(T_1/T_c)**4) # Calculation of penetration depth at 3.5K\n", - "print \" Penetration depth at 0 K is %f angstrom.\"%(lambda_0)\n", - "\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.10\n", - "Smath.radians(numpy.arcmath.tan(ard formula used lambda_0 = lambda_t*math.sqrt1-T_1/T_c**4 \n", - " Penetration depth at 0 K is 527.960928 angstrom.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch8_3.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch8_3.ipynb deleted file mode 100644 index 31c79ed3..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch8_3.ipynb +++ /dev/null @@ -1,448 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:497ece365171c1444da23921d16f63e0c1b4e77fbb7335ff607952ae9c43252d" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 8 : Superconductivity" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.1 Page No251" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# Given that\n", - "H_c_0= 0.0306# Critical Field in tesla\n", - "T_c = 3.7 # Critical temperature in kelvin\n", - "T = 2 # Temperature in kelvin\n", - "print \"Example 8.1\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "H_c = H_c_0*(1-(T/T_c)**2) # Calculation of critical field\n", - "\n", - "print \"Magnetic Field at %d K is %f tesla.\"%(T,H_c)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.1\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Magnetic Field at 2 K is 0.021659 tesla.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.2 Page No255" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "H_c= 3.3e4 # # Magnetic field in A/m\n", - "T_c = 7.2 # Critical temperature in kelvin\n", - "T = 5 # Temperature in kelvin\n", - "print \"Example 8.2\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "H_c_0 = H_c*(1-(T/T_c)**2)**(-1) # Calculation of critical field\n", - "print \"Magnetic Field at %d K is %e A/m\"%(T,H_c_0)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.2\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Magnetic Field at 5 K is 6.373770e+04 A/m\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.3 Page No257" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "H_c_0= 1 # Let \n", - "H_c= 0.1 * H_c_0 # Magnetic field in A/m\n", - "T_c = 7.2 # Critical temperature in kelvin\n", - "\n", - "print \"Example 8.3\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T = T_c*math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n", - "\n", - "print \"Required temperature is %f K.\"%(T)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.3\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Required temperature is 6.830520 K.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.4 Page No257" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "# Given that\n", - "H_c_0= 0.0803# Critical Field in tesla\n", - "T_c = 7.2 # Critical temperature in kelvin\n", - "T = 4.2 # Temperature in kelvin\n", - "print \"Example 8.4\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "H_c = H_c_0*(1-(T/T_c)**2) # Calculation of critical field\n", - "\n", - "print \"Magnetic Field at %d K is %f tesla.\"%(T,H_c)\n", - "# Answer in book is 0.0548 tesla\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.4\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Magnetic Field at 4 K is 0.052976 tesla.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.5 Page No261" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "# Given that\n", - "H_c_0= 1.5e5# Critical field in A/m \n", - "H_c= 1.05e5 # Magnetic field in A/m\n", - "T_c = 9.2 # Critical temperature in kelvin\n", - "\n", - "print \"Example 8.5\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T = T_c*math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n", - "\n", - "print \"Required temperature is %f K.\"%(T)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.5\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Required temperature is 5.039048 K.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.6 Page No264" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "# Given that\n", - "H_c_0= 2e5# Critical field in A/m \n", - "H_c= 1e5 # Magnetic field in A/m\n", - "T_c = 8 # Critical temperature in kelvin\n", - "\n", - "print \"Example 8.6\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T = T_c/math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n", - "\n", - "print \"Required temperature is %f K.\"%(T)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.6\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Required temperature is 11.313708 K.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.7 Page No265" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "H_c_0= 8e5# Critical field in A/m \n", - "H_c= 4e4 # Magnetic field in A/m\n", - "T_c = 7.26 # Critical temperature in kelvin\n", - "\n", - "print \"Example 8.7\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T = T_c*math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n", - "\n", - "print \"Required temperature is %f K.\"%(T)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.7\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Required temperature is 7.076173 K.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.8 Page No268" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "T1 = 14 # Temp in K\n", - "T2 = 13 # Temp in K\n", - "T = 4.2 # Temp in K\n", - "Hc_T1 = 0.176 # Critical field at Temp T1\n", - "Hc_T2 = 0.528 # Critical field at Temp T2\n", - "\n", - "print \"Example 8.8\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T_c = math.sqrt((T1**2*(Hc_T2/Hc_T1)- T2**2) /(Hc_T2/Hc_T1 - 1)) # Calculation of transition temperature\n", - "t_c = (T_c*10)/10 # Rounding off two two decimal places\n", - "Hc_0 = Hc_T1/(1-(T1/t_c)**2) # Calculation of critical field\n", - "Hc_T = Hc_0*(1-(T/t_c)**2) # Calculation of critical field \n", - "\n", - "print \" Transition temperature is %f K.\"%(t_c)\n", - "print \"Critical field at %f K is %fT.\"%(T,Hc_0)\n", - "print \"Critical field at 0 K is %fT.\"%(Hc_T)\n", - "# Answer in book is 2.588 T for 0 K and 2.37 for 4.2 K\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.8\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - " Transition temperature is 14.474115 K.\n", - "Critical field at 4.200000 K is 2.731259T.\n", - "Critical field at 0 K is 2.501286T.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.9 Page No273" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "m_0 = 9.1e-31 # Mass of electron in kg\n", - "mu_0 = 1.256e-6# SI\n", - "e = 1.6e-19 # Charge on electron in coulomb\n", - "eta_s = 1e28 # superelectron density in no. per cube\n", - "T_1 = 0 # First temp in kelvin\n", - "T_2 = 1 # Second temp in kelvin\n", - "T_c = 3 # Critical temp in kelvin\n", - "\n", - "print \"Example 8.9\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tlambda_0 = math.sqrtm_0/mu_0*eta_s*e**2\"\n", - "lambda_0 = math.sqrt(m_0/(mu_0*eta_s*e**2))# Calculation of penetration depth at 0K\n", - "lambda_t = lambda_0/math.sqrt(1-(T_2/T_c)**4) # Calculation of penetration depth at 2K\n", - "\n", - "print \"Penetration depth at %d K is %d angestrom.\"%(T_1,lambda_0*1e10)\n", - "print \"Penetration depth at %d K is %f angestrom.\"%(T_2,lambda_t*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.9\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tlambda_0 = math.sqrtm_0/mu_0*eta_s*e**2\n", - "Penetration depth at 0 K is 531 angestrom.\n", - "Penetration depth at 1 K is 531.992971 angestrom.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.10 Page No277" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "T_1 = 3.5 # Temperature in kelvin\n", - "T_c = 4.153 # Critical temp in kelvin\n", - "lambda_t = 750 # Penetration depth at T_1 in angstrom\n", - "print \"Example 8.10\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used lambda_0 = lambda_t*math.sqrt1-T_1/T_c**4 \"\n", - "\n", - "lambda_0 = lambda_t*math.sqrt(1-(T_1/T_c)**4) # Calculation of penetration depth at 3.5K\n", - "print \" Penetration depth at 0 K is %f angstrom.\"%(lambda_0)\n", - "\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.10\n", - "Smath.radians(numpy.arcmath.tan(ard formula used lambda_0 = lambda_t*math.sqrt1-T_1/T_c**4 \n", - " Penetration depth at 0 K is 527.960928 angstrom.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch8_4.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch8_4.ipynb deleted file mode 100644 index 31c79ed3..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch8_4.ipynb +++ /dev/null @@ -1,448 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:497ece365171c1444da23921d16f63e0c1b4e77fbb7335ff607952ae9c43252d" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 8 : Superconductivity" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.1 Page No251" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# Given that\n", - "H_c_0= 0.0306# Critical Field in tesla\n", - "T_c = 3.7 # Critical temperature in kelvin\n", - "T = 2 # Temperature in kelvin\n", - "print \"Example 8.1\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "H_c = H_c_0*(1-(T/T_c)**2) # Calculation of critical field\n", - "\n", - "print \"Magnetic Field at %d K is %f tesla.\"%(T,H_c)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.1\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Magnetic Field at 2 K is 0.021659 tesla.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.2 Page No255" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "H_c= 3.3e4 # # Magnetic field in A/m\n", - "T_c = 7.2 # Critical temperature in kelvin\n", - "T = 5 # Temperature in kelvin\n", - "print \"Example 8.2\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "H_c_0 = H_c*(1-(T/T_c)**2)**(-1) # Calculation of critical field\n", - "print \"Magnetic Field at %d K is %e A/m\"%(T,H_c_0)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.2\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Magnetic Field at 5 K is 6.373770e+04 A/m\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.3 Page No257" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "H_c_0= 1 # Let \n", - "H_c= 0.1 * H_c_0 # Magnetic field in A/m\n", - "T_c = 7.2 # Critical temperature in kelvin\n", - "\n", - "print \"Example 8.3\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T = T_c*math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n", - "\n", - "print \"Required temperature is %f K.\"%(T)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.3\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Required temperature is 6.830520 K.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.4 Page No257" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "# Given that\n", - "H_c_0= 0.0803# Critical Field in tesla\n", - "T_c = 7.2 # Critical temperature in kelvin\n", - "T = 4.2 # Temperature in kelvin\n", - "print \"Example 8.4\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "H_c = H_c_0*(1-(T/T_c)**2) # Calculation of critical field\n", - "\n", - "print \"Magnetic Field at %d K is %f tesla.\"%(T,H_c)\n", - "# Answer in book is 0.0548 tesla\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.4\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Magnetic Field at 4 K is 0.052976 tesla.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.5 Page No261" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "# Given that\n", - "H_c_0= 1.5e5# Critical field in A/m \n", - "H_c= 1.05e5 # Magnetic field in A/m\n", - "T_c = 9.2 # Critical temperature in kelvin\n", - "\n", - "print \"Example 8.5\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T = T_c*math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n", - "\n", - "print \"Required temperature is %f K.\"%(T)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.5\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Required temperature is 5.039048 K.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.6 Page No264" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "# Given that\n", - "H_c_0= 2e5# Critical field in A/m \n", - "H_c= 1e5 # Magnetic field in A/m\n", - "T_c = 8 # Critical temperature in kelvin\n", - "\n", - "print \"Example 8.6\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T = T_c/math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n", - "\n", - "print \"Required temperature is %f K.\"%(T)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.6\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Required temperature is 11.313708 K.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.7 Page No265" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "H_c_0= 8e5# Critical field in A/m \n", - "H_c= 4e4 # Magnetic field in A/m\n", - "T_c = 7.26 # Critical temperature in kelvin\n", - "\n", - "print \"Example 8.7\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T = T_c*math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n", - "\n", - "print \"Required temperature is %f K.\"%(T)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.7\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Required temperature is 7.076173 K.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.8 Page No268" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "T1 = 14 # Temp in K\n", - "T2 = 13 # Temp in K\n", - "T = 4.2 # Temp in K\n", - "Hc_T1 = 0.176 # Critical field at Temp T1\n", - "Hc_T2 = 0.528 # Critical field at Temp T2\n", - "\n", - "print \"Example 8.8\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T_c = math.sqrt((T1**2*(Hc_T2/Hc_T1)- T2**2) /(Hc_T2/Hc_T1 - 1)) # Calculation of transition temperature\n", - "t_c = (T_c*10)/10 # Rounding off two two decimal places\n", - "Hc_0 = Hc_T1/(1-(T1/t_c)**2) # Calculation of critical field\n", - "Hc_T = Hc_0*(1-(T/t_c)**2) # Calculation of critical field \n", - "\n", - "print \" Transition temperature is %f K.\"%(t_c)\n", - "print \"Critical field at %f K is %fT.\"%(T,Hc_0)\n", - "print \"Critical field at 0 K is %fT.\"%(Hc_T)\n", - "# Answer in book is 2.588 T for 0 K and 2.37 for 4.2 K\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.8\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - " Transition temperature is 14.474115 K.\n", - "Critical field at 4.200000 K is 2.731259T.\n", - "Critical field at 0 K is 2.501286T.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.9 Page No273" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "m_0 = 9.1e-31 # Mass of electron in kg\n", - "mu_0 = 1.256e-6# SI\n", - "e = 1.6e-19 # Charge on electron in coulomb\n", - "eta_s = 1e28 # superelectron density in no. per cube\n", - "T_1 = 0 # First temp in kelvin\n", - "T_2 = 1 # Second temp in kelvin\n", - "T_c = 3 # Critical temp in kelvin\n", - "\n", - "print \"Example 8.9\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tlambda_0 = math.sqrtm_0/mu_0*eta_s*e**2\"\n", - "lambda_0 = math.sqrt(m_0/(mu_0*eta_s*e**2))# Calculation of penetration depth at 0K\n", - "lambda_t = lambda_0/math.sqrt(1-(T_2/T_c)**4) # Calculation of penetration depth at 2K\n", - "\n", - "print \"Penetration depth at %d K is %d angestrom.\"%(T_1,lambda_0*1e10)\n", - "print \"Penetration depth at %d K is %f angestrom.\"%(T_2,lambda_t*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.9\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tlambda_0 = math.sqrtm_0/mu_0*eta_s*e**2\n", - "Penetration depth at 0 K is 531 angestrom.\n", - "Penetration depth at 1 K is 531.992971 angestrom.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.10 Page No277" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "T_1 = 3.5 # Temperature in kelvin\n", - "T_c = 4.153 # Critical temp in kelvin\n", - "lambda_t = 750 # Penetration depth at T_1 in angstrom\n", - "print \"Example 8.10\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used lambda_0 = lambda_t*math.sqrt1-T_1/T_c**4 \"\n", - "\n", - "lambda_0 = lambda_t*math.sqrt(1-(T_1/T_c)**4) # Calculation of penetration depth at 3.5K\n", - "print \" Penetration depth at 0 K is %f angstrom.\"%(lambda_0)\n", - "\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.10\n", - "Smath.radians(numpy.arcmath.tan(ard formula used lambda_0 = lambda_t*math.sqrt1-T_1/T_c**4 \n", - " Penetration depth at 0 K is 527.960928 angstrom.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch8_5.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch8_5.ipynb deleted file mode 100644 index 31c79ed3..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/ch8_5.ipynb +++ /dev/null @@ -1,448 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:497ece365171c1444da23921d16f63e0c1b4e77fbb7335ff607952ae9c43252d" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 8 : Superconductivity" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.1 Page No251" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "# Given that\n", - "H_c_0= 0.0306# Critical Field in tesla\n", - "T_c = 3.7 # Critical temperature in kelvin\n", - "T = 2 # Temperature in kelvin\n", - "print \"Example 8.1\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "H_c = H_c_0*(1-(T/T_c)**2) # Calculation of critical field\n", - "\n", - "print \"Magnetic Field at %d K is %f tesla.\"%(T,H_c)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.1\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Magnetic Field at 2 K is 0.021659 tesla.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.2 Page No255" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "H_c= 3.3e4 # # Magnetic field in A/m\n", - "T_c = 7.2 # Critical temperature in kelvin\n", - "T = 5 # Temperature in kelvin\n", - "print \"Example 8.2\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "H_c_0 = H_c*(1-(T/T_c)**2)**(-1) # Calculation of critical field\n", - "print \"Magnetic Field at %d K is %e A/m\"%(T,H_c_0)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.2\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Magnetic Field at 5 K is 6.373770e+04 A/m\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.3 Page No257" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "H_c_0= 1 # Let \n", - "H_c= 0.1 * H_c_0 # Magnetic field in A/m\n", - "T_c = 7.2 # Critical temperature in kelvin\n", - "\n", - "print \"Example 8.3\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T = T_c*math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n", - "\n", - "print \"Required temperature is %f K.\"%(T)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.3\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Required temperature is 6.830520 K.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.4 Page No257" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "# Given that\n", - "H_c_0= 0.0803# Critical Field in tesla\n", - "T_c = 7.2 # Critical temperature in kelvin\n", - "T = 4.2 # Temperature in kelvin\n", - "print \"Example 8.4\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "H_c = H_c_0*(1-(T/T_c)**2) # Calculation of critical field\n", - "\n", - "print \"Magnetic Field at %d K is %f tesla.\"%(T,H_c)\n", - "# Answer in book is 0.0548 tesla\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.4\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Magnetic Field at 4 K is 0.052976 tesla.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.5 Page No261" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "# Given that\n", - "H_c_0= 1.5e5# Critical field in A/m \n", - "H_c= 1.05e5 # Magnetic field in A/m\n", - "T_c = 9.2 # Critical temperature in kelvin\n", - "\n", - "print \"Example 8.5\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T = T_c*math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n", - "\n", - "print \"Required temperature is %f K.\"%(T)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.5\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Required temperature is 5.039048 K.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.6 Page No264" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "# Given that\n", - "H_c_0= 2e5# Critical field in A/m \n", - "H_c= 1e5 # Magnetic field in A/m\n", - "T_c = 8 # Critical temperature in kelvin\n", - "\n", - "print \"Example 8.6\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T = T_c/math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n", - "\n", - "print \"Required temperature is %f K.\"%(T)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.6\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Required temperature is 11.313708 K.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.7 Page No265" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "H_c_0= 8e5# Critical field in A/m \n", - "H_c= 4e4 # Magnetic field in A/m\n", - "T_c = 7.26 # Critical temperature in kelvin\n", - "\n", - "print \"Example 8.7\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T = T_c*math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n", - "\n", - "print \"Required temperature is %f K.\"%(T)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.7\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - "Required temperature is 7.076173 K.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.8 Page No268" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "T1 = 14 # Temp in K\n", - "T2 = 13 # Temp in K\n", - "T = 4.2 # Temp in K\n", - "Hc_T1 = 0.176 # Critical field at Temp T1\n", - "Hc_T2 = 0.528 # Critical field at Temp T2\n", - "\n", - "print \"Example 8.8\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tH_c = H_c_0*1-T/T_c**2 \"\n", - "T_c = math.sqrt((T1**2*(Hc_T2/Hc_T1)- T2**2) /(Hc_T2/Hc_T1 - 1)) # Calculation of transition temperature\n", - "t_c = (T_c*10)/10 # Rounding off two two decimal places\n", - "Hc_0 = Hc_T1/(1-(T1/t_c)**2) # Calculation of critical field\n", - "Hc_T = Hc_0*(1-(T/t_c)**2) # Calculation of critical field \n", - "\n", - "print \" Transition temperature is %f K.\"%(t_c)\n", - "print \"Critical field at %f K is %fT.\"%(T,Hc_0)\n", - "print \"Critical field at 0 K is %fT.\"%(Hc_T)\n", - "# Answer in book is 2.588 T for 0 K and 2.37 for 4.2 K\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.8\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tH_c = H_c_0*1-T/T_c**2 \n", - " Transition temperature is 14.474115 K.\n", - "Critical field at 4.200000 K is 2.731259T.\n", - "Critical field at 0 K is 2.501286T.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.9 Page No273" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "m_0 = 9.1e-31 # Mass of electron in kg\n", - "mu_0 = 1.256e-6# SI\n", - "e = 1.6e-19 # Charge on electron in coulomb\n", - "eta_s = 1e28 # superelectron density in no. per cube\n", - "T_1 = 0 # First temp in kelvin\n", - "T_2 = 1 # Second temp in kelvin\n", - "T_c = 3 # Critical temp in kelvin\n", - "\n", - "print \"Example 8.9\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used \\tlambda_0 = math.sqrtm_0/mu_0*eta_s*e**2\"\n", - "lambda_0 = math.sqrt(m_0/(mu_0*eta_s*e**2))# Calculation of penetration depth at 0K\n", - "lambda_t = lambda_0/math.sqrt(1-(T_2/T_c)**4) # Calculation of penetration depth at 2K\n", - "\n", - "print \"Penetration depth at %d K is %d angestrom.\"%(T_1,lambda_0*1e10)\n", - "print \"Penetration depth at %d K is %f angestrom.\"%(T_2,lambda_t*1e10)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.9\n", - "Smath.radians(numpy.arcmath.tan(ard formula used \tlambda_0 = math.sqrtm_0/mu_0*eta_s*e**2\n", - "Penetration depth at 0 K is 531 angestrom.\n", - "Penetration depth at 1 K is 531.992971 angestrom.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.10 Page No277" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# Given that\n", - "T_1 = 3.5 # Temperature in kelvin\n", - "T_c = 4.153 # Critical temp in kelvin\n", - "lambda_t = 750 # Penetration depth at T_1 in angstrom\n", - "print \"Example 8.10\"\n", - "print \"Smath.radians(numpy.arcmath.tan(ard formula used lambda_0 = lambda_t*math.sqrt1-T_1/T_c**4 \"\n", - "\n", - "lambda_0 = lambda_t*math.sqrt(1-(T_1/T_c)**4) # Calculation of penetration depth at 3.5K\n", - "print \" Penetration depth at 0 K is %f angstrom.\"%(lambda_0)\n", - "\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Example 8.10\n", - "Smath.radians(numpy.arcmath.tan(ard formula used lambda_0 = lambda_t*math.sqrt1-T_1/T_c**4 \n", - " Penetration depth at 0 K is 527.960928 angstrom.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter1_1.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter1_1.ipynb deleted file mode 100755 index 9eda293b..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter1_1.ipynb +++ /dev/null @@ -1,1147 +0,0 @@ -{
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "\n",
- "# chapter1: De Broglie Matter Waves"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "##example 1.1;page no:10"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 3,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.1,page no:10\n",
- "\n",
- " de Broglie wavelength of earth in metres is= 3.68e-63\n"
- ]
- }
- ],
- "source": [
- "# cal of de brogle wavelength of earth\n",
- "#intiation of all variables \n",
- "#given that\n",
- "M = 6.*10**24 # Mass of earth in Kg\n",
- "v = 3.*10**4 # Orbital velocity of earth in m/s\n",
- "h = 6.625*10**-34 # Plank constant\n",
- "print(\"Example 1.1,page no:10\")\n",
- "lamda=h/(M*v) \n",
- "print(\"\\n de Broglie wavelength of earth in metres is=\"),round(lamda,65)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.2;page no:10"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 4,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.2,page no:11\n",
- "\n",
- " de Broglie wavelength of body in metres is= 6.625e-35\n"
- ]
- }
- ],
- "source": [
- "#cal of de Broglie wavelength of body\n",
- "#intiation of all variables\n",
- "#given that\n",
- "M = 1 # Mass of object in Kg\n",
- "v = 10 # velocity of object in m/s\n",
- "h = 6.625*10**-34 # Plank constant\n",
- "print(\"Example 1.2,page no:11\");\n",
- "lamda=h/(M*v)#calculation of de Broglie wavelength\n",
- "print(\"\\n de Broglie wavelength of body in metres is=\"),round(lamda,38)\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "##example 1.3;page no:11"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 5,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.3,page no:11\n",
- "\n",
- " de Broglie wavelength of body in metres is= 6.625e-09\n"
- ]
- }
- ],
- "source": [
- "#cal of de brogle wavelength of body\n",
- "#intiation of all variables \n",
- "# Given that\n",
- "m = 1e-30 # Mass of any object in Kg\n",
- "v = 1e5 # velocity of object in m/s\n",
- "h = 6.625e-34 # Plank constant\n",
- "print(\"Example 1.3,page no:11\")\n",
- "lamda=h/(m*v) # calculation of de Broglie wavelength in metres\n",
- "print(\"\\n de Broglie wavelength of body in metres is=\"),round(lamda,12)\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "##example 1.4;page no:15"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 6,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.4,page no:15\n",
- "velocity of electron in m/s: 1000.0\n",
- "momentum of electron in Kgm/s: 9.1e-28\n",
- "de Broglie wavelength of electron is: 7.27e-07\n",
- "Note:The value given in the book for lamda is wrong hence corrected above\n"
- ]
- }
- ],
- "source": [
- "#cal of velocity,momenteum and wave lenght of electron\n",
- "#intiation of all variables \n",
- "# Given that\n",
- "import math\n",
- "KE = 4.55e-25 # Kinetic energy of an electron in Joule\n",
- "m = 9.1e-31 # Mass of any object in Kg\n",
- "h = 6.62e-34 # Plank constant\n",
- "print(\"Example 1.4,page no:15\")\n",
- "v = math.sqrt(2*KE/m) # Calculation of velocity of moving electron\n",
- "p = m*v #Calculation of momentum of moving electron\n",
- "lamda= h/p # calculation of de Broglie wavelength\n",
- "print(\"velocity of electron in m/s:\"),round(v)\n",
- "print(\"momentum of electron in Kgm/s:\"),round(p,29)\n",
- "print(\"de Broglie wavelength of electron is:\"),round(lamda,9)\n",
- "print(\"Note:The value given in the book for lamda is wrong hence corrected above\")\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.5;page no:16"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 7,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.5,page no:16\n",
- "de Broglie wavelength of proton is: 2.645e-14\n"
- ]
- }
- ],
- "source": [
- "#cal of de brogle wavelength of proton\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "c = 3e8 # speed of light in m/s\n",
- "v = c/20 # Speed of proton in m/s\n",
- "m = 1.67e-27 # Mass of proton in Kg\n",
- "h = 6.625e-34 # Plank constant\n",
- "print(\"Example 1.5,page no:16\")\n",
- "lamda= h/(m*v) # calculation of de Broglie wavelength\n",
- "print(\"de Broglie wavelength of proton is:\"),round(lamda,17)\n",
- "# Answer in book is 6.645e-14m which is a calculation mistake\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.6;page no:16"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 1,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.6,page no:16\n",
- "\n",
- " de Broglie wavelength of neutron in angstrom= 7.99e-05\n"
- ]
- }
- ],
- "source": [
- "#cal of de brogle wavelength of neutron\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "e = 12.8 # Energy of neutron in MeV\n",
- "c = 3.e8 # speed of light in m/s\n",
- "m = 1.675e-27 # Mass of neutron in Kg\n",
- "h = 6.62e-34 # Plank constant\n",
- "print(\"Example 1.6,page no:16\")\n",
- "rest_e = m*c**2/(1e6*1.6e-19)# rest mass energy of neutron in MeV\n",
- "if e/rest_e < 0.015:\n",
- "\tE = e\n",
- "else:\n",
- "\tE = rest_e +e\n",
- "lamda = h/(math.sqrt(2*m*e*1e6*1.6e-19)) # calculation of de Broglie wavelength\n",
- "print(\"\\n de Broglie wavelength of neutron in angstrom=\"),round(lamda*1e10,7)\n",
- "# Answer in book is 8.04e-5 angstrom which is misprinted\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.7;page no:17"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 7,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.7,page no:17\n",
- "\n",
- " de Broglie wavelength of neutron in angstrom= 1.734\n"
- ]
- }
- ],
- "source": [
- "#cal of de brogle wavelength of neutron\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "import math\n",
- "e = 1.602e-19 # charge on electron in coulomb\n",
- "V = 50. # Applied voltage in volts\n",
- "m = 9.1e-31 # Mass of electron in Kg\n",
- "h = 6.62e-34 # Plank constant\n",
- "print(\"Example 1.7,page no:17\")\n",
- "lamda= h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength\n",
- "print(\"\\n de Broglie wavelength of neutron in angstrom=\"),round(lamda*1e10,3)\n",
- "\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.9;page no:18"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 8,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.9,page no:18\n",
- "de Broglie wavelength associated with the electron in angstrom= 1.67\n"
- ]
- }
- ],
- "source": [
- "#cal of de brogle wavelength associated with the electron\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "import math\n",
- "e = 1.6e-19 # charge on electron in coulomb\n",
- "V = 54 # Applied voltage in volts\n",
- "m = 9.1e-31 # Mass of electron in Kg\n",
- "h = 6.63e-34 # Plank constant\n",
- "print(\"Example 1.9,page no:18\")\n",
- "lamda = h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength\n",
- "print(\"de Broglie wavelength associated with the electron in angstrom=\"),round(lamda*1e10,2)\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.10;page no:19"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 9,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.10,page no:19\n",
- "velocity of electron in m/s: 59299945.33\n",
- "momentum of electron in Kgm/s: 5.3963e-23\n",
- "de Broglie wavelength of electron in angstrom= 0.123\n"
- ]
- }
- ],
- "source": [
- "#cal of velocity of electron,momentum of electron,de Broglie wavelength of electron\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "import math\n",
- "E = 10. # Energy of electron in KeV\n",
- "me = 9.1e-31 # Mass of electron in Kg\n",
- "h = 6.63e-34 # Plank constant\n",
- "print(\"Example 1.10,page no:19\")\n",
- "v = math.sqrt(2*E*1.6e-16/me) # Calculation of velocity of moving electron\n",
- "p = me*v #Calculation of momentum of moving electron\n",
- "lamda = h/p # calculation of de Broglie wavelength\n",
- "print(\"velocity of electron in m/s:\"),round(v,2)\n",
- "print(\"momentum of electron in Kgm/s:\"),round(p,27)\n",
- "print(\"de Broglie wavelength of electron in angstrom=\"),round(lamda*1e10,3)\n",
- "# Answers in book are v = 5.93e6 m/s, p = 5.397e-24 kgm/s, lambda = 1.23 angstrom\n",
- "# Which is due to wrong calculation"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.11;page no:20"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 10,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.11,page no:20\n",
- " velocity of neutron in m/s: 3964.072\n",
- " Kinetic energy of neutron in eV= 0.082\n"
- ]
- }
- ],
- "source": [
- "#cal of velocity and kinetic energy of neutron\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "lamda= 1 # de Broglie wavelength of neutron in angstrom\n",
- "m = 1.67e-27 # Mass of electron in Kg\n",
- "h = 6.62e-34 # Plank constant\n",
- "print(\"Example 1.11,page no:20\")\n",
- "v = h/(m*lamda*1e-10) # Calculation of velocity of moving neutron\n",
- "print(\" velocity of neutron in m/s:\"),round(v,3)\n",
- "E = 1./2.*m*v**2 # Calculation of kinetic energy of moving neutron\n",
- "print(\" Kinetic energy of neutron in eV=\"),round(E/1.6e-19,3)\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.12;page no:20"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 12,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.12,page no:20\n",
- "Wavelength of electron in metres= 2.74e-11\n"
- ]
- }
- ],
- "source": [
- "#cal of de brogle wavelength of electron\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "E = 2 # Energy of accelerated electron in KeV\n",
- "m = 9.1e-31 # Mass of electron in Kg\n",
- "h = 6.62e-34 # Plank constant\n",
- "print(\"Example 1.12,page no:20\")\n",
- "lamda = h/math.sqrt(2*m*E*1e3*1.6e-19) # Calculation of velocity of moving electron\n",
- "print(\"Wavelength of electron in metres=\"),round(lamda,13)\n",
- "# Answer in book is 2.74e-12m\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.13;page no:21"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 13,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.13,page no:21\n",
- "Wavelength of matter wave in angstrom= 1.48e-05\n"
- ]
- }
- ],
- "source": [
- "#cal of de brogle wavelength of matter wave\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "v = 2e8 # speed of moving proton in m/s\n",
- "c = 3e8 # speed of light in m/s\n",
- "m = 1.67e-27 # Mass of proton in Kg\n",
- "h = 6.62e-34 # Plank constant\n",
- "print(\"Example 1.13,page no:21\")\n",
- "lamda = h/(m*v/math.sqrt(1-(v/c)**2)) # Calculation of velocity of moving electron\n",
- "print(\"Wavelength of matter wave in angstrom=\"),round(lamda*1e10,7)\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.14;page no:22"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 31,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.14,page no:22\n",
- "Momentum of photon in Kgm/s while Momentum of electron in Kgm/s which are equal: 6.63e-24 6.63e-24\n",
- "Total Energy of photon in joule while Total Energy of electron in MeV: 1.989e-15 2.42e-17\n",
- "Ratio of kinetic energies in: 0.0121\n"
- ]
- }
- ],
- "source": [
- "#cal of momentum,total energy and ratio of kinetic energy of photon\n",
- "#intiation of all variables \n",
- "#given that\n",
- "lamda = 1# wavelength in m/s\n",
- "m_e = 9.1e-31 # Mass of electron in Kg\n",
- "m_p = 1.67e-27 # Mass of proton in kg\n",
- "c = 3e8 # speed of light in m/s\n",
- "h = 6.63e-34 # Plank constant\n",
- "print(\"Example 1.14,page no:22\")\n",
- "p_e = h/(lamda*1e-10) # Momentum of electron\n",
- "p_p = h/(lamda*1e-10) # Momentum of photon\n",
- "print(\"Momentum of photon in Kgm/s while Momentum of electron in Kgm/s which are equal:\"),round(p_p,26),round(p_e,26)\n",
- "E_e = p_e**2/(2*m_e) +m_e*c**2 # Total energy of electron\n",
- "E_e1=(2.42*10**-17)+(m_e*c**2/1.6*10**-19)\n",
- "E_p = h*c/(lamda*1e-10) # Total energy of photon\n",
- "print(\"Total Energy of photon in joule while Total Energy of electron in MeV:\"),round(E_p,18),E_e1\n",
- "K_e = p_e**2/(2*m_e) # Kinetic energy of electron \n",
- "K_p = h*c/(lamda*1e-10)# Kinetic energy of photon\n",
- "r_K = K_e/K_p # Ratio of kinetic energies\n",
- "print(\"Ratio of kinetic energies in:\"),round(r_K,4)\n",
- "\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.15;page no:24"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 2,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.15,page no:24\n",
- "de Broglie wavelength of neutron in angstrom: 0.0573\n"
- ]
- }
- ],
- "source": [
- "#cal of de brogle wavelength of neutron\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "e = 25 # Energy of neutron in eV\n",
