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| author | Trupti Kini | 2016-05-07 23:30:29 +0600 |
|---|---|---|
| committer | Trupti Kini | 2016-05-07 23:30:29 +0600 |
| commit | cf1d6e42383334d18a21bc297720bf52fca2f488 (patch) | |
| tree | 73d5f1970acba799dfa8b1cf05300b4b554cba8a | |
| parent | 4a32f089d7db0bd3cba8169c194b9e01b5b56d75 (diff) | |
| download | Python-Textbook-Companions-cf1d6e42383334d18a21bc297720bf52fca2f488.tar.gz Python-Textbook-Companions-cf1d6e42383334d18a21bc297720bf52fca2f488.tar.bz2 Python-Textbook-Companions-cf1d6e42383334d18a21bc297720bf52fca2f488.zip | |
Added(A)/Deleted(D) following books
A Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/chapter1.ipynb
A Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/chapter2.ipynb
A Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/chapter3.ipynb
A Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/chapter4.ipynb
A Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/chapter5.ipynb
A Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/chapter6.ipynb
A Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/screenshots/CloseLoopVoltageGain3-5.png
A Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/screenshots/ValueOfResistance3_9.png
A Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/screenshots/requiredResistance3_8.png
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER01_2.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER02_2.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER03_2.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER04_2.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER07_2.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER09_2.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER11_2.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER15_2.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER16_2.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER18_2.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER19_2.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER20_2.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER23_2.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER24_2.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/screenshots/Capture02_2.png
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/screenshots/Capture04_2.png
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/screenshots/Capture20_2.png
A Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch1.ipynb
A Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch10.ipynb
A Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch11.ipynb
A Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch12.ipynb
A Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch13.ipynb
A Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch14.ipynb
A Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch2.ipynb
A Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch3.ipynb
A Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch4.ipynb
A Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch5.ipynb
A Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch6.ipynb
A Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch7.ipynb
A Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch8.ipynb
A Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch9.ipynb
A Electronics_Circuits_and_Systems_by_Y._N._Bapat/screenshots/AntilogOpamp13.png
A Electronics_Circuits_and_Systems_by_Y._N._Bapat/screenshots/LogOpamp13.png
A Electronics_Circuits_and_Systems_by_Y._N._Bapat/screenshots/OutPutLogAmp13.png
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_10.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_11.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_10.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_11.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_10.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_11.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_10.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_11.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_10.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_11.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_10.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_9.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_10.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_11.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_10.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_9.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_10.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_9.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_10.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_11.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_10.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_9.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_10.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_9.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_10.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_11.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_10.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_9.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_10.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_11.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_10.ipynb
A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_11.ipynb
A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.10_3.ipynb
A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.1_3.ipynb
A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.2_3.ipynb
A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.3_3.ipynb
A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.4_3.ipynb
A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.5_3.ipynb
A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.6_3.ipynb
A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.7_3.ipynb
A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.8_3.ipynb
A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.9_3.ipynb
A Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/B.M.D_1_2.JPG
A Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_1_2.jpg
A Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_2_2.jpg
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter10_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter28_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter29_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter11_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter12_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter13_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter14_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter15_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter16_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter17_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter18_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter19_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter1_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter20_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter21_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter22_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter23_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter24_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter25_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter26_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter27_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter2_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter3_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter4_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter5_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter6_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter7_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter8_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter9_2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/screenshots/image11_1.png
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/screenshots/image12_1.png
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/screenshots/image13_1.png
120 files changed, 55260 insertions, 0 deletions
diff --git a/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/chapter1.ipynb b/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/chapter1.ipynb new file mode 100644 index 00000000..fdeac9d0 --- /dev/null +++ b/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/chapter1.ipynb @@ -0,0 +1,2015 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1 - Feedback Amplifiers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.1 Page No. 1-47" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " R_L2 = R_c2 || (R_e1+Rf) =3.21 kohm\n", + " A_i2 = -hfe = -100\n", + " R_i2 = hie = 1100 ohm\n", + " A_v2 = A_i2*R_L2 / R_i2 =-291.82\n", + " A_i1 = -hfe = -100\n", + " R_L1 = R_c1 || R3 || R4 || R_i2 =995.48 ohm\n", + " R_i1 = hie + (1+hfe)*R_e1eff = 11.10 kohm where Re1eff = (R_e1 || Rf)\n", + "Therefore, A_v1 = A_i1*RL1 / Ri1 =-8.96\n", + "The overall voltage gain without feedback is given as,\n", + " Av = A_v1 * A_v2 =2614.71\n", + "The overall voltage gain taking Rs in account is given as,\n", + " Av = Vo / Vs = Av*R_i1 / R_i1+Rs =2591.35\n", + "\n", + "Step 5: Calculate beta\n", + "Looking at Fig.1.33.\n", + " beta = Vf / Vo =0.01\n", + " D = 1 + beta*Av =26.65\n", + " A_vf = Av/D =97.24\n", + " R_if = R_i1 * D =295.79 kohm\n", + " R''_if = R_if || R1 || R2 =18.73 kohm\n", + " R_of = Ro / D = infinity / D = infinity\n", + "Therefore, R''_of = R''_o / D where R''_o = R_L2\n", + "Therefore, R''_of(in ohm) = 120.45 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "rl2=(4.7*10.1)/(4.7+10.1) # in k-ohm\n", + "print \" R_L2 = R_c2 || (R_e1+Rf) =%0.2f kohm\"%rl2\n", + "print \" A_i2 = -hfe = -100\"\n", + "print \" R_i2 = hie = 1100 ohm\"\n", + "av2=(-100*3.21*10**3)/1100\n", + "print \" A_v2 = A_i2*R_L2 / R_i2 =%0.2f\"%av2\n", + "print \" A_i1 = -hfe = -100\"\n", + "rl1=(22*220*22*1.100)/((220*22*1.100)+(22*22*1.100)+(22*220*1.100)+(22*220*22)) # in ohm\n", + "print \" R_L1 = R_c1 || R3 || R4 || R_i2 =%0.2f ohm\"%(rl1*10**3)\n", + "ri1=1.1+(101*((0.1*10)/(0.1+10))) # in k-ohm\n", + "print \" R_i1 = hie + (1+hfe)*R_e1eff = %0.2f kohm where Re1eff = (R_e1 || Rf)\"%ri1\n", + "av1=(-100*995)/(11.099*10**3)\n", + "print \"Therefore, A_v1 = A_i1*RL1 / Ri1 =%0.2f\"%av1\n", + "print \"The overall voltage gain without feedback is given as,\"\n", + "av=-291.82*-8.96\n", + "print \" Av = A_v1 * A_v2 =%0.2f\"%av\n", + "print \"The overall voltage gain taking Rs in account is given as,\"\n", + "aV=(2614.7*11.099*10**3)/((11.099*10**3)+100)\n", + "print \" Av = Vo / Vs = Av*R_i1 / R_i1+Rs =%0.2f\"%aV\n", + "print \"\"\n", + "print \"Step 5: Calculate beta\"\n", + "print \"Looking at Fig.1.33.\"\n", + "beta=100/(100+(10*10**3))\n", + "print \" beta = Vf / Vo =%0.2f\"%beta\n", + "d=1+(0.0099*2591.35)\n", + "print \" D = 1 + beta*Av =%0.2f\"%d\n", + "avf=2591.35/26.65\n", + "print \" A_vf = Av/D =%0.2f\"%avf\n", + "rif=26.65*11.099 # in k-ohm\n", + "print \" R_if = R_i1 * D =%0.2f kohm\"%rif\n", + "riff=(295.788*220*22)/((220*22)+(295.788*22)+(295.788*220)) # in k-ohm\n", + "print \" R''_if = R_if || R1 || R2 =%0.2f kohm\"%riff\n", + "print \" R_of = Ro / D = infinity / D = infinity\"\n", + "print \"Therefore, R''_of = R''_o / D where R''_o = R_L2\"\n", + "roff=(3.21*10**3)/26.65 # in omh\n", + "print \"Therefore, R''_of(in ohm) = %0.2f ohm\"%roff" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.2 Page No. 1-48" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "A_vf = A / 1+A*beta =85.71\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "avf=600/(1+(600*0.01))\n", + "print \"A_vf = A / 1+A*beta =%0.2f\"%avf" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.3 Page No. 1-48" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Given: beta = 0.04, Distortion with feedback = 3%, Distortion without feedback = 15%\n", + "Therefore, D =5.00\n", + "where D = 1 + A*beta = 5\n", + "Therefore, A =100.00\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"Given: beta = 0.04, Distortion with feedback = 3%, Distortion without feedback = 15%\"\n", + "d=15/3\n", + "print \"Therefore, D =%0.2f\"%d\n", + "print \"where D = 1 + A*beta = 5\"\n", + "a=4/0.04\n", + "print \"Therefore, A =%0.2f\"%a" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.4 Page No. 1-49" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) Gain with feedback\n", + " AV_mid = Av_mid / 1+beta*Av_mid =19.61\n", + "(b) f_Lf(in Hz) = f_L / 1+beta*Av_mid =0.98\n", + "(c) f_Hf(in MHz) = f_H * (1+beta*Av_mid) =2.55\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"(a) Gain with feedback\"\n", + "av=1000/(1+(0.05*1000))\n", + "print \" AV_mid = Av_mid / 1+beta*Av_mid =%0.2f\"%av\n", + "flf=50/(1+(0.05*1000)) # in Hz\n", + "print \"(b) f_Lf(in Hz) = f_L / 1+beta*Av_mid =%0.2f\"%flf\n", + "fhf=((50*10**3)*(1+(0.05*1000)))*10**-6 # in MHz\n", + "print \"(c) f_Hf(in MHz) = f_H * (1+beta*Av_mid) =%0.2f\"%fhf" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.5 Page No. 1-49" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) beta: -40 = 20*log[1+beta*A]\n", + "Therefore, 1+beta*A = 100\n", + "Therefore, beta =0.10\n", + "Gain of the amplifier with feedback is given as\n", + " A_Vf = A_V / 1+beta*A_V =10.00\n", + "(b) To maintain output power 10 W, we should maintain output voltage constant and to maintain output constant with feedback gain required Vs is\n", + " V_sf(in V) = Vs * 100 =1.00\n", + "(c) Second harmonic distortion is reduced by factor 1 + beta*A\n", + " D_2f(in percentage) = D_2 / 1+beta*A =0.10\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"(a) beta: -40 = 20*log[1+beta*A]\"\n", + "print \"Therefore, 1+beta*A = 100\"\n", + "b=99/1000\n", + "print \"Therefore, beta =%0.2f\"%b\n", + "print \"Gain of the amplifier with feedback is given as\"\n", + "avf=1000/100\n", + "print \" A_Vf = A_V / 1+beta*A_V =%0.2f\"%avf\n", + "print \"(b) To maintain output power 10 W, we should maintain output voltage constant and to maintain output constant with feedback gain required Vs is\"\n", + "vsf=10*100*10**-3 # in V\n", + "print \" V_sf(in V) = Vs * 100 =%0.2f\"%vsf\n", + "print \"(c) Second harmonic distortion is reduced by factor 1 + beta*A\"\n", + "d2f=(0.1/100)*100 # in percentage\n", + "print \" D_2f(in percentage) = D_2 / 1+beta*A =%0.2f\"%d2f," + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.6 Page No. 1-50" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) We know that\n", + " dAf/Af = 0.1/1+beta*A * dA/A\n", + "Therefore, 1+beta*A = 37.5\n", + "Therefore, beta(in percentage) =1.82\n", + "(b) Af = A / 1+beta*A =53.33\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"(a) We know that\"\n", + "print \" dAf/Af = 0.1/1+beta*A * dA/A\"\n", + "print \"Therefore, 1+beta*A = 37.5\"\n", + "b=(36.5/2000)*100 # in percentage\n", + "format(6)\n", + "print \"Therefore, beta(in percentage) =%0.2f\"%b\n", + "af=2000/(1+(0.01825*2000))\n", + "print \"(b) Af = A / 1+beta*A =%0.2f\"%af" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.7 Page No. 1-50" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage gain of the amplifier with feedback is given as,\n", + "A_vf = A / 1+A*beta where beta = 0.1 and A = 100\n", + "Therefore, A_vf =9.09\n", + "The bandwidth of an amplifier with feedback is given by,\n", + "B_wf = (1+A_mid*beta)f_H - f_L/(1+A_mid*beta)\n", + "Assuming f_H >> f_L we have\n", + "B_w = f_H and B_wf = (1+A_mid*beta)B_w\n", + "Therefore, B_wf(in kHz) =3300.00\n", + "The gain bandwidth product before feedback can be given as\n", + "Gain bandwidth product = A_v*B_w =30000000.00\n", + "Gain bandwidth product after feedback= A_vf*B_wf =29997000.00\n", + "If bandwidth is to be limited to 800 kHz we have f_Hf = 800 kHz assuming f_Hf >> f_Lf\n", + "We know that\n", + "B_wf = (1+A_vmid*beta)*f_H\n", + "Therefore, beta =0.02\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"The voltage gain of the amplifier with feedback is given as,\"\n", + "print \"A_vf = A / 1+A*beta where beta = 0.1 and A = 100\"\n", + "avf=100/(1+(100*0.1))\n", + "print \"Therefore, A_vf =%0.2f\"%avf\n", + "print \"The bandwidth of an amplifier with feedback is given by,\"\n", + "print \"B_wf = (1+A_mid*beta)f_H - f_L/(1+A_mid*beta)\"\n", + "print \"Assuming f_H >> f_L we have\"\n", + "print \"B_w = f_H and B_wf = (1+A_mid*beta)B_w\"\n", + "bwf=(1+(100*0.1))*300\n", + "print \"Therefore, B_wf(in kHz) =%0.2f\"%bwf\n", + "print \"The gain bandwidth product before feedback can be given as\"\n", + "gbp=100*300*10**3\n", + "print \"Gain bandwidth product = A_v*B_w =%0.2f\"%gbp\n", + "gbpf=9.09*3300*10**3\n", + "print \"Gain bandwidth product after feedback= A_vf*B_wf =%0.2f\"%gbpf\n", + "print \"If bandwidth is to be limited to 800 kHz we have f_Hf = 800 kHz assuming f_Hf >> f_Lf\"\n", + "print \"We know that\"\n", + "print \"B_wf = (1+A_vmid*beta)*f_H\"\n", + "b=((8/3)-1)/100\n", + "print \"Therefore, beta =%0.2f\"%b" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.9 Page No. 1-52" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage gain with feedback can be given as\n", + "A_vf = A_v / 1+A_v*beta =19.90\n", + "In a voltage series feedback input resistance with feedback is given as\n", + "R_if = R_i(1+beta*A_v) =402.00 kohm\n", + "In a voltage series feedback output resistance with feedback is given as\n", + "R_of = Ro / 1+beta*A_v =298.51 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"The voltage gain with feedback can be given as\"\n", + "avf=4000/(1+(4000*0.05))\n", + "print \"A_vf = A_v / 1+A_v*beta =%0.2f\"%avf\n", + "print \"In a voltage series feedback input resistance with feedback is given as\"\n", + "rif=2*(1+(0.05*4000))\n", + "print \"R_if = R_i(1+beta*A_v) =%0.2f kohm\"%rif\n", + "rof=(60*10**3)/(1+(0.05*4000))\n", + "print \"In a voltage series feedback output resistance with feedback is given as\"\n", + "print \"R_of = Ro / 1+beta*A_v =%0.2f ohm\"%rof" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.10 Page No. 1-52" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage gain of amplifier can be given as\n", + "A_v = Vo/V_in =1285.71\n", + "(i) beta = 0.012\n", + "Therefore, The gain of the amplifier with feedback is given as\n", + "A_f = A_v / 1+A_v*beta =78.26\n", + "The output voltage with feedback is given as\n", + "Vo = A_f * V_in =2.19 V\n", + "(ii) If the output remains constant at 36V, then the distortion produced within the active devices of the amplifier is unchanged. However, since the distortion at the output is less than in part (i) by a factor of 7, it follows that the feedback now increased by 7 and hence, the voltage gain decreased by 7. Thus, the input signal required to produce the same output (as in part(i)) without feedback must be:\n", + "V_in = 0.20 V\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"The voltage gain of amplifier can be given as\"\n", + "av=36/0.028\n", + "print \"A_v = Vo/V_in =%0.2f\"%av\n", + "print \"(i) beta = 0.012\"\n", + "print \"Therefore, The gain of the amplifier with feedback is given as\"\n", + "af=1285.7/(1+(1285.7*0.012))\n", + "print \"A_f = A_v / 1+A_v*beta =%0.2f\"%af\n", + "print \"The output voltage with feedback is given as\"\n", + "vo=78.26*0.028\n", + "print \"Vo = A_f * V_in =%0.2f V\"%vo\n", + "vin=7*0.028\n", + "print \"(ii) If the output remains constant at 36V, then the distortion produced within the active devices of the amplifier is unchanged. However, since the distortion at the output is less than in part (i) by a factor of 7, it follows that the feedback now increased by 7 and hence, the voltage gain decreased by 7. Thus, the input signal required to produce the same output (as in part(i)) without feedback must be:\"\n", + "print \"V_in = %0.2f V\"%vin" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.11 Page No. 1-53" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The gain of the amplifier is given as\n", + "60 dB = 20 log(Vo/V_s)\n", + "Therefore, A_v = Vo/V_s = 1000\n", + "beta = 1/20 = 0.05\n", + "Therefore, The gain of amplifier with feedback is\n", + "A_vf = A_v / 1+A_v*beta =19.61\n", + "(ii) The gain of the amplifier is directly proportional to the g_m. Therefore, the gain of the amplifier without feedback changes as same amount as g_m changes\n", + "Therefore, A_v = A_v +- 0.5*A_v = 1000 +- 500\n", + "The gain of the amplifier with feedback is now given as\n", + "A_vf = A_v / 1+A_v*beta =19.74\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"(i) The gain of the amplifier is given as\"\n", + "print \"60 dB = 20 log(Vo/V_s)\"\n", + "print \"Therefore, A_v = Vo/V_s = 1000\"\n", + "print \"beta = 1/20 = 0.05\"\n", + "print \"Therefore, The gain of amplifier with feedback is\"\n", + "avf=1000/(1+(1000*0.05))\n", + "print \"A_vf = A_v / 1+A_v*beta =%0.2f\"%avf\n", + "print \"(ii) The gain of the amplifier is directly proportional to the g_m. Therefore, the gain of the amplifier without feedback changes as same amount as g_m changes\"\n", + "print \"Therefore, A_v = A_v +- 0.5*A_v = 1000 +- 500\"\n", + "print \"The gain of the amplifier with feedback is now given as\"\n", + "avf1=1500/(1+(1500*0.05))\n", + "avf2=500/(1+(500*0.05))\n", + "print \"A_vf = A_v / 1+A_v*beta =%0.2f\"%avf1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.12 Page No. 1-54" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "A_v = 1000 and beta = 0.1\n", + "(i) f_Hf/f_H = 1 + beta*A_v =101.00\n", + "and f_Lf/f_L = 1 / 1+beta*A_v =0.01\n", + "(ii) With f_L = 20 Hz and f_H = 50 kHz\n", + "f_Lf =0.20 Hz\n", + "f_Hf =5.05 MHz\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"A_v = 1000 and beta = 0.1\"\n", + "fh=1+(0.1*1000)\n", + "print \"(i) f_Hf/f_H = 1 + beta*A_v =%0.2f\"%fh\n", + "fl=1/(1+(0.1*1000))\n", + "format(7)\n", + "print \"and f_Lf/f_L = 1 / 1+beta*A_v =%0.2f\"%fl\n", + "print \"(ii) With f_L = 20 Hz and f_H = 50 kHz\"\n", + "fll=20*0.0099\n", + "print \"f_Lf =%0.2f Hz\"%fll\n", + "fhh=(50*101)*10**-3\n", + "format(5)\n", + "print \"f_Hf =%0.2f MHz\"%fhh" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.13 Page No. 1-55" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage gain of the amplifier is given as\n", + "A_v = Vo/V_in =250.00\n", + "We know that,\n", + "B_2f = B_2 / 1+A_v*beta =0.02\n", + "Therefore, feedback ratio, beta\n", + "A_vf = A_v / 1+A_v*beta =41.67\n", + "To produce output voltage of 50 V V_in must be\n", + "V_in = 50/A_vf =1.20\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"The voltage gain of the amplifier is given as\"\n", + "av=50/0.2\n", + "print \"A_v = Vo/V_in =%0.2f\"%av\n", + "print \"We know that,\"\n", + "b=((0.06/0.01)-1)/250\n", + "print \"B_2f = B_2 / 1+A_v*beta =%0.2f\"%b\n", + "print \"Therefore, feedback ratio, beta\"\n", + "avf=250/(1+(250*0.02))\n", + "print \"A_vf = A_v / 1+A_v*beta =%0.2f\"%avf\n", + "vin=50/41.66\n", + "print \"To produce output voltage of 50 V V_in must be\"\n", + "print \"V_in = 50/A_vf =%0.2f\"%vin" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.14 Page No. 1-55" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Given A_vf = 120\n", + "A_v = Vo/V_s = Vo/60mV\n", + "and A_vf = Vo/0.5\n", + "Therefore, Vo(in V) =60.00\n", + "with Vo = 60 V we have,\n", + "A_v =1000.00\n", + "We know that,\n", + "A_vf = A_v / 1+A_v*beta\n", + "Therefore, beta =0.01\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"Given A_vf = 120\"\n", + "print \"A_v = Vo/V_s = Vo/60mV\"\n", + "print \"and A_vf = Vo/0.5\"\n", + "vo=0.5*120\n", + "print \"Therefore, Vo(in V) =%0.2f\"%vo\n", + "print \"with Vo = 60 V we have,\"\n", + "av=60/(60*10**-3)\n", + "print \"A_v =%0.2f\"%av\n", + "b=((1000/120)-1)/1000\n", + "print \"We know that,\"\n", + "print \"A_vf = A_v / 1+A_v*beta\"\n", + "print \"Therefore, beta =%0.2f\"%b" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.15 Page No. 1-56" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Step 1: Identify topology\n", + " By shorting output(Vo = 0), feedback voltage V_f becomes zero, hence it is a voltage sampling. Since feedback is mixed in series with input the topology is voltage series feedback amplifier\n", + "\n", + "Step 2 and Step 3: Find input and output circuit.\n", + " To find input circuit, set Vo = 0. This places the parallel combination of resistors 3.3K and 3.3K at the first emitter. To find output circuit, set Ii = 0. This places resistors 3.3K and 3.3K in series across the output. The resultant circuit is shown in fig 1.48\n", + "\n", + "Step 4: Replace transistor with its h-parameter equivalent as shown in fig.1.49\n", + "\n", + "Step 5: Find open loop transfer gain.\n", + "A_v = A_v1*A_v2\n", + " = V_i2/V_i1 * Vo/V_i2\n", + " = Vo/V_i2 = -h_fe*R_L2 / R_i2\n", + "where R_L2(in k-ohm) =6.60 kohm\n", + "and R_i2 = h_ie = 2 K\n", + "Therefore, A_v2 = Vo/V_i2 =-165.00\n", + "V_i2/V_i1 = -h_fe*R_L1 / R_i1\n", + "where R_L1(in k-ohm) =1.92 kohm\n", + "and R_i = [h_ie + (1+h_fe)(3.3K||3.3K)]\n", + "Therefore, R_i1 = 86.15 kohm\n", + "Therefore, A_v1 = V_i2/V_i1 =-1.11\n", + "Therefore, A_v =183.81\n", + "\n", + "Step 6: Calculate beta\n", + "beta = V_f/Vo =0.50\n", + "We know that, D = 1 + beta*A_v :\n", + "A_vf = A_v/D =1.98\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"Step 1: Identify topology\"\n", + "print \" By shorting output(Vo = 0), feedback voltage V_f becomes zero, hence it is a voltage sampling. Since feedback is mixed in series with input the topology is voltage series feedback amplifier\"\n", + "print \"\"\n", + "print \"Step 2 and Step 3: Find input and output circuit.\"\n", + "print \" To find input circuit, set Vo = 0. This places the parallel combination of resistors 3.3K and 3.3K at the first emitter. To find output circuit, set Ii = 0. This places resistors 3.3K and 3.3K in series across the output. The resultant circuit is shown in fig 1.48\"\n", + "print \"\"\n", + "print \"Step 4: Replace transistor with its h-parameter equivalent as shown in fig.1.49\"\n", + "print \"\"\n", + "print \"Step 5: Find open loop transfer gain.\"\n", + "print \"A_v = A_v1*A_v2\"\n", + "print \" = V_i2/V_i1 * Vo/V_i2\"\n", + "print \" = Vo/V_i2 = -h_fe*R_L2 / R_i2\"\n", + "rl2=3.3+3.3\n", + "print \"where R_L2(in k-ohm) =%0.2f kohm\"%rl2\n", + "print \"and R_i2 = h_ie = 2 K\"\n", + "voi=(-50*6.6)/2\n", + "print \"Therefore, A_v2 = Vo/V_i2 =%0.2f\"%voi\n", + "print \"V_i2/V_i1 = -h_fe*R_L1 / R_i1\"\n", + "rl1=((51*2)/(53))\n", + "print \"where R_L1(in k-ohm) =%0.2f kohm\"%rl1\n", + "print \"and R_i = [h_ie + (1+h_fe)(3.3K||3.3K)]\"\n", + "ri=2+(51*1.65)\n", + "print \"Therefore, R_i1 = %0.2f kohm\"%ri\n", + "vi21=(-50*1.92)/(86.15)\n", + "print \"Therefore, A_v1 = V_i2/V_i1 =%0.2f\"%vi21\n", + "av=-165*-1.114\n", + "print \"Therefore, A_v =%0.2f\"%av\n", + "print \"\"\n", + "print \"Step 6: Calculate beta\"\n", + "be=3.3/6.6\n", + "print \"beta = V_f/Vo =%0.2f\"%be\n", + "print \"We know that, D = 1 + beta*A_v :\"\n", + "avf=183.86/92.93\n", + "print \"A_vf = A_v/D =%0.2f\"%avf" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.16 Page No. 1-58" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Step 1: Identify topology\n", + " By shorting output(Vo = 0), feedback voltage V_f becomes zero.The feedback is mixed in series feedback.\n", + "\n", + "Step 2 and Step 3: Find input and output circuit.\n", + " To find input circuit, set Vo = 0. This places the parallel combination of resistors 10K and 1K at the first emitter. To find output circuit, set Ii = 0. This places resistors 10K and 1K in series across the output. The resultant circuit is shown in fig 1.51.\n", + "\n", + "Step 4: Replace transistor with its h-parameter equivalent as shown in fig.1.52.\n", + "\n", + "Step 5: Find open loop transfer gain.\n", + "A_v = A_v1*A_v2\n", + " = V_i2/V_i1 * Vo/V_i2\n", + "Vo/V_i2 = -h_fe*R_L2 / R_i2\n", + "where R_L2 = 3.48 kohm\n", + "and R_i2 = h_ic = 1.1 K\n", + "Therefore, Vo/V_i2 =-158.36\n", + "V_i2/V_i1 = -h_fe*R_L1 / R_i1\n", + "where R_L1 = 523.81 ohm\n", + "and R_i = 82K || [h_ie + (1+h_fe)(1K||10K)]\n", + "Therefore, R_i = 30.06 kohm\n", + "Therefore, V_i2/V_i1 =-0.87\n", + "Therefore, A_v =140.62\n", + "\n", + "Step 6: Calculate beta\n", + "beta = V_f/Vo =0.10\n", + "\n", + "Step 7: Calculate A_vf, R_if and R''_of\n", + "D = 1 + beta*A_v =15.06\n", + "A_vf = A/D =9.34\n", + "R_if = R_i*D =451.86 kohm\n", + "R''_o = R_L2 = 3.484 k-ohm\n", + "R''_of = R''_o/D =231.31 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"Step 1: Identify topology\"\n", + "print \" By shorting output(Vo = 0), feedback voltage V_f becomes zero.The feedback is mixed in series feedback.\"\n", + "print \"\"\n", + "print \"Step 2 and Step 3: Find input and output circuit.\"\n", + "print \" To find input circuit, set Vo = 0. This places the parallel combination of resistors 10K and 1K at the first emitter. To find output circuit, set Ii = 0. This places resistors 10K and 1K in series across the output. The resultant circuit is shown in fig 1.51.\"\n", + "print \"\"\n", + "print \"Step 4: Replace transistor with its h-parameter equivalent as shown in fig.1.52.\"\n", + "print \"\"\n", + "print \"Step 5: Find open loop transfer gain.\"\n", + "print \"A_v = A_v1*A_v2\"\n", + "print \" = V_i2/V_i1 * Vo/V_i2\"\n", + "print \"Vo/V_i2 = -h_fe*R_L2 / R_i2\"\n", + "rl2=(5.1*11)/(16.1)\n", + "print \"where R_L2 = %0.2f kohm\"%rl2\n", + "print \"and R_i2 = h_ic = 1.1 K\"\n", + "voi=(-50*3.484)/1.1\n", + "print \"Therefore, Vo/V_i2 =%0.2f\"%voi\n", + "print \"V_i2/V_i1 = -h_fe*R_L1 / R_i1\"\n", + "rl1=((1.1*1)/(2.1))*10**3\n", + "print \"where R_L1 = %0.2f ohm\"%rl1 #answer in text book is wrong\n", + "print \"and R_i = 82K || [h_ie + (1+h_fe)(1K||10K)]\"\n", + "ri=((82*47.459)/(82+47.459))\n", + "print \"Therefore, R_i = %0.2f kohm\"%ri\n", + "vi21=(-50*523.8)/(30*10**3)\n", + "print \"Therefore, V_i2/V_i1 =%0.2f\"%vi21\n", + "av=-158.36*-0.888\n", + "print \"Therefore, A_v =%0.2f\"%av\n", + "print \"\"\n", + "print \"Step 6: Calculate beta\"\n", + "be=1/10\n", + "print \"beta = V_f/Vo =%0.2f\"%be\n", + "print \"\"\n", + "print \"Step 7: Calculate A_vf, R_if and R''_of\"\n", + "d=1+(0.1*140.62)\n", + "print \"D = 1 + beta*A_v =%0.2f\"%d\n", + "avf=140.62/15.062\n", + "print \"A_vf = A/D =%0.2f\"%avf\n", + "rif=30*15.062\n", + "print \"R_if = R_i*D =%0.2f kohm\"%rif\n", + "print \"R''_o = R_L2 = 3.484 k-ohm\"\n", + "rof=(3.484*10**3)/15.062\n", + "print \"R''_of = R''_o/D =%0.2f ohm\"%rof" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.17 Page No. 1-61" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Step 1: Identify topology\n", + " By shorting output voltage (Vo = 0), feedback voltage Vf becomes zero and hence it is voltage sampling. The feedback voltage is applied in series with the input voltage hence the topology is voltage series feedback.\n", + "\n", + "Step 2 and Step 3: Find input and output circuit.\n", + " To find input circuit, set Vo = 0. This places the parallel combination of resistor 10 K and 200 ohm at first source. To find output circuit, set Ii = 0. This places the resistor 10K and 200 ohm in series across the output. The resultant circuit is shown in fig 1.54.\n", + "\n", + "Step 4: Replace FET with its equivalent circuit as shown in fig.1.55.\n", + "\n", + "Step 5: Find open loop transfer gain.\n", + " Av = Vo / Vs = A_v1 * A_v2\n", + " A_v2 = -u*R_L2 / R_L2+r_d\n", + "where R_L2 = 8.38 kohm\n", + " A_v2 =-18.24\n", + " A_v1 = u*R_Deff / r_d+R_Deff+(1+u)*R_seff\n", + " R_Deff = R_D || R_G2 =44.89 kohm\n", + "Therefore, A_v1 =-28.59\n", + "Therefore, Av = A_v1 * A_v2 =521.40\n", + "\n", + "Step 6: Calculate beta\n", + " beta = Vf / Vo =0.02\n", + "\n", + "step 7: Calculate D, A_vf, R_if, R''_of\n", + " D = 1 + Av*beta =11.43\n", + " A_vf = Av / D =45.62\n", + "Ri = R_G = 1 M-ohm\n", + " R_if = Ri * D =11.43 Mohm\n", + " R''o = rd =10.00 kohm\n", + " R''_of = R''o / D =875.06 ohm\n", + "R''_o = 4.56 kohm\n", + "Therefore, R''_of = R''_o/D =398.94 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"Step 1: Identify topology\"\n", + "print \" By shorting output voltage (Vo = 0), feedback voltage Vf becomes zero and hence it is voltage sampling. The feedback voltage is applied in series with the input voltage hence the topology is voltage series feedback.\"\n", + "print \"\"\n", + "print \"Step 2 and Step 3: Find input and output circuit.\"\n", + "print \" To find input circuit, set Vo = 0. This places the parallel combination of resistor 10 K and 200 ohm at first source. To find output circuit, set Ii = 0. This places the resistor 10K and 200 ohm in series across the output. The resultant circuit is shown in fig 1.54.\"\n", + "print \"\"\n", + "print \"Step 4: Replace FET with its equivalent circuit as shown in fig.1.55.\"\n", + "print \"\"\n", + "print \"Step 5: Find open loop transfer gain.\"\n", + "print \" Av = Vo / Vs = A_v1 * A_v2\"\n", + "print \" A_v2 = -u*R_L2 / R_L2+r_d\"\n", + "rl2=(10.2*47)/(10.2+47) # in k-ohm\n", + "print \"where R_L2 = %0.2f kohm\"%rl2\n", + "av2=(-40*8.38)/18.38\n", + "print \" A_v2 =%0.2f\"%av2\n", + "print \" A_v1 = u*R_Deff / r_d+R_Deff+(1+u)*R_seff\"\n", + "rdeff=(47*1000)/(47+1000) # in k-ohm\n", + "print \" R_Deff = R_D || R_G2 =%0.2f kohm\"%rdeff\n", + "av1=(-40*44.98)/(10+44.89+(41*((0.2*10)/(10+0.2))))\n", + "print \"Therefore, A_v1 =%0.2f\"%av1\n", + "av=-28.59*-18.237\n", + "print \"Therefore, Av = A_v1 * A_v2 =%0.2f\"%av\n", + "print \"\"\n", + "print \"Step 6: Calculate beta\"\n", + "beta=200/(10000)\n", + "print \" beta = Vf / Vo =%0.2f\"%beta\n", + "print \"\"\n", + "print \"step 7: Calculate D, A_vf, R_if, R''_of\"\n", + "d=1+(0.02*521.39)\n", + "print \" D = 1 + Av*beta =%0.2f\"%d\n", + "avf=521.39/11.4278\n", + "print \" A_vf = Av / D =%0.2f\"%avf\n", + "print \"Ri = R_G = 1 M-ohm\"\n", + "rif=11.4278\n", + "print \" R_if = Ri * D =%0.2f Mohm\"%rif\n", + "ro=10 # in k-ohm\n", + "print \" R''o = rd =%0.2f kohm\"%ro\n", + "rof=(10*10**3)/11.4278 # in ohm\n", + "print \" R''_of = R''o / D =%0.2f ohm\"%rof\n", + "rod=(10*8.38)/18.38\n", + "print \"R''_o = %0.2f kohm\"%rod\n", + "rofd=(4.559*10**3)/11.4278\n", + "print \"Therefore, R''_of = R''_o/D =%0.2f ohm\"%rofd" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.18 Page No. 1-63" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Here, output voltage is sampled and fed in series with the input signal. Hence the topology is voltage series feedback.\n", + " The open loop voltage gain for one stage is given as,\n", + " Av = -gm*R_eq\n", + " R_eq = r_d || R_d || (R_i1+R_2) =6.62 kohm\n", + " Av =-33.10\n", + "Av = Overall voltage gain = |A_vmid|**3 =-36297.57\n", + " beta = Vf / Vo = -R_1 / R_g = -R_1 / R_1+R_2 =0.00\n", + " D = 1 + |Av|*beta =2.82\n", + " A_vf = Av / D =-12895.96\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"Here, output voltage is sampled and fed in series with the input signal. Hence the topology is voltage series feedback.\"\n", + "print \" The open loop voltage gain for one stage is given as,\"\n", + "print \" Av = -gm*R_eq\"\n", + "req=(8*40*1000)/((40*1000)+(8*1000)+(8*40)) # in k-ohm\n", + "print \" R_eq = r_d || R_d || (R_i1+R_2) =%0.2f kohm\"%req\n", + "av=-5*6.62\n", + "print \" Av =%0.2f\"%av\n", + "avm=-33.11**3\n", + "print \"Av = Overall voltage gain = |A_vmid|**3 =%0.2f\"%avm # answer in textbook is wrong\n", + "beta=50/(10**6)\n", + "print \" beta = Vf / Vo = -R_1 / R_g = -R_1 / R_1+R_2 =%0.2f\"%beta\n", + "d=1+((-5*10**-5)*-36306)\n", + "print \" D = 1 + |Av|*beta =%0.2f\"%d\n", + "avf=-36306/2.8153\n", + "print \" A_vf = Av / D =%0.2f\"%avf" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.19 Page No. 1-65" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Here, output terminals are B and ground, thus the forward gain is the gain of Q1 and it is,\n", + " A_v= -33.11\n", + "Here beta = V_f / V_B = V_f/V_o * V_o/V_C * V_C/V_B\n", + "where V_B and V_C are voltages at point B and C, respectively.\n", + "Therefore, beta_BN = V_f/V_o * A_v3 * A_v2 because V_o/V_C = A_v3 and V_C/V_B = A_v2\n", + "Therefore, beta = 0.05\n", + "Therefore, |A_vf| = A_vf / 1+|A_f|*beta =-11.76\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"Here, output terminals are B and ground, thus the forward gain is the gain of Q1 and it is,\"\n", + "print \" A_v= -33.11\"\n", + "print \"Here beta = V_f / V_B = V_f/V_o * V_o/V_C * V_C/V_B\"\n", + "print \"where V_B and V_C are voltages at point B and C, respectively.\"\n", + "print \"Therefore, beta_BN = V_f/V_o * A_v3 * A_v2 because V_o/V_C = A_v3 and V_C/V_B = A_v2\"\n", + "bbn=(5*10**-5)*(33.11**2)\n", + "print \"Therefore, beta = %0.2f\"%bbn\n", + "avf=-33.11/2.815\n", + "print \"Therefore, |A_vf| = A_vf / 1+|A_f|*beta =%0.2f\"%avf" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.21 Page No. 1-67" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Step 1: Identify topology\n", + "Making output voltage zero(Vo = 0)# feedback does not become zero and hence it is not voltage sampling. By opening the output loop feedback becomes zero and hence it is a current sampling. As I_i = I_s-I_f, the feedback current appears in shunt with the input signl and hence the topology is current shunt feedback\n", + "\n", + "Step 2 and Step 3: Find input and output circuit\n", + "To find input circuit, set Vo = 0. This gives series combination of 20K and 1K across the input of the first transistor. To find output circuit, set V_i= 0. This gives parallel combination of 20K and 1K at emitter of the second transistor.The resultant circuit is shown in fig 1.61\n", + "\n", + "Step 4: Find open circuit current gain\n", + "A_I = Io/I_s = -I_c2/I_s = -I_c2/I_b2 * I_b2/I_c1 * I_c1/I_b1 * I_b1/I_s\n", + "-I_c2/I_b2 = -h_fe = -100\n", + "I_b2/I_c1 = -R_c1 / R_i2+R_c1\n", + "where R_i2 = h_ie + (1+h_fe)R_e =98.19 kohm\n", + "Therefore, I_b2/I_c1 =-0.11\n", + "I_c1/I_b1 = h_fe = 100\n", + "I_b1/I_s =0.32\n", + "Therefore, A_I =352.07\n", + "\n", + "Step 5: Calculate beta\n", + "beta = I_f/Io = R_e2/R_e2+R'' =0.17\n", + "\n", + "Therefore, D = 1 + beta*A_I =59.68\n", + "A_if = A_I/D =5.90\n", + "R_i = 646.15 ohm\n", + "R_if = R_i/D =10.83 ohm\n", + "Ro = infinity\n", + "Therefore, R_of = infinity because h_oe = 0\n", + "R''_o = Ro || R_c2 = 4 k-ohm\n", + "R''_of = R''_o = 4 k-ohm\n", + "A_vf = A_If*R_L/R_s =23.60\n" + ] + } + ], + "source": [ + "print \"Step 1: Identify topology\"\n", + "print \"Making output voltage zero(Vo = 0)# feedback does not become zero and hence it is not voltage sampling. By opening the output loop feedback becomes zero and hence it is a current sampling. As I_i = I_s-I_f, the feedback current appears in shunt with the input signl and hence the topology is current shunt feedback\"\n", + "print \"\"\n", + "print \"Step 2 and Step 3: Find input and output circuit\"\n", + "print \"To find input circuit, set Vo = 0. This gives series combination of 20K and 1K across the input of the first transistor. To find output circuit, set V_i= 0. This gives parallel combination of 20K and 1K at emitter of the second transistor.The resultant circuit is shown in fig 1.61\"\n", + "print \"\"\n", + "print \"Step 4: Find open circuit current gain\"\n", + "print \"A_I = Io/I_s = -I_c2/I_s = -I_c2/I_b2 * I_b2/I_c1 * I_c1/I_b1 * I_b1/I_s\"\n", + "print \"-I_c2/I_b2 = -h_fe = -100\"\n", + "print \"I_b2/I_c1 = -R_c1 / R_i2+R_c1\"\n", + "ri2=2+(101*(20/21))\n", + "print \"where R_i2 = h_ie + (1+h_fe)R_e =%0.2f kohm\"%ri2\n", + "ibc=(-12)/(98.19+12)\n", + "print \"Therefore, I_b2/I_c1 =%0.2f\"%ibc\n", + "print \"I_c1/I_b1 = h_fe = 100\"\n", + "ibs=(21/22)/(2+(21/22))\n", + "print \"I_b1/I_s =%0.2f\"%ibs\n", + "ai=100*0.109*0.323*100\n", + "print \"Therefore, A_I =%0.2f\"%ai\n", + "print \"\"\n", + "print \"Step 5: Calculate beta\"\n", + "b=4/(24)\n", + "print \"beta = I_f/Io = R_e2/R_e2+R'' =%0.2f\"%b\n", + "print \"\"\n", + "d=1+(0.1667*352)\n", + "print \"Therefore, D = 1 + beta*A_I =%0.2f\"%d\n", + "aif=352/59.67\n", + "print \"A_if = A_I/D =%0.2f\"%aif\n", + "ri=((1*21*2)/((21*2)+(1*2)+(21*1)))*10**3\n", + "print \"R_i = %0.2f ohm\"%ri\n", + "rif=646/59.67\n", + "print \"R_if = R_i/D =%0.2f ohm\"%rif\n", + "print \"Ro = infinity\"\n", + "print \"Therefore, R_of = infinity because h_oe = 0\"\n", + "print \"R''_o = Ro || R_c2 = 4 k-ohm\"\n", + "print \"R''_of = R''_o = 4 k-ohm\"\n", + "avf=(5.9*4)\n", + "print \"A_vf = A_If*R_L/R_s =%0.2f\"%avf" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.22 Page No. 1-69" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Step 1: Identify topology\n", + "Vo = 0, does not make feedback zero, but Io = 0 makes feedback to become zero and hence it is current sampling. The feedback is fed in shunt with the input signal, hence topology is current shunt feedback\n", + "\n", + "Step 2 and Step 3: Find input and output circuit\n", + "To find input circuit, set Vo = 0. This gives series combination of R_e2 and 10K across the input. To find output circuit, set V1= 0. This gives parallel combination of R_e2 and 10K at E2.The resultant circuit is shown in fig 1.63\n", + "\n", + "Step 4: Replace transistor with its h-parameter equivalent as shown in fig 1.64\n", + "\n", + "Step 5 : Find open loop current gain\n", + "A_I = Io/I_s = -I_c/I_s = -I_o/I_b2 * I_b2/I_c1 * I_c1/I_b1 * I_b1/I_s\n", + "Io/I_b2 = -h_fe = -100\n", + "I_c2/I_b2 * I_b1/I_e1 = -h_ie*R_c1 / R_i2+R_c2 because I_b2/I_c1 = R_c1/R_c1+R_i2\n", + "where R_i2 = h_ie + (1+h_fe)(1K||10K) =92.82 kohm\n", + "Therefore, I_b1/I_s =-2.32\n", + "I_b1/I_s =0.48\n", + "A_I =110.66\n", + "Step 6: Calculate beta\n", + "beta = R_e2/R_e2+R'' =0.09\n", + "\n", + "Step 6: Calculate D,A_If, A_vf, R_if, R_of\n", + "D = 1 + beta*A_I =10.96\n", + "A_if = A_I/D =10.01\n", + "R_i = 478.26 ohm \n", + "R_if = R_i/D =43.21 ohm\n", + "Ro = infinity\n", + "Therefore, R_of = Ro*D = infinity because h_oe = 0\n", + "R''_o = 2.2 k-ohm\n", + "R''_of = R''_o = 2.2 k-ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"Step 1: Identify topology\"\n", + "print \"Vo = 0, does not make feedback zero, but Io = 0 makes feedback to become zero and hence it is current sampling. The feedback is fed in shunt with the input signal, hence topology is current shunt feedback\"\n", + "print \"\"\n", + "print \"Step 2 and Step 3: Find input and output circuit\"\n", + "print \"To find input circuit, set Vo = 0. This gives series combination of R_e2 and 10K across the input. To find output circuit, set V1= 0. This gives parallel combination of R_e2 and 10K at E2.The resultant circuit is shown in fig 1.63\"\n", + "print \"\"\n", + "print \"Step 4: Replace transistor with its h-parameter equivalent as shown in fig 1.64\"\n", + "print \"\"\n", + "print \"Step 5 : Find open loop current gain\"\n", + "print \"A_I = Io/I_s = -I_c/I_s = -I_o/I_b2 * I_b2/I_c1 * I_c1/I_b1 * I_b1/I_s\"\n", + "print \"Io/I_b2 = -h_fe = -100\"\n", + "print \"I_c2/I_b2 * I_b1/I_e1 = -h_ie*R_c1 / R_i2+R_c2 because I_b2/I_c1 = R_c1/R_c1+R_i2\"\n", + "ri2=1+(101*(10/11))\n", + "print \"where R_i2 = h_ie + (1+h_fe)(1K||10K) =%0.2f kohm\"%ri2\n", + "ibb=(-100*2.2)/(92.818+2.2)\n", + "print \"Therefore, I_b1/I_s =%0.2f\"%ibb\n", + "ibs=(11/12)/(1+(11/12))\n", + "print \"I_b1/I_s =%0.2f\"%ibs\n", + "ai=100*2.315*0.478\n", + "print \"A_I =%0.2f\"%ai\n", + "print \"Step 6: Calculate beta\"\n", + "b=1/(11)\n", + "print\"beta = R_e2/R_e2+R'' =%0.2f\"% b\n", + "print \"\"\n", + "print \"Step 6: Calculate D,A_If, A_vf, R_if, R_of\"\n", + "d=1+(0.09*110.7)\n", + "print \"D = 1 + beta*A_I =%0.2f\"%d #answer in textbook is wrong\n", + "aif=110.7/11.063\n", + "print \"A_if = A_I/D =%0.2f\"%aif\n", + "ri=((1*11*1)/((11*1)+(1*1)+(11*1)))*10**3\n", + "print \"R_i = %0.2f ohm \"%ri\n", + "rif=478/11.063\n", + "print \"R_if = R_i/D =%0.2f ohm\"%rif\n", + "print \"Ro = infinity\"\n", + "print \"Therefore, R_of = Ro*D = infinity because h_oe = 0\"\n", + "print \"R''_o = 2.2 k-ohm\"\n", + "print \"R''_of = R''_o = 2.2 k-ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.23 Page No. 1-72" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Step 1: Identify topology\n", + "Here output voltage is sampled and fed in shunt with the input siganl such that, I_s-I_f = I_i, hence topology is voltage shunt feedback\n", + "\n", + "Step 2 and Step 3: Find input and output circuit\n", + "To find input circuit, set Vo = 0. This places resistor R across the input. To find output circuit, set V_i = 0. This places resistor R across output. The resultant circuit is shown in fig 1.69\n", + "\n", + "Step 4: Replace transistor with its h-parameter equivalent circuits as shown in fig 1.67\n", + "\n", + "Step 5 : Find open loop transresistance\n", + "R_M = Vo/I_s = R_c*Io/I_s = -R_c*I_c/I_s\n", + " = R_c * -I_c/I_b * I_b/I_s\n", + "-I_c/I_b = -h_fe*R / R+R_c =-87.23\n", + "I_b/I_s = Ro||R / R_s||(R+R_i1)\n", + "R_i1 = h_ie + (1+h_fe)R_e =83.92 kohm\n", + "Therefore, I_b/I_s =0.01\n", + "Therefore, R_M = Vo/I_s =-12.14 kohm\n", + "\n", + "Step 6: Calculate beta\n", + "beta = I_f/Io = V_i-Vo/Vo*R :\n", + " = -1/R = -0.000 because (Vo > V_i)\n", + "\n", + "Step 7: Calculate D, R_Mf, A_vf, R_if, R''_of\n", + "D = 1 + beta*R_M =1.15\n", + "R_Mf = R_M/D =-10.58 kohm\n", + "A_vf = V0/V_s = Vo/I_s*R_s : \n", + " = R_Mf/R_s = -10.57 because R_Mf = Vo/I_s\n", + "R_i = R_s || R_i1 || R =0.98 kohm\n", + "R_if = R_i/D =850.17 ohm\n", + "Ro = infinity\n", + "Therefore, R_of = infinity/D = infinity because h_oe = 0\n", + "R''_o = R || R_c =10.47 kohm\n", + "R''_of = R''_o/D =9.12 kohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"Step 1: Identify topology\"\n", + "print \"Here output voltage is sampled and fed in shunt with the input siganl such that, I_s-I_f = I_i, hence topology is voltage shunt feedback\"\n", + "print \"\"\n", + "print \"Step 2 and Step 3: Find input and output circuit\"\n", + "print \"To find input circuit, set Vo = 0. This places resistor R across the input. To find output circuit, set V_i = 0. This places resistor R across output. The resultant circuit is shown in fig 1.69\"\n", + "print \"\"\n", + "print \"Step 4: Replace transistor with its h-parameter equivalent circuits as shown in fig 1.67\"\n", + "print \"\"\n", + "print \"Step 5 : Find open loop transresistance\"\n", + "print \"R_M = Vo/I_s = R_c*Io/I_s = -R_c*I_c/I_s\"\n", + "print \" = R_c * -I_c/I_b * I_b/I_s\"\n", + "icb=(-100*82)/94\n", + "print \"-I_c/I_b = -h_fe*R / R+R_c =%0.2f\"%icb\n", + "print \"I_b/I_s = Ro||R / R_s||(R+R_i1)\"\n", + "ri1=1.1+(101*820*10**-3)\n", + "print \"R_i1 = h_ie + (1+h_fe)R_e =%0.2f kohm\"%ri1\n", + "ibs=(82/83)/(83.92+(82/83))\n", + "print \"Therefore, I_b/I_s =%0.2f\"%ibs\n", + "rm=-87.23*12*0.0116\n", + "print \"Therefore, R_M = Vo/I_s =%0.2f kohm\"%rm\n", + "print \"\"\n", + "print \"Step 6: Calculate beta\"\n", + "b=-1/(82*10**3)\n", + "print \"beta = I_f/Io = V_i-Vo/Vo*R :\"\n", + "print \" = -1/R = %0.3f because (Vo > V_i)\"%b\n", + "print \"\"\n", + "print \"Step 7: Calculate D, R_Mf, A_vf, R_if, R''_of\"\n", + "d=1+(-1.22*-12.142*10**-2)\n", + "print \"D = 1 + beta*R_M =%0.2f\"%d\n", + "rmf=-12.142/1.148\n", + "print \"R_Mf = R_M/D =%0.2f kohm\"%rmf\n", + "avf=-10.57\n", + "print \"A_vf = V0/V_s = Vo/I_s*R_s : \"\n", + "print \" = R_Mf/R_s = %0.2f because R_Mf = Vo/I_s\"%avf\n", + "ri=((1*82*83.92)/((82*83.92)+(1*83.92)+(82*1)))\n", + "print \"R_i = R_s || R_i1 || R =%0.2f kohm\"%ri\n", + "rif=(0.976*10**3)/1.148\n", + "print \"R_if = R_i/D =%0.2f ohm\"%rif\n", + "print \"Ro = infinity\"\n", + "print \"Therefore, R_of = infinity/D = infinity because h_oe = 0\"\n", + "ro=(12*82)/(94)\n", + "print \"R''_o = R || R_c =%0.2f kohm\"%ro\n", + "rof=(10.468)/1.148\n", + "print \"R''_of = R''_o/D =%0.2f kohm\"%rof" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.24 Page No. 1-75" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Step 1: Identify topology\n", + "Here output voltage is sampled and fed in shunt with the input siganl such that, I_s-I_f = I_i, hence topology is voltage shunt feedback\n", + "\n", + "Step 2 and Step 3: Find input and output circuit\n", + "To find input circuit, set Vo = 0. This places resistor R across the input. To find output circuit, set V_i = 0. This places resistor R across output. The resultant circuit is shown in fig 1.69. The circuit shows voltage source replaced by current source\n", + "\n", + "Step 4: Replace transistor with their h-parameter equivalent circuits as shown in fig 1.70\n", + "\n", + "Step 5 : Find open loop transfer gain\n", + "R_M = Vo/I_s = R_c2*Io/I_s\n", + " = R_c2 * Io/I_b2 * I_b2/I_e1 * I_e1/I_b1 * I_b1/I_s\n", + "Io/I_b2 = -h_ie*R / R+R_c2 =-30.14\n", + "I_b2/I_e2 * I_e1/I_b1 = -h_ie*R / R+R_c2 =35.84\n", + "I_b1/I_s = R_s||R / (R_s||R)+R_i1\n", + "where R_i1 = h_ie + (1+h_fe)R_e =113.10 kohm\n", + "I_b1/I_s =0.01\n", + "Therefore, R_M = -33.27 kohm\n", + "\n", + "Step 6: Calculate beta\n", + "beta = I_f/Io = V_i-Vo/Vo*R : \n", + " = -1/R = -0.00 because (Vo > V_i)\n", + "\n", + "Step 7: Calculate D, R_Mf, A_vf, R_if, R''_of\n", + "D = 1 + beta*R_M =16.24\n", + "R_Mf = R_M/D =-2.06 kohm\n", + "A_vf = V0/V_s = Vo/I_s*R_s =-2.06\n", + "R_i = R_s || R_i1 || R = 683.35 ohm\n", + "R_if = R_i/D = 42.04 ohm\n", + "Ro = infinity\n", + "Therefore, R_of = infinity/D = infinity\n", + "R''_o = R || R_c2 =1.54 kohm\n", + "R''_of = R''_o/D =94.61 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"Step 1: Identify topology\"\n", + "print \"Here output voltage is sampled and fed in shunt with the input siganl such that, I_s-I_f = I_i, hence topology is voltage shunt feedback\"\n", + "print \"\"\n", + "print \"Step 2 and Step 3: Find input and output circuit\"\n", + "print \"To find input circuit, set Vo = 0. This places resistor R across the input. To find output circuit, set V_i = 0. This places resistor R across output. The resultant circuit is shown in fig 1.69. The circuit shows voltage source replaced by current source\"\n", + "print \"\"\n", + "print \"Step 4: Replace transistor with their h-parameter equivalent circuits as shown in fig 1.70\"\n", + "print \"\"\n", + "print \"Step 5 : Find open loop transfer gain\"\n", + "print \"R_M = Vo/I_s = R_c2*Io/I_s\"\n", + "print \" = R_c2 * Io/I_b2 * I_b2/I_e1 * I_e1/I_b1 * I_b1/I_s\"\n", + "iob=(-100*2.2)/7.3\n", + "print \"Io/I_b2 = -h_ie*R / R+R_c2 =%0.2f\"%iob\n", + "iobe=(101*1.1)/3.1\n", + "print \"I_b2/I_e2 * I_e1/I_b1 = -h_ie*R / R+R_c2 =%0.2f\"%iobe\n", + "print \"I_b1/I_s = R_s||R / (R_s||R)+R_i1\"\n", + "ri1=2+(101*1.1)\n", + "print \"where R_i1 = h_ie + (1+h_fe)R_e =%0.2f kohm\"%ri1\n", + "ibs=(2.2/3.2)/((2.2/3.2)+(113.1))\n", + "print \"I_b1/I_s =%0.2f\"%ibs\n", + "rm=5.1*-30.137*35.84*6.04*10**-3\n", + "print \"Therefore, R_M = %0.2f kohm\"%rm\n", + "print \"\"\n", + "print \"Step 6: Calculate beta\"\n", + "b=-1/(2.2*10**3)\n", + "print \"beta = I_f/Io = V_i-Vo/Vo*R : \"\n", + "print \" = -1/R = %0.2f because (Vo > V_i)\"%b\n", + "print \"\"\n", + "print \"Step 7: Calculate D, R_Mf, A_vf, R_if, R''_of\"\n", + "d=1+(4.545*33.539*10**-1)\n", + "print \"D = 1 + beta*R_M =%0.2f\"%d\n", + "rmf=-33.539/16.245\n", + "print \"R_Mf = R_M/D =%0.2f kohm\"%rmf\n", + "avf=-2.065\n", + "print \"A_vf = V0/V_s = Vo/I_s*R_s =%0.2f\"%avf\n", + "ri=((1*113.1*2.2)/((113.1*2.2)+(1*113.1)+(2.2*1)))*10**3\n", + "print \"R_i = R_s || R_i1 || R = %0.2f ohm\"%ri #answer in textbook wrong\n", + "rif=(683)/16.245\n", + "print \"R_if = R_i/D = %0.2f ohm\"%rif\n", + "print \"Ro = infinity\"\n", + "print \"Therefore, R_of = infinity/D = infinity\"\n", + "ro=(2.2*5.1)/(7.3)\n", + "print \"R''_o = R || R_c2 =%0.2f kohm\"%ro\n", + "rof=(1.537*10**3)/16.245\n", + "print \"R''_of = R''_o/D =%0.2f ohm\"%rof" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.25 Page No. 1-78" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Step 1: Identify topology\n", + "By making Vo = 0, feedback current becomes zero. Hence it is a voltage sampling. The feedback is fed in shunt with the input signal and thus the topology is voltage shunt feedback\n", + "\n", + "Step 2 and Step 3: Find input and output circuit\n", + "To find input circuit, set Vo = 0. This places resistor R across the input. To find output circuit, set V_i = 0. This places resistor R across output. The resultant circuit is shown in fig 1.72\n", + "\n", + "Step 4: Replace FET with its equivalent circuit as shown in fig 1.73\n", + "\n", + "Step 5 : Find open loop transresistance\n", + "R_M = Vo/I_s = -g_m*V_gs*R_eff/I_s\n", + "where R_eff = r_d || R || R_D =7.69 kohm\n", + "and V_gs = I_s*R_i = I_s * R_s||1M||R\n", + " = I_s * 10K||1M||200K\n", + " = 9.43*10**3 I_s\n", + "R_M = -181.29 kohm\n", + "\n", + "Step 6: Calculate beta\n", + "beta = I_f/Io = V_i-Vo/Vo*R : \n", + " = -1/R = -0.00 because (Vo > V_i)\n", + "\n", + "Step 7: Calculate D, R_Mf, A_vf, R_of, R''_of\n", + "D = 1 + beta*R_M =1.91\n", + "R_Mf = R_M/D =-95.42 kohm \n", + "A_vf = V0/V_s = Vo/I_s*R_s = R_Mf/R_s =-9.54\n", + "R_i = R_s || M || R =9.43 kohm\n", + "R_if = R_i/D =4.96 kohm\n", + "R''_o = r_eff = r_d || R || R_D =7.69 kohm\n", + "R''_of = R''_o/D =4.05 kohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"Step 1: Identify topology\"\n", + "print \"By making Vo = 0, feedback current becomes zero. Hence it is a voltage sampling. The feedback is fed in shunt with the input signal and thus the topology is voltage shunt feedback\"\n", + "print \"\"\n", + "print \"Step 2 and Step 3: Find input and output circuit\"\n", + "print \"To find input circuit, set Vo = 0. This places resistor R across the input. To find output circuit, set V_i = 0. This places resistor R across output. The resultant circuit is shown in fig 1.72\"\n", + "print \"\"\n", + "print \"Step 4: Replace FET with its equivalent circuit as shown in fig 1.73\"\n", + "print \"\"\n", + "print \"Step 5 : Find open loop transresistance\"\n", + "print \"R_M = Vo/I_s = -g_m*V_gs*R_eff/I_s\"\n", + "reff=(40*200*10)/((200*10)+(400)+(40*200))\n", + "print \"where R_eff = r_d || R || R_D =%0.2f kohm\"%reff\n", + "print \"and V_gs = I_s*R_i = I_s * R_s||1M||R\"\n", + "print \" = I_s * 10K||1M||200K\"\n", + "print \" = 9.43*10**3 I_s\"\n", + "rm=-2.5*9.43*7.69\n", + "print \"R_M = %0.2f kohm\"%rm\n", + "print \"\"\n", + "print \"Step 6: Calculate beta\"\n", + "b=-1/(200*10**3)\n", + "print \"beta = I_f/Io = V_i-Vo/Vo*R : \"\n", + "print \" = -1/R = %0.2f because (Vo > V_i)\"%b\n", + "print \"\"\n", + "print \"Step 7: Calculate D, R_Mf, A_vf, R_of, R''_of\"\n", + "d=1+(5*181.29*10**-3)\n", + "print \"D = 1 + beta*R_M =%0.2f\"%d\n", + "rmf=-181.29/1.9\n", + "print \"R_Mf = R_M/D =%0.2f kohm \"%rmf\n", + "avf=-95.415/10\n", + "print \"A_vf = V0/V_s = Vo/I_s*R_s = R_Mf/R_s =%0.2f\"%avf\n", + "ri=(10*1000*200)/((1000*200)+(10*200)+(1000*10))\n", + "print \"R_i = R_s || M || R =%0.2f kohm\"%ri\n", + "rif=(9.43)/1.9\n", + "print \"R_if = R_i/D =%0.2f kohm\"%rif\n", + "ro=(40*200*10)/((200*10)+(400)+(40*200))\n", + "print \"R''_o = r_eff = r_d || R || R_D =%0.2f kohm\"%ro\n", + "rof=(7.69/1.9)\n", + "print \"R''_of = R''_o/D =%0.2f kohm\"%rof" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.26 Page No. 1-80" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Step 1: Identify topology\n", + "The feedback voltage is applied across the resistance R_e1 and it is in series with input signal. Hence feedback is voltage series feedback\n", + "\n", + "Step 2 and Step 3: Find input and output circuit\n", + "To find input circuit, set Vo = 0, which gives parallel combination of R_e1 with R_f at E1 as shown in fig 1.75. To find output circuit, set I_i = 0 opening the input node E1 at emitter of Q1, which gives series combination of R_f and R_e1 across the output. The resultant circuit is shown in fig 1.75\n", + "\n", + "Step 4: Find open loop voltage gain (A_v)\n", + "R_L2 = R_c2 || (R_f+R_e1) =2.11 kohm\n", + "A_i2 = -h_fe = -50\n", + "R_i2 = h_ie = 1.2 k-ohm\n", + "A_v2 = A_i2*R_L2 / R_i2 =-87.92\n", + "R_L1 = R_c1 || R_i2 =1.19 kohm\n", + "A_i1 = -h_fe = -50\n", + "R_i1 = h_ie + (1+h_fe)R_e =75.51 kohm\n", + "Therefore, A_v1 = A_i1*R_L1 / R_i1 =-0.78\n", + "The overall gain without feedback is given as\n", + "A_v = A_v1*A_v2 =68.92\n", + "\n", + "Step 5: Calculate beta\n", + "beta = V_f/Vo =0.03\n", + "\n", + "Step 6: Calculate D,A_vf, R_if, R_of\n", + "D = 1 + beta*A_v =2.96\n", + "A_vf = A_v/D =23.25\n", + "R_i = R || R_i1 =54.83 kohm\n", + "R_if = R_i/D =162.49 kohm\n", + "Ro = infinity because h_oe = 0\n", + "R''_o = Ro || R_c2 || (R_f+R_e1) = Ro || R_L2 = infinity || 2.11 K = 2.11 K\n", + "R''_of = R''_o/D =711.88 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"Step 1: Identify topology\"\n", + "print \"The feedback voltage is applied across the resistance R_e1 and it is in series with input signal. Hence feedback is voltage series feedback\"\n", + "print \"\"\n", + "print \"Step 2 and Step 3: Find input and output circuit\"\n", + "print \"To find input circuit, set Vo = 0, which gives parallel combination of R_e1 with R_f at E1 as shown in fig 1.75. To find output circuit, set I_i = 0 opening the input node E1 at emitter of Q1, which gives series combination of R_f and R_e1 across the output. The resultant circuit is shown in fig 1.75\"\n", + "print \"\"\n", + "print \"Step 4: Find open loop voltage gain (A_v)\"\n", + "rl2=(2.2*52.5)/54.7\n", + "print \"R_L2 = R_c2 || (R_f+R_e1) =%0.2f kohm\"%rl2\n", + "print \"A_i2 = -h_fe = -50\"\n", + "print \"R_i2 = h_ie = 1.2 k-ohm\"\n", + "av2=(-50*2.11)/1.2\n", + "print \"A_v2 = A_i2*R_L2 / R_i2 =%0.2f\"%av2\n", + "rl1=(100*1.2)/101.2\n", + "print \"R_L1 = R_c1 || R_i2 =%0.2f kohm\"%rl1\n", + "print \"A_i1 = -h_fe = -50\"\n", + "ri2=1.2+(51*(51*1.5/52.5))\n", + "print \"R_i1 = h_ie + (1+h_fe)R_e =%0.2f kohm\"%ri2\n", + "av1=(-50*1.185)/75.51\n", + "print \"Therefore, A_v1 = A_i1*R_L1 / R_i1 =%0.2f\"%av1\n", + "print \"The overall gain without feedback is given as\"\n", + "av=-0.784*-87.91\n", + "print \"A_v = A_v1*A_v2 =%0.2f\"%av\n", + "print \"\"\n", + "print \"Step 5: Calculate beta\"\n", + "b=1.5/52.5\n", + "format(7)\n", + "print \"beta = V_f/Vo =%0.2f\"%b\n", + "print \"\"\n", + "print \"Step 6: Calculate D,A_vf, R_if, R_of\"\n", + "d=1+(0.0285*68.92)\n", + "format(6)\n", + "print \"D = 1 + beta*A_v =%0.2f\"%d\n", + "avf=68.92/2.964\n", + "print \"A_vf = A_v/D =%0.2f\"%avf\n", + "ri=(75.51*200.1485)/(200.1485+75.51)\n", + "print \"R_i = R || R_i1 =%0.2f kohm\"%ri\n", + "rif=54.82*2.964\n", + "print \"R_if = R_i/D =%0.2f kohm\"%rif\n", + "print \"Ro = infinity because h_oe = 0\"\n", + "print \"R''_o = Ro || R_c2 || (R_f+R_e1) = Ro || R_L2 = infinity || 2.11 K = 2.11 K\"\n", + "rof=(2.11*10**3)/2.964\n", + "print \"R''_of = R''_o/D =%0.2f ohm\"%(rof)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.27 Page No. 1-83" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Step 1: Identity topology\n", + "The feedback is given from emitter of Q2 to the base of Q2. If Io=0 then feedback current through 5K register is zero, hence it is current sampling. As feedback signal is mixed in shunt with input, the amplifier is current shunt feedback amplifier\n", + "\n", + "Step 2 and Step 3: Find input and output circuit\n", + "The input circuit of the amplifier without feedback is obatined by opening the output loop at the emitter of Q2(Io = 0). This places R''(5K) in series with R_s from base to emitter of Q1. The output circuit is found by shorting the input node, i.e. making V1=0. This places R''(5K) in parallel with R_e. The resultant equivalent circuit is shown in fig 1.78\n", + "Step 4: Find open circuit transfer gain\n", + "A_I = Io/I_s = -I_c/I_s = -I_c2/I_b2 * I_b2/I_c1 * I_c1/I_b1 * I_b1/I_s\n", + "We know that -I_c2/I_b2 = A_i2 = -h_fe = -50 and\n", + "-I_c/I_b1 = A_i1 = -h_fe = -50\n", + "Therefore, I_c1/I_b1 = 50\n", + "Looking at fig 1.77 we can write\n", + "I_b2/I_c1 = -R_c1/R_c1+R_i2\n", + "R_i2 = h_ie + (1+h_fe)(R_e2||R'') =28.82 kohm\n", + "Therefore, I_b2/I_c1 =-0.06\n", + "Looking at fig 1.78 we can write\n", + "I_b1/I_s = R/R+R_i1\n", + "where R = R3 || (R''+R_e) =848.48 ohm\n", + "and R_i1 = h_ie + (1+h_fe)R_e1 =21.90 kohm \n", + "Therefore, I_b1/I_s =0.04\n", + "Substituting the numerical values obtained in equations of A_I we get,\n", + "A_I =6.04\n", + "\n", + "Step 5: Calculate beta\n", + "beta = I_f/Io = R_e2 / R_e2+R'' =0.11\n", + "\n", + "Step 6: Calculate D,A_If, A_vI, R_sf, R_of\n", + "D = 1 + beta*A_I =1.64\n", + "A_if = A_I/D =3.65\n", + "A_vf = Vo/V_s = -I_c2/I_s * R_c2/R_s = A_if*R_c2 / R_s =43.85\n", + "R_i1 = R || R_i1 =816.39 ohm\n", + "R_i = R_i/D =497.19 ohm\n", + "Ro = infinity because h_oe = 0\n", + "Therefore, R_of = Ro*D = infinity\n", + "R''_o = Ro || R_c2 = infinity || 12 K = 12 K\n", + "R''_of = R''_o * 1+beta*A_i/1+beta*A1 = R''_o = R_c2 = 12K\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from __future__ import division\n", + "print \"Step 1: Identity topology\"\n", + "print \"The feedback is given from emitter of Q2 to the base of Q2. If Io=0 then feedback current through 5K register is zero, hence it is current sampling. As feedback signal is mixed in shunt with input, the amplifier is current shunt feedback amplifier\"\n", + "print \"\"\n", + "print \"Step 2 and Step 3: Find input and output circuit\"\n", + "print \"The input circuit of the amplifier without feedback is obatined by opening the output loop at the emitter of Q2(Io = 0). This places R''(5K) in series with R_s from base to emitter of Q1. The output circuit is found by shorting the input node, i.e. making V1=0. This places R''(5K) in parallel with R_e. The resultant equivalent circuit is shown in fig 1.78\"\n", + "print \"Step 4: Find open circuit transfer gain\"\n", + "print \"A_I = Io/I_s = -I_c/I_s = -I_c2/I_b2 * I_b2/I_c1 * I_c1/I_b1 * I_b1/I_s\"\n", + "print \"We know that -I_c2/I_b2 = A_i2 = -h_fe = -50 and\"\n", + "print \"-I_c/I_b1 = A_i1 = -h_fe = -50\"\n", + "print \"Therefore, I_c1/I_b1 = 50\"\n", + "print \"Looking at fig 1.77 we can write\"\n", + "print \"I_b2/I_c1 = -R_c1/R_c1+R_i2\"\n", + "ri2=1.5+(51*(5*0.6/5.6))\n", + "print \"R_i2 = h_ie + (1+h_fe)(R_e2||R'') =%0.2f kohm\"%ri2\n", + "ibc=-2/30.82\n", + "print \"Therefore, I_b2/I_c1 =%0.2f\"%ibc\n", + "print \"Looking at fig 1.78 we can write\"\n", + "print \"I_b1/I_s = R/R+R_i1\"\n", + "r=(5.6*10**3)/6.6\n", + "print \"where R = R3 || (R''+R_e) =%0.2f ohm\"%r\n", + "ri1=1.5+20.4\n", + "print \"and R_i1 = h_ie + (1+h_fe)R_e1 =%0.2f kohm \"%ri1\n", + "ib1=0.848/22.748\n", + "print \"Therefore, I_b1/I_s =%0.2f\"%ib1\n", + "print \"Substituting the numerical values obtained in equations of A_I we get,\"\n", + "ai=50*0.0649*50*0.0372\n", + "print \"A_I =%0.2f\"%ai\n", + "print \"\"\n", + "print \"Step 5: Calculate beta\"\n", + "b=0.6/5.6\n", + "print \"beta = I_f/Io = R_e2 / R_e2+R'' =%0.2f\"%b\n", + "print \"\"\n", + "print \"Step 6: Calculate D,A_If, A_vI, R_sf, R_of\"\n", + "d=1+(0.107*6)\n", + "print \"D = 1 + beta*A_I =%0.2f\"%d\n", + "aif=6/1.642\n", + "print \"A_if = A_I/D =%0.2f\"%aif\n", + "avf=(3.654*12)/1\n", + "print \"A_vf = Vo/V_s = -I_c2/I_s * R_c2/R_s = A_if*R_c2 / R_s =%0.2f\"%avf\n", + "ri1=(848*21900)/(21900+848)\n", + "print \"R_i1 = R || R_i1 =%0.2f ohm\"%ri1\n", + "rif=816.38/1.642\n", + "print \"R_i = R_i/D =%0.2f ohm\"%rif\n", + "print \"Ro = infinity because h_oe = 0\"\n", + "print \"Therefore, R_of = Ro*D = infinity\"\n", + "print \"R''_o = Ro || R_c2 = infinity || 12 K = 12 K\"\n", + "print \"R''_of = R''_o * 1+beta*A_i/1+beta*A1 = R''_o = R_c2 = 12K\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.28 Page No. 1-86" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Step 1: Identify topology\n", + " The feedback voltage is applied across R_e1 = 1.5 k-ohm, which is in series with input signal. Hence feedback is voltage series feedback\n", + "\n", + "Step 2 and step 3: Find input and output circuit\n", + " To find input circuit, set Vo = 0, which gives parallel combination of R_e1 with R_f at E1 as shown in fig.1.80. To find ouput circuit, set I_i = 0 by opening the input node, E1 at emitter of Q1, which gives the series combination of R_f and R_e1 across the output. The resultant circuit is shown in fig.1.80\n", + "\n", + "Step 4: Find the open loop voltage gain (Av)\n", + " R_L2 = R_c2 || (Rf + R_e1) =2.12 kohm\n", + "Since hoe*R_L2 = 10**-6*2.119 k-ohm = 0.002119 is less than 0.1 we use approximate analysis.\n", + " A_i2 = -h_fe = -200\n", + " R_i2 = hie = 2 k-ohm\n", + " A_v2 = A_i2*R_L2 / R_i2 =-211.90\n", + " R_L1 = R_C1 || R_i2 =1.97 kohm\n", + "Since hoe*R_L1 = 10**-6*1.967 = 0.001967 is less than 0.1 we use approximate analysis.\n", + " A_i1 = -hfe = -200\n", + " R_i1 = hie + (1+hfe)*Re =295.63 kohm\n", + "Therefore, A_v1 = A_i1*R_L1 / R_i1 =-1.33\n", + "The overall gain without feedback is\n", + " Av = A_v1 * A_v2 =281.83\n", + "\n", + "Step 5: Calculate beta\n", + " beta = Vf / Vo =0.03\n", + "\n", + "Step 6: calculate D, A_vf, R_if, R_of\n", + " D = 1 + Av*beta =8.33\n", + "Therefore, A_vf = Av / D =33.84\n", + " Ri = R_i1 || R =99.51 kohm\n", + " R_if = Ri *D =828.54\n", + " Ro = 1/hoe = 1 M-ohm\n", + " R_of( = Ro / D = 120.09\n", + " R''o = Ro || R_c2 || (Rf+R_e1) = Ro || R_L2 =2.11 kohm\n", + " R''_of = R''o / D =253.93 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"Step 1: Identify topology\"\n", + "print \" The feedback voltage is applied across R_e1 = 1.5 k-ohm, which is in series with input signal. Hence feedback is voltage series feedback\"\n", + "print \"\"\n", + "print \"Step 2 and step 3: Find input and output circuit\"\n", + "print \" To find input circuit, set Vo = 0, which gives parallel combination of R_e1 with R_f at E1 as shown in fig.1.80. To find ouput circuit, set I_i = 0 by opening the input node, E1 at emitter of Q1, which gives the series combination of R_f and R_e1 across the output. The resultant circuit is shown in fig.1.80\"\n", + "print \"\"\n", + "print \"Step 4: Find the open loop voltage gain (Av)\"\n", + "rl2=(2.2*57.5)/(2.2+57.5) # in k-ohm\n", + "print \" R_L2 = R_c2 || (Rf + R_e1) =%0.2f kohm\"%rl2\n", + "print \"Since hoe*R_L2 = 10**-6*2.119 k-ohm = 0.002119 is less than 0.1 we use approximate analysis.\"\n", + "print \" A_i2 = -h_fe = -200\"\n", + "print \" R_i2 = hie = 2 k-ohm\"\n", + "av2=(-200*2.119)/2\n", + "print \" A_v2 = A_i2*R_L2 / R_i2 =%0.2f\"%av2\n", + "rl1=(120*2)/(122) # in k-ohm\n", + "print \" R_L1 = R_C1 || R_i2 =%0.2f kohm\"%rl1\n", + "print \"Since hoe*R_L1 = 10**-6*1.967 = 0.001967 is less than 0.1 we use approximate analysis.\"\n", + "print \" A_i1 = -hfe = -200\"\n", + "ri1=2+(201*((1.5*56)/(57.5))) # in k-ohm\n", + "print \" R_i1 = hie + (1+hfe)*Re =%0.2f kohm\"%ri1\n", + "av1=(-200*1.967)/295.63\n", + "print \"Therefore, A_v1 = A_i1*R_L1 / R_i1 =%0.2f\"%av1\n", + "print \"The overall gain without feedback is\"\n", + "av=-1.33*-211.9\n", + "format(7)\n", + "print \" Av = A_v1 * A_v2 =%0.2f\"%av\n", + "print \"\"\n", + "print \"Step 5: Calculate beta\"\n", + "beta=1.5/57.5\n", + "print \" beta = Vf / Vo =%0.2f\"%beta\n", + "print \"\"\n", + "print \"Step 6: calculate D, A_vf, R_if, R_of\"\n", + "d=1+(0.026*281.82)\n", + "print \" D = 1 + Av*beta =%0.2f\"%d\n", + "avf=281.82/8.327\n", + "print \"Therefore, A_vf = Av / D =%0.2f\"%avf\n", + "ri=(295.63*150)/(295.63+150) # in k-ohm\n", + "print \" Ri = R_i1 || R =%0.2f kohm\"%ri\n", + "rif=99.5*8.327 # in k-ohm\n", + "print \" R_if = Ri *D =%0.2f\"%rif\n", + "print \" Ro = 1/hoe = 1 M-ohm\"\n", + "rof=((1*10**6)/8.327)*10**-3 # in k-ohm\n", + "print \" R_of( = Ro / D = %0.2f\"%rof\n", + "ro=(1000*2.119)/(2.119+1000) # in k-ohm\n", + "print \" R''o = Ro || R_c2 || (Rf+R_e1) = Ro || R_L2 =%0.2f kohm\"%ro\n", + "rof=(2.1145*10**3)/8.327 # in ohm\n", + "print \" R''_of = R''o / D =%0.2f ohm\"%rof," + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.29 Page No. 1-89" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Fig 1.83 shows current shunt feedback amplifier open circuit transfer gain\n", + "A_I = -I_c2/I_s = -I_c2/I_b2 * I_b2/I_c1 * I_c1/I_b1 * I_b1/I_s\n", + "I_c2/I_b2 = A_i2 = -h_fe = -100\n", + "-I_c1/I_b1 = 100\n", + "R_i2 = h_ie + (1+h_fe)(R_e2||R'') =11.00 kohm\n", + "I_b2/I_c1 = -R_c1 / R_c1+(R_i2+R_b2) =-0.15\n", + "I_b1/I_s = R/R+h_ie\n", + "R = R_s || (R''+R_e) =909.91 ohm\n", + "Therefore, I''_b/I_s =0.48\n", + "Therefore, A_I =737.80\n", + "Calculate of beta:\n", + "I_f = -I_o*R_e2 / R_e2+R''\n", + "beta = I_f/Io = R_e2/R_e2+R'' = 100/100+10K\n", + "D = 1 + beta*A_I =8.30\n", + "A_if = A_I/D = 88.89\n", + "A_vf = Vo/V_s = -I_c2/I_s * R_c2/R_s = A_if*R_c2 / R_s =195.56\n", + "R_i1 = R || h_ie =476.41 ohm\n", + "R_if = R_i/D =57.35 ohm\n", + "R_of = R_c2 = 2.2 k-ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"Fig 1.83 shows current shunt feedback amplifier open circuit transfer gain\"\n", + "print \"A_I = -I_c2/I_s = -I_c2/I_b2 * I_b2/I_c1 * I_c1/I_b1 * I_b1/I_s\"\n", + "print \"I_c2/I_b2 = A_i2 = -h_fe = -100\"\n", + "print \"-I_c1/I_b1 = 100\"\n", + "ri2=1+(101*(1/10.1))\n", + "print \"R_i2 = h_ie + (1+h_fe)(R_e2||R'') =%0.2f kohm\"%ri2\n", + "ibc=-2.2/14.2\n", + "print \"I_b2/I_c1 = -R_c1 / R_c1+(R_i2+R_b2) =%0.2f\"%ibc\n", + "print \"I_b1/I_s = R/R+h_ie\"\n", + "r=(10.1*10**3)/11.1\n", + "print \"R = R_s || (R''+R_e) =%0.2f ohm\"%r\n", + "ibs=0.9099/1.9099\n", + "print \"Therefore, I''_b/I_s =%0.2f\"%ibs\n", + "ai=100*0.155*100*0.476\n", + "print \"Therefore, A_I =%0.2f\"%ai\n", + "print \"Calculate of beta:\"\n", + "print \"I_f = -I_o*R_e2 / R_e2+R''\"\n", + "print \"beta = I_f/Io = R_e2/R_e2+R'' = 100/100+10K\"\n", + "d=1+(9.9*737.8*10**-3)\n", + "print \"D = 1 + beta*A_I =%0.2f\"%d\n", + "print \"A_if = A_I/D = 88.89\"\n", + "avf=88.89*2.2\n", + "print \"A_vf = Vo/V_s = -I_c2/I_s * R_c2/R_s = A_if*R_c2 / R_s =%0.2f\"%avf\n", + "ri1=(909.9*1000)/1909.9\n", + "print \"R_i1 = R || h_ie =%0.2f ohm\"%ri1\n", + "rif=476/8.3\n", + "print \"R_if = R_i/D =%0.2f ohm\"%rif\n", + "print \"R_of = R_c2 = 2.2 k-ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.30 Page No. 1-91" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Voltage gain with feedback A_f = A_v/D\n", + "Where, D = 1 + beta*A_v =3.00\n", + "Therefore, A_vf =33.33\n", + "(ii) Feedback voltage V_f = beta*Vo = beta*A_vf*V_i =26.66 mV\n", + "(iii) Output voltage Vo(in V) = A_vi*V_i =1.33\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"(i) Voltage gain with feedback A_f = A_v/D\"\n", + "d=1+(0.02*100)\n", + "print \"Where, D = 1 + beta*A_v =%0.2f\"%d\n", + "avf=100/3\n", + "print \"Therefore, A_vf =%0.2f\"%avf\n", + "vf=0.02*33.33*40\n", + "print \"(ii) Feedback voltage V_f = beta*Vo = beta*A_vf*V_i =%0.2f mV\"%vf\n", + "vo=33.33*40*10**-3\n", + "print \"(iii) Output voltage Vo(in V) = A_vi*V_i =%0.2f\"%vo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.31 Page No. 1-92" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "For beta = -0.01, D = 1+beta*A_v = 11\n", + "(i) Voltage gain A_vf = A_v/D =-9.09\n", + "(ii) Input impedance R_if(in k-ohm) = R_i*D =110.00\n", + "(iii) Output impedance R_of(in k-ohm) = Ro/D =1.82\n", + "For beta = -0.01, D = 1+beta*A_v = 51\n", + "(i) Voltage gain A_vf = A_v/D =-1.96\n", + "(ii) Input impedance R_if(in k-ohm) = R_i*D =510.00\n", + "(iii) Output impedance R_of(in k-ohm) = Ro/D =0.39\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"For beta = -0.01, D = 1+beta*A_v = 11\"\n", + "avf=-100/11\n", + "print \"(i) Voltage gain A_vf = A_v/D =%0.2f\"%avf\n", + "rif=10*11\n", + "print \"(ii) Input impedance R_if(in k-ohm) = R_i*D =%0.2f\"%rif\n", + "rof=20/11\n", + "print \"(iii) Output impedance R_of(in k-ohm) = Ro/D =%0.2f\"%rof\n", + "print \"For beta = -0.01, D = 1+beta*A_v = 51\"\n", + "avf=-100/51\n", + "print \"(i) Voltage gain A_vf = A_v/D =%0.2f\"%avf\n", + "rif=10*51\n", + "print \"(ii) Input impedance R_if(in k-ohm) = R_i*D =%0.2f\"%rif\n", + "rof=20/51\n", + "print \"(iii) Output impedance R_of(in k-ohm) = Ro/D =%0.2f\"%rof" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.33 Page No. 1-92" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "We know that,\n", + "A_vf = A_v / 1+beta*A_v\n", + "Therefore, A_vf + beta*A_v*A_vf = A_v\n", + "Therefore, beta = A_v-A_vf / A_v*A_vf =0.01\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"We know that,\"\n", + "print \"A_vf = A_v / 1+beta*A_v\"\n", + "print \"Therefore, A_vf + beta*A_v*A_vf = A_v\"\n", + "b=20/2400\n", + "print \"Therefore, beta = A_v-A_vf / A_v*A_vf =%0.2f\"%b" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.34 Page No. 1-93" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Given: A_v mid = 40, f_L = 100 Hz, f_H = 15 kHz and beta = 0.01\n", + "(i) A_vf = A_v mid / 1+beta*A_v mid =80.00\n", + "(ii) f_Lf = f_L / 1+beta*A_v mid =20.00\n", + "(iii) f_Hf = f_H * (1+beta*A_v mid) =75.00 kHz\n", + "(iv) New Bandwidth = f_Hf - f_Lf =74.98 kHz\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"Given: A_v mid = 40, f_L = 100 Hz, f_H = 15 kHz and beta = 0.01\"\n", + "avf=400/(1+(0.01*400))\n", + "print \"(i) A_vf = A_v mid / 1+beta*A_v mid =%0.2f\"%avf\n", + "flf=100/(1+(0.01*400))\n", + "print \"(ii) f_Lf = f_L / 1+beta*A_v mid =%0.2f\"%flf\n", + "fhf=(15)*(1+(0.01*400)) # in kHz\n", + "print \"(iii) f_Hf = f_H * (1+beta*A_v mid) =%0.2f kHz\"%fhf\n", + "bw=75-0.02 # in kHz\n", + "format(6)\n", + "print \"(iv) New Bandwidth = f_Hf - f_Lf =%0.2f kHz\"%bw" + ] + } + ], + "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/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/chapter2.ipynb b/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/chapter2.ipynb new file mode 100644 index 00000000..971dc16f --- /dev/null +++ b/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/chapter2.ipynb @@ -0,0 +1,1494 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2 - Oscillators" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.1 Page No. 2-70" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Therefore, f = omega / 2pi = 318.31 Hz\n" + ] + } + ], + "source": [ + "from math import pi\n", + "omega = 2.0*10**3 # rad/sec\n", + "f=(omega)/(2.0*pi) # in Hz)\n", + "print \"Therefore, f = omega / 2pi = %0.2f Hz\"%f" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.2 Page No. 2-70" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Refering to equation(1),\n", + " R''_i = R1 || R2 || h_ie =1.63 kohm\n", + "Now R''_i + R3 = R\n", + "Therefore, R3 = R - R''_i =5.47 kohm\n", + " K = R_C / R =2.82\n", + "Now f = 1 / 2*pi*R*C*sqrt(6+4K)\n", + "Therefore, C = 539.50 pF\n", + " h_fe >= 4K + 23 + 29/K\n", + " h_fe >=44.56\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"Refering to equation(1),\"\n", + "ri=(25*57*1.8)/((57*1.8)+(25*1.8)+(25*57)) # in k-ohm\n", + "print \" R''_i = R1 || R2 || h_ie =%0.2f kohm\"%ri\n", + "print \"Now R''_i + R3 = R\"\n", + "r3=7.1-1.631 # in k-ohm\n", + "print \"Therefore, R3 = R - R''_i =%0.2f kohm\"%r3\n", + "k=20/7.1\n", + "print \" K = R_C / R =%0.2f\"%k\n", + "print \"Now f = 1 / 2*pi*R*C*sqrt(6+4K)\"\n", + "c=(1/(sqrt(6+(4*2.816))*2*pi*7.1*10*10**6))*10**12 # in pF\n", + "print \"Therefore, C = %0.2f pF\"%c\n", + "print \" h_fe >= 4K + 23 + 29/K\"\n", + "hfe=(4*2.816)+23+(29/2.816)\n", + "print \" h_fe >=%0.2f\"%hfe" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.3 Page No. 2-71" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The given values are, R = 4.7 k-ohm and C = 0.47 uF\n", + " f = 1 / 2*pi*sqrt(6)*R*C =29.41 Hz\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "from math import sqrt,pi\n", + "print \"The given values are, R = 4.7 k-ohm and C = 0.47 uF\"\n", + "f=1/(2*pi*sqrt(6)*(4.7*10**3)*(0.47*10**-6)) # in Hz\n", + "print \" f = 1 / 2*pi*sqrt(6)*R*C =%0.2f Hz\"%f" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.4 Page No. 2-71" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "f = 1 kHz\n", + "Now f = 1 / 2*pi*sqrt(6)*R*C\n", + "Choose C = 0.1 uF\n", + "Therefore, R = 649.75 ohm\n", + "Choose R = 680 ohm standard value\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"f = 1 kHz\"\n", + "print \"Now f = 1 / 2*pi*sqrt(6)*R*C\"\n", + "print \"Choose C = 0.1 uF\"\n", + "r=1/(sqrt(6)*2*pi*0.1*1*10**-3) # in ohm\n", + "print \"Therefore, R = %0.2f ohm\"%r\n", + "print \"Choose R = 680 ohm standard value\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.5 Page No. 2-72" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Using the expression for the frequency\n", + "Now, f = 1 / 2*pi*R*C*sqrt(6)\n", + "Therefore, C = 1.34 nF\n", + "Now using the equation(27)\n", + " |A| = g_m * R_L\n", + "Therefore, |A| >= 29\n", + "Therefore, g_m * R_L >= 29\n", + "Therefore, R_L >= 29 / g_m = 5.80 kohm\n", + " R_L = R_D*r_d / R_D+r_d\n", + " Therefore, R_D = 8.19 k-ohm\n", + "While for minimum value of R_L = 5.8 k-ohm\n", + " R_D = 6.78 k-ohm\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"Using the expression for the frequency\"\n", + "print \"Now, f = 1 / 2*pi*R*C*sqrt(6)\"\n", + "f=(1/(sqrt(6)*2*pi*9.7*5*10**6))*10**9 # in nF\n", + "print \"Therefore, C = %0.2f nF\"%f\n", + "print \"Now using the equation(27)\"\n", + "print \" |A| = g_m * R_L\"\n", + "print \"Therefore, |A| >= 29\"\n", + "print \"Therefore, g_m * R_L >= 29\"\n", + "rl=(29/(5000*10**-6))*10**-3 # in k-ohm\n", + "print \"Therefore, R_L >= 29 / g_m = %0.2f kohm\"%rl\n", + "print \" R_L = R_D*r_d / R_D+r_d\"\n", + "rd=(40)/4.8823\n", + "print \" Therefore, R_D = %0.2f k-ohm\"%rd\n", + "print \"While for minimum value of R_L = 5.8 k-ohm\"\n", + "print \" R_D = 6.78 k-ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Exmaple 2.6 Page No. 2-73" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The circuit is Wien bridge oscillator using op-amp. The gain of the op-amp is\n", + "A = 1 + R3/R4 =4.00\n", + "So A > 3\n", + "This satisfies the required oscillating condition. The feedback is given to non-inverting terminal ensuring the zero phase shift. Hence the circuit will work as the oscillator.\n", + "f = 1 / 2*pi*R*C = 31.21 kHz\n", + "This will be the frequency of oscillations\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "from math import sqrt,pi\n", + "print \"The circuit is Wien bridge oscillator using op-amp. The gain of the op-amp is\"\n", + "a=1+3\n", + "print \"A = 1 + R3/R4 =%0.2f\"%a\n", + "print \"So A > 3\"\n", + "print \"This satisfies the required oscillating condition. The feedback is given to non-inverting terminal ensuring the zero phase shift. Hence the circuit will work as the oscillator.\"\n", + "f=1/(2*pi*5.1*0.001)\n", + "print \"f = 1 / 2*pi*R*C = %0.2f kHz\"%f\n", + "print \"This will be the frequency of oscillations\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.7 Page No. 2-73" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The frequency of the oscillator is given by,\n", + " f = 1 / 2*pi*sqrt(R1*R2*C1*C2)\n", + "For f = 10 kHz,\n", + "Therefore, R2 = 25.33 kohm\n", + "For f = 50 kHz,\n", + "Therefore, R2 = 1.01 k-ohm\n", + "So minimum value of R2 is 1.013 k-ohm while the maximum value of R2 is 25.33 k-ohm\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "from math import sqrt,pi\n", + "print \"The frequency of the oscillator is given by,\"\n", + "print \" f = 1 / 2*pi*sqrt(R1*R2*C1*C2)\"\n", + "print \"For f = 10 kHz,\"\n", + "r2=(1/(4*(pi**2)*(100*10**6)*(10*10**3)*(0.001*10**-12))) # in k-ohm\n", + "print \"Therefore, R2 = %0.2f kohm\"%r2\n", + "print \"For f = 50 kHz,\"\n", + "r2=(1/(4*(pi**2)*(2500*10**6)*(10*10**3)*(0.001*10**-12))) # in k-ohm\n", + "print \"Therefore, R2 = %0.2f k-ohm\"%r2\n", + "print \"So minimum value of R2 is 1.013 k-ohm while the maximum value of R2 is 25.33 k-ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.8 Page No. 2-74" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The frequency is given by,\n", + " f = 1 / 2*pi*sqrt(C*L_eq)\n", + "where L_eq = L1 + L2 =0.00 kHz\n", + "For f = f_max = 2050 kHz\n", + "Therefore, C = 2.98 pF\n", + "For f = f_min = 950 kHz\n", + "Therefore, C = 13.89 pF\n", + "Hence C must be varied from 2.98 pF to 13.89 pF, to get the required frequency variation.\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"The frequency is given by,\"\n", + "print \" f = 1 / 2*pi*sqrt(C*L_eq)\"\n", + "leq=(2*10**-3)+(20*10**-6)\n", + "print \"where L_eq = L1 + L2 =%0.2f kHz\"%leq\n", + "print \"For f = f_max = 2050 kHz\"\n", + "c=(1/(4*(pi**2)*((2050*10**3)**2)*0.00202))*10**12 # in pF\n", + "print \"Therefore, C = %0.2f pF\"%c\n", + "print \"For f = f_min = 950 kHz\"\n", + "c=(1/(4*(pi**2)*((950*10**3)**2)*0.00202))*10**12 # in pF\n", + "print \"Therefore, C = %0.2f pF\"%c\n", + "print \"Hence C must be varied from 2.98 pF to 13.89 pF, to get the required frequency variation.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.9 Page No. 2-75" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The given values are,\n", + "L1 = 0.5 mH, L2 = 1 mH, C = 0.2 uF\n", + "Now f = 1 / 2*pi*sqrt(C*L_eq)\n", + "and L_eq = L1 + L2 =1.50 mH\n", + "Therefore, f =9.19 kHz\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"The given values are,\"\n", + "print \"L1 = 0.5 mH, L2 = 1 mH, C = 0.2 uF\"\n", + "print \"Now f = 1 / 2*pi*sqrt(C*L_eq)\"\n", + "leq=0.5+1 # in mH\n", + "print \"and L_eq = L1 + L2 =%0.2f mH\"%leq\n", + "f=(1/(2*pi*sqrt(1.5*0.2*10**-9)))*10**-3 # in kHz\n", + "format(5)\n", + "print \"Therefore, f =%0.2f kHz\"%f" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.10 Page No. 2-75" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "L1 = 20 uH, L2 = 2 mH\n", + "Therefore, L_eq = L1 + L2 =0.00 H\n", + "For f = f_max = 2.5 MHz\n", + "f = 1 / 2*pi*sqrt(C*L_eq)\n", + "Therefore, C = 2.02 pF\n", + "For f = f_min = 1 MHz\n", + "Therefore, C = 12.65 pF\n", + "This C must be varied from 2.0244 pF to 12.6525 pF\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"L1 = 20 uH, L2 = 2 mH\"\n", + "leq=(20*10**-6)+(2*10**-3)\n", + "print \"Therefore, L_eq = L1 + L2 =%0.2f H\"%leq\n", + "print \"For f = f_max = 2.5 MHz\"\n", + "print \"f = 1 / 2*pi*sqrt(C*L_eq)\"\n", + "c=(1/(((2*pi*2.5*10**6)**2)*(2.002*10**-3)))*10**12\n", + "print \"Therefore, C = %0.2f pF\"%c\n", + "print \"For f = f_min = 1 MHz\"\n", + "c=(1/(((2*pi*1*10**6)**2)*(2.002*10**-3)))*10**12\n", + "print \"Therefore, C = %0.2f pF\"%c\n", + "print \"This C must be varied from 2.0244 pF to 12.6525 pF\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.11 Page No. 2-76" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The equivalent capacitance is given by,\n", + "C_eq = C1*C2 / C1+C2 =0.00 F\n", + "Now, f = 1 / 2*pi*sqrt(L*C_eq)\n", + "f = 1.93 MHz\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"The equivalent capacitance is given by,\"\n", + "ceq=(150*1.5*10**-21)/((150*10**-12)+(1.5*10**-9)) # in F\n", + "print \"C_eq = C1*C2 / C1+C2 =%0.2f F\"%ceq\n", + "print \"Now, f = 1 / 2*pi*sqrt(L*C_eq)\"\n", + "f=(1/(2*pi*sqrt(50*136.363*10**-18)))*10**-6 # in MHz\n", + "print \"f = %0.2f MHz\"%f" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.12 Page No. 2-77" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The given values are,\n", + " L = 100 uH, C1 = C2 = C and f = 500 kHz\n", + "Now, f = 1 / 2*pi*sqrt(L*C_eq)\n", + "Therefore, C_eq = 0.00 F\n", + "but C_eq = C1*C2 / C1+C2 and C1 = C2 = C\n", + "Therefore, C_eq = C / 2\n", + "Therefore, C = 2.03 nF\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"The given values are,\"\n", + "print \" L = 100 uH, C1 = C2 = C and f = 500 kHz\"\n", + "print \"Now, f = 1 / 2*pi*sqrt(L*C_eq)\"\n", + "ceq=1/(4*(pi**2)*(100*10**-6)*((500*10**3)**2)) # in F\n", + "print \"Therefore, C_eq = %0.2f F\"%ceq\n", + "print \"but C_eq = C1*C2 / C1+C2 and C1 = C2 = C\"\n", + "print \"Therefore, C_eq = C / 2\"\n", + "c=1.0132*2\n", + "print \"Therefore, C = %0.2f nF\"%c" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.13 Page No. 2-78" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Given, C1 = 100 pF, C2 = 50 pF, f = 10 MHz, L = ?\n", + "C_eq = C1*C2 / C1+C2 = 0.00 F \n", + "f = 1 / 2*pi*sqrt(L*C_eq)\n", + "Therefore, L = 7.60 uH\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"Given, C1 = 100 pF, C2 = 50 pF, f = 10 MHz, L = ?\"\n", + "ceq=(5000*10**-24)/(150*10**-12)\n", + "print \"C_eq = C1*C2 / C1+C2 = %0.2f F \"%ceq\n", + "print \"f = 1 / 2*pi*sqrt(L*C_eq)\"\n", + "l=(1/(4*(pi**2)*(33.33*10**-12)*((10*10**6)**2)))*10**6 # in F\n", + "print \"Therefore, L = %0.2f uH\"%l" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.14 Page No. 2-79" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "For a tuned collector oscillator,\n", + "f_r = 1 / 2*pi*sqrt(L*C)\n", + "where L = 30 uH and f_r to be varied 300 kHz to 1.5 MHz\n", + "For f_r = 300 kHz\n", + "Therefore, C1 =9.38 nF\n", + "For f_r = 1.5 MHz\n", + "Therefore, C2 = 375.26 pF\n", + "Hence C must be varied over 375.264 pF to 9.3816 nF, to achieve frequency variations\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"For a tuned collector oscillator,\"\n", + "print \"f_r = 1 / 2*pi*sqrt(L*C)\"\n", + "print \"where L = 30 uH and f_r to be varied 300 kHz to 1.5 MHz\"\n", + "print \"For f_r = 300 kHz\"\n", + "c1=(1/(4*(pi**2)*(30*10**-6)*((300*10**3)**2)))*10**9 # in nF\n", + "print \"Therefore, C1 =%0.2f nF\"%c1\n", + "print \"For f_r = 1.5 MHz\"\n", + "c2=(1/(4*(pi**2)*(30*10**-6)*((1.5*10**6)**2)))*10**12 # in pF\n", + "print \"Therefore, C2 = %0.2f pF\"%c2\n", + "print \"Hence C must be varied over 375.264 pF to 9.3816 nF, to achieve frequency variations\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.15 Page No. 2-79" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) f_s(in MHz) = 1 / 2*pi*sqrt(L*C) =0.86 MHz\n", + "(ii) C_eq = C*C_M / C+C_M =0.08 pF\n", + "Therefore, f_p = 1 / 2*pi*sqrt(L*C_eq) =0.90 MHz\n", + "(iii) % increase =5.02\n", + "(iv) Q = omega_s*L / R = 2*pi*f_s*L / R =430.27\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "fs=(1/(2*pi*sqrt(0.4*0.085*10**-12)))*10**-6 # in MHz\n", + "print \"(i) f_s(in MHz) = 1 / 2*pi*sqrt(L*C) =%0.2f MHz\"%fs\n", + "ceq=0.085/1.085 # in pF\n", + "print \"(ii) C_eq = C*C_M / C+C_M =%0.2f pF\"%ceq\n", + "fp=(1/(2*pi*sqrt(0.4*0.078*10**-12)))*10**-6 # in MHz (the answer in textbook is wrong)\n", + "print \"Therefore, f_p = 1 / 2*pi*sqrt(L*C_eq) =%0.2f MHz\"%fp\n", + "inc=((0.899-0.856)/0.856)*100 # in percentage\n", + "print \"(iii) %% increase =%0.2f\"%inc\n", + "q=(2*pi*0.4*0.856*10**6)/(5*10**3)\n", + "print \"(iv) Q = omega_s*L / R = 2*pi*f_s*L / R =%0.2f\"%q" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.16 Page No. 2-80" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " C_M = 2 pF\n", + "Now f_s = 1 / 2*pi*sqrt(L*C) =1.13 MHz\n", + " C_eq = C_M*C / C_M+C =0.00 F\n", + " f_p = 1 / 2*pi*sqrt(L*C_eq) =1.13\n", + "So f_s and f_p values are almost same.\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \" C_M = 2 pF\"\n", + "fs=(1/(2*pi*sqrt(2*0.01*10**-12)))*10**-6 # in MHz\n", + "print \"Now f_s = 1 / 2*pi*sqrt(L*C) =%0.2f MHz\"%fs\n", + "ceq=(2*0.01*10**-24)/(2.01*10**-12) # in F\n", + "print \" C_eq = C_M*C / C_M+C =%0.2f F\"%ceq\n", + "fp=(1/(2*pi*sqrt(2*9.95*10**-15)))*10**-6 # in MHz\n", + "print \" f_p = 1 / 2*pi*sqrt(L*C_eq) =%0.2f\"%fp\n", + "print \"So f_s and f_p values are almost same.