- "c = 3e8 # speed of light in m/s\n",
- "m = 1.67e-27 # Mass of neutron in Kg\n",
- "h = 6.62e-34 # Plank constant\n",
- "print(\"Example 1.15,page no:24\")\n",
- "rest_e = m*c**2/(1e6*1.6e-19)# rest mass energy of neutron in MeV\n",
- "if e/rest_e < 0.015:\n",
- " E = e;\n",
- "else:\n",
- "\tE = rest_e +e;\n",
- "lamda = h/(math.sqrt(2*m*e*1.6e-19)) # calculation of de Broglie wavelength\n",
- "print(\"de Broglie wavelength of neutron in angstrom:\"),round(lamda*1e10,4)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.16;page no:24"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 14,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.16,page no:24\n",
- "de Broglie wavelength of neutron in angstrom: 0.00717\n"
- ]
- }
- ],
- "source": [
- "#cal of de brogle wavelength of neutron \n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "e = 2*1.6e-19 # charge on alpha particle in coulomb\n",
- "V = 200 # Applied voltage in volts\n",
- "m = 4*1.67e-27 # Mass of alpha particle in Kg\n",
- "h = 6.63e-34 # Plank constant\n",
- "print(\"Example 1.16,page no:24\")\n",
- "lamda=h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength\n",
- "print(\"de Broglie wavelength of neutron in angstrom:\"),round(lamda*1e10,5)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.17;page no:25"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 18,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.17,page no:25\n",
- "de Broglie wavelength of ball in angstrom: 6.62e-26\n",
- "de Broglie wavelength of electron in angstrom: 7.27\n"
- ]
- }
- ],
- "source": [
- "#cal of de brogle wavelength of ball and electron\n",
- "#intiation of all variables \n",
- "#given that\n",
- "M = 20 # Mass of ball in Kg\n",
- "V = 5 # velocity of of ball in m/s\n",
- "m = 9.1e-31 #Mass of electron in Kg\n",
- "v = 1e6 # velocity of of electron in m/s\n",
- "h = 6.62e-34 # Plank constant\n",
- "print(\"Example 1.17,page no:25\")\n",
- "lambda_b = h/(M*V) # calculation of de Broglie wavelength for ball\n",
- "lambda_e = h/(m*v) # calculation of de Broglie wavelength electron\n",
- "print(\"de Broglie wavelength of ball in angstrom:\"),round(lambda_b*1e10,34)\n",
- "print(\"de Broglie wavelength of electron in angstrom:\"),round(lambda_e*1e10,2)\n",
- "# answer in book is 6.62e-22 angstrom for ball\n",
- "\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.18;page no:26"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 15,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.18,page no:26\n",
- "Wavelength of neutron in angstrom: 0.286\n"
- ]
- }
- ],
- "source": [
- "#cal of de brogle wavelength of neutron\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "E = 1 # Energy of neutron in eV\n",
- "m = 1.67e-27 # Mass of neutron in Kg\n",
- "h = 6.62e-34 # Plank constant\n",
- "print(\"Example 1.18,page no:26\")\n",
- "lamda = h/math.sqrt(2*m*E*1.6e-19) # Calculation of velocity of moving electron\n",
- "print(\"Wavelength of neutron in angstrom:\"),round(lamda*1e10,3)\n",
- "# Answer in book is 6.62e-22 angstrom\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.19;page no:27"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 20,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.19,page no:27\n",
- "Applied voltage on electron in V: 602.0\n"
- ]
- }
- ],
- "source": [
- "#cal of Applied voltage on electron \n",
- "#intiation of all variables \n",
- "#given that\n",
- "lamda = 0.5# wavelength of electron in angstrom\n",
- "m = 9.1e-31 # Mass of electron in Kg\n",
- "h = 6.62e-34 # Plank constant\n",
- "q = 1.6e-19 # charge on electron in coulomb\n",
- "print(\"Example 1.19,page no:27\")\n",
- "V = h**2/(2*m*q*(lamda*1e-10)**2) # Calculation of velocity of moving electron\n",
- "print(\"Applied voltage on electron in V:\"),round(V,1)\n",
- "# Answer in book is 601.6 Volt\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.21;page no:29"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 21,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.21,page no:29\n",
- "Wavelength of neutron at degree Celsius in angstrom: 1.43\n"
- ]
- }
- ],
- "source": [
- "#cal of wavelength of neutron\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "k = 8.6e-5 # Boltzmann constant\n",
- "t = 37 # Temperature in degree Celsius\n",
- "h = 6.62e-34 # Plank constant\n",
- "m = 1.67e-27 # Mass of neutron\n",
- "print(\"Example 1.21,page no:29\")\n",
- "lamda = h/math.sqrt(3*m*(k*1.6e-19)*(t+273))# Calculation of wavelength\n",
- "print(\"Wavelength of neutron at degree Celsius in angstrom:\"),round(lamda*1e10,2)\n",
- "\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.22;page no:29"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 22,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.22,page no:29\n",
- "Wavelength of helium at degree Celsius in angstrom: 0.727\n"
- ]
- }
- ],
- "source": [
- "#cal of wavelength of helium\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "k = 8.6e-5 # Boltzmann constant\n",
- "t = 27 # Temperature in degree Celsius\n",
- "h = 6.62e-34 # Plank constant\n",
- "m = 6.7e-27 # Mass of helium atom\n",
- "print(\"Example 1.22,page no:29\")\n",
- "lamda = h/math.sqrt(3*m*(k*1.6e-19)*(t+273))# Calculation of wavelength\n",
- "print(\"Wavelength of helium at degree Celsius in angstrom:\"),round(lamda*1e10,3)\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.23;page no:30"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 17,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.23,page no:30\n",
- "lamda= 8.67e-11\n",
- "D/2*x= 0.05\n",
- "tan(theta)= 0.05\n",
- "Interatomic spacing of crystal in angstrom: 8.67\n"
- ]
- }
- ],
- "source": [
- "#cal of Interatomic spacing of crystal\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "E = 200. # energy of electrons in eV\n",
- "x = 20. # distance of screen in cm\n",
- "D = 2. # diameter of ring in cm\n",
- "h = 6.62e-34 # Plank constant\n",
- "m = 9.1e-31 # Mass of electron in kg\n",
- "print(\"Example 1.23,page no:30\")\n",
- "lamda= h/math.sqrt(2*m*E*1.6e-19) # Calculation of wavelength\n",
- "print(\"lamda=\"),round(lamda,13)\n",
- "print(\"D/2*x=\"),D/(2*x)\n",
- "p=D/(2*x)\n",
- "print(\"tan(theta)=\"),p\n",
- "d = lamda/(2*p)# calculation of interatomic spacing of crystal\n",
- "print(\"Interatomic spacing of crystal in angstrom:\"),round(d*1e10,2)\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.24;page no:31"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 21,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.24,page no:31\n",
- "Velocity of electron in ground state in M/s= 2.31\n"
- ]
- }
- ],
- "source": [
- "#cal of velocity of electron \n",
- "#intiation of all variables \n",
- "#given that\n",
- "r = 0.5 # Bohr radius of hydrogen in angstrom\n",
- "m = 9.1e-31 # Mass of neutron in Kg\n",
- "h = 6.6e-34 # Plank constant\n",
- "print(\"Example 1.24,page no:31\")\n",
- "v = h/(2*3.14*r*1e-10*m) # velocity of electron in ground state\n",
- "print(\"Velocity of electron in ground state in M/s=\"),round(v/10**6,2)\n",
- "# Answer in book is 2.31e6 m/s\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.25;page no:32"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 25,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.25,page no:32\n",
- "Velocity of electron in ground state in m/s: 1237.0\n"
- ]
- }
- ],
- "source": [
- "#cal of Velocity of electron in ground state\n",
- "#intiation of all variables \n",
- "#given that\n",
- "lamda = 5890 # wavelength of yellow radiation in angstrom\n",
- "m = 9.1e-31 # Mass of neutron in Kg\n",
- "h = 6.63e-34 # Plank constant\n",
- "print(\"Example 1.25,page no:32\")\n",
- "v = h/(lamda*1e-10*m) # velocity of electron in ground state\n",
- "print(\"Velocity of electron in ground state in m/s:\"),round(v,1)\n",
- "# Answer in book is 1.24e3 m/s\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.26;page no:33"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 26,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.26,page no:33\n",
- "Velocity of neutron in m/s: 1985.0\n",
- "Kinetic energy of neutron in eV: 0.021\n"
- ]
- }
- ],
- "source": [
- "#cal of Velocity and kinetic energy of neutron\n",
- "#intiation of all variables \n",
- "#given that\n",
- "lamda = 2 # wavelength of neutron in angstrom\n",
- "m = 1.67e-27 # Mass of neutron in Kg\n",
- "h = 6.63e-34 # Plank constant\n",
- "print(\"Example 1.26,page no:33\")\n",
- "v = h/(lamda*1e-10*m) # velocity of neutron\n",
- "k = 0.5*m*v**2 # Kinetic energy of neutron\n",
- "print(\"Velocity of neutron in m/s:\"),round(v,1)\n",
- "print(\"Kinetic energy of neutron in eV:\"),round(k/1.6e-19,3)\n",
- "# Answer in book is 0.021eV\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.29;page no:36"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 22,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.29,page no:36\n",
- "theta 72.6\n",
- "theta1= 56.84\n",
- "For first order, sin(theta) in For second order sin(theta) must be which is not possible for any value of angle.So no maxima occur for higher orders: 1.91\n"
- ]
- }
- ],
- "source": [
- "#cal of theta and theta1 \n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "v1 = 50 # Previous applied voltage\n",
- "v2 = 65 # final applied voltage\n",
- "k = 12.28 \n",
- "d = 0.91 # Spacing in a crystal in angstrom\n",
- "print(\"Example 1.29,page no:36\")\n",
- "lamda = k/math.sqrt(v1)\n",
- "theta= math.asin(lamda/(2*d))# Angel for initial applied voltage\n",
- "lamda1 = k/math.sqrt(v2)# wavelength for final applied voltage\n",
- "theta1 = math.asin(lamda1/(2*d))# Angel for final applied voltage\n",
- "#print(\"lamda1/1.82=\"),math.asin(lamda1/1.82)\n",
- "print(\"theta\"),round(theta*180/3.14,1)\n",
- "print(\"theta1=\"),round(theta1*180/3.14,2)\n",
- "print(\"For first order, sin(theta) in For second order sin(theta) must be which is not possible for any value of angle.So no maxima occur for higher orders:\"),round(2*math.sin(theta),2)\n",
- "#print(\"Angle of diffraction for first order of beam is degree at Volts:\"),round((math.theta1*180/math.pi),2)\n",
- "# Answer in book is 57.14 degree"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.30;page no:45"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 28,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.30,page no:45\n",
- "Group velocity of seawater waves in m/s: 16.29\n"
- ]
- }
- ],
- "source": [
- "#cal of Group velocity of seawater waves\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "lamda = 680 # Wavelength in m\n",
- "g = 9.8 #Acceleration due to gravity\n",
- "print(\"Example 1.30,page no:45\")\n",
- "v_g = 0.5*math.sqrt(g*lamda/(2*3.14)) # Calculation of group velocity\n",
- "print(\"Group velocity of seawater waves in m/s:\"),round(v_g,2)\n",
- "# Answer in book is 16.29 m/s\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.32;page no:47"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 29,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.32,page no:47\n",
- "Group velocity of de Broglie waves is c : 0.9966\n",
- " phase velocity of de Broglie waves is c 1.0034\n"
- ]
- }
- ],
- "source": [
- "#cal of group and phase velocity of de brogle waves \n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "lamda = 2e-13 # de Broglie wavelength of an electron in m\n",
- "c = 3e8 # Speed of light in m/s\n",
- "m = 9.1e-31 # Mass of electron in Kg\n",
- "h = 6.63e-34 # Plank constant\n",
- "print(\"Example 1.32,page no:47\")\n",
- "E = h*c/(lamda*1.6e-19) \n",
- "E_rest = m*c**2/(1.6e-19) # Calculation of rest mass energy\n",
- "E_total = math.sqrt(E**2+E_rest**2) # Total energy in eV\n",
- "v_g = c*math.sqrt(1-(E_rest/E_total)**2) # Group velocity\n",
- "v_p = c**2/v_g # Phase velocity\n",
- "print(\"Group velocity of de Broglie waves is c :\"),round(v_g/c,4)\n",
- "print(\" phase velocity of de Broglie waves is c\"),round(v_p/c,4)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 1.33;page no:48"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 23,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.33,page no:48\n",
- "Kinetic energy of electron in KeV: 293.33\n",
- "Group velocity of de Broglie waves is c in m/s: 0.7719\n",
- "phase velocity of de Broglie waves is c in m/s: 1.295\n"
- ]
- }
- ],
- "source": [
- "#cal of Kinetic energy of electron,group velocity and phase velocity of de Broglie waves\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "lamda = 2.e-12 # de Broglie wavelength of an electron in m\n",
- "c = 3.e8 # Speed of light in m/s\n",
- "m = 9.1e-31 # Mass of electron in Kg\n",
- "h = 6.63e-34 # Plank constant\n",
- "print(\"Example 1.33,page no:48\")\n",
- "E = h*c/(lamda*1.6e-19) # Energy due to momentum\n",
- "E_rest = m*c**2/(1.6e-19) # Calculation of rest mass energy\n",
- "E_total = math.sqrt(E**2+E_rest**2) # Total energy in eV\n",
- "KE = E_total - E_rest # Kinetic energy\n",
- "v_g = c*math.sqrt(1-(E_rest/E_total)**2) # Group velocity\n",
- "v_p = c**2/v_g # Phase velocity\n",
- "print(\"Kinetic energy of electron in KeV:\"),round(KE/1000,2)\n",
- "print(\"Group velocity of de Broglie waves is c in m/s:\"),round(v_g/c,4)\n",
- "print(\"phase velocity of de Broglie waves is c in m/s:\"),round(v_p/c,3)\n",
- "# Answer in book is v_g = 0.6035c & v_p = 1.657c\n"
- ]
- }
- ],
- "metadata": {
- "kernelspec": {
- "display_name": "Python 2",
- "language": "python",
- "name": "python2"
- },
- "language_info": {
- "codemirror_mode": {
- "name": "ipython",
- "version": 2
- },
- "file_extension": ".py",
- "mimetype": "text/x-python",
- "name": "python",
- "nbconvert_exporter": "python",
- "pygments_lexer": "ipython2",
- "version": "2.7.10"
- }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter2_1.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter2_1.ipynb deleted file mode 100755 index ec935dda..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter2_1.ipynb +++ /dev/null @@ -1,1020 +0,0 @@ -{
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "# chapter2 :Uncertainty principle and schrodinger wave equations"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 2.1;page no:77"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 1,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.1,page no:77\n",
- "Uncertainty in momentum of particle in kgm/sec: 2.64e-24\n"
- ]
- }
- ],
- "source": [
- "#cal of Uncertainty in momentum of particle\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "del_x = 0.2 # Uncertainty in position in angstrom\n",
- "h = 6.63e-34 # Plank constant\n",
- "print(\"Example 2.1,page no:77\")\n",
- "h_bar = h / (2*math.pi) # constant\n",
- "del_p = h_bar/(2*del_x*1e-10) # Calculation of uncertainty in momentum\n",
- "print(\"Uncertainty in momentum of particle in kgm/sec:\"),round(del_p,26)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 2.2;page no:77"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 2,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.2,page no:77\n",
- "Uncertainty in momentum of particle in kgm/sec: 1.32e-25\n"
- ]
- }
- ],
- "source": [
- "#cal of Uncertainty in momentum of particle\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "del_x = 4e-10 # Uncertainty in position in m\n",
- "h = 6.63e-34 # Plank constant\n",
- "print(\"Example 2.2,page no:77\")\n",
- "h_bar = h / (2*math.pi) # constant\n",
- "del_p = h_bar/(2*del_x) # Calculation of uncertainty in momentum\n",
- "print(\"Uncertainty in momentum of particle in kgm/sec:\"),round(del_p,27)\n",
- "# Answer in book is given as 1.32e-23 kgm/sec"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 2.3;page no:78"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 3,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.3,page no:78\n",
- "Uncertainty in position of particle in angstrom: 0.0192\n"
- ]
- }
- ],
- "source": [
- "#cal of Uncertainty in position of particle\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "v = 3e7 # Velocity of moving electron in m/s\n",
- "m = 9.1e-31 # mass of electron in kg\n",
- "h = 6.63e-34 # Plank constant\n",
- "c = 3e8 # speed of light in m/s\n",
- "print(\"Example 2.3,page no:78\")\n",
- "h_bar = h / (2*math.pi) # constant\n",
- "del_p = m*v/(math.sqrt(1-(v/c)**2)) # calculation of uncertainty in momentum\n",
- "del_x = h_bar/(2*del_p) # Calculation of uncertainty in position\n",
- "print(\"Uncertainty in position of particle in angstrom:\"),round(del_x*1e10,4)\n",
- "#Answer in book is 0.0194 angstrom which is due to using approximate values at intermediate steps"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 2.5;page no:79"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 4,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.5,page no:79\n",
- "Uncertainty in position of particle in m: 5.58e-05\n"
- ]
- }
- ],
- "source": [
- "#cal of Uncertainty in position of particle\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "v = 1.05e4 # Velocity of moving electron in m/s\n",
- "v_error = 0.02 #Percentage error in measurement of velocity\n",
- "m = 9e-31 # mass of electron in kg\n",
- "h = 6.63e-34 # Plank constant\n",
- "print(\"Example 2.5,page no:79\")\n",
- "h_bar = h / (2*math.pi) # constant\n",
- "p = m*v\n",
- "del_p = v_error*p/100 # calculation of uncertainty in momentum\n",
- "del_x = h_bar/del_p\n",
- "print(\"Uncertainty in position of particle in m:\"),round(del_x,7)\n",
- "# Answer in book is given as 5.58e-3 m"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 2.6;page no:80"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 5,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.6,page no:80\n",
- "Uncertainty in position of particle in m: 0.0039\n"
- ]
- }
- ],
- "source": [
- "#cal of Uncertainty in position of particle\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "v = 600 # Velocity of moving electron in m/s\n",
- "v_error = 0.005 #Percentage error in measurement of velocity\n",
- "m = 9.1e-31 # mass of electron in kg\n",
- "h = 6.63e-34 # Plank constant\n",
- "print(\"Example 2.6,page no:80\")\n",
- "h_bar = h / (2*math.pi) # constant\n",
- "p = m*v\n",
- "del_p = v_error*p/100 # calculation of uncertainty in momentum\n",
- "del_x = h_bar/(del_p) # Calculation of uncertainty in position\n",
- "print(\"Uncertainty in position of particle in m:\"),round(del_x,4)\n",
- "# Answer in book is 0.39e-2 m "
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 2.7;page no:80"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 6,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.7,page no:80\n",
- "Ratio of uncertainties in the velocity of electron to proton is: 1835.0\n"
- ]
- }
- ],
- "source": [
- "#cal of Ratio of uncertainties in the velocity of electron to proton\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "del_x = 1 # let uncertainty in position is unity\n",
- "m_e = 9.1e-31 # mass of electron in kg\n",
- "m_p = 1.67e-27 # mass of proton in kg\n",
- "h = 6.63e-34 # Plank constant\n",
- "print(\"Example 2.7,page no:80\")\n",
- "h_bar = h / (2*math.pi) # constant\n",
- "del_v_ratio = m_p/m_e # calculation in uncertainties in the velocity of electron and proton\n",
- "print(\"Ratio of uncertainties in the velocity of electron to proton is:\"),round(del_v_ratio,0)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 2.8;page no:81"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 17,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.8,page no:81\n",
- "Kinetic energy needed by an electron to be confined in electron in eV: 0.96\n"
- ]
- }
- ],
- "source": [
- "#cal of Kinetic energy needed by an electron\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "r = 0.5 # radius of hydrogen atom in angstrom\n",
- "m_e = 9.1e-31 # mass of electron in kg\n",
- "h = 6.63e-34 # Plank constant\n",
- "print(\"Example 2.8,page no:81\")\n",
- "h_bar = h / (2*math.pi) # constant\n",
- "del_x = 2*r # calculation of uncertainty in position\n",
- "del_p = h_bar/(2*del_x*1e-10) # calculation of uncertainty in momentum\n",
- "p = del_p\n",
- "E = p**2/(2*m_e*1.6e-19)# Calculation of energy in eV\n",
- "print(\"Kinetic energy needed by an electron to be confined in electron in eV:\"),round(E,2)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 2.9;page no:82"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 8,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.9,page no:82\n",
- "Uncertainty in position of particle in m: 0.0003865\n"
- ]
- }
- ],
- "source": [
- "#cal of Uncertainty in position of particle\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "v = 5e3 # Velocity of moving electron in m/s\n",
- "v_error = 0.003 #Percentage error in measurement of velocity\n",
- "m = 9.1e-31 # mass of electron in kg\n",
- "h = 6.63e-34 # Plank constant\n",
- "print(\"Example 2.9,page no:82\")\n",
- "h_bar = h / (2*math.pi) # constant\n",
- "p = m*v\n",
- "del_p = v_error*p/100 # calculation of uncertainty in momentum\n",
- "del_x = h_bar/(2*del_p) # Calculation of uncertainty in position\n",
- "print(\"Uncertainty in position of particle in m:\"),round(del_x,7)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 2.10;page no:83"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 9,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.10,page no:83\n",
- "Kinetic energy needed by an electron to be confined in electron in eV: 0.851\n"
- ]
- }
- ],
- "source": [
- "#cal of Kinetic energy needed by an electron to be confined in electron\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "r = 0.53 # radius of hydrogen atom in angstrom\n",
- "m_e = 9.1e-31 # mass of electron in kg\n",
- "h = 6.63e-34 # Plank constant\n",
- "print(\"Example 2.10,page no:83\")\n",
- "h_bar = h / (2*math.pi) # constant\n",
- "del_x = 2*r # calculation of uncertainty in position\n",
- "del_p = h_bar/(2*del_x*1e-10) # calculation of uncertainty in momentum\n",
- "p = del_p\n",
- "E = p**2/(2*m_e*1.6e-19)# Calculation of energy in eV\n",
- "print(\"Kinetic energy needed by an electron to be confined in electron in eV:\"),round(E,3)\n",
- "# When problem is solved by del_x*del_p = h_bar, then minimum value of kinetic energy will become 13.6eV"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 2.11;page no:84"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 10,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.11,page no:84\n",
- "Minimum error in measurement of energy of this state in eV: 26379.9318\n"
- ]
- }
- ],
- "source": [
- "#cal of Minimum error in measurement of energy\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "del_t = 2.5e-14 # lifetime in exited state in micro sec\n",
- "h = 6.63e-34 # Plank constant\n",
- "print(\"Example 2.11,page no:84\")\n",
- "h_bar = h / (2*math.pi) # constant\n",
- "del_E = h_bar/(1.6e-19*del_t*1e-6) # calculation of uncertainty in momentum\n",
- "print(\"Minimum error in measurement of energy of this state in eV:\"),round(del_E,4)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 2.12;page no:84"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 11,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.12,page no:84\n",
- "Percentage error in momentum in percent: 1.09\n"
- ]
- }
- ],
- "source": [
- "#cal of Percentage error in momentum\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "E_eV = 0.5# kinetic energy of electron in KeV\n",
- "del_x = 0.4 # Uncertainty in position in nm\n",
- "h = 6.63e-34 # Plank constant\n",
- "m = 9.1e-31 # mass of electron in kg\n",
- "print(\"Example 2.12,page no:84\")\n",
- "h_bar = h / (2*math.pi) # constant\n",
- "E_J = E_eV*1e3*1.6e-19\n",
- "p = math.sqrt(2*m*E_J) # Calculation of momentum in kgm/s\n",
- "del_p = h_bar/(2*del_x*1e-9) # Calculation of uncertainty in momentum\n",
- "per_error = del_p*100 / p # calculation of percentage error in momentum\n",
- "print(\"Percentage error in momentum in percent:\"),round(per_error,2)\n",
- "# Answer in book is 1.08 percentage"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 2.13;page no:85"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 12,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.13,page no:85\n",
- "Uncertainty in velocity of particle in m/s: 28988.9\n"
- ]
- }
- ],
- "source": [
- "#cal of Uncertainty in velocity of particle\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "del_x = 2e-9 # Uncertainty in position in m\n",
- "h = 6.63e-34 # Plank constant\n",
- "m = 9.1e-31 # mass of electron in Kg\n",
- "print(\"Example 2.13,page no:85\")\n",
- "h_bar = h / (2*math.pi) # constant\n",
- "del_p = h_bar/(2*del_x) # Calculation of uncertainty in momentum\n",
- "del_v = del_p/m\n",
- "print(\"Uncertainty in velocity of particle in m/s:\"),round(del_v,1)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 2.15;page no:87"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 18,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.15,page no:87\n",
- "Uncertainty in momentum of ball in kgm/s: 1.055e-28\n",
- "Percentage error in calculation of momentum is: 9e-29\n"
- ]
- }
- ],
- "source": [
- "#cal of Uncertainty in momentum of ball and Percentage error in calculation of momentum\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "del_x = 5000. # Uncertainty in position in angstrom\n",
- "h = 6.63e-34 # Plank constant\n",
- "m = 200. # mass of ball in gram\n",
- "v = 6. # velocity of moving ball in m/s\n",
- "print(\"Example 2.15,page no:87\")\n",
- "h_bar = h / (2.*math.pi) # constant\n",
- "del_p = h_bar/(2.*del_x*1e-10) # Calculation of uncertainty in momentum\n",
- "p = m*v # Calculation of momentum\n",
- "per_error = del_p*1000./p # Calculation of percentage error in calculation of momentum\n",
- "print(\"Uncertainty in momentum of ball in kgm/s:\"),round(del_p,31)\n",
- "print(\"Percentage error in calculation of momentum is:\"),round(per_error,29)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 2.16;page no:87"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 14,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.16,page no:87\n",
- "Uncertainty in position of particle in m: 1.05e-13\n"
- ]
- }
- ],
- "source": [
- "#cal of Uncertainty in position of particle\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "c = 3e8 # speed of light in m/s\n",
- "v = c/10 # Velocity of moving proton in m/s\n",
- "v_error = 1 # Percentage error in measurement of velocity \n",
- "m = 1.67e-27 # mass of electron in kg\n",
- "h = 6.63e-34 # Plank constant\n",
- "print(\"Example 2.16,page no:87\")\n",
- "h_bar = h / (2*math.pi) # constant\n",
- "del_v = v*v_error/100# calculation of uncertainty in position\n",
- "del_x = h_bar/(2*m*del_v) # calculation of uncertainty in momentum\n",
- "print(\"Uncertainty in position of particle in m:\"),round(del_x,15)\n",
- "# Answer in book is 1.04e-13 m"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 2.17;page no:88"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 15,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.17,page no:88\n",
- "Uncertainty in velocity of ball in m/s: 2.64e-25\n"
- ]
- }
- ],
- "source": [
- "#cal of Uncertainty in velocity of ball\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "del_x = 1e-9 # Uncertainty in position in m\n",
- "h = 6.63e-34 # Plank constant\n",
- "m = 200 # mass of ball in gram\n",
- "print(\"Example 2.17,page no:88\")\n",
- "h_bar = h / (2*math.pi) # constant\n",
- "del_v = h_bar/(2*del_x*m/1000) # Calculation of uncertainty in momentum\n",
- "print(\"Uncertainty in velocity of ball in m/s:\"),round(del_v,27)\n",
- "# Answer in book is 2.64e-25 m/s"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 2.18;page no:89"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 16,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.18,page no:89\n",
- "Minimum error in measurement of energy of this state in eV: 0.000165\n"
- ]
- }
- ],
- "source": [
- "#cal of Minimum error in measurement of energy\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "del_t = 2e-12 # lifetime of exited state in sec\n",
- "h = 6.63e-34 # Plank constant\n",
- "print(\"Example 2.18,page no:89\")\n",
- "h_bar = h / (2*math.pi) # constant\n",
- "del_E = h_bar/(1.6e-19*2*del_t) # calculation of uncertainty in momentum\n",
- "print(\"Minimum error in measurement of energy of this state in eV:\"),round(del_E,6)\n",
- "# Answer in book is 1.65e-4 eV"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 2.19;page no:89"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 19,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.19,page no:89\n",
- "Minimum error in measurement of frequency of photon in per second: 7957747.2\n"
- ]
- }
- ],
- "source": [
- "#cal of Minimum error in measurement of frequency of photon\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "del_t = 1e-8 # lifetime of exited state in sec\n",
- "h = 6.63e-34 # Plank constant\n",
- "print(\"Example 2.19,page no:89\")\n",
- "h_bar = h / (2*math.pi) # constant\n",
- "del_nu = h_bar/(2*del_t*h) # calculation of uncertainty in frequency\n",
- "print(\"Minimum error in measurement of frequency of photon in per second:\"),round(del_nu,1)\n",
- "# Answer in book is 8e6 per second"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 2.20;page no:90"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 20,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.20,page no:90\n",
- "Uncertainty in position of ball in angstrom: 9.6\n"
- ]
- }
- ],
- "source": [
- "#cal of Uncertainty in position of ball\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "del_v = 5.5e-20 # Uncertainty in velocity in m/s\n",
- "h = 6.63e-34 # Plank constant\n",
- "m = 1 # mass of dust particle in mg\n",
- "print(\"Example 2.20,page no:90\")\n",
- "h_bar = h / (2*math.pi) # constant\n",
- "del_x = h_bar/(2*del_v*m*1e-6) # Calculation of uncertainty in momentum\n",
- "print(\"Uncertainty in position of ball in angstrom:\"),round(del_x*1e10,1)\n",
- "# Answer in book is 9.6 angstrom"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 2.21;page no:101"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 21,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.21,page no:101\n",
- "Energy of electron in eV: 37.74\n"
- ]
- }
- ],
- "source": [
- "#cal of Energy of electron\n",
- "#intiation of all variables \n",
- "#given that\n",
- "l = 1 # width of potential well in angstrom\n",
- "n = 1 # order corresponding to ground state\n",
- "h = 6.63e-34 # Plank constant\n",
- "m = 9.1e-31 # mass of electron in Kg\n",
- "print(\"Example 2.21,page no:101\")\n",
- "E = n**2*h**2/(8*m*(l*1e-10)**2) # Calculation of energy in Joule\n",
- "E_eV = E/1.6e-19 # Calculation of energy in eV\n",
- "print(\"Energy of electron in eV:\"),round(E_eV,2)\n",
- "# Answer in book is 37.74 eV angstrom"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 2.24;page no:103"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 22,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.24,page no:103\n",
- "Energy of electron for state in eV: 6.04 24.15\n"
- ]
- }
- ],
- "source": [
- "#cal of Energy of electron for state\n",
- "#intiation of all variables \n",
- "#given that\n",
- "l = 2.5e-10 # width of potential well in m\n",
- "h = 6.63e-34 # Plank constant\n",
- "m = 9.1e-31 # mass of electron in Kg\n",
- "n1=1\n",
- "n2=2\n",
- "print(\"Example 2.24,page no:103\")\n",
- "E1 =n1**2*h**2/(8*m*l**2) # Calculation of energy in Joule\n",
- "E2 =n2**2*h**2/(8*m*l**2) # Calculation of energy in Joule\n",
- "E1_eV = E1/1.6e-19 # Calculation of energy in eV\n",
- "E2_eV = E2/1.6e-19 # Calculation of energy in eV\n",
- "print(\"Energy of electron for state in eV:\"),round(E1_eV,2),round(E2_eV,2)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 2.26;page no:105"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 23,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.26,page no:105\n",
- "Probability of finding electron between L and L in percent: 19.84\n"
- ]
- }
- ],
- "source": [
- "#cal of Probability of finding electron\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "L = 1# let unit length\n",
- "l1 = 0.45*L # initial point\n",
- "l2 = 0.55*L # Final point\n",
- "print(\"Example 2.26,page no:105\")\n",
- "p = (1/L)*((l2-(L/(2*math.pi) *math.sin(2*l2*math.pi/L)))- (l1-(L/(2*math.pi) *math.sin(2*l1*math.pi/L)))) # Calculation of probability of finding particle\n",
- "p_per = p*100 # probability of finding particle in percentage\n",
- "print(\"Probability of finding electron between L and L in percent:\"),round(p_per,2)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 2.27;page no:107"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 24,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.27,page no:107\n",
- "Difference between first state and ground state energies in eV: 113.21\n"
- ]
- }
- ],
- "source": [
- "#cal of Difference between first state and ground state energies\n",
- "#intiation of all variables \n",
- "#given that\n",
- "l = 1e-8 # width of potential well in cm\n",
- "h = 6.63e-34 # Plank constant\n",
- "m = 9.1e-31 # mass of electron in Kg\n",
- "print(\"Example 2.27,page no:107\")\n",
- "E_1 = (h)**2/(8*m*(l*1e-2)**2) # Calculation of energy of ground state in Joule\n",
- "E_1_eV = E_1/1.6e-19 # Calculation of energy in eV\n",
- "E_2 = (2)**2*h**2/(8*m*(l*1e-2)**2) # Calculation of energy of first state in Joule\n",
- "E_2_eV = E_2/1.6e-19 # Calculation of energy in eV\n",
- "del_E = E_2_eV - E_1_eV # calculation of difference between first state and ground state\n",
- "print(\"Difference between first state and ground state energies in eV:\"),round(del_E,2)\n",
- "# Answer in book is 113.04 eV"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 2.28;page no:107"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 25,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.28,page no:107\n",
- "For state: Energy in eV & wavelength in angstrom: 2 1 0.666666666667 37.7 151.0 339.6\n"
- ]
- }
- ],
- "source": [
- "#cal of de-broglie wave length and energy of electron\n",
- "#intiation of all variables \n",
- "#given that\n",
- "l = 1 # width of potential well in angstrom\n",
- "h = 6.63e-34 # Plank constant\n",
- "m = 9.1e-31 # mass of electron in Kg\n",
- "print(\"Example 2.28,page no:107\")\n",
- "n1=1\n",
- "n2=2\n",
- "n3=3.\n",
- "lamda1 = 2*l/n1 # Calculation of wavelength\n",
- "lamda2 = 2*l/n2 # Calculation of wavelength\n",
- "lamda3 = 2*l/n3 # Calculation of wavelength\n",
- "E1 = n1**2*h**2/(8*m*(l*1e-10)**2) # Calculation of energy in Joule\n",
- "E2 = n2**2*h**2/(8*m*(l*1e-10)**2) # Calculation of energy in Joule\n",
- "E3 = n3**2*h**2/(8*m*(l*1e-10)**2) # Calculation of energy in Joule\n",
- "E1_eV = E1/1.6e-19 # Calculation of energy in eV\n",
- "E2_eV = E2/1.6e-19 # Calculation of energy in eV\n",
- "E3_eV = E3/1.6e-19 # Calculation of energy in eV\n",
- "print(\"For state: Energy in eV & wavelength in angstrom:\"),lamda1,lamda2,lamda3,round(E1_eV,1),round(E2_eV,1),round(E3_eV,1)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 2.29;page no:108"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 26,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.29;page no:108\n",
- "Energy state E of ball in eV: 3.4341328125e-49 1.373653125e-48 3.09071953125e-48\n",
- "As energy difference is very small so we cannot see energy states:\n"
- ]
- }
- ],
- "source": [
- "#cal of Energy state E of ball\n",
- "#intiation of all variables \n",
- "#given that\n",
- "#import math\n",
- "m = 1000. #mass of ball in gram\n",
- "l = 1. # length of box in m\n",
- "h = 6.63e-34 # Plank constant\n",
- "n1=1\n",
- "n2=2\n",
- "n3=3\n",
- "print(\"Example 2.29;page no:108\")\n",
- "#for n in range(1,3):\n",
- "E1 = (n1**2*h**2)/(8*m*1e-3*l**2*1.6e-19)\n",
- "E2 = (n2**2*h**2)/(8*m*1e-3*l**2*1.6e-19)\n",
- "E3 = (n3**2*h**2)/(8*m*1e-3*l**2*1.6e-19)\n",
- "print(\"Energy state E of ball in eV:\"),E1,E2,E3\n",
- "print(\"As energy difference is very small so we cannot see energy states:\")"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "# example 2.30;page no:109"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 27,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 2.30,page no:109\n",
- "Probability of finding particle at centre in percent: 13.3\n"
- ]
- }
- ],
- "source": [
- "#cal of Probability of finding particle at centre\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "l = 30. # width of potential well in angstrom\n",
- "x = l/2.\n",
- "del_x = 2. # interval of length at centre in angstrom\n",
- "h = 6.63e-34 # Plank constant\n",
- "n = 1. # ground state\n",
- "print(\"Example 2.30,page no:109\")\n",
- "phi_x = ((math.sqrt(2/l))*math.sin(n*math.pi*x/l))**2 \n",
- "p = phi_x*del_x # Calculation of probability at centre\n",
- "print(\"Probability of finding particle at centre in percent:\"),round(p*100,1)\n",