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.17 Page No. 2-80" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R = 6 k-ohm, C = 1500 pF, R_C = 18 k-ohm\n", + "Now K = R_C / R =3.00\n", + "Therefore, f = 1 / 2*pi*R*C*sqrt(6+4K)\n", + " f = 4.17 kHz\n", + " (h_fe)min = 4K + 23 + 29/K =44.67\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"R = 6 k-ohm, C = 1500 pF, R_C = 18 k-ohm\"\n", + "k=18/6\n", + "print \"Now K = R_C / R =%0.2f\"%k\n", + "print \"Therefore, f = 1 / 2*pi*R*C*sqrt(6+4K)\"\n", + "f=(1/(2*pi*(6*10**3)*(1500*10**-12)*sqrt(6+12)))*10**-3 # in kHZ\n", + "print \" f = %0.2f kHz\"%f\n", + "hfe=(4*3)+23+(29/3)\n", + "print \" (h_fe)min = 4K + 23 + 29/K =%0.2f\"%hfe" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.18 Page No. 2-81" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Refering to equation(1) of section 4.5.3, the input impedance is given by,\n", + "R''_i = R1 || R2 || h_ie\n", + "Now R1 = 25 k-ohm, R2 = 47 k-ohm, and h_ie = 2 k-ohm\n", + "Therefore, R''_i = 1.78 kohm\n", + " K = R_C / R\n", + "Now R_C = 10 k-ohm ...given\n", + "Now f = 1 / 2*pi*R*C*sqrt(6+4K)\n", + "Therefore, R*sqrt(6+4K) = 31830.989\n", + "Now K = R_C / R = 10*10**3 / R\n", + "Therefore, R*sqrt(6+(40*10*10**3/R)) = 31830.989\n", + "Therefore, R**2*(6+(40*10*10**3/R)) = (31830.989)**2\n", + "Therefore, R = -10.08 kohm Neglecting negative value\n", + "Therefore, K = R_C / R =0.60\n", + "Therefore, h_fe >= 4K + 23 + 29/K\n", + " h_fe >=73.94\n" + ] + } + ], + "source": [ + "from sympy import symbols,solve\n", + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"Refering to equation(1) of section 4.5.3, the input impedance is given by,\"\n", + "print \"R''_i = R1 || R2 || h_ie\"\n", + "print \"Now R1 = 25 k-ohm, R2 = 47 k-ohm, and h_ie = 2 k-ohm\"\n", + "ri=(25*47*2)/((47*2)+(25*2)+(25*47)) # in k-ohm\n", + "print \"Therefore, R''_i = %0.2f kohm\"%ri\n", + "print \" K = R_C / R\"\n", + "print \"Now R_C = 10 k-ohm ...given\"\n", + "print \"Now f = 1 / 2*pi*R*C*sqrt(6+4K)\"\n", + "print \"Therefore, R*sqrt(6+4K) = 31830.989\"\n", + "print \"Now K = R_C / R = 10*10**3 / R\"\n", + "print \"Therefore, R*sqrt(6+(40*10*10**3/R)) = 31830.989\"\n", + "print \"Therefore, R**2*(6+(40*10*10**3/R)) = (31830.989)**2\"\n", + "R=symbols('R')\n", + "p1=6*R**2+(40*10**3)*R-(31830.989)**2\n", + "t1=solve(p1,R)[1]\n", + "ans1=t1\n", + "print \"Therefore, R = %0.2f kohm Neglecting negative value\"%((-ans1)*10**-3)\n", + "k=10/16.74\n", + "print \"Therefore, K = R_C / R =%0.2f\"%k\n", + "print \"Therefore, h_fe >= 4K + 23 + 29/K\"\n", + "hfe=(4*0.5973)+23+(29/0.5973)\n", + "print \" h_fe >=%0.2f\"%hfe" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.19 Page No. 2-82" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " C_eq = C1*C2 / C1+C2 =0.02 pF\n", + "Therefore, f_s(in MHz) = 1 / 2*pi*sqrt(L*C1) =5.03 MHz\n", + "Therefore, f_p = 1 / 2*pi*sqrt(L*C_eq) =5.04 MHz\n", + "Let C_s = 5 pF connected across the crystal\n", + "Therefore, C''2 = C2 + C_x =17.00 pF\n", + "Therefore, C''_eq = C1*C''2 / C1+C''2 =0.02 pF\n", + "Therefore, f''_p = 1 / 2*pi*sqrt(L*C_eq) =5.04 MHz\n", + "New C_x = 6 pF is connected then,\n", + " C''''2 = C2 + C_x =18.00 pF\n", + "Therefore, C''''_eq = C1*C''''2 / C1+C''''2 = 0.02 pF\n", + "Therefore, f''''_p = 1 / 2*pi*sqrt(L*C''''_eq) =5.04 MHz\n", + "Therefore, Change = f''_p - f''''_p =164.00 Hz\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "ceq=((0.02*12*10**-24)/(12.02*10**-12))*10**12 # in pF\n", + "print \" C_eq = C1*C2 / C1+C2 =%0.2f pF\"%ceq\n", + "fs=(1/(2*pi*sqrt(50*0.02*10**-15)))*10**-6 # in MHz\n", + "print \"Therefore, f_s(in MHz) = 1 / 2*pi*sqrt(L*C1) =%0.2f MHz\"%fs\n", + "fp=(1/(2*pi*sqrt(50*0.01996*10**-15)))*10**-6 # in MHz\n", + "print \"Therefore, f_p = 1 / 2*pi*sqrt(L*C_eq) =%0.2f MHz\"%fp\n", + "print \"Let C_s = 5 pF connected across the crystal\"\n", + "c2=12+5\n", + "print \"Therefore, C''2 = C2 + C_x =%0.2f pF\"%c2\n", + "ceq1=0.019976\n", + "print \"Therefore, C''_eq = C1*C''2 / C1+C''2 =%0.2f pF\"%ceq1\n", + "fp1=5.03588\n", + "print \"Therefore, f''_p = 1 / 2*pi*sqrt(L*C_eq) =%0.2f MHz\"%fp1\n", + "print \"New C_x = 6 pF is connected then,\"\n", + "c21=12+6\n", + "print \" C''''2 = C2 + C_x =%0.2f pF\"%c21\n", + "ceq2=0.0199778\n", + "print \"Therefore, C''''_eq = C1*C''''2 / C1+C''''2 = %0.2f pF\"%ceq2\n", + "fp2=5.035716\n", + "print \"Therefore, f''''_p = 1 / 2*pi*sqrt(L*C''''_eq) =%0.2f MHz\"%fp2\n", + "c=(5.03588-5.035716)*10**6\n", + "print \"Therefore, Change = f''_p - f''''_p =%0.2f Hz\"%c" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.20 Page No. 2-82" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R''_i = R1 || R2 || h_fe =1.79 kohm\n", + "Now R''_i + R3 = R\n", + "Therefore, R3 = R - R''_i =4.98 kohm\n", + "K = R_C / R = 2.94\n", + "Therefore, f = 1 / 2*pi*RC*sqrt(6+4K)\n", + "Therefore, C = 111.06 pF\n", + "And h_fe >= 4 K + 23 + 29/K >=44.62\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "ri=(22*68*2)/((68*2)+(22*2)+(22*68))\n", + "print \"R''_i = R1 || R2 || h_fe =%0.2f kohm\"%ri #answer in textbook is wrong\n", + "print \"Now R''_i + R3 = R\"\n", + "r3=6.8-1.8243\n", + "print \"Therefore, R3 = R - R''_i =%0.2f kohm\"%r3\n", + "k=20/6.8\n", + "print \"K = R_C / R = %0.2f\"%k\n", + "print \"Therefore, f = 1 / 2*pi*RC*sqrt(6+4K)\"\n", + "c=(1/(2*pi*6.8*50*sqrt(6+(4*2.9411))*10**6))*10**12\n", + "print \"Therefore, C = %0.2f pF\"%c\n", + "hfe=(4*2.9411)+23+(29/2.9411)\n", + "print \"And h_fe >= 4 K + 23 + 29/K >=%0.2f\"%hfe" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.21 Page No. 2-82" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The frequency of the oscillator is given by,\n", + " f = 1 / 2*pi*sqrt(R1*R2*C1*C2)\n", + "For f = 20 kHz,\n", + "Therefore, R2 = 6.33 kohm\n", + "For f = 70 kHz,\n", + "Therefore, R2 = 0.52 kohm\n", + "So minimum value of R2 is 0.517 k-ohm while the maximum value of R2 is 6.33 k-ohm\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "print \"The frequency of the oscillator is given by,\"\n", + "print \" f = 1 / 2*pi*sqrt(R1*R2*C1*C2)\"\n", + "print \"For f = 20 kHz,\"\n", + "r2=(1/(4*(pi**2)*((20*10**3)**2)*(10*10**3)*((0.001*10**-6)**2)))*10**-3\n", + "print \"Therefore, R2 = %0.2f kohm\"%r2\n", + "print \"For f = 70 kHz,\"\n", + "r2=(1/(4*(pi**2)*((70*10**3)**2)*(10*10**3)*((0.001*10**-6)**2)))*10**-3\n", + "print \"Therefore, R2 = %0.2f kohm\"%r2\n", + "print \"So minimum value of R2 is 0.517 k-ohm while the maximum value of R2 is 6.33 k-ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.22 Page No. 2-83" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "For a Hartley oscillator,\n", + " f = 1 / 2*pi*sqrt(L_eq*C) where L_eq = L1 + L2 + 2M\n", + "Therefore, L_eq =17.95 mH\n", + "Therefore, L2 = 2.95 mH\n", + "Now h_fe = L1+M / L2+M =5.09\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "print \"For a Hartley oscillator,\"\n", + "print \" f = 1 / 2*pi*sqrt(L_eq*C) where L_eq = L1 + L2 + 2M\"\n", + "leq=(1/(4*(pi**2)*((168*10**3)**2)*(50*10**-12)))*10**3 # in mH\n", + "print \"Therefore, L_eq =%0.2f mH\"%leq\n", + "l2=((17.95*10**-3)-(15*10**-3)-(5*10**-6))*10**3 # in mH\n", + "print \"Therefore, L2 = %0.2f mH\"%l2\n", + "hfe=((15*10**-3)+(5*10**-6))/((2.945*10**-3)+(5*10**-6))\n", + "print \"Now h_fe = L1+M / L2+M =%0.2f\"%hfe" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.24 Page No. 2-84" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Assume one perticular coupling direction for which,\n", + " L_eq = L1 + L2 + 2M = 0.25 mH\n", + "Therefore, f = 1 / 2*pi*sqrt(L_eq*C) =1.01 MHz\n", + "Let the direction of coupling is reversed,\n", + " L_eq = L1 + L2 - 2M = 0.15 mH\n", + "Therefore, f'' = 1 / 2*pi*sqrt(L_eq*C) =1.30 MHz\n", + "Therefore, % change = f''-f/f * 100 = 29.09 \n", + "(ii) Let us assume direction of coupling such that,\n", + " L_eq = L1 + L2 + 2M = 0.25 mH\n", + " C_t = Trim capacitor = 100 pF\n", + "Therefore, C_eq = C*C_t / C+C_t = 50 pF\n", + "Therefore, f = 1 / 2*pi*sqrt(L_eq*C_eq) =1.42 MHz\n", + "If now direction of coupling is reversed,\n", + " L_eq = L1 + L2 - 2M = 0.15 mH\n", + "Therefore, f'' = 1 / 2*pi*sqrt(L_eq*C_eq) =1.84 MHz\n", + "Therefore, % change = f''-f/f * 100 =29.10\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "print \"(i) Assume one perticular coupling direction for which,\"\n", + "print \" L_eq = L1 + L2 + 2M = 0.25 mH\"\n", + "f=(1/(2*pi*sqrt(0.25*100*10**-15)))*10**-6 # in MHz\n", + "print \"Therefore, f = 1 / 2*pi*sqrt(L_eq*C) =%0.2f MHz\"%f\n", + "print \"Let the direction of coupling is reversed,\"\n", + "print \" L_eq = L1 + L2 - 2M = 0.15 mH\"\n", + "fd=(1/(2*pi*sqrt(0.15*100*10**-15)))*10**-6 # in MHz\n", + "print \"Therefore, f'' = 1 / 2*pi*sqrt(L_eq*C) =%0.2f MHz\"%fd\n", + "pc=((1.2994-1.00658)/1.00658)*100 # in percentage\n", + "print \"Therefore, %% change = f''-f/f * 100 = %0.2f \"%pc\n", + "print \"(ii) Let us assume direction of coupling such that,\"\n", + "print \" L_eq = L1 + L2 + 2M = 0.25 mH\"\n", + "print \" C_t = Trim capacitor = 100 pF\"\n", + "print \"Therefore, C_eq = C*C_t / C+C_t = 50 pF\"\n", + "f1=(1/(2*pi*sqrt(0.25*50*10**-15)))*10**-6 # in MHz\n", + "print \"Therefore, f = 1 / 2*pi*sqrt(L_eq*C_eq) =%0.2f MHz\"%f1\n", + "print \"If now direction of coupling is reversed,\"\n", + "print \" L_eq = L1 + L2 - 2M = 0.15 mH\"\n", + "f2=(1/(2*pi*sqrt(0.15*50*10**-15)))*10**-6 # in MHz\n", + "print \"Therefore, f'' = 1 / 2*pi*sqrt(L_eq*C_eq) =%0.2f MHz\"%f2\n", + "pc1=((1.83776-1.4235)/1.4235)*100\n", + "print \"Therefore, %% change = f''-f/f * 100 =%0.2f\"%pc1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.25 Page No. 2-84" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "For RC phase shhift oscillator,\n", + " h_fe = 4K + 23 + 29/K ...given h_fe = 150\n", + "Therefore, 150 = 4K + 23 + 29/K\n", + "Therefore, 4K**2 - 127K + 29 = 0\n", + "Therefore, K =0.23\n", + " f = 1 / 2*pi*R*C*sqrt(6+4K) ...given f = 5 kHz\n", + "Therefore,Choose C = 100 pF\n", + "Therefore, R = 12.10 kohm\n", + " K = R_C / R i.e. R_C = KR = 2.7 k-ohm\n", + "Neglecting effect of biasing resistances assuming them to be large and selecting transistor with h_ie = 2 k-ohm\n", + " R''_i = h_ie = 2 k-ohm\n", + "Therefore,Last resistance in phase network\n", + " R3 = R - R''_i =10.00\n", + "Using the back to back connected zener diodes of 9.3 V (Vz) each at the output of emitter follower and using this at the output of the oscillator, the output amplitude can be controlled to 10 V i.e. 20 V peak to peak. The zener diode 9.3V and forward biased diode of 0.7 V gives total 10 V\n", + "The designed circuit is shown in fig.2.58\n" + ] + } + ], + "source": [ + "from sympy import symbols,solve\n", + "from __future__ import division \n", + "print \"For RC phase shhift oscillator,\"\n", + "print \" h_fe = 4K + 23 + 29/K ...given h_fe = 150\"\n", + "print \"Therefore, 150 = 4K + 23 + 29/K\"\n", + "print \"Therefore, 4K**2 - 127K + 29 = 0\"\n", + "K=symbols('K')\n", + "p1=4*K**2-127*K+29\n", + "t1=solve(p1,K)[0]\n", + "print \"Therefore, K =%0.2f\"%t1\n", + "print \" f = 1 / 2*pi*R*C*sqrt(6+4K) ...given f = 5 kHz\"\n", + "print \"Therefore,Choose C = 100 pF\"\n", + "r=(1/(2*pi*(1000*10**-12)*(5*10**3)*sqrt(6+(4*0.23))))*10**-3 # in k-ohm\n", + "print \"Therefore, R = %0.2f kohm\"%r\n", + "print \" K = R_C / R i.e. R_C = KR = 2.7 k-ohm\"\n", + "print \"Neglecting effect of biasing resistances assuming them to be large and selecting transistor with h_ie = 2 k-ohm\"\n", + "print \" R''_i = h_ie = 2 k-ohm\"\n", + "print \"Therefore,Last resistance in phase network\"\n", + "r3=12-2\n", + "print \" R3 = R - R''_i =%0.2f\"%r3\n", + "print \"Using the back to back connected zener diodes of 9.3 V (Vz) each at the output of emitter follower and using this at the output of the oscillator, the output amplitude can be controlled to 10 V i.e. 20 V peak to peak. The zener diode 9.3V and forward biased diode of 0.7 V gives total 10 V\"\n", + "print \"The designed circuit is shown in fig.2.58\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.26 Page No. 2-85" + ] + }, + { + "cell_type": "code", + "execution_count": 36, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(1) f = 1 / 2*pi*sqrt(L*C_eq)\n", + "Where C_eq = C1*C2 / C1+C2 = 83.33 pF\n", + "f = 87.17 kHz\n", + "(2) The input voltage is not required for the oscillator. The feedback voltage, which is the part of the output voltage is enough to drive the oscillator\n", + "V0 = 10 V\n", + "For Colpitts oscillator, gain = C2 / C1\n", + "Therefore, Gain =5.00\n", + "Therefore, Feedback voltage(in V) = V0 / Gain =2.00 V\n", + "(3) Minimum gain = C2/C1 = 5\n", + "h_fe(min) = C2/C1 = 5\n", + "(4) Gain = 10 = C2/C1\n", + "Therefore, C1 = 50.00 pF\n", + "(5) For C1 = 50 pF and C2 = 500 pF\n", + "Where C_eq = C1*C2 / C1+C2 =45.45 pF\n", + "f = 1 / 2*pi*sqrt(L*C_eq) = 118.03 kHz\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "print \"(1) f = 1 / 2*pi*sqrt(L*C_eq)\"\n", + "ceq=(100*500)/600\n", + "print \"Where C_eq = C1*C2 / C1+C2 = %0.2f pF\"%ceq\n", + "f=(1/(2*pi*sqrt(40*83.333*10**-15)))*10**-3\n", + "print \"f = %0.2f kHz\"%f\n", + "print \"(2) The input voltage is not required for the oscillator. The feedback voltage, which is the part of the output voltage is enough to drive the oscillator\"\n", + "print \"V0 = 10 V\"\n", + "print \"For Colpitts oscillator, gain = C2 / C1\"\n", + "gain=500/100\n", + "print \"Therefore, Gain =%0.2f\"%gain\n", + "fv=10/5\n", + "print \"Therefore, Feedback voltage(in V) = V0 / Gain =%0.2f V\"%fv\n", + "print \"(3) Minimum gain = C2/C1 = 5\"\n", + "print \"h_fe(min) = C2/C1 = 5\"\n", + "print \"(4) Gain = 10 = C2/C1\"\n", + "c1=500/10\n", + "print \"Therefore, C1 = %0.2f pF\"%c1\n", + "print \"(5) For C1 = 50 pF and C2 = 500 pF\"\n", + "ceq=(50*500)/550\n", + "print \"Where C_eq = C1*C2 / C1+C2 =%0.2f pF\"%ceq\n", + "f=(1/(2*pi*sqrt(40*45.4545*10**-15)))*10**-3\n", + "print \"f = 1 / 2*pi*sqrt(L*C_eq) = %0.2f kHz\"%f" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.27 Page No. 2-86" + ] + }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The frequency required is, f = 1 MHz and for FET, u = 20\n", + "Now u = C2/C1 for oscillations\n", + "Therefore, 20 = C2/C1\n", + "Therefore, C2 = 20*C1 ....(1)\n", + "Let C1 = 0.01 uF hence C2 = 0.2 uF\n", + "Therefore, C_eq = C1*C2 / C1+C2 =9.52 nF\n", + "Now f = 1 / 2*pi*sqrt(L*C_eq)\n", + "Therefore, L( = 2.66 uH\n", + "The baising resistances can be selected as,\n", + "R1 = 12 M-ohm and R2 = 8 M-ohm\n", + "These resistances must be large\n", + "The designed circuit is shown in the fig 2.59\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "print \"The frequency required is, f = 1 MHz and for FET, u = 20\"\n", + "print \"Now u = C2/C1 for oscillations\"\n", + "print \"Therefore, 20 = C2/C1\"\n", + "print \"Therefore, C2 = 20*C1 ....(1)\"\n", + "print \"Let C1 = 0.01 uF hence C2 = 0.2 uF\"\n", + "ceq=((0.01*0.2)/(0.21))*10**3\n", + "print \"Therefore, C_eq = C1*C2 / C1+C2 =%0.2f nF\"%ceq\n", + "print \"Now f = 1 / 2*pi*sqrt(L*C_eq)\"\n", + "l=(1/(((2*pi*1*10**6)**2)*(9.5238*10**-9)))*10**6\n", + "print \"Therefore, L( = %0.2f uH\"%l\n", + "print \"The baising resistances can be selected as,\"\n", + "print \"R1 = 12 M-ohm and R2 = 8 M-ohm\"\n", + "print \"These resistances must be large\"\n", + "print \"The designed circuit is shown in the fig 2.59\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.28 Page No. 2-87" + ] + }, + { + "cell_type": "code", + "execution_count": 38, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "L_eq = L1 + L2 + 2M =6100.00 uH\n", + "f = 1 / 2*pi*sqrt(C*L_eq) =166.38 kHz\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "leq=500+5000+600\n", + "print \"L_eq = L1 + L2 + 2M =%0.2f uH\"%leq\n", + "f=(1/(2*pi*sqrt(150*6100*10**-18)))*10**-3\n", + "print \"f = 1 / 2*pi*sqrt(C*L_eq) =%0.2f kHz\"%f" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.29 Page No. 2-87" + ] + }, + { + "cell_type": "code", + "execution_count": 39, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "L_s = 0.8 H, C_s = 0.08 pF, R_s = 5 k-ohm, C_M = 1 pF\n", + "f_s = 1 / 2*pi*sqrt(C_s*L_s) =629.12 kHz\n", + "C_eq = C_M*C_s / C_M+C_s =0.00 F\n", + "Therefore, f_p = 1 / 2*pi*sqrt(C*L_eq) =653.80 kHz\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "print \"L_s = 0.8 H, C_s = 0.08 pF, R_s = 5 k-ohm, C_M = 1 pF\"\n", + "fs=(1/(2*pi*sqrt(0.8*0.08*10**-12)))*10**-3\n", + "print \"f_s = 1 / 2*pi*sqrt(C_s*L_s) =%0.2f kHz\"%fs\n", + "ceq=(0.08*10**-12)/1.08\n", + "print \"C_eq = C_M*C_s / C_M+C_s =%0.2f F\"%ceq\n", + "fp=(1/(2*pi*sqrt(0.8*7.4074*10**-14)))*10**-3\n", + "print \"Therefore, f_p = 1 / 2*pi*sqrt(C*L_eq) =%0.2f kHz\"%fp" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.31 Page No. 2-88" + ] + }, + { + "cell_type": "code", + "execution_count": 40, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The series and parallel resonating frequencies are,\n", + "f_s = 1 / 2*pi*sqrt(C*L) while f_p = 1 / 2*pi*sqrt(L*C_eq)\n", + "f_p/f_s = 1/2*pi*sqrt(L*C_eq) * 2*pi*sqrt(LC) = sqrt(c/C_eq) but C_eq = C*C_M/C+C_M\n", + "f_p/f_s = sqrt(C/(C*C_M/C+C_M)) = sqrt(C*(C+C_M)/C*C_M) = sqrt(1+(C/C_M)) =1.01\n", + "f_p = 1.00995*f_s\n", + "Therefore, % increase = (1.00995*f_s-f_s / f_s)*100 =1.00\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "print \"The series and parallel resonating frequencies are,\"\n", + "print \"f_s = 1 / 2*pi*sqrt(C*L) while f_p = 1 / 2*pi*sqrt(L*C_eq)\"\n", + "print \"f_p/f_s = 1/2*pi*sqrt(L*C_eq) * 2*pi*sqrt(LC) = sqrt(c/C_eq) but C_eq = C*C_M/C+C_M\"\n", + "fp=sqrt(1+(0.04/2))\n", + "print \"f_p/f_s = sqrt(C/(C*C_M/C+C_M)) = sqrt(C*(C+C_M)/C*C_M) = sqrt(1+(C/C_M)) =%0.2f\"%fp\n", + "print \"f_p = 1.00995*f_s\"\n", + "inc=0.00995*100\n", + "print \"Therefore, %% increase = (1.00995*f_s-f_s / f_s)*100 =%0.2f\"%inc," + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.32 Page No. 2-88" + ] + }, + { + "cell_type": "code", + "execution_count": 41, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "C = 20 pF, L2 = 1000 uH, L1 = 100 uH, M = 20 uH\n", + "Therefore, L_eq = L1 + L2 + 2M =1140.00 uH\n", + "Therefore, f = 1 / 2*pi*sqrt(L_eq*C) =1.05 MHz\n", + "The feedback fraction beta is given by,\n", + "beta = V_f/V0 = X_L1 / X_L1+X_L2 = L1 / L1+L2 =0.09\n", + "It is a Hartley oscillator\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "print \"C = 20 pF, L2 = 1000 uH, L1 = 100 uH, M = 20 uH\"\n", + "leq=100+1000+40\n", + "print \"Therefore, L_eq = L1 + L2 + 2M =%0.2f uH\"%leq\n", + "f=(1/(2*pi*sqrt(1140*20*10**-18)))*10**-6\n", + "print \"Therefore, f = 1 / 2*pi*sqrt(L_eq*C) =%0.2f MHz\"%f\n", + "print \"The feedback fraction beta is given by,\"\n", + "b=100/1100\n", + "print \"beta = V_f/V0 = X_L1 / X_L1+X_L2 = L1 / L1+L2 =%0.2f\"%b\n", + "print \"It is a Hartley oscillator\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.33 Page No. 2-89" + ] + }, + { + "cell_type": "code", + "execution_count": 42, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Using the expression of the frequency,\n", + "f = 1 / 2*pi*RC*sqrt(6)\n", + "Therefore, C = 6.50 nF\n", + "For FET phase shift oscillator,\n", + "|A| = g_m*R_L and |A| >= 29\n", + "Therefore, g_m*R_L >= 29 i.e. R_L(in k-ohm) >=5.80\n", + "With R_L = 5.8 k-ohm,\n", + "R_L = R_D*r_d / R_D+r_d\n", + "Therefore, R_D =6.78 kohm\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "print \"Using the expression of the frequency,\"\n", + "print \"f = 1 / 2*pi*RC*sqrt(6)\"\n", + "c=(1/(2*pi*10*sqrt(6)*10**6))*10**9\n", + "print \"Therefore, C = %0.2f nF\"%c\n", + "print \"For FET phase shift oscillator,\"\n", + "print \"|A| = g_m*R_L and |A| >= 29\"\n", + "rl=(29/5000)*10**3\n", + "print \"Therefore, g_m*R_L >= 29 i.e. R_L(in k-ohm) >=%0.2f\"%rl\n", + "print \"With R_L = 5.8 k-ohm,\"\n", + "print \"R_L = R_D*r_d / R_D+r_d\"\n", + "rd=40/5.8965\n", + "format(7)\n", + "print \"Therefore, R_D =%0.2f kohm\"%rd" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.34 Page No. 2-89" + ] + }, + { + "cell_type": "code", + "execution_count": 43, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The name of the oscillator is Pierce oscillator\n", + "C1 = 1000 pF, C2 = 100 pF, f_s = 1 MHz\n", + "C_eq = C1*C2 / C1+C2 =0.00 F\n", + "At resonance, X_L = X_Ceq i.e. 2*pi*f*L = 1 / 2*pi*f*C_eq\n", + "Therefore, L = 1/(2*pi*f)**2*C_eq = 278.63 uH\n", + "The fig 2.61(a) shows the electrical equivalent of the crystal\n", + "At series resonance,\n", + "X_L = X_C for crystal\n", + "Therefore, C = 90.909 pF for crystal\n", + "The mounting capacitance is about 1 to 2 pF\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "print \"The name of the oscillator is Pierce oscillator\"\n", + "print \"C1 = 1000 pF, C2 = 100 pF, f_s = 1 MHz\"\n", + "ceq=(1000*100*10**-12)/1100\n", + "print \"C_eq = C1*C2 / C1+C2 =%0.2f F\"%ceq\n", + "print \"At resonance, X_L = X_Ceq i.e. 2*pi*f*L = 1 / 2*pi*f*C_eq\"\n", + "l=(1/(((2*pi*10**6)**2)*(90.909*10**-12)))*10**6\n", + "print \"Therefore, L = 1/(2*pi*f)**2*C_eq = %0.2f uH\"%l\n", + "print \"The fig 2.61(a) shows the electrical equivalent of the crystal\"\n", + "print \"At series resonance,\"\n", + "print \"X_L = X_C for crystal\"\n", + "print \"Therefore, C = 90.909 pF for crystal\"\n", + "print \"The mounting capacitance is about 1 to 2 pF\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.36 Page No. 2-90" + ] + }, + { + "cell_type": "code", + "execution_count": 44, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "f = 2 kHz\n", + "f = 1/ 2*pi*R*c*sqrt(6) ...For phase shift oscillator\n", + "Choose C = 1 nF\n", + "Therefore, r = 32.49 kohm\n", + "Select FET with g_m = 5000 us and r_d = 50 k-ohm\n", + "For phase shift oscillator, |A| >= 29 and |A| = g_m*R_L\n", + "Therefore, g_m*R_L >= 29\n", + "i.e. R_L(in k-ohm) >= 29/g_m >=5.80\n", + "Select R_L = 6.8 k-ohm\n", + "But R_L = R_D*r_d / R_D+r_d\n", + "Therefore, R_D = 7.87 kohm\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "print \"f = 2 kHz\"\n", + "print \"f = 1/ 2*pi*R*c*sqrt(6) ...For phase shift oscillator\"\n", + "print \"Choose C = 1 nF\"\n", + "r=(1/(2*pi*2*sqrt(6)*10**-6))*10**-3\n", + "print \"Therefore, r = %0.2f kohm\"%r\n", + "print \"Select FET with g_m = 5000 us and r_d = 50 k-ohm\"\n", + "print \"For phase shift oscillator, |A| >= 29 and |A| = g_m*R_L\"\n", + "print \"Therefore, g_m*R_L >= 29\"\n", + "rl=(29/(5000*10**-6))*10**-3\n", + "print \"i.e. R_L(in k-ohm) >= 29/g_m >=%0.2f\"%rl\n", + "print \"Select R_L = 6.8 k-ohm\"\n", + "print \"But R_L = R_D*r_d / R_D+r_d\"\n", + "rd=7.87\n", + "print \"Therefore, R_D = %0.2f kohm\"%rd" + ] + } + ], + "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/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/chapter3.ipynb b/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/chapter3.ipynb new file mode 100644 index 00000000..b6d65281 --- /dev/null +++ b/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/chapter3.ipynb @@ -0,0 +1,390 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3 - Multivibrator and Blocking Oscillators" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.1 : Page No. 3-101" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Close loop voltage gain AF = 11.00 \n", + "Value of RiF = 1.82e+10 ohm\n", + "Value of RoF = 4.12e-03 ohm\n", + "Value of fF = 91 kHz\n" + ] + } + ], + "source": [ + "# given data\n", + "R1=1.5 #in kohm\n", + "RF=15 #in kohm\n", + "A=2*10**5 #unitless\n", + "Ri=1 #in Mohm\n", + "Ro=75 #in ohm\n", + "fo=5 #in Hz\n", + "AF=1+RF/R1 #unitless\n", + "B=1/AF #unitless\n", + "RiF=(1+A*B)*Ri*10**6 #in ohm\n", + "RoF=Ro/(1+A*B) #in ohm\n", + "fF=fo*(1+A*B) #in ohm\n", + "print \"Close loop voltage gain AF = %0.2f \" %AF\n", + "print \"Value of RiF = %0.2e ohm\"%RiF\n", + "print \"Value of RoF = %0.2e ohm\"%RoF\n", + "print \"Value of fF = %0.f kHz\"%(fF/1000)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.2 : Page No. 3-101" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Value of RiF = 99.00 kohm\n", + "Value of RiF = 2.00e+09 ohm\n", + "Value of RoF = 3.75e-02 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# given data\n", + "AF=100 #unitless\n", + "A=2*10**5 #unitless\n", + "Ri=1 #in Mohm\n", + "Ro=75 #in ohm\n", + "#let R1 =1 ohm\n", + "R1=1 #in ohm\n", + "#formula : AF=1+RF/R1\n", + "RF=(AF-1)*R1 #in kohm\n", + "B=1/AF #unitless\n", + "RiF=(1+A*B)*Ri*10**6 #in ohm\n", + "RoF=Ro/(1+A*B) #in ohm\n", + "print \"Value of RF = %0.2f kohm\"%RF\n", + "print \"Value of RiF = %0.2e ohm\"%RiF\n", + "print \"Value of RoF = %0.2e ohm\"%RoF" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.3 : Page No. 3-102" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Value of RiF = 2.00e+11 ohm\n", + "Value of RoF = 3.75e-04 ohm\n", + "Value of fF = 1.00 MHz\n" + ] + } + ], + "source": [ + "# given data\n", + "AF=1 #unitless\n", + "B=1 #unitless\n", + "A=2*10**5 #unitless\n", + "fo=5 #in Hz\n", + "Ri=1 #in Mohm\n", + "Ro=75 #in ohm\n", + "#let 1+AB=A as A>>>1\n", + "RiF=A*Ri*10**6 #in ohm\n", + "RoF=Ro/A #in ohm\n", + "fF=fo*A #in ohm\n", + "print \"Value of RiF = %0.2e ohm\"%RiF\n", + "print \"Value of RoF = %0.2e ohm\"%RoF\n", + "print \"Value of fF = %0.2f MHz\"%(fF/10**6)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.4 : Page No. 3-103" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Value of RF = 300 kohm\n" + ] + } + ], + "source": [ + " # given data\n", + "R1=50 #in Kohm\n", + "AF=-6 #unitless\n", + "# here AF=-RF/R1\n", + "RF=-AF*R1 #in kohm\n", + "print \"Value of RF = %0.f kohm\" %RF" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.5 : Page No. 3-103" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Close loop voltage gain AF = -6.00\n", + "Value of RiF = 50.00 kohm\n", + "Value of RoF = 2.62e-03 ohm\n", + "Value of fF = 143 kHz\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# given data\n", + "A=2*10**5 #unitless\n", + "Ri=1 #in Mohm\n", + "Ro=75 #in ohm\n", + "fo=5 #in Hz\n", + "R1=50 #in kohm\n", + "RF=300 #in kohm\n", + "K=RF/(R1+RF) #unitless\n", + "B=R1/(R1+RF) #unitless\n", + "AF=-(A*K/(1+A*B)) #unitless\n", + "RiF=R1 #in kohm ideal\n", + "RoF=Ro/(1+A*B) #in ohm\n", + "fF=-(A*K*fo/AF) #in Hz\n", + "print \"Close loop voltage gain AF = %0.2f\"%AF\n", + "print \"Value of RiF = %0.2f kohm\"%RiF\n", + "print \"Value of RoF = %0.2e ohm\"%RoF\n", + "print \"Value of fF = %0.f kHz\"%(fF/1000)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.6 : Page No. 3-104" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Value of RF = 30 kohm\n" + ] + } + ], + "source": [ + "# given data\n", + "# let R1=R2=R3=R=10kohm\n", + "R=10 #in kohm\n", + "R1=R #in kohm\n", + "R2=R #in kohm\n", + "R3=R #in kohm\n", + "RF=3*R #in Kohm\n", + "print \"Value of RF = %0.0f kohm\"%RF" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.7 : Page No. 3-104" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Value of R1 = 33.33 kohm\n", + "Value of RiF = 33.33 kohm\n" + ] + } + ], + "source": [ + " # given data\n", + "RF=1 #in Mohm\n", + "AV=-30 #unitless\n", + "#AV=-RF/R1=Vo/V1\n", + "R1=-RF*10**6/AV #in ohm\n", + "#for an inverting amplifier RiF=R1\n", + "RiF=R1 #in ohm\n", + "print \"Value of R1 = %0.2f kohm\"%(R1/1000)\n", + "print \"Value of RiF = %0.2f kohm\"%(RiF/1000)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.8 : Page No. 3-105" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Required value of R2 = 533.33 kohm\n", + "Required value of R1 = 66.67 kohm\n" + ] + } + ], + "source": [ + " # given data\n", + "AV=-8 #unitless\n", + "Vin=-1 #in Volts\n", + "Imax=15 #in uA\n", + "Vo=AV*Vin #in Volts\n", + "#Formula : Vo=Imax*R2min\n", + "R2min=Vo/(Imax*10**-6) #in kohm\n", + "R1min=-Vin/(Imax*10**-6) #in kohm\n", + "print \"Required value of R2 = %0.2f kohm\"%(R2min/1000)\n", + "print \"Required value of R1 = %0.2f kohm\"%(R1min/1000)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.9 : Page No. 3-106" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of RF = 75.00 Mohm\n", + "Value of R1 = 98.80 kohm\n", + "Value of R2 = 494.00 kohm\n", + "Value of R3 = 494.00 kohm\n", + "Value of R4 = 17.64 kohm\n" + ] + } + ], + "source": [ + " # given data\n", + "Vo=1.5 #in Volts\n", + "Vin=10 #in mVolts\n", + "RiF=500 #in kohm\n", + "R1=500 #in kohm\n", + "AF=Vo/(Vin*10**-3) #unitless\n", + "RF=AF*R1 #in Kohm\n", + "print \"The value of RF = %0.2f Mohm\"%(RF/1000)\n", + "#AF=-R2/R1*(1+R3/R2+R3/R4)\n", + "#Microphone resistance is Rm=1.2 Kohm\n", + "R1eff=100 #in Kohm\n", + "Rm=1.2 #in Kohm\n", + "R1=R1eff-Rm\n", + "R3=5*R1 \n", + "R2=R3 #in Kohm\n", + "R4=R3/28 #in Kohm\n", + "print \"Value of R1 = %0.2f kohm\" %R1\n", + "print \"Value of R2 = %0.2f kohm\" %R2\n", + "print \"Value of R3 = %0.2f kohm\" %R3\n", + "print \"Value of R4 = %0.2f kohm\" %R4" + ] + } + ], + "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/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/chapter4.ipynb b/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/chapter4.ipynb new file mode 100644 index 00000000..c3968d86 --- /dev/null +++ b/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/chapter4.ipynb @@ -0,0 +1,382 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4 - High Frequency Amplifiers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.1 Page No. 4-34" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) g_m = I_C / V_T =38.46 mA/V\n", + "(ii) r_b''e = h_fe / g_m =5.20 kohm\n", + "(iii) (C_e + C_C) = g_m / 2*pi*f_T = g_m / omega_T =-76.92 pF\n", + "Therefore, C_b''e = C_e =73.92 pF\n", + "(iv) We know that,\n", + "f_T = h_fe*f_beta\n", + "Therefore, 2*pi*f_T = h_fe*2*pi*f_beta\n", + "omega_T = h_fe*omega_beta\n", + "omega_beta = omega_T / h_fe =2500.00 rad/s\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "gm=(1/26)*10**3\n", + "print \"(i) g_m = I_C / V_T =%0.2f mA/V\"%gm\n", + "rbe=200/(38.46)\n", + "print \"(ii) r_b''e = h_fe / g_m =%0.2f kohm\"%rbe\n", + "cc=((38.46*10**-3)/(500*10**6))*10**12\n", + "print \"(iii) (C_e + C_C) = g_m / 2*pi*f_T = g_m / omega_T =%0.2f pF\"%-cc\n", + "cbe=76.92-3\n", + "print \"Therefore, C_b''e = C_e =%0.2f pF\"%cbe\n", + "print \"(iv) We know that,\"\n", + "print \"f_T = h_fe*f_beta\"\n", + "print \"Therefore, 2*pi*f_T = h_fe*2*pi*f_beta\"\n", + "print \"omega_T = h_fe*omega_beta\"\n", + "ob=((500*10**6)/200)*10**-3\n", + "print \"omega_beta = omega_T / h_fe =%0.2f rad/s\"%ob" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.2 Page No. 4-35" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) f_T = |A_i|*f =50.00 MHz\n", + "(ii) h_fe(in kHz) = f_T / f_beta =250.00 kHz\n", + "(iii) |A_i| = h_fe / sqrt(1+((f/f_beta)**2)) :\n", + "At f = 10 MHz\n", + "|A_i| =5.00\n", + "At f = 100 MHz\n", + "|A_i| =0.50\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "from __future__ import division \n", + "ft=25*2\n", + "print \"(i) f_T = |A_i|*f =%0.2f MHz\"%ft\n", + "hfe=50000/200\n", + "print \"(ii) h_fe(in kHz) = f_T / f_beta =%0.2f kHz\"%hfe\n", + "print \"(iii) |A_i| = h_fe / sqrt(1+((f/f_beta)**2)) :\"\n", + "print \"At f = 10 MHz\"\n", + "ai=250/sqrt(1+(((10*10**6)/(200*10**3))**2))\n", + "print \"|A_i| =%0.2f\"%ai\n", + "print \"At f = 100 MHz\"\n", + "ai=250/sqrt(1+(((100*10**6)/(200*10**3))**2))\n", + "print \"|A_i| =%0.2f\"%ai" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.10 Page No. 4-39" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) The 3dB frequency for circuit gain and voltage gain is given as,\n", + "(f_H)=1/(2*pi*R_eq*C_eq)\n", + "where R_eq =(R_s+r_bb'')parallel to r+b''e =166.67 ohm\n", + "and C_eq =(C_b''e)+(1+(g_m*R_L)*C_b''c)= 0.00 F\n", + "f_H = 3774350.04 Hz\n", + "b)Voltage gain is given as,\n", + "A=(-g_m*R_L)=-50.00\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"a) The 3dB frequency for circuit gain and voltage gain is given as,\"\n", + "print \"(f_H)=1/(2*pi*R_eq*C_eq)\"\n", + "r=(200*1000)/(200+1000)\n", + "print \"where R_eq =(R_s+r_bb'')parallel to r+b''e =%0.2f ohm\"%r\n", + "c=(100*10**-12)+((1+50)*3*10**-12)\n", + "print \"and C_eq =(C_b''e)+(1+(g_m*R_L)*C_b''c)= %0.2f F\"%c\n", + "f=1/((2*pi*166.67*253*10**-12))\n", + "print \"f_H = %0.2f Hz\"%f\n", + "print \"b)Voltage gain is given as,\"\n", + "a=(-50*1)\n", + "print \"A=(-g_m*R_L)=%0.2f\"%a" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.12 Page No. 4-40" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "f_H=1/(2*pi*R_eq*C_eq)\n", + "and f_H''=2(f_H)\n", + "1/(2*pi*R_eq*C_eq) = 2/(2*pi*R_eq*C_eq)\n", + "R_eq'' = R_eq/2\n", + "R_eq=(r_b''e)parallel to (r_bb''+R_s)\n", + "= (r_b''e)=1000 ohm\n", + "Therefore R_eq'' =500 ohm\n", + "Therefore 500=((r_b''e)*(r_bb''+R_s))/((r_b''e)+(r_bb'')+R_s)\n", + " = 1000(100+R_s)/(1000+100+R_s)\n", + "R_s = 900.00 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "print \"f_H=1/(2*pi*R_eq*C_eq)\"\n", + "print \"and f_H''=2(f_H)\"\n", + "print \"1/(2*pi*R_eq*C_eq) = 2/(2*pi*R_eq*C_eq)\"\n", + "print \"R_eq'' = R_eq/2\"\n", + "print \"R_eq=(r_b''e)parallel to (r_bb''+R_s)\"\n", + "print \"= (r_b''e)=1000 ohm\"\n", + "print \"Therefore R_eq'' =500 ohm\"\n", + "print \"Therefore 500=((r_b''e)*(r_bb''+R_s))/((r_b''e)+(r_bb'')+R_s)\"\n", + "print \" = 1000(100+R_s)/(1000+100+R_s)\"\n", + "r=(4.5*10**5)/500\n", + "print \"R_s = %0.2f ohm\"%r" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.16 Page No. 4-44" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Hybrid-pi Equivalent is as shown in fig.4.29\n", + "(i) Mid frequency voltage gain :\n", + "V_o / V_s = -h_fe*R_L / R_s+h_ie\n", + "h_ie = r_bb'' + r_b''e =1.10 kohm\n", + "h_fe = g_m * r_b''e =200.00\n", + "Therefore, V_o / V_s =-100.00\n", + "(ii) f_beta = 1 / 2*pi*r_b''e*(C_e+C_C) =780.17 kHz\n", + "f_beta = 780.17 kHz\n", + "(iii) f_T = h_fe * f_beta =156.00 kHz\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"Hybrid-pi Equivalent is as shown in fig.4.29\"\n", + "print \"(i) Mid frequency voltage gain :\"\n", + "print \"V_o / V_s = -h_fe*R_L / R_s+h_ie\"\n", + "hie=(100+1000)*10**-3\n", + "print \"h_ie = r_bb'' + r_b''e =%0.2f kohm\"%hie\n", + "hfe=0.2*1000\n", + "print \"h_fe = g_m * r_b''e =%0.2f\"%hfe\n", + "vo=-200/2\n", + "print \"Therefore, V_o / V_s =%0.2f\"%vo\n", + "fb=(1/(2*pi*1000*(204*10**-12)))*10**-3\n", + "print \"(ii) f_beta = 1 / 2*pi*r_b''e*(C_e+C_C) =%0.2f kHz\"%fb\n", + "print \"f_beta = %0.2f kHz\"%fb\n", + "ft=(200*780)*10**-3\n", + "print \"(iii) f_T = h_fe * f_beta =%0.2f kHz\"%ft" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.17 Page No. 4-44" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) We know that,\n", + " f_H = 1 / 2*pi*R_eq*C_eq\n", + "where R_eq = (R_s+r_bb'')*r_b''e / R_s+r_bb''+r_b''e\n", + "and C_eq = C_e + C_C*[1+g_m*R_L]\n", + " r_b''e(in k-ohm) = h_fo / g_m = 1.00 kohm\n", + "C_eq = C_e + C_C*[1+g_m*R_L] = C_e + C_C[1+100*10**-3*500]\n", + " = C_e + 51 pF\n", + "C_e = g_m / 2*pi*f_T =39.79 pF\n", + "Therefore, C_eq =90.79 pF\n", + "R_eq = 1 / 2*pi*f_H*C_eq =350.60 ohm\n", + "Therefore, 350.6 = (R_s+100)*1000 / R_s+1100\n", + "Therefore, R_s = 439.88 ohm\n", + "(ii) The mid-band voltage gain V_o/V_s is given as\n", + " V_o/V_s = -h_fe*R_L / R_s+h_ie\n", + "where h_ie = r_bb'' + r_b''e =1.10 K\n", + "Therefore, V_o/V_s =-32.47\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"(i) We know that,\"\n", + "print \" f_H = 1 / 2*pi*R_eq*C_eq\"\n", + "print \"where R_eq = (R_s+r_bb'')*r_b''e / R_s+r_bb''+r_b''e\"\n", + "print \"and C_eq = C_e + C_C*[1+g_m*R_L]\"\n", + "rbe=100/100\n", + "print \" r_b''e(in k-ohm) = h_fo / g_m = %0.2f kohm\"%rbe\n", + "print \"C_eq = C_e + C_C*[1+g_m*R_L] = C_e + C_C[1+100*10**-3*500]\"\n", + "print \" = C_e + 51 pF\"\n", + "ce=((100*10**-3)/(2*pi*(400*10**6)))*10**12\n", + "print \"C_e = g_m / 2*pi*f_T =%0.2f pF\"%ce\n", + "ceq=39.79+51\n", + "print \"Therefore, C_eq =%0.2f pF\"%ceq\n", + "req=1/(2*pi*5*90.79*10**-6)\n", + "print \"R_eq = 1 / 2*pi*f_H*C_eq =%0.2f ohm\"%req\n", + "print \"Therefore, 350.6 = (R_s+100)*1000 / R_s+1100\"\n", + "rs=(285.66*10**3)/649.4\n", + "print \"Therefore, R_s = %0.2f ohm\"%rs\n", + "print \"(ii) The mid-band voltage gain V_o/V_s is given as\"\n", + "print \" V_o/V_s = -h_fe*R_L / R_s+h_ie\"\n", + "hie=(100+1000)*10**-3\n", + "print \"where h_ie = r_bb'' + r_b''e =%0.2f K\"%hie\n", + "vo=(-100*500)/(439.88+1100)\n", + "print \"Therefore, V_o/V_s =%0.2f\"%vo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.20 Page No. 4-47" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Assume that the output time-constant is negligible as compared to the time consedtant. When this is the case\n", + "A_vs = V_o/V_s = -g_m*R''_L*G''_s / G''_s+g_b''e+sC\n", + "where G''_s = 1 / (R_s||R_b)+r_bb'' =0.01\n", + "g_b''e = 1 / r_b''e =0.00\n", + "R''_L = R_L || R_C =333.33 ohm\n", + " sC = admittance of C\n", + "where C = C_e + C_C*(1+g_m*R''_L) =153.00\n", + "At 10 kHz,\n", + "sC = 2*pi*f*C =0.00\n", + "Therefore, At 10kHz signal frequency\n", + "A_vs = V_o / V_s =-14.47\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"Assume that the output time-constant is negligible as compared to the time consedtant. When this is the case\"\n", + "print \"A_vs = V_o/V_s = -g_m*R''_L*G''_s / G''_s+g_b''e+sC\"\n", + "gs=6.66*10**-3\n", + "print \"where G''_s = 1 / (R_s||R_b)+r_bb'' =%0.2f\"%gs\n", + "gbe=1/1000\n", + "print \"g_b''e = 1 / r_b''e =%0.2f\"%gbe\n", + "rl=(0.5/1.5)*10**3\n", + "print \"R''_L = R_L || R_C =%0.2f ohm\"%rl\n", + "print \" sC = admittance of C\"\n", + "c=100+(3*(1+(50*333.33*10**-3)))\n", + "print \"where C = C_e + C_C*(1+g_m*R''_L) =%0.2f\"%c\n", + "print \"At 10 kHz,\"\n", + "sc=2*pi*10*153*10**-9\n", + "print \"sC = 2*pi*f*C =%0.2f\"%sc\n", + "print \"Therefore, At 10kHz signal frequency\"\n", + "avs=(-50*333.33*6.66*10**-6)/((6.66*10**-3)+(10**-3)+(9.613*10**-6))\n", + "print \"A_vs = V_o / V_s =%0.2f\"%avs" + ] + } + ], + "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/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/chapter5.ipynb b/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/chapter5.ipynb new file mode 100644 index 00000000..69dd062d --- /dev/null +++ b/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/chapter5.ipynb @@ -0,0 +1,500 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5 - Tuned Amplifiers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.1 Page No. 5-45" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R_p = omega_0 * L * Q =471.24 kohm\n", + "R_s = omega_0*L / Q =5.24 ohm\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "rp=2*pi*10**6*250*300*10**-9\n", + "print \"R_p = omega_0 * L * Q =%0.2f kohm\"%rp\n", + "rs=(2*pi*250)/300\n", + "print \"R_s = omega_0*L / Q =%0.2f ohm\"%rs" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.2 Page No. 5-45" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "From equation 9 we have\n", + " BW = 1 / 2*pi*R*C\n", + "Therefore, R*C = 1 / 2*pi*BW =0.00\n", + "From equation 3 we have\n", + " R = r_i || R_p || r_b''e\n", + "where r_i = 4 k-ohm\n", + "r_b''e = h_fe / g_m =2500.00 ohm\n", + "R_p = Q_c * omega_0 * L = Q_c / omega_0*C\n", + "Therefore, R = 4*10**3 || 2500 || Q_c/omega_0*C\n", + "C = 1 / 2*pi*10*10**3*R\n", + "Therefore, C = 1 / 2*pi*10*10**3*[4*10**3 || 2500 || Q_c/2*pi*500*10**3*C]\n", + "The typical range for Q_c is 10 to 150. However, we have to assume Q such that value of C_p should be positive. Let us assume Q = 100\n", + "Therefore, C = 1 / 2*pi*10*10**3*[1538.5 || 1/2*pi*5000*C]\n", + " = 1 / 2*pi*10*10**3*[1 / 1/1538.5+2*pi*5000*C]\n", + "Solving for C we get\n", + " C = 0.02 uF\n", + "We have\n", + " C = C'' + C_b''e + (1+g_m*R_L)*C_b''e\n", + "Therefore, C'' = C - [C_b''e + (1+g_m*R_L)*C_b''e]\n", + "Therefore, C'' = 0.02 uF\n", + "We have,\n", + "omega_0**2 = 1 / L*C\n", + "Therefore, L = 1 / omega_0**2*C =5.07 uH\n", + "From equation 2 we have,\n", + "R_p = omega*L*Q_c =1570.80 ohm\n", + "Therefore, R = r_i || R_p || r_b''e =777.04 ohm\n", + "We have mid frequency gain as\n", + "A_v(max) = -g_m*R =-31.08\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"From equation 9 we have\"\n", + "print \" BW = 1 / 2*pi*R*C\"\n", + "rc=1/(2*pi*10*10**3)\n", + "print \"Therefore, R*C = 1 / 2*pi*BW =%0.2f\"%rc\n", + "print \"From equation 3 we have\"\n", + "print \" R = r_i || R_p || r_b''e\"\n", + "print \"where r_i = 4 k-ohm\"\n", + "rbe=100/0.04\n", + "print \"r_b''e = h_fe / g_m =%0.2f ohm\"%rbe\n", + "print \"R_p = Q_c * omega_0 * L = Q_c / omega_0*C\"\n", + "print \"Therefore, R = 4*10**3 || 2500 || Q_c/omega_0*C\"\n", + "print \"C = 1 / 2*pi*10*10**3*R\"\n", + "print \"Therefore, C = 1 / 2*pi*10*10**3*[4*10**3 || 2500 || Q_c/2*pi*500*10**3*C]\"\n", + "print \"The typical range for Q_c is 10 to 150. However, we have to assume Q such that value of C_p should be positive. Let us assume Q = 100\"\n", + "print \"Therefore, C = 1 / 2*pi*10*10**3*[1538.5 || 1/2*pi*5000*C]\"\n", + "print \" = 1 / 2*pi*10*10**3*[1 / 1/1538.5+2*pi*5000*C]\"\n", + "print \"Solving for C we get\"\n", + "print \" C = 0.02 uF\"\n", + "print \"We have\"\n", + "print \" C = C'' + C_b''e + (1+g_m*R_L)*C_b''e\"\n", + "print \"Therefore, C'' = C - [C_b''e + (1+g_m*R_L)*C_b''e]\"\n", + "c=((0.02*10**-6)-((1000*10**-12)+((1+(0.04*510))*100*10**-12)))*10**6\n", + "print \"Therefore, C'' = %0.2f uF\"%c\n", + "print \"We have,\"\n", + "print \"omega_0**2 = 1 / L*C\"\n", + "l=(1/(((2*pi*500*10**3)**2)*(0.02*10**-6)))*10**6\n", + "print \"Therefore, L = 1 / omega_0**2*C =%0.2f uH\"%l\n", + "print \"From equation 2 we have,\"\n", + "rp=2*pi*500*5*100*10**-3\n", + "print \"R_p = omega*L*Q_c =%0.2f ohm\"%rp\n", + "r=(4000*1570*2500)/((1570*2500)+(4000*2500)+(4000*1570))\n", + "print \"Therefore, R = r_i || R_p || r_b''e =%0.2f ohm\"%r\n", + "print \"We have mid frequency gain as\"\n", + "av=-0.04*777\n", + "print \"A_v(max) = -g_m*R =%0.2f\"%av" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.3 Page No. 5-46" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) We know that,\n", + "BW_n = BW_1 * sqrt(2**1/n - 1) =10.20 kHz\n", + "(ii) BW_n = BW_1 * sqrt(2**1/n - 1) =8.70 kHz\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"(i) We know that,\"\n", + "bw=((20*10**3)*sqrt(((2)**(1/3))-1))*10**-3\n", + "print \"BW_n = BW_1 * sqrt(2**1/n - 1) =%0.2f kHz\"%bw\n", + "bw1=((20*10**3)*sqrt(((2)**(1/4))-1))*10**-3\n", + "print \"(ii) BW_n = BW_1 * sqrt(2**1/n - 1) =%0.2f kHz\"%bw1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.6 Page No. 5-48" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) We have,\n", + "A_vmid=(-g_m*R)= -15\n", + "Therefore R(in ohms)=(-15)/(-5*10**-3)= 3000.00\n", + "b) The Miller effect capacitance is given by \n", + "C_d = C_gs+(1+g_m*R)*(C_g*d)\n", + " = (1*10**-12)+(1+15)*(3*10**-12)=0.00 F\n", + "c) The limit frequency of the uncompemsated amplifier is \n", + "f2 =1/(2*pi*C_d*R)= 1082686.69 Hz\n", + "d) L =q*C_d*R**2= 0.00 H\n", + "e) Possible extension of frequency range\n", + "f''2 =1.72*f2= 1857600.00 Hz\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"a) We have,\"\n", + "print \"A_vmid=(-g_m*R)= -15\"\n", + "r=15/(5*10**-3)\n", + "print \"Therefore R(in ohms)=(-15)/(-5*10**-3)= %0.2f\"%r\n", + "print \"b) The Miller effect capacitance is given by \"\n", + "print \"C_d = C_gs+(1+g_m*R)*(C_g*d)\"\n", + "c=(10**-12)+((1+15)*(3*10**-12))\n", + "print \" = (1*10**-12)+(1+15)*(3*10**-12)=%0.2f F\"%c\n", + "print \"c) The limit frequency of the uncompemsated amplifier is \"\n", + "f=1/(2*pi*49*3*10**-9)\n", + "print \"f2 =1/(2*pi*C_d*R)= %0.2f Hz\"%f\n", + "l=0.414*((3*10**3)**2)*(49*10**-12)\n", + "print \"d) L =q*C_d*R**2= %0.2f H\"%l\n", + "print \"e) Possible extension of frequency range\"\n", + "e=1.72*1.08*10**6\n", + "print \"f''2 =1.72*f2= %0.2f Hz\"%e" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.8 Page No. 5-49" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Resonant frequency:\n", + "f_r = 1 / 2*pi*sqrt(LC) =1.59 MHz\n", + "(ii) We know that\n", + "Q_r = R_p / omega_r*L\n", + "Therefore, Impedance of tuned circuit R_p = Q_r * omega_r * L =5994.16\n", + "(iii) Voltage gain of stage A_v,\n", + "A_v = A_I*R''_L / R''_i =-299.94\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"(i) Resonant frequency:\"\n", + "fr=(1/(2*pi*sqrt(20*500*10**-18)))*10**-6\n", + "print \"f_r = 1 / 2*pi*sqrt(LC) =%0.2f MHz\"%fr\n", + "print \"(ii) We know that\"\n", + "print \"Q_r = R_p / omega_r*L\"\n", + "rp=30*2*pi*1.59*20\n", + "print \"Therefore, Impedance of tuned circuit R_p = Q_r * omega_r * L =%0.2f\"%rp\n", + "print \"(iii) Voltage gain of stage A_v,\"\n", + "av=(-50*((5994*1500)/(5994+1500)))/200\n", + "print \"A_v = A_I*R''_L / R''_i =%0.2f\"%av" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.10 Page No. 5-50" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "i) f_r=Resonant frequency\n", + "= 1/(2*pi*sqrt(L*C))= 159154.94\n", + "ii) Tuned circuit dynamic resistance=R_p=L/CR\n", + "= (400 microH)/(2500pF)*(5ohm)= 32000.00\n", + "iii) Gain at resonance=A_v=(-g_m*R_L)=(-g_m*R_p)\n", + " = 6mA/V * 32kohm = -192.00\n", + "iv) The signal bandwidth =BW=(f_r)/Q\n", + "Q=(omega_r*L)/R= 79.92\n", + "BW =(f_r)/Q= 1989.49 Hz\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"i) f_r=Resonant frequency\"\n", + "f=1/((2*pi)*sqrt(0.0004*2500*10**-12))\n", + "print \"= 1/(2*pi*sqrt(L*C))= %0.2f\"%f\n", + "print \"ii) Tuned circuit dynamic resistance=R_p=L/CR\"\n", + "r=(80*10**6)/2500\n", + "print \"= (400 microH)/(2500pF)*(5ohm)= %0.2f\"%r\n", + "print \"iii) Gain at resonance=A_v=(-g_m*R_L)=(-g_m*R_p)\"\n", + "a=-6*32\n", + "print \" = 6mA/V * 32kohm = %0.2f\"%a\n", + "print \"iv) The signal bandwidth =BW=(f_r)/Q\"\n", + "q=(2*pi*0.159*400)/5\n", + "print \"Q=(omega_r*L)/R= %0.2f\"%q\n", + "b=159000/79.92\n", + "print \"BW =(f_r)/Q= %0.2f Hz\"%b" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.11 Page No. 5-51" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) R_L = r_d || R_p\n", + "R_p = Tank circuit impedance at resonance = L / CR\n", + "f_r = 1 / 2*pi*sqrt(L*C)\n", + "Therefore, C = 1 / 4*pi**2*f_r**2*L =50.10 pF\n", + "Q = omega_r*L / R = 2*pi*f_r*L / R\n", + "Therefore, R = 2*pi*f_r*L / Q =39.96 ohm\n", + "R_F = L / C*R =100.00 kohm\n", + "R_L = r_d*R_p / r_d+R_p =83.33 kohm\n", + "A_v = -g_m*R_L = 416.65 at resonance frequency omega_r\n", + "(ii) At f = f_r+10 kHz = 1.6 MHz\n", + "|A_v / A_v(at resonance)| = 1 / sqrt(1+(f/f_r)**2)\n", + "Therefore, |A_v| = |A_v(at resonance| / sqrt(1+(f/f_r)**2) =293.71\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"(i) R_L = r_d || R_p\"\n", + "print \"R_p = Tank circuit impedance at resonance = L / CR\"\n", + "print \"f_r = 1 / 2*pi*sqrt(L*C)\"\n", + "c=(1/(4*pi**2*200*1.59**2*10**6))*10**12\n", + "print \"Therefore, C = 1 / 4*pi**2*f_r**2*L =%0.2f pF\"%c\n", + "print \"Q = omega_r*L / R = 2*pi*f_r*L / R\"\n", + "r=(2*pi*200*1.59)/50\n", + "print \"Therefore, R = 2*pi*f_r*L / Q =%0.2f ohm\"%r\n", + "rf=((200*10**-6)/(50*40*10**-12))*10**-3\n", + "print \"R_F = L / C*R =%0.2f kohm\"%rf\n", + "rl=(500*100)/600\n", + "print \"R_L = r_d*R_p / r_d+R_p =%0.2f kohm\"%rl\n", + "av=5*83.33\n", + "print \"A_v = -g_m*R_L = %0.2f at resonance frequency omega_r\"%av\n", + "print \"(ii) At f = f_r+10 kHz = 1.6 MHz\"\n", + "print \"|A_v / A_v(at resonance)| = 1 / sqrt(1+(f/f_r)**2)\"\n", + "ava=416.67/sqrt(1+((1.6/1.59)**2))\n", + "print \"Therefore, |A_v| = |A_v(at resonance| / sqrt(1+(f/f_r)**2) =%0.2f\"%ava" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.12 Page No. 5-52" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Resonant frequency f_r = 1 / 2*pi*sqrt(L*C) =503292.12 kHz\n", + "(ii) Tank circuit impedance at resonance can be given as\n", + "R_P = L / C*R =20.00 kohm\n", + "(iii) A_v = -g_m*R_L = -g_m*(r_d||R_P) =-96.15\n", + "(iv) BW = f_r/Q\n", + " BW = f_r*R / omega_r*L Therefore, Q = omega_r*L / R\n", + " BW = R / 2*pi*L =7.96 kHz\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "fr=1/(2*pi*sqrt(100*1000*10**-18))\n", + "print \"(i) Resonant frequency f_r = 1 / 2*pi*sqrt(L*C) =%0.2f kHz\"%fr\n", + "print \"(ii) Tank circuit impedance at resonance can be given as\"\n", + "rp=((100*10**6)/5000)*10**-3\n", + "print \"R_P = L / C*R =%0.2f kohm\"%rp\n", + "av=(-5*10**-3)*((500*20*10**3)/(520))\n", + "print \"(iii) A_v = -g_m*R_L = -g_m*(r_d||R_P) =%0.2f\"%av\n", + "bw=(5/(2*pi*100*10**-6))*10**-3\n", + "print \"(iv) BW = f_r/Q\"\n", + "print \" BW = f_r*R / omega_r*L Therefore, Q = omega_r*L / R\"\n", + "print \" BW = R / 2*pi*L =%0.2f kHz\"%bw" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.13 Page No. 5-53" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "BW = f_r / Q\n", + "Therefore, Q = f_r / BW =53.50\n", + "Q = omega_r*L / R = 2*pi*f_r*L / R\n", + "Therefore, L/R = Q / 2*pi*f_r =0.00\n", + "|A_v| = g_m*R_L = 30\n", + "Therefore, R_L = (r_d || R_p) =6.00 kohm\n", + "Therefore, R_p = 6383 ohm\n", + "We know that\n", + "R_p = L/C*R\n", + "Therefore, C = 124.55 pF\n", + "We know that\n", + "f_r = 1 / 2*pi*sqrt(L*C)\n", + "Therefore, L = 1.78 uH\n", + "We have\n", + "R_p = L / C*R\n", + "Therefore, R = L / C*R_p =2.24 ohm\n", + "Therefore, elements of tank circuit are:\n", + "L = 1.777 uH, C = 124.5 pF and R = 2.236 ohm\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"BW = f_r / Q\"\n", + "q=10700/200\n", + "print \"Therefore, Q = f_r / BW =%0.2f\"%q\n", + "print \"Q = omega_r*L / R = 2*pi*f_r*L / R\"\n", + "lr=53.5/(2*pi*10.7*10**6)\n", + "print \"Therefore, L/R = Q / 2*pi*f_r =%0.2f\"%lr\n", + "print \"|A_v| = g_m*R_L = 30\"\n", + "rl=(30/5)\n", + "print \"Therefore, R_L = (r_d || R_p) =%0.2f kohm\"%rl\n", + "print \"Therefore, R_p = 6383 ohm\"\n", + "print \"We know that\"\n", + "print \"R_p = L/C*R\"\n", + "c=((795*10**-9)/6383)*10**12\n", + "print \"Therefore, C = %0.2f pF\"%c\n", + "print \"We know that\"\n", + "l=(1/(4*pi**2*((10.7*10**6)**2)*124.5*10**-12))*10**6\n", + "print \"f_r = 1 / 2*pi*sqrt(L*C)\"\n", + "print \"Therefore, L = %0.2f uH\"%l\n", + "print \"We have\"\n", + "print \"R_p = L / C*R\"\n", + "r=(1.777*10**-6)/(6383*124.5*10**-12)\n", + "print \"Therefore, R = L / C*R_p =%0.2f ohm\"%r\n", + "print \"Therefore, elements of tank circuit are:\"\n", + "print \"L = 1.777 uH, C = 124.5 pF and R = 2.236 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/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/chapter6.ipynb b/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/chapter6.ipynb new file mode 100644 index 00000000..f718c0b0 --- /dev/null +++ b/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/chapter6.ipynb @@ -0,0 +1,1273 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6 - Power Amplifiers [Large Signal Amplifiers]" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.1 Page No. 6-64" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) I_BQ = V_CC-V_BE / R_B =12.87 mA\n", + "I_CQ = beta * I_BQ =643.50 mA\n", + "(ii) V_CC = I_CQ*R_L + V_CEQ\n", + "Therefore, V_CEQ(in V) = V_CC - I_CQ*R_L =9.70 V\n", + "(iii) P_DC = V_CC * I_CQ =12.87 W\n", + "(iv) P_ac Peak current i_b = 9 mA\n", + "i_c = beta * i_b =450.00 mA\n", + "Therefore, i_c(rms) = I_rms(in mA) = i_c(peak) / sqrt(2) =318.20\n", + "Therefore, P_ac = (I_rms)**2 * R_L =1.62 W\n", + "(v) Efficiency eta(in percentage) = P_ac/P_DC * 100 =12.58\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "from __future__ import division \n", + "ibq=(20-0.7)/1.5\n", + "print \"(i) I_BQ = V_CC-V_BE / R_B =%0.2f mA\"%ibq\n", + "icq=50*12.87\n", + "print \"I_CQ = beta * I_BQ =%0.2f mA\"%icq\n", + "print \"(ii) V_CC = I_CQ*R_L + V_CEQ\"\n", + "vceq=20-(643.5*16*10**-3)\n", + "print \"Therefore, V_CEQ(in V) = V_CC - I_CQ*R_L =%0.2f V\"%vceq\n", + "pdc=20*643.5*10**-3\n", + "print \"(iii) P_DC = V_CC * I_CQ =%0.2f W\"%pdc\n", + "print \"(iv) P_ac Peak current i_b = 9 mA\"\n", + "ic=50*9\n", + "print \"i_c = beta * i_b =%0.2f mA\"%ic\n", + "icm=450/sqrt(2)\n", + "print \"Therefore, i_c(rms) = I_rms(in mA) = i_c(peak) / sqrt(2) =%0.2f\"%icm\n", + "pac=318.19**2*16*10**-6\n", + "print \"Therefore, P_ac = (I_rms)**2 * R_L =%0.2f W\"%pac\n", + "n=(1.619*100)/12.87\n", + "print \"(v) Efficiency eta(in percentage) = P_ac/P_DC * 100 =%0.2f\"%n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.2 Page No. 6-64" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R_L = 4 ohm, N1 = 200, N2 = 20\n", + "Therefore, n = N2 / N1 =0.10\n", + "Therefore, R''_L = R1 / n**2 =400.00 ohm\n", + "As N2 < N1, the transformer is step down and hence R''_L > R_L, as the primary winding is high voltage winding.\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "print \"R_L = 4 ohm, N1 = 200, N2 = 20\"\n", + "n=20/200\n", + "print \"Therefore, n = N2 / N1 =%0.2f\"%n\n", + "rl=4/(0.1**2)\n", + "print \"Therefore, R''_L = R1 / n**2 =%0.2f ohm\"%rl\n", + "print \"As N2 < N1, the transformer is step down and hence R''_L > R_L, as the primary winding is high voltage winding.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.3 Page No. 6-65" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R_L = 8 ohm, R''_L = 648 ohm\n", + "Now R''_L = R_L / n**2\n", + "Therefore, n**2 = R_L / R''_L =0.01\n", + "Therefore, n = 0.1111 = Turn ratio\n", + "But, n = N2 / N1 = 0.1111\n", + "Therefore, N1/N2 = 9\n", + "Generally the turns ratio is specified as Ni/N2 : 1 i.e. for this transformer it is 9:1\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "print \"R_L = 8 ohm, R''_L = 648 ohm\"\n", + "print \"Now R''_L = R_L / n**2\"\n", + "n=8/648\n", + "print \"Therefore, n**2 = R_L / R''_L =%0.2f\"%n\n", + "print \"Therefore, n = 0.1111 = Turn ratio\"\n", + "print \"But, n = N2 / N1 = 0.1111\"\n", + "print \"Therefore, N1/N2 = 9\"\n", + "print \"Generally the turns ratio is specified as Ni/N2 : 1 i.e. for this transformer it is 9:1\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example6.4 Page No. 6-66" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R_L=8ohm, I_CQ=140 mA, V_CC=10V\n", + "P_ac= 0.48 W\n", + "The turns ratio are specified as N1/N2:1 i.e 3:1\n", + "Therefore N1/N2=3\n", + "n=N2/N1=1/3=0.33\n", + "Therefore R''_L=R_L/n**2=72.14\n", + "1. As the transformer is ideal, whatever is the power delivered to the load,same is the power developed across primary.\n", + "Therefore P_ac(across primary)=0.48W\n", + "2. Using equation (9),\n", + "we get, P_ac=(V_1rms**2)/(R''_L)\n", + "Therefore 0.48=(V_1rms**2)/72\n", + "Therefore V_1rms(in V)=5.88 But rms value of the load voltage is V_2rms\n", + "So (V1_rms)/(V2_rms)=N1/N2=3/1\n", + "Therefore (V2_rms)(in V)=(V1_rms)/3=1.96\n", + "This is the rms value of the load voltage.\n", + "3. The rms value of the primary voltage is (V1_rms) as calculated above.\n", + "Therefore (V1_rms)=5.8787 V\n", + "4. The power delivered to the load = (I_2rms**2)*R_L ..Refer equation 13.\n", + "0.48=(I_2rms**2)*8\n", + "(I_2rms)[in A]=0.24\n", + "This is the rms value of the load current as the resistance value used is R_L and not R''_L\n", + "5. The rms values of primary and secondary are related through the transformation ratio.\n", + "Therefore (I_1rms)/(I_2rms)=N2/N1=n=0.333\n", + "Thererfore (I_1rms)[in A]=0.2449*0.333= 0.08\n", + "6. The dc power input is,\n", + "P_DC =(V_CC)*(I_CQ)=1.40 W\n", + "7. %eta=(P_ac *100)/(P_dc)=34.29\n", + "P_d = 0.92 W\n", + "This is the power dissipation.\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "from __future__ import division \n", + "print \"R_L=8ohm, I_CQ=140 mA, V_CC=10V\"\n", + "print \"P_ac= 0.48 W\"\n", + "print \"The turns ratio are specified as N1/N2:1 i.e 3:1\"\n", + "print \"Therefore N1/N2=3\"\n", + "n=1/3\n", + "print \"n=N2/N1=1/3=%0.2f\"%n\n", + "r=8/(0.333)**2\n", + "print \"Therefore R''_L=R_L/n**2=%0.2f\"%r\n", + "print \"1. As the transformer is ideal, whatever is the power delivered to the load,same is the power developed across primary.\"\n", + "print \"Therefore P_ac(across primary)=0.48W\"\n", + "print \"2. Using equation (9),\"\n", + "print \"we get, P_ac=(V_1rms**2)/(R''_L)\"\n", + "print \"Therefore 0.48=(V_1rms**2)/72\"\n", + "v=sqrt(34.56)\n", + "print \"Therefore V_1rms(in V)=%0.2f\"%v,\n", + "print \"But rms value of the load voltage is V_2rms\"\n", + "print \"So (V1_rms)/(V2_rms)=N1/N2=3/1\"\n", + "v=5.8787/3\n", + "print \"Therefore (V2_rms)(in V)=(V1_rms)/3=%0.2f\"%v\n", + "print \"This is the rms value of the load voltage.\"\n", + "print \"3. The rms value of the primary voltage is (V1_rms) as calculated above.\"\n", + "print \"Therefore (V1_rms)=5.8787 V\"\n", + "print \"4. The power delivered to the load = (I_2rms**2)*R_L ..Refer equation 13.\"\n", + "print \"0.48=(I_2rms**2)*8\"\n", + "i=sqrt(0.06)\n", + "print \"(I_2rms)[in A]=%0.2f\"%i\n", + "print \"This is the rms value of the load current as the resistance value used is R_L and not R''_L\"\n", + "print \"5. The rms values of primary and secondary are related through the transformation ratio.\"\n", + "print \"Therefore (I_1rms)/(I_2rms)=N2/N1=n=0.333\"\n", + "i=0.2449*0.333\n", + "print \"Thererfore (I_1rms)[in A]=0.2449*0.333= %0.2f\"%i\n", + "print \"6. The dc power input is,\"\n", + "p=140*10**-2\n", + "print \"P_DC =(V_CC)*(I_CQ)=%0.2f W\"%p\n", + "n=(0.48*100)/1.4\n", + "print \"7. %%eta=(P_ac *100)/(P_dc)=%0.2f\"%n\n", + "d=1.4-0.48\n", + "print \"P_d = %0.2f W\"%d\n", + "print \"This is the power dissipation.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.6 Page No. 6-67" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R_L = 4 k-ohm, (P_ac)_D = 0.85 W\n", + "The current without signal is I_CQ = 31 mA\n", + "The current with signal is I_CQ + B0 = 34 mA\n", + "The increase is due to harmonic content in the signal\n", + "Therefore, B0 = 34 - 31 = 3 mA\n", + "But, B2 = B0 = 3 mA\n", + "Now (P_ac)_D = P_ac * [1+D2**2] ... Assuming only second harmonic\n", + "Therefore, (P_ac)_D = 1/2*B1**2*R_L * [1 + B2**2/B1**2]\n", + "Therefore, (P_ac)_D = 1/2*B1**2*R_L + 1/2*B2**2*R_L\n", + "0.85 = 1/2*B1**2*(4*10**3) + 1/2*(9*10**-6)*(4*10**3)\n", + "Therefore, B1 = 20.396 mA\n", + "Therefore, D2(in percentage) = |B2|/|B1| * 100 =14.71\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "print \"R_L = 4 k-ohm, (P_ac)_D = 0.85 W\"\n", + "print \"The current without signal is I_CQ = 31 mA\"\n", + "print \"The current with signal is I_CQ + B0 = 34 mA\"\n", + "print \"The increase is due to harmonic content in the signal\"\n", + "print \"Therefore, B0 = 34 - 31 = 3 mA\"\n", + "print \"But, B2 = B0 = 3 mA\"\n", + "print \"Now (P_ac)_D = P_ac * [1+D2**2] ... Assuming only second harmonic\"\n", + "print \"Therefore, (P_ac)_D = 1/2*B1**2*R_L * [1 + B2**2/B1**2]\"\n", + "print \"Therefore, (P_ac)_D = 1/2*B1**2*R_L + 1/2*B2**2*R_L\"\n", + "print \"0.85 = 1/2*B1**2*(4*10**3) + 1/2*(9*10**-6)*(4*10**3)\"\n", + "print \"Therefore, B1 = 20.396 mA\"\n", + "d2=300/20.396\n", + "print \"Therefore, D2(in percentage) = |B2|/|B1| * 100 =%0.2f\"%d2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.8 Page No. 6-68" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R_L = 12 ohm, n = N2/N1 = 1/3 = 0.333, eta_trans = 78.5%\n", + "Therefore, R''_L = R_L / n**2 =108.22\n", + "(i) For P_max, V_m = V_CC\n", + "Threfore, (P_ac)_max = 1/2 * V_CC**2/R''_L =1.85 W\n", + "But eta_trans = 78.5%\n", + "Therefore, P_L = eta_trans * (P_ac)_max =1.45 W\n", + "(ii) Condition for (P_d)_max is V_m(in V) = 2*V_CC/pi =12.73\n", + "Therefore, (P_d)_max(in W) = 2*V_CC**2 / pi**2*R''_L =0.75 W\n", + "Therefore, (P_d)_max per transistor =0.38 W\n", + "(iii) (P_ac)_max = V_rms * I_rms = V_m/sqrt(2) * I_m/sqrt(2) = V_m*I_m / 2 and V_m = V_CC\n", + "Therefore, 1.8518 = 20*I_m / 2\n", + "Therefore, I_m = (I_c)_max =0.19 A\n", + "and (i_b)_max = (i_c))max / h_fe =7.40 mA\n" + ] + } + ], + "source": [ + "from math import pi,sqrt\n", + "from __future__ import division \n", + "print \"R_L = 12 ohm, n = N2/N1 = 1/3 = 0.333, eta_trans = 78.5%\"\n", + "rl=12/(0.333**2)\n", + "print \"Therefore, R''_L = R_L / n**2 =%0.2f\"%rl\n", + "pac=(0.5*20**2)/108\n", + "print \"(i) For P_max, V_m = V_CC\"\n", + "print \"Threfore, (P_ac)_max = 1/2 * V_CC**2/R''_L =%0.2f W\"%pac\n", + "print \"But eta_trans = 78.5%\"\n", + "pl=0.785*1.8518\n", + "print \"Therefore, P_L = eta_trans * (P_ac)_max =%0.2f W\"%pl\n", + "vm=(2*20)/pi\n", + "print \"(ii) Condition for (P_d)_max is V_m(in V) = 2*V_CC/pi =%0.2f\"%vm\n", + "pd=(2*20**2)/(108*pi**2)\n", + "print \"Therefore, (P_d)_max(in W) = 2*V_CC**2 / pi**2*R''_L =%0.2f W\"%pd\n", + "pdm=0.7505/2\n", + "print \"Therefore, (P_d)_max per transistor =%0.2f W\"%pdm\n", + "print \"(iii) (P_ac)_max = V_rms * I_rms = V_m/sqrt(2) * I_m/sqrt(2) = V_m*I_m / 2 and V_m = V_CC\"\n", + "print \"Therefore, 1.8518 = 20*I_m / 2\"\n", + "im=(2*1.8518)/20\n", + "print \"Therefore, I_m = (I_c)_max =%0.2f A\"%im\n", + "ibm=(0.1851/25)*10**3\n", + "print \"and (i_b)_max = (i_c))max / h_fe =%0.2f mA\"%ibm" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example6.9 Page No. 6-69" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R_L=16 ohm, V_CC=25 V\n", + "Now 2N1=200, N2=50\n", + "Therefore N1=100.00\n", + "Therefore n=N2/N1=0.50\n", + "Therefore R''_L =(R_L)/(n**2)=64.00\n", + "For maximum power output, V_m=V_CC\n", + "i) (P_ac)_max =(V_CC**2)/(2*R_L)=4.88 W\n", + "ii) (P_dc)=(2*V_CC*I_m)/pi\n", + "Now (V_m)/(I_m)=(R''_L)\n", + "and V_m=V_CC\n", + "Therefore (I_m)=(V_CC)/(R''_L)=0.39\n", + "Therefore (P_DC) = 6.22 W\n", + "iii) %eta=(P_ac*100)/(P_DC)=78.57\n", + "iv) (P_d)_max[in W]=(2*(P_ac)_max)/(pi**2)=0.99 W\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"R_L=16 ohm, V_CC=25 V\"\n", + "print \"Now 2N1=200, N2=50\"\n", + "n=200/2\n", + "print \"Therefore N1=%0.2f\"%n\n", + "n=50/100\n", + "print \"Therefore n=N2/N1=%0.2f\"%n\n", + "r=16/(0.5**2)\n", + "print \"Therefore R''_L =(R_L)/(n**2)=%0.2f\"%r\n", + "print \"For maximum power output, V_m=V_CC\"\n", + "p=(25**2)/(2*64)\n", + "print \"i) (P_ac)_max =(V_CC**2)/(2*R_L)=%0.2f W\"%p\n", + "print \"ii) (P_dc)=(2*V_CC*I_m)/pi\"\n", + "print \"Now (V_m)/(I_m)=(R''_L)\"\n", + "print \"and V_m=V_CC\"\n", + "i=25/64\n", + "print \"Therefore (I_m)=(V_CC)/(R''_L)=%0.2f\"%i\n", + "p=(2*25*0.3906)/(pi)\n", + "print \"Therefore (P_DC) = %0.2f W\"%p\n", + "n=(4.8848*100)/6.2169\n", + "print \"iii) %%eta=(P_ac*100)/(P_DC)=%0.2f\"%n\n", + "p=(2*4.8828)/(pi**2)\n", + "print \"iv) (P_d)_max[in W]=(2*(P_ac)_max)/(pi**2)=%0.2f W\"%p" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.12 Page No. 6-70" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R_L = 8 ohm, V_CC = +-12 V hence dual supply version\n", + "(1) (P_ac)_max = 1/2 * V_CC**2/R_L =9.00 W\n", + "(2) P_DC = V_CC*I_DC but I_DC = 2*I_m / pi\n", + " = V_CC * (2*I_m/pi)\n", + "Now R_L = V_m/I_m i.e. I_m = V_m/R_L and V_m = V_CC\n", + "Therefore, P_DC = V_CC * 2 * V_CC/R_L * 1/pi =11.46 W\n", + "Therefore, Total P_D = P_DC - P_ac =2.46 W\n", + "Therefore, P_D per transistor =1.23 W\n", + "(3) %eta(in percentage) = P_ac/P_DC * 100 =78.54\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"R_L = 8 ohm, V_CC = +-12 V hence dual supply version\"\n", + "pac=0.5*(12**2/8)\n", + "print \"(1) (P_ac)_max = 1/2 * V_CC**2/R_L =%0.2f W\"%pac\n", + "print \"(2) P_DC = V_CC*I_DC but I_DC = 2*I_m / pi\"\n", + "print \" = V_CC * (2*I_m/pi)\"\n", + "print \"Now R_L = V_m/I_m i.e. I_m = V_m/R_L and V_m = V_CC\"\n", + "pdc=(12**2*2)/(8*pi)\n", + "print \"Therefore, P_DC = V_CC * 2 * V_CC/R_L * 1/pi =%0.2f W\"%pdc\n", + "pdt=11.4591-9\n", + "print \"Therefore, Total P_D = P_DC - P_ac =%0.2f W\"%pdt\n", + "pd=2.4591/2\n", + "print \"Therefore, P_D per transistor =%0.2f W\"%pd\n", + "n=900/11.4591\n", + "print \"(3) %%eta(in percentage) = P_ac/P_DC * 100 =%0.2f\"%n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example6.13 Page No. 6-71" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "V_CC=10V ,R_L=5 ohm\n", + "i) (P_ac)_max = (V_CC**2)/(2*R_L)=(10**2)/(2*5)=10.00 W\n", + "ii) To decide Power rating of transistors means to find (P_D)_max\n", + "V_m = 6.37 V\n", + "Now, R_L=(V_m)/(I_m)\n", + "Therefore (I_m) = 1.27 A\n", + "Therefore (P_DC)=(V_CC)*(I_DC)=(V_CC)*(2*I_m)/pi (I_DC)=(2*I_m)/pi\n", + " =(10*2*1.2732)/(pi) =8.11\n", + "and (P_ac)[in W]=(V_m*I_m)/2=4.05 W\n", + "(P_D)_max = (P_DC)-(P_ac)=3.60 W\n", + "Therefore P_D rating fora each transistor =(P_D)_max/2=2.03\n", + "iii) For (P_ac)_max, V_m=V_CC=10 V\n", + "I_m = (V_m)/R_L=2.00 A\n", + "P_DC = 12.73 W\n" + ] + } + ], + "source": [ + "from math import pi\n", + "from __future__ import division \n", + "print \"V_CC=10V ,R_L=5 ohm\"\n", + "p=100/10\n", + "print \"i) (P_ac)_max = (V_CC**2)/(2*R_L)=(10**2)/(2*5)=%0.2f W\"%p\n", + "print \"ii) To decide Power rating of transistors means to find (P_D)_max\"\n", + "v=(2*10)/(pi)\n", + "print \"V_m = %0.2f V\"%v\n", + "print \"Now, R_L=(V_m)/(I_m)\"\n", + "i=6.3662/5 \n", + "print \"Therefore (I_m) = %0.2f A\"%i\n", + "print \"Therefore (P_DC)=(V_CC)*(I_DC)=(V_CC)*(2*I_m)/pi (I_DC)=(2*I_m)/pi\"\n", + "p=(10*2*1.2732)/(pi)\n", + "print \" =(10*2*1.2732)/(pi) =%0.2f\"%p\n", + "p=(6.3662*1.2732)/2\n", + "print \"and (P_ac)[in W]=(V_m*I_m)/2=%0.2f W\"%p\n", + "p=8.1056-4.5027\n", + "print \"(P_D)_max = (P_DC)-(P_ac)=%0.2f W\"%p\n", + "p=4.0528/2\n", + "print \"Therefore P_D rating fora each transistor =(P_D)_max/2=%0.2f\"%p\n", + "print \"iii) For (P_ac)_max, V_m=V_CC=10 V\"\n", + "i=10/5\n", + "print \"I_m = (V_m)/R_L=%0.2f A\"%i\n", + "p=(10*2*2)/pi\n", + "print \"P_DC = %0.2f W\"%p" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.15 Page No. 6-72" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "V_CC = 20 V, R_L = 4 ohm\n", + "For (P_d)max, V_m(in V) = 2/pi * V_CC = 12.73\n", + "R_L = V_m / I_m\n", + "Therefore, I_m = V_m / R_L =3.18 A\n", + "Therefore, I_dc = 2*I_m / pi =2.03 A\n", + "Therefore, P_ac = 1/2 * V_m**2/R_L =20.26 W\n", + "and P_dc = V_CC * I_DC =40.51 W\n", + "Therefore, Total (P_d)max = P_dc - P_ac =20.25 W\n", + "Therefore, (P_d)max per transistor =10.13 W\n" + ] + } + ], + "source": [ + "from math import pi\n", + "from __future__ import division \n", + "print \"V_CC = 20 V, R_L = 4 ohm\"\n", + "vm=(2*20)/pi\n", + "print \"For (P_d)max, V_m(in V) = 2/pi * V_CC = %0.2f\"%vm\n", + "print \"R_L = V_m / I_m\"\n", + "im=12.7324/4\n", + "print \"Therefore, I_m = V_m / R_L =%0.2f A\"%im\n", + "idc=(2*3.183)/pi\n", + "print \"Therefore, I_dc = 2*I_m / pi =%0.2f A\"%idc\n", + "pac=(0.5*12.7324**2)/4\n", + "print \"Therefore, P_ac = 1/2 * V_m**2/R_L =%0.2f W\"%pac\n", + "pdc=20*2.0254\n", + "print \"and P_dc = V_CC * I_DC =%0.2f W\"%pdc\n", + "pdm=40.508-20.2542\n", + "print \"Therefore, Total (P_d)max = P_dc - P_ac =%0.2f W\"%pdm\n", + "pdma=20.2538/2\n", + "print \"Therefore, (P_d)max per transistor =%0.2f W\"%pdma" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example6.16 Page No. 6-73" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "For a given transistor,\n", + "Maximum collector current =I_cm =1A\n", + "Maximum power dissipation =P_d=10W\n", + "Maximum V_CEO =40V\n", + "For maximum output power,\n", + "I_cm=2*I_CQ\n", + "I_CQ=1/2=0.50\n", + "and V_CEO=2*V_CC\n", + "V_CC = V_CEO/2= 20.00 V\n", + "and V_cc=V_m=20V for (P_ac)_max\n", + "(P_ac)_max=(V_cc**2)/(2*R_L)\n", + "R''_L=(V_m)/I_m and I_m=I_CQ=0.5 A\n", + "R''_L =40.00 ohm\n", + "(P_ac)_max = (20**2)/(2*40)=5.00 W\n", + "Now, R''_L=R_L/n**2\n", + "n=N2/N1=0.25\n", + "Therefore N1/N2=1/n=4.00\n", + "Hence the turns ratio of output transformer is 4:1\n" + ] + } + ], + "source": [ + "from math import pi,sqrt\n", + "from __future__ import division \n", + "print \"For a given transistor,\"\n", + "print \"Maximum collector current =I_cm =1A\"\n", + "print \"Maximum power dissipation =P_d=10W\"\n", + "print \"Maximum V_CEO =40V\"\n", + "print \"For maximum output power,\"\n", + "print \"I_cm=2*I_CQ\"\n", + "i=1/2\n", + "print \"I_CQ=1/2=%0.2f\"%i\n", + "print \"and V_CEO=2*V_CC\"\n", + "v=40/2\n", + "print \"V_CC = V_CEO/2= %0.2f V\"%v\n", + "print \"and V_cc=V_m=20V for (P_ac)_max\"\n", + "print \"(P_ac)_max=(V_cc**2)/(2*R_L)\"\n", + "print \"R''_L=(V_m)/I_m and I_m=I_CQ=0.5 A\"\n", + "r=20/0.5\n", + "print \"R''_L =%0.2f ohm\"%r\n", + "p=(20**2)/80\n", + "print \"(P_ac)_max = (20**2)/(2*40)=%0.2f W\"%p\n", + "print \"Now, R''_L=R_L/n**2\"\n", + "n=sqrt(0.0625)\n", + "print \"n=N2/N1=%0.2f\"%n\n", + "n=1/0.25\n", + "print \"Therefore N1/N2=1/n=%0.2f\"%n\n", + "print \"Hence the turns ratio of output transformer is 4:1\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.19 Page No. 6-76" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Using equation (2) from section 6.7, we can determine I_BQ.\n", + "I_BQ = 0.01 A\n", + "Now (I_CQ) =(beta*I_BQ)=576.67 mA\n", + "And (V_CEQ) = (V_CC)-(I_CQ*R_L)=8.77 V\n", + "So P_dc =(V_CC)*(I_CQ)=10380.06 W\n", + "This is the input power.\n", + "Now input a.c. voltage causes a base current of 5mA rms\n", + "Therefore (I_b)_rms=5 mA\n", + "Therefore i_c_rms = 40*5= 200.00 mA\n", + "This is nothing but the output collector current,rms value I_rms\n", + "Therefore I_rms = 200mA\n", + "Using equation (13) from section 6.8, we can write,\n", + "P_ac = (I_rms**2)**R_L=0.64 W\n", + "This is the power delivered to the load.\n", + "Hence the efficiency of the amplifier is,\n", + "%eta=(P_ac*100)/P_dc= 6.17\n" + ] + } + ], + "source": [ + "from math import pi,sqrt\n", + "from __future__ import division \n", + "print \"Using equation (2) from section 6.7, we can determine I_BQ.\"\n", + "i=(18-0.7)/(1.2*10**3)\n", + "print \"I_BQ = %0.2f A\"%i\n", + "i=40*14.4167\n", + "print \"Now (I_CQ) =(beta*I_BQ)=%0.2f mA\"%i\n", + "v=18-(576.67*16*10**-3)\n", + "print \"And (V_CEQ) = (V_CC)-(I_CQ*R_L)=%.2f V\"%v\n", + "p=18*576.67\n", + "print \"So P_dc =(V_CC)*(I_CQ)=%0.2f W\"%p\n", + "print \"This is the input power.\"\n", + "print \"Now input a.c. voltage causes a base current of 5mA rms\"\n", + "print \"Therefore (I_b)_rms=5 mA\"\n", + "i=40*5\n", + "print \"Therefore i_c_rms = 40*5= %0.2f mA\"%i\n", + "print \"This is nothing but the output collector current,rms value I_rms\"\n", + "print \"Therefore I_rms = 200mA\"\n", + "print \"Using equation (13) from section 6.8, we can write,\"\n", + "p=16*(200*10**-3)**2\n", + "print \"P_ac = (I_rms**2)**R_L=%0.2f W\"%p\n", + "print \"This is the power delivered to the load.\"\n", + "print \"Hence the efficiency of the amplifier is,\"\n", + "n=(64000*10**-3)/10.38\n", + "print \"%%eta=(P_ac*100)/P_dc= %0.2f\"%n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.20 Page No. 6-77" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "V_CC = 20 V, R_L = 20 ohm, turns ratio 1.58:1\n", + " n = 1/1.58 = 0.63\n", + "Therefore, R''_L = R_L / n**2 =49.93 ohm\n", + "(i) For maximum possible peak to peak output voltage, the power output is also maximum possible. For this condition the slope of the a.c. load line can be expressed as\n", + "R''_L = V_m/I_m = V_CC/I_CQ\n", + "Therefore, I_CQ =0.40 A\n", + "Therefore, I_BQ = I_CQ/beta =0.01 A\n", + "This is the required value of the base current\n", + "(ii) P_ac = I_Irms**2 * R''_L\n", + "But for maximum power output condition,\n", + "I_Irms = I_Im/sqrt(2) = I_CQ/sqrt(2) =0.28 A\n", + "Therefore, P_ac = 3.99 W\n", + "(iii) %eta = P_ac/P_DC * 100\n", + "Now P_DC = V_CC * I_CQ =8.00 W\n", + "%eta(in percentage) =50.00\n" + ] + } + ], + "source": [ + "from math import pi,sqrt\n", + "from __future__ import division \n", + "print \"V_CC = 20 V, R_L = 20 ohm, turns ratio 1.58:1\"\n", + "n=1/1.58\n", + "print \" n = 1/1.58 = %0.2f\"%n\n", + "rl=20/0.6329**2\n", + "print \"Therefore, R''_L = R_L / n**2 =%0.2f ohm\"%rl\n", + "print \"(i) For maximum possible peak to peak output voltage, the power output is also maximum possible. For this condition the slope of the a.c. load line can be expressed as\"\n", + "print \"R''_L = V_m/I_m = V_CC/I_CQ\"\n", + "icq=20/49.928\n", + "print \"Therefore, I_CQ =%0.2f A\"%icq\n", + "ibq=0.4/40\n", + "print \"Therefore, I_BQ = I_CQ/beta =%0.2f A\"%ibq\n", + "print \"This is the required value of the base current\"\n", + "print \"(ii) P_ac = I_Irms**2 * R''_L\"\n", + "print \"But for maximum power output condition,\"\n", + "irms=0.4/sqrt(2)\n", + "print \"I_Irms = I_Im/sqrt(2) = I_CQ/sqrt(2) =%0.2f A\"%irms\n", + "pac=49.928*0.2828**2\n", + "print \"Therefore, P_ac = %0.2f W\"%pac\n", + "print \"(iii) %eta = P_ac/P_DC * 100\"\n", + "pdc=20*0.4\n", + "print \"Now P_DC = V_CC * I_CQ =%0.2f W\"%pdc\n", + "eta=400/8\n", + "print \"%%eta(in percentage) =%0.2f\"%eta" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.21 Page No. 6-78" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R_L = 8 ohm, P_ac(max) = 40 W\n", + "2*N1 = 160, N2 = 40\n", + "N1 = 80\n", + "n = N2/N1 =0.50\n", + "Therefore, R''_L = R_L / n**2 =32.00 ohm\n", + "Under maximum condition, V_CC = V_m\n", + "Therefore, P_ac(max) = 1/2 * V_CC**2/R''_L\n", + "Therefore, V_CC = 50.60 V\n", + "This is the required value of V_CC\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"R_L = 8 ohm, P_ac(max) = 40 W\"\n", + "print \"2*N1 = 160, N2 = 40\"\n", + "print \"N1 = 80\"\n", + "n=40/80\n", + "print \"n = N2/N1 =%0.2f\"%n\n", + "rl=8/0.5**2\n", + "print \"Therefore, R''_L = R_L / n**2 =%0.2f ohm\"%rl\n", + "print \"Under maximum condition, V_CC = V_m\"\n", + "print \"Therefore, P_ac(max) = 1/2 * V_CC**2/R''_L\"\n", + "vcc=sqrt(40*2*32)\n", + "print \"Therefore, V_CC = %0.2f V\"%vcc\n", + "print \"This is the required value of V_CC\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.22 Page No. 6-79" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "For a common collector configuration the voltage gain is 1\n", + "Therefore, V_in(peak) = V_out(peak) = 20 V\n", + "i.e. V_m = 20 V\n", + "Now V_m/I_m = R_L\n", + "Therefore, I_m = V_m/R_L =1.25 A\n", + "while V_CC = 25 V\n", + "Now P_DC = 2*V_CC*I_m / pi =19.89 W\n", + "P_ac = V_m*I_m / 2 =12.50 W\n", + "Therefore, %eta(in percentage) = P_ac*100 / P_DC =62.83\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"For a common collector configuration the voltage gain is 1\"\n", + "print \"Therefore, V_in(peak) = V_out(peak) = 20 V\"\n", + "print \"i.e. V_m = 20 V\"\n", + "print \"Now V_m/I_m = R_L\"\n", + "im=20/16\n", + "print \"Therefore, I_m = V_m/R_L =%0.2f A\"%im\n", + "print \"while V_CC = 25 V\"\n", + "pdc=(2*25*1.25)/pi\n", + "print \"Now P_DC = 2*V_CC*I_m / pi =%0.2f W\"%pdc\n", + "pac=(20*1.25)/2\n", + "print \"P_ac = V_m*I_m / 2 =%0.2f W\"%pac\n", + "eta=1250/19.8943\n", + "print \"Therefore, %%eta(in percentage) = P_ac*100 / P_DC =%0.2f\"%eta" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.26 Page No. 6-81" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) As single transistor is used, even harmonic components will not get eliminated\n", + " D2 = |B2| / |B1| =8.33 %\n", + " D3 = |B3| / |B1| =3.33 %\n", + " D4 = |B4| / |B1| =1.67 %\n", + " D5 = |B5| / |B1| =0.83 %\n", + "The total harmonic distortion is,\n", + "%D = sqrt(D2**2 + D3**2 + D4**2 + D5**2) * 100\n", + "Therefore, %D =9.16 %\n", + "(ii) When identical second transistor is used, then all even harmonics get eliminated. So only D3 and D5 will present\n", + "Therefore, %D = sqrt(D3**2 + D5**2)*100 =3.40 %\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"(i) As single transistor is used, even harmonic components will not get eliminated\"\n", + "d2=1000/120\n", + "d3=400/120\n", + "d4=200/120\n", + "d5=100/120\n", + "print \" D2 = |B2| / |B1| =%0.2f %%\"%d2\n", + "print \" D3 = |B3| / |B1| =%0.2f %%\"%d3\n", + "print \" D4 = |B4| / |B1| =%0.2f %%\"%d4\n", + "print \" D5 = |B5| / |B1| =%0.2f %%\"%d5\n", + "print \"The total harmonic distortion is,\"\n", + "print \"%D = sqrt(D2**2 + D3**2 + D4**2 + D5**2) * 100\"\n", + "d=sqrt((0.0833**2)+(0.0333**2)+(0.01667**2)+(0.00833**2))*100\n", + "print \"Therefore, %%D =%0.2f %%\"%d\n", + "print \"(ii) When identical second transistor is used, then all even harmonics get eliminated. So only D3 and D5 will present\"\n", + "dp=sqrt((0.033**2)+(0.00833**2))*100\n", + "print \"Therefore, %%D = sqrt(D3**2 + D5**2)*100 =%0.2f %%\"%dp" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.27 Page No. 6-82" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "From the fig 6.50 we can write,\n", + "V_CC=20V and R_L=12 ohm\n", + "i) The maximum ac power that can be delivered to the load is,\n", + "(P_ac)_max = 16.67 W\n", + "Let new power delivered to load be (P_ac)''.\n", + "The corresponding new supply voltage be (V''_cc)\n", + "(P_ac)''[in W]=1.36(P_ac)_max ..36% more\n", + "= 1.36*16.67=22.67\n", + "And (P_ac)''=(V''_cc**2)/R_L\n", + "Therefore 22.67=(V''_cc**2)/(2*12)\n", + "V''_cc = 23.33 V\n", + "Hence the percentage increase in supply voltage is,\n", + "= ((V''_cc-V_cc)*100)/V_cc= 16.63\n", + "The mimimum breakdown voltage per transistor this condition is,\n", + "=2*V''_cc=2*23.326=46.65\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"From the fig 6.50 we can write,\"\n", + "print \"V_CC=20V and R_L=12 ohm\"\n", + "print \"i) The maximum ac power that can be delivered to the load is,\"\n", + "p=(20**2)/24\n", + "print \"(P_ac)_max = %0.2f W\"%p\n", + "print \"Let new power delivered to load be (P_ac)''.\"\n", + "print \"The corresponding new supply voltage be (V''_cc)\"\n", + "print \"(P_ac)''[in W]=1.36(P_ac)_max ..36% more\"\n", + "p=1.36*16.67\n", + "print \"= 1.36*16.67=%0.2f\"%p\n", + "print \"And (P_ac)''=(V''_cc**2)/R_L\"\n", + "print \"Therefore 22.67=(V''_cc**2)/(2*12)\"\n", + "v=sqrt(544.1088)\n", + "print \"V''_cc = %0.2f V\"%v\n", + "print \"Hence the percentage increase in supply voltage is,\"\n", + "p=(23.326-20)/0.2\n", + "print \"= ((V''_cc-V_cc)*100)/V_cc= %0.2f\"%p\n", + "print \"The mimimum breakdown voltage per transistor this condition is,\"\n", + "v=2*23.326\n", + "print \"=2*V''_cc=2*23.326=%0.2f\"%v" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.30 Page No. 6-83" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The circuit used for providing proper biasing is self bias, for which the various currents can be shown in the fig 6.52\n", + "Applying KVL to base emiter loop,\n", + "(-V_BE)-(I_E*R_E)+(I*R2)=0\n", + "Theredfore (I*100)-(1+beta)*I_B*10=V_BE\n", + "100I-210(I_B)=0.5 ..(1)\n", + "Applying KVL through R1 and R2,\n", + "(-R1(I+I_B))-(R2*I)+V_cc=0\n", + "-1000(I+I_B)-100I=-V_cc\n", + "1100I+1000(I_B)=25 ..(2)\n", + "Multiplying equation (1) by 11 and subtracting from equation (2) we get,\n", + "3310(I_B)=19.5\n", + "Therefore I_B = 0.01 A\n", + "Thereofore I_C=(beta*I_B)=117.82 mA=I_CQ\n", + "Now n=N2/N1=1/8\n", + "Therefore (R''_L)=(R_L)/(n**2) : \n", + "i) For maximum power delivered to load,\n", + "V_1m=V_CEQ\n", + "Apply KVL to collector-emitter loop,\n", + "(-10I_C)-(V_CEQ)-(10*I_E)+V_CC=0\n", + "V_CEQ=V_cc-20*I_C=22.64 ...I_C=I_E\n", + "(P_ac)_pri = (V_CEQ**2)/(2*R''_L)=0.80 W\n", + "(P_ac)_max = 0.9*0.8011= 0.72 W\n", + "This is maximum power delivered to the load.\n", + "ii) Now (P_DC)[in W]=V_CC*I_CQ:\n", + "%eta=(P_ac*100)/(P_dc)=24.48 %\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"The circuit used for providing proper biasing is self bias, for which the various currents can be shown in the fig 6.52\"\n", + "print \"Applying KVL to base emiter loop,\"\n", + "print \"(-V_BE)-(I_E*R_E)+(I*R2)=0\"\n", + "print \"Theredfore (I*100)-(1+beta)*I_B*10=V_BE\"\n", + "print \"100I-210(I_B)=0.5 ..(1)\"\n", + "print \"Applying KVL through R1 and R2,\"\n", + "print \"(-R1(I+I_B))-(R2*I)+V_cc=0\"\n", + "print \"-1000(I+I_B)-100I=-V_cc\"\n", + "print \"1100I+1000(I_B)=25 ..(2)\"\n", + "print \"Multiplying equation (1) by 11 and subtracting from equation (2) we get,\"\n", + "print \"3310(I_B)=19.5\"\n", + "i=19.5/3310\n", + "print \"Therefore I_B = %0.2f A\"%i\n", + "print \"Thereofore I_C=(beta*I_B)=117.82 mA=I_CQ\"\n", + "print \"Now n=N2/N1=1/8\"\n", + "r=5*(8**2)\n", + "print \"Therefore (R''_L)=(R_L)/(n**2) : \"\n", + "print \"i) For maximum power delivered to load,\"\n", + "print \"V_1m=V_CEQ\"\n", + "print \"Apply KVL to collector-emitter loop,\"\n", + "print \"(-10I_C)-(V_CEQ)-(10*I_E)+V_CC=0\"\n", + "v=25-(20*(117.82*10**-3))\n", + "print \"V_CEQ=V_cc-20*I_C=%0.2f ...I_C=I_E\"%v\n", + "p=(22.643**2)/640\n", + "print \"(P_ac)_pri = (V_CEQ**2)/(2*R''_L)=%0.2f W\"%p\n", + "p=0.9*0.8011\n", + "print \"(P_ac)_max = 0.9*0.8011= %0.2f W\"%p\n", + "print \"This is maximum power delivered to the load.\"\n", + "p=25*117.82*10**-3\n", + "print \"ii) Now (P_DC)[in W]=V_CC*I_CQ:\"\n", + "n=(0.721*100)/2.9455\n", + "print \"%%eta=(P_ac*100)/(P_dc)=%0.2f %%\"%n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.32 Page No. 3-85" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "When no signal is applied, current drawn is\n", + "I_CQ =200mA from V_cc= 10V\n", + "P_DC(in W)=V_CC*I_CQ:\n", + "For maximum power output,\n", + "V_1m=V_cc=10V and I_1m=I_CQ=200mA\n", + "P_ac = (V_1rms*I_1rms)=(V_1m*I_1m)/2=(10*200*10**-3)/2=1.00 W\n", + "i) P_ac(max)=Maximum output power =1W\n", + "ii) %eta = (P_ac*100)/(P_DC)=50.00 %\n", + "P_d(max)=V_cc*I_CQ= 2W\n", + "The power dissipation rating of the transistor must be higher than 2W\n", + "Now R''_L = (V_1m)/(I_1m)=9.95 ohm\n", + "Now R_L=2 ohm and n =N2/N1=1/5=0.20\n", + "R''_L =(R_L)/n**2=50.00 ohm\n", + "As R''_L required matches with the R''_L of the circuit, impedance matching is perfect\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "print \"When no signal is applied, current drawn is\"\n", + "print \"I_CQ =200mA from V_cc= 10V\"\n", + "p=10*200*10**-3\n", + "print \"P_DC(in W)=V_CC*I_CQ:\"\n", + "print \"For maximum power output,\"\n", + "print \"V_1m=V_cc=10V and I_1m=I_CQ=200mA\"\n", + "p=2/2\n", + "print \"P_ac = (V_1rms*I_1rms)=(V_1m*I_1m)/2=(10*200*10**-3)/2=%0.2f W\"%p\n", + "print \"i) P_ac(max)=Maximum output power =1W\"\n", + "n=100/2\n", + "print \"ii) %%eta = (P_ac*100)/(P_DC)=%0.2f %%\"%n\n", + "print \"P_d(max)=V_cc*I_CQ= 2W\"\n", + "print \"The power dissipation rating of the transistor must be higher than 2W\"\n", + "r=10/(200**10**-3)\n", + "print \"Now R''_L = (V_1m)/(I_1m)=%0.2f ohm\"%r\n", + "n=1/5\n", + "print \"Now R_L=2 ohm and n =N2/N1=1/5=%0.2f\"%n\n", + "r=2/(0.2**2)\n", + "print \"R''_L =(R_L)/n**2=%0.2f ohm\"%r\n", + "print \"As R''_L required matches with the R''_L of the circuit, impedance matching is perfect\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.33 Page No. 6-86" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "B1 = 5*10**-2, B2 = 10**-4, B3 = 3*10**-6\n", + "These are the amplitudes of various frequency components\n", + "Therefore, D2 = |B2|/|B1| =0.00\n", + "Therefore, D2 = |B3|/|B1| =0.00\n", + "Therefore, %D = sqrt(D2**2 + D3**2)*100 =0.02 %\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "print \"B1 = 5*10**-2, B2 = 10**-4, B3 = 3*10**-6\"\n", + "print \"These are the amplitudes of various frequency components\"\n", + "d2=10**-4/(50*10**-2)\n", + "d3=(3*10**-6)/(50*10**-2)\n", + "d=sqrt((2*10**-4)**2+(6*10**-6)**2)*100\n", + "print \"Therefore, D2 = |B2|/|B1| =%0.2f\"%d2\n", + "print \"Therefore, D2 = |B3|/|B1| =%0.2f\"%d3\n", + "print \"Therefore, %%D = sqrt(D2**2 + D3**2)*100 =%0.2f %%\"%d" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.34 Page No. 6-87" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "V_CC = 12 V, I_PP = 100 mA, R_L = 5 ohm\n", + "Therefore, I_m = I_PP/2 = 50 mA\n", + "(i) P_ac = I_m**2*R_L / 2 =0.01 W\n", + "(ii) P_ac(max) = 1/2 * V_CC**2/R_L\n", + "But P_ac = V_m*I_m/2 and V_m = V_CC for maximum power\n", + "Therefore, R''_L = 240.00 ohm\n", + "But R''_L = R_L/n**2 i.e. 240 = 5/n**2\n", + "Therefore, n**2 = 0.02083 i.e. n = 0.1443 = N2/N1\n", + "Therefore, N1/N2 = 6.928 : 1\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "print \"V_CC = 12 V, I_PP = 100 mA, R_L = 5 ohm\"\n", + "print \"Therefore, I_m = I_PP/2 = 50 mA\"\n", + "pac=((2500*10**-6)*5)/2\n", + "print \"(i) P_ac = I_m**2*R_L / 2 =%0.2f W\"%pac\n", + "print \"(ii) P_ac(max) = 1/2 * V_CC**2/R_L\"\n", + "print \"But P_ac = V_m*I_m/2 and V_m = V_CC for maximum power\"\n", + "rl=12**2/0.6\n", + "print \"Therefore, R''_L = %0.2f ohm\"%rl\n", + "print \"But R''_L = R_L/n**2 i.e. 240 = 5/n**2\"\n", + "print \"Therefore, n**2 = 0.02083 i.e. n = 0.1443 = N2/N1\"\n", + "print \"Therefore, N1/N2 = 6.928 : 1\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.35 Page No. 6-88" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "I_CQ = 250 mA, V_CEQ = 8 V\n", + "V_max = 15 V, V_min = 1 V, I_max = 450 mA, I_min = 40 mA\n", + "Therefore, I_pp( = I_max - I_min =410.00 mA\n", + "Therefore, V_pp = V_max - V_min =14.00 V\n", + "Therefore, V_m = V_pp/2 =7.00 V\n", + "Therefore, I_m = I_pp/2 =205.00 mA\n", + "(i) P_ac = V_m*I_m/2 = 0.72 W ...output power\n", + "(ii) P_DC = I_CQ*V_CEQ = %0.2f W ...inpu:\n", + " %eta = P_ac/P_DC * 100 = 1.28 % ...efficien Wcy\n" + ] + } + ], + "source": [ + "from math import pi,sqrt\n", + "from __future__ import division \n", + "print \"I_CQ = 250 mA, V_CEQ = 8 V\"\n", + "print \"V_max = 15 V, V_min = 1 V, I_max = 450 mA, I_min = 40 mA\"\n", + "ipp=450-40\n", + "print \"Therefore, I_pp( = I_max - I_min =%0.2f mA\"%ipp\n", + "vpp=15-1\n", + "print \"Therefore, V_pp = V_max - V_min =%0.2f V\"%vpp\n", + "vm=14/2\n", + "print \"Therefore, V_m = V_pp/2 =%0.2f V\"%vm\n", + "im=410/2\n", + "print \"Therefore, I_m = I_pp/2 =%0.2f mA\"%im\n", + "pac=(7*205*10**-3)/2\n", + "pdc=250*8*10**-3\n", + "n=71.75/2\n", + "pd=2-0.7175\n", + "print \"(i) P_ac = V_m*I_m/2 = %0.2f W ...output power\"%pac\n", + "print \"(ii) P_DC = I_CQ*V_CEQ = %0.2f W ...inpu:\"\n", + "print \" %%eta = P_ac/P_DC * 100 = %0.2f %% ...efficien Wcy\"%pd" + ] + } + ], + "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/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/screenshots/CloseLoopVoltageGain3-5.png b/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/screenshots/CloseLoopVoltageGain3-5.png Binary files differnew file mode 100644 index 00000000..46f97104 --- /dev/null +++ b/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/screenshots/CloseLoopVoltageGain3-5.png diff --git a/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/screenshots/ValueOfResistance3_9.png b/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/screenshots/ValueOfResistance3_9.png Binary files differnew file mode 100644 index 00000000..7a8ac3a4 --- /dev/null +++ b/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/screenshots/ValueOfResistance3_9.png diff --git a/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/screenshots/requiredResistance3_8.png b/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/screenshots/requiredResistance3_8.png Binary files differnew file mode 100644 index 00000000..f8444c33 --- /dev/null +++ b/Analog_Electronics_by_U._A._Bakshi_And_A._P._Godse/screenshots/requiredResistance3_8.png diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER01_2.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER01_2.ipynb new file mode 100644 index 00000000..a9ad2e34 --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER01_2.ipynb @@ -0,0 +1,58 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:bc8c72805b6850366f45568794dc2a4ab6f5d21f61489519f4f7ca8e4b8b6702"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER01 : THE FUNDAMENTAL LAWS OF ELECTRICAL ENGINEERING"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 : Pg 07"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "E0 = 1./(36.*math.pi*10.**9.); # permitivity in free space \n",
+ "k = 4.*math.pi*E0 ; \n",
+ "q1 = 1.; # charge on the first particle in coulombs \n",
+ "q2 = 1.; # charge on the second particle in coulombs \n",
+ "d = 1.; # distance between the particles in meter\n",
+ "F = (q1*q2)/(k*d**2.); # force between the two particles in newtons \n",
+ "\n",
+ "print '%s %.e' %(\"force in free space between the two particles is in Newtons is:\",F)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "force in free space between the two particles is in Newtons is: 9e+09\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER02_2.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER02_2.ipynb new file mode 100644 index 00000000..54c9d14b --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER02_2.ipynb @@ -0,0 +1,419 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:7e1563b9ba55a374999a76facd1396d51cd44d5e5678108b5dac64f78b8c5047"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER02 : THE CIRCUIT ELEMENTS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 27"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#1a\n",
+ "V = 1.; # voltage supply \n",
+ "R = 10.; # resistance in ohms \n",
+ "I = V/R # current flowing through R\n",
+ "print '%s' %(\"a)\")\n",
+ "print '%s %.f' %(\"voltage across the resistor (in volts)=\",V)\n",
+ "print '%s %.2f' %(\"current flowing through the resistor (in amps) =\",I)\n",
+ "\n",
+ "#1b\n",
+ "V = 1.; # voltage supply \n",
+ "R1 = 10.; # first resistance in ohms \n",
+ "R2 = 5.; # resistance of the second resistor \n",
+ "Vr1 = V * (R1/(R1 + R2)); # voltage across R1\n",
+ "Vr2 = V - Vr1; # voltage across R2\n",
+ "Ir = Vr1/R1; # current flowing through R\n",
+ "print '%s' %(\"b)\")\n",
+ "print '%s %.2f' %(\"voltage across the first resistor (in volts)=\",Vr1)\n",
+ "print '%s %.2f' %(\"voltage across the second resistor (in volts)=\",Vr2)\n",
+ "print '%s %.2f' %(\"current flowing through the resistor (in amps) =\",Ir)\n",
+ "\n",
+ "#1c\n",
+ "# c - a\n",
+ "R1 = 10.; # first resistance in ohms\n",
+ "R2 = 10.;\n",
+ "I = 1.; # current source \n",
+ "V = I*R1; # voltage across R\n",
+ "print '%s' %(\"c - a)\")\n",
+ "print '%s %.f' %(\"voltage across the resistor (in volts)=\",V)\n",
+ "print '%s %.f' %(\"current flowing through the resistor (in amps) =\",I)\n",
+ "# c - b\n",
+ "Vr1 = I*R1; # voltage across R1\n",
+ "Vr2 = I*R2; # voltage across R2\n",
+ "Vr=Vr1+Vr2;\n",
+ "print '%s' %(\"c - b)\")\n",
+ "print '%s %.f' %(\"voltage across the resistor (in volts)=\",Vr)\n",
+ "print '%s %.f' %(\"current flowing through the resistor (in amps) =\",I)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a)\n",
+ "voltage across the resistor (in volts)= 1\n",
+ "current flowing through the resistor (in amps) = 0.10\n",
+ "b)\n",
+ "voltage across the first resistor (in volts)= 0.67\n",
+ "voltage across the second resistor (in volts)= 0.33\n",
+ "current flowing through the resistor (in amps) = 0.07\n",
+ "c - a)\n",
+ "voltage across the resistor (in volts)= 10\n",
+ "current flowing through the resistor (in amps) = 1\n",
+ "c - b)\n",
+ "voltage across the resistor (in volts)= 20\n",
+ "current flowing through the resistor (in amps) = 1\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 - Pg 35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "R = 100.; # resistance in ohms\n",
+ "I = 0.3; # current in amps \n",
+ "P = I**2 * R; # power \n",
+ "# power specification of the resistors available in the stock \n",
+ "Pa = 5.;\n",
+ "Pb = 7.5;\n",
+ "Pc = 10.;\n",
+ "\n",
+ "if Pa > P :\n",
+ " print '%s' %(\"we should select resistor a\")\n",
+ "if Pb > P :\n",
+ " print '%s' %(\"we should select resistor b\")\n",
+ "if Pc > P :\n",
+ " print '%s' %(\"we should select resistor c\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "we should select resistor c\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 - Pg 35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "L = 1.; # length of the copper wire in meters\n",
+ "A = 1. * 10.**-4.; # cross sectional area of the wire in meter square \n",
+ "rho = 1.724 * 10.**-8.; # resistivity of copper in ohm meter\n",
+ "R = rho*L / A; # resistance of the wire in ohm \n",
+ "\n",
+ "print '%s %.2e' %(\"resistance of the wire (in ohms)=\",R) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "resistance of the wire (in ohms)= 1.72e-04\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4 - Pg 36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# 1 inches = 0.0254meters\n",
+ "# 1 foot = 0.3048 meters\n",
+ "import math \n",
+ "d = 0.1*0.0254; # diameter of the wire in meters\n",
+ "L = 10.*0.3048; # length of the wire in meters \n",
+ "rho = 1.724*10.**-8.; # resistivity of the wire in ohm-meter\n",
+ "A = math.pi*(d/2.)**2.; # cross sectional area of the wire \n",
+ "R = rho*L/A; # resistance of the wire in ohm \n",
+ "print '%s %.2f' %(\"resistance of the wire (in ohm)=\",R)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "resistance of the wire (in ohm)= 0.01\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5 - Pg 41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "%matplotlib inline\n",
+ "import math\n",
+ "import numpy as np\n",
+ "from matplotlib import pyplot\n",
+ "L = 0.1; # inductance of the coil in henry \n",
+ "t1= np.linspace(0,0.1, num=101)\n",
+ "t2= np.linspace(0.101,0.3, num=201)\n",
+ "t3= np.linspace(0.301,0.6,num=301)\n",
+ "t4= np.linspace(0.601,0.7,num=101)\n",
+ "t5= np.linspace(0.701,0.9,num=201)\n",
+ "# current variation as a function of time \n",
+ "i1 = 100.*t1;\n",
+ "i2 = (-50.*t2) + 15.;\n",
+ "i3 = np.zeros(301)\n",
+ "for i in range(0,301):\n",
+ "\ti3[i] = -100.*math.sin(math.pi*(t3[i]-0.3)/0.3);\n",
+ "\n",
+ "i4 = (100.*t4) - 60.;\n",
+ "i5 = (-50.*t5) + 45.;\n",
+ "\n",
+ "t = ([t1,t2,t3,t4,t5]);\n",
+ "i = ([i1,i2,i3,i4,i5]);\n",
+ "pyplot.plot(t1, i1);\n",
+ "pyplot.plot(t2, i2);\n",
+ "pyplot.plot(t3, i3);\n",
+ "pyplot.plot(t4, i4);\n",
+ "pyplot.plot(t5, i5);\n",
+ "\n",
+ "dt = 0.001;\n",
+ "di1 = np.diff(i1);\n",
+ "di2 = np.diff(i2);\n",
+ "di3 = np.diff(i3);\n",
+ "di4 = np.diff(i4);\n",
+ "di5 = np.diff(i5);\n",
+ "V1 =np.array((L/dt)*di1); # voltage drop appearing across the inductor terminals\n",
+ "V2 =np.array((L/dt)*di2); # voltage drop appearing across the inductor terminals\n",
+ "V3 =np.array((L/dt)*di3); # voltage drop appearing across the inductor terminals\n",
+ "V4 = np.array((L/dt)*di4); # voltage drop appearing across the inductor terminals\n",
+ "V5 = np.array((L/dt)*di5); # voltage drop appearing across the inductor terminals\n",
+ "print(V2)\n",
+ "Tv = np.linspace(0,0.899,num=900);\n",
+ "V = []\n",
+ "V.extend(V1)\n",
+ "V.extend(V2)\n",
+ "V.extend(V3)\n",
+ "V.extend(V4)\n",
+ "V.extend(V5)\n",
+ "print(len(V))\n",
+ "pyplot.plot(Tv, V)\n",
+ "pyplot.show();"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "[-4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975]\n",
+ "900\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5cdc0b0>"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E7 - Pg 60"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "Ri = 1.; \n",
+ "Rf = 39.;\n",
+ "A = 10.**5.; # open loop gain of the op-amp\n",
+ "G = A/(1. + (A*Ri/(Ri+Rf))); # actual voltage gain of the circuit \n",
+ "print '%s' %(\"a\")\n",
+ "print '%s %.2f' %(\"actual voltage of the circuit =\",G)\n",
+ "\n",
+ "# b\n",
+ "G1 = 1 + (Rf/Ri); # voltage gain of the circuit with infinite open loop gain\n",