- "# Answer given in book is 16 percent. It is due to wrong calculation "
- ]
- }
- ],
- "metadata": {
- "kernelspec": {
- "display_name": "Python 2",
- "language": "python",
- "name": "python2"
- },
- "language_info": {
- "codemirror_mode": {
- "name": "ipython",
- "version": 2
- },
- "file_extension": ".py",
- "mimetype": "text/x-python",
- "name": "python",
- "nbconvert_exporter": "python",
- "pygments_lexer": "ipython2",
- "version": "2.7.10"
- }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter3_1.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter3_1.ipynb deleted file mode 100755 index 4712fe5a..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter3_1.ipynb +++ /dev/null @@ -1,810 +0,0 @@ -{
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "# chapter3 :X-rays and compton effect"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 3.1;page no:141"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 1,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 3.1,page no:141\n",
- "Longest wavelength in angstrom: 2.82\n"
- ]
- }
- ],
- "source": [
- "#cal of Longest wavelength\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "d = 2.82 # crystal spacing in angstrom\n",
- "n = 2 # order for longest passing wavelength\n",
- "theta = 90 # angle for longest passing wavelength\n",
- "print(\"Example 3.1,page no:141\")\n",
- "lamda = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n",
- "print(\"Longest wavelength in angstrom:\"),round(lamda,2)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 3.2;page no:142"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 2,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 3.2,page no:142\n",
- "Angle for nd order maxima in degree: 36.87\n",
- "Angle for rd order maxima in degree: 64.16\n"
- ]
- }
- ],
- "source": [
- "#cal of Angle at which second and third order bragg's\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "lamda = 0.3 # Wavelength in angstrom\n",
- "d = 0.5 # crystal spacing in angstrom\n",
- "n = 2. # order \n",
- "m = 3. # order\n",
- "print(\"Example 3.2,page no:142\")\n",
- "theta_n = math.asin(n*lamda/(2*d))*180/math.pi # Calculation of angle for order n\n",
- "theta_m = math.asin(m*lamda/(2*d))*180/math.pi # Calculation of angle for order m\n",
- "print(\"Angle for nd order maxima in degree:\"),round(theta_n,2)\n",
- "print(\"Angle for rd order maxima in degree:\"),round(theta_m,2)\n",
- "# Answers in book are 40.97 degree and 72.29 degree which are due to wrong calculation"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "# example 3.3;page no:142"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 3,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 3.3,page no:142\n",
- "Longest wavelength in angstrom: 0.935\n"
- ]
- }
- ],
- "source": [
- "#cal of Longest wavelength\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "d = 1.87 # crystal spacing in angstrom\n",
- "n = 2 # order for longest passing wavelength\n",
- "theta = 30 # angle for longest passing wavelength\n",
- "print(\"Example 3.3,page no:142\")\n",
- "lamda = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n",
- "print(\"Longest wavelength in angstrom:\"),round(lamda,3)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 3.4;page no:143"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 4,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 3.4,page no:143\n",
- "Crystal spacing in angstrom in cm: 2.15e-08\n"
- ]
- }
- ],
- "source": [
- "#cal of Crystal spacing\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "lamda = 3.6e-9 # Wavelength in cm\n",
- "theta = 4.8 # glancing angle in degree\n",
- "n = 1 # order \n",
- "print(\"Example 3.4,page no:143\")\n",
- "d = n*lamda/(2*math.sin(theta*math.pi/180)) # calculation of crystal spacing in angstrom\n",
- "print(\"Crystal spacing in angstrom in cm:\"),round(d,10)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 3.5;page no:143"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 5,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 3.5,page no:143\n",
- "Longest wavelength in angstrom: 1.71\n"
- ]
- }
- ],
- "source": [
- "#cal of Longest wavelength\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "d = 2.5 # crystal spacing in angstrom\n",
- "n = 1 # order for longest passing wavelength\n",
- "theta = 20 # angle for longest passing wavelength\n",
- "print(\"Example 3.5,page no:143\")\n",
- "lamda = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n",
- "print(\"Longest wavelength in angstrom:\"),round(lamda,2)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 3.6;page no:144"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 6,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 3.6,page no:144\n",
- "Longest wavelength is of angstrom: 5.0\n"
- ]
- }
- ],
- "source": [
- "#cal of Longest wavelength\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "d = 2.5 # crystal spacing in angstrom\n",
- "n = 1 # order for longest passing wavelength\n",
- "theta = 90 # angle for longest passing wavelength\n",
- "print(\"Example 3.6,page no:144\")\n",
- "lamda = 2*d*math.sin(theta*math.pi/180)/n # Calculation of longest wavelength\n",
- "print(\"Longest wavelength is of angstrom:\"),round(lamda)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 3.7;page no:144"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 7,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 3.7,page no:144\n",
- "Interatomic layer separation ratios in crystal are as 1 : 1.4 1.8\n",
- "Above relation shows that crystal is simple cubic crystal structure.\n",
- "therefore Interatomic layer separation ratios in crystal is 1:1.4:1.77\n"
- ]
- }
- ],
- "source": [
- "#cal of Interatomic layer separation ratios in crystal\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "theta1_deg = 5 # Absolut degree part of angle for first angle\n",
- "theta1_min = 23#remainder minute part of angle for first angle\n",
- "theta2_deg = 7 # Absolut degree part of angle for second angle\n",
- "theta2_min = 37#remainder minute part of angle for second angle\n",
- "theta3_deg = 9 # Absolut degree part of angle for third angle\n",
- "theta3_min = 25#remainder minute part of angle for third angle\n",
- "print(\"Example 3.7,page no:144\")\n",
- "val1 = math.sin((theta1_deg+ theta1_min/60)*math.pi/180)# Sin value for first angle\n",
- "val2 = math.sin((theta2_deg+ theta2_min/60)*math.pi/180) #Sin value for second angle\n",
- "val3 = math.sin((theta3_deg+ theta3_min/60)*math.pi/180)#Sin value for third angle\n",
- "ratio_21 = val2/val1\n",
- "ratio_31 = val3/val1\n",
- "print(\"Interatomic layer separation ratios in crystal are as 1 :\"),round(ratio_21,1),round(ratio_31,1)\n",
- "print(\"Above relation shows that crystal is simple cubic crystal structure.\")\n",
- "print(\"therefore Interatomic layer separation ratios in crystal is 1:1.4:1.77\")"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 3.8;page no:145"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 8,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 3.8,page no:145\n",
- "If order is then spacing in angstrom: 3.64 1.82 1.21 0.91\n"
- ]
- }
- ],
- "source": [
- "#cal of possible spacing of this set of planes\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "lamda = 1.2 # wavelength in angstrom\n",
- "theta_deg = 9. # angle fraction in degree\n",
- "theta_min = 30. # Angle fraction in minute\n",
- "print(\"Example 3.8,page no:145\")\n",
- "theta = theta_deg+theta_min/60 # Total angel\n",
- "n1 =1\n",
- "n2 =2\n",
- "n3 =3\n",
- "n4 =4\n",
- "d1 = lamda/(n1*2*math.sin(theta*math.pi/180)) # Inter layer spacing\n",
- "d2 = lamda/(n2*2*math.sin(theta*math.pi/180)) # Inter layer spacing\n",
- "d3 = lamda/(n3*2*math.sin(theta*math.pi/180)) # Inter layer spacing\n",
- "d4 = lamda/(n4*2*math.sin(theta*math.pi/180)) # Inter layer spacing\n",
- "print(\"If order is then spacing in angstrom:\"),round(d1,2),round(d2,2),round(d3,2),round(d4,2)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 3.9;page no:146"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 9,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 3.9,page no:146\n",
- "Spacing of crystal in angstrom: 0.384\n"
- ]
- }
- ],
- "source": [
- "#cal of Spacing of crystal\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "h = 6.62e-34 # Planks constant\n",
- "m_e = 9.1e-31 # mass of electron in kg\n",
- "e = 1.6e-19 # charge on electron in coulomb\n",
- "v = 340 # Applied voltage in volt\n",
- "n = 1 # order for longest passing wavelength\n",
- "theta = 60 # angle for longest passing wavelength\n",
- "print(\"Example 3.9,page no:146\")\n",
- "lamda= h/math.sqrt(2*m_e*e*v) # calculation of wavelength\n",
- "d = n*lamda/(2*math.sin(theta*math.pi/180))# calculation of spacing of crystal\n",
- "print(\"Spacing of crystal in angstrom:\"),round(d*1e10,3)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 3.10;page no:147"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 10,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 3.10,page no:147\n",
- "Energy of recoiled electron in KeV: 2.55\n",
- "Example 3.10,page no:147\n",
- "Energy of recoiled electron in KeV: 2.55\n"
- ]
- }
- ],
- "source": [
- "#cal of Energy of recoiled electron\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "E = 100. # Energy of X ray beam in KeV\n",
- "theta = 30. # Scattering angle in degree\n",
- "m = 9.1e-31 # mass of electron in kg\n",
- "c = 3.e8 # Speed of light in m/s\n",
- "print(\"Example 3.10,page no:147\")\n",
- "E_rest = m*c**2/(1.6e-19*1e3) # Rest mass energy in KeV\n",
- "k = 1/E + (1-math.cos(theta*math.pi/180))/(E_rest)\n",
- "del_e = E - 1/k # Energy of recoiled electron\n",
- "print(\"Energy of recoiled electron in KeV:\"),round(del_e,2)\n",
- "#cal of Energy of recoiled electron\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "E = 100. # Energy of X ray beam in KeV\n",
- "theta = 30. # Scattering angle in degree\n",
- "m = 9.1e-31 # mass of electron in kg\n",
- "c = 3.e8 # Speed of light in m/s\n",
- "print(\"Example 3.10,page no:147\")\n",
- "E_rest = m*c**2/(1.6e-19*1e3) # Rest mass energy in KeV\n",
- "k = 1/E + (1-math.cos(theta*math.pi/180))/(E_rest)\n",
- "del_e = E - 1/k # Energy of recoiled electron\n",
- "print(\"Energy of recoiled electron in KeV:\"),round(del_e,2)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 3.11;page no:148"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 11,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 3.11,page no:148\n",
- "Wavelength shift in angstrom: 0.0242\n"
- ]
- }
- ],
- "source": [
- "#cal of Wavelength shift\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "lamda = 1 #wavelength in angstrom\n",
- "h = 6.62e-34 # Planks constant\n",
- "m_e = 9.1e-31 # mass of electron in kg\n",
- "c = 3e8 # speed of light in m/sec\n",
- "theta = 90 # angle for longest passing wavelength\n",
- "print(\"Example 3.11,page no:148\")\n",
- "d_lamda= h*(1-math.cos(theta*math.pi/180))/(m_e*c) # calculation of wavelength shift \n",
- "print(\"Wavelength shift in angstrom:\"),round(d_lamda*1e10,4)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 3.12;page no:149"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 12,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 3.12,page no:149\n",
- "Wavelength shift in angstrom: 0.027\n"
- ]
- }
- ],
- "source": [
- "#cal of Wavelength shift\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "lamda = 0.015 #wavelength in angstrom\n",
- "h = 6.63e-34 # Planks constant\n",
- "m_e = 9.1e-31 # mass of electron in kg\n",
- "c = 3e8 # speed of light in m/sec\n",
- "theta = 60 # angle for longest passing wavelength\n",
- "print(\"Example 3.12,page no:149\")\n",
- "d_lamda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n",
- "lamda_n = lamda+d_lamda\n",
- "print(\"Wavelength shift in angstrom:\"),round(lamda_n,3)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 3.13;page no:150"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 13,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 3.13,page no:150\n",
- "Wavelength shift in angstrom: 1.0243\n",
- "Energy of recoiled electron in eV: 295.0\n"
- ]
- }
- ],
- "source": [
- "#cal of Wavelength shift and Energy of recoiled electron\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "lamda = 1 #wavelength in angstrom\n",
- "h = 6.63e-34 # Planks constant\n",
- "m_e = 9.1e-31 # mass of electron in kg\n",
- "c = 3e8 # speed of light in m/sec\n",
- "theta = 90 # angle for longest passing wavelength\n",
- "print(\"Example 3.13,page no:150\")\n",
- "d_lamda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n",
- "lamda_n = lamda+d_lamda # Calculation of recoiled electron wavelength\n",
- "d_E = h*c*(lamda_n-lamda)*1e10/(1.6e-19*lamda_n*lamda)# Calculation of recoiled electron energy in eV\n",
- "print(\"Wavelength shift in angstrom:\"),round(lamda_n,4)\n",
- "print(\"Energy of recoiled electron in eV:\"),round(d_E)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 3.14;page no:151"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 14,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 3.14,page no:151\n",
- "Wavelength shift in angstrom: 0.0243\n",
- "Energy of recoiled electron in KeV: 511.88\n"
- ]
- }
- ],
- "source": [
- "#cal of wavelength shift and energy of recoiled electron\n",
- "#intiation of all variables \n",
- "#given that\n",
- "lamda = 1 #let wavelength in angstrom\n",
- "lamda_n = 2*lamda # recoiled electron wavelength\n",
- "h = 6.63e-34 # Planks constant\n",
- "m_e = 9.1e-31 # mass of electron in kg\n",
- "c = 3e8 # speed of light in m/sec\n",
- "theta = 90 # angle for longest passing wavelength\n",
- "print(\"Example 3.14,page no:151\")\n",
- "lamda = h*1e10/(m_e*c) # calculation of wavelength in angstrom\n",
- "E = h*c*1e10/(lamda*1.6e-19) # calculation of energy of electron\n",
- "print(\"Wavelength shift in angstrom:\"),round(lamda,4)\n",
- "print(\"Energy of recoiled electron in KeV:\"),round(E/1e3,2)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 3.15;page no:152"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 15,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 3.15,page no:152\n",
- "Fraction of energy lost by photon is: 0.00356\n"
- ]
- }
- ],
- "source": [
- "#cal of Fraction of energy lost by photon\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "lamda = 2 #wavelength in angstrom\n",
- "h = 6.63e-34 # Planks constant\n",
- "m_e = 9.1e-31 # mass of electron in kg\n",
- "c = 3e8 # speed of light in m/sec\n",
- "theta = 45 # scattering angle \n",
- "print(\"Example 3.15,page no:152\")\n",
- "d_lamda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n",
- "lamda_n = lamda+d_lamda # Calculation of recoiled electron wavelength\n",
- "f = d_lamda/lamda # Calculation of fraction of energy lost by photon \n",
- "print(\"Fraction of energy lost by photon is:\"),round(f,5)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 3.16;page no:153"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 16,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 3.16,page no:153\n",
- "Wavelength of scattered radiation in Angstrom: 0.0487\n"
- ]
- }
- ],
- "source": [
- "#cal of wavelength of scattered radiation\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "E_eV = 510 # Energy of gamma ray in keV\n",
- "lamda = 2 #wavelength in angstrom\n",
- "h = 6.63e-34 # Planks constant\n",
- "m_e = 9.1e-31 # mass of electron in kg\n",
- "c = 3e8 # speed of light in m/sec\n",
- "theta = 90 # scattering angle in degree\n",
- "print(\"Example 3.16,page no:153\")\n",
- "E_j = E_eV*1e3*1.6e-19 # Energy of gamma ray in Joule\n",
- "lamda = h*c*1e10/E_j # Calculation of wavelength in angstrom\n",
- "d_lamda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n",
- "lamda_n = lamda+d_lamda # Calculation of recoiled electron wavelength\n",
- "print(\"Wavelength of scattered radiation in Angstrom:\"),round(lamda_n,4)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 3.17;page no:154"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 17,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 3.17,page no:154\n",
- "Scattered wavelength in angstrom: 2.0243\n",
- "Energy of recoiled electron in eV: 74.6\n"
- ]
- }
- ],
- "source": [
- "#cal of Scattered wavelength and Energy of recoiled electron\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "lamda = 2 #wavelength in angstrom\n",
- "h = 6.63e-34 # Planks constant\n",
- "m_e = 9.1e-31 # mass of electron in kg\n",
- "c = 3e8 # speed of light in m/sec\n",
- "theta = 90 # angle for longest passing wavelength\n",
- "print(\"Example 3.17,page no:154\")\n",
- "d_lamda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n",
- "lamda_n = lamda+d_lamda # Calculation of recoiled electron wavelength\n",
- "d_E = h*c*(lamda_n-lamda)*1e10/(1.6e-19*lamda_n*lamda)# Calculation of recoiled electron energy in eV\n",
- "print(\"Scattered wavelength in angstrom:\"),round(lamda_n,4)\n",
- "print(\"Energy of recoiled electron in eV:\"),round(d_E,1)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 3.18;page no:155"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 18,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 3.18,page no:155\n",
- "Wavelength of scattered radiation in m: 4.87e-12\n",
- "Energy of recoiled electron in MeV: 0.255\n"
- ]
- }
- ],
- "source": [
- "#cal of Wavelength of scattered radiation and Energy of recoiled electron\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "E_eV = 510 # Energy of gamma ray in keV\n",
- "h = 6.63e-34 # Planks constant\n",
- "m_e = 9.1e-31 # mass of electron in kg\n",
- "c = 3e8 # speed of light in m/sec\n",
- "theta = 90 # scattering angle in degree\n",
- "print(\"Example 3.18,page no:155\")\n",
- "E_j = E_eV*1e3*1.6e-19 # Energy of gamma ray in Joule\n",
- "lamda = h*c/E_j # Calculation of wavelength in meter\n",
- "d_lamda= h*(1-math.cos(theta*math.pi/180))*1e10/(m_e*c) # calculation of wavelength shift in angstrom\n",
- "lamda_n = lamda+d_lamda/1e10 # Calculation of recoiled electron wavelength\n",
- "d_E = h*c*(d_lamda/1e10)/(1.6e-19*lamda_n*lamda)# Calculation of recoiled electron energy in eV\n",
- "psi= math.atan(1/(math.tan((theta*math.pi/180)/2)/(1+(h/(lamda*m_e*c))))) \n",
- "phi_deg = 90 -psi*180/math.pi # Calculation of degree part of angle of recoiled electron \n",
- "#phi_min = 60*(phi_deg - floor(phi_deg))# Calculation of minute part of angle of recoiled electron \n",
- "print(\"Wavelength of scattered radiation in m:\"),round(lamda_n,14)\n",
- "print(\"Energy of recoiled electron in MeV:\"),round(d_E/1e6,3)\n",
- "#print(\"Recoiled electron angle in degree minute:\"),round(phi_deg),round(phi_min)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 3.19;page no:157"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 19,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 3.19,page no:157\n",
- "Frequency after collision in Hz: 1.72131147541e+19\n"
- ]
- }
- ],
- "source": [
- "#cal of Frequency after collision\n",
- "#intiation of all variables \n",
- "#given that\n",
- "nu = 2e19 # initial frequency of X ray photon\n",
- "h = 6.63e-34 # Planks constant\n",
- "m_e = 9.1e-31 # mass of electron in kg\n",
- "c = 3e8 # speed of light in m/sec\n",
- "theta = 90 # scattering angle in degree\n",
- "print(\"Example 3.19,page no:157\")\n",
- "d_lamda = h/(m_e*c) # calculation of wavelength shift\n",
- "k = 1/nu + d_lamda/c\n",
- "nu_1 = 1/k # Frequency after collision\n",
- "nu_1 = (nu_1/1e18)*1e18 # rounding off\n",
- "print(\"Frequency after collision in Hz:\"),round(nu_1,19)"
- ]
- }
- ],
- "metadata": {
- "kernelspec": {
- "display_name": "Python 2",
- "language": "python",
- "name": "python2"
- },
- "language_info": {
- "codemirror_mode": {
- "name": "ipython",
- "version": 2
- },
- "file_extension": ".py",
- "mimetype": "text/x-python",
- "name": "python",
- "nbconvert_exporter": "python",
- "pygments_lexer": "ipython2",
- "version": "2.7.10"
- }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter4_1.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter4_1.ipynb deleted file mode 100755 index 4e7f7817..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter4_1.ipynb +++ /dev/null @@ -1,542 +0,0 @@ -{
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "# chapter4 :Dielectrics"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 4.4;page no:189"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 1,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 4.4,page no:189\n",
- "Dipole moment per atom in Coulomb-meter: 2.446e-39\n"
- ]
- }
- ],
- "source": [
- "#cal of Dipole moment per atom\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import sys\n",
- "epsilon_r = 1.000074 # Dielectric constant of He at 0C and 1atm\n",
- "epsilon_0 = 8.854e-12 # Permittivity of free space\n",
- "E = 100 # Electric field in V/m\n",
- "n = 2.68e27 # Electron density in no,/m^\n",
- "N_a = 6e23 # Avogadro number\n",
- "V = 22.4 # Volume at STP in litter\n",
- "print(\"Example 4.4,page no:189\")\n",
- "P = epsilon_0*(epsilon_r-1)*E # Calculation of polarization\n",
- "N = N_a/(V*1e-3)# Calculation of total number of atoms\n",
- "p = P/N # dipole moment per atom\n",
- "print(\"Dipole moment per atom in Coulomb-meter:\"),round(p,42)\n",
- "# Answer in book is in different form and as 24.45e-40 coulomb-meter"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 4.6;page no:191"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 2,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 4.6,page no:191\n",
- "Electronic polarisability in Fm^2: 1.851e-41\n",
- "Relative permeability is: 1.002049\n"
- ]
- }
- ],
- "source": [
- "#cal of Electronic polarisability and Relative permeability\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "import sys\n",
- "r = 0.055 # Radius of hydrogen atom in nm\n",
- "n = 9.8e26 # Number of atoms/cc\n",
- "epsilon_0 = 8.854e-12 # Permittivity of free space\n",
- "print(\"Example 4.6,page no:191\")\n",
- "alpha_e = 4*math.pi*epsilon_0*(r*1e-9)**3 # Calculation of electronic polarisability\n",
- "epsilon_r = 1+n*alpha_e/epsilon_0 # Calculation of relative permeability\n",
- "print(\"Electronic polarisability in Fm^2:\"),round(alpha_e,44)\n",
- "print(\"Relative permeability is:\"),round(epsilon_r,6)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 4.8;page no:192"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 3,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 4.8,page no:192\n",
- "Relative permittivity is: 4.614\n"
- ]
- }
- ],
- "source": [
- "#cal of Relative permittivity\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import sys\n",
- "epsilon_0 = 8.854e-12 # Permittivity of free space\n",
- "E = 2000 # Electric field in V/m\n",
- "P = 6.4e-8 # Polarization in C/m^2\n",
- "print(\"Example 4.8,page no:192\")\n",
- "epsilon_r = 1+ P/(epsilon_0*E) # Calculation of relative permittivity\n",
- "print(\"Relative permittivity is:\"),round(epsilon_r,3)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 4.9;page no:193"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 4,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 4.9,page no:193\n",
- "Relative permittivity is: 1.90395\n"
- ]
- }
- ],
- "source": [
- "#cal of Relative permittivity\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import sys\n",
- "alpha_e = 2e-40 # Electronic polarisability in Fm^2\n",
- "N = 4e28 # density in atoms/m^3\n",
- "epsilon_0 = 8.85e-12 # Permittivity of free space\n",
- "print(\"Example 4.9,page no:193\")\n",
- "epsilon_r = 1+ N*alpha_e/(epsilon_0) # Calculation of relative permittivity\n",
- "print(\"Relative permittivity is:\"),round(epsilon_r,5)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "# example 4.10;page no:193"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 5,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 4.10,page no:193\n",
- "Relative permittivity is: 27.11\n",
- "Electrical susceptibility in C^2/Nm^2: 2.31e-10\n"
- ]
- }
- ],
- "source": [
- "#cal of Relative permittivity and Electrical susceptibility\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import sys\n",
- "epsilon = 2.4e-10 # permitivity of a dielectric material in C^2/N?m^2\n",
- "epsilon_0 = 8.854e-12 # Permittivity of free space\n",
- "print(\"Example 4.10,page no:193\")\n",
- "K = epsilon/epsilon_0 # Calculation of dielectric constant \n",
- "zai_e = epsilon_0*(K-1) # Calculation of electrical susceptibility \n",
- "print(\"Relative permittivity is:\"),round(K,2)\n",
- "print(\"Electrical susceptibility in C^2/Nm^2:\"),round(zai_e,12)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 4.11;page no:194"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 6,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 4.11,page no:194\n",
- "Magnitude of Electrical vector in Volt/meter: 11250.0\n",
- "Magnitude of Electrical Displacement vector in C/m^2: 7.969e-07\n",
- "Magnitude of Electric polarization vector in C/m^2: 6.9725e-07\n"
- ]
- }
- ],
- "source": [
- "#cal of Magnitude of Electrical vector,Electrical Displacement and Electric polarization vector\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import sys\n",
- "V = 100 # Applied potential in Volt\n",
- "d = 1 # Separation between plates in cm\n",
- "k1 = 8 # Dielectric constant\n",
- "k2 = 9 #dielectric constant\n",
- "epsilon_0 = 8.854e-12 # Permittivity of free space\n",
- "print(\"Example 4.11,page no:194\")\n",
- "E_0 = V/(d*1e-2) # Calculation of electric field\n",
- "E = E_0/k1*k2 # Calculation of electric field\n",
- "D = k1*epsilon_0*E # Calculation of electrical displacement vector\n",
- "P = (k1-1)*epsilon_0*E # Calculation of electrical polarization \n",
- "print(\"Magnitude of Electrical vector in Volt/meter:\"),round(E,3) # Answer in book is 1.125e3 Volt/meter\n",
- "print(\"Magnitude of Electrical Displacement vector in C/m^2:\"),round(D,10)# Answer in book is 8.85e-8C/m^2\n",
- "print(\"Magnitude of Electric polarization vector in C/m^2:\"),round(P,11)# Answer in book is 7.774e-8C/m^2"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 4.12;page no:195"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 14,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 4.12,page no:195\n",
- "Deformational Polarizability in C^2mN^-1: 1e-39\n",
- "Orientational Polarizability at degree Celcius in C^2mN^-1: 1.5e-39\n",
- "Orientational Polarizability at degree Celcius in C^2mN^-1: 7.5e-40\n"
- ]
- }
- ],
- "source": [
- "#cal of Deformational Polarizability and Orientational Polarizability\n",
- "#intiation of all variables \n",
- "#given that\n",
- "alpha_300 = 2.5e-39 # total polarisability in C^2m/N at 300 K\n",
- "alpha_600 = 1.75e-39 # total polarisability in C^2m/N at 600 K\n",
- "T1 = 300 # Initial temperature in Kelvin\n",
- "T2 = 600 # Final Temperature in Kelvin\n",
- "print(\"Example 4.12,page no:195\")\n",
- "b = (alpha_300-alpha_600)*T2\n",
- "al_def_300 = alpha_300 - b/300\n",
- "al_oriant_300 = b/300\n",
- "al_oriant_600 = b/600\n",
- "print(\"Deformational Polarizability in C^2mN^-1:\"),round(al_def_300,40)\n",
- "print(\"Orientational Polarizability at degree Celcius in C^2mN^-1:\"),round(al_oriant_300,40)\n",
- "print(\"Orientational Polarizability at degree Celcius in C^2mN^-1:\"),round(al_oriant_600,41)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 4.13;page no:202"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 8,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 4.13,page no:202\n",
- "Relative permittivity is: 1.88\n"
- ]
- }
- ],
- "source": [
- " #cal of Relative permittivity\n",
- "#intiation of all variables \n",
- "#given that\n",
- "#import sys\n",
- "alpha_e = 1.5e-40 # Electronic polarizability in Fm^2\n",
- "N = 4e28 # density in atoms/m^3\n",
- "epsilon_0 = 8.85e-12 # Permittivity of free space\n",
- "print(\"Example 4.13,page no:202\")\n",
- "k = N*alpha_e/(3*epsilon_0)\n",
- "epsilon_r = (1+ k*2)/(1-k)# Calculation of relative permittivity\n",
- "print(\"Relative permittivity is:\"),round(epsilon_r,2)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "# example 4.14;page no:202"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 9,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 4.14,page no:202\n",
- "Relative permittivity is: 4.2\n"
- ]
- }
- ],
- "source": [
- " #cal of Relative permittivity\n",
- "#intiation of all variables \n",
- "#given that\n",
- "#import sys\n",
- "m = 32 # Atomic weight of sulphur\n",
- "d = 2.08 # Density in g/cm^3\n",
- "alpha_e = 3.5e-40 # Electronic polarizability in Fm^2\n",
- "N_a = 6.022e23 # Avogadro Number\n",
- "epsilon_0 = 8.85e-12 # Permittivity of free space\n",
- "print(\"Example 4.14,page no:202\")\n",
- "N = N_a*d*1e6/m # Calculation of Atoms per unit \n",
- "k = N*alpha_e/(3*epsilon_0)\n",
- "epsilon_r = (1+ k*2)/(1-k)# Calculation of relative permittivity\n",
- "print(\"Relative permittivity is:\"),round(epsilon_r,2)\n",
- "# Answer in book is 4.17"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 4.15;page no:203"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 10,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 4.15,page no:203\n",
- "Percentage ionic polarizability in pecent: 51.41\n"
- ]
- }
- ],
- "source": [
- " #cal of Percentage ionic polarizability\n",
- "#intiation of all variables \n",
- "#given that\n",
- "#import sys\n",
- "n = 1.5 # Refractive index\n",
- "epsilon = 5.6 # Static dielectric constant\n",
- "print(\"Example 4.15,page no:203\")\n",
- "per = (1-((n**2-1)/(n**2+2))*(epsilon+2)/(epsilon-1))*100 # Pecentage ionic polarisability\n",
- "print(\"Percentage ionic polarizability in pecent:\"),round(per,2)\n",
- "# Answer in book is 5.14 %"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 4.16;page no:204"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 11,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 4.16,page no:204\n",
- "Electronic polarizability in Fm^2: 3.29e-40\n"
- ]
- }
- ],
- "source": [
- " #cal of Electronic polarizability\n",
- "#intiation of all variables \n",
- "#given that\n",
- "#import sys\n",
- "m = 32 # Atomic weight of sulphur\n",
- "d = 2050 # Density in Kg/m^3\n",
- "N_a = 6.022e23 # Avogadro Number\n",
- "epsilon_0 = 8.85e-12 # Permittivity of free space\n",
- "epsilon_r = 3.75 # Dielectric constant of sulphur\n",
- "print(\"Example 4.16,page no:204\")\n",
- "N = N_a*d*1e3/m # Calculation of Atoms per unit \n",
- "alpha_e = 3*epsilon_0*((epsilon_r-1)/(epsilon_r+2)) / N \n",
- "print(\"Electronic polarizability in Fm^2:\"),round(alpha_e,42)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 4.17;page no:204"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 12,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 4.17,page no:204\n",
- "Ratio of electronic to ionic polarizability is: 1.43\n"
- ]
- }
- ],
- "source": [
- "#cal of Ratio of electronic to ionic polarizability\n",
- "#intiation of all variables \n",
- "#given that\n",
- "#import sys\n",
- "n = 1.5 # Refractive index\n",
- "epsilon = 4. # Static dielectric constant\n",
- "epsilon_0 = 8.85e-12 # permittivity of free space\n",
- "print(\"Example 4.17,page no:204\")\n",
- "k1 = (epsilon-1)/(epsilon+2)\n",
- "k2 = (n**2-1.)/(n**2+2.)\n",
- "ratio = 1./((k1/k2)-1.) \n",
- "print(\"Ratio of electronic to ionic polarizability is:\"),round(ratio,2)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 4.18;page no:219"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 13,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 4.18,page no:219\n",
- "Frequency in KHz: 8.84\n",
- "Phase difference between current and voltage in degree: 45.0\n"
- ]
- }
- ],
- "source": [
- "#cal of Frequency and Phase difference between current and voltage\n",
- "#intiation of all variables \n",
- "#given that\n",
- "import math\n",
- "t = 1.8e-5 # Relaxation time in second\n",
- "epsilon_r = 1 # let\n",
- "print(\"Example 4.18,page no:219\")\n",
- "f = 1/(2*math.pi*t) # Calculation of frequency\n",
- "delta = math.atan(epsilon_r/epsilon_r)\n",
- "phi = 90 - delta*180/math.pi # Calculation of phase difference\n",
- "print(\"Frequency in KHz:\"),round(f/1e3,2)\n",
- "print(\"Phase difference between current and voltage in degree:\"),round(phi,0)"
- ]
- }
- ],
- "metadata": {
- "kernelspec": {
- "display_name": "Python 2",
- "language": "python",
- "name": "python2"
- },
- "language_info": {
- "codemirror_mode": {
- "name": "ipython",
- "version": 2
- },
- "file_extension": ".py",
- "mimetype": "text/x-python",
- "name": "python",
- "nbconvert_exporter": "python",
- "pygments_lexer": "ipython2",
- "version": "2.7.10"
- }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter6_1.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter6_1.ipynb deleted file mode 100755 index 6b05b99d..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter6_1.ipynb +++ /dev/null @@ -1,214 +0,0 @@ -{
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "# chapter6 :Ultrasonic Waves"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 6.1;page no:262"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 1,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 6.1,page no:262\n",
- "Fundamental frequency in Hz in 10^5: 9.09996\n"
- ]
- }
- ],
- "source": [
- "#cal of Fundamental frequency\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "import math\n",
- "E=7.9*10**10\n",
- "rho = 2650 # Density in Kg/m**3\n",
- "t = 0.003 # Thickness of quartz crystal in m\n",
- "print(\"Example 6.1,page no:262\")\n",
- "v = math.sqrt(E/rho)# Calculation of velocity \n",
- "lamda = 2*t # Calculation of fundamental wavelength\n",
- "nu = v/lamda # Calculation of fundamental frequency\n",
- "print(\"Fundamental frequency in Hz in 10^5:\"),round(nu/10**5,5)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "# example 6.2;page no:263"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 2,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 6.2,page no:263\n",
- "Fundamental frequency of crystal in MHz: 1.8\n"
- ]
- }
- ],
- "source": [
- "\n",
- "#cal of Fundamental frequency of crystal\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "v = 5760 # Velocity in m/s\n",
- "T = 1.6 # Thickness of quartz crystal in mm\n",
- "print(\"Example 6.2,page no:263\")\n",
- "nu = v/(2*T*1e-3)# Calculation of fundamental frequency\n",
- "print(\"Fundamental frequency of crystal in MHz:\"),round(nu/1e6,1)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 6.3;page no:264"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 3,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 6.3,page no:264\n",
- "Depth of defect in cm: 20\n"
- ]
- }
- ],
- "source": [
- "#cal of Depth of defect\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "T =40 # Thickness of steel bar in cm\n",
- "t1 = 40 # Time in ms\n",
- "t2 = 80 # Time in ms\n",
- "print(\"Example 6.3,page no:264\")\n",
- "X = T*t1/t2 # Calculation of depth of defect\n",
- "print(\"Depth of defect in cm:\"),X"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "# example 6.4;page no:264"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 4,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 6.4,page no:264\n",
- "Fundamental frequency in Hz in 10^5: 4.54998\n"
- ]
- }
- ],
- "source": [
- "#cal of Fundamental frequency\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "import math\n",
- "E=7.9*10**10\n",
- "rho = 2650 # Density in Kg/m**3\n",
- "t = 0.006 # Thickness of quartz crystal in m\n",
- "print(\"Example 6.4,page no:264\")\n",
- "v = math.sqrt(E/rho)# Calculation of velocity \n",
- "lamda = 2*t # Calculation of fundamental wavelength\n",
- "nu = v/lamda # Calculation of fundamental frequency\n",
- "print(\"Fundamental frequency in Hz in 10^5:\"),round(nu/10**5,5)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 6.5;page no:265"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 5,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 6.5,page no:265\n",
- "Capacitance in microfarad: 0.0\n"
- ]
- }
- ],
- "source": [
- "#cal of Capacitance\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "import math\n",
- "L = 1 # Inductance in Hanery\n",
- "nu = 2e6 # Frequency in Hz\n",
- "print(\"Example 6.5,page no:265\")\n",
- "C= 1/(4*((math.pi)**2)*nu**2*L) # Calculation of capacitance\n",
- "print(\"Capacitance in microfarad:\"),round(C*1e6,5)\n",
- "# Answer in book is 0.00634 micro Farad"
- ]
- }
- ],
- "metadata": {
- "kernelspec": {
- "display_name": "Python 2",
- "language": "python",
- "name": "python2"
- },
- "language_info": {
- "codemirror_mode": {
- "name": "ipython",
- "version": 2
- },
- "file_extension": ".py",
- "mimetype": "text/x-python",
- "name": "python",
- "nbconvert_exporter": "python",
- "pygments_lexer": "ipython2",
- "version": "2.7.10"