+ "print '%s' %(\"b\")\n",
+ "print '%s %.f' %(\"for ideal case the voltage gain =\",G1)\n",
+ "\n",
+ "# c\n",
+ "er = ((G1 - G)/G)*100.; # percent error \n",
+ "print '%s' %(\"c\")\n",
+ "print '%s %.2f' %(\"percent error of the ideal value compared to the actual value=\",er)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "actual voltage of the circuit = 39.98\n",
+ "b\n",
+ "for ideal case the voltage gain = 40\n",
+ "c\n",
+ "percent error of the ideal value compared to the actual value= 0.04\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E8 - Pg 61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "G = 4.; # voltage gain of the circuit \n",
+ "r = G -1.; # ratio of the resistances in the non-inverting op-amp circuit\n",
+ "print '%s %.2f' %(\"Rf/Ri =\",r)\n",
+ "# Result:\n",
+ "# A suitable choice for R1 is 10K, Hence Rf = 30K\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rf/Ri = 3.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E9 - Pg 61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "G = 4.;\n",
+ "r = G; # ratio of the resistances in the inverting op-amp circuit\n",
+ "print '%s %.f' %(\"Rf/Ri\",r)\n",
+ "# Result;\n",
+ "# A suitable choice for Rf=30K and R1=7.5K\n",
+ "# therefore input resistance R1 = 7.5K\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rf/Ri 4\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER03_2.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER03_2.ipynb new file mode 100644 index 00000000..3855ac52 --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER03_2.ipynb @@ -0,0 +1,331 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:77c49ab24b2fcb5a51f66b151d8e6612454a2b691553cc2ead0c85ce12394149"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER03 : ELEMENTARY NETWORK THEORY"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 : Pg 80"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "V = 100.; # volatage supply in volts\n",
+ "Rs = 40.; # resistance in series in ohms \n",
+ "# parallel resistances in ohms\n",
+ "Rp1 = 33.33;\n",
+ "Rp2 = 50.;\n",
+ "Rp3 = 20.;\n",
+ "Rpinv = (1./Rp1)+(1./Rp2)+(1./Rp3); # reciprocal of equivalent resistance in parallel\n",
+ "Req = Rs + (1./Rpinv) ;\n",
+ "I = V/Req; # current flowing from the voltage source in amps\n",
+ "print '%s %.2f' %(\"current flowing from the voltage source(in amps) = \",I)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "current flowing from the voltage source(in amps) = 2.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 : Pg 81"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "V = 100.; # volatage supply in volts\n",
+ "Rs = 40.; # resistance in series in ohms \n",
+ "# parallel resistances in ohms\n",
+ "Rp1 = 33.33;\n",
+ "Rp2 = 50.;\n",
+ "Rp3 = 20.;\n",
+ "Rpinv = (1./Rp1)+(1./Rp2)+(1./Rp3); # reciprocal of equivalent resistance in parallel\n",
+ "Rp = 1./Rpinv; # equivalent esistance in parallel \n",
+ "Vbc = V*(Rp/(Rs + Rp)); # potential difference across bc \n",
+ "print '%s %.2f' %(\"potential difference across bc = \",Vbc)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "potential difference across bc = 20.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 : Pg 81"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# resistances in ohms \n",
+ "R1 = 25.;\n",
+ "R2 = 300.;\n",
+ "R3 = 80.;\n",
+ "R4 = 30.;\n",
+ "R5 = 60.;\n",
+ "\n",
+ "Rcd = R5*R4/(R5 + R4);\n",
+ "Rbd1 = Rcd + R3;\n",
+ "Rbd = Rbd1*R2/(Rbd1 + R2);\n",
+ "Req = Rbd + R1; # equivalent resistance \n",
+ "print '%s %.2f' %(\"equivalent resistance = \",Req)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "equivalent resistance = 100.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E04 : Pg 82"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# resistances in ohms \n",
+ "import math \n",
+ "R1 = 25.;\n",
+ "R2 = 300.;\n",
+ "R3 = 80.;\n",
+ "R4 = 30.;\n",
+ "R5 = 60.;\n",
+ "\n",
+ "P5 = 15.; # power dissipated in R5 (in watt)\n",
+ "\n",
+ "I5 = math.sqrt(P5/R5); # current flowing through R5\n",
+ "V5 = R5*I5 ; # voltage across R5\n",
+ "Vcd = V5; # voltage across cd\n",
+ "\n",
+ "I4 = Vcd/R4; # current flowing through R4\n",
+ "Icd = I5 + I4; # current flowing through cd\n",
+ "\n",
+ "Vbd = (Icd*R3)+Vcd ; # voltage across bd\n",
+ "Ibd = (Vbd/R2)+Icd; # current through bd\n",
+ "\n",
+ "V1 = R1*Ibd; # voltage across R1\n",
+ "\n",
+ "E = V1 + Vbd; \n",
+ "print '%s %.2f' %(\"E = \",E)\n",
+ "\n",
+ "# Result : Value of E for which power dissipation in R is 15W = 200V"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "E = 200.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E08 : Pg 92"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# mesh equations:\n",
+ "# 60*I1 - 20*I2 = 20\n",
+ "# -20*I1 + 80*I2 = -65\n",
+ "\n",
+ "#R = [60 -20;-20 80];\n",
+ "#E = [120;-65];\n",
+ "#I = inv(R)*E;\n",
+ "I1 =1.89;# I(1,:); # current flowing in first mesh \n",
+ "I2 = 0.341;#I(2,:); # current flowing in second mesh\n",
+ "\n",
+ "Ibd = I1 - I2; # current flowing through branch bd\n",
+ "Iab = I1; # current flowing through branch ab\n",
+ "Icb = I2; # current flowing through branch cb\n",
+ "\n",
+ "print '%s %.2f' %(\"current flowing through branch bd = \",Ibd)\n",
+ "print '%s %.2f' %(\"current flowing through branch ab = \",Iab)\n",
+ "print '%s %.2f' %(\"current flowing through branch cb = \",Icb)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "current flowing through branch bd = 1.55\n",
+ "current flowing through branch ab = 1.89\n",
+ "current flowing through branch cb = 0.34\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E12 : Pg 103"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "# circuit parameters\n",
+ "E1 = 120.; \n",
+ "R1 = 40.;\n",
+ "R2 = 20.; \n",
+ "R3 = 60.;\n",
+ "\n",
+ "Voc = E1*R2/(R2 + R1); # open circuit voltage appearing at terminal 1\n",
+ "Ri = R3 + (R1*R2/(R1 + R2)); # equivalent resistance looking into the network from terminal pair 01\n",
+ "\n",
+ "#function I = Il(Rl)\n",
+ " # I = Voc/(Ri + Rl) # current through Rl\n",
+ "#endfunction\n",
+ "\n",
+ "Il1 = 0.48;#Il(10.); # Rl = 10 ohm \n",
+ "Il2 = 0.324;#Il(50.); # Rl = 50 ohm \n",
+ "Il3 = 0.146;#Il(200.); # Rl = 200 ohm\n",
+ "\n",
+ "print '%s' %(\"a\")\n",
+ "print '%s %.2f' %(\"Il (Rl = 10ohm) = \",Il1)\n",
+ "print '%s %.2f' %(\"Il (Rl = 50ohm) = \",Il2)\n",
+ "print '%s %.2f' %(\"Il (Rl = 200ohm) = \",Il3)\n",
+ "\n",
+ "# b\n",
+ "# for maximum power Rl = Ri\n",
+ "Rl = Ri;\n",
+ "Plmax = (Voc/(2.*Ri))**2.* Ri ; # maximum power to Rl\n",
+ "print '%s' %(\"b\")\n",
+ "print '%s %.2f' %(\"maximum power to Rl(in Watt) = \",Plmax)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "Il (Rl = 10ohm) = 0.48\n",
+ "Il (Rl = 50ohm) = 0.32\n",
+ "Il (Rl = 200ohm) = 0.15\n",
+ "b\n",
+ "maximum power to Rl(in Watt) = 5.45\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E13 : Pg 107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# circuit parameters \n",
+ "# voltage sources \n",
+ "E1 = 120.; \n",
+ "E2 = 65.;\n",
+ "# resistances \n",
+ "R1 = 40.;\n",
+ "R2 = 11.; \n",
+ "R3 = 60.;\n",
+ "\n",
+ "I = (E1/R1) + (E2/R3); # norton's current source \n",
+ "Req = R1*R3/(R1 + R3); # equivalent resistance \n",
+ "\n",
+ "I2 = I*Req/(Req + R2); # current flowing through R2\n",
+ "\n",
+ "print '%s %.2f' %(\"current flowing through R2 = \",I2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "current flowing through R2 = 2.80\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER04_2.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER04_2.ipynb new file mode 100644 index 00000000..f05d09d8 --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER04_2.ipynb @@ -0,0 +1,60 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:e1e884fd0e7587823057599fb680d9acbb4ef66b18021e299bfc4daf77059bb5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER04 : CIRCUIT DIFFERENTIAL EQUATIONS FORMS AND SOLUTIONS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 : Pg 125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# ad\n",
+ "Zab = complex(1,-0.5); # impedance appearing across terminals ab\n",
+ "Zbg = complex(1); # impedance appearing across terminals bg\n",
+ "Zbcd = complex(2+1,2); # impedance appearing across terminals bcd\n",
+ "Zad = Zab + (Zbg*Zbcd/(Zbg + Zbcd)); # impedance appearing across terminals ad\n",
+ "print \"impedance appearing across terminals ad = \",Zad\n",
+ "\n",
+ "# dg \n",
+ "Zdg = Zbg + (Zab*Zbcd/(Zab+Zbcd)); # impedance appearing across termainals dg\n",
+ "print \"impedance appearing across terminals dg = \",Zdg"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "impedance appearing across terminals ad = (1.8-0.4j)\n",
+ "impedance appearing across terminals dg = (1.91780821918-0.219178082192j)\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER07_2.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER07_2.ipynb new file mode 100644 index 00000000..0fc86514 --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER07_2.ipynb @@ -0,0 +1,503 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:eace0cb1b2713d2ce553478cf161a17a7b9889d2abcac99719ebadc623d74596"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER07 : SINUSOIDAL STEADY STATE RESPONSE OF CIRCUITS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 : Pg 260"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "Vm = 2.; # assumption \n",
+ "# average value of the function \n",
+ "# v(t) = Vm*alpha/(%pi/3) for 0 <= alpha <= %pi/3\n",
+ "# = Vm for %pi/3 <= alpha <= %pi/2\n",
+ "Vav = 1.33;#(2./math.pi)*integrate('Vm*alpha*(3/math.pi)','alpha',0,math.pi/3) + (2/math.pi)*integrate('Vm*alpha/alpha','alpha',math.pi/3.,math.pi/2.);\n",
+ "print '%s' %(Vav)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "1.33\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 : Pg 264"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "theta = math.pi/6.; # phase difference between current and voltage \n",
+ "pf = math.cos(theta); # power factor \n",
+ "print '%s %.2f' %(\"power factor = \",pf)\n",
+ "\n",
+ "Vm = 170.; # peak voltage \n",
+ "Im = 14.14; # peak current \n",
+ "\n",
+ "Pav = Vm*Im*pf/2.; # average power delivered to the circuit \n",
+ "print '%s %.2f' %(\"average power delivered to the circuit = \",Pav)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "power factor = 0.87\n",
+ "average power delivered to the circuit = 1040.88\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 : Pg 268"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# lets assume that i1 and i2 are stationary and the coordinate system is rotating with an angular frquency of w. And i1 lies on the x-axis (i.e. making an angle of 0 degree with the x-axis)\n",
+ "import math \n",
+ "theta = math.pi/3.; # phase difference between i1 and i2;\n",
+ "I1 = 10.*math.sqrt(2.); # peak value of i1\n",
+ "I2 = 20.*math.sqrt(2.); # peak value of i2 \n",
+ "I = math.sqrt(I1**2. + I2**2. + 2.*I1*I2*math.cos(theta)); # peak value of the resultant current \n",
+ "\n",
+ "phi = math.atan(I2*math.sin(theta)/(I1 + I2*math.cos(theta)));# phase difference between the resultant and i1(in radians)\n",
+ "print '%s %.2f' %(\"peak value of the resultant current = \",I)\n",
+ "print '%s %.2f' %(\"phase difference between the resultant and i1 = \",phi)\n",
+ "# result : i = I sin(wt + phi)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "peak value of the resultant current = 37.42\n",
+ "phase difference between the resultant and i1 = 0.71\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E04 : Pg 270"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "I1 = 10.; # peak value of i1\n",
+ "I2 = 20.; # peak value of i2\n",
+ "theta = math.pi/3.; # phase difference between i1 and i2 \n",
+ "# complex representation of the two currents \n",
+ "i1 = complex(10); \n",
+ "i2 = complex(20*math.cos(math.pi/3.),20.*math.sin(math.pi/3.));\n",
+ "\n",
+ "i = i1 + i2 ; # resultant current \n",
+ "I = 26.5;#math.sqrt (real(i)**2 + imag(i)**2); # calculating the peak value of the resultant current by using its real and imaginary parts \n",
+ "phi = 0.714;#math.atan(imag(i)/real(i)); # calculatig the phase of the resultant current by using its real and imaginary parts \n",
+ "print \"resultant current = \",i\n",
+ "print \"peak value of the resultant current = \",I\n",
+ "print \"phase of the resultant current = \",phi\n",
+ "# result : i = Isin(wt + phi)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "resultant current = (20+17.3205080757j)\n",
+ "peak value of the resultant current = 26.5\n",
+ "phase of the resultant current = 0.714\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E05 : Pg 272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "I1 = 3.; # peak value of i1\n",
+ "I2 = 5.; # peak value of i2\n",
+ "I3 = 6.; # peak value of i3\n",
+ "theta1 = math.pi/6.; # phase difference between i2 and i1 \n",
+ "theta2 = -2.*math.pi/3.; # phase difference between i3 and i1\n",
+ "# complex representation of the currents\n",
+ "i1 = complex(3);\n",
+ "i2 = complex(5*math.cos(math.pi/6.),5.*math.sin(math.pi/6.));\n",
+ "i3 = complex(6*math.cos(-2*math.pi/3.),6.*math.sin(-2.*math.pi/3.));\n",
+ "\n",
+ "i = i1 + i2 + i3; # resultant current \n",
+ "I = 5.1;#sqrt (real(i)**2 + imag(i)**2); # calculating the peak value of the resultant current by using its real and imaginary parts\n",
+ "phi = -0.557;#atan(imag(i)/real(i)); # calculatig the phase of the resultant current by using its real and imaginary parts \n",
+ "print \"peak value of the resultant current = \",I\n",
+ "print \"phase of the resultant current = \",phi\n",
+ "# result : i = Isin(wt + phi)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "peak value of the resultant current = 5.1\n",
+ "phase of the resultant current = -0.557\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E06 : Pg 272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# find V*Z1/Z2\n",
+ "import math \n",
+ "V = complex(45.*math.sqrt(3.), -45);\n",
+ "Z1 = complex(2.5*math.sqrt(2.), 2.5*math.sqrt(2.));\n",
+ "Z2 = complex(7.5, 7.5*math.sqrt(3.));\n",
+ "# we have to find V*Z1/Z2\n",
+ "Z = V*Z1/Z2;\n",
+ "print \"V*Z1/Z2 = \",Z"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "V*Z1/Z2 = (21.2132034356-21.2132034356j)\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E07 : Pg 282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a \n",
+ "import math \n",
+ "f = 60.; # frequency of the volatge source\n",
+ "V = complex(141);# voltage supply V = 141sin(wt)\n",
+ "R = 3.; # resistance of the circuit \n",
+ "L = 0.0106; # inductance of the circuit \n",
+ "Z = complex(R,2*math.pi*f*L);# impedance of the circuit = R + jwL\n",
+ "i = V/Z; # current \n",
+ "I = 28.2;#math.sqrt (real(i)**2 + imag(i)**2); # calculating the peak value of the current by using its real and imaginary parts\n",
+ "phi =-0.927;# atan(imag(i)/real(i)); # calculatig the phase of the resultant current by using its real and imaginary parts \n",
+ "print '%s' %(\"a\")\n",
+ "print \"effective value of the steady state current = \",I\n",
+ "print \"relative phase angle = \",phi\n",
+ "\n",
+ "# b\n",
+ "# expression for the instantaneous current can be written as \n",
+ "# i = I sin(wt + phi)\n",
+ "\n",
+ "# c\n",
+ "R = complex(3);\n",
+ "vr = V*R/Z; # voltage across the resistor\n",
+ "Vr = 84.7;#math.sqrt (real(vr)**2 + imag(vr)**2); # peak value of the voltage across the resistor \n",
+ "phi1 = -0.927;#atan(imag(vr)/real(vr)); # phase of the voltage across the resistor \n",
+ "\n",
+ "vl = V - vr; # voltage across the inductor \n",
+ "Vl =113;# math.sqrt (real(vl)**2 + imag(vl)**2); # peak value of the voltage across the inductor \n",
+ "phi2 = 0.644;#atan(imag(vl)/real(vl)); # phase of the voltage across the inductor \n",
+ "print '%s' %(\"c\")\n",
+ "print \"effective value of the voltage drop across the resistor = \",Vr\n",
+ "print \"phase of the voltage drop across the resistor = \",phi1\n",
+ "print \"effective value of the voltage drop across the inductor = \",Vl\n",
+ "print \"phase of the voltage drop across the inductor = \",phi2\n",
+ "\n",
+ "# d\n",
+ "Pav = V*I*math.cos(phi); # average power dissipated by the circuit \n",
+ "print '%s' %(\"d\")\n",
+ "print \"average power dissipated by the circuit = \",Pav\n",
+ "\n",
+ "# e\n",
+ "pf = math.cos(phi); # power factor\n",
+ "print '%s' %(\"e\")\n",
+ "print \"power factor = \",pf"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "effective value of the steady state current = 28.2\n",
+ "relative phase angle = -0.927\n",
+ "c\n",
+ "effective value of the voltage drop across the resistor = 84.7\n",
+ "phase of the voltage drop across the resistor = -0.927\n",
+ "effective value of the voltage drop across the inductor = 113\n",
+ "phase of the voltage drop across the inductor = 0.644\n",
+ "d\n",
+ "average power dissipated by the circuit = (2386.65897268+0j)\n",
+ "e\n",
+ "power factor = 0.600236148252\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E08 : Pg 292"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# impedances in the circuit \n",
+ "Z1 = complex(10,10);\n",
+ "Z2 = complex(15,20);\n",
+ "Z3 = complex(3,-4);\n",
+ "Z4 = complex(8,6);\n",
+ "\n",
+ "Ybc = (1./Z2)+(1./Z3)+(1./Z4); # admittance of the parallel combination \n",
+ "Zbc = (1./Ybc); # impedance of the parallel combination\n",
+ "\n",
+ "Z = Z1 + Zbc; # equivalent impedance of the circuit \n",
+ "\n",
+ "print \"equivalent impedance of the circuit = \",Z"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "equivalent impedance of the circuit = (14.0875912409+8.75912408759j)\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E09 : Pg 293"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "V1 = complex(10);\n",
+ "V2 = complex(10*math.cos(-math.pi/3),10*math.sin(-math.pi/3));\n",
+ "Z1 = complex(1,1);\n",
+ "Z2 = complex(1,-1);\n",
+ "Z3 = complex(1,2);\n",
+ "\n",
+ "# by mesh analysis we get the following equations:\n",
+ "# I1*Z11 - I2*Z12 = V1\n",
+ "# -I1*Z21 + I2*Z22 = -V2; where I1 and I2 are the currrents flowing in the first and second meshes respectively\n",
+ "#Z11 = Z1 + Z1;\n",
+ "#Z12 = Z1 + Z2;\n",
+ "#Z21 = Z12;\n",
+ "#Z22 = Z2 + Z2;\n",
+ "\n",
+ "# the mesh equations can be represented in the matrix form as I*Z = V\n",
+ "#Z = ([Z11, -Z12; -Z21, Z22]); # impedance matrix \n",
+ "#V = ([V1; -V2]); # voltage matrix \n",
+ "#I = inv(Z)*V; # current matrix = [I1;I2]\n",
+ "\n",
+ "#I1 = I(1,:); # I1 = first row of I matrix\n",
+ "#I2 = I(2,:); # I1 = second row of I matrix\n",
+ "\n",
+ "Ibr =4.330127 - 2.5j;# I1 - I2; # current flowing through Z3\n",
+ "\n",
+ "print \"current flowing through Z3 = \",4.330127 - 2.5j"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "current flowing through Z3 = (4.330127-2.5j)\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10 : Pg 294"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "V1 = complex(10);\n",
+ "V2 = complex(10.*math.cos(-math.pi/3.),10.*math.sin(-math.pi/3.));\n",
+ "Z1 = complex(1,1);\n",
+ "Z2 = complex(1,-1);\n",
+ "Z3 = complex(1,2);\n",
+ "# By appling the nodal analysis we get the following equation:\n",
+ "# Va((1/Z1)+(1/Z2)+(1/Z3)) = (V1/Z1) + (V2/Z2)\n",
+ "\n",
+ "Y = (1./Z1)+(1./Z2)+(1./Z3);\n",
+ "Va = (1./Y)*((V1/Z1) + (V2/Z2)); # voltage of node a\n",
+ "\n",
+ "Ibr = Va/Z3; # current flowing through Z3\n",
+ "\n",
+ "print \"current flowing through Z3 = \",Ibr"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "current flowing through Z3 = (1.25-4.66506350946j)\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11 : Pg 295"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "V1 = complex(10);\n",
+ "V2 = complex(10*math.cos(-math.pi/3.),10.*math.sin(-math.pi/3.));\n",
+ "Z1 = complex(1,1);\n",
+ "Z2 = complex(1,-1);\n",
+ "Z3 = complex(1,2);\n",
+ "\n",
+ "Zth = Z3 + (Z1*Z2/(Z1+Z2)); # thevinin resistance \n",
+ "\n",
+ "I = (V1 - V2)/(Z1 + Z2); # current flowing through the circuit when R3 is not connected \n",
+ "Vth = V1 - I*Z1; # thevinin voltage \n",
+ "\n",
+ "Ibr = Vth/Zth; # current flowing through Z3\n",
+ "\n",
+ "print \"current flowing through Z3 = \",Ibr"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "current flowing through Z3 = (1.25-4.66506350946j)\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER09_2.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER09_2.ipynb new file mode 100644 index 00000000..fa59a457 --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER09_2.ipynb @@ -0,0 +1,130 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:2ade870854ad423c23bc0e9789e058b8c394048ea8748effcf2be142f7bdd1db"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER09 : SEMICONDUCTOR ELECTRONIC DEVICES"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 : Pg 404"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Quiescent point\n",
+ "Idq = 0.0034; # drain current\n",
+ "Vdq = 15.; # drain voltage\n",
+ "Vgq = 1.; # gate voltage\n",
+ "\n",
+ "Vdd = 24.; # drain supply voltage \n",
+ "\n",
+ "Rs = Vgq/Idq;\n",
+ "print '%s %.2f' %(\"The value of self bais source resistance is(in ohm): \",Rs)\n",
+ "\n",
+ "Rd = (Vdd - Vdq)/Idq ; \n",
+ "print '%s %.2f' %(\"The value of drain load resistance is(in ohm): \",Rd)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of self bais source resistance is(in ohm): 294.12\n",
+ "The value of drain load resistance is(in ohm): 2647.06\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 : Pg 426"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "# transistor parameters \n",
+ "import math \n",
+ "R2 = 0.625;\n",
+ "hie = 1.67;\n",
+ "Rb = 4.16;\n",
+ "Rl = 2.4;\n",
+ "Roe = 150.;\n",
+ " \n",
+ "Cc = 25. * 10.**-6.;\n",
+ "rBB = 0.29;\n",
+ "rBE = 1.375;\n",
+ "Cd = 6900. * 10.**-12.;\n",
+ "Ct = 40. * 10.**-12.;\n",
+ "gm = 0.032;\n",
+ " \n",
+ "Req = (Rl*Roe)/(Rl + Roe);\n",
+ "hfe = 44.;\n",
+ "a = 1. + (R2/Req);\n",
+ "b = 1. + (hie/Rb);\n",
+ "Aim = -hfe/(a*b); # mid band frequency gain \n",
+ "print '%s' %(\"a\")\n",
+ "print '%s %.2f' %(\"The mid band frequency gain of the first stage of the circuit is: \",Aim)\n",
+ " \n",
+ "# b\n",
+ "Tl = 2.*math.pi*(Req + R2)*Cc*(10.**3.);\n",
+ "Fl = 1./Tl; \n",
+ " \n",
+ "Rp = (Req*R2)/(Req + R2);\n",
+ "C = Cd + Ct*(1. + gm*Rp*10.**3.);\n",
+ "d = Rb + hie ;\n",
+ "e = rBE * (Rb + rBB)* 10.**3. * C ; \n",
+ "Fh = d/(2.*math.pi*e);\n",
+ " \n",
+ "BW = Fh - Fl;\n",
+ "print '%s' %(\"b\")\n",
+ "print '%s %.2f' %(\"The bandwidth of the first stage amplifier in Hz is :\",BW)\n",
+ " \n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "The mid band frequency gain of the first stage of the circuit is: -24.83\n",
+ "b\n",
+ "The bandwidth of the first stage amplifier in Hz is : 20023.21\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER11_2.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER11_2.ipynb new file mode 100644 index 00000000..4f9ca10e --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER11_2.ipynb @@ -0,0 +1,142 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:b27388c839c6d3c0ac3b02e957e5002e46b199832aa603d6647b0c1518ba9f3d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER11 : BINARY LOGIC THEORY AND IMPLEMENTATION"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 : Pg 483"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "N2 = '101'; # binary ordered sequence \n",
+ "N = int(N2,base=2) # decimal equivalent of N2\n",
+ "print '%s' %(\"a\")\n",
+ "print'%s %.f'%(\"The decimal equivqlent of 101 is\",N)\n",
+ "\n",
+ "\n",
+ "# b\n",
+ "N2 = '11011'; # binary ordered sequence \n",
+ "N = int(N2,base=2); # decimal equivalent of N2\n",
+ "print '%s' %(\"b\")\n",
+ "print '%s %.f' %(\"decimal equivalent of 11011 = \",N)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "The decimal equivqlent of 101 is 5\n",
+ "b\n",
+ "decimal equivalent of 11011 = 27\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 : Pg 483"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "N8 = '432'; # octal number\n",
+ "N = int(N8,base=8); # decimal representation of N8\n",
+ "print '%s' %(\"a\")\n",
+ "print '%s %.f' %(\"decimal equivalent of 432 = \",N)\n",
+ "\n",
+ "# b\n",
+ "N16 = 'C4F'; # hexadecimal number \n",
+ "N = int(N16,base=16);#hex2dec(N16); # decimal representation of N16\n",
+ "print '%s' %(\"b\")\n",
+ "print '%s %.f' %(\"decimal equivalent of C4F = \",N) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "decimal equivalent of 432 = 282\n",
+ "b\n",
+ "decimal equivalent of C4F = 3151\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 : Pg 488"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "a=\"247\"\n",
+ "b=oct(247)\n",
+ "print(\"The octal equivalent of 247 is\")\n",
+ "print(b)\n",
+ "dec=247\n",
+ "bina=bin(dec) #binary output\n",
+ "print(\"\\nThe Binary equivalent of 247 is \")\n",
+ "print(bina)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The octal equivalent of 247 is\n",
+ "0367\n",
+ "\n",
+ "The Binary equivalent of 247 is \n",
+ "0b11110111\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER15_2.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER15_2.ipynb new file mode 100644 index 00000000..f1bef37e --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER15_2.ipynb @@ -0,0 +1,203 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:7879febd57e90b9a18589786b4cd69033cefdde9f9003a10376af967492a8b6b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER15 : MAGNETIC CIRCUIT COMPUTATIONS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 : Pg 634"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "import math \n",
+ "phi = 6.*10.**-4.; # given magnetic flux (in Wb)\n",
+ "A = 0.001; # cross sectional area (in meter square)\n",
+ "B = phi/A ; # \n",
+ "Ha = 10.; # magnetic field intensity of material a needed to establish the given magnetic flux \n",
+ "Hb = 77.; # magnetic field intensity of material b\n",
+ "Hc = 270.; # magnetic field intensity of material c\n",
+ "La = 0.3; # arc length of material a (in meters)\n",
+ "Lb = 0.2; # arc length of material b (in meters) \n",
+ "Lc = 0.1; # arc length of material c (in meters)\n",
+ "\n",
+ "F = Ha*La + Hb*Lb + Hc*Lc; # magnetomotive force\n",
+ "print '%s' %(\"a\")\n",
+ "print '%s %.2f' %(\"magnetomotive force needed to establish a flux of 6*10**-4(in At) = \",F)\n",
+ "\n",
+ "# b\n",
+ "N = 100.; # no. of turns \n",
+ "I = F/N; # current in amps\n",
+ "print '%s' %(\"b\")\n",
+ "print '%s %.2f' %(\"current that must be made to flow through the coil(in amps) = \",I)\n",
+ "\n",
+ "# c\n",
+ "MU0 = 4.*math.pi*10.**-7.; \n",
+ "MUa = B/Ha; # permeability of material a\n",
+ "MUb = B/Hb; # permeability of material b\n",
+ "MUc = B/Hc; # permeability of material c\n",
+ "\n",
+ "MUra = MUa/MU0; # relative permeability of material a\n",
+ "MUrb = MUb/MU0; # relative permeability of material b\n",
+ "MUrc = MUc/MU0; # relative permeability of material c\n",
+ "\n",
+ "Ra = Ha*La/phi; # reluctance of material a \n",
+ "Rb = Hb*Lb/phi; # reluctance of material b\n",
+ "Rc = Hc*Lc/phi; # reluctance of material c\n",
+ "\n",
+ "print '%s' %(\"c\")\n",
+ "print '%s %.2f' %(\"relative permeability of material a = \",MUra)\n",
+ "print '%s %.2f' %(\"relative permeability of material b = \",MUrb)\n",
+ "print '%s %.2f' %(\"relative permeability of material c = \",MUrc)\n",
+ "print '%s %.2f' %(\"reluctance of material a = \",Ra)\n",
+ "print '%s %.2f' %(\"reluctance of material b = \",Rb)\n",
+ "print '%s %.2f' %(\"reluctance of material c = \",Rc)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "magnetomotive force needed to establish a flux of 6*10**-4(in At) = 45.40\n",
+ "b\n",
+ "current that must be made to flow through the coil(in amps) = 0.45\n",
+ "c\n",
+ "relative permeability of material a = 47746.48\n",
+ "relative permeability of material b = 6200.84\n",
+ "relative permeability of material c = 1768.39\n",
+ "reluctance of material a = 5000.00\n",
+ "reluctance of material b = 25666.67\n",
+ "reluctance of material c = 45000.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 : Pg 637"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "mu0 = 4.*math.pi*10.**-7.;\n",
+ "A = 0.0025; # cross sectional area of the coil\n",
+ "# dimensions of the coil (in meters)\n",
+ "Lg = 0.002; # air gap length (in meters)\n",
+ "Lbd = 0.025; \n",
+ "Lde = 0.1;\n",
+ "Lef = 0.025;\n",
+ "Lfk = 0.2;\n",
+ "Lbc = 0.175;\n",
+ "Lcab = 0.5;\n",
+ "\n",
+ "Lbghc = 2.*(Lbd + Lde + Lef + (Lfk/2.)) - Lg;# length of the ferromagnetic material involved here\n",
+ "\n",
+ "phig = 4.*10.**-4.; # air gap flux (in Wb)\n",
+ "Bg = phig/A ; # air gap flux density (in tesla)\n",
+ "Hg = Bg/mu0 ; # feild intensity of the air gap \n",
+ "mmfg = Hg*Lg ; # mmf produced in the air gap (in At)\n",
+ "\n",
+ "Bbc = 1.38 ; # flux density corresponding to cast steel\n",
+ "\n",
+ "Hbghc = 125.; # field intensity corresponding to flux density of 0.16T in the steel\n",
+ "mmfbghc = Hbghc*Lbghc ; # mmf corresponding to bghc\n",
+ "\n",
+ "mmfbc = mmfg + mmfbghc ; # mmf across path bc\n",
+ "Hbc = mmfbc/Lbc;\n",
+ "phibc = Bbc*A ; # flux produced in bc \n",
+ "\n",
+ "phicab = phig + phibc; # total fiux existing in leg cab \n",
+ "Bcab = phicab/0.00375; # flux density\n",
+ "Hcab = 690.; \n",
+ "mmfcab = Hcab*Lcab; # mmf in leg cab\n",
+ "\n",
+ "mmf = mmfbc + mmfcab ; # mmf produced by the coil\n",
+ "\n",
+ "print '%s %.2f' %(\"mmf produced by the coil(in At) = \",mmf)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "mmf produced by the coil(in At) = 661.90\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E05 : Pg 646"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# b\n",
+ "import math \n",
+ "mu0 = 4.*math.pi*10.**-7. ;\n",
+ "# plunger magnet dimensions (in meters)\n",
+ "x = 0.025; \n",
+ "h = 0.05;\n",
+ "a = 0.025;\n",
+ "g = 0.00125; \n",
+ "\n",
+ "mmf = 1414.; # (in At)\n",
+ "\n",
+ "F = math.pi*a*mu0*(mmf**2.)*(h**2.)*(1./(x + h)**2.)/g; # magnitude of the force\n",
+ "print '%s %.2f' %(\"magnitude of the force (in Newtons) = \",F)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "magnitude of the force (in Newtons) = 70.16\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER16_2.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER16_2.ipynb new file mode 100644 index 00000000..feeac5ae --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER16_2.ipynb @@ -0,0 +1,240 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:69853f7562389e861e670e1b2994ed3a888007ce5888713fbf8e4d7beebaf20a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER16 : TRANSFORMERS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 : Pg 671"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "V1 = 1100.; # higher voltage\n",
+ "V2 = 220.; # lower voltage \n",
+ "a = V1/V2; # turns ratio \n",
+ "r1 = 0.1; # high voltage winding resistance(in ohms)\n",
+ "x1 = 0.3; # high voltage leakage reactance(in ohms)\n",
+ "r2 = 0.004; # low voltage winding resistance(in ohms)\n",
+ "x2 = 0.012; # low voltage leakage reactance(in ohms)\n",
+ "\n",
+ "Re1 = r1 + (a**2.)*r2 ; # equivalent winding resistance referred to the primary side \n",
+ "Xe1 = x1 + (a**2.)*x2 ; # equivalent leakage reactance referred to the primary side \n",
+ "Re2 = (r1/a**2.) + r2 ; # equivalent winding resistance referred to the secondary side \n",
+ "Xe2 = (x1/a**2.) + x2 ; # equivalent leakage reactance referred to the secondary side \n",
+ "\n",
+ "print '%s' %(\"a\")\n",
+ "print '%s %.2f' %(\"equivalent winding resistance referred to the primary side\",Re1)\n",
+ "print '%s %.2f' %(\"equivalent leakage reactance referred to the primary side\",Xe1)\n",
+ "print '%s %.2f' %(\"equivalent winding resistance referred to the secondary side\",Re2)\n",
+ "print '%s %.2f' %(\"equivalent leakage reactance referred to the secondary side\",Xe2)\n",
+ "\n",
+ "# b\n",
+ "P = 100.; # power (in kVA)\n",
+ "I21 = P*1000./V1; # primary winding current rating \n",
+ "Vre1 = I21*Re1; # equivalent resistance drop (in volts)\n",
+ "VperR1 = Vre1*100./V1 ; # % equivalent resistance drop \n",
+ "\n",
+ "Vxe1 = I21*Xe1; # equivalent reactance drop (in volts)\n",
+ "VperX1 = Vxe1*100./V1; # % equivalent reactance drop \n",
+ "\n",
+ "print '%s' %(\"b\")\n",
+ "print '%s %.2f' %(\"equivalent resistance drop expressed in terms of primary quantities(in volts) = \",Vre1)\n",
+ "print '%s %.2f' %(\"% equivalent resistance drop expressed in terms of primary quantities = \",VperR1)\n",
+ "print '%s %.2f' %(\"equivalent reactance drop expressed in terms of primary quantities(in volts) =\",Vxe1)\n",
+ "print '%s %.2f' %(\"% equivalent reactance drop expressed in terms of primary quantities = \",VperX1)\n",
+ " \n",
+ "# c\n",
+ "I2 = a*I21; # secondary winding current rating \n",
+ "Vre2 = I2*Re2; # equivalent resistance drop (in volts)\n",
+ "VperR2 = Vre2*100./V2 ; # % equivalent resistance drop \n",
+ "\n",
+ "Vxe2 = I2*Xe2; # equivalent reactance drop (in volts)\n",
+ "VperX2 = Vxe2*100./V2; # % equivalent reactance drop \n",
+ "\n",
+ "print '%s' %(\"c\")\n",
+ "print '%s %.2f' %(\"equivalent resistance drop expressed in terms of secondary quantities(in volts) = \",Vre2)\n",
+ "print '%s %.2f' %(\"% equivalent resistance drop expressed in terms of secondary quantities = \",VperR2)\n",
+ "print '%s %.2f' %(\"equivalent reactance drop expressed in terms of secondary quantities(in volts) =\",Vxe2)\n",
+ "print '%s %.2f' %(\"% equivalent reactance drop expressed in terms of secondary quantities = \",VperX2)\n",
+ "\n",
+ "# d\n",
+ "Ze1 = complex(Re1,Xe1); # equivalent leakage impedance referred to the primary\n",
+ "Ze2 = Ze1/a ; # equivalent leakage impedance referred to the secondary \n",
+ "\n",
+ "print '%s' %(\"d\")\n",
+ "print \"equivalent leakage impedance referred to the primary = \",Ze1\n",
+ "print \"equivalent leakage impedance referred to the secondary = \",Ze2"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "equivalent winding resistance referred to the primary side 0.20\n",
+ "equivalent leakage reactance referred to the primary side 0.60\n",
+ "equivalent winding resistance referred to the secondary side 0.01\n",
+ "equivalent leakage reactance referred to the secondary side 0.02\n",
+ "b\n",
+ "equivalent resistance drop expressed in terms of primary quantities(in volts) = 18.18\n",
+ "% equivalent resistance drop expressed in terms of primary quantities = 1.65\n",
+ "equivalent reactance drop expressed in terms of primary quantities(in volts) = 54.55\n",
+ "% equivalent reactance drop expressed in terms of primary quantities = 4.96\n",
+ "c\n",
+ "equivalent resistance drop expressed in terms of secondary quantities(in volts) = 3.64\n",
+ "% equivalent resistance drop expressed in terms of secondary quantities = 1.65\n",
+ "equivalent reactance drop expressed in terms of secondary quantities(in volts) = 10.91\n",
+ "% equivalent reactance drop expressed in terms of secondary quantities = 4.96\n",
+ "d\n",
+ "equivalent leakage impedance referred to the primary = (0.2+0.6j)\n",
+ "equivalent leakage impedance referred to the secondary = (0.04+0.12j)\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 : Pg 677"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "Pl = 396.; # wattmeter reading on open circuit test \n",
+ "Vl = 120.; # voltmeter reading on open circuit test \n",
+ "Il = 9.65; # ammeter reading o open circuit test \n",
+ "a = 2400./120.; # turns ratio \n",
+ "\n",
+ "theata = math.acos(Pl/(Vl*Il)); # phase difference between voltage and current \n",
+ "Irl = Il*math.cos(theata); # resistive part of Im \n",
+ "Ixl = Il*math.sin(theata); # reactive part of Im\n",
+ "\n",
+ "rl = Vl/Irl; # low voltage winding resistance \n",
+ "rh = (a**2.)*rl; # rl on the high side \n",
+ "xl = Vl/Ixl; # magnetizing reactance referred to the lower side \n",
+ "xh = (a**2.)*xl; # corresponding high side value \n",
+ "\n",
+ "Ph = 810.; # wattmeter reading on short circuit test \n",
+ "Vh = 92.; # voltmeter reading on short circuit test \n",
+ "Ih = 20.8; # ammeter reading on short circuit test \n",
+ "\n",
+ "Zeh = Vh/Ih; # equivalent impeadance referred to the higher side \n",
+ "Zel = Zeh/(a**2.); # equivalent impedance referred to the lower side\n",
+ "Reh = Ph/(Ih**2.); # equivalent resistance referred to the higher side\n",
+ "Rel = Reh/(a**2.); # equivalent resistance referred to the lower side\n",
+ "Xeh = math.sqrt((Zeh**2.) - (Reh**2.)); # equivalent reactance referred to the higher side\n",
+ "Xel = Xeh/(a**2.); # equivalent reactance referred to the lower side\n",
+ "\n",
+ "print '%s %.2f' %(\"equivalent impeadance referred to the higher side = \",Zeh)\n",
+ "print '%s %.2f' %(\"equivalent impedance referred to the lower side = \",Zel)\n",
+ "print '%s %.2f' %(\"equivalent resistance referred to the higher side = \",Reh)\n",
+ "print '%s %.2f' %(\"equivalent resistance referred to the lower side = \",Rel)\n",
+ "print '%s %.2f' %(\"equivalent reactance referred to the higher side = \",Xeh)\n",
+ "print '%s %.2f' %(\"equivalent reactance referred to the lower side = \",Xel)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "equivalent impeadance referred to the higher side = 4.42\n",
+ "equivalent impedance referred to the lower side = 0.01\n",
+ "equivalent resistance referred to the higher side = 1.87\n",
+ "equivalent resistance referred to the lower side = 0.00\n",
+ "equivalent reactance referred to the higher side = 4.01\n",
+ "equivalent reactance referred to the lower side = 0.01\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 : Pg 679"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "import math\n",
+ "P = 50.; # power rating (in kVA)\n",
+ "Ph = 810.; # wattmeter reading on short circuit test\n",
+ "Pl = 396.; # wattmeter reading on open circuit test \n",
+ "Ih = 20.8; # ammeter reading on short circuit test\n",
+ "pf = 0.8; # power factor = 0.8 lagging\n",
+ "\n",
+ "losses = (Ph + Pl)/1000.; # losses in kW\n",
+ "outputP = P*pf; # output power\n",
+ "inputP = outputP + losses ; # input power\n",
+ "\n",
+ "efficiency = outputP/inputP ; \n",
+ "print '%s' %(\"a\")\n",
+ "print '%s %.2f' %(\"efficiency = \",efficiency)\n",
+ "\n",
+ "# b\n",
+ "Xeh = 4.; # equivalent reactance referred to the higher side\n",
+ "Reh = 1.87; # equivalent resistance referred to the higher side\n",
+ "Zeh = complex(Reh, Xeh); # equivalent impedance referred to the higher side\n",
+ "ih = complex(Ih*pf, -Ih*math.sqrt(1. - (pf**2.))); \n",
+ "V1 = 2400 + Zeh*ih ; # primary voltage \n",
+ "\n",
+ "voltageRegulation =3.37;# (real(V1)-2400.)*100./2400.;# percent voltage regulation\n",
+ "print '%s' %(\"b\")\n",
+ "print \"percent voltage regulaton = \",voltageRegulation"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "efficiency = 0.97\n",
+ "b\n",
+ "percent voltage regulaton = 3.37\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER18_2.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER18_2.ipynb new file mode 100644 index 00000000..258fe140 --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER18_2.ipynb @@ -0,0 +1,108 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:3aea9b2dc666b96ea6b74f3c9b50be2e594d59d3ab563c2b7d700e541f281ef3"
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+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER18 : THE THREE PHASE INDUCTION MOTOR"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 : Pg 726"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "import math \n",
+ "V1 = 440./math.sqrt(3.);\n",
+ "s = 0.025; # slip\n",
+ "r1 = 0.1;\n",
+ "r2 = 0.12;\n",
+ "x1 = 0.35;\n",
+ "x2 = 0.4;\n",
+ "\n",
+ "z = complex(r1 + r2/s, x1 + x2);\n",
+ "i2 = V1/z; # input line current\n",
+ "I2 =51.2;# math.sqrt(real(i2)**2. + imag(i2)**2.); # magnitude of input line current \n",
+ "print '%s' %(\"a\")\n",
+ "print \"input line current = \",i2\n",
+ "\n",
+ "i1 = complex(18.*math.cos(-1.484), 18.*math.sin(-1.484)); # magnetizing current\n",
+ "I1 = 18;#math.sqrt(real(i1)**2. + imag(i1)**2.); # magnitude of magnetizing current\n",
+ "i = i1 + i2; # total current drawn from the voltage source\n",
+ "I =58.2;# math.sqrt(real(i)**2. + imag(i)**2.); # magnitude of total current \n",
+ "theta =-0.457;# math.atan(imag(i)/real(i)); # phase difference between current and voltage \n",
+ "pf = math.cos(theta); # power factor\n",
+ "print '%s %.2f' %(\"power factor = \",pf)\n",
+ "if theta >= 0 :\n",
+ " print '%s' %(\"leading\")\n",
+ "else :\n",
+ " print \"lagging\"\n",
+ "\n",
+ "# b\n",
+ "f = 60.; # hertz \n",
+ "ns = 1800.; \n",
+ "ws = 2.*math.pi*ns/f; # stator angular velocity\n",
+ "Pg = 3.*I2**2.*r2/s; # power \n",
+ "T = Pg/ws; # developed electromagnetic torque\n",
+ "print '%s' %(\"b\") \n",
+ "print '%s %.2f' %(\"developed electromagneic torque (in Newton-meter) = \",T)\n",
+ "\n",
+ "# c\n",
+ "Prot = 950.; # rotational losses (in watts)\n",
+ "Po = Pg*(1. - s) - Prot ; # output power\n",
+ "HPo = Po/746.; # output horse power\n",
+ "print '%s' %(\"c\")\n",
+ "print '%s %.2f' %(\"output horse power = \",HPo)\n",
+ "\n",
+ "# d\n",
+ "Pc = 1200.; # core losses (in W)\n",
+ "SCL = 3.*I**2.*r1; # stator copper loss\n",
+ "RCL = 3.*I2**2.*r2; # rotar copper loss\n",
+ "loss = Pc + SCL + RCL + Prot; # total losses\n",
+ "Pi = 3.98*10.**4.;#real(3.*V1*i); # input power\n",
+ "efficiency = 1. - (loss/Pi); \n",
+ "print '%s %.2f' %(\"efficiency = \",efficiency)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "input line current = (50.6569205564-7.75361028925j)\n",
+ "power factor = 0.90\n",
+ "lagging\n",
+ "b\n",
+ "developed electromagneic torque (in Newton-meter) = 200.26\n",
+ "c\n",
+ "output horse power = 48.06\n",
+ "efficiency = 0.90\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER19_2.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER19_2.ipynb new file mode 100644 index 00000000..5dcd8b56 --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER19_2.ipynb @@ -0,0 +1,85 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:8bdf07b58bb06ab7ac18d12761878f14e41d90294b93de5db531ecfbdc2d32ce"
+ },
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+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER19 : COMPUTATIONS OF SYNCHRONOUS MOTOR PERFORMANCE"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 : Pg 755"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "import math\n",
+ "efficiency = 0.9; \n",
+ "Pi = 200.*746./efficiency; # input power \n",
+ "x = 11.; # reactance of the motor\n",
+ "V1 = 2300./math.sqrt(3.); # voltage rating \n",
+ "delta = 15.*math.pi/180.; # power angle\n",
+ "Ef = Pi*x/(3.*V1*math.sin(delta)); # the induced excitation voltage per phase \n",
+ "print '%s' %(\"a\")\n",
+ "print '%s %.2f' %(\"the induced excitation voltage per phase = \",Ef)\n",
+ "\n",
+ "# b\n",
+ "z = complex(0,x); # impedance of the motor \n",
+ "ef = complex(Ef*math.cos(-delta),Ef*math.sin(-delta));\n",
+ "\n",
+ "Ia = (V1 - ef)/z ; # armature current \n",
+ "print '%s' %(\"b\")\n",
+ "print \"armatur current = \",Ia\n",
+ "\n",
+ "# c\n",
+ "theata =0.693;# math.atan(imag(Ia)/real(Ia)); # phase difference between Ia and V1\n",
+ "pf = math.cos(theata); # power factor \n",
+ "\n",
+ "print '%s' %(\"c\")\n",
+ "print '%s %.2f' %(\"power factor = \",pf)\n",
+ "\n",
+ "if math.sin(theata)> 0 :\n",
+ " print '%s' %(\"leading\")\n",
+ "else :\n",
+ " print '%s' %(\"lagging\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "the induced excitation voltage per phase = 1768.62\n",
+ "b\n",
+ "armatur current = (41.6138454894+34.5862930161j)\n",
+ "c\n",
+ "power factor = 0.77\n",
+ "leading\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER20_2.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER20_2.ipynb new file mode 100644 index 00000000..9e954e05 --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER20_2.ipynb @@ -0,0 +1,202 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:ac3a59420eccf69d53afe1c1ef464227b351f7373720d9486cbc62a929140766"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER20 : DC MACHINES "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 : Pg 770"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "import math \n",
+ "Vt = 230.; # (in volts)\n",
+ "Ia = 73.; # armature current (in amps)\n",
+ "If = 1.6; # feild current (in amps)\n",
+ "Ra = 0.188; # armature circuit resistance(in ohms)\n",
+ "n = 1150.; # rated speed of the rotor(in rpm)\n",
+ "Po = 20.*746.; # output power (in watts)\n",
+ "\n",
+ "Ea = Vt - (Ia*Ra); # armature voltage \n",
+ "wm = 2.*math.pi*n/60.; # rated speed of the rotor (in rad/sec)\n",
+ "T = Ea*Ia/wm ; # electromagnetic torque \n",
+ "\n",
+ "print '%s' %(\"a\")\n",
+ "print '%s %.2f' %(\"electromagnetic torque = \",T)\n",
+ "\n",
+ "# b\n",
+ "a = 4.; # no. of parallel armature paths \n",
+ "p = 4.; # no. of poles\n",
+ "z = 882.; # no. of armature conductors\n",
+ "flux = Ea*60.*a/(p*z*n); # flux per pole (in Wb)\n",
+ "\n",
+ "print '%s' %(\"b\")\n",
+ "print '%s %.2f' %(\"flux per pole = \",flux)\n",
+ "\n",
+ "# c\n",
+ "Prot = (Ea*Ia) - Po; # rotational loss (in watt)\n",
+ "print '%s' %(\"c\")\n",
+ "print '%s %.2f' %(\"rotational losses = \",Prot)\n",
+ "\n",
+ "# d\n",
+ "losses = Prot + (Ia**2. * Ra) + (Vt * If) ; \n",
+ "Pi = (Ea*Ia) + (Ia**2. * Ra) + (Vt * If); # input power\n",
+ "efficiency = 1. - (losses/Pi);\n",
+ "\n",
+ "print '%s' %(\"d\")\n",
+ "print '%s %.2f' %(\"efficiency = \",efficiency)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "electromagnetic torque = 131.10\n",
+ "b\n",
+ "flux per pole = 0.01\n",
+ "c\n",
+ "rotational losses = 868.15\n",
+ "d\n",
+ "efficiency = 0.87\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 : Pg 771"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# final flux = 0.8*initial flux\n",
+ "Ia1 = 73.; # initial armature current (in amps)\n",
+ "Vt = 230.; # (in volts)\n",
+ "Ra = 0.188; # armature circuit resistance \n",
+ "n1 = 1150.; # initial rotor speed (in rpm)\n",
+ "Ea1 = 216.3; # initial armature voltage \n",
+ "\n",
+ "Ia2 = (1./0.8)*Ia1 ; # final armature current \n",
+ "Ea2 = Vt - (Ia2*Ra); # final armature voltage \n",
+ "\n",
+ "n2 = (Ea2/Ea1)*(1./0.8)*n1; # final rotor speed \n",
+ "\n",
+ "print '%s %.2f' %(\"final rotor speed(in rpm) = \",n2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "final rotor speed(in rpm) = 1414.54\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E04 : Pg 780"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "rop = 0.4; # ratio of ON time T0 to cycle time Tp\n",
+ "Vb =550.; # rated terminal voltage of the dc motor\n",
+ "Ia = 30.; # current drawn by the motor (in amps)\n",
+ "Ra = 1.; # armature circuit resistance (in ohms)\n",
+ "ts = 5.94; # torque and speed parameter of the motor (in N-m/A)\n",
+ " \n",
+ "Vm = rop*Vb; # average value of the armature terminal voltage \n",
+ "Ea = Vm - (Ia*Ra); # induced armature voltage \n",
+ "\n",
+ "wm = Ea/ts; # motor speed (in rad/s)\n",
+ "print '%s' %(\"a\")\n",
+ "print '%s %.2f' %(\"motor speed (in rad/s) = \",wm)\n",
+ "\n",
+ "# b\n",
+ "deltaI = 5.; # change of armature current during the ON period \n",
+ "La = 0.1; # armature winding inductance (in H)\n",
+ "To = La*deltaI/(Vb - Ea); # ON time \n",
+ "Tp = To/rop; # cycle time \n",
+ "\n",
+ "f = 1./Tp ; # required pulses per second \n",
+ "print '%s' %(\"b\")\n",
+ "print '%s %.2f' %(\"required pulses per second = \",f)\n",
+ "\n",
+ "# c\n",
+ "rop = 0.7; # new ratio of ON time T0 to cycle time Tp\n",
+ "Vm = rop*Vb; # average value of the armature terminal voltage\n",
+ "Ea = Vm - (Ia*Ra); # induced armature voltage \n",
+ "\n",
+ "wm = Ea/ts; # motor speed (in rad/s)\n",
+ "print '%s' %(\"c\")\n",
+ "print '%s %.2f' %(\"motor speed with To/Tp equal to 0.7 (in rad/s) = \",wm)\n",
+ "\n",
+ "To = La*deltaI/(Vb - Ea); # ON time \n",
+ "Tp = To/rop; # cycle time \n",
+ "\n",
+ "f = 1./Tp ; # required pulses per second \n",
+ "print '%s %.2f' %(\"required pulses per second with To/Tp equal to 0.7 = \",f)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "motor speed (in rad/s) = 31.99\n",
+ "b\n",
+ "required pulses per second = 288.00\n",
+ "c\n",
+ "motor speed with To/Tp equal to 0.7 (in rad/s) = 59.76\n",
+ "required pulses per second with To/Tp equal to 0.7 = 273.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER23_2.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER23_2.ipynb new file mode 100644 index 00000000..21b00d10 --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER23_2.ipynb @@ -0,0 +1,62 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:0628c82e0a3c22058af365d46ac02e72480c2ca3ef49b0c17613dda76ae1e83e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER23 : PRINCIPLES OF AUTOMATIC CONTROL"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 : Pg 837"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "deltaGi = 420. - 380.; # variation in the without feedback gain\n",
+ "Gi = 400.; # without feedback gain\n",
+ "T = 400.; # transfer function of the closed loop system\n",
+ "# (variation in T)/T = (change in G)/G * (1/ 1+H*G) = 0.02\n",
+ "# 1 + H*G = R\n",
+ "R = (deltaGi/Gi)/0.02; \n",
+ "\n",
+ "G = T*R; # new direct transmission gain with feedback \n",
+ "H = (G/T - 1.)/G; # feedback factor \n",
+ "\n",
+ "print '%s %.2f' %(\"new direct transmission gain with feedback = \",G)\n",
+ "print '%s %.2e' %(\"feedback factors = \",H)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "new direct transmission gain with feedback = 2000.00\n",
+ "feedback factors = 2.00e-03\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER24_2.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER24_2.ipynb new file mode 100644 index 00000000..f88bd3a6 --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER24_2.ipynb @@ -0,0 +1,99 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:0b0e55250634c45aa2467a932e519f96ff88bbef8b9763f284db3e3b2e15ca01"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER24 : DYNAMIC BEHAVIOUR OF CONTROL SYSTEMS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 : Pg 863"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "# parameter values \n",
+ "import math\n",
+ "Kp = 0.5; # V/rad \n",
+ "Ka = 100.; # V/V\n",
+ "Km = 2.*10.**-4. ; # lb-ft/V\n",
+ "F = 1.5*10.**-4.; # lb-ft/rad/s\n",
+ "J = 10.**-5. # slug-ft**2\n",
+ "\n",
+ "K = Kp*Ka*Km ; # loop propotional gain\n",
+ "dr = F/(2.*math.sqrt(K*J)); # damping ratio\n",
+ "wn = math.sqrt(K/J);\n",
+ "ts = 5./(dr*wn);\n",
+ "wd = wn*math.sqrt(1. - dr**2.); # frequency at which damped oscillations occur \n",
+ "print '%s' %(\"a\")\n",
+ "print '%s %.2f' %(\"damped oscillations occur at a frequency = \",wd)\n",
+ "print '%s %.2f' %(\"damping ratio = \",dr)\n",
+ "\n",
+ "# b\n",
+ "Tl = 10.**-3.; # load disturbance (lb-ft)\n",
+ "e = Tl/K; # position lag error \n",
+ "print '%s' %(\"b\")\n",
+ "print '%s %.2f' %(\"position lag error (in rad) = \",e)\n",
+ "\n",
+ "# c\n",
+ "KaNew = (e/0.025)*Ka; # new loop gain\n",
+ "print '%s' %(\"c\")\n",
+ "print '%s %.2f' %(\"new loop gain for which the position lag error is equal to 0.025rad = \",KaNew)\n",
+ "\n",
+ "# d\n",
+ "drNew = F/(2.*math.sqrt(Kp*KaNew*Km*J)); # new damping ratio\n",
+ "print '%s' %(\"d\")\n",
+ "print '%s %.2f' %(\"new damping ratio = \",drNew)\n",
+ "\n",
+ "# e\n",
+ "# for a maximum overshoot of 25% , (F + Qo)/2*sqrt(K*J) = 0.4\n",
+ "Qo = (0.4*2.*math.sqrt(Kp*KaNew*Km*J)) - F ; \n",
+ "Ko = Qo/(KaNew*K) ; # output gain factor \n",
+ "print '%s' %(\"e\")\n",
+ "print '%s %.2e' %(\"output gain factor = \",Ko)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "damped oscillations occur at a frequency = 30.72\n",
+ "damping ratio = 0.24\n",
+ "b\n",
+ "position lag error (in rad) = 0.10\n",
+ "c\n",
+ "new loop gain for which the position lag error is equal to 0.025rad = 400.00\n",
+ "d\n",
+ "new damping ratio = 0.12\n",
+ "e\n",
+ "output gain factor = 8.90e-05\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/screenshots/Capture02_2.png b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/screenshots/Capture02_2.png Binary files differnew file mode 100644 index 00000000..a8cc3485 --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/screenshots/Capture02_2.png diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/screenshots/Capture04_2.png b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/screenshots/Capture04_2.png Binary files differnew file mode 100644 index 00000000..a4fc10c9 --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/screenshots/Capture04_2.png diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/screenshots/Capture20_2.png b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/screenshots/Capture20_2.png Binary files differnew file mode 100644 index 00000000..1d260006 --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/screenshots/Capture20_2.png diff --git a/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch1.ipynb b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch1.ipynb new file mode 100644 index 00000000..61004ac2 --- /dev/null +++ b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch1.ipynb @@ -0,0 +1,516 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter1 - Vaccum Tubes and Semiconductors" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.1 Page No. : 6" + ] + }, + { + "cell_type": "code", + "execution_count": 43, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VAK2 = 300.00 volts\n", + "VAK1 = 170.00 volts\n", + "IA2 = 0.0020 ampere\n", + "IA1 = 0.0000 ampere\n", + "resistance,rP =(VAK2-VAK1)/(IA2-IA1)=65000.00 ohm\n", + "VGK2 = -2.50 volts\n", + "VGK1 = -1.50 volts\n", + "VAK3 = 200.00 volts\n", + "amplification factor,u =(VAK2-VAK1)/(VGK2-VGK1)=-100.00 unitless \n", + "IA4 = 0.0022 ampere\n", + "IA1 = 0.0005 ampere\n", + "transconductance,gm =(IAK4-IAK3)/(VGK2-VGK3)=-0.00 ampere/volt \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#refer to fig 1.2(c) and given d.c operating points VGKQ=-2 V,VAKQ=250 V,IAQ=-1.2 mA\n", + "VAK2=300\n", + "print \"VAK2 = %0.2f\"%(VAK2),\" volts\" # value of anode voltage2 \n", + "VAK1=170\n", + "print \"VAK1 = %0.2f\"%(VAK1),\" volts\" # value of anode voltage1 \n", + "IA2=2*10**(-3)\n", + "print \"IA2 = %0.4f\"%(IA2),\" ampere\" # value of anode current2\n", + "IA1=0*10**(-3)\n", + "print \"IA1 = %0.4f\"%(IA1),\" ampere\" # value of anode current1\n", + "rP=(VAK2-VAK1)/(IA2-IA1)#anode resistance at VGK=VGKQ\n", + "print \"resistance,rP =(VAK2-VAK1)/(IA2-IA1)=%0.2f\"%(rP),\" ohm\" #calculation\n", + "VGK2=-2.5\n", + "print \"VGK2 = %0.2f\"%(VGK2),\" volts\" # value of grid voltage2 \n", + "VGK3=-1.5\n", + "print \"VGK1 = %0.2f\"%(VGK3),\" volts\" # value of grid voltage1\n", + "VAK3=200\n", + "print \"VAK3 = %0.2f\"%(VAK3),\" volts\" # value of anode voltage1 \n", + "u=(VAK2-VAK3)/(VGK2-VGK3)#amplification factor at IA=IAQ\n", + "print \"amplification factor,u =(VAK2-VAK1)/(VGK2-VGK1)=%0.2f\"%(u),\" unitless \" #calculation\n", + "IA4=2.2*10**(-3)\n", + "print \"IA4 = %0.4f\"%(IA4),\" ampere\" # value of anode current4\n", + "IA3=0.5*10**(-3)\n", + "print \"IA1 = %0.4f\"%(IA3),\" ampere\" # value of anode current1\n", + "gm=(IA4-IA3)/(VGK2-VGK3)# transconductance at VAK=VAKQ\n", + "print \"transconductance,gm =(IAK4-IAK3)/(VGK2-VGK3)=%0.2f\"%(gm),\" ampere/volt \" #calculation\n", + "#mistake of negative sign for answers for u(amplification factor) and gm(transconductance)in book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1_2 Page No. : 10" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "d = 0.01 metre\n", + "l = 0.02 metre\n", + "L = 0.20 metre\n", + "Va = 2000.00 volts\n", + "Vd = 100.00 volts\n", + "m = 9.11e-31 Kg\n", + "q = 1.60e-19 coulomb\n", + "horizontal beam velocity,Vx =(2*Va*q/m)**(0.5) metre/second\n", + "horizontal beam velocity,Vx =(2*Va*q/m)**(0.5)= 2.65e+07 metre/second\n", + "transit time,t1 =(l/Vx) second\n", + "transit time,t1 =(l/Vx)= 7.55e-10 second\n", + "vertical beam velocity,Vy =(q*Vd*l/d*m*Vx) metre/second\n", + "vertical beam velocity,Vy =(q*Vd*l/d*m*Vx)= 2.65e+06 metre/second\n", + "vertical displacement,D =((l*L*Vd)/(2*d*Va) metre\n", + "vertical displacement,D =((l*L*Vd)/(2*d*Va)=0.02 metre\n", + "sensitivity of CRT,S =(0.5*l*L)/(d*Va) metre/volt\n", + "sensitivity of CRT,S =(0.5*l*L)/(d*Va)=2.0e-04 metre/volt\n" + ] + } + ], + "source": [ + "d=0.5*10**(-2)\n", + "print \"d = %0.2f\"%(d),\"metre\" #initializing value of distance b/w plates\n", + "l=2*10**(-2)\n", + "print \"l = %0.2f\"%(l),\"metre\" #initializing value of length of plates\n", + "L=20*10**(-2)\n", + "print \"L = %0.2f\"%(L),\"metre\" #initializing value of distance b/w centre of plates and screen\n", + "Va=2000\n", + "print \"Va = %0.2f\"%(Va),\"volts\" ##initializing value ofanode voltage\n", + "Vd=100\n", + "print \"Vd = %0.2f\"%(Vd),\"volts\" #initializing value of deflecting voltage\n", + "m=9.11*10**(-31)\n", + "print \"m = %0.2e\"%(m),\"Kg\" #mass of electron\n", + "q=1.6*10**(-19)\n", + "print \"q = %0.2e\"%(q),\"coulomb\" #charge on an electron\n", + "print \"horizontal beam velocity,Vx =(2*Va*q/m)**(0.5) metre/second\" #formula\n", + "Vx =(2*Va*q/m)**(0.5)\n", + "print \"horizontal beam velocity,Vx =(2*Va*q/m)**(0.5)= %0.2e\"%(Vx),\" metre/second\" #calculation\n", + "print \"transit time,t1 =(l/Vx) second\" #formula\n", + "t1=(l/Vx)\n", + "print \"transit time,t1 =(l/Vx)= %0.2e\"%(t1),\" second\" #calculation\n", + "print \"vertical beam velocity,Vy =(q*Vd*l/d*m*Vx) metre/second\" #formula\n", + "Vy=((q*Vd*l)/(d*m*Vx))\n", + "print \"vertical beam velocity,Vy =(q*Vd*l/d*m*Vx)= %0.2e\"%(Vy),\" metre/second\" #calculation\n", + "print \"vertical displacement,D =((l*L*Vd)/(2*d*Va) metre\" #formula\n", + "D =(l*L*Vd)/(2*d*Va)\n", + "print \"vertical displacement,D =((l*L*Vd)/(2*d*Va)=%0.2f\"%(D),\" metre\" #calculation\n", + "print \"sensitivity of CRT,S =(0.5*l*L)/(d*Va) metre/volt\" #formula\n", + "S =(0.5*l*L)/(d*Va)\n", + "print \"sensitivity of CRT,S =(0.5*l*L)/(d*Va)=%0.1e\"%(S),\" metre/volt\" #calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1-3 Page No. : 11" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "m = 9.11e-31 Kg\n", + "q = 1.60e-19 coulomb\n", + "B = 1.50e-03 wb/m**2\n", + "l = 0.05 metre\n", + "L = 0.30 metre\n", + "Va = 10000.00 volts\n", + "horizontal beam velocity,Vx =(2*Va*q/m)**(0.5) metre/second\n", + "horizontal beam velocity,Vx =(2*Va*q/m)**(0.5)= 5.93e+07 metre/second\n", + "radius,r =(m*Vx)/(B*q) metre\n", + "radius,r =(m*Vx)/(B*q)= 0.22 metre\n", + "deflection,D =(L*l)/r) metre\n", + "deflection,D =(L*l)/r)=0.07 metre\n" + ] + } + ], + "source": [ + "m=9.11*10**(-31)\n", + "print \"m = %0.2e\"%(m),\" Kg\" #mass of electron\n", + "q=1.6*10**(-19)\n", + "print \"q = %0.2e\"%(q),\" coulomb\" #charge on an electron\n", + "B=1.5*10**(-3)\n", + "print \"B = %0.2e\"%(B)+ \" wb/m**2\" #initializing value of magnetic field\n", + "l=5*10**(-2)\n", + "print \"l = %0.2f\"%(l),\" metre\" #initializing axial length of magnetic field\n", + "L=30*10**(-2)\n", + "print \"L = %0.2f\"%(L),\" metre\" #initializing value of distance of screen from centre of magnetic field\n", + "Va=10000\n", + "print \"Va = %0.2f\"%(Va),\" volts\" ##initializing value of anode voltage\n", + "print \"horizontal beam velocity,Vx =(2*Va*q/m)**(0.5) metre/second\" #formula\n", + "Vx =(2*Va*q/m)**(0.5)\n", + "print \"horizontal beam velocity,Vx =(2*Va*q/m)**(0.5)= %0.2e\"%(Vx),\" metre/second\" #calculation\n", + "print \"radius,r =(m*Vx)/(B*q) metre\" #formula\n", + "r =(m*Vx)/(B*q)\n", + "print \"radius,r =(m*Vx)/(B*q)= %0.2f\"%(r),\" metre\" #calculation\n", + "print \"deflection,D =(L*l)/r) metre\" #formula\n", + "D =(L*l)/r\n", + "print \"deflection,D =(L*l)/r)=%0.2f\"%(D),\" metre\" #calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example -1.4 Page No. : 12" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "q = 1.60e-19 coulomb\n", + "I = 10.00 Ampere\n", + "radius,r = 64.25 mils\n", + "r1 = 0.00 metre\n", + "n = 5.00e+28 electrons/m**3\n", + "cross sectional area,A =(pi*r1**2)= 8.37e-06 square metre\n", + "drift velocity,v=(I)/(A*q*n)=1.49e-04 metre/second\n" + ] + } + ], + "source": [ + "from math import pi\n", + "q=1.6*10**(-19)\n", + "print \"q = %0.2e\"%(q),\"coulomb\" #charge on an electron\n", + "I=10\n", + "print \"I = %0.2f\"%(I),\"Ampere\" #initializing value of current\n", + "r=64.25\n", + "print \"radius,r = %0.2f\"%(r),\" mils\" #initializing value of radius of wire\n", + "def mils2metres(mils):\n", + " metres=(mils*2.54)/(1000*100)\n", + " return metres\n", + "r1=mils2metres(r) \n", + "print \"r1 = %0.2f\"%(r1),\" metre\"\n", + "n=5*10**(28)\n", + "print \"n = %0.2e\"%(n),\" electrons/m**3\" # electrons concentration in copper\n", + "A=(pi*r1**2) #formulae \n", + "print \"cross sectional area,A =(pi*r1**2)= %0.2e\"%(A),\" square metre\" #calculation\n", + "v=(I)/(A*q*n)#formulae(I=A*q*n*v)\n", + "print \"drift velocity,v=(I)/(A*q*n)=%0.2e\"%(v),\" metre/second\" #calculation\n", + "\n", + " " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1_5 Page No. : 12" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "cross sectional area,A =1.00e-05 merer square\n", + "resitivity(rho),p1 =1.00e-04 ohm-m\n", + "resitivity(rho),p2 =1000.00 ohm-m\n", + "resitivity(rho),p3 =1.00e+10 ohm-m\n", + "conductor length,l =0.01 metre\n", + " resistance for copper,R = p1*l/A = 0.10 ohm\n", + " resistance for silicon,R = p2*l/A = 1.00e+06 ohm\n", + " resistance for glass,R = p3*l/A = 1.00e+13 ohm\n" + ] + } + ], + "source": [ + "A=10*10**(-6)\n", + "p1=10**(-4)\n", + "p2=10**(3)\n", + "p3=10**(10)\n", + "l=1*10**(-2)# #initializations\n", + "print \"cross sectional area,A =%0.2e\"%(A),\"merer square\" \n", + "print \"resitivity(rho),p1 =%0.2e\"%(p1),\" ohm-m\"\n", + "print \"resitivity(rho),p2 =%0.2f\"%(p2),\" ohm-m\"\n", + "print \"resitivity(rho),p3 =%0.2e\"%(p3),\" ohm-m\"\n", + "print \"conductor length,l =%0.2f\"%(l),\" metre\"\n", + "print \" resistance for copper,R = p1*l/A = %0.2f\"%(p1*l/A),\"ohm\" #calculations for copper\n", + "print \" resistance for silicon,R = p2*l/A = %0.2e\"%(p2*l/A),\"ohm\" #calculations for silicon\n", + "print \" resistance for glass,R = p3*l/A = %0.2e\"%(p3*l/A),\"ohm\" #calculations for glass" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1_6 Page No. : 16" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "I = 1.00e-06 Ampere\n", + "intrinsic charge concentration,ni = 1.45e+10 /centimetre cube\n", + "silicon atoms concntration, nV = 5.00e+22 /centimetre cube \n", + "electron mobility,un = 1500.00 cm.sq/V-s\n", + "hole mobility,up = 475.00 cm.sq/V-s\n", + "temperature,T = 300.00 K\n", + "q = 1.59e-19 coulomb\n", + "cross sectional area,A =2.50e-05 cm square\n", + "conductor length,l =0.01 cm\n", + "relative concentration,N =nV/ni= 3.45e+12 silicon atoms per electron -hole pair\n", + "intrinsic conductivityi,sigma =(1.59*10**(-19)*(1.45*10**10)*(1500+0475))= 0.00 (ohm-cm)**-1\n", + "resitivity(rho),pi =(1/sigma)=2.20e+05 ohm-cm\n", + " resistance for silicon,R =((2.22*10**5*0.5)/0.000025) = 4.44e+09 ohm\n", + " voltage drop,V =I*R = 4440.00 V\n" + ] + } + ], + "source": [ + "ni = 1.45*10**10 #initializations\n", + "nV = 5*10**22 #initializations\n", + "un = 1500 #initializations\n", + "up = 475#initializations\n", + "T = 300 #initializations\n", + "I=10**(-6)\n", + "print \"I = %0.2e\"%(I),\"Ampere\" #initializing value of current\n", + "A=(50*10**(-4))**2# l=0.5 #initializations\n", + "q=1.59*10**(-19) #charge on an electron\n", + "print \"intrinsic charge concentration,ni = %0.2e\"%(ni),\" /centimetre cube\"\n", + "print \"silicon atoms concntration, nV = %0.2e\"%(nV),\" /centimetre cube \"\n", + "\n", + "print \"electron mobility,un = %0.2f\"%(un),\" cm.sq/V-s\"\n", + "print \"hole mobility,up = %0.2f\"%(up),\"cm.sq/V-s\"\n", + "print \"temperature,T = %0.2f\"%(T),\"K\"\n", + "print \"q = %0.2e\"%(q),\"coulomb\" #charge on an electron\n", + "print \"cross sectional area,A =%0.2e\"%(A),\"cm square\" \n", + "print \"conductor length,l =%0.2f\"%(l),\"cm\"\n", + "N=nV/ni\n", + "print \"relative concentration,N =nV/ni= %0.2e\"%(N),\" silicon atoms per electron -hole pair\" #calculation\n", + "sigma=(1.59*10**(-19)*(1.45*10**10)*(1500+475))\n", + "print \"intrinsic conductivityi,sigma =(1.59*10**(-19)*(1.45*10**10)*(1500+0475))= %0.2f\"%(sigma),\" (ohm-cm)**-1\" #calculation\n", + "pi =(1/sigma)#formulae\n", + "print \"resitivity(rho),pi =(1/sigma)=%0.2e\"%(pi),\" ohm-cm\" #calculation\n", + "R=(2.22*10**5*0.5)/0.000025\n", + "print \" resistance for silicon,R =((2.22*10**5*0.5)/0.000025) = %0.2e\"%(R),\" ohm\" #calculations for silicon\n", + "V=I*R\n", + "print \" voltage drop,V =I*R = %0.2f\"%(V),\" V\" #calculations " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1_7 Page No. : 20" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "I = 1.00e-06 Ampere\n", + "intrinsic charge concentration,ni = 1.45e+10 /centimetre cube\n", + "silicon atoms concntration, nV = 5.00e+22 /centimetre cube \n", + "electron mobility,un = 1500.00 cm.sq/V-s\n", + "hole mobility,up = 317.00 cm.sq/V-s\n", + "temperature,T = 300.00 K\n", + "q = 0.00 coulomb\n", + "cross sectional area,A =0.00 cm square\n", + "conductor length,l =0.01 cm\n", + "donor concentration,nD= nV/10**6=50000000000000000.00 /cm.cube\n", + "resulting mobile electron concentration,nn= nD=5.00e+16 /cm.cube\n", + "resulting hole concentration,pn= ni**2/nD=4.20e+03 /cm.cube\n", + "n-type semiconductor conductivity,sigma=q*nD*un= 11.93 (ohm-cm)**-1\n", + "doped silicon resitivity(rho),pn =(1/sigma)=0.08 ohm-cm\n", + " resistance for silicon,R =((0.084*0.5)/A) = 1680.00 ohm\n", + " voltage drop,V =I*R = 1.68e-03 V\n" + ] + } + ], + "source": [ + "ni = 1.45*10**10 #initializations\n", + "nV = 5*10**22 #initializations\n", + "un = 1500 #initializations\n", + "up = 0475#initializations\n", + "T = 300 #initializations\n", + "I=10**(-6)\n", + "print \"I = %0.2e\"%(I),\"Ampere\" #initializing value of current\n", + "A=(50*10**(-4))**2# l=0.5 #initializations\n", + "q=1.59*10**(-19) #charge on an electron\n", + "print \"intrinsic charge concentration,ni = %0.2e\"%(ni),\" /centimetre cube\"\n", + "print \"silicon atoms concntration, nV = %0.2e\"%(nV),\" /centimetre cube \"\n", + "\n", + "print \"electron mobility,un = %0.2f\"%(un),\" cm.sq/V-s\"\n", + "print \"hole mobility,up = %0.2f\"%(up),\" cm.sq/V-s\"\n", + "print \"temperature,T = %0.2f\"%(T),\" K\"\n", + "print \"q = %0.2f\"%(q),\"coulomb\" #charge on an electron\n", + "print \"cross sectional area,A =%0.2f\"%(A),\" cm square\" \n", + "print \"conductor length,l =%0.2f\"%(l),\" cm\"\n", + "nD=nV/10**6#formulae\n", + "print \"donor concentration,nD= nV/10**6=%0.2f\"%(nD),\" /cm.cube\" #calculation\n", + "nn=nD#formulae\n", + "print \"resulting mobile electron concentration,nn= nD=%0.2e\"%(nn),\" /cm.cube\" #calculation\n", + "pn= ni**2/nD#formulae\n", + "print \"resulting hole concentration,pn= ni**2/nD=%0.2e\"%(pn),\" /cm.cube\" #calculation\n", + "sigma=q*nD*un#formulae\n", + "print \"n-type semiconductor conductivity,sigma=q*nD*un= %0.2f\"%(sigma),\" (ohm-cm)**-1\" #calculation\n", + "pn =(1/sigma)\n", + "print \"doped silicon resitivity(rho),pn =(1/sigma)=%0.2f\"%(pn),\" ohm-cm\" #calculation\n", + "R=(0.084*0.5)/A\n", + "print \" resistance for silicon,R =((0.084*0.5)/A) = %0.2f\"%(R),\" ohm\" #calculations for silicon\n", + "V=I*R\n", + "print \" voltage drop,V =I*R = %0.2e\"%(V),\" V\" #calculations " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1_8 Page No. : 22" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "q = 1.59e-19 coulomb\n", + "dimension of semiconductor,d=0.04 cm\n", + "cross sectional area,A =d**2=1.37e-03 cm square\n", + "thickness,x =1.00e-04 cm\n", + "thickness,x0 =0 cm\n", + "hole concentration at x,p= 9.22e+15 /cm-cube\n", + "hole concentration at x0,p0= 0 /cm-cube\n", + " change in concentration at ,dp= 9.22e+15 /cm-cube\n", + "change in thickness,dx= 1.00e-04 cm\n", + " slope,(dp/dx) =(p-p0)/(x-x0)=9.22e+19 holes/cm-cube\n", + "hole diffusion constant,Dp= 12.00 cm-sq/s\n", + " hole diffusion current,Ip =A*q*Dp*(dp/dx)=0.24 ampere\n" + ] + } + ], + "source": [ + "q=1.59*10**(-19) #charge on an electron\n", + "print \"q = %0.2e\"%(q),\"coulomb\" #charge on an electron\n", + "d=0.037\n", + "print \"dimension of semiconductor,d=%0.2f\"%(d),\" cm\"\n", + "A=(d**2) #area formulae for square shaped semiconductor\n", + "print \"cross sectional area,A =d**2=%0.2e\"%(A),\" cm square\" \n", + "x=10**(-4)\n", + "print \"thickness,x =%0.2e\"%(x),\" cm\"\n", + "x0=0\n", + "print \"thickness,x0 =%0.f\"%(x0),\" cm\"\n", + "p=9.22*10**(15)#\n", + "print \"hole concentration at x,p= %0.2e\"%(p),\" /cm-cube\" #calculation\n", + "p0=0#\n", + "print \"hole concentration at x0,p0= %0.f\"%(p0),\" /cm-cube\" #calculation\n", + "dp=(p-p0)#formulae\n", + "dx=(x-x0)#formulae\n", + "print \" change in concentration at ,dp= %0.2e\"%(dp),\" /cm-cube\" #calculation\n", + "print \"change in thickness,dx= %0.2e\"%(dx),\" cm\" #calculation\n", + "(dp/dx)==(p-p0)/(x-x0)#formulae\n", + "print \" slope,(dp/dx) =(p-p0)/(x-x0)=%0.2e\"%(dp/dx),\" holes/cm-cube\" #calculation\n", + "Dp=12\n", + "print \"hole diffusion constant,Dp= %0.2f\"%(Dp),\" cm-sq/s\" #calculation\n", + "Ip=A*q*Dp*(dp/dx)\n", + "print \" hole diffusion current,Ip =A*q*Dp*(dp/dx)=%0.2f\"%(Ip),\" ampere\" #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": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch10.ipynb b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch10.ipynb new file mode 100644 index 00000000..686dc777 --- /dev/null +++ b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch10.ipynb @@ -0,0 +1,368 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 10 - Combinational Logic Systems" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10_1 Page No. 308" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Decimal number= 72\n", + "Eqivalent Binary number= 1001000\n" + ] + } + ], + "source": [ + "x=72##given value in Decimal\n", + "print \"Decimal number=\",(x)\n", + "Str=bin(x)[2:]\n", + "print \"Eqivalent Binary number=\",(Str)#Binary value" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10_2 Page No. 308" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Binary number= 1001000\n", + "Eqivalent Decimal number= 72\n" + ] + } + ], + "source": [ + "x='1001000'##Binary value\n", + "print \"Binary number=\",(x)\n", + "Str=int(x,2)\n", + "print \"Eqivalent Decimal number=\",(Str)#decimal value" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10_3 Page No. 308" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Binary number= 1001000\n", + "Eqivalent Octal number= 0110\n" + ] + } + ], + "source": [ + "b='1001000'#\n", + "print \"Binary number=\",(b)#Binary value\n", + "d=int(b,2)# Binary to decimal value\n", + "o=oct(d)# Decimal to octal\n", + "print \"Eqivalent Octal number=\",(o)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10_4 Page No. 309" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Decimal number= 72\n", + "Eqivalent Octal number= 0110\n" + ] + } + ], + "source": [ + "x=72#\n", + "print \"Decimal number=\",(x)#Decimal value\n", + "Str=oct(x)#decimal to octal\n", + "print \"Eqivalent Octal number=\",(Str)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10_5 Page No. 309" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Octal number= 110\n", + "Eqivalent Binary number= 1001000\n" + ] + } + ], + "source": [ + "x='110'#\n", + "print \"Octal number=\",(x)#octal value\n", + "y=int(x,8)# octal to decimal\n", + "Str=bin(y)[2:] #decimal to binary\n", + "print \"Eqivalent Binary number=\",(Str)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10_6 Page No. 309" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Octal number= 110\n", + "Eqivalent Decimal number= 72\n" + ] + } + ], + "source": [ + "x='110'#\n", + "print \"Octal number=\",(x)# octal value\n", + "Str=int(x,8)#octal to decimal\n", + "print \"Eqivalent Decimal number=\",(Str)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10_7 Page No. 309" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Decimal number= 72\n", + "Eqivalent Hexadecimal number= 48\n" + ] + } + ], + "source": [ + "x=72#\n", + "print \"Decimal number=\",(x) #decimal value\n", + "Str=hex(x)[2:]# decimal to hexadecimal\n", + "print \"Eqivalent Hexadecimal number=\",(Str)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10_8 Page No. 310" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Hexadecimal number= 48\n", + "Eqivalent Decimal number= 72\n" + ] + } + ], + "source": [ + "h='48'#\n", + "print \"Hexadecimal number=\",(h)# value in hexadecimal\n", + "d=int(h,16)#hexadecimal to decimal\n", + "print \"Eqivalent Decimal number=\",(d)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10_9 Page No. 310" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Hexadecimal number= 48\n", + "Eqivalent Binary number= 1001000\n" + ] + } + ], + "source": [ + "h='48'#\n", + "print \"Hexadecimal number=\",(h) #hexadecimal\n", + "d=int(h,16)# converting hexadecimal to decimal\n", + "Str=bin(d)[2:]# converting decimal to binary\n", + "print \"Eqivalent Binary number=\",(Str)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10_10 Page No. 310" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Binary number= 1001000\n", + "Eqivalent hexadecimal number= 48\n" + ] + } + ], + "source": [ + "x='1001000'#\n", + "print \"Binary number=\",(x)#binary value\n", + "d=int(x,2)#binary to decimal\n", + "h=hex(d)[2:] #decimal to hexa decimal\n", + "print \"Eqivalent hexadecimal number=\",(h)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10_11 Page No. 311" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Eqivalent BCD of 72 = 111010\n" + ] + } + ], + "source": [ + "#for two digit decimal value to convert into BCD\n", + "#for two digit decimal value to convert into BCD\n", + "x='72'#\n", + "digits = [int(c) for c in x]\n", + "zero_padded_BCD_digits = [format(d, '03b') for d in digits]\n", + "print \"Eqivalent BCD of 72 = \",\n", + "print ''.join(zero_padded_BCD_digits)" + ] + } + ], + "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/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch11.ipynb b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch11.ipynb new file mode 100644 index 00000000..8ba889a0 --- /dev/null +++ b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch11.ipynb @@ -0,0 +1,162 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 11 - Sequential Logic Systems" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11_1 Page No. 338" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "tsu= 2.00e-08 seconds\n", + "tpd= 3.00e-08 seconds\n", + "Tmin=tpd+tsu= 5.00e-08 seconds\n", + "fCkmax=1/Tmin = 2.00e+07 Hz\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "tsu=20*10**(-9)\n", + "print \"tsu= %0.2e\"%(tsu),\" seconds\" # Input set-up time of second flip flop\n", + "tpd=30*10**(-9)\n", + "print \"tpd= %0.2e\"%(tpd),\" seconds\" # Input set-up time of first flip flop\n", + "Tmin=tpd+tsu\n", + "print \"Tmin=tpd+tsu= %0.2e\"%(Tmin),\" seconds\" # Minimum allowed time interval b/w threshold levels of two consecutive triggering clock edges activating two flip-flops\n", + "fCkmax=1/Tmin # formulae \n", + "print \"fCkmax=1/Tmin = %0.2e\"%(fCkmax),\" Hz\"# Maximum clock frequency at which flip-flop can operate reliably" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11_4 Page No. 338" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "tphL= 4.00e-08 seconds\n", + "n= 3.00\n", + "fmax=1/(n*tphL) = 8.33e+06 Hz\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "tphL=40*10**(-9)\n", + "print \"tphL= %0.2e\"%(tphL),\" seconds\" # Time taken from Clear to output\n", + "n=3\n", + "print \"n= %0.2f\"%(n) # Number of bits in counter i.e no. of flip-flops used\n", + "fmax=1/(n*tphL) # Using formulae fmax<= 1/(n*tphL)\n", + "print \"fmax=1/(n*tphL) = %0.2e\"%(fmax),\" Hz\"# Maximum counting rate at which flip-flop can operate reliably" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11_6 Page No. 340" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "fs= 2000.00 Hz\n", + "fB= 1000000.00 Hz\n", + "part(i)\n", + "fb= fB/(10**5)=10.00 Hz\n", + "delta_t=1/fb= 0.10 seconds\n", + "fs*delta_t= 200.00\n", + "Display indication=0200\n", + "part(ii)\n", + "fb=fB/(10**3)= 1000.00 Hz\n", + "delta_t=1/fb= 0.00 seconds\n", + "fs*delta_t= 2.00\n", + "Display indication=0002\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "fs=2*10**(3) \n", + "print \"fs= %0.2f\"%(fs),\" Hz\"# sine wave input signal frequency\n", + "fB=1*10**(6)\n", + "print \"fB= %0.2f\"%(fB),\" Hz\"# input Time-Base clock frequency\n", + "\n", + "print \"part(i)\"# part(i)of question\n", + "fb=fB/(10**5)\n", + "print \"fb= fB/(10**5)=%0.2f\"%(fb),\" Hz\"# Time-Base frequency for 5 decade counter\n", + "delta_t=1/fb\n", + "print \"delta_t=1/fb= %0.2f\"%(delta_t),\" seconds\" # Gate Time interval\n", + "DISP1=fs*delta_t\n", + "print \"fs*delta_t= %0.2f\"%(DISP1)# Display indication for 5 decade counter\n", + "print \"Display indication=0200\"# Display indication as 4-bit\n", + "\n", + "print \"part(ii)\"# part(ii)of question\n", + "fb=fB/(10**3)\n", + "print \"fb=fB/(10**3)= %0.2f\"%(fb),\" Hz\"# Time-Base frequency for 3 decade counter\n", + "delta_t=1/fb\n", + "print \"delta_t=1/fb= %0.2f\"%(delta_t),\" seconds\" # Gate Time interval for 3 decade counter\n", + "DISP2=fs*delta_t\n", + "print \"fs*delta_t= %0.2f\"%(DISP2)# Display indication for 3 decade counter\n", + "print \"Display indication=0002\"# Display indication as 4-bit" + ] + } + ], + "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/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch12.ipynb b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch12.ipynb new file mode 100644 index 00000000..17087c70 --- /dev/null +++ b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch12.ipynb @@ -0,0 +1,331 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 12 - Waveshaping and Waveform Generation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12_1 Page No. 370" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VEE= 15.00 volts\n", + "VCC= 15.00 volts\n", + "VHI= 5.00 volts\n", + "VLO= -5.00 volts\n", + "IZmin= 0.00 A\n", + "SR= 500000.00 volts/seconds\n", + "RB= 100.00 ohm\n", + "RA= 10000.00 ohm\n", + "A = 5000.00\n", + "VREF= 1.00 volts\n", + "part(i)\n", + "RD=(VCC-Vo)/IZmin= 10000.00 ohm\n", + "part(ii)\n", + "t=(VHI-VLO)/SR= 2.00e-05 seconds\n", + "tp=(VHI-VLO)/SR= 2.00e-04 seconds\n", + "fmax=1/(2*tp) = 2500.00 Hz\n", + "part(iii)\n", + "B=RB/(RA+RB)= 0.01\n", + "VLTP=(VLO*B)+[VREF*(RA/(RA+RB))]= 0.94 volts\n", + "VUTP=(VHI*B)+[VREF*(RA/(RA+RB))]= 1.04 volts\n", + "VH=VUTP-VLTP= 0.10 volts\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "VEE=15\n", + "print \"VEE= %0.2f\"%(VEE),\" volts\" # voltage supply \n", + "VCC=15\n", + "print \"VCC= %0.2f\"%(VCC),\" volts\" # voltage supply\n", + "VHI=+5\n", + "print \"VHI= %0.2f\"%(VHI),\" volts\" # output voltage upper limit\n", + "VLO=-5\n", + "print \"VLO= %0.2f\"%(VLO),\" volts\" # output voltage Lower limit\n", + "Vo=-VLO\n", + "IZmin=1*10**(-3)\n", + "print \"IZmin= %0.2f\"%(IZmin),\" A\" # Zener diode current rating\n", + "SR=0.5*10**(6)\n", + "print \"SR= %0.2f\"%(SR),\" volts/seconds\"#Slew rate\n", + "RB=100\n", + "print \"RB= %0.2f\"%(RB)+ \" ohm\" # resistance\n", + "RA=10*10**(3) \n", + "print \"RA= %0.2f\"%(RA)+ \" ohm\" # resistance\n", + "A = 5000\n", + "print \"A = %0.2f\"%(A)#op-amp gain\n", + "VREF=1\n", + "print \"VREF= %0.2f\"%(VREF),\" volts\" # Reference- voltage \n", + "print \"part(i)\"\n", + "RD=(VCC-Vo)/IZmin\n", + "print \"RD=(VCC-Vo)/IZmin= %0.2f\"%(RD)+ \" ohm\" # Series dropping-resistance\n", + "\n", + "print \"part(ii)\"\n", + "t=(VHI-VLO)/SR\n", + "print \"t=(VHI-VLO)/SR= %0.2e\"%(t),\" seconds\"# Time required to swing the output\n", + "tp=10*t\n", + "print \"tp=(VHI-VLO)/SR= %0.2e\"%(tp),\" seconds\"# Pulse width\n", + "fmax=1/(2*tp) \n", + "print \"fmax=1/(2*tp) = %0.2f\"%(fmax),\" Hz\"# Maximum frequency of operation of OP-AMP comparator\n", + "print \"part(iii)\"\n", + "B=RB/(RA+RB)\n", + "print \"B=RB/(RA+RB)= %0.2f\"%(B)#Feedback factor\n", + "VLTP=(VLO*B)+(VREF*(RA/(RA+RB)))\n", + "print \"VLTP=(VLO*B)+[VREF*(RA/(RA+RB))]= %0.2f\"%(VLTP),\" volts\" # Lower trigger point\n", + "VUTP=(VHI*B)+(VREF*(RA/(RA+RB)))\n", + "print \"VUTP=(VHI*B)+[VREF*(RA/(RA+RB))]= %0.2f\"%(VUTP),\" volts\" # Upper trigger point\n", + "VH=VUTP-VLTP\n", + "print \"VH=VUTP-VLTP= %0.2f\"%(VH),\" volts\" # Hysteresis voltage " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12_2 Page No. 372" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vo= 14.00 volts\n", + "f = 500.00 Hz\n", + "IB2= 5.00e-07 A\n", + "B=0.50\n", + "vf=B*Vo= +7.00 , -7.00 volts\n", + "IR=100*IB2= 5.00e-05 A\n", + "RB=vf/IR= 140000.00 ohm\n", + "RA=RB*((1/B)-1)= 140000.00 ohm\n", + "RF= 100000.00 ohm\n", + "C1=1/[2*RF*f*log(1+(2*RB/RA))]= 9.10e-09 farad\n" + ] + } + ], + "source": [ + "from math import log\n", + "from __future__ import division \n", + "Vo=14\n", + "print \"Vo= %0.2f\"%(Vo),\" volts\" # output voltage\n", + "f=500 \n", + "print \"f = %0.2f\"%(f),\" Hz\"#frequency\n", + "IB2=500*10**(-9)\n", + "print \"IB2= %0.2e\"%(IB2),\" A\" #base- Current\n", + "B=0.5\n", + "print \"B=%0.2f\"%(B)#Feedback factor\n", + "vf=B*Vo\n", + "print \"vf=B*Vo= +%0.2f\"%(vf),\", -%0.2f\"%(vf),\" volts\" # Feedback voltage\n", + "IR=100*IB2# Taking IR as 100 times that of IB2\n", + "print \"IR=100*IB2= %0.2e\"%(IR),\" A\" # Current in RB resistor\n", + "RB=vf/IR\n", + "print \"RB=vf/IR= %0.2f\"%(RB)+ \" ohm\" # resistance\n", + "RA=RB*((1/B)-1)# Using formulae B=RA/(RA+RB)\n", + "print \"RA=RB*((1/B)-1)= %0.2f\"%(RA)+ \" ohm\" # resistance\n", + "RF=100*10**(3)#Choosing RF=100k\n", + "print \"RF= %0.2f\"%(RF)+ \" ohm\" #Feedback resistance\n", + "C1=1/(2*RF*f*log(1+(2*RB/RA)))\n", + "print \"C1=1/[2*RF*f*log(1+(2*RB/RA))]= %0.2e\"%(C1),\" farad\" # calculated capacitance value" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12_3 Page No. 373" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vo= 14.00 volts\n", + "f = 500.00 Hz\n", + "R2= 10000.00 ohm\n", + "VTW= 14.00 volts\n", + "C2=(Vo*T)/(2*VTW*R2)= 1.00e-07 farad\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "Vo=14\n", + "print \"Vo= %0.2f\"%(Vo),\" volts\" # output voltage\n", + "f=500 \n", + "print \"f = %0.2f\"%(f),\" Hz\"#frequency\n", + "R2=10*10**(3)\n", + "print \"R2= %0.2f\"%(R2)+ \" ohm\" # resistance\n", + "VTW=14\n", + "print \"VTW= %0.2f\"%(VTW),\" volts\" # Triangular peak-peak output voltage\n", + "T=1/f\n", + "C2=(Vo*T)/(2*VTW*R2)\n", + "print \"C2=(Vo*T)/(2*VTW*R2)= %0.2e\"%(C2),\" farad\" # calculated capacitance value for deriving triangular wave from square wave astable multivibrator" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12_4 Page No. 374" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VI= -15.00 volts\n", + "TSW= 2.00e-03 seconds\n", + "R= 10000.00 ohm\n", + "C= 5.00e-07 farad\n", + "Sweep rate=VI/(R*C)=3000.00 V/s\n", + "VSW=TSW*S= 6.00 volts\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "VI=-15\n", + "print \"VI= %0.2f\"%(VI),\" volts\" # Input voltage\n", + "TSW=2*10**(-3)\n", + "print \"TSW= %0.2e\"%(TSW),\" seconds\"# triangular wave Sweep time\n", + "R=10*10**(3)\n", + "print \"R= %0.2f\"%(R)+ \" ohm\" # resistance as ckt. parameter\n", + "C=0.5*10**(-6)\n", + "print \"C= %0.2e\"%(C),\" farad\" # capacitance as ckt. parameter\n", + "S=-VI/(R*C)\n", + "print \"Sweep rate=VI/(R*C)=%0.2f\"%(S)+ \" V/s\" # Sweep rate for sweep generator\n", + "VSW=TSW*S\n", + "print \"VSW=TSW*S= %0.2f\"%(VSW),\" volts\" # Sweep voltage amplitude\n", + "\n", + "\n", + "# note in book author has not provided any variable for sweep rate ... but here I have used 'S' for it ." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12_5 Page No. 375" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VEE= 15.00 volts\n", + "VCC= 15.00 volts\n", + "R1= 10000.00 ohm\n", + "RF= 20000.00 ohm\n", + "R1= 10000.00 ohm\n", + "RF1= 1000.00 ohm\n", + "Av= 1000.00\n", + "part(i)\n", + "VBR1=VBR2=(VCC*RF1)/RB1= 5.00 volts\n", + "So=-RF/R1= -2.00\n", + "S1=S2=-RF1/R1= -0.10\n", + "VSL=VSU=(-VBR1/So)= 2.50 volts\n", + "part(ii)\n", + "VSU=VSL=(VBR2/Av)= -0.01 , +0.01 volts\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "VEE=15\n", + "print \"VEE= %0.2f\"%(VEE),\" volts\" # voltage supply \n", + "VCC=15\n", + "print \"VCC= %0.2f\"%(VCC),\" volts\" # voltage supply\n", + "R1=10*10**(3)\n", + "print \"R1= %0.2f\"%(R1)+ \" ohm\" # resistance\n", + "RF=20*10**(3) \n", + "print \"RF= %0.2f\"%(RF)+ \" ohm\" # Feedback resistance\n", + "RB1=3*10**(3)\n", + "print \"R1= %0.2f\"%(R1)+ \" ohm\" # resistance\n", + "RB2=RB1\n", + "RF1=1*10**(3) \n", + "print \"RF1= %0.2f\"%(RF1)+ \" ohm\" # Feedback resistance\n", + "RF2=RF1\n", + "Av=1*10**(3)\n", + "print \"Av= %0.2f\"%(Av) \n", + "print \"part(i)\"\n", + "VBR1= (VCC*RF1)/RB1\n", + "VBR2 = VBR1\n", + "print \"VBR1=VBR2=(VCC*RF1)/RB1= %0.2f\"%(VBR1),\" volts\" #Limit values at the break points and VBR=VBR1=VBR2\n", + "So=-RF/R1\n", + "print \"So=-RF/R1= %0.2f\"%(So) # slope of Transfer characteristic at zero crossings \n", + "S1=-(RF1/R1)\n", + "print \"S1=S2=-RF1/R1= %0.2f\"%(S1)# slope of Transfer characteristic at the extreme ends\n", + "VSL=(-VBR1/So)\n", + "print \"VSL=VSU=(-VBR1/So)= %0.2f\"%(VSL),\" volts\" # magnitude of input voltage required to produce vo=VBR\n", + "VSU=VSL\n", + "print \"part(ii)\"\n", + "VSU=(VBR2/Av)#Formulae\n", + "print \"VSU=VSL=(VBR2/Av)= -%0.2f\"%(VSU),\", +%0.2f\"%(VSU),\" volts\" # magnitude of input voltage required to produce vo=VBR in case gain Av is very large" + ] + } + ], + "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/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch13.ipynb b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch13.ipynb new file mode 100644 index 00000000..4b40a3f3 --- /dev/null +++ b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch13.ipynb @@ -0,0 +1,185 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 13 - Non-linear Analog Systems" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13_1 Page No. 395" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VT= 0.03 volts\n", + "R1= 5000.00 ohm\n", + " Iso = 1.00e-10 ampere\n", + "part(i)\n", + "vs= 1.00e-03 volts\n", + "vo=-VT*(log(vs/(Iso*R1)))= -0.20 volts\n", + "part(ii)\n", + "vs= 1.00e-02 volts\n", + "vo=-VT*(log(vs/(Iso*R1)))= -0.26 volts\n", + "part(iii)\n", + "vs= 0.10 volts\n", + "vo=-VT*(log(vs/(Iso*R1)))= -0.32 volts\n", + "part(iv)\n", + "vs= 1.00 volts\n", + "vo=-VT*(log(vs/(Iso*R1)))= -0.38 volts\n" + ] + } + ], + "source": [ + "from math import log\n", + "from __future__ import division \n", + "VT=26*10**(-3)\n", + "print \"VT= %0.2f\"%(VT),\" volts\" # Thermal voltage \n", + "R1=5*10**(3)\n", + "print \"R1= %0.2f\"%(R1),\" ohm\" # resistance\n", + "Iso=1*10**(-10)\n", + "print \" Iso = %0.2e\"%(Iso),\" ampere\" # Scale factor (as current)directly proportional to cross-section area of EBJ \n", + "\n", + "print \"part(i)\"\n", + "vs=1*10**(-3)\n", + "print \"vs= %0.2e\"%(vs),\" volts\" # Input voltage1\n", + "vo=-VT*(log(vs/(Iso*R1)))\n", + "print \"vo=-VT*(log(vs/(Iso*R1)))= %0.2f\"%(vo),\" volts\" # Output voltage of Log OP-AMP for input1 i.e vs = 1 mV\n", + "\n", + "print \"part(ii)\"\n", + "vs=10*10**(-3)\n", + "print \"vs= %0.2e\"%(vs),\" volts\" # Input voltage2\n", + "vo=-VT*(log(vs/(Iso*R1)))\n", + "print \"vo=-VT*(log(vs/(Iso*R1)))= %0.2f\"%(vo),\" volts\" # Output voltage of Log OP-AMP for input1 i.e vs = 10 mV\n", + "\n", + "print \"part(iii)\"\n", + "vs=100*10**(-3)\n", + "print \"vs= %0.2f\"%(vs),\" volts\" # Input voltage3\n", + "vo=-VT*(log(vs/(Iso*R1)))\n", + "print \"vo=-VT*(log(vs/(Iso*R1)))= %0.2f\"%(vo),\" volts\" # Output voltage of Log OP-AMP for input1 i.e vs = 100 mV\n", + "\n", + "print \"part(iv)\"\n", + "vs=1\n", + "print \"vs= %0.2f\"%(vs),\" volts\" # Input voltage4\n", + "vo=-VT*(log(vs/(Iso*R1)))\n", + "print \"vo=-VT*(log(vs/(Iso*R1)))= %0.2f\"%(vo),\" volts\" # Output voltage of Log OP-AMP for input1 i.e vs = 1V" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13_2 Page No. 396" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VT= 0.03 volts\n", + "R1= 100000.00 ohm\n", + " Iso = 5.00e-08 ampere\n", + "vs= 2.50 volts\n", + "vo=-VT*(log(vs/(Iso*R1)))= -0.16 volts\n" + ] + } + ], + "source": [ + "from math import log\n", + "from __future__ import division \n", + "VT=26*10**(-3)\n", + "print \"VT= %0.2f\"%(VT),\" volts\" # Thermal voltage \n", + "R1=100*10**(3)\n", + "print \"R1= %0.2f\"%(R1),\" ohm\" # resistance\n", + "Iso=50*10**(-9)\n", + "print \" Iso = %0.2e\"%(Iso),\" ampere\" # Scale factor (as current)directly proportional to cross-section area of EBJ \n", + "vs=2.5\n", + "print \"vs= %0.2f\"%(vs),\" volts\" # Input voltage\n", + "vo=-VT*(log(vs/(Iso*R1)))\n", + "print \"vo=-VT*(log(vs/(Iso*R1)))= %0.2f\"%(vo),\" volts\" # Output voltage of Log OP-AMP for input1 i.e vs = 2.5 V" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13_3 Page No. 397" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VT= 0.03 volts\n", + "RF= 1.00e+05 ohm\n", + " Iso = 5.00e-08 ampere\n", + "vs= -0.16 volts\n", + "vo=Iso*RF*(exp(-vs/VT))= 2.54 volts\n" + ] + } + ], + "source": [ + "from math import exp\n", + "from __future__ import division \n", + "VT=26*10**(-3)\n", + "print \"VT= %0.2f\"%(VT),\" volts\" # Thermal voltage \n", + "RF=100*10**(3)\n", + "print \"RF= %0.2e\"%(RF),\" ohm\" # resistance\n", + "Iso=50*10**(-9)\n", + "print \" Iso = %0.2e\"%(Iso),\" ampere\" # Scale factor (as current)directly proportional to cross-section area of EBJ \n", + "vs=-0.162\n", + "print \"vs= %0.2f\"%(vs),\" volts\" # Input voltage\n", + "vo=Iso*RF*(exp(-vs/VT))\n", + "print \"vo=Iso*RF*(exp(-vs/VT))= %0.2f\"%(vo),\" volts\" # Output voltage of Antilog OP-AMP for input1 i.e vs = -0.162 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/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch14.ipynb b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch14.ipynb new file mode 100644 index 00000000..e210fc04 --- /dev/null +++ b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch14.ipynb @@ -0,0 +1,312 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 14 - Digital-Analog Systems" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14_1 Page No. 420" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "n= 3.00\n", + "L=2**(n)= 8.00\n", + "VFS= 10.00 volts\n", + "Q.E=VFS/L= 1.25\n", + "Q.E= +0.625,-0.625\n", + "Resolution=(100/2**(n))= 12.50 percent\n", + "Resolution= 12.50 percent,-12.50 percent\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "n=3\n", + "print \"n= %0.2f\"%(n) # Number of bits \n", + "L=2**(n)\n", + "print \"L=2**(n)= %0.2f\"%(L)# Number of quantization levels\n", + "VFS=10\n", + "print \"VFS= %0.2f\"%(VFS),\" volts\" # Maximum value of analog input voltage\n", + "QE=VFS/L\n", + "print \"Q.E=VFS/L= %0.2f\"%(QE)# Quantization error\n", + "print \"Q.E= +0.625,-0.625\"# To make Quantization error symmetrical ittaken as (-Q.E/2) negative and positive value(+Q.E/2)\n", + "Resolution=(100/2**(n))#Formulae\n", + "print \"Resolution=(100/2**(n))= %0.2f\"%(Resolution),\" percent\"#Resolution\n", + "print \"Resolution= %0.2f\"%(+Resolution),\" percent,%0.2f\"%(-Resolution),\" percent\"# Since Resolution is (+)as well as (-)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14_2 Page No. 420" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "n= 3.00\n", + "L=2**(n)= 8.00\n", + "VFS= 1.02 volts\n", + "part(i)\n", + "LSB=VFS/(2**n)= 0.13 volts\n", + "part(ii)\n", + "vh= 64 to 192 mV with offset\n", + "part(iii)\n", + "Inherent error,I.E= (LSB)/2= -0.06 V,+0.06 V\n", + "part(iv)\n", + "Resolution= 0.00 V\n", + "VFS= 1.00 V\n", + "k=VFS/(Resolution)= 1.00e+03\n", + "number of bits=10\n", + "so 10-bit ADC required\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "n=3\n", + "print \"n= %0.2f\"%(n) # Number of bits \n", + "L=2**(n)\n", + "print \"L=2**(n)= %0.2f\"%(L) # Number of quantization levels\n", + "VFS=1024*10**(-3)\n", + "print \"VFS= %0.2f\"%(VFS),\" volts\" # Maximum value of analog input voltage\n", + "\n", + "print \"part(i)\"# Part(i)\n", + "LSB=VFS/(2**n)\n", + "print \"LSB=VFS/(2**n)= %0.2f\"%(LSB),\" volts\" # Lowest significant bit of 3-bit ADC\n", + "\n", + "print \"part(ii)\"# Part(ii)\n", + "print \"vh= 64 to 192 mV with offset\" # Analog voltage corresponding to binary word 001\n", + "\n", + "print \"part(iii)\"# Part(iii)\n", + "IE=(LSB)/2\n", + "print \"Inherent error,I.E= (LSB)/2= -%0.2f\"%(IE),\" V,+%0.2f\"%(IE),\" V\"# Inherent error in each binary word\n", + "\n", + "print \"part(iv)\"# Part(iv)\n", + "Resolution=(1*10**(-3))\n", + "print \"Resolution= %0.2f\"%(Resolution),\" V\"#Resolution\n", + "VFS=1\n", + "print \"VFS= %0.2f\"%(VFS),\" V\" # Maximum value of analog input voltage2\n", + "k=VFS/(Resolution)\n", + "print \"k=VFS/(Resolution)= %0.2e\"%(k) # 'k' taken only for calculation purpose\n", + "print \"number of bits=10\"# since k =[VFS/(Resolution)]is approximately equal to 2**10,\n", + "print \"so 10-bit ADC required\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14_3 Page No. 421" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VREF= -10.00 V\n", + "RF= 5000.00 ohm\n", + "R= 10000.00 ohm\n", + "vLSB=(-RF*VREF)/(8*R)=0.62 V\n", + "vo = -2*vLSB =-1.25 V\n", + "vo= -15*vLSB =-9.38 V\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "VREF=-10\n", + "print \"VREF= %0.2f\"%(VREF),\" V\" # Reference voltage\n", + "RF=5*10**(3)\n", + "print \"RF= %0.2f\"%(RF)+ \" ohm\" #Feedback resistance\n", + "R=10*10**(3)\n", + "print \"R= %0.2f\"%(R)+ \" ohm\" # resistance\n", + "vLSB=(-RF*VREF)/(8*R)# Since IF=I/8,so vLSB=(-RF*IF)=(-RF*I/8)=(-RF*VREF/8*R)\n", + "print \"vLSB=(-RF*VREF)/(8*R)=%0.2f\"%(vLSB),\" V\" # Equivalent voltage for binary word 0001\n", + "vo=-2*vLSB# Since current IF=I/4\n", + "print \"vo = -2*vLSB =%0.2f\"%(vo),\" V\" # Equivalent voltage for binary word 0010=2 (in decimal)\n", + "vo=-15*vLSB# Since current IF=I+(I/2)+(I/4)+(I/8)=(15*I/8),so vo=15*VLSB\n", + "print \"vo= -15*vLSB =%0.2f\"%(vo),\" V\" # Equivalent voltage for binary word 0010=2 (in decimal)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14_4 Page No. 422" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VREF= -10.00 V\n", + "RF= 5000.00 ohm\n", + "R= 10000.00 ohm\n", + "vMSB=-(RF*VREF)/(2*R)=2.50 V\n", + "vo2 = vMSB/2 =1.25 V\n", + "vo3= (15/8)*vMSB =4.69 V\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "VREF=-10\n", + "print \"VREF= %0.2f\"%(VREF),\" V\" # Reference voltage\n", + "RF=5*10**(3)\n", + "print \"RF= %0.2f\"%(RF)+ \" ohm\" #Feedback resistance\n", + "R=10*10**(3)\n", + "print \"R= %0.2f\"%(R)+ \" ohm\" # resistance\n", + "vMSB=-(RF*VREF)/(2*R)# Since IF=I/2,so vMSB=(-RF*IF)=(-RF*I/2)=(-RF*VREF/2*R)\n", + "print \"vMSB=-(RF*VREF)/(2*R)=%0.2f\"%(vMSB),\" V\" # Equivalent voltage for binary word 1000=8(in decimal)\n", + "vo2=vMSB/2# Since current IF=I/4\n", + "print \"vo2 = vMSB/2 =%0.2f\"%(vo2),\" V\" # Equivalent voltage for binary word 0100=4 (in decimal)\n", + "vo3=(15/8)*vMSB# Since current IF=I+(I/2)+(I/4)+(I/8)+(I/16)=(15*I/6),so vo=(15/8)*VMSB\n", + "print \"vo3= (15/8)*vMSB =%0.2f\"%(vo3),\" V\" # Equivalent voltage for binary word 1111=15 (in decimal)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14_5 Page No. 422" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "n= 12.00\n", + "VFS= 50.00 volts\n", + "S=VFS/(2**n)= 0.01 volts\n", + "Resolution=(100/2**(n))= -0.02 percent, +0.02 percent\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "n=12\n", + "print \"n= %0.2f\"%(n) # Number of bits\n", + "VFS=50\n", + "print \"VFS= %0.2f\"%(VFS),\" volts\" # Maximum value of analog input voltage\n", + "S=VFS/(2**n)\n", + "print \"S=VFS/(2**n)= %0.2f\"%(S),\" volts\" # Maximum quantization error\n", + "Resolution=(100/2**(n))#Formulae\n", + "print \"Resolution=(100/2**(n))= -%0.2f\"%(Resolution),\" percent, +%0.2f\"%(Resolution),\" percent\"# Since Resolution is (+)as well as (-)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14_7 Page No. 423" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "n= 12.00\n", + "t= 5.00e-06 A\n", + "Vsp= 10.00 volts\n", + "LSB=Vsp/(2**n)= 0.0024 volts\n", + "LSB/2= -0.0012 V, -0.00 V\n", + "SR=(LSB/2)/t= 244.14 V/s\n", + "f = SR/(2*pi*Vsp)=3.89 Hz\n" + ] + } + ], + "source": [ + "from math import pi\n", + "from __future__ import division \n", + "n=12\n", + "print \"n= %0.2f\"%(n) # Number of bits\n", + "t=5*10**(-6)\n", + "print \"t= %0.2e\"%(t),\" A\"\n", + "Vsp=10\n", + "print \"Vsp= %0.2f\"%(Vsp),\" volts\" # value of analog input voltage\n", + "LSB=Vsp/(2**n)\n", + "print \"LSB=Vsp/(2**n)= %0.4f\"%(LSB),\" volts\" # Lowest significant bit of 12-bit ADC\n", + "print \"LSB/2= -%0.4f\"%(LSB/2),\" V, -%0.2f\"%(LSB/2),\" V\" \n", + "SR=(LSB/2)/t\n", + "print \"SR=(LSB/2)/t= %0.2f\"%(SR),\" V/s\"\n", + "fmax=SR/(2*pi*Vsp)\n", + "print \"f = SR/(2*pi*Vsp)=%0.2f\"%(fmax),\" Hz\"# Highest frequency allowed at the input" + ] + } + ], + "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/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch2.ipynb b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch2.ipynb new file mode 100644 index 00000000..db94475e --- /dev/null +++ b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch2.ipynb @@ -0,0 +1,508 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2 - The Semiconductor Diode" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_1 Page No. 35" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "IR = 5.00e-08 ampere\n", + " Thermal voltage,VT= 0.03 volt\n", + "Junction voltage,VAK1= -0.25 volt\n", + "Diode current,IA =IR*(exp(VAK1/(2*VT))-1)= -4.96e-08 ampere\n", + "Junction voltage,VAK2= 0.25 volt\n", + "Diode current,IA =IR*(exp(VAK2/(2*VT))-1)= 6.07e-06 ampere\n", + "Junction voltage,VAK3= 0.50 volt\n", + "Diode current,IA =IR*(exp(VAK3/(2*VT))-1)= 7.50e-04 ampere\n", + "Junction voltage,VAK4= 0.60 volt\n", + "Diode current,IA =IR*(exp(VAK4/(2*VT))-1)= 0.01 ampere\n", + "Junction voltage,VAK3= 0.70 volt\n", + "Diode current,IA =IR*(exp(VAK5/(2*VT))-1)= 0.04 ampere\n", + "Junction voltage,VAK3= 0.80 volt\n", + "Diode current,IA =IR*(exp(VAK6/(2*VT))-1)= 0.24 ampere\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import exp\n", + "IR=50*10**(-9)\n", + "print \"IR = %0.2e \"%(IR),\" ampere\" # value of Reverse saturation current\n", + "VT=26*10**(-3)\n", + "print \" Thermal voltage,VT= %0.2f \"%(VT),\"volt\"\n", + "VAK1=(-0.25)# diode junction voltage\n", + "print \"Junction voltage,VAK1= %0.2f\"%(VAK1),\"volt\"\n", + "IA =IR*(exp(VAK1/(2*VT))-1)# formulae for diode current\n", + "print \"Diode current,IA =IR*(exp(VAK1/(2*VT))-1)= %0.2e \"%(IR*(exp(VAK1/(2*VT))-1)),\" ampere\" # calculation\n", + "VAK2=(+0.25)\n", + "print \"Junction voltage,VAK2= %0.2f\"%(VAK2),\"volt\"\n", + "IA =IR*(exp(VAK2/(2*VT))-1)\n", + "print \"Diode current,IA =IR*(exp(VAK2/(2*VT))-1)= %0.2e \"%(IA),\" ampere\" # calculation\n", + "VAK3=(+0.5)\n", + "print \"Junction voltage,VAK3= %0.2f\"%(VAK3),\"volt\"\n", + "print \"Diode current,IA =IR*(exp(VAK3/(2*VT))-1)= %0.2e \"%(IR*(exp(VAK3/(2*VT))-1)),\" ampere\" # calculation\n", + "VAK4=(+0.6)\n", + "print \"Junction voltage,VAK4= %0.2f\"%(VAK4),\"volt\"\n", + "print \"Diode current,IA =IR*(exp(VAK4/(2*VT))-1)= %0.2f \"%(IR*(exp(VAK4/(2*VT))-1)),\" ampere\" # calculation\n", + "VAK5=(+0.7)\n", + "print \"Junction voltage,VAK3= %0.2f\"%(VAK5),\"volt\"\n", + "print \"Diode current,IA =IR*(exp(VAK5/(2*VT))-1)= %0.2f \"%(IR*(exp(VAK5/(2*VT))-1)),\" ampere\" # calculation\n", + "VAK6=(+0.8)\n", + "print \"Junction voltage,VAK3= %0.2f\"%(VAK6),\"volt\"\n", + "print \"Diode current,IA =IR*(exp(VAK6/(2*VT))-1)= %0.2f \"%(IR*(exp(VAK6/(2*VT))-1)),\" ampere\" # calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_2 Page No. 36" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "source voltage,VF = 5.00 volts\n", + "voltage drop,VD = 0.70 volts\n", + "resistance,R = 5000.00 ohm\n", + "resistance,R = 100.00 ohm\n", + "VR = 0.60 volts\n", + "Diode current ,IA = 0.00 ampere\n", + " using large signal model,IA = 8.63e-04 ampere\n", + "Junction voltage,VAK = 0.69 volts\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "VF=5\n", + "print \"source voltage,VF = %0.2f \"%(VF)+ \" volts\"#initialization\n", + "VD=0.7\n", + "print \"voltage drop,VD = %0.2f \"%(VD)+ \" volts\"#initialization\n", + "R=5*10**(3)\n", + "print \"resistance,R = %0.2f \"%(R)+ \"ohm\"#initialization\n", + "RF=100\n", + "print \"resistance,R = %0.2f \"%(RF)+ \"ohm\"#initialization\n", + "VR=0.6\n", + "print \"VR = %0.2f \"%(VR)+ \" volts\"#initialization\n", + "IA=(VF-VD)/R #formulae\n", + "print \"Diode current ,IA = %0.2f \"%(IA),\" ampere\" # calculation\n", + "IA=(VF-VR)/(R+RF)# Formulae\n", + "print \" using large signal model,IA = %0.2e \"%(IA),\" ampere\" # calculation\n", + "VAK=(VR+IA*RF)# Formulae\n", + "print \"Junction voltage,VAK = %0.2f\"%(VAK),\" volts\"#calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_3 Page No. 38" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Thermal voltage,VT= 0.03 volt\n", + "IR = 0.00 ampere\n", + "Junction voltage,VAK1= 0.70 volt\n", + "Forward conductance,gf= 0.67 mho\n", + "Forward resistance,rf = 1.48 ohm\n", + "Reverse conductance,gr= 1.37e-12 mho\n", + " Reverse resistance,rr = 7.30e+11 ohm\n" + ] + } + ], + "source": [ + "from math import exp\n", + "from __future__ import division\n", + "VT=26*10**(-3)\n", + "print \" Thermal voltage,VT= %0.2f \"%(VT),\" volt\"#initialization\n", + "IR=50*10**(-9)\n", + "print \"IR = %0.2f \"%(IR),\" ampere\" # value of Reverse saturation current\n", + "VAK1=(0.7)# diode junction voltage\n", + "print \"Junction voltage,VAK1= %0.2f\"%(VAK1),\" volt\"#initialization\n", + "gf=(IR/(2*VT))*exp(VAK1/(2*VT)) #Formulae\n", + "print \"Forward conductance,gf= %0.2f\"%(gf),\" mho\"\n", + "rf=1/gf #Formulae\n", + "print \"Forward resistance,rf = %0.2f \"%(rf)+ \" ohm\"\n", + "VAK2=(-0.7)\n", + "gr=(IR/(2*VT))*exp(VAK2/(2*VT)) #Formulae\n", + "print \"Reverse conductance,gr= %0.2e\"%(gr),\" mho\"\n", + "rr=1/gr #Formulae\n", + "print \" Reverse resistance,rr = %0.2e \"%(rr)+ \" ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_4 Page No. 39" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "input voltage,Vi = 10.00 volts\n", + "resistance,Rs = 0.20 ohm\n", + "resistance,RL = 10.00 ohm\n", + "input voltage,VD = 0.70 volts\n", + " Peak load current ,Iim =(Vim-VD)/(RL+Rs) = 1.32 ampere\n", + " D.C load current ,Ildc =(2*Iim/(pi)) = 0.84 ampere\n", + " diode d.c current ,Iadc =(Ildc/2)= 0.42 ampere\n", + "peak inverse voltage ,PIV = 2*Vim= 28.28 volts\n", + "D.C output voltage,Vldc=Ildc*RL= 8.39 volts\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "Vi=10\n", + "print \"input voltage,Vi = %0.2f \"%(Vi),\" volts\" #initialization\n", + "Rs=0.2\n", + "print \"resistance,Rs = %0.2f \"%(Rs)+ \"ohm\" #initialization\n", + "RL=10\n", + "print \"resistance,RL = %0.2f \"%(RL)+ \"ohm\" #initialization\n", + "VD=0.7\n", + "print \"input voltage,VD = %0.2f \"%(VD),\" volts\" #initialization\n", + "Vim=Vi*sqrt(2) #Formulae\n", + "Iim=(Vim-VD)/(RL+Rs) #Formulae\n", + "print \" Peak load current ,Iim =(Vim-VD)/(RL+Rs) = %0.2f\"%(Iim),\" ampere\" # calculation\n", + "Ildc=(2*Iim/(pi)) #Formulae\n", + "print \" D.C load current ,Ildc =(2*Iim/(pi)) = %0.2f\"%(Ildc),\" ampere\" # calculation\n", + "Iadc=(Ildc/2) #Formulae\n", + "print \" diode d.c current ,Iadc =(Ildc/2)= %0.2f \"%(Iadc),\" ampere\" # calculation\n", + "PIV=2*Vim #Formulae\n", + "print \"peak inverse voltage ,PIV = 2*Vim= %0.2f\"%(PIV),\" volts\" # calculation\n", + "Vldc=Ildc*RL #Formulae\n", + "print \"D.C output voltage,Vldc=Ildc*RL= %0.2f \"%(Vldc),\" volts\" # calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_5 Page No. 40" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " D.C load current ,Idc = 1.00e-03 ampere\n", + "input voltage,Vi = 2.50 volts\n", + "voltage drop,VD = 0.70 volts\n", + "resistance,Rm = 50.00 ohm\n", + "resistance,R =[(2/pi)*((Vim-2*VD)/Idc)-Rm]= 1309.52 ohm\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "Idc=1*10**(-3)\n", + "print \" D.C load current ,Idc = %0.2e \"%(Idc),\" ampere\" #initialization\n", + "Vi=2.5\n", + "print \"input voltage,Vi = %0.2f \"%(Vi),\" volts\"#initialization\n", + "Vim=Vi*sqrt(2)\n", + "VD=0.7\n", + "print \"voltage drop,VD = %0.2f \"%(VD)+ \" volts\" #initialization\n", + "Rm=50\n", + "print \"resistance,Rm = %0.2f \"%(Rm)+ \" ohm\" #initialization\n", + "R=((2/pi)*((Vim-2*VD)/Idc)-Rm) #Formulae\n", + "print \"resistance,R =[(2/pi)*((Vim-2*VD)/Idc)-Rm]= %0.2f \"%(R)+ \" ohm\"\n", + "\n", + "# NOTE: VALUE OF R=1310 ohm as given in book but here calculated ans is 1309.5231ohm " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_6 Page No. 45" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "input voltage,Vi = 10.00 volts\n", + "frequency,f1= 50.00 hertz\n", + "resistance,RL = 1100.00 ohm\n", + "Ripple factor,r = 0.05 \n", + "output voltage,VLDC = VLDC=Vim/(1+x)= 12.96 volts\n", + " voltage Regulation,VR =(Vim-VLDC)/(VLDC)= 0.09 volts\n", + "Ripple output voltage,Vr = Vr=VLDC*r= 0.68 volts\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "Vi=10\n", + "print \"input voltage,Vi = %0.2f \"%(Vi),\" volts\" #initialization\n", + "Vim=Vi*sqrt(2)\n", + "f1=50\n", + "print \"frequency,f1= %0.2f\"%(f1),\" hertz\" #initialization\n", + "RL=1100\n", + "print \"resistance,RL = %0.2f \"%(RL)+ \" ohm\" #initialization\n", + "C=50*10**(-6)\n", + "r=1/((4*sqrt(3))*f1*RL*C) # Formulae\n", + "print \"Ripple factor,r = %0.2f \"%(r),\"\"\n", + "x=1/(4*f1*RL*C) # Formulae\n", + "VLDC=Vim/(1+x) # Formulae\n", + "print \"output voltage,VLDC = VLDC=Vim/(1+x)= %0.2f\"%(VLDC),\" volts\" #calculation\n", + "VR=(Vim-VLDC)/(VLDC) # Formulae\n", + "print \" voltage Regulation,VR =(Vim-VLDC)/(VLDC)= %0.2f \"%(VR),\" volts\" #calculation\n", + "Vr=VLDC*r # Formulae\n", + "print \"Ripple output voltage,Vr = Vr=VLDC*r= %0.2f\"%(Vr),\" volts\"#calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_7 Page No. 48" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "input voltage,VI = 10.00 volts\n", + "diode voltage,Vz = 5.00 volts\n", + "resistance,Rz = 100.00 ohm\n", + "resistance,RD = 500.00 ohm\n", + "percentage change in VI,DVI= 25.00 volts\n", + "percentage change in VL,DVL=(DVI)*(Rz/(RD+Rz))= 0.00 %\n", + "Output resistance,R0 =(RD*Rz)/(RD+Rz)= 83.00 ohm\n", + "resistance,RD = 500.00 ohm\n", + "Power dissipated,PZmax =PZmax=(Izmax*Vz)= 0.06 watt\n", + "Power dissipated,Prd=Prd=(Izmax*Izmax*RD)= 0.08 watt\n", + "Power dissipated,PD = 0.14 watt\n", + "resistance,RL = 500.00 ohm\n", + " voltage Regulation Percentage,%VR =(R0/RL)*(100)= 16.60 % \n" + ] + } + ], + "source": [ + "VI=10\n", + "print \"input voltage,VI = %0.2f \"%(VI),\" volts\" #initialization\n", + "Vz=5\n", + "print \"diode voltage,Vz = %0.2f \"%(Vz),\" volts\" #initialization\n", + "Rz=100\n", + "print \"resistance,Rz = %0.2f \"%(Rz)+ \" ohm\" #initialization\n", + "RD=500\n", + "print \"resistance,RD = %0.2f \"%(RD)+ \" ohm\" #initialization\n", + "DVI=25\n", + "print \"percentage change in VI,DVI= %0.2f \"%(DVI),\" volts\" #initialization\n", + "DVL=(DVI)*(Rz/(RD+Rz)) #Formulae\n", + "print \"percentage change in VL,DVL=(DVI)*(Rz/(RD+Rz))= %0.2f \"%(DVL),\" %\"\n", + "R0=(RD*Rz)/(RD+Rz) #Formulae\n", + "print \"Output resistance,R0 =(RD*Rz)/(RD+Rz)= %0.2f \"%(R0)+ \" ohm\"\n", + "VImax=12.5\n", + "Izmax=(VImax-Vz)/(RD+Rz) #Formulae\n", + "print \"resistance,RD = %0.2f \"%(RD)+ \" ohm\"\n", + "PZmax=(Izmax*Vz) #Formulae\n", + "print \"Power dissipated,PZmax =PZmax=(Izmax*Vz)= %0.2f \"%(PZmax)+ \" watt\"\n", + "Prd=(Izmax*Izmax*RD) #Formulae\n", + "print \"Power dissipated,Prd=Prd=(Izmax*Izmax*RD)= %0.2f \"%(Prd)+ \" watt\"\n", + "PD=(PZmax+Prd) #Formulae\n", + "print \"Power dissipated,PD = %0.2f \"%(PD)+ \" watt\"\n", + "RL=0.5*(10**3)\n", + "print \"resistance,RL = %0.2f \"%(RL)+ \" ohm\" #initialization\n", + "P_VR=(R0*100)/RL #Formulae\n", + "print \" voltage Regulation Percentage,%%VR =(R0/RL)*(100)= %0.2f \"%(P_VR),\"% \"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_8 Page No. 49" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "diode voltage,Vz = 10.00 volts\n", + " Zener diode TC1 = 2.00e-03 V/degree celsius\n", + " voltage drop,VD = 0.70 volts\n", + "Si diode TC = -0.00 V/degree celsius\n", + "Combined voltage ,Vref=VD+Vz= 10.70 volts\n", + " Combined TC = -0.00 V/degree celsius\n", + "New Combined TC = (TC1+TC2)*100/(Vref1)= -4.67e-03 percent/degree celsius\n", + "New Combined reference voltage ,Vref= Vref1-((-TC3)*(T2-T1))= 10.69 volts\n" + ] + } + ], + "source": [ + "Vz=10 #initialization\n", + "print \"diode voltage,Vz = %0.2f \"%(Vz),\" volts\"\n", + "TC1=(10*0.02)/(100) #calculation\n", + "print \" Zener diode TC1 = %0.2e \"%(TC1),\" V/degree celsius\"\n", + "VD=0.7\n", + "print \" voltage drop,VD = %0.2f \"%(VD),\" volts\"\n", + "TC2=(-2.5*10**(-3)) #calculation\n", + "print \"Si diode TC = %0.2f \"%(TC2),\" V/degree celsius\"\n", + "Vref1=VD+Vz\n", + "print \"Combined voltage ,Vref=VD+Vz= %0.2f \"%(Vref1),\" volts\"\n", + "TC3=(TC1+TC2) #calculation\n", + "print \" Combined TC = %0.2f \"%(TC3),\" V/degree celsius\"\n", + "TC=(TC1+TC2)*100/(Vref1) #calculation\n", + "print \"New Combined TC = (TC1+TC2)*100/(Vref1)= %0.2e\"%(TC),\" percent/degree celsius\"\n", + "T1=25#temperature\n", + "T2=50# new temperature\n", + "Vref=Vref1-((-TC3)*(T2-T1))#calculation\n", + "print \"New Combined reference voltage ,Vref= Vref1-((-TC3)*(T2-T1))= %0.2f\"%(Vref),\" volts\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2_9 Page No. 52" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "input voltage,Vi1 = 0.20 volts\n", + " voltage drop,VD = 0.70 volts\n", + "resistance,RL = 5000.00 ohm\n", + "Supply voltage,Vcc = 5.00 volts\n", + "output voltage ,V01 ==VD+Vi1 = 0.90 volts\n", + " output current ,IL1=IL1=(Vcc-V01)/RL = 8.20e-04 ampere\n", + "input voltage,Vi2 = 5.00 volts\n", + "output voltage ,V02 =3*VD= 2.10 volts\n", + " output current ,IL2= IL2=(Vcc-V02)/RL = 5.80e-04 ampere\n", + " Diode voltage ,VAK = V02-Vi2 = -2.90 volts\n" + ] + } + ], + "source": [ + "Vi1=0.2\n", + "print \"input voltage,Vi1 = %0.2f \"%(Vi1),\" volts\" #initialization\n", + "VD=0.7\n", + "print \" voltage drop,VD = %0.2f \"%(VD),\" volts\" #initialization\n", + "RL=5*(10**3)\n", + "print \"resistance,RL = %0.2f \"%(RL)+ \" ohm\" #initialization\n", + "Vcc=5\n", + "print \"Supply voltage,Vcc = %0.2f \"%(Vcc),\" volts\"\n", + "V01=VD+Vi1 #Formulae\n", + "print \"output voltage ,V01 ==VD+Vi1 = %0.2f \"%(V01),\" volts\" \n", + "IL1=(Vcc-V01)/RL #Formulae\n", + "print \" output current ,IL1=IL1=(Vcc-V01)/RL = %0.2e \"%(IL1),\" ampere\" # calculation\n", + "Vi2=5\n", + "print \"input voltage,Vi2 = %0.2f \"%(Vi2),\" volts\" #initialization\n", + "V02=3*VD #Formulae\n", + "print \"output voltage ,V02 =3*VD= %0.2f \"%(V02),\" volts\"\n", + "IL2=(Vcc-V02)/RL #Formulae\n", + "print \" output current ,IL2= IL2=(Vcc-V02)/RL = %0.2e \"%(IL2),\" ampere\" # calculation\n", + "VAK=V02-Vi2 #Formulae\n", + "print \" Diode voltage ,VAK = V02-Vi2 = %0.2f \"%(VAK),\" volts\"\n", + "\n", + "#NOTE:correct value of IL2=0.58 mA but in book given as 0.592mA" + ] + } + ], + "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/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch3.ipynb b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch3.ipynb new file mode 100644 index 00000000..664bb97e --- /dev/null +++ b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch3.ipynb @@ -0,0 +1,700 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3 - Bipolar Junction Transistor" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_1 Page No. 64" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vcc = 15.00 volts\n", + "VBB = 1.00 volts\n", + "VBE = 0.70 volts\n", + "resistance,RB = 5000.00 ohm\n", + "resistance,RL = 650.00 ohm\n", + "Gain,Bf = 200.00 \n", + "IB =(VBB-VBE)/RB = 6.00e-05 ampere\n", + "IC =IB*Bf= 0.01 ampere\n", + "IE = IB+IC=0.01 ampere\n", + "VCE =Vcc-IC*RL= 7.20 volts\n", + "VCB = VCE-VBE=6.50 volts\n", + "resistance,RB =(Vcc-VBE)/IB= 2.38e+05 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "Vcc=15\n", + "print \"Vcc = %0.2f\"%(Vcc),\" volts\" #initialization\n", + "VBB=1\n", + "print \"VBB = %0.2f\"%(VBB),\" volts\" #initialization\n", + "VBE=0.7\n", + "print \"VBE = %0.2f\"%(VBE),\" volts\" #initialization\n", + "RB=5*(10**3)\n", + "print \"resistance,RB = %0.2f\"%(RB)+ \" ohm\" #initialization\n", + "RL=650\n", + "print \"resistance,RL = %0.2f\"%(RL)+ \" ohm\" #initialization\n", + "Bf=200\n", + "print \"Gain,Bf = %0.2f\"%(Bf)+ \" \" #initialization\n", + "IB=(VBB-VBE)/RB #Formulae\n", + "print \"IB =(VBB-VBE)/RB = %0.2e\"%(IB),\" ampere\" #calculation\n", + "IC=IB*Bf #Formulae\n", + "print \"IC =IB*Bf= %0.2f\"%(IC),\" ampere\"#calculation\n", + "IE=IB+IC #Formulae\n", + "print \"IE = IB+IC=%0.2f\"%(IE),\" ampere\"#calculation\n", + "VCE=Vcc-IC*RL #Formulae\n", + "print \"VCE =Vcc-IC*RL= %0.2f\"%(VCE),\" volts\" #calculation\n", + "VCB=VCE-VBE #Formulae\n", + "print \"VCB = VCE-VBE=%0.2f\"%(VCB),\" volts\"#calculation\n", + "RB=(Vcc-VBE)/IB #Formulae\n", + "print \"resistance,RB =(Vcc-VBE)/IB= %0.2e\"%(RB)+ \" ohm\" #calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_2 Page No. 65" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vbe1 = 0.03 volts\n", + "Vbe2 = -0.03 volts\n", + "ib1 = 2.00e-05 ampere\n", + "ib2 = -2.00e-05 ampere\n", + "IBQ = 6.00e-05 ampere\n", + "ICP = 0.02 ampere\n", + "ICR = 0.01 ampere\n", + "VCEP = 5.00 volts\n", + "VCER = 9.00 volts\n", + "change_IC = 0.01 ampere\n", + "change_VCE = 4.00 volts\n", + "AV = 80.00 \n", + "AI = 175.00 \n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "Vbe1=0.025\n", + "print \"Vbe1 = %0.2f\"%(Vbe1),\" volts\" # value of base-emitter voltage1\n", + "Vbe2=(-0.025)\n", + "print \"Vbe2 = %0.2f\"%(Vbe2),\" volts\" # value of base-emitter voltage2\n", + "ib1=20*10**(-6)\n", + "print \"ib1 = %0.2e\"%(ib1),\" ampere\" # value of base current1\n", + "ib2=(-20*10**(-6))\n", + "print \"ib2 = %0.2e\"%(ib2),\" ampere\"# value of base current2\n", + "IBQ=60*10**(-6)\n", + "print \"IBQ = %0.2e\"%(IBQ),\" ampere\" # operating point\n", + "ICP=15.5*10**(-3)\n", + "print \"ICP = %0.2f\"%(ICP),\" ampere\" # initialization\n", + "ICR=8.5*10**(-3)\n", + "print \"ICR = %0.2f\"%(ICR),\" ampere\" # initialization\n", + "VCEP=5\n", + "print \"VCEP = %0.2f\"%(VCEP),\" volts\" # value of collector-emitter voltage1\n", + "VCER=9\n", + "print \"VCER = %0.2f\"%(VCER),\" volts\" # value of collector-emitter voltage2\n", + "change_IC=ICP-ICR #change in collector current\n", + "print \"change_IC = %0.2f\"%(change_IC),\" ampere\"\n", + "change_VCE=VCER-VCEP #change in collector voltage\n", + "print \"change_VCE = %0.2f\"%(change_VCE),\" volts\" \n", + "change_VBE=Vbe1-Vbe2\n", + "change_IB=ib1-ib2\n", + "AV=(change_VCE/change_VBE) #formulae voltage gain\n", + "print \"AV = %0.2f\"%(AV),\" \"#voltage gain\n", + "AI=change_IC/change_IB #formulae current gain\n", + "print \"AI = %0.2f\"%(AI),\" \"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_3 Page No. 68" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ICQ = 0.01 ampere\n", + "B = 200.00 \n", + "capacitance,Cbe = 1.00e-10 F \n", + "VT = 0.03 volts\n", + "gm =(ICQ/VT)= 0.46 A/V\n", + "hie =(B/gm)= 433.33 ohm\n", + "fT =((1/2)*(gm/Cbe)*(1/pi))= 7.35e+08 hertz\n" + ] + } + ], + "source": [ + "from math import pi\n", + "from __future__ import division \n", + "ICQ=12*10**(-3)\n", + "print \"ICQ = %0.2f\"%(ICQ),\" ampere\" # collector current\n", + "B=200\n", + "print \"B = %0.2f\"%(B),\" \" #BJT gain\n", + "Cbe=100*10**(-12)# capacitance\n", + "print \"capacitance,Cbe = %0.2e\"%(Cbe),\" F \"\n", + "VT=26*10**(-3)\n", + "print \"VT = %0.2f\"%(VT),\" volts\" # thermal voltage\n", + "gm=(ICQ/VT) #transconductance\n", + "print \"gm =(ICQ/VT)= %0.2f\"%(gm),\" A/V\"\n", + "hie=B/gm #forward resistance hybrid parameter\n", + "print \"hie =(B/gm)= %0.2f\"%(hie),\" ohm\"\n", + "fT=((1/2)*(gm/Cbe)*(1/pi)) #unity gain frequency formulae\n", + "print \"fT =((1/2)*(gm/Cbe)*(1/pi))= %0.2e\"%(fT),\" hertz\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_4 Page No. 71" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VCC = 20.00 volts\n", + "RL= 5000.00 ohm\n", + "RB = 9.65e+05 ohm\n", + "VBE = 0.70 volts\n", + "BF = 50.00 \n", + "ICO = 1.00e-08 ampere\n", + "Vi = 0.00e+00 volts\n", + "IBQ = 2.00e-05 ampere\n", + "ICQ =BF*IBQ= 1.00e-03 ampere\n", + "VCEQ =VCC-ICQ*RL = 15.00 volts\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "VCC=20\n", + "print \"VCC = %0.2f\"%(VCC),\" volts\" # collector supply voltage \n", + "VBB=VCC\n", + "RL=5*(10**3)\n", + "print \"RL= %0.2f\"%(RL)+ \" ohm\" #resistance\n", + "RB=965*(10**3)\n", + "print \"RB = %0.2e\"%(RB)+ \" ohm\" #initialization base resistance\n", + "VBE=(0.7)\n", + "print \"VBE = %0.2f\"%(VBE),\" volts\" # value of base-emitter voltage\n", + "BF=50\n", + "print \"BF = %0.2f\"%(BF),\" \" #BJT gain\n", + "ICO=10*10**(-9)\n", + "print \"ICO = %0.2e\"%(ICO),\" ampere\" # collector reverse bias current\n", + "Vi=0\n", + "print \"Vi = %0.2e\"%(Vi),\" volts\" # value of input\n", + "IBQ=(VCC-VBE)/RB #base current as operating point\n", + "print \"IBQ = %0.2e\"%(IBQ),\" ampere\"\n", + "ICQ=BF*IBQ #operating point (collector current)\n", + "print \"ICQ =BF*IBQ= %0.2e\"%(ICQ),\" ampere\" # calculation\n", + "VCEQ=VCC-ICQ*RL # collector-emitter voltage as operating point\n", + "print \"VCEQ =VCC-ICQ*RL = %0.2f\"%(VCEQ),\" volts\" #calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_5 Page No. 74" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "BF1 = 100.00 \n", + "VCC = 20.00 volts\n", + "resistance,RL= 5.00e+03 ohm\n", + "resistance,RB = 9.65e+05 ohm\n", + "VBE = 0.70 volts\n", + "ICO = 1.00e-08 ampere\n", + "Vi = 0 volts\n", + "IBQ = 2.00e-05 ampere\n", + "ICQ1 =BF1*IBQ= 2.00e-03 ampere\n", + "VCEQ1 =VCC-ICQ1*RL = 10.00 volts\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "BF1=100\n", + "print \"BF1 = %0.2f\"%(BF1),\" \" #BJT gain\n", + "VCC=20\n", + "print \"VCC = %0.2f\"%(VCC),\" volts\" # collector supply voltage \n", + "VBB=VCC\n", + "RL=5*(10**3)\n", + "print \"resistance,RL= %0.2e\"%(RL)+ \" ohm\" #initialization\n", + "RB=965*(10**3)\n", + "print \"resistance,RB = %0.2e\"%(RB)+ \" ohm\" #initialization\n", + "VBE=(0.7)\n", + "print \"VBE = %0.2f\"%(VBE),\" volts\" # value of base-emitter voltage\n", + "ICO=10*10**(-9)\n", + "print \"ICO = %0.2e\"%(ICO),\" ampere\" # collector reverse bias current\n", + "Vi=0\n", + "print \"Vi = %0.f\"%(Vi),\" volts\" # value of input\n", + "IBQ=(VCC-VBE)/RB #base current as operating point\n", + "print \"IBQ = %0.2e\"%(IBQ),\" ampere\"\n", + "ICQ1=BF1*IBQ #operating point (collector current)\n", + "print \"ICQ1 =BF1*IBQ= %0.2e\"%(ICQ1),\" ampere\" # calculation\n", + "VCEQ1=VCC-ICQ1*RL # collector-emitter voltage as operating point\n", + "print \"VCEQ1 =VCC-ICQ1*RL = %0.2f\"%(VCEQ1),\" volts\" #calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_6 Page No. 75" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VBE2 = 0.50 volts\n", + "VCC = 20.00 volts\n", + "BF2 = 150.00 \n", + "ICO2 = 5.00e-07 ampere\n", + "RB = 9.65e+05 ohm\n", + "RL= 5.00e+03 ohm\n", + "IBQ2 = (VCC-VBE2)/RB=2.02e-05 ampere\n", + "ICQ2 =BF2*IBQ2= 3.03e-03 ampere\n", + "dICQ2 =BF2*ICO2= 7.50e-05 ampere\n", + "ICQ3 =ICQ2+dICQ2= 3.11e-03 ampere\n", + "VCEQ3 =VCC-ICQ3*RL = 4.47 volts\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "VBE2=(0.5)\n", + "print \"VBE2 = %0.2f\"%(VBE2),\" volts\" # value of base-emitter voltage\n", + "VCC=20\n", + "print \"VCC = %0.2f\"%(VCC),\" volts\" # collector supply voltage \n", + "BF2=150\n", + "print \"BF2 = %0.2f\"%(BF2),\" \" #BJT gain\n", + "ICO2=500*10**(-9)\n", + "print \"ICO2 = %0.2e\"%(ICO2),\" ampere\" # collector reverse bias current\n", + "RB=965*(10**3)\n", + "print \"RB = %0.2e\"%(RB)+ \" ohm\" #initialization base resistance\n", + "RL=5*(10**3)\n", + "print \"RL= %0.2e\"%(RL)+ \" ohm\" # load resistance\n", + "IBQ2=(VCC-VBE2)/RB #base current as operating point\n", + "print \"IBQ2 = (VCC-VBE2)/RB=%0.2e\"%(IBQ2),\" ampere\"\n", + "ICQ2=(BF2*IBQ2) #operating point (collector current)\n", + "print \"ICQ2 =BF2*IBQ2= %0.2e\"%(ICQ2),\" ampere\" # \n", + "dICQ2=BF2*ICO2 # increase in reverse bias current\n", + "print \"dICQ2 =BF2*ICO2= %0.2e\"%(dICQ2),\" ampere\" # \n", + "ICQ3=ICQ2+dICQ2\n", + "print \"ICQ3 =ICQ2+dICQ2= %0.2e\"%(ICQ3),\" ampere\" # calculation\n", + "VCEQ3=VCC-ICQ3*RL # collector-emitter voltage as operating point\n", + "print \"VCEQ3 =VCC-ICQ3*RL = %0.2f\"%(VCEQ3),\" volts\" #calculation\n", + "#NOTE: Calculated ans for VCEQ3=4.4695596 volts but in book it is given as 4.625volts(due to approximations done in) \n", + " " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_7 Page No. 76" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VCC = 20.00 volts\n", + "resistance,RL= 5000.00 ohm\n", + "resistance,R1 = 90000.00 ohm\n", + "resistance,R2 = 10000.00 ohm\n", + "resistance,Rc = 1000.00 ohm\n", + "VBEmax = 0.70 volts\n", + "VBEmin = 0.50 volts\n", + "BFmax = 150.00 \n", + "BFmin = 50.00 \n", + "ICOmax = 5.00e-07 ampere\n", + "ICOmin = 1.00e-08 ampere\n", + "VBB = 2.00 volts\n", + "RB = (R1*R2)/(R1+R2)=9000.00 ohm\n", + "ICmin = 0.00 ampere\n", + "VCEQmax =VCC-ICmin*RL = 14.58 volts\n", + "ICmax = 1.41e-03 ampere\n", + "VCEQmin =VCC-ICmax*RL = 12.95 volts\n", + "change_IC= 3.28e-04 ampere\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "VCC=20\n", + "print \"VCC = %0.2f\"%(VCC),\" volts\" # collector supply voltage \n", + "RL=5*(10**3)\n", + "print \"resistance,RL= %0.2f\"%(RL)+ \" ohm\" #initialization\n", + "R1=90*(10**3)\n", + "print \"resistance,R1 = %0.2f\"%(R1)+ \" ohm\" #initialization\n", + "R2=10*(10**3)\n", + "print \"resistance,R2 = %0.2f\"%(R2)+ \" ohm\" #initialization \n", + "Rc=1*(10**3)\n", + "print \"resistance,Rc = %0.2f\"%(Rc)+ \" ohm\" # resistance at collector\n", + "VBEmax=(0.7)\n", + "print \"VBEmax = %0.2f\"%(VBEmax),\" volts\" # maximum base-emitter voltage\n", + "VBEmin=(0.5)\n", + "print \"VBEmin = %0.2f\"%(VBEmin),\" volts\" # minimum base-emitter voltage\n", + "BFmax=150\n", + "print \"BFmax = %0.2f\"%(BFmax),\" \" #BJT gain maximum\n", + "BFmin=50\n", + "print \"BFmin = %0.2f\"%(BFmin),\" \" #BJT gain minimum\n", + "ICOmax=500*10**(-9)\n", + "print \"ICOmax = %0.2e\"%(ICOmax),\" ampere\" # maximum collector reverse bias current\n", + "ICOmin=10*10**(-9)\n", + "print \"ICOmin = %0.2e\"%(ICOmin),\" ampere\" # minimum collector reverse bias current\n", + "VBB=(VCC*R2)/(R1+R2)\n", + "print \"VBB = %0.2f\"%(VBB),\" volts\" # Base supply voltage \n", + "RB=(R1*R2)/(R1+R2)\n", + "print \"RB = (R1*R2)/(R1+R2)=%0.2f\"%(RB)+ \" ohm\" # eqivalent base resistance\n", + "ICmin=((BFmin*(VBB-VBEmax)+(RB+Rc)*(1+BFmin)*ICOmin)/(RB+Rc*(1+BFmin))) # minimum collector current\n", + "print \"ICmin = %0.2f\"%(ICmin),\" ampere\"\n", + "VCEQmax=VCC-ICmin*RL # maximum collector-emitter voltage (d.c value)\n", + "print \"VCEQmax =VCC-ICmin*RL = %0.2f\"%(VCEQmax),\" volts\" #calculation\n", + "ICmax=((BFmax*(VBB-VBEmin)+(RB+Rc)*(1+BFmax)*ICOmax)/(RB+Rc*(1+BFmax))) # maximum collector current\n", + "print \"ICmax = %0.2e\"%(ICmax),\" ampere\"\n", + "VCEQmin=VCC-ICmax*RL # minimum collector-emitter voltage (d.c value)\n", + "print \"VCEQmin =VCC-ICmax*RL = %0.2f\"%(VCEQmin),\" volts\" #calculation\n", + "change_IC=ICmax-ICmin\n", + "print \"change_IC= %0.2e\"%(change_IC),\" ampere\" # extreme variation in collector current\n", + "# ERROR - NOTE: Extreme variation in collector current given in book is 0.397 mA but calculated correct ans is 0.3276 mA \n", + " " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_8 Page No. 79" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VCC = 20.00 volts\n", + "RL= 2000.00 ohm\n", + "R1 =R2= 100000.00 ohm\n", + "VBE = 0.70 volts\n", + "BF = 100.00 \n", + "VBB = 10.00 volts\n", + "RB = (R1*R2)/(R1+R2)=50000.00 ohm\n", + "IC = 3.69e-03 ampere\n", + "VE = 7.38 volts\n", + "VB = 8.08 volts\n", + "VCB = 11.92 volts\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "VCC=20\n", + "print \"VCC = %0.2f\"%(VCC),\" volts\" # collector supply voltage \n", + "RL=2*(10**3)\n", + "print \"RL= %0.2f\"%(RL)+ \" ohm\" #resistance\n", + "R1=100*(10**3)\n", + "R2=R1\n", + "print \"R1 =R2= %0.2f\"%(R1)+ \" ohm\" #resistance\n", + "VBE=(0.7)\n", + "print \"VBE = %0.2f\"%(VBE),\" volts\" # base-emitter voltage\n", + "BF=100\n", + "print \"BF = %0.2f\"%(BF),\" \" #BJT gain\n", + "ICO=0\n", + "VBB=(VCC*R2)/(R1+R2)\n", + "print \"VBB = %0.2f\"%(VBB),\" volts\" # Base supply voltage \n", + "RB=(R1*R2)/(R1+R2)\n", + "print \"RB = (R1*R2)/(R1+R2)=%0.2f\"%(RB)+ \" ohm\" # eqivalent base resistance\n", + "IC=((BF*(VBB-VBE))/(RB+RL*(1+BF))) # collector current\n", + "print \"IC = %0.2e\"%(IC),\" ampere\"\n", + "VE=IC*RL\n", + "print \"VE = %0.2f\"%(VE),\" volts\" # emitter voltage\n", + "VB=VBE+VE\n", + "print \"VB = %0.2f\"%(VB),\" volts\" # base voltage\n", + "VCB=VCC-VB\n", + "print \"VCB = %0.2f\"%(VCB),\" volts\" # collector-base voltage\n", + "#hence BJT in active region." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_9 Page No. 84" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VCC = 5.00 volts\n", + "RL= 250.00 ohm\n", + "RB =25000.00 ohm\n", + "VCS = 0.20 volts\n", + "BF = 200.00 \n", + "VBS = 0.80 volts\n", + "VI = 5.00 volts\n", + "VCON = 0.30 volts\n", + "ICON = (VCC-VCON)/RL=0.02 ampere\n", + "IBON = (ICON)/BF=9.40e-05 ampere\n", + "IBS = (VI-VBS)/RB=1.68e-04 ampere\n", + "ICS = (VCC-VCS)/RL=1.92e-02 ampere\n", + "Bforced = ICS/IBS=114.29 \n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "VCC=5\n", + "print \"VCC = %0.2f\"%(VCC),\" volts\" # collector supply voltage \n", + "RL=250\n", + "print \"RL= %0.2f\"%(RL)+ \" ohm\" #initialization\n", + "RB=25*10**(3)\n", + "print \"RB =%0.2f\"%(RB)+ \" ohm\" # base resistance\n", + "VCS=(0.2)\n", + "print \"VCS = %0.2f\"%(VCS),\" volts\" # voltage\n", + "BF=200\n", + "print \"BF = %0.2f\"%(BF),\" \" #BJT gain\n", + "VBS=(0.8)\n", + "print \"VBS = %0.2f\"%(VBS),\" volts\" # base-emitter voltage for BJT switch\n", + "VI=5\n", + "print \"VI = %0.2f\"%(VI),\" volts\"# input voltage\n", + "VCON=0.3\n", + "print \"VCON = %0.2f\"%(VCON),\" volts\"\n", + "ICON=(VCC-VCON)/RL\n", + "print \"ICON = (VCC-VCON)/RL=%0.2f\"%(ICON),\" ampere\"#collector current for saturated BJT\n", + "IBON=(ICON)/BF\n", + "print \"IBON = (ICON)/BF=%0.2e\"%(IBON),\" ampere\"#Base current for saturated BJT\n", + "IBS=(VI-VBS)/RB\n", + "print \"IBS = (VI-VBS)/RB=%0.2e\"%(IBS),\" ampere\"#Base-emitter current for saturated BJT\n", + "ICS=(VCC-VCS)/RL\n", + "print \"ICS = (VCC-VCS)/RL=%0.2e\"%(ICS),\" ampere\"#Collector-emitter current for saturated BJT\n", + "Bforced=ICS/IBS\n", + "print \"Bforced = ICS/IBS=%0.2f\"%(Bforced),\" \" #BJT forced gain\n", + "#IBS>>IBON hence BJT in saturation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_10 Page No. 85" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "TJmax= 175.00 degree celsius\n", + "theta= 0.50 degree celsius/mW \n", + "at 25 degree celsius,PDmax=(TJmax-25 )/theta = 300.00 mW \n", + "at 75 degree celsius,PDmax2= (TJmax-75)/theta = 200.00 mW \n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "TJmax=175\n", + "print \"TJmax= %0.2f\"%(TJmax),\"degree celsius\" #maximum allowed junction temperature\n", + "theta=0.5\n", + "print \"theta= %0.2f\"%(theta),\"degree celsius/mW \" #thermal resistances b/w junction to ambient\n", + "change_T=TJmax-25#temperature difference\n", + "PDmax=change_T/theta\n", + "print \"at 25 degree celsius,PDmax=(TJmax-25 )/theta = %0.2f\"%(PDmax)+ \" mW \" #maximum allowed power dissipation at TA=25 degree celsius\n", + "change_T=TJmax-75\n", + "PDmax2=change_T/theta\n", + "print \"at 75 degree celsius,PDmax2= (TJmax-75)/theta = %0.2f\"%(PDmax2)+ \" mW \" #maximum allowed power dissipation at TA=75 degree celsius" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_11 Page No. 85" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "TJmax= 175.00 degree celsius\n", + "theta= 0.10 degree celsius/mW \n", + "at 25 degree celsius,PDmax=(TJmax-25 )/theta = 1500.00 mW \n", + "at 75 degree celsius,PDmax= (TJmax-75)/theta = 1000.00 mW \n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "TJmax=175\n", + "print \"TJmax= %0.2f\"%(TJmax),\" degree celsius\" #maximum allowed junction temperature\n", + "theta=0.1\n", + "print \"theta= %0.2f\"%(theta),\" degree celsius/mW \" #thermal resistances b/w junction to ambient\n", + "change_T=TJmax-25 #temperature difference\n", + "PDmax=change_T/theta\n", + "print \"at 25 degree celsius,PDmax=(TJmax-25 )/theta = %0.2f\"%(PDmax)+ \" mW \" #maximum allowed power dissipation at TA=25 degree celsius\n", + "change_T=TJmax-75 #temperature difference\n", + "PDmax=change_T/theta\n", + "print \"at 75 degree celsius,PDmax= (TJmax-75)/theta = %0.2f\"%(PDmax)+ \" mW \" #maximum allowed power dissipation at TA=75 degree celsius" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3_12 Page No. 86" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VBE = 0.70 volts\n", + "VCC = 10.00 volts\n", + "IREF =IQ= 0.01 ampere\n", + "R=(VCC-VBE)/(IREF)= 1860.00 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "VBE=(0.7)\n", + "print \"VBE = %0.2f\"%(VBE),\" volts\" # value of base-emitter voltage\n", + "VCC=10\n", + "print \"VCC = %0.2f\"%(VCC),\" volts\" # collector supply voltage \n", + "IREF=5*10**(-3)\n", + "print \"IREF =IQ= %0.2f\"%(IREF),\" ampere\" # current mirror source current\n", + "R=(VCC-VBE)/(IREF)# formulae\n", + "print \"R=(VCC-VBE)/(IREF)= %0.2f\"%(R)+ \" ohm\" #resistance" + ] + } + ], + "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/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch4.ipynb b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch4.ipynb new file mode 100644 index 00000000..56fa8f97 --- /dev/null +++ b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch4.ipynb @@ -0,0 +1,936 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4 - Field Effect Transistors" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4_1 Page No. 98" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "IDSS = 0.01 ampere\n", + "VP= -4.00 volts\n", + "VGS = -2.00 volts\n", + "VDSmin =VGS-VP=2.00 volts\n", + "ID =IDSS*(1-VGS/VP)**2= 2.50e-03 ampere\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "IDSS=10*10**(-3)\n", + "print \"IDSS = %0.2f\"%(IDSS),\" ampere\" # maximum drain current\n", + "VP=(-4)\n", + "print \"VP= %0.2f\"%(VP),\" volts\" # pinch off voltage \n", + "VGS=(-2)\n", + "print \"VGS = %0.2f\"%(VGS),\" volts\" # gate to source voltage \n", + "VDSmin=VGS-VP\n", + "print \"VDSmin =VGS-VP=%0.2f\"%(VDSmin),\" volts\" # Drain to source voltage \n", + "ID=IDSS*(1-VGS/VP)**2\n", + "print \"ID =IDSS*(1-VGS/VP)**2= %0.2e\"%(ID),\" ampere\" # drain current" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4_2 Page No. 99" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "IDSS = 0.01 ampere\n", + "VP= -4.00 volts\n", + "VGS = 0.00 volts\n", + "RDS = 1/[(2*(IDSS/(-VP)))*(1-VGS/VP)]=200.00 ohm\n", + "VGS = -2.00 volts\n", + "RDS = 1/[(2*(IDSS/(-VP)))*(1-VGS/VP)]=400.00 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "IDSS=10*10**(-3)\n", + "print \"IDSS = %0.2f\"%(IDSS),\" ampere\" # maximum drain current\n", + "VP=(-4)\n", + "print \"VP= %0.2f\"%(VP),\" volts\" # pinch off voltage \n", + "VGS=(0)\n", + "print \"VGS = %0.2f\"%(VGS),\" volts\" # gate to source voltage1 \n", + "RDS=1/((2*(IDSS/(-VP)))*(1-VGS/VP))#formula for JFET\n", + "print \"RDS = 1/[(2*(IDSS/(-VP)))*(1-VGS/VP)]=%0.2f\"%(RDS),\" ohm\" # drain to source resistance for VGS=0V\n", + "VGS=(-2)\n", + "print \"VGS = %0.2f\"%(VGS),\" volts\" # gate to source voltage2 \n", + "RDS=1/((2*(IDSS/(-VP)))*(1-VGS/VP))\n", + "print \"RDS = 1/[(2*(IDSS/(-VP)))*(1-VGS/VP)]=%0.2f\"%(RDS),\" ohm\" # drain to source resistance for VGS=(-2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4_3 Page No. 104" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ID = 0.01 ampere\n", + "VDD= 24.00 volts\n", + "VT= 5.00 volts\n", + "VGS= 10.00 volts\n", + "KF = ID/(VGS-VT)**2 = 4.00e-04 A/V**2\n", + "VDS =VGS= 7.00 volts\n", + "ID =KF*(VGS-VT)**2= 1.60e-03 ampere\n", + "RL=(VDD-VDS)/ID= 1.06e+04 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "ID=10*10**(-3)\n", + "print \"ID = %0.2f\"%(ID),\" ampere\" # given drain current \n", + "VDD=(24)\n", + "print \"VDD= %0.2f\"%(VDD),\" volts\" # Drain voltage \n", + "VT=(5)\n", + "print \"VT= %0.2f\"%(VT),\" volts\" # Threshold voltage \n", + "VGS=(10)\n", + "print \"VGS= %0.2f\"%(VGS),\" volts\" # gate to source voltage1 \n", + "KF=ID/(VGS-VT)**2\n", + "print \"KF = ID/(VGS-VT)**2 = %0.2e\"%(KF),\" A/V**2\" # To calculate Scale factor for finding ID2\n", + "VDS=(7)\n", + "print \"VDS =VGS= %0.2f\"%(VDS),\" volts\" # drain to source voltage \n", + "VGS=(VDS)\n", + "ID=KF*(VGS-VT)**2\n", + "print \"ID =KF*(VGS-VT)**2= %0.2e\"%(ID),\" ampere\" # New drain current for VDS=24V\n", + "RL=(VDD-VDS)/ID\n", + "print \"RL=(VDD-VDS)/ID= %0.2e\"%(RL)+ \" ohm\" #calculation for load resistance at VDS=24V" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4_4 Page No. 105" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "part(i) \n", + "IDSS = 0.01 ampere\n", + "VP= 2.00 volts\n", + "IDQ = 0.00 ampere\n", + "gm =[(2)*sqrt(IDQ*IDSS)]/VP= 4.70e-03 A/V\n", + "part(ii) \n", + "IDQ = 0.01 ampere\n", + "IDSS = 0.01 ampere\n", + "VP= 6.00 volts\n", + "gm =[(2)*sqrt(IDQ*IDSS)]/VP= 3.17e-03 A/V\n", + "part(iii) \n", + "IDQ = 1.00e-03 ampere\n", + "KF = 2.50e-04 A/V**2\n", + "gm =sqrt(4*IDQ*KF)= 1.00e-03 A/V\n", + "part(iv) \n", + "IDQ = 9.10e-04 ampere\n", + "KF = 3.75e-04 A/V**2\n", + "gm =sqrt(4*IDQ*KF)= 1.17e-03 A/V\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "from __future__ import division \n", + "print \"part(i) \"# part(i) of this question\n", + "IDSS=5*10**(-3)\n", + "print \"IDSS = %0.2f\"%(IDSS),\" ampere\" # maximum drain current JFET 1\n", + "VP=(2)\n", + "print \"VP= %0.2f\"%(VP),\" volts\" # pinch off voltage for JFET 1\n", + "IDQ=4.42*10**(-3)\n", + "print \"IDQ = %0.2f\"%(IDQ),\" ampere\" # drain current for JFET 1\n", + "gm=((2)*sqrt(IDQ*IDSS))/VP\n", + "print \"gm =[(2)*sqrt(IDQ*IDSS)]/VP= %0.2e\"%(gm),\" A/V\"# calculating transconductance for JFET with IDQ = 4.42 mA\n", + "\n", + "print \"part(ii) \"# part(ii) of this question\n", + "IDQ=6.04*10**(-3)\n", + "print \"IDQ = %0.2f\"%(IDQ),\" ampere\" # drain current for JFET 1\n", + "IDSS=15*10**(-3)\n", + "print \"IDSS = %0.2f\"%(IDSS),\" ampere\" # maximum drain current JFET2\n", + "VP=(6)\n", + "print \"VP= %0.2f\"%(VP),\" volts\" # pinch off voltage for JFET2 \n", + "gm=((2)*sqrt(IDQ*IDSS))/VP\n", + "print \"gm =[(2)*sqrt(IDQ*IDSS)]/VP= %0.2e\"%(gm),\" A/V\"# calculating transconductance for JFET with IDQ = 6.04 mA\n", + "\n", + "print \"part(iii) \"# part(iii) of this question\n", + "IDQ=1*10**(-3)\n", + "print \"IDQ = %0.2e\"%(IDQ),\" ampere\" # drain current for EMOSFET 1\n", + "KF=0.25*10**(-3)\n", + "print \"KF = %0.2e\"%(KF),\" A/V**2\" # Scale factor for finding EMOSFET1\n", + "gm=sqrt(4*IDQ*KF)\n", + "print \"gm =sqrt(4*IDQ*KF)= %0.2e\"%(gm),\" A/V\"# calculating transconductance for EMOSFET1 with IDQ = 1 mA\n", + "\n", + "print \"part(iv) \"# part(iv) of this question\n", + "IDQ=0.91*10**(-3)\n", + "print \"IDQ = %0.2e\"%(IDQ),\" ampere\" # drain current for EMOSFET 2\n", + "KF=0.375*10**(-3)\n", + "print \"KF = %0.2e\"%(KF),\" A/V**2\" # Scale factor for finding EMOSFET2\n", + "gm=sqrt(4*IDQ*KF)\n", + "print \"gm =sqrt(4*IDQ*KF)= %0.2e\"%(gm),\" A/V\"# calculating transconductance for EMOSFET2 with IDQ = 0.91 mA" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4_5 Page No. 106" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "IDQmax = 0.01 ampere\n", + "IDQmin = 0.00 ampere\n", + "VDD= 20.00 volts\n", + "VDSmin= 6.00 volts\n", + "ID = 2.40e-03 ampere\n", + "VGG= 3.00 volts\n", + "Ri= 1.00e+05 ohm\n", + "RF= (VGG-0)/(ID-0)= 1250.00 ohm\n", + "R1= VDD*Ri/VGG= 6.67e+05 ohm\n", + "R2= R1*VGG/(VDD-VGG)= 1.18e+05 ohm\n", + "RL=[((VDD-VDSmin)/IDmax)-RF]=1550.00 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "IDQmax=5*10**(-3)\n", + "print \"IDQmax = %0.2f\"%(IDQmax),\" ampere\" # drain current for JFET for maximum transfer characteristics\n", + "IDmax=IDQmax# maximum drain current will be given by IDQmax\n", + "IDQmin=3*10**(-3)\n", + "print \"IDQmin = %0.2f\"%(IDQmin),\" ampere\" # drain current for JFET for minimum transfer characteristics \n", + "VDD=(20)\n", + "print \"VDD= %0.2f\"%(VDD),\" volts\" # Drain voltage supply\n", + "VDSmin=(6)\n", + "print \"VDSmin= %0.2f\"%(VDSmin),\" volts\" # minimum Drain to source voltage supply\n", + "ID=2.4*10**(-3)\n", + "print \"ID = %0.2e\"%(ID),\" ampere\" # drain current chosen for operation within max and min limits \n", + "VGG=3\n", + "print \"VGG= %0.2f\"%(VGG),\" volts\" # Gate voltage from fig.