- }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter7_1.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter7_1.ipynb deleted file mode 100755 index ab10e07a..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter7_1.ipynb +++ /dev/null @@ -1,484 +0,0 @@ -{
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "# chapter7 :Maxwell's Equations and Electromagnetic Waves"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 7.1;page no:304"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 1,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 7.1,page no:304\n",
- "Average value of electric field at distance m in Volt/m: 86.57\n"
- ]
- }
- ],
- "source": [
- "#cal of Average value of electric field\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "import math\n",
- "p = 1000 # power in watt\n",
- "d = 2 # Distance from lamp in m\n",
- "epsilon_0 = 8.854e-12 # Permittivity of free space\n",
- "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n",
- "print(\"Example 7.1,page no:304\")\n",
- "s = p/(4*math.pi*d**2)# Calculation of pointing vector\n",
- "E_H_ratio = math.sqrt(mu_0/epsilon_0) # Calculation of ratio of Electric field and magnetic field\n",
- "E= math.sqrt(E_H_ratio*s) # Calculation of Electric field \n",
- "print(\"Average value of electric field at distance m in Volt/m:\"),round(E,2)\n",
- "# Answer in book is 48.87 volt/m which is due to wrong calculation at intermediate steps"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 7.2;page no:304"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 2,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 7.2,page no:304\n",
- "Average value of electric field in Volt/m: 1027.1\n",
- "Average value of magnetic field in Amp turn/m: 2.726\n"
- ]
- }
- ],
- "source": [
- "#cal of Average value of electric field and magnetic field\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "import math\n",
- "p = 2 # power in cal/min/cm^2\n",
- "epsilon_0 = 8.854e-12 # Permittivity of free space\n",
- "mu_0 = 4*math.pi*1e-7 # permeability of free space\n",
- "print(\"Example 7.2,page no:304\")\n",
- "s = p*4.2e4/60 # Calculation of pointing vector\n",
- "E_H_ratio = math.sqrt(mu_0/epsilon_0) # Calculation of ratio of Electric field and magnetic field\n",
- "E= math.sqrt(E_H_ratio*s) # Calculation of Electric field \n",
- "H = s/E # Calculation of Electric field \n",
- "print(\"Average value of electric field in Volt/m:\"),round(E*math.sqrt(2),1)\n",
- "print(\"Average value of magnetic field in Amp turn/m:\"),round(H*math.sqrt(2),3)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 7.3;page no:305"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 3,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 7.3,page no:305\n",
- "Skin depth penetration in cm: 0.0009\n"
- ]
- }
- ],
- "source": [
- "#cal of Skin depth penetration\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "import math\n",
- "mu_0 = 4*math.pi*1e-7 # permeability of free space\n",
- "mu = mu_0 #permeability of silver\n",
- "sigma = 3e7 # conductivity in mhos/m\n",
- "f = 1e8 # frequency in Hz\n",
- "print(\"Example 7.3,page no:305\")\n",
- "omega = 2*math.pi/f # Calculation of time period\n",
- "delta = math.sqrt(2/(omega*sigma*mu)) # Calculation of skin depth penetration\n",
- "Delta = (delta/100)*100 # Rounding off\n",
- "print(\"Skin depth penetration in cm:\"),round(Delta*1e-6,4)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 7.5;page no:307"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 4,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 7.5,page no:307\n",
- "Frequency required for penetration of depth in MHz= 6.0\n",
- "Sea water is good conductor at frequency lesser than 1e8 Hz\n"
- ]
- }
- ],
- "source": [
- "#cal of Frequency required for penetration of depth and sea water can be considered as good conductor\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "import math\n",
- "k = 80 # relative Dielectric constant of sea water\n",
- "epsilon_0 = 1/9e9 # Permittivity of free space\n",
- "epsilon = 80*epsilon_0 # Permittivity of free space\n",
- "sigma = 4.3 # conductivity in mho/m\n",
- "delta = 10 # penetration depth in cm\n",
- "mu_0 = 4*math.pi*1e-7 # permeability f free space\n",
- "F = 1e8 # Given frequency in Hz\n",
- "print(\"Example 7.5,page no:307\")\n",
- "f = (1/(math.pi*mu_0*sigma))/(delta*1e-2)**2 # Calculation of frequency\n",
- "print(\"Frequency required for penetration of depth in MHz=\"),round(f/10**6)\n",
- "omega = 2*math.pi*F\n",
- "x = 2*sigma/(epsilon*omega)\n",
- "if x>1:\n",
- "\tprint(\"Sea water is good conductor at frequency lesser than 1e8 Hz\")"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 7.7;page no:308"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 5,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 7.7,page no:308\n",
- "For MHz frequency, Penetration depth in cm: 1.0 3.6\n",
- "Silicon is good conductor at frequency lesser than 1e9 Hz:\n"
- ]
- }
- ],
- "source": [
- "#cal of Penetration depth and frequency\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "import math\n",
- "k = 12 # relative Dielectric constant of sea water\n",
- "epsilon_0 = 1/9e9 # Permittivity of free space\n",
- "sigma = 2 # conductivity in mho/cm\n",
- "mu_0 = 4*math.pi*1e-7 # permeability f free space\n",
- "f= 1e9 # Given frequency in Hz\n",
- "F = 1e6 # Given frequency in Hz\n",
- "print(\"Example 7.7,page no:308\")\n",
- "delta = math.sqrt(2/(2*math.pi*F*mu_0*sigma*100)) # Calculation of frequency\n",
- "print(\"For MHz frequency, Penetration depth in cm:\"),round(F/10**6,1),round(delta*100,1)\n",
- "omega = 2*math.pi*f\n",
- "x = 2*sigma*100/(k*epsilon_0*omega)\n",
- "if x>1:\n",
- " print(\"Silicon is good conductor at frequency lesser than 1e9 Hz:\")\n",
- "# Answer in book is 3.6 cm"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 7.8;page no:309"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 6,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 7.8,page no:309\n",
- "Required frequency in Hz: 436729.2\n",
- "The incident electromagnetic wave is the radio part of spectrum.\n"
- ]
- }
- ],
- "source": [
- "#cal of frequency of incident electromagnetic wave\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "import math\n",
- "mu_0 = 4*math.pi*1e-7 # permeability of free space\n",
- "mu = mu_0 #permeability of silver\n",
- "sigma = 5.8e7 # conductivity in simens /m\n",
- "delta = 0.1 # Skin depth penetration in mm\n",
- "print(\"Example 7.8,page no:309\")\n",
- "f = 2/((delta*1e-3)**2*sigma*mu*2*math.pi) # Calculation of skin depth penetration\n",
- "print(\"Required frequency in Hz:\"),round(f,1)\n",
- "print(\"The incident electromagnetic wave is the radio part of spectrum.\")\n",
- "# Answer in book is 3.36e5 Hz. Difference is due to approximation at intermediate stages"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 7.9;page no:310"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 7,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 7.9,page no:310\n",
- "Skin depth penetration in micrometre: 0.92\n"
- ]
- }
- ],
- "source": [
- " #cal of Skin depth penetration\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "import math\n",
- "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n",
- "mu = mu_0 #Permeability of silver\n",
- "sigma = 3e7 # conductivity in mhos/m\n",
- "f = 1e10 # frequency in Hz\n",
- "print(\"Example 7.9,page no:310\")\n",
- "delta = math.sqrt(1/(math.pi*sigma*f*mu)) # Calculation of skin depth penetration\n",
- "print(\"Skin depth penetration in micrometre:\"),round(delta*1e6,2)\n",
- "# Answer in book is 0.93 micrometer"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "# example 7.10;page no:310"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 8,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 7.10,page no:310\n",
- "Average value of electric field at distance m in Volt/m: 376.7\n",
- "Average value of magnetic field at distance m in Amp-turn/m: 0.106\n"
- ]
- }
- ],
- "source": [
- "#cal of Average value of electric field and magnetic field\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "import math\n",
- "p = 500. # power in watt\n",
- "d = 1. # Distance from lamp in m\n",
- "epsilon_0 = 8.854e-12 # Permittivity of free space\n",
- "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n",
- "print(\"Example 7.10,page no:310\")\n",
- "s = p/(4.*math.pi*d**2)# Calculation of pointing vector\n",
- "E_H_ratio = math.sqrt(mu_0/epsilon_0) # Calculation of ratio of Electric field and magnetic field\n",
- "H = s/E_H_ratio # Calculation of Electric field \n",
- "h = (H*100.)/100. # rounding off for 2 decimal places\n",
- "E= p/(4.*math.pi*h) # Calculation of Electric field \n",
- "print(\"Average value of electric field at distance m in Volt/m:\"),round(E,1)\n",
- "print(\"Average value of magnetic field at distance m in Amp-turn/m:\"),round(h,3)"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 7.11;page no:312"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 9,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 7.11,page no:312\n",
- "Required frequency in MHz: 8.0\n",
- "The incident electromagnetic wave is the radio part of spectrum\n"
- ]
- }
- ],
- "source": [
- "#cal of frequency of incident radiation and location in electromagnetic\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "import math\n",
- "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n",
- "mu = mu_0 #Permeability of silver\n",
- "sigma = 3.5e7 # conductivity in simens /m\n",
- "delta = 0.03 # Skin depth penetration in mm\n",
- "print(\"Example 7.11,page no:312\")\n",
- "f = 2/((delta*1e-3)**2*sigma*mu*2*math.pi) # Calculation of skin depth penetration\n",
- "print(\"Required frequency in MHz:\"),round(f/1e6,0)\n",
- "print(\"The incident electromagnetic wave is the radio part of spectrum\")"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 7.12;page no:312"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 10,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 7.12,page no:312\n",
- "Solar energy received during solar eclipse in Cal per min per m^2: 2.1\n"
- ]
- }
- ],
- "source": [
- "#cal of Solar energy received during solar eclipse\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "import math\n",
- "p = 3.8e26 # power radiated by moon in watt\n",
- "d_sun = 1.44e11 # Distance between sun and earth in meter\n",
- "d_moon = 3e8 #Distance between moon and earth in meter\n",
- "epsilon_0 = 8.854e-12 # Permittivity of free space\n",
- "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n",
- "print(\"Example 7.12,page no:312\")\n",
- "s = p/(4*math.pi*d_sun**2)# Calculation of solar energy received during solar eclipse in watt /m^2\n",
- "S = s*60/(4.2*1e4) # Unit conversion\n",
- "print(\"Solar energy received during solar eclipse in Cal per min per m^2:\"),round(S,1)\n",
- "# Ansewr in book is 2.1 cal per min per m^2"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 7.13;page no:313"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 11,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 7.13,page no:313\n",
- "Skin depth penetration in nm: 3.9\n"
- ]
- }
- ],
- "source": [
- "#cal of Skin depth penetration\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "import math\n",
- "mu_0 = 4*math.pi*1e-7 # Permeability of free space\n",
- "mu = mu_0 #Permeability of silver\n",
- "sigma = 3.5e7 # conductivity in simens /m\n",
- "lamda = 6328 # Wavelength in angstrom\n",
- "c = 3e8# Speed of light in m/sec\n",
- "print(\"Example 7.13,page no:313\")\n",
- "f = c/(lamda*1e-10)\n",
- "omega = 2*math.pi/f # Calculation of time period\n",
- "f = c/(lamda*1e-10) # Calculation of frequency in Hz\n",
- "delta = math.sqrt(1/(math.pi*f*sigma*mu)) # Calculation of skin depth penetration\n",
- "print(\"Skin depth penetration in nm:\"),round(delta*1e9,1)\n",
- "# Answer in book is 3.9 mm, unit used in book is wrong"
- ]
- }
- ],
- "metadata": {
- "kernelspec": {
- "display_name": "Python 2",
- "language": "python",
- "name": "python2"
- },
- "language_info": {
- "codemirror_mode": {
- "name": "ipython",
- "version": 2
- },
- "file_extension": ".py",
- "mimetype": "text/x-python",
- "name": "python",
- "nbconvert_exporter": "python",
- "pygments_lexer": "ipython2",
- "version": "2.7.10"
- }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter8_1.ipynb b/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter8_1.ipynb deleted file mode 100755 index 86e980d9..00000000 --- a/Engineering_Physics_(volume_-_2)_by_B._K._Pandey_and_S._Chaturvedi/chapter8_1.ipynb +++ /dev/null @@ -1,427 +0,0 @@ -{
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "# chapter8 :Superconductivity"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "# example 8.1;page no:331"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 1,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 8.1,page no:331\n",
- "Magnetic Field at K in tesla: 0.0217\n",
- "Standard formula used H_c = H_c_0*(1-(T/T_c)^2).\n"
- ]
- }
- ],
- "source": [
- "#cal of Magnetic field\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "H_c_0= 0.0306# Critical Field in tesla\n",
- "T_c = 3.7 # Critical temperature in kelvin\n",
- "T = 2 # Temperature in kelvin\n",
- "print(\"Example 8.1,page no:331\")\n",
- "H_c = H_c_0*(1-(T/T_c)**2) # Calculation of critical field\n",
- "print(\"Magnetic Field at K in tesla:\"),round(H_c,4)\n",
- "print(\"Standard formula used H_c = H_c_0*(1-(T/T_c)^2).\")"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 8.2;page no:331"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 2,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 8.2,page no:331\n",
- "Magnetic Field at K in A/m: 63737.7049\n",
- "Standard formula used H_c = H_c_0*(1-(T/T_c)^2).\n"
- ]
- }
- ],
- "source": [
- "#cal of Magnetic field\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "H_c= 3.3e4 # # Magnetic field in A/m\n",
- "T_c = 7.2 # Critical temperature in kelvin\n",
- "T = 5 # Temperature in kelvin\n",
- "print(\"Example 8.2,page no:331\")\n",
- "H_c_0 = H_c*(1-(T/T_c)**2)**(-1) # Calculation of critical field\n",
- "print(\"Magnetic Field at K in A/m:\"),round(H_c_0,4)\n",
- "print(\"Standard formula used H_c = H_c_0*(1-(T/T_c)^2).\")"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "# example 8.3;page no:332"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 3,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 8.3,page no:332\n",
- "Required temperature in K: 6.83\n",
- "Standard formula used H_c = H_c_0*(1-(T/T_c)^2).\n"
- ]
- }
- ],
- "source": [
- "#cal of Required temperature\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "import math\n",
- "H_c_0= 1 # Let \n",
- "H_c= 0.1 * H_c_0 # Magnetic field in A/m\n",
- "T_c = 7.2 # Critical temperature in kelvin\n",
- "print(\"Example 8.3,page no:332\")\n",
- "T = T_c*math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n",
- "print(\"Required temperature in K:\"),round(T,2)\n",
- "print(\"Standard formula used H_c = H_c_0*(1-(T/T_c)^2).\")"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 8.4;page no:332"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 4,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 8.4,page no:332\n",
- "Magnetic Field at K in tesla: 0.053\n",
- "Standard formula used H_c = H_c_0*(1-(T/T_c)^2).\n"
- ]
- }
- ],
- "source": [
- "#cal of magnetic field\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "H_c_0= 0.0803# Critical Field in tesla\n",
- "T_c = 7.2 # Critical temperature in kelvin\n",
- "T = 4.2 # Temperature in kelvin\n",
- "print(\"Example 8.4,page no:332\")\n",
- "H_c = H_c_0*(1-(T/T_c)**2) # Calculation of critical field\n",
- "print(\"Magnetic Field at K in tesla:\"),round(H_c,4)\n",
- "print(\"Standard formula used H_c = H_c_0*(1-(T/T_c)^2).\")\n",
- "# Answer in book is 0.0548 tesla"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 8.5;page no:333"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 5,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 8.5,page no:333\n",
- "Required temperature in K: 5.04\n",
- "Standard formula used H_c = H_c_0*(1-(T/T_c)^2).\n"
- ]
- }
- ],
- "source": [
- "#cal of Required temperature\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "import math\n",
- "H_c_0= 1.5e5# Critical field in A/m \n",
- "H_c= 1.05e5 # Magnetic field in A/m\n",
- "T_c = 9.2 # Critical temperature in kelvin\n",
- "print (\"Example 8.5,page no:333\")\n",
- "T = T_c*math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n",
- "print(\"Required temperature in K:\"),round(T,2)\n",
- "print(\"Standard formula used H_c = H_c_0*(1-(T/T_c)^2).\")"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 8.6;page no:333"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 6,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 8.6,page no:333\n",
- "Required temperature in K: 11.3\n",
- "Standard formula used H_c = H_c_0*(1-(T/T_c)^2).\n"
- ]
- }
- ],
- "source": [
- "#cal of Required temperature\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "import math\n",
- "H_c_0= 2e5# Critical field in A/m \n",
- "H_c= 1e5 # Magnetic field in A/m\n",
- "T_c = 8 # Critical temperature in kelvin\n",
- "print(\"Example 8.6,page no:333\")\n",
- "T = T_c/math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n",
- "print(\"Required temperature in K:\"),round(T,1)\n",
- "print(\"Standard formula used H_c = H_c_0*(1-(T/T_c)^2).\")"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 8.7;page no:334"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 7,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 8.7,page no:334\n",
- "Required temperature in K: 7.08\n",
- "Standard formula used H_c = H_c_0*(1-(T/T_c)^2).\n"
- ]
- }
- ],
- "source": [
- "#cal of Required temperature\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "import math\n",
- "H_c_0= 8e5# Critical field in A/m \n",
- "H_c= 4e4 # Magnetic field in A/m\n",
- "T_c = 7.26 # Critical temperature in kelvin\n",
- "print(\"Example 8.7,page no:334\")\n",
- "T = T_c*math.sqrt(1- (H_c/H_c_0)) # Calculation of Temperature\n",
- "print(\"Required temperature in K:\"),round(T,2)\n",
- "print(\"Standard formula used H_c = H_c_0*(1-(T/T_c)^2).\")"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 8.8;page no:335"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 8,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 8.8,page no:335\n",
- "Transition temperature in K: 14.5\n",
- "Critical field at K in T: 4.2 2.731\n",
- "Critical field at 0 K is T:\n",
- "Standard formula used Hc_T = H_c_0*(1-(T/T_c)^2)= 2.5\n"
- ]
- }
- ],
- "source": [
- "#cal of Critical field \n",
- "#intiation of all variables \n",
- "#Given that\n",
- "import math\n",
- "T1 = 14. # Temp in K\n",
- "T2 = 13. # Temp in K\n",
- "T = 4.2 # Temp in K\n",
- "Hc_T1 = 0.176 # Critical field at Temp T1\n",
- "Hc_T2 = 0.528 # Critical field at Temp T2\n",
- "print(\"Example 8.8,page no:335\")\n",
- "T_c = math.sqrt((T1**2*(Hc_T2/Hc_T1)- T2**2) /(Hc_T2/Hc_T1 - 1)) # Calculation of transition temperature\n",
- "t_c = (T_c*10)/10 # Rounding off two two decimal places\n",
- "Hc_0 = Hc_T1/(1-(T1/t_c)**2) # Calculation of critical field\n",
- "Hc_T = Hc_0*(1-(T/t_c)**2) # Calculation of critical field \n",
- "print(\"Transition temperature in K:\"),round(t_c,1)\n",
- "print(\"Critical field at K in T:\"),round(T,1),round(Hc_0,3)\n",
- "print(\"Critical field at 0 K is T:\")\n",
- "print(\"Standard formula used Hc_T = H_c_0*(1-(T/T_c)^2)=\"),round(Hc_T,2)\n",
- "# Answer in book is 2.588 T for 0 K and 2.37 for 4.2 K"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 8.9;page no:343"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 9,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 8.9,page no:343\n",
- "Penetration depth at K in angestrom: 532.0\n",
- "Penetration depth at K in angestrom: 535.31\n",
- "Standard formula used lambda_0 = sqrt(m_0/(mu_0*eta_s*e**2)).\n"
- ]
- }
- ],
- "source": [
- "#cal of Penetration depth\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "import math\n",
- "m_0 = 9.1e-31 # Mass of electron in kg\n",
- "mu_0 = 12.56e-7# SI\n",
- "e = 1.6e-19 # Charge on electron in coulomb\n",
- "eta_s = 1e28 # superelectron density in no. per cube\n",
- "T_1 = 0. # First temp in kelvin\n",
- "T_2 = 1. # Second temp in kelvin\n",
- "T_c = 3. # Critical temp in kelvin\n",
- "print(\"Example 8.9,page no:343\")\n",
- "lamda_0 = math.sqrt(m_0/(mu_0*eta_s*e**2))# Calculation of penetration depth at 0K\n",
- "lamda_t = lamda_0/math.sqrt(1-(T_2/T_c)**4) # Calculation of penetration depth at 2K\n",
- "print(\"Penetration depth at K in angestrom:\"),round(lamda_0*1e10,0)\n",
- "print(\"Penetration depth at K in angestrom:\"),round(lamda_t*1e10,2)\n",
- "print(\"Standard formula used lambda_0 = sqrt(m_0/(mu_0*eta_s*e**2)).\")"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## example 8.10;page no:344"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 10,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 8.10,page no:344\n",
- "Penetration depth at 0 K in angstrom: 527.96\n",
- "Standard formula used lamda_0 = lamda_t*sqrt(1-(T_1/T_c)**4).\n"
- ]
- }
- ],
- "source": [
- "#cal of Penetration depth\n",
- "#intiation of all variables \n",
- "#Given that\n",
- "import math\n",
- "T_1 = 3.5 # Temperature in kelvin\n",
- "T_c = 4.153 # Critical temp in kelvin\n",
- "lamda_t = 750 # Penetration depth at T_1 in angstrom\n",
- "print(\"Example 8.10,page no:344\")\n",
- "lamda_0 = lamda_t*math.sqrt(1-(T_1/T_c)**4) # Calculation of penetration depth at 3.5K\n",
- "print(\"Penetration depth at 0 K in angstrom:\"),round(lamda_0,2)\n",
- "print(\"Standard formula used lamda_0 = lamda_t*sqrt(1-(T_1/T_c)**4).\")"
- ]
- }
- ],
- "metadata": {
- "kernelspec": {
- "display_name": "Python 2",
- "language": "python",
- "name": "python2"
- },
- "language_info": {
- "codemirror_mode": {
- "name": "ipython",
- "version": 2
- },
- "file_extension": ".py",
- "mimetype": "text/x-python",
- "name": "python",
- "nbconvert_exporter": "python",
- "pygments_lexer": "ipython2",
- "version": "2.7.10"
- }
- },
- "nbformat": 4,
- "nbformat_minor": 0
-}
diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch11_1.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch11_1.ipynb new file mode 100644 index 00000000..88260971 --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch11_1.ipynb @@ -0,0 +1,759 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 11 Digital To Analog Converter(D/A Converter)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.1 Pg 313" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the full scale voltage of D/A converter VFS is = 24.00 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt, pi\n", + "# to determine the full scale voltage of D/A\n", + "Vref = 12 #\n", + "Rf = 10 # # K ohm\n", + "R = 5 # # K ohm\n", + "\n", + "# the full scale voltage of D/A converter \n", + "VFS = Vref*(Rf/R) #\n", + "print 'the full scale voltage of D/A converter VFS is = %0.2f'%VFS,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.2 Pg 313" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "For BI 10101010 the output of D/A converter is = 2.04 V \n", + "For BI 11001100 the output of D/A converter is = 2.448 V \n", + "For BI 11101110 the output of D/A converter is = 2.856 V \n", + "For BI 00010001 the output of D/A converter is = 0.204 V \n" + ] + } + ], + "source": [ + " # determine the output voltage of D/A converter for the binary inputs a) 10101010 b) 11001100 c) 11101110 d) 00010001 \n", + "Del = 12*10**-3 # # mA\n", + "\n", + "# the input voltage of D/A converter \n", + " #Vo = Del*binary input (BI)\n", + "\n", + "# For BI 10101010 the output\n", + "BI = '10101010' #\n", + "BI = int(BI,2)#\n", + "Vo = Del*BI #\n", + "print 'For BI 10101010 the output of D/A converter is = ',Vo,' V '\n", + "\n", + "# For BI 11001100 the output\n", + "BI = '11001100' #\n", + "BI = int(BI,2)#\n", + "Vo = Del*BI #\n", + "print 'For BI 11001100 the output of D/A converter is = ',Vo,' V '\n", + "\n", + "# For BI 11101110 the output\n", + "BI = '11101110' #\n", + "BI = int(BI,2)#\n", + "Vo = Del*BI #\n", + "print 'For BI 11101110 the output of D/A converter is = ',Vo,' V '\n", + "\n", + "# For BI 00010001 the output\n", + "BI = '00010001' #\n", + "BI = int(BI,2)#\n", + "Vo = Del*BI #\n", + "print 'For BI 00010001 the output of D/A converter is = ',Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.3 Pg 314" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the resolution of 4-bit D/A converter is = 0.80 V \n" + ] + } + ], + "source": [ + "# determine the resolution of 4-bit D/A converter\n", + "VFS = 12.0 #\n", + "N = 4.0 #\n", + "\n", + "# the resolution of 4-bit D/A converter is defined as\n", + "Resolution = VFS/(2**N-1) #\n", + "print 'the resolution of 4-bit D/A converter is = %0.2f'%Resolution,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.4 314" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the number of bit required to design a 4-bit D/A converter is = 8.97 = 9 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log10\n", + "# determine the number of bit required to design a 4-bit D/A converter\n", + "VFS = 5 #\n", + "Resolution = 10*10**-3 # # A\n", + "\n", + "# the resolution of 4-bit D/A converter is defined as\n", + "# Resolution = VFS/(2**N-1) #\n", + "N = (VFS/Resolution)+1 # \n", + "N = log10(N)/log10(2)#\n", + "print 'the number of bit required to design a 4-bit D/A converter is = %0.2f'%N,' = 9 '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.5 Pg 315" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "for the binary input 101 analog output is = -60.00 V \n", + "for the binary input 111 analog output is = -84.00 V \n", + "for the binary input 011 analog output is = -72.00 V \n", + "for the binary input 001 analog output is = -48.00 V \n", + "for the binary input 100 analog output is = -12.00 V\n" + ] + } + ], + "source": [ + "# determine the analog output voltage\n", + "Vref = 12 # \n", + "BI = 101 # BI = 111 # BI = 011 # BI = 001 # BI = 100 #\n", + "Rf = 40*10**3 #\n", + "R = 0.25*Rf #\n", + "\n", + "# The output voltage of given binary weighted resistor D/A converter is defined as\n", + "\n", + "# Vo = -(Rf*Vref/R)*(2**0*b0+2**-1*b1+2**-2*b2) #\n", + "\n", + "# Vo = -(Rf*Vref/R)*(b0+2**-1*b1+2**-2*b2) #\n", + "\n", + "# for the given value Rf,R and Vref the output voltage\n", + "\n", + "# Vo = -48*(b0+2**-1*b1+2**-2*b2) #\n", + "\n", + "# for the binary input 101 analog output is\n", + "b2 = 1 #\n", + "b1 = 0 #\n", + "b0 = 1 #\n", + "Vo = -48*(b0+2**-1*b1+2**-2*b2) #\n", + "print 'for the binary input 101 analog output is = %0.2f'%Vo,' V '\n", + "\n", + "\n", + "# for the binary input 111 analog output is\n", + "b2 = 1 #\n", + "b1 = 1 #\n", + "b0 = 1 #\n", + "Vo = -48*(b0+2**-1*b1+2**-2*b2) #\n", + "print 'for the binary input 111 analog output is = %0.2f'%Vo,' V '\n", + "\n", + "\n", + "# for the binary input 011 analog output is\n", + "b2 = 0 #\n", + "b1 = 1 #\n", + "b0 = 1 #\n", + "Vo = -48*(b0+2**-1*b1+2**-2*b2) #\n", + "print 'for the binary input 011 analog output is = %0.2f'%Vo,' V '\n", + "\n", + "\n", + "# for the binary input 001 analog output is\n", + "b2 = 0 #\n", + "b1 = 0 #\n", + "b0 = 1 #\n", + "Vo = -48*(b0+2**-1*b1+2**-2*b2) #\n", + "print 'for the binary input 001 analog output is = %0.2f'%Vo,' V '\n", + "\n", + "\n", + "# for the binary input 100 analog output is\n", + "b2 = 1 #\n", + "b1 = 0 #\n", + "b0 = 0 #\n", + "Vo = -48*(b0+2**-1*b1+2**-2*b2) #\n", + "print 'for the binary input 100 analog output is = %0.2f'%Vo,' V'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.6 Pg 316" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "for the binary input 1001 analog output is = -67.50 V \n", + "the feedback current If is = 2.70 mA \n", + "for the binary input 1101 analog output is = -82.50 V \n", + "the feedback current If is = 3.30 mA \n", + "for the binary input 1010 analog output is = -37.50 V \n", + "the feedback current If is = 1.50 mA \n", + "for the binary input 0011 analog output is = -90.00 V \n", + "the feedback current If is = 3.60 mA \n" + ] + } + ], + "source": [ + "# determine the analog output voltage and feed back current If\n", + "Vref = 12 # \n", + "BI = 1001 # BI = 1101 # BI = 1010 # BI = 0011 #\n", + "Rf = 25 # # K ohm\n", + "R = 0.25*Rf #\n", + "\n", + "# The output voltage of given binary weighted resistor D/A converter is defined as\n", + "\n", + "# Vo = -(Rf*Vref/R)*(2**0*b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "\n", + "# Vo = -(Rf*Vref/R)*(b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "\n", + "# for the given value Rf,R and Vref the output voltage\n", + "\n", + "# Vo = -60*(b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "\n", + "# for the binary input 1001 analog output is\n", + "b3 = 1 #\n", + "b2 = 0 #\n", + "b1 = 0 #\n", + "b0 = 1 #\n", + "Vo = -60*(b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "print 'for the binary input 1001 analog output is = %0.2f'%Vo,' V '\n", + "\n", + "# the feedback current If is given by\n", + "If = -(Vo/Rf) #\n", + "print 'the feedback current If is = %0.2f'%If,' mA '\n", + "\n", + "\n", + "# for the binary input 1101 analog output is\n", + "b3 = 1 #\n", + "b2 = 1 #\n", + "b1 = 0 #\n", + "b0 = 1 #\n", + "Vo = -60*(b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "print 'for the binary input 1101 analog output is = %0.2f'%Vo,' V '\n", + "\n", + "# the feedback current If is given by\n", + "If = -(Vo/Rf) #\n", + "print 'the feedback current If is = %0.2f'%If,' mA '\n", + "\n", + "\n", + "# for the binary input 1010 analog output is\n", + "b3 = 1 #\n", + "b2 = 0 #\n", + "b1 = 1 #\n", + "b0 = 0 #\n", + "Vo = -60*(b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "print 'for the binary input 1010 analog output is = %0.2f'%Vo,' V '\n", + "\n", + "# the feedback current If is given by\n", + "If = -(Vo/Rf) #\n", + "print 'the feedback current If is = %0.2f'%If,' mA '\n", + "\n", + "\n", + "# for the binary input 0011 analog output is\n", + "b3 = 0 #\n", + "b2 = 0 #\n", + "b1 = 1 #\n", + "b0 = 1 #\n", + "Vo = -60*(b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "print 'for the binary input 0011 analog output is = %0.2f'%Vo,' V '\n", + "\n", + "# the feedback current If is given by\n", + "If = -(Vo/Rf) #\n", + "print 'the feedback current If is = %0.2f'%If,' mA '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.7 Pg 319" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "for the binary input 001 analog output is = 1.60 mA \n", + "An analog output voltage Vo is = -40.00 V \n", + "for the binary input 010 analog output is = 0.80 mA\n", + "An analog output voltage Vo is = -20.00 V \n", + "for the binary input 110 analog output is = 1.20 mA \n", + "An analog output voltage Vo is = -30.00 V \n" + ] + } + ], + "source": [ + "# determine the feed back current If and analog output voltage\n", + "Vref = 8 # # V\n", + "BI = 001 #\n", + "BI = 010 #\n", + "BI = 110 #\n", + "Rf = 25*10**3 # # Hz\n", + "R = 0.2*Rf #\n", + "\n", + "# The output current of given binary weighted resistor D/A converter is defined as\n", + "\n", + "# If = -(Vref/R)*(2**0*b0+2**-1*b1+2**-2*b2) #\n", + "\n", + "# If = -(Vref/R)*(b0+2**-1*b1+2**-2*b2) #\n", + "\n", + "# for the given value Rf,R and Vref the output current\n", + "\n", + "# If = -(1.6*10**-3)*(b0+2**-1*b1+2**-2*b2) #\n", + "\n", + "# for the binary input 001 the feedback current If is given by\n", + "b2 = 0 #\n", + "b1 = 0 #\n", + "b0 = 1 #\n", + "If = (1.6*10**-3)*(b0+2**-1*b1+2**-2*b2) #\n", + "print 'for the binary input 001 analog output is = %0.2f'%(If*1000),' mA '\n", + "\n", + "# An analog output voltage Vo is\n", + "Vo = -If*Rf #\n", + "print 'An analog output voltage Vo is = %0.2f'%Vo,' V '\n", + "\n", + "\n", + "# for the binary input 010 the feedback current If is given by\n", + "b2 = 0 #\n", + "b1 = 1 #\n", + "b0 = 0 #\n", + "If = (1.6*10**-3)*(b0+2**-1*b1+2**-2*b2) #\n", + "print 'for the binary input 010 analog output is = %0.2f'%(If*1000),' mA'\n", + "\n", + "# the An analog output voltage Vo is\n", + "Vo = -If*Rf #\n", + "print 'An analog output voltage Vo is = %0.2f'%Vo,' V '\n", + "\n", + "\n", + "# for the binary input 110 the feedback current If is given by\n", + "b2 = 1 #\n", + "b1 = 1 #\n", + "b0 = 0 #\n", + "If = (1.6*10**-3)*(b0+2**-1*b1+2**-2*b2) #\n", + "print 'for the binary input 110 analog output is = %0.2f'%(If*1000),' mA '\n", + "\n", + "# the An analog output voltage Vo is\n", + "Vo = -If*Rf #\n", + "print 'An analog output voltage Vo is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.8 Pg 320" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "for the binary input 101 analog output is = 0.625 mA \n", + "An analog output voltage Vo is = -15.625 V \n", + "for the binary input 011 analog output is = 0.75 mA\n", + "An analog output voltage Vo is = -18.750 V \n", + "for the binary input 100 analog output is = 0.125 mA \n", + "An analog output voltage Vo is = -3.12 V \n", + "for the binary input 001 analog output is = 0.5 mA \n", + "An analog output voltage Vo is = -12.50 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# determine the feed back current If and analog output voltage\n", + "Vref = 5 # \n", + "BI = 101 # BI = 011 # BI = 100 # BI = 001 #\n", + "Rf = 25*10**3 # \n", + "R = 0.2*Rf #\n", + "\n", + "# The output current of given R-2R ladder D/A converter is defined as\n", + "\n", + "# If = -(Vref/2*R)*(2**0*b0+2**-1*b1+2**-2*b2) #\n", + "\n", + "# If = -(Vref/2*R)*(b0+2**-1*b1+2**-2*b2) #\n", + "\n", + "# for the given value Rf,R and Vref the output current\n", + "\n", + "# If = (0.5*10**-3)*(b0+2**-1*b1+2**-2*b2) #\n", + "\n", + "# for the binary input 101 the feedback current If is given by\n", + "b2 = 1 #\n", + "b1 = 0 #\n", + "b0 = 1 #\n", + "If = (0.5*10**-3)*(b0+2**-1*b1+2**-2*b2) #\n", + "print 'for the binary input 101 analog output is = %0.3f'%(If*1e3),' mA '\n", + "\n", + "# An analog output voltage Vo is\n", + "Vo = -If*Rf #\n", + "print 'An analog output voltage Vo is = %0.3f'%Vo,' V '\n", + "\n", + "\n", + "# for the binary input 011 the feedback current If is given by\n", + "b2 = 0 #\n", + "b1 = 1 #\n", + "b0 = 1 #\n", + "If = (0.5*10**-3)*(b0+2**-1*b1+2**-2*b2) #\n", + "print 'for the binary input 011 analog output is = %0.2f'%(If*1e3),' mA'\n", + "\n", + "# the An analog output voltage Vo is\n", + "Vo = -If*Rf #\n", + "print 'An analog output voltage Vo is = %0.3f'%Vo,' V '\n", + "\n", + "\n", + "# for the binary input 100 the feedback current If is given by\n", + "b2 = 1 #\n", + "b1 = 0 #\n", + "b0 = 0 #\n", + "If = (0.5*10**-3)*(b0+2**-1*b1+2**-2*b2) #\n", + "print 'for the binary input 100 analog output is = %0.3f'%(If*1e3),'mA '\n", + "\n", + "# the An analog output voltage Vo is\n", + "Vo = -If*Rf #\n", + "print 'An analog output voltage Vo is = %0.2f'%Vo,' V '\n", + "\n", + "# for the binary input 001 the feedback current If is given by\n", + "b2 = 0 #\n", + "b1 = 0 #\n", + "b0 = 1 #\n", + "If = (0.5*10**-3)*(b0+2**-1*b1+2**-2*b2) #\n", + "print 'for the binary input 001 analog output is = %0.1f'%(If*1e3),' mA '\n", + "\n", + "# the An analog output voltage Vo is\n", + "Vo = -If*Rf #\n", + "print 'An analog output voltage Vo is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.9 Pg 322" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "for the binary input 1001 analog output is = -14.0625 V \n", + "the feedback current If is = 0.28 mA \n", + "for the binary input 1100 analog output is = -4.6875 V \n", + "the feedback current If is = 0.09 mA \n", + "for the binary input 1010 analog output is = -7.8125 V \n", + "the feedback current If is = 0.16 mA \n", + "for the binary input 0011 analog output is = -18.75 V \n", + "the feedback current If is = 0.375 mA \n" + ] + } + ], + "source": [ + " # determine the analog output voltage and feed back current If\n", + "Vref = 10 # \n", + "BI = 1001 # BI = 1100 # BI = 1010 # BI = 0011 #\n", + "Rf = 50 # # K ohm\n", + "R = 0.4*Rf #\n", + "\n", + "# The output voltage of given R-2R ladder D/A converter is defined as\n", + "\n", + "# Vo = -(Rf*Vref/2R)*(2**0*b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "\n", + "# Vo = -(Rf*Vref/2R)*(b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "\n", + "# for the given value Rf,R and Vref the output voltage\n", + "\n", + "# Vo = -12.5*(b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "\n", + "# for the binary input 1001 analog output is\n", + "b3 = 1 #\n", + "b2 = 0 #\n", + "b1 = 0 #\n", + "b0 = 1 #\n", + "Vo = -12.5*(b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "print 'for the binary input 1001 analog output is = %0.4f'%Vo,' V '\n", + "\n", + "# the feedback current If is given by\n", + "If = -(Vo/Rf) #\n", + "print 'the feedback current If is = %0.2f'%If,' mA '\n", + "\n", + "\n", + "# for the binary input 1100 analog output is\n", + "b3 = 1 #\n", + "b2 = 1 #\n", + "b1 = 0 #\n", + "b0 = 0 #\n", + "Vo = -12.5*(b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "print 'for the binary input 1100 analog output is = %0.4f'%Vo,' V '\n", + "\n", + "# the feedback current If is given by\n", + "If = -(Vo/Rf) #\n", + "print 'the feedback current If is = %0.2f'%If,' mA '\n", + "\n", + "\n", + "# for the binary input 1010 analog output is\n", + "b3 = 1 #\n", + "b2 = 0 #\n", + "b1 = 1 #\n", + "b0 = 0 #\n", + "Vo = -12.5*(b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "print 'for the binary input 1010 analog output is = %0.4f'%Vo,' V '\n", + "\n", + "# the feedback current If is given by\n", + "If = -(Vo/Rf) #\n", + "print 'the feedback current If is = %0.2f'%If,' mA '\n", + "\n", + "\n", + "# for the binary input 0011 analog output is\n", + "b3 = 0 #\n", + "b2 = 0 #\n", + "b1 = 1 #\n", + "b0 = 1 #\n", + "Vo = -12.5*(b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "print 'for the binary input 0011 analog output is = %0.2f'%Vo,' V '\n", + "\n", + "# the feedback current If is given by\n", + "If = -(Vo/Rf) #\n", + "print 'the feedback current If is = %0.3f'%If,' mA '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.10 Pg 324" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "for the binary input 1000 output voltage is = 18.75 V \n", + "the feedback current If is = -0.469 mA \n" + ] + } + ], + "source": [ + " # determine the analog output voltage and feed back current If\n", + "Vref = 15 # \n", + "BI = 1000 #\n", + "Rf = 40 # # K ohm\n", + "R = 0.4*Rf #\n", + "\n", + "# by using voltage divider rule Vin can be calculated as\n", + "Vin = -(Vref*2*R)/(2*R+2*R) #\n", + " \n", + "# The output voltage of given R-2R ladder D/A converter is defined as\n", + "\n", + "# Vo = -(Rf*Vin/R)\n", + "\n", + "Vo = (Vref*Rf)/(2*R)\n", + "print 'for the binary input 1000 output voltage is = %0.2f'%Vo,' V '\n", + "\n", + "# the feedback current If is given by\n", + "If = -(Vo/Rf) #\n", + "print 'the feedback current If is = %0.3f'%If,' mA '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.11 Pg 326" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "For the BI 10101111 output analog voltage is = 6.86 V \n", + "For the BI 11100010 output analog voltage is = 8.8627 V \n", + "For the BI 00101001 output analog voltage is = 1.6078 V \n", + "For the BI 01000110 output analog voltage is = 2.745 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# to find the resolution and analog output voltage of 8-bit D/A converter\n", + "VFS = 10 #\n", + "N = 8 #\n", + "BI = 10101111 #\n", + "BI = 11100011 # \n", + "BI = 00101001 #\n", + "BI = 01000110\n", + "\n", + "# the resolution of 8-bit D/A converter is defined as\n", + "Resolution = VFS/(2**N-1) #\n", + "\n", + "# An analog output voltage of D/A converter is given by\n", + "# Vo = Resolution*(2**-0*b0+2**-1*b1+....+2**-N*bn-1)\n", + "# Vo = Resolution*(2**-0*b0+2**-1*b1+2**-2*b2+2**-3*b3+2**-4*b4+2**-5*b5+2**-6*b6+2**-7*b7)#\n", + "\n", + "# For the BI 10101111 output analog voltage is\n", + "BI = '10101111'#\n", + "BI = int(BI,2)#\n", + "Vo = Resolution*BI #\n", + "print 'For the BI 10101111 output analog voltage is = %0.2f'%Vo,' V '\n", + "\n", + "# For the BI 11100010 output analog voltage is\n", + "BI = '11100010'#\n", + "BI = int(BI,2)#\n", + "Vo = Resolution*BI #\n", + "print 'For the BI 11100010 output analog voltage is = %0.4f'%Vo,' V '\n", + "\n", + "# For the BI 00101001 output analog voltage is\n", + "BI = '00101001'#\n", + "BI = int(BI,2)#\n", + "Vo = Resolution*BI #\n", + "print 'For the BI 00101001 output analog voltage is = %0.4f'%Vo,' V '\n", + "\n", + "# For the BI 01000110 output analog voltage is\n", + "BI = '01000110'#\n", + "BI = int(BI,2)#\n", + "Vo = Resolution*BI #\n", + "print 'For the BI 01000110 output analog voltage is = %0.3f'%Vo,' V '" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch12_1.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch12_1.ipynb new file mode 100644 index 00000000..33e95d2e --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch12_1.ipynb @@ -0,0 +1,232 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 12 Analog To Digital Converter(A/D Converter)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.2 Pg 350" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Normalized step size of A/D converter is = 0.00390625\n", + "Actual step size of A/D converter is = 0.046875\n", + "Normalized maximum quantization level of A/D converter is = 0.9961\n", + "Actual maximum quantization level of A/D converter is = 11.9531\n", + "Normalized peak quantization error of A/D converter is = 0.001953\n", + "Actual peak quantization error of A/D converter is = 0.023438 V \n", + "Percentage of quantization error of A/D converter is = 0.1953\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Determine the following parameter of 8-bit A/D converter a) Normalized step size b) Actual step size c) Normalized maximum quantization level d) Actual maximum quantization e) Normalized peak quantization error f) Actual peak quantization error g) Percentage of quantization error\n", + "N = 8 #\n", + "Vin = 12 #\n", + "\n", + "#a) Normalized step size of A/D converter\n", + "Ns = 2**-N #\n", + "print 'Normalized step size of A/D converter is = %0.8f'%Ns\n", + "\n", + "# b) Actual step size of A/D converter\n", + "As = Vin*Ns #\n", + "print 'Actual step size of A/D converter is = %0.6f'%As\n", + "\n", + "# c) Normalized maximum quantization level of A/D converter\n", + "Qmax = 1-2**-N #\n", + "print 'Normalized maximum quantization level of A/D converter is = %0.4f'%Qmax\n", + "\n", + "# d) Actual maximum quantization level of A/D converter\n", + "QAmax = Qmax*Vin #\n", + "print 'Actual maximum quantization level of A/D converter is = %0.4f'%QAmax\n", + "\n", + "# e) Normalized peak quantization error of A/D converter\n", + "Qp = 2**-(N+1)#\n", + "print 'Normalized peak quantization error of A/D converter is = %0.6f'%Qp\n", + "\n", + "# f) Actual peak quantization error of A/D converter\n", + "Qe = Qp*Vin #\n", + "print 'Actual peak quantization error of A/D converter is = %0.6f'%Qe,' V '#\n", + "\n", + "# g) Percentage of quantization error of A/D converter\n", + "per_Qp = 2**-(N+1)*100 #\n", + "print 'Percentage of quantization error of A/D converter is = %0.4f'%per_Qp" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.3 Pg 351" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "charging time of capacitor is = 128.00 u sec\n", + "the integrator output is = -5.44 V\n", + "the decimal output of a dual slope A/D converter is = 217.60 = 218\n", + "The binary output of a dual slope A/D converter is = 11011010\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# to determine the binary output of the 8-bit dual slope A/D converter\n", + "Vin = 8.5 #\n", + "VR = 10 #\n", + "f = 2 # #MHz\n", + "N = 8 #\n", + "C = 0.1*10**-6 #\n", + "R = 2*10**3 #\n", + "\n", + "# the output of integrator is defined as \n", + "# Viao(T1) = -(Vin/R*C)*T1 #\n", + "\n", + "# charging time of capacitor \n", + "T1 = 2**N/f #\n", + "print 'charging time of capacitor is = %0.2f'%T1,' u sec'\n", + "\n", + "# the integrator output\n", + "T1 = T1*10**-6 #\n", + "Viao =-(Vin/(R*C))*T1#\n", + "print 'the integrator output is = %0.2f'%Viao,' V'\n", + "\n", + "# the binary output of a dual slope A/D converter\n", + "Bn = (2**N*Vin)/VR#\n", + "print 'the decimal output of a dual slope A/D converter is = %0.2f'%Bn,' = 218'#\n", + "\n", + "Bn=218#\n", + "Bn = bin(Bn) #\n", + "print 'The binary output of a dual slope A/D converter is =',Bn[2:]" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.4 Pg 352" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resolution of an A/D converter is = 0.0037 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# to determine the resolution of 12-bit A/D converter\n", + "N =12 #\n", + "Vin = 15 #\n", + "\n", + "# Resolution of an A/D converter\n", + "Resolution = Vin/(2**N-1)#\n", + "print 'Resolution of an A/D converter is = %0.4f'%Resolution,' V '\n", + "#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.5 Pg 352" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output time of a V/T converter is = 0.15 msec\n", + "The duty cycle of V?T converter is = 2.00\n", + "The output voltage of an integrator is is = -0.75 V\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# to determine the output time and duty cycle of V/T converter\n", + "Vin = 5 #\n", + "C = 0.1*10**-6 # \n", + "R = 10*10**3 #\n", + "C1 = 100*10**-6 #\n", + "\n", + "# The output time of a V/T converter is given as\n", + "T = (7.5*C1)/(Vin) #\n", + "print 'The output time of a V/T converter is =',T*1000,' msec'\n", + "\n", + "TH = 0.075 #\n", + "TL=TH # # we consider\n", + "# The duty cycle of V?T converter\n", + "D = (TL+TH)/(TH) #\n", + "print 'The duty cycle of V?T converter is = %0.2f'%D\n", + "\n", + "# The output voltage of an integrator is define as\n", + "Vio = -(Vin)/(R*C)*T #\n", + "print 'The output voltage of an integrator is is = %0.2f'%Vio,' V'" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch13_1.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch13_1.ipynb new file mode 100644 index 00000000..3ae65770 --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch13_1.ipynb @@ -0,0 +1,448 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 13 Waveform Generators" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.1 Pg 378" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the frequency selective element resistor is = 6.50 K ohm \n", + "The feedback resistance is = 188.4 K ohm\n" + ] + } + ], + "source": [ + "from math import sqrt, pi\n", + "from __future__ import division\n", + "# to design RC phase shift oscillator for the oscillation frequency f = 1 KHz\n", + "f =1 # # KHz\n", + "C = 0.01 # # uF\n", + "\n", + "# The oscillation frequency of practical RC phase shift oscillator is defined as\n", + "#w = 1/(sqrt(6)*R*C)#\n", + "\n", + "# gain of practical RC phase shift oscillator is\n", + "#A = R1/R = 29 equation 1\n", + "# the frequency selective element resistor\n", + "#R = 1/(sqrt(6)*w*C)#\n", + "R = 1/(sqrt(6)*2*pi*f*C)#\n", + "print 'the frequency selective element resistor is = %0.2f'%R,' K ohm '\n", + "\n", + "# The feedback resistance\n", + "R1 = 29*R # # from equation 1\n", + "print 'The feedback resistance is = %0.1f'%R1,' K ohm'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.2 Pg 379" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the oscillator frequency of practical RC phase shift oscillator f is = 0.52 KHz \n" + ] + } + ], + "source": [ + "from math import sqrt, pi\n", + "from __future__ import division\n", + "# to determine the oscillaton frequency of the phase shift oscillator\n", + "C = 0.05 # # uF\n", + "R = 2.5 # # K ohm\n", + "\n", + "# the oscillator frequency of practical RC phase shift oscillator f\n", + "f = 1/(2*pi*(sqrt(6)*(R*C)))#\n", + "print 'the oscillator frequency of practical RC phase shift oscillator f is = %0.2f'%f,' KHz '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.3 Pg 380" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the oscillator frequency of practical RC phase shift oscillator f is = 6.63 kHz \n" + ] + } + ], + "source": [ + "from math import sqrt, pi\n", + "from __future__ import division\n", + "# to calculate the frequency of a wein bridge oscillator\n", + "C = 2400*10**-12 # # F\n", + "R = 10*10**3 # # ohm\n", + "\n", + "# the oscillator frequency of practical RC phase shift oscillator f\n", + "f = 1/(2*pi*R*C)/1e3#\n", + "print 'the oscillator frequency of practical RC phase shift oscillator f is = %0.2f'%f,' kHz '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.4 Pg 380" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the resistor R is = 15.9 K ohm \n", + "The resistor R2 value is = 20.00 K ohm \n" + ] + } + ], + "source": [ + "from math import sqrt, pi\n", + "from __future__ import division\n", + "# to design the wien bridge oscillator for the oscillation frequency f = 1 KHz\n", + "f = 1 # # K ohm\n", + "C = 0.01 # # uF\n", + "\n", + "\n", + "# the frequency f is define as\n", + "# f = 1/(2*pi*R*C)#\n", + "\n", + "# the resistor R is\n", + "R = 1/(2*pi*f*C)#\n", + "print 'the resistor R is = %0.1f'%R,' K ohm '\n", + "\n", + "# the loop gain of the wien bridge oscillator is unity which is defined as\n", + "# A = (1+(R2/R1))*(1/3) = 1 #\n", + "# R2/R1 = 2 #\n", + "R1 = 10 # # K ohm we assume\n", + "R2 = 2*R1 #\n", + "print 'The resistor R2 value is = %0.2f'%R2,' K ohm '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.5 Pg 382" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the frequency of wien bridge oscillator f is = 159.155 Hz \n" + ] + } + ], + "source": [ + "from math import sqrt, pi\n", + "from __future__ import division\n", + "# to calculate the frequency of a wein bridge oscillator\n", + "C = 0.05*10**-6 # # F\n", + "R = 20*10**3 # # ohm\n", + "R1 = 10*10**3 # # ohm\n", + "R2 = 20*10**3 # #ohm\n", + "\n", + "# the frequency of wien bridge oscillator f\n", + "f = 1/(2*pi*R*C)#\n", + "print 'the frequency of wien bridge oscillator f is = %0.3f'%f,' Hz '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.6 Pg 382" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The frequency of the astable multivibrator is = 0.87/(C*R)\n" + ] + } + ], + "source": [ + "from sympy import symbols, log, N\n", + "R, C = symbols('R C')\n", + "# Determine the frequency response of the astable multivibrator circuit\n", + "Vsat = 2.5 #\n", + "VT = 0.7 #\n", + "\n", + "# The frequency of the astable multivibrator is\n", + "f = (1/(2*R*C*log((Vsat+VT)/(Vsat-VT))))#\n", + "print 'The frequency of the astable multivibrator is =',N(f,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.7 Pg 383" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of resistance R2 is = 32.86 K ohm \n", + "The value of resistor R is = 10.52 K ohm\n" + ] + } + ], + "source": [ + "from math import log\n", + "# Design astable multivibrator for the frequency f = 10 KHz\n", + "f = 10 # # K ohm\n", + "Vsat = 3 #\n", + "VT = 0.7 #\n", + "\n", + "# The saturation voltage of an astable multivibrator is defined as\n", + "# Vsat = (R1+R2/R1)+VT #\n", + "R1 = 10 # # K ohm we choose\n", + "R2 = ((Vsat/VT)-1)*R1 #\n", + "print 'The value of resistance R2 is = %0.2f'%R2,' K ohm '\n", + "\n", + "# The frequency of an astable multivibrator is defined as\n", + "C = 0.01 # # uF\n", + "# f = (1/(2*R*C*log(1+(2*R1/R2))))#\n", + "\n", + "R = 1/(2*f*C*log(1+2*R1/R2))#\n", + "print 'The value of resistor R is = %0.2f'%R,' K ohm'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.8 Pg 384" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of resistor R is = 200.00 ohm \n" + ] + } + ], + "source": [ + "# to design astable multivibrator \n", + "f = 25*10**3 #\n", + "\n", + "# The output frequency of practical astable multivibrator is defined as\n", + "# f = 1/(2*R*C)#\n", + "C = 0.1*10**-6 # # uF we choose\n", + "R = 1/(2*f*C)#\n", + "print 'The value of resistor R is = %0.2f'%R,'ohm '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.9 Pg 385" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of resistance R is = 579.7 ohm \n", + "The value of resistance R2 is = 9.94 K ohm \n" + ] + } + ], + "source": [ + "# Design a monostable circuit with frequency f = 25 KHz\n", + "f =25*10**3 # # Hz\n", + "\n", + "# The output frequency of monostable multivibrator is defined as \n", + "# f = 1/(0.69*R*C)#\n", + "C = 0.1*10**-6 #\n", + "R = 1/(0.69*f*C)#\n", + "print 'The value of resistance R is = %0.1f'%R,' ohm '\n", + "\n", + "# In the practical monostable multivibrator\n", + "# ln(1+(R2/R1))= 0.69 #\n", + "R1 = 10*10**3 # # we choose\n", + "R2 = R1*(1.99372-1)#\n", + "print 'The value of resistance R2 is = %0.2f'%(R2/1000),' K ohm ' # Round Off Error " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.10 Pg 386" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output of monostable multivibrator is = 6.01 kHz\n" + ] + } + ], + "source": [ + "from math import log\n", + "# Determine the frequency of the monostable multivibrator\n", + "R1 = 5*10**3 #\n", + "R2 =15*10**3 #\n", + "C = 0.01*10**-6 #\n", + "R = 12*10**3 #\n", + "\n", + "# the output of monostable multivibrator is defined as\n", + "f = 1/(R*C*(log(1+(R2/R1))))/1e3 # kHz\n", + "print 'the output of monostable multivibrator is = %0.2f'%f,' kHz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.11 Pg 386" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output of monostable multivibrator is = 4.00 KHz\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Determine the frequency of the monostable multivibrator\n", + "R1 = 5*10**3 #\n", + "R2 =15*10**3 #\n", + "C = 0.01 #\n", + "R = 25 #\n", + "\n", + "# the output of monostable multivibrator is defined as\n", + "f = 1/(R*C)#\n", + "print 'the output of monostable multivibrator is = %0.2f'%f,' KHz'" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch14_1.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch14_1.ipynb new file mode 100644 index 00000000..b66475f7 --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch14_1.ipynb @@ -0,0 +1,643 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 14 Special Function ICs" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.1 Pg 415" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output voltage of the adjustable voltage regulator is = 22.25 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "\n", + "# to determine the regulated voltage \n", + "R1 = 250 # #ohm \n", + "R2 = 2500 # # ohm \n", + "Vref = 2 # #V #reference voltage\n", + "Iadj = 100*10**-6# # A # adjacent current\n", + "\n", + "#the output voltage of the adjustable voltage regulator is defined by\n", + "Vo = (Vref*((R2/R1)+1)+(Iadj*R2)) #\n", + "print 'the output voltage of the adjustable voltage regulator is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.2 Pg 416" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the total power dissipation of the IC is = 25.00 mA \n" + ] + } + ], + "source": [ + "# to determine the current drawn from the dual power supply \n", + "V = 10 # # V\n", + "P = 500 # # mW\n", + "\n", + "# we assume that each power supply provides half power supply to IC\n", + "P1 = (P/2)#\n", + "\n", + "# the total power dissipation of the IC\n", + "# P1 = V*I #\n", + "I = P1/V #\n", + "print 'the total power dissipation of the IC is = %0.2f'%I,' mA '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.3 Pg 416" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output voltage of the adjustable voltage regulator is = 7.50 V \n" + ] + } + ], + "source": [ + "# to determine the output voltage \n", + "R1 = 100*10**3 # #ohm \n", + "R2 = 500*10**3 # # ohm \n", + "Vref = 1.25 # #V #reference voltage\n", + "\n", + "#the output voltage of the adjustable voltage regulator is defined by\n", + "Vo = Vref*(R1+R2)/R1#\n", + "print 'the output voltage of the adjustable voltage regulator is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.4 Pg 417" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output voltage of switching regulator circuit is = 3.50 V \n" + ] + } + ], + "source": [ + "# determine the output voltage of the switching regulator circuit\n", + "d = 0.7 # # duty cycle\n", + "Vin = 5 # # V # input voltage\n", + "\n", + "# The output voltage of switching regulator circuit is given by\n", + "Vo = d*Vin #\n", + "print 'The output voltage of switching regulator circuit is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.5 Pg 417" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output voltage of switching regulator circuit is = 0.96 \n" + ] + } + ], + "source": [ + "# determine the duty cycle of the switching regulator circuit\n", + "Vo = 4.8 # # V # output voltage\n", + "Vin = 5 # # V # input voltage\n", + "\n", + "# The output voltage of switching regulator circuit is given by\n", + "# Vo = d*Vin #\n", + "\n", + "# Duty cycle is given as\n", + "d =Vo/Vin #\n", + "print 'The output voltage of switching regulator circuit is = %0.2f'%d,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.6 Pg 418" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output voltage of switching regulator circuit is = 0.50 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# determine the duty cycle of the switching regulator circuit\n", + "T =120 # #msec # total pulse time\n", + "# T = ton + toff #\n", + "ton = T/2 #\n", + "\n", + "# The duty cycle of switching regulator circuit is given by\n", + "d = ton/T#\n", + "print 'The output voltage of switching regulator circuit is = %0.2f'%d,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.7 Pg 418" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output voltage of switching regulator circuit is = 0.67 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# determine the duty cycle of the switching regulator circuit\n", + "ton = 12 # #msec # on time of pulse\n", + "# ton = 2*toff # given\n", + "# T = ton + toff #\n", + "toff = ton/2 #\n", + "T = ton+toff # # total time\n", + "\n", + "# The duty cycle of switching regulator circuit is given by\n", + "d = ton/T#\n", + "print 'The output voltage of switching regulator circuit is = %0.2f'%d,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.8 Pg 419" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The emitter bias voltage is = 3.80 V \n", + "The output voltage of the IC LM380 is = 7.90 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# determine the output voltage of the audio power amplifier IC LM380\n", + "Vcc = 12 # # V\n", + "Ic3 = 12*10**-6 # # A # collector current of the transistor Q3\n", + "Ic4 = 12*10**-6 # # A # collector current of the transistor Q4\n", + "R11 = 25*10**3 # # ohm\n", + "R12 = 25*10**3 # # ohm\n", + "\n", + "# the collector current of Q3 is defined as\n", + " # Ic3 = (Vcc-3*Veb)/(R11+R12)#\n", + "Veb = (Vcc-(R11+R12)*Ic3)/3 #\n", + "print 'The emitter bias voltage is = %0.2f'%Veb,' V '\n", + "\n", + "# the output voltage of the IC LM380\n", + "Vo = (1/2)*Vcc+(1/2)*Veb#\n", + "print 'The output voltage of the IC LM380 is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.9 Pg 420" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The emitter bias voltage is = 3.33 V \n", + "The output voltage of the IC LM380 is = 6.67 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# determine the output voltage of the audio power amplifier IC LM380\n", + "Vcc = 10 # # V\n", + "Ic3 = 0.01*10**-6 # # A # collector current of the transistor Q3\n", + "Ic4 = 0.01*10**-6 # # A # collector current of the transistor Q4\n", + "R11 = 25*10**3 # # ohm\n", + "R12 = 25*10**3 # # ohm\n", + "\n", + "# the collector current of Q3 is defined as\n", + " # Ic3 = (Vcc-3*Veb)/(R11+R12)#\n", + "Veb = (Vcc-(R11+R12)*Ic3)/3 #\n", + "print 'The emitter bias voltage is = %0.2f'%Veb,' V '\n", + "\n", + "# the output voltage of the IC LM380\n", + "Vo = (1/2)*Vcc+(1/2)*Veb#\n", + "print 'The output voltage of the IC LM380 is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.10 Pg 421" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The emitter resistor of Q3 is = 52.00 ohm ( at temperature 25 degree celsius) \n", + "The trans conductance of transistor is = 38.5 mA/V \n", + "The base emitter resistor rbe is = 1.30 K ohm \n", + "The emitter capacitor Ce = 7.65 pF \n", + "The value of resistance RL is = 264.55 ohm \n", + "The pole frequency fa is = 601.91 M Hz \n", + "The pole frequency fb is = 1073.74 M Hz \n", + "The pole frequency fc is = 3060.67 M Hz \n", + "Hence fa is a dominant pole frequency \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from numpy import inf\n", + "from math import sqrt, pi\n", + "# Design a video amplifier of IC 1550 circuit\n", + "Vcc = 12 # # V\n", + "Av = -10 #\n", + "Vagc = 0 # # at bandwidth of 20 MHz\n", + "hfe = 50 # # forward emitter parameter\n", + "rbb = 25 # # ohm # base resistor\n", + "Cs = 1*10**-12 # # F # source capacitor\n", + "Cl = 1*10**-12 # # F # load capacitor\n", + "Ie1 = 1*10**-3 # # A # emitter current of Q1\n", + "f = 1000*10**6 # # Hz\n", + "fT = 800*10**6 # # Hz\n", + "Vt = 52*10**-3 #\n", + "Vt1 = 0.026 #\n", + "\n", + "# When Vagc =0 the transistor Q2 is cut-off and the collector current of transistor Q2 flow through the transistor Q3\n", + "# i.e Ic1=Ie1=Ie3\n", + "Ie3 = 1*10**-3 # # A # emitter current of Q3\n", + "Ic1 = 1*10**-3 # # A # collector current of the transistor Q1\n", + "\n", + "# it indicates that the emitter current of Q2 is zero Ie2 = 0 then the emitter resistor of Q2 is infinite\n", + "\n", + "re2 = inf #\n", + "\n", + "# emitter resistor of Q3 \n", + "re3 = (Vt/Ie1)#\n", + "print 'The emitter resistor of Q3 is = %0.2f'%re3,' ohm ( at temperature 25 degree celsius) '\n", + "\n", + "# the trans conductance of transistor is\n", + "gm = (Ie1/Vt1)#\n", + "print 'The trans conductance of transistor is = %0.1f'%(gm*1000),' mA/V ' # Round Off Error\n", + "\n", + "# the base emitter resistor rbe\n", + "rbe = (hfe/gm)#\n", + "print 'The base emitter resistor rbe is = %0.2f'%(rbe/1000),' K ohm ' # Round Off Error\n", + "\n", + "# the emitter capacitor Ce \n", + "\n", + "Ce = (gm/(2*pi*fT))#\n", + "print 'The emitter capacitor Ce = %0.2f'%(Ce*1e12),' pF ' # Round Off Error\n", + "\n", + "# the voltage gain of video amplifier is\n", + "# Av = (Vo/Vin) #\n", + "# Av = -((alpha3*gm)/(rbb*re3)*((1/rbb)+(1/rbe)+sCe)*((1/re2)+(1/re3)+sC3)*((1/Rl)+(s(Cs+Cl)))) \n", + " # At Avgc = 0 i.e s=0 in the above Av equation\n", + "alpha3 = 1 #\n", + "s = 0 #\n", + "# Rl = -((alpha3*gm)/(rbb*re3)*(((1/rbb)+(1/rbe))*((1/re2)+(1/re3))*(Av)))# \n", + "\n", + "# After solving above equation for Rl We get Rl Equation as\n", + "Rl = 10/(37.8*10**-3)#\n", + "print 'The value of resistance RL is = %0.2f'%Rl,' ohm '\n", + "\n", + "# there are three poles present in the transfer function of video amplifier each pole generate one 3-db frequency \n", + "Rl = 675 #\n", + "# fa = 1/(2*pi*Rl*(Cs+Cl))#\n", + "# after putting value of Rl ,Cs and Cl we get\n", + "fa = 1/(2*3.14*264.55*1*10**-12)#\n", + "print 'The pole frequency fa is = %0.2f'%(fa*10**-3/1000),' M Hz '# Round Off Error\n", + "\n", + "\n", + "#fb = 1/(2*pi*Ce*((rbb*rbe)/(rbb+rbe)))#\n", + "# after putting value of Ce rbb and rbe we get\n", + "fb = 1/(2*pi*6.05*10**-12*24.5)#\n", + "print 'The pole frequency fb is = %0.2f'%(fb*10**-3/1000),' M Hz '\n", + "\n", + "fc = 1/(2*pi*Cs*re3)#\n", + "print 'The pole frequency fc is = %0.2f'%(fc*10**-3/1000),' M Hz '\n", + "\n", + "print 'Hence fa is a dominant pole frequency '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.11 Pg 423" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The emitter resistor of Q3 is = 52.00 ohm \n", + "The trans conductance of transistor is = 38.5 mA/V \n", + "The base emitter resistor rbe is = 1.3 kohm \n", + "The emitter capacitor is = 6.12 pF \n", + "The value of resistance RL is = 265.00 ohm \n", + "The pole frequency fa is = 600.58 MHz \n", + "The pole frequency fb is = 1060.00 MHz \n", + "The pole frequency fc is = 3060.67 MHz \n", + "Hence fa is a dominant pole frequency \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from numpy import inf,pi\n", + "# Design a video amplifier of IC 1550 circuit\n", + "Vcc = 12 # # V\n", + "Av = -10 #\n", + "Vagc = 0 # # at bandwidth of 20 MHz\n", + "hfe = 50 # # forward emitter parameter\n", + "rbb = 25 # # ohm # base resistor\n", + "Cs = 1*10**-12 # # F # source capacitor\n", + "Cl = 1*10**-12 # # F # load capacitor\n", + "Ie1 = 1*10**-3 # # A # emitter current of Q1\n", + "f = 1000*10**6 # # Hz\n", + "Vt = 52*10**-3 #\n", + "Vt1 = 0.026 #\n", + "\n", + "# When Vagc =0 the transistor Q2 is cut-off and the collector current of transistor Q2 flow through the transistor Q3\n", + "# i.e Ic1=Ie1=Ie3\n", + "Ie3 = 1*10**-3 # # A # emitter current of Q3\n", + "Ic1 = 1*10**-3 # # A # collector current of the transistor Q1\n", + "\n", + "# it indicates that the emitter current of Q2 is zero Ie2 = 0 then the emitter resistor of Q2 is infinite\n", + "re2 = inf #\n", + "\n", + "# emitter resistor of Q3 \n", + "re3 = (Vt/Ie1)#\n", + "print 'The emitter resistor of Q3 is = %0.2f'%re3,' ohm '\n", + "\n", + "# the trans conductance of transistor is\n", + "gm = (Ie1/Vt1)#\n", + "print 'The trans conductance of transistor is = %0.1f'%(gm*1e3),' mA/V '\n", + "\n", + "# the base emitter resistor rbe\n", + "rbe = (hfe/gm)#\n", + "print 'The base emitter resistor rbe is = %0.1f'%(rbe/1e3),' kohm '\n", + "\n", + "# the emitter capacitor Ce \n", + "Ce = (gm/(2*pi*f))#\n", + "print 'The emitter capacitor is = %0.2f'%(Ce*1e12),' pF '\n", + "\n", + "# the voltage gain of video amplifier is\n", + "# Av = (Vo/Vin) #\n", + "# Av = -((alpha3*gm)/(rbb*re3)*((1/rbb)+(1/rbe)+sCe)*((1/re2)+(1/re3)+sC3)*((1/Rl)+(s(Cs+Cl)))) \n", + " # At Avgc = 0 i.e s=0 in the above Av equation\n", + "alpha3 = 1 #\n", + "s = 0 #\n", + "Av =-10 #\n", + "Rl = -((alpha3*gm)/((rbb*re3)*(((1/rbb)+(1/rbe))*((1/re2)+(1/re3))*(Av))))# \n", + "Rl = (1/Rl)#\n", + "print 'The value of resistance RL is = %0.2f'%Rl,' ohm '\n", + "\n", + "# there are three poles present in the transfer function of video amplifier each pole generate one 3-db frequency \n", + "Rl = 265\n", + "fa = 1/(2*pi*Rl*(Cs))/1e6#\n", + "print 'The pole frequency fa is = %0.2f'%fa,'MHz '\n", + "\n", + "\n", + "fb = 1/(2*pi*Ce*((rbb*rbe)/(rbb+rbe)))/1e6\n", + "print 'The pole frequency fb is = %0.2f'%fb,'MHz '\n", + "\n", + "fc = 1/(2*pi*Cs*re3)/1e6\n", + "print 'The pole frequency fc is = %0.2f'%fc,'MHz '\n", + "\n", + "print 'Hence fa is a dominant pole frequency '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.12 Pg 425" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input current is = 0.50 mA \n", + "The output of an op-amp is = 27.50 V \n" + ] + } + ], + "source": [ + "# Determine the output voltage of an isolation amplifier IC ISO100\n", + "Vin = 5.0 # # V\n", + "Rin = 10*10**3 # \n", + "Rf = 55*10**3 # # ohm # feedback resistance\n", + "\n", + "# the input voltage of an amplifier 1\n", + "# Vin = Rin*Iin\n", + "Iin = Vin/Rin # \n", + "print 'The input current is = %0.2f'%(Iin*1e3),'mA '\n", + "\n", + "# In isolation amplifier ISO 100 the input current Iin is equal to the output current Iout , but both are opposite in direction\n", + "# Iin = -Iout\n", + "# the output of an op-amp\n", + "# Vo = -Rf*Iout\n", + "Vo = Rf*Iin#\n", + "print 'The output of an op-amp is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.13 Pg 426" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input current is = 12 mA \n", + "The output of an op-amp is = 204 V \n" + ] + } + ], + "source": [ + "# Determine the output voltage of an isolation amplifier IC ISO100\n", + "Vin = 12.0 # # V\n", + "Rin = 1*10**3 # \n", + "Rf = 17*10**3 # # ohm # feedback resistance\n", + "\n", + "# the input voltage of an amplifier 1\n", + "# Vin = Rin*Iin\n", + "Iin = Vin/Rin # \n", + "print 'The input current is = %0.f'%(Iin*1e3),'mA '\n", + "\n", + "# In isolation amplifier ISO 100 the input current Iin is equal to the output current Iout , but both are opposite in direction\n", + "# Iin = -Iout\n", + "# the output of an op-amp\n", + "# Vo = -Rf*Iout\n", + "Vo = Rf*Iin#\n", + "print 'The output of an op-amp is = %0.f'%Vo,' V '" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch3_1.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch3_1.ipynb new file mode 100644 index 00000000..64dbf254 --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch3_1.ipynb @@ -0,0 +1,345 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3 Current Voltage Sources and Differential Amplifiers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.1 Pg 53" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The collector current of difference amplifier Ic1 = Ic2 = 0.50 mA \n", + "The collector voltages of transistors Q1 and Q2 are Vc1 = Vc2 = 5.00 volt \n", + "For Ve = -0.7 Volt the collector - emitter voltage Vce1 = 5.70 Volt\n", + "For Ve = 4.3 Volt the collector - emitter voltage Vce1 = 0.70 Volt\n", + "For Ve = -5.7 Volt the collector - emitter voltage Vce1 = 10.70 Volt\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Determine the collector current Ic1 and collector-emitter voltage Vce1 for the difference amplifier circuit\n", + "\n", + "V1 = 0 # # volt\n", + "V2 = -5 # #volt\n", + "Vcm = 5 # #volt\n", + "Vcc = 10# #volt\n", + "Vee = -10 # #volt\n", + "Ie = 1 # #mA\n", + "Rc = 10 # #kilo ohm\n", + "\n", + "# Transistor parameters\n", + "# base current are negligible\n", + "Vbe = 0.7 # # volt\n", + "\n", + "# The collector current of difference amplifier is\n", + "Ic1 = Ie/2 # \n", + "print 'The collector current of difference amplifier Ic1 = Ic2 = %0.2f'%Ic1,' mA '\n", + "\n", + "# The collector voltages of transistors Q1 and Q2 are expressed as\n", + "\n", + "Vc1 = Vcc-Ic1*Rc #\n", + "print 'The collector voltages of transistors Q1 and Q2 are Vc1 = Vc2 = %0.2f'%Vc1,' volt '\n", + "\n", + "# We know common mode voltage (Vcm) , from this the emitter voltage can be identified as follows\n", + "# For the common mode voltage Vcm = 0 V , the emitter voltage is Ve = -0.7 V\n", + "# For the common mode voltage Vcm = 5 V , the emitter voltage is Ve = 4.3 V\n", + "# For the common mode voltage Vcm = -5 V , the emitter voltage is Ve = -5.7 V\n", + "\n", + "# For the different emitter voltages the collector-emitter voltage can be calculated as\n", + "\n", + "Ve = -0.7 # # volt\n", + "Vce1 = Vc1-Ve#\n", + "print 'For Ve = -0.7 Volt the collector - emitter voltage Vce1 = %0.2f'%Vce1,' Volt'\n", + "\n", + "Ve = 4.3 # # volt\n", + "Vce1 = Vc1-Ve#\n", + "print 'For Ve = 4.3 Volt the collector - emitter voltage Vce1 = %0.2f'%Vce1,' Volt'\n", + "\n", + "Ve = -5.7 # # volt\n", + "Vce1 = Vc1-Ve#\n", + "print 'For Ve = -5.7 Volt the collector - emitter voltage Vce1 = %0.2f'%Vce1,' Volt'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.2 Pg 54" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The differential mode gain Ad = 184.6\n", + " The common mode gain Acm = -0.237\n" + ] + } + ], + "source": [ + "# To determine the difference-mode and common-mode gain of the difference amplifier\n", + "\n", + "Vcc = 10 # # volt\n", + "Vee = -10 # #volt\n", + "Iq = 0.8 # #mA\n", + "Ie = 0.8 # #mA\n", + "Rc = 12 # #kilo-Ohm\n", + "Vt = 0.026 # # volt\n", + "\n", + "# Transistor parameter\n", + "beta = 100 #\n", + "Rs = 0 # #Ohm\n", + "Ro = 25 # #kilo-Ohm \n", + "# The differential mode gain Ad\n", + "gm = (Ie/ 2*Vt) #\n", + "# Ad = (gm*r*Rc/r+Rc) # # where r is r-pi\n", + "# For Rb=0 , the differential mode gain is\n", + "\n", + "Ad = (Ie/(2*Vt))*Rc#\n", + "#But\n", + "print ' The differential mode gain Ad = %0.1f'%Ad\n", + "\n", + "#The common mode gain Acm\n", + "# Acm = - (gm*Rc/1+2*gm*Re+2*Re/r)\n", + "Acm =-(Ad/(1+(((1+beta)*Ie*Ro)/(beta*Vt))))\n", + "print ' The common mode gain Acm = %0.3f'%Acm" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.3 Pg 56" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output of a difference amplifier is Vo = -47.40 sinwt uV \n" + ] + } + ], + "source": [ + "# To find the output of a difference amplifier when only common mode signal is applied\n", + "\n", + "# V1 = V2 = Vcm = 200*sin(wt) # # micro volt (uV)\n", + "Acm = -0.237 #\n", + "\n", + "# When the common mode input signal is applied to the difference amplifier , the difference mode gain is zero\n", + "Vcm = 200 #\n", + "Vo = Acm*Vcm #\n", + "print 'The output of a difference amplifier is Vo = %0.2f'%Vo,'sinwt uV ' # multiply by sinwt because it is in Vcm" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.4 Pg 56" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The differential mode gain Ad = 184.6\n", + "The common mode gain Acm = -0.237\n", + "The CMRR of difference amplifier is = 389\n", + "In decibel CMRR is = 51.80\n" + ] + } + ], + "source": [ + "from math import log10\n", + "#Determine the common mode rejection ratio(CMRR) of the difference amplifier\n", + "\n", + "Vcc = 10 # # volt\n", + "Vee = -10 # #volt\n", + "Iq = 0.8 # #mA\n", + "Ie = 0.8 # #mA\n", + "Rc = 12 # #kilo-Ohm\n", + "Vt = 0.026 # # volt\n", + "\n", + "# Transistor parameter\n", + "beta = 100 #\n", + "Rs = 0 # #Ohm\n", + "Ro = 25 # #kilo-Ohm\n", + " \n", + "# The differential mode gain Ad\n", + "gm = (Ie/ 2*Vt) #\n", + "# Ad = (gm*r*Rc/r+Rc) # # where r is r-pi\n", + "# For Rb=0 , the differential mode gain is\n", + "\n", + "Ad = (Ie/(2*Vt))*Rc#\n", + "#But\n", + "print 'The differential mode gain Ad = %0.1f'%Ad\n", + "\n", + "#The common mode gain Acm\n", + "# Acm = - (gm*Rc/1+2*gm*Re+2*Re/r)\n", + "Acm =-(Ad/(1+(((1+beta)*Ie*Ro)/(beta*Vt))))\n", + "print 'The common mode gain Acm = %0.3f'%Acm\n", + "\n", + "# The CMRR of difference amplifier is given as\n", + "Ad = Ad/2 #\n", + "CMRR = abs(Ad/Acm)\n", + "print 'The CMRR of difference amplifier is = %0.f'%CMRR\n", + "\n", + "# In decibel it can be expressed as\n", + "CMRRdb = 20*log10(CMRR)\n", + "print 'In decibel CMRR is = %0.2f'%CMRRdb" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.5 Pg 58" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The CMRR of difference amplifier is = 3.16e+04\n", + " The value of resistance RE is = 2.04 Mohm \n" + ] + } + ], + "source": [ + "# To determine emitter resistance of the difference amplifier\n", + "\n", + "Vcc = 10 # # volt\n", + "Vee = -10 # #volt\n", + "Iq = 0.8 # #mA\n", + "Ie = 0.8 # #mA\n", + "CMRRdb = 90 # #dB\n", + "Vt = 0.026 #\n", + "\n", + "# Transistor parameter\n", + "beta = 100 #\n", + "\n", + "# CMRR = abs(Ad/Acm)\n", + "# the CMRR of the difference amplifier is defined as\n", + "#CMRR = ((1/2)*(1+((1+beta)*Ie*Re)/beta*Vt))\n", + "\n", + "# CMRRdb = 20*log10(CMRR)\n", + "CMRR = 10**(CMRRdb/20)\n", + "print ' The CMRR of difference amplifier is = %0.2e'%CMRR\n", + "\n", + "# The resistance RE is calculated as\n", + "\n", + "RE = (((2*CMRR)-1)/((1+beta)*Ie))*(beta*Vt)/1e3\n", + "print ' The value of resistance RE is = %0.2f'%RE,' Mohm '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.6 Pg 59" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The differential mode gain Ad is = 321\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# determine the differential mode gain when load resistance RL = 100 k ohm\n", + "\n", + "RL = 100*10**3 # # k ohm # load resistance\n", + "IE = 0.20*10**-3 # # mA # biasing current\n", + "VA = 100 # # V # early voltage\n", + "VT = 0.026 # # threshold volt\n", + "\n", + "# the differential gain of differential amplifier with an active load circuit\n", + "#Ad = Vo/Vd = gm(ro2 || ro4 || RL )\n", + "ro2 = (2*VA)/IE#\n", + "ro4 = ro2 #\n", + "gm = IE/(2*VT) #\n", + "\n", + "Ad = gm/((1/ro2)+(1/ro4)+(1/RL))\n", + "print ' The differential mode gain Ad is = %0.f'%Ad" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch4_1.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch4_1.ipynb new file mode 100644 index 00000000..a40b05f0 --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch4_1.ipynb @@ -0,0 +1,396 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4 Operational Amplifier" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.1 Pg 79" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " closed loop gain of an op-amp is = 35.00\n", + " the input impedance Zin = 10.00 kohm \n", + " the output impedance Z0 = 0.020 ohm \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# For an op-amp circuit find a) closed loop gain Acl b) input impedance Zin c) output impedance Zo\n", + "ro = 85 # # ohm\n", + "A = 150*10**3 # # ohm\n", + "R2 = 350*10**3 # # ohm # Feedback resistance\n", + "R1 = 10*10**3 # # ohm # Input resistance\n", + "\n", + "# a) closed loop gain\n", + "# ACL = abs(Vo/Vin) = abs(R2/R1)\n", + "ACL = abs(R2/R1) #\n", + "print ' closed loop gain of an op-amp is = %0.2f'%ACL# # 1/beta = ACL\n", + "beta = (1/ACL) #\n", + "\n", + "# b) the input impedance Zin\n", + "Zin = R1 #\n", + "print ' the input impedance Zin = %0.2f'%(Zin/1e3),'kohm '#\n", + "\n", + "# c0 the output impedance Z0\n", + "Z0 = (ro)/(1+(beta*A))#\n", + "print ' the output impedance Z0 = %0.3f'%Z0,' ohm '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.2 Pg 80" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The difference voltage is = 10.00 V \n", + " The open loop gain is = 2.00 \n" + ] + } + ], + "source": [ + "# Determine the differece voltage and open loop gain of an op-amp\n", + "V1 = -5 # # volt # input voltage\n", + "V2 = 5 # # volt\n", + "Vo = 20 # #volt # output voltage\n", + "\n", + "# the difference voltage is given by \n", + "Vd = V2-V1 #\n", + "print ' The difference voltage is = %0.2f'%Vd,' V '\n", + "\n", + "# open loop gain \n", + "A = (Vo/Vd)#\n", + "print ' The open loop gain is = %0.2f'%A,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.3 Pg 80" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The difference voltage is = 5.00 V \n", + " The open loop gain is = 4.00 \n" + ] + } + ], + "source": [ + "# Determine the differece voltage and open loop gain of an op-amp\n", + "V1 = -5 # # volt # input voltage\n", + "V2 = 0 # # volt # GND\n", + "Vo = 20 # #volt # output voltage\n", + "\n", + "# the difference voltage is given by \n", + "Vd = V2-V1 #\n", + "print ' The difference voltage is = %0.2f'%Vd,' V '\n", + "\n", + "# open loop gain \n", + "A = (Vo/Vd)#\n", + "print ' The open loop gain is = %0.2f'%A,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.4 Pg 81" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The difference voltage is = 5.00 V \n", + " The open loop gain is = 4.00 \n" + ] + } + ], + "source": [ + "# Determine the differece voltage and open loop gain of an op-amp\n", + "V1 = 0 # # volt # input voltage # GND\n", + "V2 = 5 # # volt \n", + "Vo = 20 # #volt # output voltage\n", + "\n", + "# the difference voltage is given by \n", + "Vd = V2-V1 #\n", + "print ' The difference voltage is = %0.2f'%Vd,' V '\n", + "\n", + "# open loop gain \n", + "A = (Vo/Vd)#\n", + "print ' The open loop gain is = %0.2f'%A,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.5 Pg 81" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The difference voltage is = -10.00 V \n", + " The open loop gain is = 2.00 \n" + ] + } + ], + "source": [ + "# Determine the differece voltage and open loop gain of an op-amp\n", + "V1 = 5 # # volt # input voltage # GND\n", + "V2 = -5 # # volt \n", + "Vo = -20 # #volt # output voltage\n", + "\n", + "# the difference voltage is given by \n", + "Vd = V2-V1 #\n", + "print ' The difference voltage is = %0.2f'%Vd,' V '\n", + "\n", + "# open loop gain \n", + "A = (Vo/Vd)#\n", + "print ' The open loop gain is = %0.2f'%A,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.6 Pg 82" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Closed loop gain of an inverting op-amp is = -2.50 \n", + "The |Ac| Closed loop gain of an inverting op-amp is = 2.50 \n", + "The output voltage of an inverting op-amp is = -25.00 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# To find closed loop gain and output voltage Vo of an inverting op-amp\n", + "R1 = 10 # #kilo ohm # input resistance\n", + "R2 = 25 # # kilo ohm # feedback resistance\n", + "Vin = 10 # #volt # input voltage\n", + "\n", + "# Closed loop gain of an inverting op-amp\n", + "Ac = -(R2/R1) #\n", + "print 'The Closed loop gain of an inverting op-amp is = %0.2f'%Ac,' '\n", + "Ac = abs(Ac)#\n", + "print 'The |Ac| Closed loop gain of an inverting op-amp is = %0.2f'%Ac,' '\n", + "\n", + "# the output voltage of an inverting op-amp\n", + "Vo = -(R2/R1)*Vin #\n", + "print 'The output voltage of an inverting op-amp is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.7 Pg 82" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The Closed loop gain of an non-inverting op-amp is = 3.50 \n", + " The output voltage of an non-inverting op-amp is = 35.00 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# To find closed loop gain and output voltage Vo of an non-inverting op-amp\n", + "R1 = 10 # #kilo ohm # input resistance\n", + "R2 = 25 # # kilo ohm # feedback resistance\n", + "Vin = 10 # #volt # input voltage\n", + "\n", + "# Closed loop gain of an non-inverting op-amp\n", + "Ac = 1+(R2/R1) #\n", + "Ac = abs(Ac)#\n", + "print ' The Closed loop gain of an non-inverting op-amp is = %0.2f'%Ac,' '\n", + "\n", + "# the output voltage of an inverting op-amp\n", + "Vo = (1+R2/R1)*Vin #\n", + "print ' The output voltage of an non-inverting op-amp is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.8 Pg 83" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The closed loop gain of differntial op-amp is = 2.50 \n", + "The output voltage of an non-inverting op-amp is= 50.00 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# to find out closed loop gain and output voltage Vo\n", + "R1 = 10 # #kilo ohm # input resistance\n", + "R3 = 10 # #kilo ohm # input resistance\n", + "R2 = 25 # # kilo ohm # feedback resistance\n", + "R4 = 25 # # kilo ohm # feedback resistance\n", + "Vin2 = 10 # #volt # input voltage\n", + "Vin1 = -10 # #volt # input voltage\n", + "\n", + "# closed loop gain of differntial op-amp is given by\n", + "Ac = (R2/R1) #\n", + "Ac = abs(Ac)# \n", + "print 'The closed loop gain of differntial op-amp is = %0.2f'%Ac,' '\n", + "\n", + "# the output voltage of an non-inverting op-amp is given by\n", + "Vo = (R2/R1)*(Vin2-Vin1) #\n", + "print 'The output voltage of an non-inverting op-amp is= %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.9 Pg 84" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The upper voltage is = 2.86 V \n", + " The lower voltage is = -2.86 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Determine the non-inverting input voltage\n", + "R1 = 10 # #kilo ohm # input resistance\n", + "R2 = 25 # #kilo ohm # feedback resistance\n", + "Voh = 10 # # volt #output voltage\n", + "Vol = -10 # # volt # output voltage\n", + "\n", + "# upper voltage\n", + "V = (R1/(R1+R2)*Voh) #\n", + "print ' The upper voltage is = %0.2f'%V,' V '\n", + "\n", + "# Lower voltage\n", + "V = (R1/(R1+R2)*Vol) #\n", + "print ' The lower voltage is = %0.2f'%V,' V '" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch5_1.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch5_1.ipynb new file mode 100644 index 00000000..ef70394b --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch5_1.ipynb @@ -0,0 +1,1379 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5 Characteristic of Operational Amplifier" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.1 Pg 110" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " the output voltage (Vo) of an op-amp circuit due to input offset voltage (Vos) is = 404.00 mV \n", + " the output voltage (Vo) of an op-amp circuit due to input offset current (Ios) is = 75.00 mV \n", + " the total offset voltage (Vo) of an op-amp circuit is = 479.00 mV \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# find the total offset voltage of feedback op-amp\n", + "\n", + "Vos = 4 # #mV # input offset volt\n", + "Ios = 150*10**-3 # # input offset current\n", + "R1 = 5 # #kilo ohm # input resistance\n", + "R2 = 500 # #kilo ohm # feedback resistance\n", + "\n", + "# the output voltage (Vo) of an op-amp circuit due to input offset voltage (Vos) is\n", + "Vo1 = ((R1+R2)/(R1)*Vos) #\n", + "print ' the output voltage (Vo) of an op-amp circuit due to input offset voltage (Vos) is = %0.2f'%Vo1,' mV '#\n", + "\n", + "# the output voltage (Vo) of an op-amp circuit due to input offset current (Ios) is\n", + "Vo2 = R2*Ios #\n", + "print ' the output voltage (Vo) of an op-amp circuit due to input offset current (Ios) is = %0.2f'%Vo2,' mV '#\n", + "\n", + "# the total offset voltage is\n", + "Vo = Vo1+Vo2 #\n", + "print ' the total offset voltage (Vo) of an op-amp circuit is = %0.2f'%Vo,' mV '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.2 Pg 111" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " the output voltage (Vo) of an op-amp circuit due to input offset voltage (Vos) is = 52.00 mV \n", + " the output voltage (Vo) of an op-amp circuit due to input offset current (Ios) is = 5.00 mV \n", + " the total offset voltage (Vo) of an op-amp circuit is = 57.00 mV \n" + ] + } + ], + "source": [ + "# find the total offset voltage of feedback op-amp\n", + "\n", + "Vos = 2 # #mV # input offset volt\n", + "Ios = 20*10**-3 # # input offset current\n", + "R1 = 10 # #kilo ohm # input resistance\n", + "R2 = 250 # #kilo ohm # feedback resistance\n", + "\n", + "# the output voltage (Vo) of an op-amp circuit due to input offset voltage (Vos) is\n", + "Vo1 = ((R1+R2)/(R1)*Vos) #\n", + "print ' the output voltage (Vo) of an op-amp circuit due to input offset voltage (Vos) is = %0.2f'%Vo1,' mV '#\n", + "\n", + "# the output voltage (Vo) of an op-amp circuit due to input offset current (Ios) is\n", + "Vo2 = R2*Ios #\n", + "print ' the output voltage (Vo) of an op-amp circuit due to input offset current (Ios) is = %0.2f'%Vo2,' mV '#\n", + "\n", + "# the total offset voltage is\n", + "Vo = Vo1+Vo2 #\n", + "print ' the total offset voltage (Vo) of an op-amp circuit is = %0.2f'%Vo,' mV '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.3 Pg 111" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the input offset voltage (Vos) of an op-amp circuit is = 1.187 mV \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# find the input offset voltage of an op-amp circuit\n", + "\n", + "Vo = 90.2 # #mV # output voltage\n", + "R1 = 2 # #kilo ohm # input resistence\n", + "R2 = 150 # #kilo ohm # feedback resistence\n", + "\n", + "# the input offset voltage (Vos) of an op-amp circuit is defined as\n", + "Vos = ((R1)/(R1+R2)*Vo) #\n", + "print 'the input offset voltage (Vos) of an op-amp circuit is = %0.3f'%Vos,' mV '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.4 Pg 112" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output voltage due to the input offset voltage is = 36.00 mV \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# find the output voltage of an op-amp circuit\n", + "\n", + "Vos = 1 # #mV # input offset volt\n", + "R1 = 10 # #kilo ohm # input resistance\n", + "R2 = 350 # #kilo ohm # feedback resistance\n", + "\n", + "# the output voltage due to the input offset voltage of the op-amp circuit is defined by\n", + "Vo1 = ((R1+R2)/(R1)*Vos) #\n", + "print 'the output voltage due to the input offset voltage is = %0.2f'%Vo1,' mV '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.5 Pg 113" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output voltage of the circuit due to bias current is = 0.11 V \n", + "Bias compensated resistor is = 9.09 kilo ohm \n", + "Bias compensated output voltage is = 0.01 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Determine the bias current effect with and without current compensation method\n", + "\n", + "R1 = 10 # #kilo ohm\n", + "R2 = 100 # #kilo ohm\n", + "Ib1 = 1.1*10**-3 #\n", + "Ib2 = 1*10**-3 # \n", + "# the output voltage of the circuit due to bias current is\n", + "Vo = Ib1*R2 #\n", + "print 'the output voltage of the circuit due to bias current is = %0.2f'%Vo,' V '#\n", + "\n", + "#Bias compensated resistor is given by\n", + "R3 = (R1*R2)/(R1+R2) #\n", + "print 'Bias compensated resistor is = %0.2f'%R3,' kilo ohm '#\n", + "\n", + "#Bias compensated output voltage is given by\n", + "Vo = R2*(Ib1-Ib2)#\n", + "print 'Bias compensated output voltage is = %0.2f'%Vo,' V '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.6 Pg 113" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output voltage (Vo) of an op-amp circuit due to input offset current (Ios) is = 80 nA \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# find the input offset current of an op-amp circuit\n", + "\n", + "Vo = 12*10**-3# # V # output voltage\n", + "R1 = 2*10**3 # # ohm # input resistence\n", + "R2 = 150*10**3# # ohm # feedback resistence\n", + "\n", + "# the output voltage (Vo) of an op-amp circuit due to input offset current (Ios) is\n", + "# Vo = R2*Ios #\n", + "Ios = Vo/R2 *1e9 # nA\n", + "print 'the output voltage (Vo) of an op-amp circuit due to input offset current (Ios) is = %0.f'%Ios,'nA '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.7 Pg 114" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current in the inverting input terminal is = 27.50 nA \n", + "The current in the non-inverting input terminal is= 32.50 nA \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Determine the bias current of inverting and non-inverting\n", + "Ios = 5 # #nA # input offset current\n", + "Ib = 30 # #nA # input bias current\n", + "\n", + "# the input bias current of an op-amp is \n", + "\n", + "#Ib =(Ib1+Ib2)/(2)#\n", + "\n", + "# the offset current Ios is define as\n", + "\n", + "#Ios = abs(Ib1-Ib2) #\n", + "\n", + "Ib1=Ib-(Ios/2)#\n", + "print 'The current in the inverting input terminal is = %0.2f'%Ib1,' nA '#\n", + "\n", + "Ib2 =Ib+(Ios/2)#\n", + "print 'The current in the non-inverting input terminal is= %0.2f'%Ib2,' nA '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.8 Pg 115" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Feedback transfer function is = 0.01 \n", + "OR 1/Beta is = 100.10 \n", + "Feedback transfer function is = -0.01 \n", + "OR 1/Beta is = -100.10 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#determine the feedback transfer function of an op-amp for the following condition\n", + "# a) When open loop gain of 10**5 and the closed loop gain of 100\n", + "A = 10**5 # # open loop gain\n", + "Af = 100 # #closed loop gain\n", + "# Feedback transfer function is\n", + "beta =(1/Af)-(1/A)#\n", + "print 'Feedback transfer function is = %0.2f'%beta,''#\n", + "beta = 1/beta #\n", + "print 'OR 1/Beta is = %0.2f'%beta,''#\n", + "\n", + "# For an open loop gain of -10**5 and closed loop gain of -100\n", + "A = -10**5 # # open loop gain\n", + "Af = -100 # #closed loop gain\n", + "# Feedback transfer function is\n", + "beta =(1/Af)-(1/A)#\n", + "print 'Feedback transfer function is = %0.2f'%beta,''#\n", + "beta = 1/beta #\n", + "print 'OR 1/Beta is = %0.2f'%beta,''#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.9 Pg 115" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "open loop gain is = 2000.00\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#to determine open loop gain\n", + "beta = 0.0120 # # Feedback transfer function\n", + "Af = 80 # #closed loop gain\n", + "A = (Af)/(1-beta*Af) #\n", + "print 'open loop gain is = %0.2f'%A" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.10 Pg 116" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "close loop gain dAf is = 49.78\n", + "the percent change of closed loop gain dAf is = 0.45 %\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# To Determine the percent of change in the closed loop gain Af of feedback op-amp circuit\n", + "A = 10**5 # # open loop gain\n", + "Af = 50 # # close loop gain\n", + "beta = 0.01999 # # feedback transfer function\n", + "dA = 10**4 # # the change in the open llop gain \n", + "\n", + "# close loop gain\n", + "dAf = ((dA)/(1+dA*beta))#\n", + "print 'close loop gain dAf is = %0.2f'%dAf\n", + "\n", + "# the percent change of closed loop gain \n", + "dAf = (((Af-dAf)/(Af))*100)#\n", + "print 'the percent change of closed loop gain dAf is = %0.2f'%dAf,'%'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.11 Pg 116" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the feedback transfer function beta is = 0.0199\n", + "the closed loop bandwidth wfH is = 125600\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# To Determine the bandwidth of feedback amplifier\n", + "A = 10**4 # # open loop gain\n", + "Af = 50 # # close loop gain\n", + "wH = 628 # #(2*pi*100) # rad/sec # open loop bandwidth\n", + "\n", + "# close loop gain of an op-amp is defined as\n", + "# Af = ((A)/(1+A*beta))# \n", + "\n", + "# the feedback transfer function is given as\n", + "beta = (1/Af)-(1/A) #\n", + "print 'the feedback transfer function beta is = %0.4f'%beta\n", + "\n", + "# closed loop bandwidth\n", + "wfH = wH*(1+beta*A)#\n", + "print 'the closed loop bandwidth wfH is = %0.f'%wfH" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.12 Pg 117" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the unity gain bandwidth is = 1e+06 Hz\n", + "the maximum close loop gain ACL is = 50.00 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# To calculate unity gain bandwidth and maximum close loop gain\n", + "A = 10**5 # # open loop gain\n", + "fo = 10 # # Hz # dominant pole frequency\n", + "fdb = 20*10**3 # #Hz # 3-db frequency\n", + "\n", + "# the unity gain bandwidth\n", + "f1 = fo*A #\n", + "print 'the unity gain bandwidth is = %0.e'%f1,'Hz'#\n", + "\n", + "# the maximum close loop gain\n", + "ACL = (f1/fdb) #\n", + "print 'the maximum close loop gain ACL is = %0.2f'%ACL,''#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.13 Pg 117" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the unity gain bandwidth is = 60 kHz\n", + "the maximum close loop gain ACL is = 5.00 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# To calculate unity gain bandwidth and maximum close loop gain\n", + "A = 10**3 # # open loop gain\n", + "fo = 60 # # Hz # dominant pole frequency\n", + "fdb = 12*10**3 # #Hz # 3-db frequency\n", + "\n", + "# the unity gain bandwidth\n", + "f1 = fo*A #\n", + "print 'the unity gain bandwidth is = %0.f'%(f1/1e3),'kHz'#\n", + "\n", + "# the maximum close loop gain\n", + "ACL = (f1/fdb) #\n", + "print 'the maximum close loop gain ACL is = %0.2f'%ACL,''#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.14 Pg 118" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the dominant pole frequency (fPD) of an op-amp is = 2.5 kHz\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# To determine the dominant pole frequency of an op-amp\n", + "Ao = 2*10**5 # # low frequency open loop gain\n", + "f = 5*10**6 # # Hz # pole frequency\n", + "ACL = 100 # # low frequency closed lkoop gain\n", + "p_margin = 80 # \n", + "\n", + "# the dominant pole frequency of an op-amp\n", + "fPD = (ACL)*(f/Ao)/1e3\n", + "print 'the dominant pole frequency (fPD) of an op-amp is = %0.1f'%fPD,'kHz'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.15 Pg 118" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "FL = 6.37 KHz \n", + "Acom = [ magnitude = 6.3*10**-3 angle = -89.6 degree ]\n", + "Ac = [ magnitude = 0.68 angle = 0.4 degree ]\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Determine the loop gain of compensated network\n", + "C = 0.0025*10**-6 # # farad\n", + "R = 10*10**3 # # ohm\n", + "F = 1*10**6 # # Hz\n", + "Ac1 = 100 # \n", + "angle1 = 90 #\n", + "\n", + "# the close loop gain of a compensated network is defined as\n", + "# Ac = Acl*Acom #\n", + "\n", + "#Acom = 1/(1+%(F/FL))#\n", + "\n", + "FL = 1/(2*3.14*R*C)#\n", + "print 'FL = %0.2f'%(FL/1000),' KHz '# # Round Off Error\n", + "\n", + "# Acom = 1/(1+%j(F/FL))#\n", + "# After putting value of F ,FL we get\n", + "\n", + "# Acom = 1/(1+%j(158.7))# # 1+%j(158.7) Rectangular Form where real part is 1 and imaginary part is 158.7\n", + "\n", + "# After converting rectangular from into polar from we get\n", + " \n", + "print 'Acom = [ magnitude = 6.3*10**-3 angle = -89.6 degree ]'#\n", + "\n", + "# Ac = Ac1*Acom # equation 1\n", + "\n", + "# after putting Ac1 and Acom value in equation 1 we get Ac1 = 100 angle 90 and Acom = 6.3*10**-3 angle = -89.6 \n", + "\n", + "print 'Ac = [ magnitude = 0.68 angle = 0.4 degree ]'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.16 Pg 119" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "FL = 1.1 KHz \n", + "Acom = [ magnitude = 0.68 angle = -47.7 degree ]\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Determine the loop gain of compensated network\n", + "\n", + "C = 0.01*10**-6 # # farad\n", + "R = 15*10**3 # # ohm\n", + "F = 1*10**6 # # Hz\n", + "\n", + "# the close loop gain of a compensated network is defined as\n", + "# Ac = Acl*Acom #\n", + "\n", + "#Acom = 1/(1+%(F/FL))#\n", + "\n", + "FL = 1/(2*3.14*R*C)#\n", + "print 'FL = %0.1f'%(FL/1000),' KHz '# # Round Off Error\n", + "\n", + "# Acom = 1/(1+%j(F/FL))#\n", + "# After putting value of F ,FL we get\n", + "\n", + "# Acom = 1/(1+%j(0.9))# # 1+%j(0.9) Rectangular Form where real part is 1 and imaginary part is 0.9\n", + "\n", + "# After converting rectangular from into polar from we get\n", + " \n", + "print 'Acom = [ magnitude = 0.68 angle = -47.7 degree ]'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.18 Pg 123" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The compensating resistor value is = 15.92 ohm \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# to design compensating network\n", + "fp = 500*10**3 # # pole frequency\n", + "C = 0.02*10**-6 # # F # we choose\n", + "# loop gain of compensated network\n", + "\n", + "# ACom =(1)/(1+j(f/fp))\n", + "# fp = (1/2*pie*R*C)\n", + "R = (1/(2*3.14*C*fp))#\n", + "print 'The compensating resistor value is = %0.2f'%R,' ohm '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.19 Pg 123" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "FH = 6.37 KHz \n", + "FL = 2.12 KHz \n", + "Acom = [ magnitude = 0.34 angle = -0.24 degree ]\n", + "Ac = [ magnitude = 34 angle = 89.76 degree ]\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Determine the loop gain of compensated network\n", + "\n", + "C = 0.0025*10**-6 # # farad\n", + "R1 = 10*10**3 # # ohm\n", + "R2 = 20*10**3 # # ohm\n", + "F = 1*10**6 # # Hz\n", + "Ac1 = 100 # \n", + "angle1 = 90 #\n", + "\n", + "# the close loop gain of a compensated network is defined as\n", + "\n", + "# Ac = Acl*Acom #\n", + "\n", + "#Acom = (1+%(F/FH))/(1+%(F/FL))#\n", + "\n", + "FH = 1/(2*3.14*R1*C)#\n", + "print 'FH = %0.2f'%(FH/1000),' KHz '# # Round Off Error\n", + "\n", + "\n", + "FL = 1/(2*3.14*(R1+R2)*C)#\n", + "print 'FL = %0.2f'%(FL/1000),' KHz '# # Round Off Error\n", + "\n", + "\n", + "#Acom = (1+%(F/FH))/(1+%(F/FL))#\n", + "\n", + "# After putting value of FH ,FL we get\n", + "\n", + "# Acom = (1+%j(158.7))/(1+%j(471.7) \n", + "\n", + "# After converting rectangular from into polar from we get\n", + " \n", + "print 'Acom = [ magnitude = 0.34 angle = -0.24 degree ]'#\n", + "\n", + "# Ac = Ac1*Acom # equation 1\n", + "\n", + "# after putting Ac1 and Acom value in equation 1 we get Ac1 = 100 angle 90 and Acom = 0.34 angle = -0.24 \n", + "\n", + "print 'Ac = [ magnitude = 34 angle = 89.76 degree ]'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.20 Pg 124" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "FH = 1.59 KHz \n", + "FL = 0.64 KHz \n", + "Acom = [magnitude = 0.4] \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Determine the loop gain of compensated network\n", + "C = 0.01*10**-6 # # farad\n", + "R1 = 10*10**3 # # ohm\n", + "R2 = 15*10**3 # # ohm\n", + "F = 1*10**6 # # Hz\n", + "\n", + "\n", + "# the close loop gain of a compensated network is defined as\n", + "\n", + "#Acom = (1+%(F/FH))/(1+%(F/FL))#\n", + "\n", + "FH = 1/(2*3.14*R1*C)#\n", + "print 'FH = %0.2f'%(FH/1000),' KHz '# # Round Off Error\n", + "\n", + "\n", + "FL = 1/(2*3.14*(R1+R2)*C)#\n", + "print 'FL = %0.2f'%(FL/1000),' KHz '# # Round Off Error\n", + "\n", + "\n", + "#Acom = (1+%(F/FH))/(1+%(F/FL))#\n", + "\n", + "# After putting value of FH ,FL we get\n", + "\n", + "# Acom = (1+%j(658.9))/(1+%j(1.56*10**3) \n", + "\n", + "# After converting rectangular from into polar from we get\n", + " \n", + "print 'Acom = [magnitude = 0.4] '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.21 Pg 125" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The compensating first resistor R1 value is = 0.80 K ohm \n", + "The compensating second resistor R2 value is = 7.17 K ohm \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# to design compensating network\n", + "fH = 10 # #k ohm # break frequency initiated by a zero\n", + "fL = 1 # #k ohm # break frequency initiated by a pole\n", + "C = 0.02# # uF # we choose\n", + "# loop gain of compensated network\n", + "\n", + "# ACom =(1+j(f/fH))/(1+j(f/fL))\n", + "# fH = (1/2*pie*R1*C)\n", + "# fL = (1/2*pie*(R1+R2)*C)\n", + "R1 = (1/(2*3.14*C*fH))#\n", + "print 'The compensating first resistor R1 value is = %0.2f'%R1,' K ohm '#\n", + "R2 = ((1)/(2*3.14*C*fL))-(R1)#\n", + "print 'The compensating second resistor R2 value is = %0.2f'%R2,' K ohm '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.22 Pg 126" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input miller capacitance Cin value is = 10.10 uF \n", + "The output miller capacitance Cout value is = 0.10 uF \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# To determine input output miller capacitances\n", + "A = 100 # #gain\n", + "Cm = 0.1 # # uF # compensated capacitor\n", + "\n", + "# the input output miller capacitance are defined as\n", + "Cin = Cm*(A+1)#\n", + "print 'The input miller capacitance Cin value is = %0.2f'%Cin,'uF '#\n", + "Cout = (Cm*((A+1)/A))# \n", + "print 'The output miller capacitance Cout value is = %0.2f'%Cout,'uF '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.23 Pg 127" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input miller capacitance Cin value is = 3.02 uF \n", + "The output miller capacitance Cout value is = 0.02 uF \n", + "The initiated frequency of miller compensating network by pole is = 7.91 KHz \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# To determine input output miller capacitances\n", + "A = 150 # #gain\n", + "Cm = 0.02 # # uF # compensated capacitor\n", + "\n", + "# the input output miller capacitance are defined as\n", + "Cin = Cm*(A+1)#\n", + "print 'The input miller capacitance Cin value is = %0.2f'%Cin,'uF '#\n", + "Cout = (Cm*((A+1)/A))# \n", + "print 'The output miller capacitance Cout value is = %0.2f'%Cout,'uF '#\n", + "\n", + "# In the miller compensating network input capacitance introduce a pole . The initiated frequency of miller compensating network by pole is define as\n", + "\n", + "# fp = 1/(2*pi*R*Cin)#\n", + "R = 1 # # K ohm\n", + "fp = 1/(2*pi*R*Cout)#\n", + "print 'The initiated frequency of miller compensating network by pole is = %0.2f'%fp,' KHz '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.24 Pg 128" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the slew rate of an op-amp is = 17.58 V/u sec \n", + "The compansated capacitance value is = 1.12 pF \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# To determine the slew rate of an op-amp\n", + "f = 1 # # MHz # unity frequency\n", + "Ic = 1*10**-6 # # uA # capacitor current\n", + "Vt = 0.7 # # V # threshold voltage\n", + "\n", + "# the slew rate of an op-amp is defined as\n", + "# Slew rate = (dVo/dt)\n", + "Slewrate = 8*3.14*Vt*f #\n", + "print 'the slew rate of an op-amp is = %0.2f'%Slewrate,' V/u sec '#\n", + "\n", + "# The compansated capacitance Cm is\n", + "gm = (Ic/Vt)#\n", + "Cm = (gm/4*3.14*f)*1e6 # pF\n", + "print 'The compansated capacitance value is = %0.2f'%Cm,'pF '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.25 Pg 129" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Cut -off frequency of an op-amp is = 5.00 Hz \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# To determine the cut off frequency of an op-amp\n", + "f = 1*10**3 # # Hz # unity frequency\n", + "Av = 200 # # V/mV # dc gain\n", + "\n", + "# the unity gain frequency of an op-amp is defined as\n", + "# f = Av*fc #\n", + "\n", + "# cut off frequency\n", + "fc = (f/Av)#\n", + "print 'Cut -off frequency of an op-amp is = %0.2f'%fc,' Hz '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.26 Pg 129" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the closed loop gain ACL is = 35.00 \n", + "The output gain factor K is = 0.88 V\n", + "The maximum frequency of an op-amp fmax = 145.59 KHz\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# To find the maximum frequency of input signal in op-amp circuit\n", + "Vin = 25*10**-3 # # V # input voltage\n", + "Slewrate = 0.8/10**-6 # # V/uV # Slew rate of an op-amp\n", + "R2 = 350*10**3 # # ohm # feedback resistance\n", + "R1 = 10*10**3 # # ohm # input resistance\n", + "\n", + "# the closed loop gain\n", + "# ACL = (mod (Vo/Vin)) = (mod(R2/R1))#\n", + "ACL = abs(R2/R1)#\n", + "print 'the closed loop gain ACL is = %0.2f'%ACL,' '#\n", + "\n", + "# the output gain factor K is given as\n", + "K = ACL*Vin #\n", + "print 'The output gain factor K is = %0.2f'%K,' V'#\n", + "\n", + "# the maximum frequency of an op-amp is\n", + "wmax = (Slewrate/K)#\n", + "fmax = wmax/(2*3.14)#\n", + "print 'The maximum frequency of an op-amp fmax = %0.2f'%(fmax/1000),' KHz'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.27 Pg 129" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the closed loop gain ACL is = 24.00 \n", + "The output gain factor K is = 0.36 V\n", + "The wmax is = 2.22 *10**6 rad/sec\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# To find the maximum frequency of op-amp circuit\n", + "Vin = 0.015 # # V # input voltage\n", + "Slewrate = 0.8 # # V/uV # Slew rate of an op-amp\n", + "R2 = 120*10**3 # # ohm # feedback resistance\n", + "R1 = 5*10**3 # # ohm # input resistance\n", + "\n", + "# the closed loop gain\n", + "# ACL = (mod (Vo/Vin)) = (mod(R2/R1))#\n", + "ACL = abs(R2/R1)#\n", + "print 'the closed loop gain ACL is = %0.2f'%ACL,' '#\n", + "\n", + "# the output gain factor K is given as\n", + "K = ACL*Vin #\n", + "print 'The output gain factor K is = %0.2f'%K,' V'#\n", + "\n", + "# the maximum frequency of an op-amp is\n", + "wmax = (Slewrate/K)#\n", + "print 'The wmax is = %0.2f'%wmax,'*10**6 rad/sec'# # *10**6 because Slewrate is V/uV \n", + "\n", + "# the signal frequency may be w = 500*10**3 rad/sec that is less than the maximum frequency value" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.28 Pg 130" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the unity frequency f is = 568.70 kHz \n", + "The compansated capacitance Cm value is = 0.2 nF \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# To determine the compensated capacitance of an op-amp\n", + "Slewrate = 10 # # V/u sec\n", + "Ic = 1*10**-3 # # mA # capacitor current\n", + "Vt = 0.7 # # V # threshold voltage\n", + "\n", + "# the slew rate of an op-amp is defined as\n", + "# Slew rate = (dVo/dt)\n", + "# the unity frequency f is\n", + "f =(Slewrate/(8*3.14*Vt))#\n", + "f = f*10**6# # *10**6 because Slew rate is V/uV \n", + "print 'the unity frequency f is = %0.2f'%(f/1e3),'kHz '#\n", + "\n", + "# The compansated capacitance Cm is\n", + "gm = (Ic/Vt)#\n", + "Cm = (gm)/(4*3.14*f)*1e9 #\n", + "print 'The compansated capacitance Cm value is = %0.1f'%Cm,'nF '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.29 Pg 131" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the Slew rate of an op-amp is = 0.50 V/u sec\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# To find Slew rate of an op-amp\n", + "Iq = 15 # # uA # bias current\n", + "Cm = 30 # # pF # internal frequency compensated capacitor\n", + "Slewrate = (Iq/Cm)\n", + "print 'the Slew rate of an op-amp is = %0.2f'%Slewrate,' V/u sec'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.30 Pg 131" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the Slew rate of an op-amp is = 0.68 V/u sec\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# To find Slew rate of an op-amp\n", + "Iq = 21 # # uA # bias current\n", + "Cm = 31 # # pF # internal frequency compensated capacitor\n", + "Slewrate = (Iq/Cm)#\n", + "print 'the Slew rate of an op-amp is = %0.2f'%Slewrate,' V/u sec'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.31 Pg 131" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The full power bandwidth FPBW is = 8.12 kHz \n", + "The 3-db frequency or small signal band width f3db is = 10 kHz \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# To determine full power and small signal bandwidth of an op-amp with unity gain\n", + "f = 100*10**6 # # Hz unity gain bandwidth\n", + "ACL = 10**4 # # maximum closed loop gain\n", + "Slewrate = 0.51 # # V/u sec\n", + "Vp = 10 # # V peak volt\n", + "\n", + "# The full power bandwidth\n", + "FPBW = (Slewrate/(2*3.14*Vp))#\n", + "FPBW = FPBW*10**6 # # *10**6 because Slew rate is V/uV \n", + "print 'The full power bandwidth FPBW is = %0.2f'%(FPBW/1e3),'kHz '#\n", + "\n", + "# the 3-db frequency or small signal band width \n", + "f3db = (f/ACL)#\n", + "print 'The 3-db frequency or small signal band width f3db is = %0.f'%(f3db/1e3),'kHz '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.32 Pg 132" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The full power bandwidth FPBW is = 8.12 kHz \n", + "The 3-db frequency or small signal band width f3db is = 10 kHz \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# To determine full power and small signal bandwidth of an op-amp with unity gain\n", + "f = 100*10**6 # # Hz unity gain bandwidth\n", + "ACL = 10**4 # # maximum closed loop gain\n", + "Slewrate = 0.51 # # V/u sec\n", + "Vp = 10 # # V peak volt\n", + "\n", + "# The full power bandwidth\n", + "FPBW = (Slewrate/(2*3.14*Vp))#\n", + "FPBW = FPBW*10**6 # # *10**6 because Slew rate is V/uV \n", + "print 'The full power bandwidth FPBW is = %0.2f'%(FPBW/1e3),'kHz '#\n", + "\n", + "# the 3-db frequency or small signal band width \n", + "f3db = (f/ACL)#\n", + "print 'The 3-db frequency or small signal band width f3db is = %0.f'%(f3db/1e3),'kHz '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.33 Pg 132" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the Slew rate of an op-amp is = 0.31 V/u sec \n", + "The closed loop gain ACL is = 83.33 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# To find Slew rate and closed loop gain of an op-amp\n", + "fu = 1*10**6 # # Hz # unity gain bandwidth\n", + "fmax = 5*10**3 # # KHz # full power bandwidth\n", + "F3db = 12*10**3 # # Hz # small signal bandwidth\n", + "Vp = 10 # # V # peak volt\n", + "\n", + "# the full power bandwidth of an op-amp\n", + "# fmax=FPBW = (Slew rate/2*3.14*Vp)#\n", + "Slewrate = 2*3.14*Vp*fmax#\n", + "Slewrate = Slewrate*(10**-6)# # *10**-6 because Slewrate is V/u \n", + "print 'the Slew rate of an op-amp is = %0.2f'%Slewrate,' V/u sec '#\n", + "\n", + "# # the 3-db frequency or small signal band width \n", + "#f3db = (f/ACL)#\n", + "#the closed loop gain ACL\n", + "ACL = fu/F3db #\n", + "print 'The closed loop gain ACL is = %0.2f'%ACL,' '#" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch6_1.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch6_1.ipynb new file mode 100644 index 00000000..1fffc432 --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch6_1.ipynb @@ -0,0 +1,2059 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6 Applications of Operational Amplifier" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.1 Pg 140" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the input resistance is = 20.00 kohm\n", + "The resistance R2 is = 100.00 kohm\n" + ] + } + ], + "source": [ + "# Design an inverting amplifier\n", + "Av = -5 #\n", + "#V1 = 0.1 sin wt #\n", + "V1 = 0.1 # # *sin wt #\n", + "i = 5*10**-6 #\n", + "\n", + "# the input resistance \n", + "R1 = V1/i / 1000 # kohm\n", + "print 'the input resistance is = %0.2f'%R1,'kohm'#\n", + "\n", + "# The resistance R2\n", + "#Av = -(R2/R1)#\n", + "R2 = -(Av*R1)#\n", + "print 'The resistance R2 is = %0.2f'%R2,'kohm'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.2 Pg 141" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the input resistance is = 20.00 kohm\n", + "The resistance R2 is = 80.00 kohm\n" + ] + } + ], + "source": [ + "# Design an non inverting amplifier\n", + "Av = 5 #\n", + "#V1 = 0.1 sin wt #\n", + "V1 = 0.1 #\n", + "i = -5*10**-6 #\n", + "\n", + "# the input resistance \n", + "R1 = -V1/i/1000 # kohm\n", + "print 'the input resistance is = %0.2f'%R1,'kohm'#\n", + "\n", + "# The resistance R2\n", + "#Av = 1+(R2/R1)#\n", + "R2 = (Av-1)*R1#\n", + "print 'The resistance R2 is = %0.2f'%R2,'kohm'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.3 Pg 146" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the cut off frequency of phase shifter is = 723.43 Hz\n", + "The phase shift is = -15.75\n" + ] + } + ], + "source": [ + "from math import atan ,degrees,pi\n", + "# To calculate phase shift between two extremes \n", + "C = 0.22*10**-6 #\n", + "R = 1*10**3 #\n", + "f = 1*10**3 #\n", + "\n", + "# the cut off frequency of phase shifter \n", + "fc = 1/(2*pi*R*C) #\n", + "print 'the cut off frequency of phase shifter is = %0.2f'%fc,'Hz'#\n", + "f\n", + "# the phase shift\n", + "f = 1 # # KHz\n", + "fc = 7.23 # # KHz \n", + "PS = -2*degrees(atan(f/fc))\n", + "print 'The phase shift is = %0.2f'%PS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.4 Pg 146" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of resistance is = 1.92 kohm\n" + ] + } + ], + "source": [ + "# To design a phase shifter\n", + "f = 2*10**3 #\n", + "PS = -135 #\n", + "# the phase shift\n", + "# PS = -2*atand(2*pi*R*C)#\n", + "#RC = 192.1*10**-6 #\n", + "C = 0.1*10**-6 #\n", + "R = (192.1*10**-6)/C/1000 # kohm\n", + "print 'The value of resistance is = %0.2f'%R,'kohm'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.5 Pg 153" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of resistance R1 is = 25.00 kohm\n", + "The value of resistance R3 is = 25.00 kohm\n", + "The value of resistance R2 is = 750.00 kohm\n", + "The value of resistance R4 is = 750.00 kohm\n" + ] + } + ], + "source": [ + "# Design a difference amplifier\n", + "Ri = 50 # kohm\n", + "Ad = 30 \n", + "\n", + "R1 = Ri/2 #\n", + "print 'The value of resistance R1 is = %0.2f'%R1,'kohm'#\n", + "R3 = R1 #\n", + "print 'The value of resistance R3 is = %0.2f'%R3,'kohm'#\n", + "\n", + "# the differential gain\n", + "#Ad = R2/R1 #\n", + "R2 = 30*R1 #\n", + "print 'The value of resistance R2 is = %0.2f'%R2,'kohm'#\n", + "\n", + "R4 = R2 #\n", + "print 'The value of resistance R4 is = %0.2f'%R4,'kohm'# " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.6 Pg 154" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The common mode rejection ratio is = 26.58 dB\n" + ] + } + ], + "source": [ + "from math import log10\n", + "# Calculate CMRR ratio\n", + "Ad = 10.24 #\n", + "Acm = 0.48 #\n", + "\n", + "# the common mode rejection ratio CMRR is defined as\n", + "CMRRdB = 20*log10(Ad/Acm)#\n", + "print 'The common mode rejection ratio is = %0.2f'%CMRRdB,' dB'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.7 Pg 156" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The feedback resistance is = 100.00 kohm\n", + " The value of resistance R1 is = 100.00 kohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Design current to voltage converter\n", + "Vo =-10 #\n", + "Is = 100*10**-6 #\n", + "\n", + "# the output voltage of current to voltage converter is defined as\n", + "#Vo =-1s*R2 \n", + "R2 = -Vo/Is/1000 #kohm\n", + "print ' The feedback resistance is = %0.2f'%R2,'kohm'#\n", + "\n", + "R1 = R2 #\n", + "print ' The value of resistance R1 is = %0.2f'%R1,'kohm'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.8 Pg 157" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of resistance R2 is = 44.78 kohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Design high sensitivity current to voltage converter\n", + "R1 = 5 # kohm\n", + "Is = 1 #\n", + "KR = 0.01/10**9 # # V / nA\n", + "\n", + "# the output voltage of high sensitivity current to voltage converter\n", + "Vo =-KR*Is #\n", + "KR = 10*10**6 #\n", + "R = 1*10**6 # #we assume then\n", + "K = 10 #\n", + "#1 + (R2/R1)+(R2/R) = 10 #\n", + "# solving above equation we get\n", + "\n", + "R2 = 9*((5*10**6)/(10**3+5))/1000 # kohm\n", + "print 'The value of resistance R2 is = %0.2f'%R2,'kohm'# " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.9 Pg 160" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The load current iL is = 5.00 mA\n", + "The current i3 is = 0.50 mA\n", + "The current iA is = 5.50 mA\n", + "The output voltage is = 6.00 V\n", + "The current i1 is = 49.40 A\n", + "The current i2 is = 49.40 A\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Determine a load current in a V to I converter\n", + "R1 = 10 # kohm\n", + "R2 = 10 # koohm\n", + "R3 = 1 # kohm\n", + "R4 = 1 # kohm\n", + "VI = -5 #\n", + "\n", + "# The Load Current\n", + "iL = -VI/R3 #\n", + "print 'The load current iL is = %0.2f'%iL,'mA'#\n", + "\n", + "VL = 0.5 #\n", + "# The Current i3 and iA\n", + "i3 = VL/R3 #\n", + "print 'The current i3 is = %0.2f'%i3,'mA'#\n", + "\n", + "iA = i3+iL #\n", + "print 'The current iA is = %0.2f'%iA,'mA'#\n", + "\n", + "# the output voltage \n", + "Vo = (iA*R3)+VL #\n", + "print 'The output voltage is = %0.2f'%Vo,' V'#\n", + "\n", + "ZL =100 #\n", + "# The current i1 and i2 \n", + "#i1 = (VI-iL*ZL)/R1 #\n", + "i1 = (iL*ZL-Vo)/R2 #\n", + "print 'The current i1 is = %0.2f'%i1,' A'#\n", + "\n", + "i2 = i1 #\n", + "print 'The current i2 is = %0.2f'%i2,' A'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.10 Pg 163" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of resistance R1f is = 0.0606 K ohm \n", + "The value of resistance R2 is = 75.5 K ohm \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Design an instrumentation amplifier\n", + "#A = 5 to 500 # adjustable gain\n", + "VR = 100*10**3 #\n", + "\n", + "# the maximum differential gain of instrumentation amplifier is 500 \n", + "#Amax = (R4/R3)*(1+(2R2/R1))#\n", + "#by solving above equation we get following equation\n", + "# 2R2 -249R1f = 0 equation 1\n", + "\n", + "# the minimum differential gain of instrumentation amplifier is 5\n", + "# Amin = (R4/R3)*(1+(2R2/R1)) #\n", + "#by solving above equation we get following equation\n", + "# 2R2 -1.5R1f = 150*10**3 equation 2\n", + "\n", + "#by solving equation 1 and 2 we get\n", + "print 'The value of resistance R1f is = 0.0606 K ohm '#\n", + "\n", + "print 'The value of resistance R2 is = 75.5 K ohm '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.11 Pg 164" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of resistane R1 is = 9.09 kohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# To find the value of resistance R1 for instrumentation amplifier\n", + "A =100 #\n", + "R2 = 450*10**3 #\n", + "R3 = 1*10**3 #\n", + "R4 = 1*10**3 #\n", + "\n", + "# The gain of differential amplifier \n", + "# A = (R4/R3)*(1+(2R2/R1)) #\n", + "#but R3 = R4 then\n", + "# A = 1+(2R2/R1) #\n", + "R1 = 2*R2/(A-1)/1000 # kohm\n", + "print 'The value of resistane R1 is = %0.2f'%R1,'kohm'# " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.12 Pg 167" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " At t = 1 msec the time constant RC is = 0.10 m sec\n", + " if C = 0.01 uF then R of RC time constant is = 10 K ohm \n", + " if C = 0.001 uF then R of RC time constant is = 100 K ohm \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# determine the time constant of an integrator\n", + "Vo = 10 # # at t= 1 m sec\n", + "t = 1 # # m sec\n", + "\n", + "# the output of integrator \n", + "#Vo = t/RC # when t is from 0 to 1\n", + "RC = t/Vo #\n", + "print ' At t = 1 msec the time constant RC is = %0.2f'%RC,' m sec'#\n", + "\n", + "print ' if C = 0.01 uF then R of RC time constant is = 10 K ohm '#\n", + "\n", + "print ' if C = 0.001 uF then R of RC time constant is = 100 K ohm '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.13 Pg 168" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The feedback resistance Rf is = 100.00 kohm\n", + " The frequency fa is = 2.00 kHz\n", + " The value of capacitor C is = 0.8 nF \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# Design an integrator circuit\n", + "A = 10 #\n", + "f =20*10**3 #\n", + "R = 10*10**3 # # we assume \n", + "Rf =10*R #\n", + "\n", + "print ' The feedback resistance Rf is = %0.2f'%(Rf/1000),'kohm'#\n", + "\n", + "# for proper integration f>= 10fa \n", + "fa = f/10/1000 #\n", + "print ' The frequency fa is = %0.2f'%fa,'kHz'#\n", + "\n", + "# in practical integrator\n", + "#fa = 1/(2*pi*Rf*C)#\n", + "\n", + "C = 1/(2*pi*Rf*fa)*1e6# nF\n", + "print ' The value of capacitor C is = %0.1f'%C,'nF '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.1 Pg 185" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the value of resistance R1 is = 0.199 Mohm\n", + "the value of resistance R2 is = 1 Mohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# design an inverting amplifier with a closed loop voltage gain of Av = -5\n", + "Av = -5 #\n", + "Is = 5*10**-6 # # A\n", + "Rs = 1*10**3 # # ohm\n", + "# input voltage source Vs = sinwt volts\n", + "\n", + "# in an inverting amplifier frequency effect is neglected then i/p volt Vin = 1 V and total resistance equal to Rs+R1\n", + "\n", + "# the input current can be written as Iin=Is\n", + "# Is = (Vin/Rs+R1)#\n", + "Iin = Is#\n", + "Vin = 1 # # V\n", + "R1 = (1-(Iin*Rs))/Iin #\n", + "print 'the value of resistance R1 is = %0.3f'%(R1/1e6),'Mohm'#\n", + "\n", + "# closed loop voltage gain of an inverting amplifier\n", + "#Av = -(R2/Rs+R1)\n", + "R2 = -(Av*(Rs+R1))#\n", + "print 'the value of resistance R2 is = %0.f'%(R2/1e6),'Mohm'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.2 Pg 186" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the value of resistance R1 is = 8 kohm\n", + "the value of resistance R2 is = 72 kohm\n" + ] + } + ], + "source": [ + " # design an inverting amplifier with a closed loop voltage gain of Av = 10\n", + "Av = 10 #\n", + "Vin = 0.8 # #V\n", + "Iin = 100*10**-6 # # A\n", + "# in an non- inverting amplifier the input voltage Vin=V1=V2 because of vortual short effect then the i/p current In = Vin/R1\n", + "R1 = Vin/Iin/1e3\n", + "print 'the value of resistance R1 is = %0.f'%R1,'kohm'#\n", + "\n", + "# closed loop voltage gain of an non-inverting amplifier\n", + "#Av = Vo/Vin = (1+R2/R1)\n", + "R2 = (Av-1)*R1 # kohm\n", + "print 'the value of resistance R2 is = %0.f'%R2,'kohm'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.3 Pg 187" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the value of resistance R1 is = 20.00 kohm\n", + "the value of resistance R2 is = 80.00 kohm\n", + "the output current I2 is = 50 uA\n" + ] + } + ], + "source": [ + "# design an non-inverting amplifier with colsed loop gain of 5 limited voltage of -5 V <= Vo <= 5 V and maximum i/p c/n 50 uA\n", + "R1 = 8*10**3 # # ohm\n", + "R2 = 72*10**3 # # ohm\n", + "Iin = 50*10**-6 # # A\n", + "Vo = 5 # # V \n", + "\n", + "# closed loop gain\n", + "#Av = Vo/Vin = (1+R2/R1)\n", + "Av = 1+(R2/R1)#\n", + "# but \n", + "Av = 5 #\n", + "# then\n", + "# (R2/R1) = 4 #\n", + "\n", + "# the output voltage of the amplifier is Vo = 5 V \n", + "#i.e\n", + "Vin = 1 # # V\n", + "# Iin = Vin/R1 #\n", + "R1 = Vin/Iin/1e3\n", + "print 'the value of resistance R1 is = %0.2f'%R1,'kohm'#\n", + "\n", + "R2 = 4*R1 #\n", + "print 'the value of resistance R2 is = %0.2f'%R2,'kohm'#\n", + "\n", + "# the output current I2 is given as\n", + "I2 = (Vo-Vin)/R2*1e3 # uA\n", + "print 'the output current I2 is = %0.f'%I2,'uA'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.4 Pg 188" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the value of resistance Ria is = 20.00 kohm\n", + "the value of resistance Rib is = 15.00 kohm\n", + "the value of resistance Ric is = 30.00 kohm\n" + ] + } + ], + "source": [ + "# Design a op-amp circuit to provide the output voltage Vo = -2(3 V1 +4 V2 +2 V3)\n", + "# Vo = -2(3 V1 + 4 V2+ 2 V3)# equation 1\n", + "# the output of the summer circuit is given as\n", + "# Vo = -R2((Via/Ria)+(Vib/Rib)+(Vic/Ric)) equation 2\n", + "\n", + "# compare equation 1 and 2 of Vo we get \n", + "\n", + "# (R2/Ria)= 6 #\n", + "# (R2/Rbi=8 #\n", + "# (R2/Ric)=4 #\n", + "\n", + "R2 = 120*10**3/1e3 # # we choose then \n", + "\n", + "Ria = R2/6 #\n", + "print 'the value of resistance Ria is = %0.2f'%Ria,'kohm'#\n", + "\n", + "Rib = R2/8 #\n", + "print 'the value of resistance Rib is = %0.2f'%Rib,'kohm'#\n", + "\n", + "Ric = R2/4 #\n", + "print 'the value of resistance Ric is = %0.2f'%Ric,'kohm'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.5 Pg 188" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the value of resistance Ria is = 90.00 kohm\n", + "the value of resistance Rib is = 42.00 kohm\n", + "the value of resistance Ric is = 63.00 kohm\n", + "the value of resistance Rid is = 210.00 kohm\n" + ] + } + ], + "source": [ + " # Design a summing amplifier circuit to provide the output voltage Vo = -(7 V11 + 15 V12 + 10 V13 + 3 V14)\n", + "R2 = 630# kohm # Assume feedback resistance\n", + "# Vo = -(7 V11 + 15 V12 + 10 V13 + 3 V14)# equation 1\n", + "# the output of the summer circuit is given as\n", + "# Vo = -R2((Via/Ria)+(Vib/Rib)+(Vic/Ric)+(Vid/Rid)) equation 2\n", + "\n", + "# compare equation 1 and 2 of Vo we get \n", + "\n", + "# (R2/Ria)= 7 #\n", + "# (R2/Rbi= 15 #\n", + "# (R2/Ric)= 10 #\n", + "# (R2/Rid)= 3 #\n", + "\n", + "Ria = R2/7 #\n", + "print 'the value of resistance Ria is = %0.2f'%Ria,' kohm'#\n", + "\n", + "Rib = R2/15 #\n", + "print 'the value of resistance Rib is = %0.2f'%Rib,' kohm'#\n", + "\n", + "Ric = R2/10 #\n", + "print 'the value of resistance Ric is = %0.2f'%Ric,' kohm'#\n", + "\n", + "Rid = R2/3 #\n", + "print 'the value of resistance Rid is = %0.2f'%Rid,'kohm'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.6 Pg 190" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the value of resistance R2 is = 300.00 kohm\n", + "the value of resistance R4 is = 33333.33 kohm\n" + ] + } + ], + "source": [ + "# Design a op-amp circuit to provide the output voltage Vo = V2 - 3 V1 with Ri1 =Ri2 = 100*10**3\n", + "Ri1 = 100 # # kohm\n", + "Ri2 = 100 # # kohm\n", + "# the i/p resistance \n", + "R1 = Ri1 #\n", + "R3 = Ri2 #\n", + "\n", + "# Vo = V2 - 3 V1# equation 1\n", + "# the output of the summer circuit is given as\n", + "# Vo = [(R4/(R3+R4)*(1+(R2/R1))*Vi2-(R2/R1)*Vi1] equation 2\n", + "\n", + "# compare equation 1 and 2 of Vo we get \n", + "# (R4/(R3+R4)*(1+(R2/R1)) = 1 # equation 3\n", + "# R2/R1 = 3 # equation 4\n", + "\n", + "# by subsituting the value of R1 and R3 in equation 3 and 4\n", + "\n", + "# from equation 4\n", + "R2 = 3*R1 #\n", + "print 'the value of resistance R2 is = %0.2f'%R2,'kohm'#\n", + "\n", + "# from equation 3\n", + "R4 = (100*10**3)/3 #\n", + "print 'the value of resistance R4 is = %0.2f'%R4,'kohm'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.7 Pg 191" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The load current iL is = 5 mA\n", + "The voltage across load VL is = 1.00 V\n", + "The non-inverting current across i3 is = 1.00 mA\n", + "The non-inverting current across i4 is = 6.00 mA\n", + "The output voltage of given voltage to current converter is = 6.00 V\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# determine the load current and output voltage\n", + "Vin = -5 # # V\n", + "ZL = 200 # # ohm\n", + "R1 = 10*10**3 # # ohm\n", + "R2 = 10*10**3 # # ohm\n", + "R3 = 1*10**3 # # ohm\n", + "R4 = 1*10**3 # # ohm\n", + "\n", + "# the load c/n of the given voltage to c/n converter circuit is given by\n", + "iL =-Vin/(R1*R4)*R2*1e3 # m\n", + "print 'The load current iL is = %0.f'%iL,'mA'#\n", + "\n", + "# the voltage across the load \n", + "VL = iL/1e3*ZL#\n", + "print 'The voltage across load VL is = %0.2f'%VL,' V'#\n", + "\n", + "# the non-inverting current across i3 and i4 are\n", + "i3 = VL/R3*1000 #mA\n", + "print 'The non-inverting current across i3 is = %0.2f'%i3,'mA'#\n", + "\n", + "i4 = iL+i3 # mA\n", + "print 'The non-inverting current across i4 is = %0.2f'%i4,'mA'#\n", + "\n", + "# the output voltage of given voltage to current converter is given by\n", + "Vo = (iL/1e3*R3)+VL #\n", + "print 'The output voltage of given voltage to current converter is = %0.2f'%Vo,' V'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.8 Pg 192" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The common mode rejection ratio CMRR is = 120.50 \n", + "The common mode rejection ratio CMRR in decibel is = 41.62 dB \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log10\n", + "# determine the common mode rejection ratio CMRR\n", + "# R2/R1 = 10 #\n", + "# R4/R3 = 11 #\n", + "\n", + "# the output of the difference amplifier is given by\n", + "# Vo = (((R4)/(R3+R4))*(((1+(R2/R1))*VI2))-((R2/R1)*VI1))#\n", + "\n", + "# putting R1 R2 R3 R4 value in above equation we get Vo as\n", + "\n", + "# Vo =(121/12)*VI2-10VI1 # equation 1\n", + "\n", + "# the differential mode input of difference amplifier is given by\n", + "# Vd = VI2-VI1 # eqution 2\n", + "\n", + "# the common mode input of difference amplifier is given by\n", + "# VCM = (VI2+VI1)/2 # equation 3\n", + "\n", + "# from equation 2 and 3 \n", + "\n", + "# VI1 = VCM-Vd/2 # equation 4\n", + "\n", + "# VI2 = VCM+Vd/2 # equation 5\n", + "\n", + "# substitute equation 4 and 5 in 1 we get \n", + "# Vo = (VCM/12)+(241Vd/24)# equation6\n", + "\n", + "# Vd = Ad*Vd+ACM*VCM # equation 7\n", + "\n", + "#equation from equation 6 and 7 we get\n", + "\n", + "Ad = 241/24 #\n", + "ACM = 1/12 #\n", + "\n", + "# the common mode rejection ratio CMRR is \n", + "CMRR = abs(Ad/ACM)#\n", + "print 'The common mode rejection ratio CMRR is = %0.2f'%CMRR,' '#\n", + "\n", + "# in decibal it can be expressed as\n", + "\n", + "CMRR = 20*log10(CMRR)#\n", + "print 'The common mode rejection ratio CMRR in decibel is = %0.2f'%CMRR,' dB '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.9 Pg 194" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output of the difference amplifier is = -8.12 V \n", + "The output of the difference amplifier is = 0.12 V \n", + "the common mode input of difference amplifier is = 2.00 \n", + "the common mode gain ACM of difference amplifier is = 0.06 \n", + "the differential gain of the difference amplifier is = 2.00 \n", + "The common mode rejection ratio CMRR is = 32.00 \n", + "The common mode rejection ratio CMRR in decibel is = 30.10 dB \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log10\n", + "# determine Vo when 1) VI1 = 2 V VI2 = -2 V and 2) VI1 = 2 V VI2 = 2 V\n", + "# and common mode rejection ratio CMRR\n", + "R1 = 10*10**3 # # ohm\n", + "R2 = 20*10**3 # # ohm\n", + "R3 = 10*10**3 # # ohm\n", + "R4 = 22*10**3 # # ohm\n", + "\n", + "\n", + "# the output of the difference amplifier is given by\n", + "# Vo = (((R4)/(R3+R4))*(((1+(R2/R1))*VI2))-((R2/R1)*VI1))#\n", + "\n", + "# Case 1 when VI1 = 2 V VI2 = -2 V\n", + "VI1 = 2 #\n", + "VI2 = -2 #\n", + "\n", + "Vo = (((R4)/(R3+R4))*(((1+(R2/R1))*VI2))-((R2/R1)*VI1))#\n", + "print 'The output of the difference amplifier is = %0.2f'%Vo,' V '#\n", + "\n", + "# case 2 when VI1 = 2 V VI2 = 2 V\n", + "VI1 = 2 #\n", + "VI2 = 2 #\n", + "\n", + "Vo = (((R4)/(R3+R4))*(((1+(R2/R1))*VI2))-((R2/R1)*VI1))#\n", + "print 'The output of the difference amplifier is = %0.2f'%Vo,' V '#\n", + "\n", + "# the common mode input of difference amplifier is given by\n", + "VCM = (VI2+VI1)/2 #\n", + "print 'the common mode input of difference amplifier is = %0.2f'%VCM,' '#\n", + "\n", + "# the common mode gain ACM of difference amplifier is given by\n", + "ACM = Vo/VCM\n", + "print 'the common mode gain ACM of difference amplifier is = %0.2f'%ACM,' '#\n", + "\n", + "# the differential gain of the difference amplifier is given \n", + "Ad = R2/R1 # \n", + "print 'the differential gain of the difference amplifier is = %0.2f'%Ad,' '#\n", + "\n", + "# the common mode rejection ratio CMRR is \n", + "CMRR = abs(Ad/ACM)#\n", + "print 'The common mode rejection ratio CMRR is = %0.2f'%CMRR,' '#\n", + "\n", + "# in decibal it can be expressed as\n", + "CMRR = 20*log10(CMRR)#\n", + "print 'The common mode rejection ratio CMRR in decibel is = %0.2f'%CMRR,' dB '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.10 Pg 195" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the maximum differential voltage gain of the instrumentation amplifier is = 101.00 \n", + "the minimum differential voltage gain of the instrumentation amplifier is = 5.00 \n", + " the range of the differential voltage gain of the instrumentation amplifier is \n", + " 5 <= Av <= 101 \n" + ] + } + ], + "source": [ + "# To determine the range of the differential voltage gain\n", + "#R1 = 1 K ohm to 25 K ohm #\n", + "R2 = 50 # # K ohm\n", + "R3 = 10 # # K ohm\n", + "R4 = 10 # # K ohm\n", + "\n", + "# the output of instrumentation amplifier is given by\n", + "#Vo = (R4/R3)*(1+(2*R2/R1))*(VI@-VI1)#\n", + "\n", + "# the differential voltage gain of the instrumentation amplifier can be written as\n", + "#Av = (Vo/(VI2-VI1)) = (R4/R3)*(1+(2R2/R1))#\n", + "\n", + "# For R1 = 1 K ohm the maximum differential voltage gain of the instrumentation amplifier is\n", + "R1 = 1 # # K ohm\n", + "Av = (R4/R3)*(1+(2*R2/R1))#\n", + "print 'the maximum differential voltage gain of the instrumentation amplifier is = %0.2f'%Av,' '#\n", + "\n", + "# For R1 = 25 K ohm the mminimum differential voltage gain of the instrumentation amplifier is\n", + "R1 = 25 # # K ohm\n", + "Av = (R4/R3)*(1+(2*R2/R1))#\n", + "print 'the minimum differential voltage gain of the instrumentation amplifier is = %0.2f'%Av,' '#\n", + "\n", + "print ' the range of the differential voltage gain of the instrumentation amplifier is '#\n", + "print ' 5 <= Av <= 101 '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.11 Pg 196" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The variable resistance R1 varies is 401 ohm <= R1 <= 100.2 K-ohm \n" + ] + } + ], + "source": [ + "# To design an instrumentation amplifier\n", + "# 4 <= Av <= 1000 # gain\n", + "Ad = 2 #\n", + "Res = 100 # # K ohm\n", + "\n", + "# we cosider the variable resistance is R1 , the maximum and the minimum range of variable resistance \n", + "# R1min = R1 # \n", + "# R1max = R1+100 #\n", + "\n", + "# the gain of difference amplifier \n", + "#A3 = Ad = Vo/(Vo2-Vo1) = (R4/R3)\n", + "\n", + "# the maximum range of differential voltage gain Avmax = 1000 when R1min = R1\n", + "#Avmax = R4/R3*(1+(2*R2/R1min))#\n", + "\n", + "# by solvin we get following equation\n", + "# 499*R1-2*R2=0 equation 1\n", + "\n", + "# the maximum range of differential voltage gain Avmin =4 when R1max = R1+100 K ohm\n", + "# Avmin = (R4/R3)*(1+(2R2/R1max))#\n", + "\n", + "# by solving above equation we get\n", + "# R1 -2 R2 = -200 K ohm equation 2\n", + "\n", + "#by solving equation 1 and 2 we get\n", + "R1 = 401 # # ohm\n", + "R2 = 100.2 # # ohm\n", + "print 'The variable resistance R1 varies is 401 ohm <= R1 <= 100.2 K-ohm ' #" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.12 Pg 198" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The time constant of the given filter is RC = 0.20 msec \n" + ] + } + ], + "source": [ + " # Determine the time constant of the integrator\n", + "Vo = 10 #\n", + "t = 2*10**-3 #\n", + "VI = -1 # # at t =0 #\n", + "\n", + "# The output voltage of an integrator is define as\n", + "RC = t/10*1e3 # ms\n", + "print ' The time constant of the given filter is RC = %0.2f'%RC,'msec '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.13 Pg 198" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The time constant of the given filter is RC = 0.1 msec \n", + "The capacitor value is = 0.1 F\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Determine the time constant of the integrator\n", + "Vo = 20 #\n", + "t = 1*10**-3 #\n", + "VI = -1 # # at t =0 #\n", + "\n", + "# The output voltage of an integrator is define as\n", + "RC = t/10 #\n", + "print ' The time constant of the given filter is RC = %0.1f'%(RC*1000),'msec '#\n", + "\n", + "R = 1*10**3 # # we assume \n", + "C = RC/R*10**6 #\n", + "print 'The capacitor value is = %0.1f'%C,'F'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.14 Pg 199" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of resistance RIa is = 90.00 K ohm \n", + "The value of resistance RIb is = 22.50 K ohm \n", + "The value of resistance RIc is = 18.00 K ohm \n", + "The value of resistance RId is = 15.00 K ohm \n" + ] + } + ], + "source": [ + " # to design a summing amplifier\n", + "\n", + "# the output of the summing amplifier is given by\n", + "#Vo = -R2*((VIa/RIa)+(VIb/RIb)+(VIc/RIc)+(VId/RId))# equation 1\n", + "\n", + "# the equation given is\n", + "#Vo = -(3*VIa+12*VIb+15*VIc+18*VId)# equation 2\n", + "\n", + "# comparing equation 1 and 2\n", + "#R2/RIa = 3 #\n", + "#R2/RIb = 12 #\n", + "#R2/RIc = 15 #\n", + "#R2/RId = 18 # \n", + "\n", + "# the feedback resistance R2= 270 K ohm \n", + "R2 = 270 # # K ohm\n", + "RIa = R2/3 #\n", + "print 'The value of resistance RIa is = %0.2f'%RIa,' K ohm '#\n", + "\n", + "RIb = R2/12 #\n", + "print 'The value of resistance RIb is = %0.2f'%RIb,' K ohm '#\n", + "\n", + "RIc = R2/15 #\n", + "print 'The value of resistance RIc is = %0.2f'%RIc,' K ohm '#\n", + "\n", + "RId = R2/18 #\n", + "print 'The value of resistance RId is = %0.2f'%RId,' K ohm '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.15 Pg 200" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output of first op-amp A1 is = -275*sin wt mV \n", + "The output of second op-amp A2 is = 275*sin wt mV \n", + "The output of third op-amp A3 is = 825*sin wt mV \n", + "current through the resistor R1 and R2 is = 5 sin wt uA \n", + "current through the non-inverting terminal resistor R3 and R4 = 5.5 sin wt uA \n", + "current through the inverting terminal resistor R3 and R4 = 22 sin wt uA \n" + ] + } + ], + "source": [ + "# for the instrumentation amplifier find Vo1 , Vo2 , Vo \n", + "# Vi1 = -25 sin wt # # mV\n", + "# Vi2 = 25 sin wt # # mV\n", + "R1 = 10*10**3 #\n", + "R2 = 20*10**3 #\n", + "R3 = 20*10**3 #\n", + "R4 = 10*10**3 #\n", + "\n", + "# the output of first op-amp A1 is given by\n", + "# Vo1 = (1+(R2/R1))*Vi1-(R2/R1)*Vi2 #\n", + "#by solving above equation we get\n", + "print 'The output of first op-amp A1 is = -275*sin wt mV '#\n", + "\n", + "# the output of second op-amp A2 is given by\n", + "# Vo2 = (1+(R2/R1))*Vi2-(R2/R1)*Vi1 #\n", + "#by solving above equation we get\n", + "print 'The output of second op-amp A2 is = 275*sin wt mV '#\n", + "\n", + "# the output of third op-amp A3 is given by\n", + "# Vo = (R4/R3)-(1+(2R2/R1)*(Vi2-Vi1) #\n", + "#by solving above equation we get\n", + "print 'The output of third op-amp A3 is = 825*sin wt mV '#\n", + "\n", + "# current through the resistor R1 and R2 is\n", + "#i = (Vi1-Vi2)/R1 #\n", + "print 'current through the resistor R1 and R2 is = 5 sin wt uA '#\n", + "\n", + "# current through the non-inverting terminal resistor R3 and R4 \n", + "#i3 = Vo2/(R3+R4)#\n", + "print 'current through the non-inverting terminal resistor R3 and R4 = 5.5 sin wt uA '#\n", + "\n", + "# current through the inverting terminal resistor R3 and R4 \n", + "#i2 = Vo1-(R3/(R3+R4))*Vo2/R3 #\n", + "print 'current through the inverting terminal resistor R3 and R4 = 22 sin wt uA '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.16 Pg 202" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input resistance Rin is = 0.00999 ohm \n", + "The value of Resistance Rs is = 1.0990 K ohm \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# for the a current to voltage converter show a) Rin = (Rf/1+Aop) b) Rf = 10 K ohm Aop = 1000 \n", + "\n", + "#a) The input resistance given as\n", + "#Rin = (Rf)/(1+Aop) #\n", + "\n", + "# The input resistance of the circuit can be written as\n", + "#Rin = (V1/i!)#\n", + "\n", + "# the feedback current of the given circuit is defined as\n", + "#i1 =(V1-Vo)/RF #\n", + "\n", + "# the feedback resistance RF is \n", + "#RF =(V1-Vo)/i1 #\n", + "\n", + "# The output voltage Vo is\n", + "#Vo = -Aop*V1 #\n", + "\n", + "#by using this output feedback currenty i1 can be reformed as\n", + "#i1 = (V1-(-Aop*V1))/RF #\n", + "\n", + "#i1 = V1*(1+Aop)/RF #\n", + "\n", + "# Then Rin Becomes \n", + "#Rin =Rf/(1+Aop)#\n", + "\n", + "Rf =10*10**3 #\n", + "Aop = 1000 #\n", + "\n", + "# the input current and output voltage of the circuit are defined as\n", + "#i1 =(Rs)/(Rs+Rin) #\n", + "# Vo = -(Aop*(RF/1+Aop))*i1 #\n", + "\n", + "#the input resistance Rin is \n", + "Rin =(Rf/(1+Aop)) #\n", + "\n", + "# subsituting the value of RF Aop Rin and Vo we get \n", + "RF = 10 #\n", + "Rin = RF/(1+Aop)\n", + "print 'The input resistance Rin is = %0.5f'%Rin,' ohm '#\n", + "\n", + "Aop = 1000 #\n", + "#(1000/1001)*(Rs/(Rs*0.00999))> 0.99 #\n", + "# by solving above equation we get \n", + "Rs = 1.099 # # K ohm \n", + "print 'The value of Resistance Rs is = %0.4f'%Rs,' K ohm '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.17 Pg 204" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "for Aop = 10**4 closed loop gain is = 0.9999 \n", + "for Aop = 10**3 closed loop gain is = 0.9990 \n", + "for Aop = 10**2 closed loop gain is = 0.9901 \n", + "for Aop = 10**1 closed loop gain is = 0.9091 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# determine the closed loop gain\n", + "\n", + "# the output of the voltage follower is given as\n", + "#Vo = Aop(V1-Vo)#\n", + "\n", + "# the closed loop gain of the voltage follower \n", + "#A = 1/(1+(1/Aop))#\n", + " \n", + "# for Aop = 10**4 closed loop gain\n", + "Aop = 10**4 #\n", + "A = 1/(1+(1/Aop))#\n", + "print 'for Aop = 10**4 closed loop gain is = %0.4f'%A,' '#\n", + "\n", + "# for Aop = 10**3 closed loop gain\n", + "Aop = 10**3 #\n", + "A = 1/(1+(1/Aop))#\n", + "print 'for Aop = 10**3 closed loop gain is = %0.4f'%A,' '#\n", + "\n", + "# for Aop = 10**2 closed loop gain\n", + "Aop = 10**2 #\n", + "A = 1/(1+(1/Aop))#\n", + "print 'for Aop = 10**2 closed loop gain is = %0.4f'%A,' '#\n", + "\n", + "# for Aop = 10**1 closed loop gain\n", + "Aop = 10**1 #\n", + "A = 1/(1+(1/Aop))#\n", + "print 'for Aop = 10**1 closed loop gain is = %0.4f'%A,' '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.18 Pg 205" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "for the frequency f = 10 Hz the output is = 106.10 V \n", + "for the frequency f = 1000 Hz the output is = 1.06 V \n", + "for the frequency f = 10000 Hz the output is = 0.106 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "# To determine the output voltage of integrator\n", + "Vin = 1 #\n", + "R = 150*10**3 ## ohm\n", + "C = 1*10**-9 # # F\n", + "\n", + "# the output voltage of an integrator is given as\n", + "#Vo = (fc/f)*Vin #\n", + "\n", + "#fc = 1/(2*pi*R*C)#\n", + "\n", + "#Vo = (1/(2*pi*R*C*f))*Vin#\n", + "\n", + "#for the frequency f = 10 Hz the output is\n", + "f = 10 # # Hz\n", + "Vo = (1/(2*pi*R*C*f))*Vin#\n", + "print 'for the frequency f = 10 Hz the output is = %0.2f'%Vo,' V '#\n", + "\n", + "#for the frequency f = 1000 Hz the output is\n", + "f = 1000 # # Hz\n", + "Vo = (1/(2*pi*R*C*f))*Vin#\n", + "print 'for the frequency f = 1000 Hz the output is = %0.2f'%Vo,' V '#\n", + "\n", + "#for the frequency f = 10000 Hz the output is\n", + "f = 10000 # # Hz\n", + "Vo = (1/(2*pi*R*C*f))*Vin#\n", + "print 'for the frequency f = 10000 Hz the output is = %0.3f'%Vo,' V '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.19 Pg 206" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The cutoff frequency of the integrator is = 13.263 kHz\n", + "The gain of the integrator is = 3.18\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt\n", + "# To determine the magnitude gain of the integrator\n", + "Vin = 1 #\n", + "f = 50*10**3 #\n", + "Rf = 120*10**3 #\n", + "R = 10*10**3 #\n", + "C = 0.1*10**-9 #\n", + "\n", + "# the magnitude gain of the integrator is given by\n", + "#A = (Rf/R)/(sqrt(1+(f/fc)**2))#\n", + "\n", + "# the cutoff frequency of the integrator \n", + "fc = 1/(2*pi*Rf*C)/1e3\n", + "print 'The cutoff frequency of the integrator is = %0.3f'%fc,'kHz'#\n", + "\n", + "\n", + "A = (Rf/R)/(sqrt(1+(f/fc)**2))*1e3#\n", + "print 'The gain of the integrator is = %0.2f'%A" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.20 Pg 207" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the break frequency fa is = 31.83 kHz \n", + "the break frequency fb is = 21.22 kHz \n", + "The gain of the differentiator is = 0.6667 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt,pi\n", + "# To determine the magnitude gain of the differentiator\n", + "Vin = 1 #\n", + "f = 50*10**3 #\n", + "R = 75*10**3 #\n", + "R1 = 50*10**3 #\n", + "C = 0.1*10**-9 #\n", + "\n", + "# the magnitude gain of the differentiator is given by\n", + "#A = (f/fa)/(sqrt(1+(f/fb)**2))#\n", + "\n", + "# the break frequency fa is defined as\n", + "fa = 1/(2*pi*R1*C) / 1e3\n", + "print 'the break frequency fa is = %0.2f'%fa,'kHz '#\n", + "\n", + "# the break frequency fb is defined as\n", + "fb = 1/(2*pi*R*C) /1e3\n", + "print 'the break frequency fb is = %0.2f'%fb,'kHz '#\n", + "\n", + "\n", + "A = (f/fa)/(sqrt(1+(f/fb)**2))#\n", + "print 'The gain of the differentiator is = %0.4f'%A,' '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.21 Pg 209" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input voltage of an op-amp is = -40 mV\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# to determine the input voltage of an op-amp\n", + "Vo = 2 # # V\n", + "R1 = 20*10**3 # # ohm\n", + "R2 = 1*10**6 # # ohm\n", + "\n", + "# the input voltage of an op-amp\n", + "Vin = -(R1/R2)*Vo *1000 # mV\n", + "print 'The input voltage of an op-amp is = %0.f'%Vin,'mV'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.22 Pg 209" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output voltage of follower Vo1 is = 2.00 V\n", + "The output voltage of an inverting amplifier is = -20.00 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# To determine the output voltage\n", + "Vin = 2 #\n", + "R2 = 20*10**3 #\n", + "R1 = 2*10**3 #\n", + "\n", + "# the output voltage of follower Vo1 is\n", + "Vo1 = Vin #\n", + "print 'the output voltage of follower Vo1 is = %0.2f'%Vo1,' V'#\n", + "# the output voltage of an inverting amplifier\n", + "Vo = -(R2/R1)*Vo1 #\n", + "print 'The output voltage of an inverting amplifier is = %0.2f'%Vo,' V '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.23 Pg 210" + ] + }, + { + "cell_type": "code", + "execution_count": 36, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output of the inverting amplifier is = -15.00 V\n", + "The output of the non-inverting amplifier is = 20.00 V\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# to determine the output voltage of an op-amp\n", + "Vin = 5 # # V\n", + "R1 = 25*10**3 # # ohm\n", + "R2 = 75*10**3 # # ohm\n", + "\n", + "# in this problem op-amp A1 perform the voltage follower and op-amp A2 perform inverting amplifier and op-amp A3 perform non-inverting amplifier\n", + "\n", + "# the output voltage of follower op-amp A1\n", + "Vo1 = Vin #\n", + "\n", + "# the output of the inverting amplifier A2\n", + "Vo2 = -((R2/R1)*Vo1) #\n", + "print 'The output of the inverting amplifier is = %0.2f'%Vo2,' V'#\n", + "\n", + "# the output of the non-inverting amplifier A3\n", + "Vo =(1+(R2/R1))*Vo1 #\n", + "print 'The output of the non-inverting amplifier is = %0.2f'%Vo,' V'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.24 Pg 211" + ] + }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output voltage of an inverting amplifier is = 27.50 V \n", + "the output voltage of follower Vo1 is = 2.50 V\n", + "the output of the inverting summing amplifier is = 101.25 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# To determine the output voltage\n", + "Vin = 2.5 #\n", + "Rf = 100*10**3 #\n", + "R1 = 10*10**3 #\n", + "RI1 = 25*10**3 #\n", + "RI2 = 10*10**3 #\n", + "R2 = 100*10**3 #\n", + "\n", + "# the output voltage of an inverting amplifier\n", + "Vo1 = (1+(R2/R1))*Vin # #\n", + "print 'The output voltage of an inverting amplifier is = %0.2f'%Vo1,' V '#\n", + "\n", + "# the output voltage of follower Vo2 is\n", + "Vo2 = Vin #\n", + "print 'the output voltage of follower Vo1 is = %0.2f'%Vo2,' V'#\n", + "\n", + "# the output of the inverting summing amplifier\n", + "R2 = 75*10**3 #\n", + "Vo = R2*((Vo1/RI1)+(Vo2/RI2))#\n", + "print 'the output of the inverting summing amplifier is = %0.2f'%Vo,' V '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.25 Pg 212" + ] + }, + { + "cell_type": "code", + "execution_count": 38, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the total gain of the circuit is = 36.00 \n", + "The output voltage of the op-amp is = 90.00 V\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# To calculate the output voltage\n", + "Vin = 2.5 # \n", + "R1 = 10*10**3 #\n", + "R2 = 10*10**3 #\n", + "R3 = 10*10**3 #\n", + "Rf = 30*10**3 #\n", + "\n", + "# the total gain of the circuit \n", + "#Av =A1v*A2v*A3v #\n", + "Av = (1+(Rf/R1))*(-Rf/R2)*(-Rf/R3)#\n", + "print 'the total gain of the circuit is = %0.2f'%Av,' '#\n", + "\n", + "# The output voltage of the op-amp \n", + "Vo = Av*Vin #\n", + "print 'The output voltage of the op-amp is = %0.2f'%Vo,' V'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.26 Pg 212" + ] + }, + { + "cell_type": "code", + "execution_count": 39, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output of op-amp A1 is = -10.00 V1\n", + "The output of op-amp A2 is Vo = 40V1 - 2V2 \n", + "The output is equal to the difference between 40V1 and 2V2 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# to calculate the output voltage of op-amp circuit\n", + "Rf = 100*10**3 # # ohm\n", + "R1 = 10*10**3 # # ohm\n", + "R2 = 25*10**3 # # ohm\n", + "R3 = 50*10**3 # # ohm\n", + "\n", + "# the output of op-amp A1 is\n", + "# VA1 = (-Rf/R1)*V1 #\n", + "VA1 = (-Rf/R1)#\n", + "print 'The output of op-amp A1 is = %0.2f'%VA1,'V1' # # *V1 because the output is come from 1 op-amp\n", + "\n", + "# the output of op-amp A2 is\n", + "# Vo = -Rf*((VA1/R2)+(V2/R3))#\n", + "#Vo = -100*(-0.4*V1+0.02V2)#\n", + "print 'The output of op-amp A2 is Vo = 40V1 - 2V2 '# \n", + "\n", + "print 'The output is equal to the difference between 40V1 and 2V2 '# " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.27 Pg 213" + ] + }, + { + "cell_type": "code", + "execution_count": 40, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the upper crossover voltage of schmitt trigger is = 1.00 V\n", + "the lower crossover voltage of schmitt trigger is = -1.00 V\n", + "the hysteresis width HW of schmitt trigger is = 2.00 V\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# to determine the hysteresis width of a schmitt trigger\n", + "R1 = 25*10**3 # # ohm\n", + "R2 = 75*10**3 # # ohm\n", + "VTH = 4 # # V\n", + "VTL = -4 # # V\n", + "\n", + "# the upper crossover voltage of schmitt trigger is defined as\n", + "VU = (R1/(R1+R2))*VTH#\n", + "print 'the upper crossover voltage of schmitt trigger is = %0.2f'%VU,' V'\n", + "\n", + "# the lower crossover voltage of schmitt trigger is defined as\n", + "VL = (R1/(R1+R2))*VTL#\n", + "print 'the lower crossover voltage of schmitt trigger is = %0.2f'%VL,' V'\n", + "\n", + "# the hysteresis width of schmitt trigger is\n", + "HW = VU-VL #\n", + "print 'the hysteresis width HW of schmitt trigger is = %0.2f'%HW,' V'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.28 Pg 214" + ] + }, + { + "cell_type": "code", + "execution_count": 41, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the upper crossover voltage of schmitt trigger is = 1.43 V\n", + "the lower crossover voltage of schmitt trigger is = -1.43 V\n", + "the hysteresis width HW of schmitt trigger is = 2.86 V\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# to determine the hysteresis width of a schmitt trigger\n", + "R1 = 15*10**3 # # ohm\n", + "R2 = 90*10**3 # # ohm\n", + "VTH = 10 # # V\n", + "VTL = -10 # # V\n", + "\n", + "# the upper crossover voltage of schmitt trigger is defined as\n", + "VU = (R1/(R1+R2))*VTH#\n", + "print 'the upper crossover voltage of schmitt trigger is = %0.2f'%VU,' V'\n", + "\n", + "# the lower crossover voltage of schmitt trigger is defined as\n", + "VL = (R1/(R1+R2))*VTL#\n", + "print 'the lower crossover voltage of schmitt trigger is = %0.2f'%VL,' V'\n", + "\n", + "# the hysteresis width of schmitt trigger is\n", + "HW = VU-VL #`\n", + "print 'the hysteresis width HW of schmitt trigger is = %0.2f'%HW,' V'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.29 Pg 214" + ] + }, + { + "cell_type": "code", + "execution_count": 42, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the value of resistance R1 is = 5.00 kohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# to determine the resistance R1 when low and high saturated output states are given\n", + "R2 = 20*10**3 # # ohm\n", + "VH = 2 # # V crossover voltage\n", + "VL = -2 # # V crossover voltage\n", + "VOH = 10 # # V saturated output states\n", + "VOL = -10 # # V saturated output states\n", + "\n", + "# the upper crossover voltage of schmitt trigger is defined as\n", + "# V = (R1/(R1+R2))*VOH#\n", + "# solving above equation we get \n", + "# 2R1+2R2 = 10R1 #\n", + "R1 = (2*R2)/8/1000 # kohm\n", + "print 'the value of resistance R1 is = %0.2f'%R1,'kohm'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.30 Pg 215" + ] + }, + { + "cell_type": "code", + "execution_count": 43, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the value of resistance R2 is = 30.00 kohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# to determine the value of resistance R1 and R2 when low and high saturated output states are given\n", + "VH = 3 # # V crossover voltage\n", + "VL = -3 # # V crossover voltage\n", + "VOH = 12 # # V saturated output states\n", + "VOL = -12 # # V saturated output states\n", + "\n", + "# the upper crossover voltage of schmitt trigger is defined as\n", + "# V = (R1/(R1+R2))*VOH#\n", + "# solving above equation we get \n", + "# 3R1+3R2 = 12R1 #\n", + "\n", + "# 3*R1 = R2 #\n", + "R1 = 10*10**3 # # ohm we assume\n", + "R2 = 3*R1 / 1e3#\n", + "print 'the value of resistance R2 is = %0.2f'%R2,'kohm'" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch7_1.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch7_1.ipynb new file mode 100644 index 00000000..c7941467 --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch7_1.ipynb @@ -0,0 +1,747 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 7 Filters and Rectifiers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.1 Pg 232" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistor value is = 1.59 k ohm \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Design active low filter with cut-off frequency 10 kHz\n", + "fc = 10 # # kHz\n", + "C = 0.01 # #uF # we assume\n", + "\n", + "# the cut-off frequency of active low pass filter is defined as\n", + "# fc = (1/2*pi*R3*C)#\n", + "\n", + "# R3 can be calculated as\n", + "R3 = (1/(2*pi*fc*C))#\n", + "print 'The resistor value is = %0.2f'%R3,' k ohm '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.2 Pg 233" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistor value is = 106 ohm \n", + "The pass band gain is = 1.50 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Design active low filter with cut-off frequency 15 kHz\n", + "fc = 15*10**3 # # Hz \n", + "C = 0.1*10**-6 # #F # we assume\n", + "\n", + "# the cut-off frequency of active low pass filter is defined as\n", + "# fc = (1/2*pi*R3*C)#\n", + "\n", + "# R3 can be calculated as\n", + "R3 = (1/(2*pi*fc*C))#\n", + "print 'The resistor value is = %0.f'%R3,' ohm '\n", + "\n", + "# the pass band gain of filter is given by\n", + "# Af = 1+(R2/R1)#\n", + "# assume that the inverting terminal resistor R2=0.5*R1#\n", + "# in Af equation if we put R2=0.5R1 in R1 R1 cancellout each other \n", + "Af = 1+(0.5)\n", + "print 'The pass band gain is = %0.2f'%Af,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.3 Pg 234" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistor value is = 159 Kohm \n", + "The resistor R2 value is = 900.00 k ohm \n", + "The magnitude of an active low pass filter is = 1.96 \n", + "The phase angle of the filter is = -78.69 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Design active low filter with cut-off frequency 20 kHz\n", + "fc = 20 # # kHz \n", + "f = 100 # # frequency of filter\n", + "Af = 10 # # desired pass band gain\n", + "C = 0.05 # #nF # we assume\n", + "\n", + "# the cut-off frequency of active low pass filter is defined as\n", + "# fc = (1/2*pi*R3*C)#\n", + "\n", + "# R3 can be calculated as\n", + "R3 = (1/(2*pi*fc*1e3*C*1e-9))/1e3 # Kohm\n", + "print 'The resistor value is = %0.f'%R3,' Kohm '\n", + "\n", + "# the pass band gain of filter is given by\n", + "# Af = 1+(R2/R1)#\n", + "# assume that the inverting terminal resistor R1= 100 k ohm#\n", + "R1 = 100 # # k ohm\n", + "R2 = (Af*R1)-R1#\n", + "print 'The resistor R2 value is = %0.2f'%R2,' k ohm '\n", + "\n", + "# the magnitude of an active low pass filter is given as\n", + "A = Af/(sqrt(1+(f/fc)**2))#\n", + "print 'The magnitude of an active low pass filter is = %0.2f'%A,' '\n", + "\n", + "#the phase angle of the filter\n", + "from math import atan , degrees\n", + "Angle = -degrees(atan(f/fc))#\n", + "print 'The phase angle of the filter is = %0.2f'%Angle,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.4 Pg 236" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The frequency of the first order low pass filter is = 2.65 kHz \n", + "The pass band gain of filter is = 13.00 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# to determine the cut-off frequency and pass band gain Af\n", + "R1 = 1 # # k ohm\n", + "R2 = 12 # # k ohm\n", + "R3 = 1.2 # # k ohm\n", + "C = 0.05 # #uF # we assume\n", + "\n", + "# the frequency of the first order low pass filter is defined as\n", + "fc = (1/(2*pi*R3*C))#\n", + "print 'The frequency of the first order low pass filter is = %0.2f'%fc,' kHz '\n", + "\n", + "# the pass band gain of filter is given by\n", + "Af =(1+R2/R1)#\n", + "print 'The pass band gain of filter is = %0.2f'%Af,''" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.5 Pg 236" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The capacitor of high pass filter is = 25.13 uF \n", + "The second resistor value is = 90.00 K ohm \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# to design a first order high pass filter with cut-off frequency 2kHz\n", + "Af = 10 #\n", + "fc = 2 # # kHz \n", + "R3 = 2 # #K ohm # we assume\n", + "R1 = 10 # # k ohm\n", + "# the capacitor of high pass filter is given by\n", + "C = 2*pi*R3*fc#\n", + "print 'The capacitor of high pass filter is = %0.2f'%C,' uF '\n", + "\n", + "# the voltage gain of the high pass filter is\n", + "# Af = 1+(R2/R1)#\n", + "R2 = R1*(Af-1)#\n", + "print 'The second resistor value is = %0.2f'%R2,' K ohm '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.6 Pg 237" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance R3 is = 1.59 K ohm \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# to design an active high pass filter with cut-off frequency 10kHz\n", + "fc = 10 # # kHz \n", + "C = 0.01 # #uF # we assume\n", + "# the cut-off frequency of active high pass filter is given by\n", + "# fc = 2*pi*R3*C#\n", + "# R3 can be calculated as\n", + "R3 = (1/(2*pi*fc*C))#\n", + "\n", + "print 'The resistance R3 is = %0.2f'%R3,' K ohm '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.7 Pg 238" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance R3 is = 64 Kohm \n", + "The pass band gain is = 1.20 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# to design an active high pass filter with cut-off frequency 25kHz\n", + "fc = 25 # # kHz \n", + "C = 0.1 # #nF # we assume\n", + "# the cut-off frequency of active high pass filter is given by\n", + "# fc = 2*pi*R3*C#\n", + "# R3 can be calculated as\n", + "R3 = (1/(2*pi*fc*1e3*C*1e-9)) / 1e3 # Kohm\n", + "print 'The resistance R3 is = %0.f'%R3,' Kohm '\n", + "\n", + "# the desire pass band gain of filter is given by \n", + "#Af = 1+(R2/R1)#\n", + "# assume that the inverting terminal resistor R2=0.2*R1#\n", + "# in Af equation if we put R2=0.2R1 in R1 R1 cancellout each other \n", + "Af = 1+(0.2)\n", + "print 'The pass band gain is = %0.2f'%Af,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.8 Pg 239" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance R3 is = 159 K ohm \n", + "The resistance R2 is = 700.00 K ohm \n", + "The magnitude of an active high pass filter is = 14.55 \n", + "The phase angle of the filter is = 14.04 degree\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "## # to design an active high pass filter with cut-off frequency 20kHz \n", + "Af = 15 #\n", + "fc = 20 # #kHz\n", + "f = 80 # # kHz the frequency of filter \n", + "C = 0.05 # #nF # we assume\n", + "# the cut-off frequency of active high pass filter is given by\n", + "# fc = 2*pi*R3*C#\n", + "# R3 can be calculated as\n", + "R3 = (1/(2*pi*fc*C))#\n", + "print 'The resistance R3 is = %0.f'%(R3*1000),' K ohm ' # Round Off Error\n", + "\n", + "# the desire pass band gain of filter is given by \n", + "#Af = 1+(R2/R1)#\n", + "# assume that the inverting terminal resistor R1=50 K ohm#\n", + "R1 = 50 # # K ohm\n", + "R2 = (R1*Af)-(R1)\n", + "print 'The resistance R2 is = %0.2f'%R2,' K ohm '\n", + "\n", + "# the magnitude of an active high pass filter is given as\n", + "A = Af*(f/fc)/(sqrt(1+(f/fc)**2))#\n", + "print 'The magnitude of an active high pass filter is = %0.2f'%A,' '\n", + "\n", + "#the phase angle of the filter\n", + "from numpy import inf\n", + "Angle = degrees(-atan(f/fc)+atan(inf))\n", + "print 'The phase angle of the filter is = %0.2f'%Angle,' degree' # Round Off Error" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.9 Pg 241" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The lower cut-off frequency FLC of band pass filter is = 159.2 Hz \n", + "The upper cut-off frequency FUC of band pass filter is = 15.92 kHz \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# to calculate upper and lower cut-off frequency of the band pass filter\n", + "R1 = 10*10**3 # #K ohm\n", + "R2 = 10 # #K ohm\n", + "C1 = 0.1*10**-6 # # uF\n", + "C2 = 0.001 # #uF\n", + "\n", + "# the lower cut-off frequency of band pass filter is\n", + "fLC = 1/(2*pi*R1*C1)#\n", + "print 'The lower cut-off frequency FLC of band pass filter is = %0.1f'%fLC,' Hz '\n", + "\n", + "# The upper cut-off frequency of band pass filter is\n", + "fUC = 1/(2*pi*R2*C2)#\n", + "print 'The upper cut-off frequency FUC of band pass filter is = %0.2f'%fUC,' kHz '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.10 Pg 242" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance R3 Value is = 7.96 M ohm \n", + "The resistance R6 value is = 1.59 M ohm \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# to design an active band pass filter with lower cut-off frequency 10 kHz an upper 50 kHz\n", + "fL = 10 # # kHz\n", + "fH = 50 # # kHz\n", + "C1 = 0.002 # # nF\n", + "C2 = 0.002 # # nF\n", + "\n", + "# the lower cut-off frequency of band pass filter is\n", + "# fL = 1/(2*pi*R3*C1)#\n", + "R3 = 1/(2*pi*fL*C1)#\n", + "print 'The resistance R3 Value is = %0.2f'%R3,' M ohm '\n", + "\n", + "# The upper cut-off frequency of band pass filter is\n", + "# fH = 1/(2*pi*R6*C2)#\n", + "R6 = 1/(2*pi*fH*C2)#\n", + "print 'The resistance R6 value is = %0.2f'%R6,' M ohm '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.11 Pg 243" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance R3 Value is = 7.96 M ohm \n", + "The resistance R6 value is = 3.98 M ohm \n", + "The desire pass band gain of filter is = 15 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# to design an active band pass filter with lower cut-off frequency 20 kHz an upper 40 kHz\n", + "fL = 20 # # kHz\n", + "fH = 40 # # kHz\n", + "# the inverting terminal resistance 2R1=R2 and 4R4=R5\n", + "C1 = 0.001 # # nF\n", + "C2 = 0.001 # # nF\n", + "\n", + "# the lower cut-off frequency of band pass filter is\n", + "# fL = 1/(2*pi*R3*C1)#\n", + "R3 = 1/(2*pi*fL*C1)#\n", + "print 'The resistance R3 Value is = %0.2f'%R3,' M ohm '\n", + "\n", + "# The upper cut-off frequency of band pass filter is\n", + "# fH = 1/(2*pi*R6*C2)#\n", + "R6 = 1/(2*pi*fH*C2)#\n", + "print 'The resistance R6 value is = %0.2f'%R6,' M ohm '\n", + "\n", + "# the desire pass band gain of filter is defined as\n", + "R1 = 1 # # M ohm we assume\n", + "#we define inverting terminal resistance 2R1=R2\n", + "R2 = 2 # # M ohm\n", + "# then\n", + "R4 = 1 # #M ohm\n", + "R5 = 4 # # M ohm\n", + "Af = (1+(R2/R1))*(1+(R5/R4))#\n", + "print 'The desire pass band gain of filter is = %d'%Af,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.12 Pg 244" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance R3 Value is = 7.96 M ohm \n", + "The resistance R6 value is = 1.99 M ohm \n", + "The desire pass band gain of filter is = 15.00 \n", + "The magnitude of gain of band pass filter is = 11.49 \n", + "The phase angle of gain of band pass filter is = 50 degree\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# to design an active band pass filter with lower cut-off frequency 20 kHz an upper 80 kHz\n", + "f = 100 # # kHz the frequency of band pass filter\n", + "fL = 20 # # kHz\n", + "fH = 80 # # kHz\n", + "# the inverting terminal resistance R1=0.5*R2 and R4=0.25*R5\n", + "C1 = 0.001 # # nF\n", + "C2 = 0.001 # # nF\n", + "\n", + "# the lower cut-off frequency of band pass filter is\n", + "# fL = 1/(2*pi*R3*C1)#\n", + "R3 = 1/(2*pi*fL*C1)#\n", + "print 'The resistance R3 Value is = %0.2f'%R3,' M ohm '\n", + "\n", + "# The upper cut-off frequency of band pass filter is\n", + "# fH = 1/(2*pi*R6*C2)#\n", + "R6 = 1/(2*pi*fH*C2)#\n", + "print 'The resistance R6 value is = %0.2f'%R6,' M ohm ' # Round Off Error\n", + "\n", + "# the desire pass band gain of filter is defined as\n", + "R1 = 1 # # M ohm we assume\n", + "#we define inverting terminal resistance R1=0.5*R2\n", + "R2 = 2 # # M ohm\n", + "# then\n", + "R4 = 1 # #M ohm\n", + "R5 = 4 # # M ohm\n", + "Af = (1+(R2/R1))*(1+(R5/R4))#\n", + "print 'The desire pass band gain of filter is = %0.2f'%Af,' '\n", + "\n", + "# the magnitude of gain of band pass filter is given as\n", + "A = Af*(f**2/(fL*fH))/((sqrt(1+(f/fL)**2))*(sqrt(1+(f/fH)**2)))#\n", + "print 'The magnitude of gain of band pass filter is = %0.2f'%A,' ' # Round Off Error\n", + "\n", + "#the phase angle of the filter\n", + "from numpy import inf\n", + "Angle = degrees(2*atan(inf)-atan(f/fL)-atan(f/fH))\n", + "print 'The phase angle of gain of band pass filter is = %0.f'%Angle,'degree' # Round Off Error" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.13 Pg 247" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output of the half wave precision rectifier Vo is = -20.00 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# to determine the output voltage of the precision rectifier circuit\n", + "Vi = 10 # #V i/p volt\n", + "R1 = 20 # # K ohm\n", + "R2 = 40 # # K ohm\n", + "Vd = 0.7 # # V the diode voltage drop\n", + "\n", + "# the output of the half wave precision rectifier is defined as\n", + "# Vo = -(R2/R1)*Vi # for Vi < 0\n", + "# = 0 otherwise\n", + "# i.e for Vi > 0\n", + "# Vo = 0\n", + "# for Vi < 0\n", + "Vo = -(R2/R1)*Vi\n", + "print 'The output of the half wave precision rectifier Vo is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.14 Pg 247" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output of the half wave precision rectifier Vo is = -15.00 V \n", + "The output of the half wave precision rectifier Vo is = 15.00 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# to determine the output voltage of the precision rectifier circuit for i/p voltage a) Vi = 5 b) Vi = -5\n", + "Vi = 5 # #V i/p volt\n", + "R1 = 5 # # K ohm\n", + "R2 = 15 # # K ohm\n", + "Vd = 0.7 # # V the diode voltage drop\n", + "\n", + "# the output of the half wave precision rectifier is defined as\n", + "# Vo = -(R2/R1)*Vi # for Vi < 0\n", + "# = 0 otherwise\n", + "\n", + "# for Vi = 5 V\n", + "# i.e for Vi > 0\n", + "# Vo = 0\n", + "# for Vi < 0\n", + "Vo = -(R2/R1)*Vi#\n", + "print 'The output of the half wave precision rectifier Vo is = %0.2f'%Vo,' V '\n", + "\n", + "# for Vi = -5 V\n", + "# i.e for Vi > 0\n", + "# Vo = 0\n", + "# for Vi < 0\n", + "Vi =-5 # # V\n", + "Vo = -(R2/R1)*Vi#\n", + "print 'The output of the half wave precision rectifier Vo is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.15 Pg 248" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The gain of precision full wave rectifier A is = 6.00 \n", + "The output voltage Vo is = 42.00 V \n", + "The output voltage Vo is = 42.00 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# to determine the output voltage of the precision rectifier circuit for i/p voltage a) Vi = 7 b) Vi = -7\n", + "Vi = 7 # #V i/p volt\n", + "R1 = 5 # # K ohm\n", + "R3 = 5 # # K ohm\n", + "R4 = 5 # # K ohm\n", + "R2 = 15 # # K ohm\n", + "R5 = 15 # # K ohm\n", + "Vd = 0.7 # # V the diode voltage drop\n", + "\n", + "# the output of the full wave precision rectifier is defined as\n", + "# Vo = -A*Vi # for Vi < 0 equation 1\n", + "# = A*Vi # otherwise equation 2\n", + "\n", + "# or Vo = abs(A*Vi) #\n", + "\n", + "# The gain of precision full wave rectifier\n", + "A = (((R2*R5)/(R1*R3))-(R5/R4)) #\n", + "print 'The gain of precision full wave rectifier A is = %0.2f'%A,' '\n", + "\n", + "\n", + "# for Vi = 7 V the output voltage is\n", + "Vi = 7 #\n", + "Vo = -A*Vi # # from equation 1\n", + "Vo = A*Vi # # from equation 2\n", + "Vo = abs(A*Vi) #\n", + "print 'The output voltage Vo is = %0.2f'%Vo,' V '\n", + "\n", + "# for Vi = -7 V the output voltage is\n", + "Vi = -7 #\n", + "Vo = -A*Vi # # from equation 1\n", + "Vo = A*Vi # # from equation 2\n", + "Vo = abs(A*Vi) #\n", + "print 'The output voltage Vo is = %0.2f'%Vo,' V '" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch8_1.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch8_1.ipynb new file mode 100644 index 00000000..3309565b --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch8_1.ipynb @@ -0,0 +1,183 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8 Analog Multiplier" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.1 Pg 267" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output voltage of inverting amplifier (V2) is = 3.00 V\n" + ] + } + ], + "source": [ + "from math import sqrt, pi\n", + "from __future__ import division\n", + "# to determine the output voltage of inverting amplifier (V2)\n", + "Vin = 18 # # V\n", + "V1 = -6 # # V\n", + "\n", + "# in the op-amp due to the infinite i/p resiostance the input current is = 0\n", + "# i1+i2 = 0\n", + "# it gives relation\n", + "Vo = -Vin #\n", + "\n", + "# the output of multiplier is defined as\n", + "#Vo = K*V1*V2\n", + "\n", + "K = 1 # # we assume\n", + "\n", + "V2 = (Vo/(K*V1))#\n", + "print 'the output voltage of inverting amplifier (V2) is = %0.2f'%(V2),'V'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.2 Pg 267" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output voltage of multiplier is = 225.00 V\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# to determine the output voltage of multiplier\n", + "Vin = 15 # # V\n", + "\n", + "# the output of multiplier is defined as\n", + "#Vo = K*V1*V2\n", + "# because of i/p terminal the circuit performs mathematical operation squaring\n", + "# i.e V1 = V2 = Vin\n", + "K = 1 # # we assume\n", + "Vo = K*(Vin)**2#\n", + "print 'the output voltage of multiplier is = %0.2f'%(Vo),'V' " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.3 Pg 268" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output voltage of inverting amplifier is = 4.00 V \n", + "the output voltage of multiplier is = 16.00 V \n" + ] + } + ], + "source": [ + "from math import sqrt, pi\n", + "from __future__ import division\n", + "# to determine the output voltage of multiplier and inverting amplifier\n", + "Vin = 16 #\n", + "# the output of the inverting amplifier\n", + "K =1 # # we assume\n", + "Vos = sqrt(abs(Vin)/K) #\n", + "print 'the output voltage of inverting amplifier is = %0.2f'%(Vos),' V '\n", + "\n", + "# the output of the multiplier\n", + "Vo = K*Vos**2 #\n", + "print 'the output voltage of multiplier is = %0.2f'%(Vo),' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.5 Pg 269" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "output voltage of RMS detector is = 10.00 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# output voltage of of RMS detector\n", + "Vin = 10 # \n", + "T = 1 # # we assume that the charging and discharging period of capacitor\n", + "\n", + "# the output voltage of RMS detector\n", + "# Vo =sqrt(1/T*)integrate(Vin**2*(t),t,0,1 [,atol [,rtol]]) #\n", + "Vo = 10 #\n", + "print 'output voltage of RMS detector is = %0.2f'%(Vo),'V '" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch9_1.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch9_1.ipynb new file mode 100644 index 00000000..155ff9ba --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch9_1.ipynb @@ -0,0 +1,413 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9 Phase Locked Loop" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.1 Pg 284" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output voltage of switching regulator circuit is = -0.30 V \n", + "The output voltage of switching regulator circuit is = 1.50 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt, pi\n", + "# to find output voltage for a constant input signal frequency of 200 KHz\n", + "fo = 2*pi*1*10**3 # # KHz/V # VCO sensitivity range 4.1\n", + "fc = 500 # # Hz a free running frequency\n", + "f1 = 200 # # Hz input frequency\n", + "f2 = 2*10**3 # # Hz input frequency\n", + "\n", + "# the output voltage of PLL is defined as\n", + "#Vo = (wo-wc)/ko\n", + "ko = fo #\n", + "# when i/p locked with o/p wo=wi\n", + "# Vo = (wi-wc)/ko #\n", + "\n", + "#for the i/p frequency fi = 200 Hz\n", + "fi = 200 # # Hz\n", + "Vo = (((2*pi*fi)-(2*pi*fc))/ko)#\n", + "print 'The output voltage of switching regulator circuit is = %0.2f'%Vo,' V '\n", + "\n", + "#for the i/p frequency fi = 200 Hz\n", + "fi = 2*10**3 # # Hz\n", + "Vo = (((2*pi*fi)-(2*pi*fc))/ko)#\n", + "print 'The output voltage of switching regulator circuit is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.2 Pg 285" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The sum frequency produce by phase detector is = 900.00 KHz \n", + "The difference frequency produce by phase detector is = 100.00 KHz \n", + "The phase detector frequencies are outside of the low pass filter\n", + "The VCO will be in its free running frequency \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# to find VCO output frequency\n", + "fc = 400 # # KHz a free running frequency\n", + "f = 10 # # KHz low pass filter bandwidth\n", + "fi = 500 # # KHz input frequency\n", + "\n", + "# In PLL a phase detector produces the sum and difference frequencies are defined as\n", + "\n", + "sum = fi+fc #\n", + "print 'The sum frequency produce by phase detector is = %0.2f'%sum,' KHz '\n", + "\n", + "difference = fi-fc #\n", + "print 'The difference frequency produce by phase detector is = %0.2f'%difference,' KHz '\n", + "\n", + "print 'The phase detector frequencies are outside of the low pass filter'#\n", + "\n", + "print 'The VCO will be in its free running frequency '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.3 Pg 286" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The sensitivity of phase detector Kd is = 0.45 \n", + "The maximum control voltage of VCO Vfmax = 1.40 V\n", + "The maximum frequency swing of VCO = 35.00 KHz\n", + "The maximum range of frequency which lock a PLL is = 15.00 KHz \n", + "The maximum range of frequency which lock a PLL is = 85.00 KHz \n", + "The maximum and minimum rage between 15 KHz to 85 KHZ \n", + "The lock range is = 70.00 KHz \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt, pi\n", + "# to determine the lock range of PLL\n", + "Ko = 25 # # KHz\n", + "fo = 50 # # KHz\n", + "A = 2 #\n", + "Vd = 0.7 #\n", + "AL = 1 #\n", + "\n", + "# the amximum output swing of phase detector \n", + "# Vd = Kd*(pi/2) #\n", + "\n", + "# the sensitivity of phase detector Kd is\n", + "Kd = Vd*(2/pi) #\n", + "print 'The sensitivity of phase detector Kd is = %0.2f'%Kd,''\n", + "\n", + "# The maximum control voltage of VCO Vfmax\n", + "Vfmax = (pi/2)*Kd*A #\n", + "print 'The maximum control voltage of VCO Vfmax = %0.2f'%Vfmax,' V'\n", + "\n", + "# the maximum frequency swing of VCO\n", + "fL = (Ko*Vfmax)#\n", + "print 'The maximum frequency swing of VCO = %0.2f'%fL,' KHz'\n", + "\n", + "# The maximum range of frequency which lock a PLL are\n", + "fi = fo-fL #\n", + "print 'The maximum range of frequency which lock a PLL is = %0.2f'%fi,' KHz '\n", + "\n", + "fi = fo+fL #\n", + "print 'The maximum range of frequency which lock a PLL is = %0.2f'%fi,' KHz '\n", + "\n", + "print 'The maximum and minimum rage between 15 KHz to 85 KHZ '\n", + "\n", + "\n", + "# the lock range is\n", + "fLock = 2*fL #\n", + "print 'The lock range is = %0.2f'%fLock,' KHz '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.4 Pg 286" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current through the control resistor R is = 0.60 mA \n", + "The charging time of capacitor is = 5.00 msec \n", + "The total time period of tringular and square wave is = 10.00 msec \n", + "The output frequency of VCO is = 0.10 KHz \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# to determine the output frequency capacitor charging time of VCO\n", + "Vcc = 12 #\n", + "Vcs = 6\n", + "R = 10 # # K ohm\n", + "C = 1 # # uF\n", + "\n", + "# the current through the control resistor R\n", + "i =(Vcc-Vcs)/R #\n", + "print 'The current through the control resistor R is = %0.2f'%i, ' mA '\n", + "\n", + "# The charging time of capacitor \n", + "t = (0.25*Vcc*C)/i #\n", + "print 'The charging time of capacitor is = %0.2f'%t, ' msec '\n", + "\n", + "# In VCO the capacitor charging and discharging time period are equal ,so the total time period of tringular and square wave forms can be written as 2*t #\n", + "t = ((0.5*Vcc*C)/i)#\n", + "print 'The total time period of tringular and square wave is = %0.2f'%t, ' msec '\n", + "\n", + "# the output frequency of VCO is\n", + "fo = 1/t #\n", + "print 'The output frequency of VCO is = %0.2f'%fo, ' KHz '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.5 Pg 287" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The charging or discharging time of capacitor is = 25.00 msec \n", + "The output frequency of VCO is is = 20.00 Hz \n", + "The output frequency of VCO is = 625.00 Kohm\n", + "The current through the control resistor R is = 1.60 uA \n", + "The capacitor charging current is = 2000.00 V = 0.33Vcc \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt, pi\n", + "# to design VCO with output square wave pulse time of 50 msec\n", + "Vcc =6 #\n", + "Vcs = 5 #\n", + "R = 22 # #K ohm\n", + "C = 0.02 # # uF\n", + "t = 50*10**-3 # # sec output square wave pluse\n", + "\n", + "# In VCO the capacitor charging and discharging time period are equal ,so the total time period of tringular and square wave forms can be written as 2*t #\n", + "\n", + "\n", + "# the charging or discharging time of capacitor \n", + "tcap = t/2*1e3 #\n", + "print 'The charging or discharging time of capacitor is = %0.2f'%tcap, ' msec '\n", + "\n", + "# the output frequency of VCO is\n", + "fo = 1/t #\n", + "print 'The output frequency of VCO is is = %0.2f'%fo, ' Hz '\n", + "\n", + "# the output frequency of VCO\n", + " # fo = (1/4*R*C)#\n", + "R = 1/(4*fo*1e3*C*1e-9)/1e3 # Kohm\n", + "print 'The output frequency of VCO is = %0.2f'%R, ' Kohm'\n", + "\n", + "# the current through the control resistor R\n", + "i =(Vcc-Vcs)/R*1e3 #\n", + "print 'The current through the control resistor R is = %0.2f'%i, ' uA '\n", + "\n", + "# the capacitor charging current \n", + "# (V/t)=(i/C) #\n", + "V = (i/C)*tcap #\n", + "print 'The capacitor charging current is = %0.2f'%V, ' V = 0.33Vcc '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.6 Pg 289" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The center frequency of VCO is is = 0.17 kHz \n", + "The lock range of PLL is = 2.67 KHz/V \n", + "The lock range of PLL is = 25.59 k Hz/V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt, pi\n", + "# to determine the center frequency of VCO lock and capture range of PLL\n", + "R = 15 # # K ohm\n", + "C = 0.12 # # uF\n", + "Vcc = 12 #\n", + "\n", + "# the center frequency of VCO fo\n", + "fo = (1.2/4/(R*1e3)/(C*1e-6))/1e3#\n", + "print 'The center frequency of VCO is is = %0.2f'%fo, ' kHz '\n", + "\n", + "fo = 4 # # KHz\n", + "# the lock range of PLL\n", + "fL = (8*fo/Vcc) #\n", + "print 'The lock range of PLL is = %0.2f'%fL, ' KHz/V '\n", + "\n", + "# the capture range of PLL\n", + "fc = ((fo-fL)/(2*pi*3.6*10**3*C*1e-6)**(1/2)) #\n", + "print 'The lock range of PLL is = %0.2f'%fc, 'k Hz/V '\n", + "# ans wrong in the textbook." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.7 Pg 290" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total time period of VCO is = 5.00 usec \n", + "The charging or discharging time of capacitor is = 2.50 usec \n", + "The voltage swing of VCO for 12 V supply is = 3.00 V \n", + "The lock range of PLL FL is = 0.955 Hz \n", + "The capture range is = 437.02 Hz \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt, pi\n", + "# determine the lock range of the FSK demodulator\n", + "Vcc = 12 #\n", + "Fvco = 0.25*Vcc #\n", + "f = 200*10**3 # # Hz\n", + "\n", + "\n", + "# the total time period of VCO \n", + "t = 1/f*1e6 #\n", + "print 'The total time period of VCO is = %0.2f'%t, ' usec '\n", + "\n", + "# In VCO the capacitor charging and discharging time period are equal ,so the total time period of tringular and square wave forms can be written as 2*t #\n", + "\n", + "\n", + "# the charging or discharging time of capacitor \n", + "tcap = t/2 #\n", + "print 'The charging or discharging time of capacitor is = %0.2f'%tcap, ' usec '\n", + "\n", + "# the voltage swing of VCO for 12 V supply\n", + "Fvco = 0.25*Vcc #\n", + "print 'The voltage swing of VCO for 12 V supply is = %0.2f'%Fvco, ' V '\n", + "\n", + "# The lock range of PLL \n", + "#FL = (1/2*pi*f)*(Fvco/tcap)#\n", + "FL = (3/(2*pi*f*tcap*1e-6))#\n", + "print 'The lock range of PLL FL is = %0.3f'%FL, ' Hz '\n", + "\n", + "# the capture range \n", + "fcap = sqrt(f*FL)#\n", + "print 'The capture range is = %0.2f'%fcap, ' Hz '" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/BiasingVoltage_1.png b/Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/BiasingVoltage_1.png Binary files differnew file mode 100644 index 00000000..3079217b --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/BiasingVoltage_1.png diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/InOutCrrnt_1.png b/Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/InOutCrrnt_1.png Binary files differnew file mode 100644 index 00000000..ebf45d77 --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/InOutCrrnt_1.png diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/inAndOutCurrnt_1.png b/Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/inAndOutCurrnt_1.png Binary files differnew file mode 100644 index 00000000..a6158aa6 --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/inAndOutCurrnt_1.png |
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