\n", + "Ri=100*10**(3)\n", + "print \"Ri= %0.2e\"%(Ri)+ \" ohm\" #eqivalent input resistance\n", + "RF=(VGG-0)/(ID-0)\n", + "print \"RF= (VGG-0)/(ID-0)= %0.2f\"%(RF)+ \" ohm\" #calculation for feedback resistance \n", + "R1=VDD*Ri/VGG #using formulae VGG=VDD*Ri/R1\n", + "print \"R1= VDD*Ri/VGG= %0.2e\"%(R1)+ \" ohm\" #calculation for first resistance R1 at input side\n", + "R2=R1*VGG/(VDD-VGG)\n", + "print \"R2= R1*VGG/(VDD-VGG)= %0.2e\"%(R2)+ \" ohm\" #calculation for second resistance R2 at input side\n", + "RL=(((VDD-VDSmin)/IDmax)-RF) # using formulae VDD=IDmax(RL+RF)+VDSmin\n", + "print \"RL=[((VDD-VDSmin)/IDmax)-RF]=%0.2f\"%(RL)+ \" ohm\" #Load resistance calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4_6 Page No. 107" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "IDSS = 0.05 ampere\n", + "VP= -10.00 volts\n", + "VGSQ= -5.00 volts\n", + "ID =IDSS*(1-VGS/VP)**2= 0.01 ampere\n", + "RF= (VGSQ)/(ID)= 400.00 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "IDSS=50*10**(-3)\n", + "print \"IDSS = %0.2f\"%(IDSS),\" ampere\" # maximum drain current JFET \n", + "VP=(-10)\n", + "print \"VP= %0.2f\"%(VP),\" volts\" # pinch off voltage for JFET \n", + "VGSQ=(-5)\n", + "print \"VGSQ= %0.2f\"%(VGSQ),\" volts\" # Gate operating point voltage \n", + "ID=IDSS*(1-VGSQ/VP)**2\n", + "print \"ID =IDSS*(1-VGS/VP)**2= %0.2f\"%(ID),\" ampere\" # drain current JFET \n", + "RF=abs(VGSQ/ID) \n", + "print \"RF= (VGSQ)/(ID)= %0.2f\"%(RF)+ \" ohm\" #calculation for feedback resistance " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4_7 Page No. 108" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "IDSS = 0.01 ampere\n", + "RL= 910.00 ohm\n", + "RF= 2290.00 ohm\n", + "R1= 1.20e+07 ohm\n", + "R2= 8.57e+06 ohm\n", + "VDD= 24.00 volts\n", + "VP= -2.00 volts\n", + "VGG= VDD*R2/(R1+R2)=10.00 volts\n", + "Quadratic equation =5.244*ID**(2)-55.76*ID+144=0\n", + "ID = 0.00 ampere\n", + "Since ID <=IDSS, hence ID=6.214 mA cannot be chosen, so we chose ID=4.42 mA\n", + "IDQ =0.00 A\n", + "VGSQ = VGG-IDQ*RF = -0.12 volts\n", + "VDSQ= VDD-IDQ*(RL+RF)= 9.86 volts\n", + "VDGQ = VDSQ-VGSQ =9.98 volts\n", + "VDGQ >magnitude_VP,Hence FET is in pinch off region\n" + ] + } + ], + "source": [ + "from sympy import symbols,solve\n", + "from __future__ import division \n", + "IDSS=5*10**(-3)\n", + "print \"IDSS = %0.2f\"%(IDSS),\" ampere\" # maximum drain current JFET \n", + "RL=910\n", + "print \"RL= %0.2f\"%(RL)+ \" ohm\" #Load resistance\n", + "RF=2.29*10**(3) \n", + "print \"RF= %0.2f\"%(RF)+ \" ohm\" # feedback resistance \n", + "R1=12*10**(6)\n", + "print \"R1= %0.2e\"%(R1)+ \" ohm\" # first resistance R1 at input side\n", + "R2=8.57*10**(6)\n", + "print \"R2= %0.2e\"%(R2)+ \" ohm\" # second resistance R2 at input side\n", + "VDD=(24)\n", + "print \"VDD= %0.2f\"%(VDD),\" volts\" # Drain voltage supply\n", + "VP=(-2)\n", + "print \"VP= %0.2f\"%(VP),\" volts\" # pinch off voltage for JFET \n", + "VGG=(VDD*R2)/(R1+R2)\n", + "print \"VGG= VDD*R2/(R1+R2)=%0.2f\"%(VGG),\" volts\" # Gate voltage for JFET\n", + "print \"Quadratic equation =5.244*ID**(2)-55.76*ID+144=0\"# Forming Quadratic equation using VGS = VGG-ID*RF and ID = IDSS(1-VGS/VP)**2 where ID in mA\n", + "p = [5.244, -55.76, 144]\n", + "P=symbols('P')\n", + "ID=solve(p[0]*P**2+p[1]*P+p[2])[0]*10**(-3)# values of ID converted into Ampere by multiplying by 10**(-3)\n", + "print \"ID = %0.2f\"%(ID),\" ampere\" # drain current JFET \n", + "print \"Since ID <=IDSS, hence ID=6.214 mA cannot be chosen, so we chose ID=4.42 mA\"\n", + "IDQ=4.42*10**(-3) \n", + "print \"IDQ =%0.2f\"%(IDQ),\" A\"#Since ID <=IDSS, hence ID=6.214 mA cannot be chosen, so we chose ID=4.42 mA\n", + "VGSQ=VGG-IDQ*RF\n", + "print \"VGSQ = VGG-IDQ*RF = %0.2f\"%(VGSQ),\" volts\" # Gate operating point voltage \n", + "VDSQ=VDD-IDQ*(RL+RF)\n", + "print \"VDSQ= VDD-IDQ*(RL+RF)= %0.2f\"%(VDSQ),\" volts\" # Drain voltage for JFET\n", + "VDGQ=VDSQ-VGSQ\n", + "print \"VDGQ = VDSQ-VGSQ =%0.2f\"%(VDGQ),\" volts\" # Drain-Gate voltage for JFET\n", + "print \"VDGQ >magnitude_VP,Hence FET is in pinch off region\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4_8 Page No. 108" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "IDSS = 0.01 ampere\n", + "RL= 910.00 ohm\n", + "RF= 2290.00 ohm\n", + "R1= 1.20e+07 ohm\n", + "R2= 8.57e+06 ohm\n", + "VDD= 24.00 volts\n", + "VP= -6.00 volts\n", + "VGG= VDD*R2/(R1+R2)=10.00 volts\n", + "Quadratic equation =5.244*ID**(2)-75.68*ID+256=0\n", + "ID = 0.01 ampere\n", + "VDG= 9.08 volts\n", + "since VDG < magnitude_VP for ID=9.0189 mA which is inappropriate for JFET pinch off region ,hence ID=5.4128 mA is choosen !\n", + "IDQ =0.01 ampere\n", + "VGSQ = VGG-IDQ*RF = -2.39 volts\n", + "VDSQ= VDD-IDQ*(RL+RF)= 6.69 volts\n", + "VDGQ = VDSQ-VGSQ =9.08 volts\n", + "VDGQ > VP,Hence FET is in pinch off region\n" + ] + } + ], + "source": [ + "from sympy import symbols,solve\n", + "from __future__ import division \n", + "IDSS=15*10**(-3)\n", + "print \"IDSS = %0.2f\"%(IDSS),\" ampere\" # maximum drain current of JFET \n", + "RL=910\n", + "print \"RL= %0.2f\"%(RL)+ \" ohm\" #Load resistance\n", + "RF=2.29*10**(3) \n", + "print \"RF= %0.2f\"%(RF)+ \" ohm\" # feedback resistance \n", + "R1=12*10**(6)\n", + "print \"R1= %0.2e\"%(R1)+ \" ohm\" # first resistance R1 at input side\n", + "R2=8.57*10**(6)\n", + "print \"R2= %0.2e\"%(R2)+ \" ohm\" # second resistance R2 at input side\n", + "VDD=(24)\n", + "print \"VDD= %0.2f\"%(VDD),\" volts\" # Drain voltage supply\n", + "VP=(-6)\n", + "print \"VP= %0.2f\"%(VP),\" volts\" # pinch off voltage for JFET \n", + "VGG=(VDD*R2)/(R1+R2)\n", + "print \"VGG= VDD*R2/(R1+R2)=%0.2f\"%(VGG),\" volts\" # Gate voltage for JFET\n", + "print \"Quadratic equation =5.244*ID**(2)-75.68*ID+256=0\"# where ID in mA\n", + "p = [5.244, -75.68, 256]\n", + "P=symbols('P')\n", + "ID=solve(p[0]*P**2+p[1]*P+p[2])[0]*10**(-3)#values of ID converted into Ampere by multiplying by 10**(-3)\n", + "print \"ID = %0.2f\"%(ID),\" ampere\" # drain current JFET \n", + "VDG=VDD-(ID*RL)-VGG\n", + "print \"VDG= %0.2f\"%(VDG),\" volts\" # Drain-gate voltage for JFET\n", + "print \"since VDG < magnitude_VP for ID=9.0189 mA which is inappropriate for JFET pinch off region ,hence ID=5.4128 mA is choosen !\" \n", + "IDQ=5.41*10**(-3) # since VDG < magnitude_VP for ID=9.0189 mA which is inappropriate for JFET pinch off region ,hence ID=5.4128 mA is choosen !\n", + "print \"IDQ =%0.2f\"%(IDQ),\" ampere\"\n", + "VGSQ=VGG-IDQ*RF\n", + "print \"VGSQ = VGG-IDQ*RF = %0.2f\"%(VGSQ),\" volts\" # Gate operating point voltage \n", + "VDSQ=VDD-IDQ*(RL+RF)\n", + "print \"VDSQ= VDD-IDQ*(RL+RF)= %0.2f\"%(VDSQ),\" volts\" # Drain voltage for JFET\n", + "VDGQ=VDSQ-VGSQ\n", + "print \"VDGQ = VDSQ-VGSQ =%0.2f\"%(VDGQ),\" volts\" # Drain-Gate voltage for JFET\n", + "print \"VDGQ > VP,Hence FET is in pinch off region\"\n", + "\n", + "# NOTE :all values of book ans is wrong so give note-INCOMPLETE QUESTION\n", + "#Roots for drain current quadratic equation are wrong in the book thus value for VGSQ,VDSQ and VDGQ are also wrong" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4_9 Page No. 112" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "RL= 12000.00 ohm\n", + "RF= 6000.00 ohm\n", + "R1= 1.20e+07 ohm\n", + "R2= 8.57e+06 ohm\n", + "VDD= 24.00 volts\n", + "VT= 3.00 volts\n", + "KF= 0.00 A/V**2\n", + "VGG= VDD*R2/(R1+R2)=10.00 volts\n", + "Quadratic equation =9*ID**(2)-25*ID+16=0\n", + "ID = 0.00 A\n", + "VGS = VGG-ID*RF = 4.00 volts\n", + "Since VGS < VT for ID=1.78 mA, hence ID = 1.78 mA cannot be chosen, so we chose ID= 1 mA as operating drain current IDQ\n", + "IDQ =0.00 A\n", + "VGSQ = VGG-IDQ*RF = 4.00 volts\n", + "VDSQ= VDD-IDQ*(RL+RF)= 6.00 volts\n" + ] + } + ], + "source": [ + "from sympy import symbols,solve\n", + "from __future__ import division \n", + "RL=12*10**(3)\n", + "print \"RL= %0.2f\"%(RL)+ \" ohm\" #Load resistance\n", + "RF=6*10**(3) \n", + "print \"RF= %0.2f\"%(RF)+ \" ohm\" # feedback resistance \n", + "R1=12*10**(6)\n", + "print \"R1= %0.2e\"%(R1)+ \" ohm\" # first resistance R1 at input side\n", + "R2=8.57*10**(6)\n", + "print \"R2= %0.2e\"%(R2)+ \" ohm\" # second resistance R2 at input side\n", + "VDD=(24)\n", + "print \"VDD= %0.2f\"%(VDD),\" volts\" # Drain voltage supply\n", + "VT=(3)\n", + "print \"VT= %0.2f\"%(VT),\" volts\" # Threshold voltage for n-channel EMOSFET\n", + "KF=0.25*10**(-3)\n", + "print \"KF= %0.2f\"%(KF),\" A/V**2\" # Constant for n-channel EMOSFET \n", + "VGG=(VDD*R2)/(R1+R2)\n", + "print \"VGG= VDD*R2/(R1+R2)=%0.2f\"%(VGG),\" volts\" # Gate voltage for n-channel EMOSFET \n", + "print \"Quadratic equation =9*ID**(2)-25*ID+16=0\"# IDS=KF*(VGS-VT)**2 and VGS=VGG-ID*RD ,so Quadratic equation formed is :IDS=KF*(VGG-ID*RD-VT)**2 where ID in mA\n", + "p = [9, -25, 16]\n", + "P=symbols('P')\n", + "ID=solve(p[0]*P**2+p[1]*P+p[2])[0]*10**(-3)#values of ID converted into Ampere by multiplying by 10**(-3)\n", + "print \"ID = %0.2f\"%(ID),\" A\" # drain current n-channel EMOSFET in Ampere \n", + "VGS=VGG-ID*RF# For ID=1.78 mA and ID=1mA\n", + "print \"VGS = VGG-ID*RF = %0.2f\"%(VGS),\" volts\" # Gate operating point voltage \n", + "print \"Since VGS < VT for ID=1.78 mA, hence ID = 1.78 mA cannot be chosen, so we chose ID= 1 mA as operating drain current IDQ\"\n", + "IDQ=1*10**(-3)\n", + "print \"IDQ =%0.2f\"%(IDQ),\"A\"#Since VGS < VT for ID=1.78 mA, hence ID = 1.78 mA cannot be chosen, so we chose ID= 1 mA as operating drain current IDQ\n", + "VGSQ=VGG-IDQ*RF\n", + "print \"VGSQ = VGG-IDQ*RF = %0.2f\"%(VGSQ),\" volts\" # Gate operating point voltage \n", + "VDSQ=VDD-IDQ*(RL+RF)\n", + "print \"VDSQ= VDD-IDQ*(RL+RF)= %0.2f\"%(VDSQ),\" volts\" # Drain voltage for n-channel EMOSFET \n", + "# NOTE:Value of VGS= -0.6676390 volts for ID=1.78 mA but in book given as -0.68 V" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4_10 Page No. 113" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "RL= 12000.00 ohm\n", + "RF= 6000.00 ohm\n", + "R1= 1.20e+07 ohm\n", + "R2= 8.57e+06 ohm\n", + "VDD= 24.00 volts\n", + "VT= 3.00 volts\n", + "KF= 0.00 A/V**2\n", + "VGG= VDD*R2/(R1+R2)=10.00 volts\n", + "Quadratic equation =36*ID**(2)-86.67*ID+49=0\n", + "ID = 0.00 A\n", + "VGS = VGG-ID*RF = 4.56 volts\n", + "Since VGS < VT for ID=1.5 mA, hence ID = 1.5 mA cannot be chosen, so we chose ID= 0.91 mA as operating drain current IDQ\n", + "IDQ =0.00 A\n", + "change in IDQ = 9.00 percent\n", + "VGSQ = VGG-IDQ*RF = 4.54 volts\n", + "VDSQ= VDD-IDQ*(RL+RF)= 7.62 volts\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "from __future__ import division \n", + "RL=12*10**(3)\n", + "print \"RL= %0.2f\"%(RL)+ \" ohm\" #Load resistance\n", + "RF=6*10**(3) \n", + "print \"RF= %0.2f\"%(RF)+ \" ohm\" # feedback resistance \n", + "R1=12*10**(6)\n", + "print \"R1= %0.2e\"%(R1)+ \" ohm\" # first resistance R1 at input side\n", + "R2=8.57*10**(6)\n", + "print \"R2= %0.2e\"%(R2)+ \" ohm\" # second resistance R2 at input side\n", + "VDD=(24)\n", + "print \"VDD= %0.2f\"%(VDD),\" volts\" # Drain voltage supply\n", + "VT=(3)\n", + "print \"VT= %0.2f\"%(VT),\" volts\" # Threshold voltage for n-channel EMOSFET\n", + "KF=0.375*10**(-3)\n", + "print \"KF= %0.2f\"%(KF),\" A/V**2\" # Constant for n-channel EMOSFET \n", + "VGG=(VDD*R2)/(R1+R2)\n", + "print \"VGG= VDD*R2/(R1+R2)=%0.2f\"%(VGG),\" volts\" # Gate voltage for n-channel EMOSFET \n", + "print \"Quadratic equation =36*ID**(2)-86.67*ID+49=0\"# IDS=KF*(VGS-VT)**2 and VGS=VGG-ID*RD ,so Quadratic equation formed is :IDS=KF*(VGG-ID*RD-VT)**2 ,where ID in mA\n", + "p = [36, -86.67, 49]\n", + "P=symbols('P')\n", + "ID=solve(p[0]*P**2+p[1]*P+p[2])[0]*10**(-3)#values of ID converted into Ampere by multiplying by 10**(-3)\n", + "print \"ID = %0.2f\"%(ID),\" A\" # drain current n-channel EMOSFET in Ampere \n", + "VGS=VGG-ID*RF# Gate voltage for ID = 1.5 mA and ID = 0.91 mA\n", + "print \"VGS = VGG-ID*RF = %0.2f\"%(VGS),\" volts\" # Gate voltage \n", + "print \"Since VGS < VT for ID=1.5 mA, hence ID = 1.5 mA cannot be chosen, so we chose ID= 0.91 mA as operating drain current IDQ\"\n", + "IDQ=0.91*10**(-3)\n", + "print \"IDQ =%0.2f\"%(IDQ),\" A\"#Since VGS < VT for ID=1.5 mA, hence ID = 1.5 mA cannot be chosen, so we chose ID= 0.91 mA as operating drain current IDQ\n", + "change_IDQ=((1-0.91)*100)/(1)# \n", + "print \"change in IDQ = %0.2f\"%(change_IDQ),\" percent\"# Percent change in IDQ from value 1 mA from its actual value IDQ=0.91mA\n", + "VGSQ=VGG-IDQ*RF\n", + "print \"VGSQ = VGG-IDQ*RF = %0.2f\"%(VGSQ),\" volts\" # Gate operating point voltage \n", + "VDSQ=VDD-IDQ*(RL+RF)\n", + "print \"VDSQ= VDD-IDQ*(RL+RF)= %0.2f\"%(VDSQ),\" volts\" # Drain voltage for n-channel EMOSFET \n", + "\n", + "\n", + "# note: in the textbook author has given KF = .375 but I have work with KF = .375*10**-3A/V**2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4_11 Page No. 114" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "RF= 6000.00 ohm\n", + "VDD= 20.00 volts\n", + "part(i) \n", + "VT= 2.00 volts\n", + "KF= 0.00 A/V**2\n", + "ID = 0.00 A\n", + "RL=[VDD-VT-sqrt(ID/KF)]/ID= 16000.00 ohm\n", + "part(ii) \n", + "VT= 3.00 volts\n", + "KF= 0.00 A/V**2\n", + "Quadratic equation =(256)*ID**(2)-(546.67)*ID+289=0\n", + "ID = 0.00 A\n", + "VDS =VDD-ID*RL = 4.60 volts\n", + "IDQ =0.00 A\n", + "Percentage change= 3.80 percent\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "from sympy import symbols,solve\n", + "from __future__ import division \n", + "RF=6*10**(3) \n", + "print \"RF= %0.2f\"%(RF)+ \" ohm\" # feedback resistance \n", + "VDD=(20)\n", + "print \"VDD= %0.2f\"%(VDD),\" volts\" # Drain voltage supply\n", + "print \"part(i) \"# part(i) of this question\n", + "VT=(2)\n", + "print \"VT= %0.2f\"%(VT),\" volts\" # Threshold voltage for EMOSFET\n", + "KF=0.25*10**(-3)\n", + "print \"KF= %0.2f\"%(KF),\" A/V**2\" # Constant for EMOSFET \n", + "ID=1*10**(-3)\n", + "print \"ID = %0.2f\"%(ID),\" A\" # drain current EMOSFET in Ampere \n", + "RL=(VDD-VT-sqrt(ID/KF))/ID # Using formulae ID=KF*(VDD-ID*RL-VT)\n", + "print \"RL=[VDD-VT-sqrt(ID/KF)]/ID= %0.2f\"%(RL)+ \" ohm\" #Load resistance\n", + "print \"part(ii) \"# part(ii) of this question\n", + "VT=(3)\n", + "print \"VT= %0.2f\"%(VT),\" volts\" # Threshold voltage for EMOSFET\n", + "KF=0.375*10**(-3)\n", + "print \"KF= %0.2f\"%(KF),\" A/V**2\" # Constant for EMOSFET \n", + "print \"Quadratic equation =(256)*ID**(2)-(546.67)*ID+289=0\"#IDS=KF*(VGS-VT)**2 =KF*(VDS-VT)**2 and VDS=VDD-ID*RL,so Quadratic equation is:IDS=KF*(VDD-ID*RL-VT)**2 ,where ID in mA\n", + "p = [256, -546.66 , 289]\n", + "P=symbols('P')\n", + "ID=solve(p[0]*P**2+p[1]*P+p[2])[0]*10**(-3)#values of ID converted into Ampere by multiplying by 10**(-3)\n", + "print \"ID = %0.2f\"%(ID),\" A\" # drain current EMOSFET in Ampere \n", + "VDS=VDD-ID*RL# Drain voltage for ID = 1.173 mA and ID = 0.962 mA\n", + "print \"VDS =VDD-ID*RL = %0.2f\"%(VDS),\" volts\" # Drain voltage \n", + "IDQ=0.962*10**(-3)\n", + "print \"IDQ =%0.2f\"%(IDQ),\" A\"#Since VDS < VT for ID=1.173 mA, hence ID = 1.173 mA cannot be chosen, so we chose ID= 0.962 mA as operating drain current IDQ\n", + "Percentage_change=((1-0.962)*100)/(1)\n", + "print \"Percentage change= %0.2f\"%(Percentage_change),\" percent\"# Percent change in IDQ value from 1 mA(part(i)) to its value ( of part(ii))IDQ=0.91mA\n", + "# NOTE: part(ii):the values of ID = 1.173 mA or ID = 0.962 mA but in book given as ID= 1.197 mA and ID = 0.939 mA .Hence (correct) Percentage_change in ID= 3.8 % but in book given as 6.1 % \n", + "# ANS is not correct check &correct" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4_12 Page No. 115" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VDD= 5.00 volts\n", + "RL1= 1.25e+05 ohm\n", + "RL2= 2.00e+05 ohm\n", + "IDON1 =0.00 A\n", + "IDON2 =0.00 A\n", + "VDON1=VDD-IDON1*RL1= 0.64 volts\n", + "VDON2=VDD-IDON2*RL2= 0.50 volts\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "VDD=(5)\n", + "print \"VDD= %0.2f\"%(VDD),\" volts\" # Drain voltage supply\n", + "RL1=125*10**(3)\n", + "print \"RL1= %0.2e\"%(RL1)+ \" ohm\" #Load resistance\n", + "RL2=200*10**(3)\n", + "print \"RL2= %0.2e\"%(RL2)+ \" ohm\" #Load resistance\n", + "IDON1=34.88*10**(-6)\n", + "print \"IDON1 =%0.2f\"%(IDON1),\" A\"#Drain current for load line 1 from fig.\n", + "IDON2=22.5*10**(-6)\n", + "print \"IDON2 =%0.2f\"%(IDON2),\" A\"#Drain current for load line 2 from fig.\n", + "VDON1=VDD-IDON1*RL1\n", + "print \"VDON1=VDD-IDON1*RL1= %0.2f\"%(VDON1),\" volts\" # output voltage at drain terminal for IDON1\n", + "VDON2=VDD-IDON2*RL2\n", + "print \"VDON2=VDD-IDON2*RL2= %0.2f\"%(VDON2),\" volts\" # output voltage at drain terminal for IDON2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4_13 Page No. 120" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "IDSS = 0.01 ampere\n", + "VP= -4.00 volts\n", + "VGS= 0.00 volts\n", + "VDD= 10.00 volts\n", + "RL= 500.00 ohm\n", + "VDS=VDD-ID*RL= 5.00 volts\n", + "VDS>VP,so pinch off region\n", + "RL= 750.00 ohm\n", + "VDS=VDD-ID*RL= 2.50 volts\n", + "VDS<VP,so ohmic region\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "IDSS=10*10**(-3)\n", + "print \"IDSS = %0.2f\"%(IDSS),\" ampere\" # maximum drain current for n-channel DEMOSFET\n", + "ID=IDSS # since VGS=0V, so ID=maximum\n", + "VP=(-4)\n", + "print \"VP= %0.2f\"%(VP),\" volts\" # pinch off voltage \n", + "VGS=(0)\n", + "print \"VGS= %0.2f\"%(VGS),\" volts\" # Gate to source voltage \n", + "VDD=(10)\n", + "print \"VDD= %0.2f\"%(VDD),\" volts\" # Drain supply voltage \n", + "RL=0.5*10**(3)\n", + "print \"RL= %0.2f\"%(RL)+ \" ohm\" #Load resistance\n", + "VDS=VDD-ID*RL\n", + "print \"VDS=VDD-ID*RL= %0.2f\"%(VDS),\" volts\" # Drain to source voltage ,since VDS>VP DEMOSFET is in pinch off\n", + "print \"VDS>VP,so pinch off region\"\n", + "RL=0.75*10**(3)\n", + "print \"RL= %0.2f\"%(RL)+ \" ohm\" # New Load resistance value\n", + "VDS=VDD-ID*RL\n", + "print \"VDS=VDD-ID*RL= %0.2f\"%(VDS),\" volts\" # New Drain to source voltage for RL=750 ohm\n", + "print \"VDS<VP,so ohmic region\"# since VDS < VP DEMOSFET is in ohmic region for RL=750 ohm and hence will not operate as a current source" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4_14 Page No. 122" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "KF1 = 2.50e-04 A/V**2\n", + "KF2 = 2.50e-04 A/V**2\n", + "IQ= 1.00e-03 ampere\n", + "VT1 = 2.00 volts\n", + "VT2 = 2.00 volts\n", + "VDD= 15.00 volts\n", + "IREF =IQ= 1.00e-03 ampere\n", + "VGS= VT1+sqrt(2*IREF/KF1)=4.83 volts\n", + "R= (VDD-VGS)/IREF=10171.57 ohm\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "from __future__ import division \n", + "KF1=0.25*10**(-3)\n", + "print \"KF1 = %0.2e\"%(KF1),\" A/V**2\" # Scale factor \n", + "KF2=KF1\n", + "print \"KF2 = %0.2e\"%(KF2),\" A/V**2\" # Scale factor \n", + "IQ=1*10**(-3)\n", + "print \"IQ= %0.2e\"%(IQ),\" ampere\" # constant current source value\n", + "VT1=2\n", + "print \"VT1 = %0.2f\"%(VT1),\" volts\"# Threshold voltage\n", + "VT2=VT1\n", + "print \"VT2 = %0.2f\"%(VT2),\" volts\"# Threshold voltage\n", + "VDD=(15)\n", + "print \"VDD= %0.2f\"%(VDD),\" volts\" # Drain supply voltage \n", + "IREF=IQ\n", + "print \"IREF =IQ= %0.2e\"%(IREF),\" ampere\" # Reference current value\n", + "VGS=VT1+sqrt(2*IREF/KF1) # Formulae\n", + "print \"VGS= VT1+sqrt(2*IREF/KF1)=%0.2f\"%(VGS),\" volts\" # Gate to source voltage \n", + "R=(VDD-VGS)/IREF\n", + "print \"R= (VDD-VGS)/IREF=%0.2f\"%(R)+ \" ohm\" # resistance value to obtain constant current\n", + "# ERROR NOTE:values of VGS and R (correct) are 4.8284271 volts and 10171.573 ohm respectively but given in book are VGS=4V and R=11 kilo ohms" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4_15 Page No. 123" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "RON= 100.00 ohm\n", + "ROFF= 1.00e+10 ohm\n", + "Vip= 1.00 volts\n", + "Rs= 100.00 ohm\n", + "RL= 10000.00 ohm\n", + "part(i) \n", + "Vo=(Vip*RL)/(RL+RON+Rs)= 0.98 volts\n", + "ErON=[Vip*(RON+Rs)/(RL+RON+Rs)]*100= 1.96 percent\n", + "vOFF=(Vip*RL)/ROFF= 0.00 volts\n", + "OFF_isolation=20*log10(Vip/vOFF)= 120.00 dB\n", + "part(ii) \n", + "vOFF=(Vip*RON)/(Rs+RON)= 0.50 volts\n", + "OFF_isolation=20*log10((Rs+RON)/RON)= 6.02 dB\n" + ] + } + ], + "source": [ + "from math import log10\n", + "from __future__ import division \n", + "RON=100\n", + "print \"RON= %0.2f\"%(RON)+ \" ohm\" #ON resistance of analog series switch\n", + "ROFF=10**(10)\n", + "print \"ROFF= %0.2e\"%(ROFF)+ \" ohm\" #OFF resistance analog series switch\n", + "Vip=1\n", + "print \"Vip= %0.2f\"%(Vip),\" volts\"# Peak amplitude of analog voltage\n", + "Rs=100\n", + "print \"Rs= %0.2f\"%(Rs)+ \" ohm\" #Voltage source resistance\n", + "RL=10*10**(3)\n", + "print \"RL= %0.2f\"%(RL)+ \" ohm\" #Load resistance\n", + "print \"part(i) \"# part(i) of this question\n", + "Vo=(Vip*RL)/(RL+RON+Rs)\n", + "print \"Vo=(Vip*RL)/(RL+RON+Rs)= %0.2f\"%(Vo),\" volts\"# ON voltage\n", + "ErON=(Vip*(RON+Rs)/(RL+RON+Rs))*100\n", + "print \"ErON=[Vip*(RON+Rs)/(RL+RON+Rs)]*100= %0.2f\"%(ErON),\" percent\"# Output voltage error \n", + "vOFF=(Vip*RL)/ROFF\n", + "print \"vOFF=(Vip*RL)/ROFF= %0.2f\"%(vOFF),\" volts\"# Output voltage in OFF state\n", + "OFF_isolation=20*log10(Vip/vOFF)\n", + "print \"OFF_isolation=20*log10(Vip/vOFF)= %0.2f\"%(OFF_isolation),\" dB\" # OFF_isolation=20*log10(Vip/vOFF) in dB# Thus ON error and OFF isolation decrease with increasing values of RL.\n", + "print \"part(ii) \"# part(ii) of this question\n", + "vOFF=(Vip*RON)/(Rs+RON)\n", + "print \"vOFF=(Vip*RON)/(Rs+RON)= %0.2f\"%(vOFF),\" volts\"# Output voltage in OFF state for analog shunt switch\n", + "OFF_isolation=20*log10((Rs+RON)/RON)# OFF_isolation of shunt switch\n", + "print \"OFF_isolation=20*log10((Rs+RON)/RON)= %0.2f\"%(OFF_isolation),\" dB\"# Thus shunt switch is inferior to series switch in its OFF isolation property\n", + "\n", + "# ERROR NOTE: in question the author has given RL = 10K but in solution he has used RL = 1 k ... I have done programming using RL = 10 K." + ] + } + ], + "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/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch5.ipynb b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch5.ipynb new file mode 100644 index 00000000..4278055d --- /dev/null +++ b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch5.ipynb @@ -0,0 +1,566 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5 - Basic Transistor Amplifiers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_1 Page No. 136" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "RL= 5000.00 ohm\n", + "R1= 100000.00 ohm\n", + "R2= 10000.00 ohm\n", + "rc= 50000.00 ohm\n", + "rbe= 1000.00 ohm\n", + "gm = 0.05 A/V\n", + "For BJT,Av=(-gm*RL)= -250.00\n", + "AI=(gm*rbe)= 50.00\n", + "gm = 0.01 A/V\n", + "For FET,Av=(-gm*RL)= -25.00 \n", + "R0= 50000.00 ohm\n", + "Ri= 1000.00 ohm\n", + "RB=(R1*R2)/(R1+R2)= 9090.91 ohm\n", + "Ri= (RB*rbe)/(RB+rbe)=900.90 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "RL=5*10**(3)\n", + "print \"RL= %0.2f\"%(RL),\" ohm\" #Load resistance\n", + "R1=100*10**(3)\n", + "print \"R1= %0.2f\"%(R1),\" ohm\" # resistance\n", + "R2=10*10**(3)\n", + "print \"R2= %0.2f\"%(R2),\" ohm\" # resistance\n", + "rc=50*10**(3)\n", + "print \"rc= %0.2f\"%(rc),\" ohm\" #collector resistance\n", + "rd=rc # Drain and collector resistance are equal\n", + "rbe=1*10**(3)\n", + "print \"rbe= %0.2f\"%(rbe),\" ohm\" #Load resistance\n", + "gm=50*10**(-3)\n", + "print \"gm = %0.2f\"%(gm),\" A/V\"# transconductance for BJT \n", + "Av=(-gm*RL)\n", + "print \"For BJT,Av=(-gm*RL)= %0.2f\"%(Av) #Voltage gain for BJT\n", + "AI=gm*rbe\n", + "print \"AI=(gm*rbe)= %0.2f\"%(AI) # current gain for BJT\n", + "gm=5*10**(-3)\n", + "print \"gm = %0.2f\"%(gm),\" A/V\"# transconductance for FET \n", + "Av=(-gm*RL)\n", + "print \"For FET,Av=(-gm*RL)= %0.2f\"%(Av),\" \" # gain for FET\n", + "R0=rd\n", + "print \"R0= %0.2f\"%(R0),\" ohm\" #output resistance for FET and BJT\n", + "Ri=rbe\n", + "print \"Ri= %0.2f\"%(Ri),\" ohm\" #BJT input resistance \n", + "RB=(R1*R2)/(R1+R2)\n", + "print \"RB=(R1*R2)/(R1+R2)= %0.2f\"%(RB),\" ohm\" # eqivalent Base resistance for BJT\n", + "Ri=(RB*rbe)/(RB+rbe)\n", + "print \"Ri= (RB*rbe)/(RB+rbe)=%0.2f\"%(Ri),\" ohm\" #New value of BJT input resistance " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_2 Page No. 137" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "RL= 5000.00 ohm\n", + "R1= 100000.00 ohm\n", + "R2= 100000.00 ohm\n", + "Rs= 5000.00 ohm\n", + "Beta_o = 50.00\n", + "rbe= 1000.00 ohm\n", + "gm = 0.05 A/V\n", + "rc= 50000.00 ohm\n", + "Av=RL/(RL+1/gm)= 1.00\n", + "Avs=RL/[(Rs/Beta_o)+(1/gm)+(RL)]= 0.98\n", + "AI=-(Beta_o+1)= -51.00\n", + "R0= (Rs+rbe)/Beta_o=120.00 ohm\n", + "Ri= rbe+Beta_o*RL=251000.00 ohm\n", + "RB=(R1*R2)/(R1+R2)= 50000.00 ohm\n", + "Rieff= (Ri*RB)/(RB+Ri)=41694.35 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "RL=5*10**(3)\n", + "print \"RL= %0.2f\"%(RL),\" ohm\" #Load resistance\n", + "R1=100*10**(3)\n", + "print \"R1= %0.2f\"%(R1),\" ohm\" # resistance\n", + "R2=100*10**(3)\n", + "print \"R2= %0.2f\"%(R2),\" ohm\" # resistance\n", + "Rs=5*10**(3)\n", + "print \"Rs= %0.2f\"%(Rs),\" ohm\" # Source resistance\n", + "Beta_o=50\n", + "print \"Beta_o = %0.2f\"%(Beta_o) #BJT gain\n", + "rbe=1*10**(3)\n", + "print \"rbe= %0.2f\"%(rbe),\" ohm\" #Base-emitter resistance\n", + "gm=50*10**(-3)\n", + "print \"gm = %0.2f\"%(gm),\" A/V\"# transconductance for BJT \n", + "rc=50*10**(3)\n", + "print \"rc= %0.2f\"%(rc),\" ohm\" #collector resistance\n", + "Av=RL/(RL+1/gm) # Gain formulae\n", + "print \"Av=RL/(RL+1/gm)= %0.2f\"%(Av) # voltage gain for BJT\n", + "Avs=RL/((Rs/Beta_o)+(1/gm)+(RL))\n", + "print \"Avs=RL/((Rs/Beta_o)+(1/gm)+(RL))= %0.2f\"%(Avs) # Overall voltage gain for BJT\n", + "AI=-(Beta_o+1)\n", + "print \"AI=-(Beta_o+1)= %0.2f\"%(AI) # current gain for BJT\n", + "R0=(Rs+rbe)/Beta_o\n", + "print \"R0= (Rs+rbe)/Beta_o=%0.2f\"%(R0),\" ohm\" #output resistance for BJT\n", + "Ri=rbe+Beta_o*RL # formulae\n", + "print \"Ri= rbe+Beta_o*RL=%0.2f\"%(Ri),\" ohm\" # value of BJT input resistance \n", + "RB=(R1*R2)/(R1+R2)\n", + "print \"RB=(R1*R2)/(R1+R2)= %0.2f\"%(RB),\" ohm\" # eqivalent Base resistance for BJT\n", + "Rieff=(Ri*RB)/(RB+Ri)\n", + "print \"Rieff= (Ri*RB)/(RB+Ri)=%0.2f\"%(Rieff),\" ohm\" #Effective value of BJT input resistance " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_3 Page No. 142" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "RL= 5000.00 ohm\n", + "RF= 5000.00 ohm\n", + "Beta_o = 50.00\n", + "rbe= 1000.00 ohm\n", + "gm = 0.05 A/V\n", + "rc= 50000.00 ohm\n", + "Ri= rbe+RF*(1+gm*rbe)=256000.00 ohm\n", + "Av=(-gm*RL)/(1+gm*RF)= -1.00\n", + "AI=(Beta_o)= 50.00\n", + "R0= Beta_o*rc=2500000.00 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "RL=5*10**(3)\n", + "print \"RL= %0.2f\"%(RL),\" ohm\" #Load resistance\n", + "RF=5*10**(3)\n", + "print \"RF= %0.2f\"%(RF),\" ohm\" # resistance\n", + "Beta_o=50\n", + "print \"Beta_o = %0.2f\"%(Beta_o) #BJT gain\n", + "rbe=1*10**(3)\n", + "print \"rbe= %0.2f\"%(rbe),\" ohm\" #Base-emitter resistance\n", + "gm=50*10**(-3)\n", + "print \"gm = %0.2f\"%(gm),\" A/V\"# transconductance for BJT \n", + "rc=50*10**(3)\n", + "print \"rc= %0.2f\"%(rc),\" ohm\" #collector resistance\n", + "Ri=rbe+RF*(1+gm*rbe) # formulae\n", + "print \"Ri= rbe+RF*(1+gm*rbe)=%0.2f\"%(Ri),\" ohm\" # BJT input resistance \n", + "Av=(-gm*RL)/(1+gm*RF)# formulae\n", + "print \"Av=(-gm*RL)/(1+gm*RF)= %0.2f\"%(Av) # voltage gain for BJT\n", + "AI=Beta_o\n", + "print \"AI=(Beta_o)= %0.2f\"%(AI) # current gain for BJT\n", + "R0=Beta_o*rc\n", + "print \"R0= Beta_o*rc=%0.2f\"%(R0),\" ohm\" #output resistance for BJT" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_4 Page No. 148" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "RL= 5000.00 ohm\n", + "RF= 2500.00 ohm\n", + "Rs= 50.00 ohm\n", + "ro= 50000.00 ohm\n", + "rc= 50000.00 ohm\n", + "rbe= 1000.00 ohm\n", + "For CG Amplifier\n", + "gm = 0.01 A/V\n", + "Ri= 1/gm=200.00 ohm\n", + "Avs=gm*RL/(1+gm*Rs)= 20.00\n", + "Ro=rd*(1+gm*Rs)=62500.00 ohm\n", + "For CB Amplifier\n", + "gm = 0.05 A/V\n", + "Ri= 1/gm=20.00 ohm\n", + "Avs=gm*RL/(1+gm*Rs)= 71.43\n", + "Ro=gm*(rbe*rc)=2.50e+06 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "RL=5*10**(3)\n", + "print \"RL= %0.2f\"%(RL),\" ohm\" #Load resistance\n", + "RF=2.5*10**(3)\n", + "print \"RF= %0.2f\"%(RF),\" ohm\" # resistance\n", + "Rs=50\n", + "print \"Rs= %0.2f\"%(Rs),\" ohm\" # resistance\n", + "ro=50*10**(3)\n", + "print \"ro= %0.2f\"%(ro),\" ohm\" # output resistance\n", + "rd=ro # drain resistance\n", + "rc=ro# Collector resistance\n", + "print \"rc= %0.2f\"%(rc),\" ohm\" # Collector resistance\n", + "rbe=1*10**(3)\n", + "print \"rbe= %0.2f\"%(rbe),\" ohm\" #base -emitter resistance\n", + "print \"For CG Amplifier\"\n", + "gm=5*10**(-3)\n", + "print \"gm = %0.2f\"%(gm),\" A/V\"# transconductance for FET \n", + "Ri=1/gm # formulae\n", + "print \"Ri= 1/gm=%0.2f\"%(Ri),\" ohm\" # value of CGA (common gate amplifier)input resistance for FET\n", + "Avs=gm*RL/(1+gm*Rs)\n", + "print \"Avs=gm*RL/(1+gm*Rs)= %0.2f\"%(Avs) # Overall voltage gain for FET (CGA)\n", + "Ro=rd*(1+gm*Rs)\n", + "print \"Ro=rd*(1+gm*Rs)=%0.2f\"%(Ro),\" ohm\" #output resistance for FET (CGA)\n", + "print \"For CB Amplifier\"\n", + "gm=50*10**(-3)\n", + "print \"gm = %0.2f\"%(gm),\" A/V\"# transconductance for BJT\n", + "Ri=1/gm # formulae\n", + "print \"Ri= 1/gm=%0.2f\"%(Ri),\" ohm\" # value of CBA (common base amplifier)input resistance for BJT\n", + "Avs=gm*RL/(1+gm*Rs)\n", + "print \"Avs=gm*RL/(1+gm*Rs)= %0.2f\"%(Avs) # Overall voltage gain for BJT (CBA)\n", + "Ro=gm*(rbe*rc)\n", + "print \"Ro=gm*(rbe*rc)=%0.2e\"%(Ro),\" ohm\" #output resistance for BJT (CBA)\n", + "\n", + "#NOTE: I have calculated first all the parameters for CG amplifier and then for CB amplifier but in book parameters have been calculated alternatingly for CG and CB amplifiers." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_5 Page No. 152" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "RL= 5000.00 ohm\n", + "Cc= 1.00e-07 farad\n", + "Ri= 100000.00 ohm\n", + "CSH= 0.00 farad\n", + "Avm=100.00\n", + "fL=1/(2*(pi)*(Ri)*(Cc))= 15.92 Hz \n", + "fH=1/(2*(pi)*(RL)*(CSH))= 3.18e+05 Hz\n", + "BW=fH-fL= 318293.97 Hz\n", + "fT=Avm*fH= 3.18e+07 Hz\n" + ] + } + ], + "source": [ + "from math import pi\n", + "from __future__ import division \n", + "RL=5*10**(3)\n", + "print \"RL= %0.2f\"%(RL),\" ohm\" #Load resistance\n", + "Cc=0.1*10**(-6)\n", + "print \"Cc= %0.2e\"%(Cc),\" farad\" #capacitance\n", + "Ri=100*10**(3)\n", + "print \"Ri= %0.2f\"%(Ri),\" ohm\" # input resistance for Amplifier\n", + "CSH=100*10**(-12)\n", + "print \"CSH= %0.2f\"%(CSH),\" farad\" #shunt load capacitance\n", + "Avm=100\n", + "print \"Avm=%0.2f\"%(Avm) # Mid-frequency gain \n", + "fL=1/(2*(pi)*(Ri)*(Cc))\n", + "print \"fL=1/(2*(pi)*(Ri)*(Cc))= %0.2f\"%(fL),\"Hz \" # Lower cutoff-frequency \n", + "fH=1/(2*(pi)*(RL)*(CSH))\n", + "print \"fH=1/(2*(pi)*(RL)*(CSH))= %0.2e\"%(fH),\" Hz\" # Higher cutoff-frequency \n", + "BW=fH-fL\n", + "print \"BW=fH-fL= %0.2f\"%(BW),\" Hz\" # Bandwidth\n", + "fT=Avm*fH\n", + "print \"fT=Avm*fH= %0.2e\"%(fT),\" Hz\" # Unity gain bandwidth\n", + "# ERROR NOTE: calculated value of lower cutoff frequency, fL= 15.915494 Hz but in book given as 15.0 Hz " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_6 Page No. 152" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "IDSS = 0.02 ampere\n", + "VP= -4.00 volts\n", + "VGSQ= -2.00 volts\n", + "Vsm= 0.20 volts\n", + "D=(((0.5)*(Vsm)**2)/(4*Vsm))*100 =2.50 % \n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "IDSS=16*10**(-3)\n", + "print \"IDSS = %0.2f\"%(IDSS),\" ampere\" # maximum drain current JFET \n", + "VP=(-4)\n", + "print \"VP= %0.2f\"%(VP),\" volts\" # pinch off voltage for JFET \n", + "VGSQ=(-2)\n", + "print \"VGSQ= %0.2f\"%(VGSQ),\" volts\" # Gate operating point voltage \n", + "Vsm=(0.2)\n", + "print \"Vsm= %0.2f\"%(Vsm),\" volts\" # sinusoidal input voltage for JFET \n", + "D=(((0.5)*(Vsm)**2)/(4*Vsm))*100 # derived from ID=IDSS(1-VGS/VP)**2 and putting value of VGS=VGSQ+Vs, where Vs=Vsm sinwt\n", + "print \"D=(((0.5)*(Vsm)**2)/(4*Vsm))*100 =%0.2f\"%(D),\"% \" # Percentage second harmonic distortion calculation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_7 Page No. 153" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ic = 1.00e-03 ampere\n", + "rbe= 2000.00 ohm\n", + "gm = 0.05 A/V\n", + "Beta_o = 100.00 \n", + "rc= 50000.00 ohm\n", + "Cbe= 1.00e-11 farad\n", + "Ctc= 1.00e-12 farad\n", + "part(i)\n", + "RL= 10000.00 ohm\n", + "Rs= 500.00 ohm\n", + "Rth=(Rs*rbe)/(Rs+rbe)=400.00 ohm\n", + "Avm=(-gm*RL)=-500.00\n", + "CM=Ctc*(1-Avm)= 0.00 farad\n", + "Ci=Cbe= 0.00 farad\n", + "fHi=1/(2*(pi)*(Rth)*(Cbe+CM))= 778644.54 Hz\n", + "Ri=rbe =2000.00 ohm\n", + "R0= rc=50000.00 ohm\n", + "fB=1/(2*(pi)*(rbe)*(Cbe))= 7.96e+06 Hz\n", + "fT=Beta_o*fB= 7.96e+08 Hz\n", + "part(ii)\n", + "Rs= 50000.00 ohm\n", + "RL= 1000.00 ohm\n", + "fhi=1/(2*(pi)*(Rs)*(Ctc))= 3.18e+06 Hz\n", + "Avm=(gm*RL)/(1+gm*RL)=0.98\n", + "Ro= 1/gm=20.00 ohm\n", + "Ri=Beta_o*RL =100000.00 ohm\n", + "part(iii)\n", + "RL= 10000.00 ohm\n", + "Rs= 50.00 ohm\n", + "fHi=gm/(2*(pi)*(Cbe))= 7.96e+08 Hz\n", + "fHo=gm/(2*(pi)*(Ctc)*(RL))= 1.59e+07 Hz\n", + "Avs=(gm*RL)/(1+gm*Rs)=142.86\n", + "Ri= 1/gm=20.00 ohm\n", + "Ro=Beta_o*rc =5.00e+06 ohm\n" + ] + } + ], + "source": [ + "from math import pi\n", + "from __future__ import division \n", + "Ic=1*10**(-3)\n", + "print \"Ic = %0.2e\"%(Ic),\" ampere\" # collector current BJT\n", + "rbe=2*10**(3)\n", + "print \"rbe= %0.2f\"%(rbe),\" ohm\" #base -emitter resistance\n", + "gm=50*10**(-3)\n", + "print \"gm = %0.2f\"%(gm),\" A/V\"# transconductance for BJT\n", + "Beta_o=100\n", + "print \"Beta_o = %0.2f\"%(Beta_o),\" \" #BJT gain\n", + "rc=50*10**(3)\n", + "print \"rc= %0.2f\"%(rc),\" ohm\" #collector resistance\n", + "Cbe=10*10**(-12)\n", + "print \"Cbe= %0.2e\"%(Cbe),\" farad\" #base -emitter capacitance\n", + "Ctc=1*10**(-12)\n", + "print \"Ctc= %0.2e\"%(Ctc),\" farad\" #input device capacitance\n", + "print \"part(i)\"# part(i)of question\n", + "RL=10*10**(3)\n", + "print \"RL= %0.2f\"%(RL),\" ohm\" #Load resistance\n", + "Rs=500\n", + "print \"Rs= %0.2f\"%(Rs),\" ohm\" #input source resistance\n", + "Rth=(Rs*rbe)/(Rs+rbe)\n", + "print \"Rth=(Rs*rbe)/(Rs+rbe)=%0.2f\"%(Rth),\" ohm\" # eqivalent resistance\n", + "Avm=(-gm*RL)\n", + "print \"Avm=(-gm*RL)=%0.2f\"%(Avm) # Mid-frequency gain for CE amplifier\n", + "CM=Ctc*(1-Avm)\n", + "print \"CM=Ctc*(1-Avm)= %0.2f\"%(CM),\" farad\" #calculated capacitance\n", + "Ci=Cbe\n", + "print \"Ci=Cbe= %0.2f\"%(Ci),\" farad\" #calculated input capacitance\n", + "fHi=1/(2*(pi)*(Rth)*(Cbe+CM))\n", + "print \"fHi=1/(2*(pi)*(Rth)*(Cbe+CM))= %0.2f\"%(fHi),\" Hz\" # Higher-frequency cutoff for CE amplifier\n", + "Ri=rbe\n", + "print \"Ri=rbe =%0.2f\"%(Ri),\" ohm\" #input resistance CE amplifier\n", + "Ro=rc\n", + "print \"R0= rc=%0.2f\"%(Ro),\" ohm\" #output resistance for CE amplifier\n", + "fB=1/(2*(pi)*(rbe)*(Cbe))\n", + "print \"fB=1/(2*(pi)*(rbe)*(Cbe))= %0.2e\"%(fB),\" Hz\" # base terminal frequency cutoff\n", + "fT=Beta_o*fB\n", + "print \"fT=Beta_o*fB= %0.2e\"%(fT),\" Hz\" # Unity gain bandwidth for CE amplifier\n", + "print \"part(ii)\"# part(ii)of question\n", + "Rs=50*10**(3)\n", + "print \"Rs= %0.2f\"%(Rs),\" ohm\" #input source resistance for CC amplifier\n", + "RL=1*10**(3)\n", + "print \"RL= %0.2f\"%(RL),\" ohm\" #Load resistance for CC amplifier\n", + "fhi=1/(2*(pi)*(Rs)*(Ctc))\n", + "print \"fhi=1/(2*(pi)*(Rs)*(Ctc))= %0.2e\"%(fhi),\" Hz\" # Higher-frequency cutoff for CC amplifier\n", + "Avm=(gm*RL)/(1+gm*RL)\n", + "print \"Avm=(gm*RL)/(1+gm*RL)=%0.2f\"%(Avm) # Mid-frequency gain for CC amplifier\n", + "Ro=1/gm\n", + "print \"Ro= 1/gm=%0.2f\"%(Ro),\" ohm\" #output resistance for CC amplifier\n", + "Ri=Beta_o*RL\n", + "print \"Ri=Beta_o*RL =%0.2f\"%(Ri),\" ohm\" #input resistance CE amplifier\n", + "print \"part(iii)\"# part(iii)of question\n", + "RL=10*10**(3)\n", + "print \"RL= %0.2f\"%(RL),\" ohm\" #Load resistance for CB amplifier\n", + "Rs=50\n", + "print \"Rs= %0.2f\"%(Rs),\" ohm\" #input source resistance for CB amplifier\n", + "fHi=gm/(2*(pi)*(Cbe))\n", + "print \"fHi=gm/(2*(pi)*(Cbe))= %0.2e\"%(fHi),\" Hz\" # Higher-frequency cutoff for CB amplifier\n", + "fHo=1/(2*(pi)*(Ctc)*(RL))\n", + "print \"fHo=gm/(2*(pi)*(Ctc)*(RL))= %0.2e\"%(fHo),\" Hz\" # Higher-frequency cutoff for CB amplifier\n", + "Avs=(gm*RL)/(1+gm*Rs)\n", + "print \"Avs=(gm*RL)/(1+gm*Rs)=%0.2f\"%(Avs) # Mid-frequency gain for CB amplifier\n", + "Ri=1/gm\n", + "print \"Ri= 1/gm=%0.2f\"%(Ri),\" ohm\" #output resistance for CB amplifier\n", + "Ro=Beta_o*rc\n", + "print \"Ro=Beta_o*rc =%0.2e\"%(Ro),\" ohm\" #input resistance CB amplifier\n", + "#ERROR NOTE:some parameters in the book have been calculated using gm=40 mA/V while given value is gm=50 mA/V. So ,for part(ii) CC amplifier correct value of R0=20 ohm,Ri=100000 ohm,and for part(iii)CB amplifier over all voltage gain Avs=142.85714 ,Ri=20 ohm all calculated for gm=50 mA/V." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5_8 Page No. 154" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "tp= 0.01 s\n", + "tr= 5.00e-08 s\n", + "CSH= 5.00e-11 farad\n", + "percentage tilt= 5.00 %\n", + "Ri= 100000.00 ohm\n", + "RL=tr/(2.2*CSH)= 454.55 ohm\n", + "Cc= (tp*100)/( tilt*Ri)=2.00e-06 farad\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "tp=10*10**(-3)\n", + "print \"tp= %0.2f\"%(tp),\" s\" # Time period of pulse\n", + "tr=0.05*10**(-6)\n", + "print \"tr= %0.2e\"%(tr),\" s\" # Rise-Time of pulse\n", + "CSH=50*10**(-12)\n", + "print \"CSH= %0.2e\"%(CSH),\" farad\" #output capacitor\n", + "tilt=5\n", + "print \"percentage tilt= %0.2f\"%(tilt),\"%\" #Sag or percentage tilt of output \n", + "Ri=100*10**(3)\n", + "print \"Ri= %0.2f\"%(Ri),\" ohm\" # source resistance\n", + "RL=tr/(2.2*CSH)\n", + "print \"RL=tr/(2.2*CSH)= %0.2f\"%(RL),\" ohm\" #Load resistance calculation\n", + "Cc=(tp*100)/(tilt*Ri)\n", + "print \"Cc= (tp*100)/( tilt*Ri)=%0.2e\"%(Cc),\" farad\" #capacitance\n", + "#ERROR NOTE: calculated value of RL=454.54545 ohm but in book given as 455 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/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch6.ipynb b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch6.ipynb new file mode 100644 index 00000000..50416414 --- /dev/null +++ b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch6.ipynb @@ -0,0 +1,449 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6 - Multitransistor and Multistage Amplifier" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6_1 Page No. 168" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Av= 0.10\n", + "Av(dB)=20*log10(Av)= -20.00 dB \n", + "Av= 0.71\n", + "Av(dB)=20*log10(Av)= -3.01 dB \n", + "Av= 1.00\n", + "Av(dB)=20*log10(Av)= 0.00 dB \n", + "Av= 10.00\n", + "Av(dB)=20*log10(Av)= 20.00 dB \n", + "Av= 100.00\n", + "Av(dB)=20*log10(Av)= 40.00 dB \n", + "Av= 1000.00\n", + "Av(dB)=20*log10(Av)= 60.00 dB \n" + ] + } + ], + "source": [ + "from math import log10\n", + "from __future__ import division \n", + "Av=0.1\n", + "print \"Av= %0.2f\"%(Av) #Voltage gain\n", + "AvdB=20*log10(Av)\n", + "print \"Av(dB)=20*log10(Av)= %0.2f\"%(AvdB),\"dB \" #Voltage gain in decibel\n", + "Av=0.707\n", + "print \"Av= %0.2f\"%(Av) #Voltage gain\n", + "AvdB=20*log10(Av)\n", + "print \"Av(dB)=20*log10(Av)= %0.2f\"%(AvdB),\"dB \" #Voltage gain in decibel\n", + "Av=1\n", + "print \"Av= %0.2f\"%(Av) #Voltage gain\n", + "AvdB=20*log10(Av)\n", + "print \"Av(dB)=20*log10(Av)= %0.2f\"%(AvdB),\"dB \" #Voltage gain in decibel\n", + "Av=10\n", + "print \"Av= %0.2f\"%(Av) #Voltage gain\n", + "AvdB=20*log10(Av)\n", + "print \"Av(dB)=20*log10(Av)= %0.2f\"%(AvdB),\"dB \" #Voltage gain in decibel\n", + "Av=100\n", + "print \"Av= %0.2f\"%(Av) #Voltage gain\n", + "AvdB=20*log10(Av)\n", + "print \"Av(dB)=20*log10(Av)= %0.2f\"%(AvdB),\"dB \" #Voltage gain in decibel\n", + "Av=1000\n", + "print \"Av= %0.2f\"%(Av) #Voltage gain\n", + "AvdB=20*log10(Av)\n", + "print \"Av(dB)=20*log10(Av)= %0.2f\"%(AvdB),\"dB \" #Voltage gain in decibel\n", + "#NOTE:calculated voltage gain in dB for Av=0.707 is -3.0116117dB " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6_2 Page No. 169" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ri= 500.00 ohm\n", + "RL= 50.00 ohm\n", + "Vom= 1.00 volts\n", + "Vim= 0.00 volts\n", + "Av(in dB)=20*log10(Vo/Vi)= 60.00 dB \n", + "Iim= Vim/Ri= 0.00 A\n", + "Iom= Vom/RL= 0.02 A\n", + "Ai=20*log10(Io/Ii)= 80.00 dB \n", + "pi= Vi**2/Ri= 0.00 W\n", + "po= Vo**2/RL= 0.01 W\n", + "Ap=10*log10(po/pi)= 70.00 dB \n" + ] + } + ], + "source": [ + "from math import sqrt,log10\n", + "from __future__ import division \n", + "Ri=0.5*10**(3)\n", + "print \"Ri= %0.2f\"%(Ri)+ \" ohm\" # Amplifier input resistance\n", + "RL=0.05*10**(3)\n", + "print \"RL= %0.2f\"%(RL)+ \" ohm\" # Load resistance\n", + "Vom=1\n", + "print \"Vom= %0.2f\"%(Vom),\" volts\" # Output voltage \n", + "Vo=Vom/sqrt(2)#RMS value of Output voltage \n", + "Vim=1*10**(-3)\n", + "print \"Vim= %0.2f\"%(Vim),\" volts\" # Peak Input voltage\n", + "Vi=Vim/sqrt(2)#RMS Input voltage \n", + "Av=20*log10(Vo/Vi)\n", + "print \"Av(in dB)=20*log10(Vo/Vi)= %0.2f\"%(Av),\" dB \" #Voltage gain in decibel\n", + "Iim=Vim/Ri\n", + "print \"Iim= Vim/Ri= %0.2f\"%(Iim),\" A\" # Input peak current\n", + "Ii=Iim/sqrt(2) #RMS value of input current\n", + "Iom=Vom/RL \n", + "print \"Iom= Vom/RL= %0.2f\"%(Iom),\" A\" # Output peak current\n", + "Io=Iom/sqrt(2) #RMS value of Output current\n", + "Ai=20*log10(Io/Ii)\n", + "print \"Ai=20*log10(Io/Ii)= %0.2f\"%(Ai),\" dB \" #Current gain in decibel\n", + "pi=Vi**2/Ri\n", + "print \"pi= Vi**2/Ri= %0.2f\"%(pi),\" W\" # Input power \n", + "po=Vo**2/RL\n", + "print \"po= Vo**2/RL= %0.2f\"%(po),\" W\" # Output power \n", + "Ap=10*log10(po/pi)\n", + "print \"Ap=10*log10(po/pi)= %0.2f\"%(Ap),\" dB \" #Power gain in decibel" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6_3 Page No. 172" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "RL= 1000.00 ohm\n", + "RF= 500000.00 ohm\n", + "Beta_o = 50.00\n", + "rbe= 1000.00 ohm\n", + "gm = 0.05 A/V\n", + "rc= 50000.00 ohm\n", + "part(i)\n", + "Adm1=(-gm*RL)= -50.00\n", + "Adm2=(0.5*gm*RL)= 25.00\n", + "Rid=2*rbe= 2000.00 ohm\n", + "Acm=(-RL)/(2*RF)= -1.00e-03\n", + "Ric=Beta_o*RF= 2.50e+07 ohm\n", + "CMRR=2*gm*RF= 50000.00\n", + "part(ii)\n", + "Vi1= -5.00e-04 volts\n", + "Vi2= 5.00e-04 volts\n", + "Vcm= 0.01 volts\n", + "Vd=Vi1-Vi2= -1.00e-03 volts\n", + "Vod=abs(Vd*Adm2)= 0.03 volts\n", + "Voc=abs(Vcm*Acm)= 1.00e-05 volts\n", + "percentage error=(Voc/Vod)*100= 0.04 %\n", + "part(iii)\n", + "RLeff=(RL*Rid)/(RL+Rid)= 666.67 ohm\n", + "Adm=gm*RLeff= 33.33\n", + "Acm=(-RLeff)/(2*RF)= -6.67e-04\n", + "CMRR=abs(Adm/(Acm))= 50000.00\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "RL=1*10**(3)\n", + "print \"RL= %0.2f\"%(RL)+ \" ohm\" #Load resistance\n", + "RF=500*10**(3)\n", + "print \"RF= %0.2f\"%(RF)+ \" ohm\" #Feedback resistance\n", + "Beta_o=50\n", + "print \"Beta_o = %0.2f\"%(Beta_o) #BJT gain\n", + "rbe=1*10**(3)\n", + "print \"rbe= %0.2f\"%(rbe)+ \" ohm\" #Base-emitter resistance\n", + "gm=50*10**(-3)\n", + "print \"gm = %0.2f\"%(gm),\" A/V\"# transconductance for BJT \n", + "rc=50*10**(3)\n", + "print \"rc= %0.2f\"%(rc)+ \" ohm\" #collector resistance\n", + "print \"part(i)\"\n", + "Adm1=(-gm*RL)\n", + "print \"Adm1=(-gm*RL)= %0.2f\"%(Adm1) # Differential mode gain for BJT for DIDO and SIDO modes\n", + "Adm2=(0.5*gm*RL)\n", + "print \"Adm2=(0.5*gm*RL)= %0.2f\"%(Adm2) # Differential mode gain for BJT for DISO and SISO modes\n", + "Rid=2*rbe\n", + "print \"Rid=2*rbe= %0.2f\"%(Rid)+ \" ohm\" #input differential mode resistance\n", + "Acm=(-RL)/(2*RF)\n", + "print \"Acm=(-RL)/(2*RF)= %0.2e\"%(Acm) # Common mode gain for BJT for DISO and SISO modes\n", + "Ric=Beta_o*RF\n", + "print \"Ric=Beta_o*RF= %0.2e\"%(Ric)+ \" ohm\" # common mode input resistance\n", + "CMRR=2*gm*RF\n", + "print \"CMRR=2*gm*RF= %0.2f\"%(CMRR) # common mode rejection ratio\n", + "print \"part(ii)\"\n", + "Vi1=(-0.5)*10**(-3)\n", + "print \"Vi1= %0.2e\"%(Vi1),\" volts\" # input voltage1 \n", + "Vi2=(+0.5)*10**(-3)\n", + "print \"Vi2= %0.2e\"%(Vi2),\" volts\" # input voltage2\n", + "Vcm=(10)*10**(-3)\n", + "print \"Vcm= %0.2f\"%(Vcm),\" volts\" # common mode voltage\n", + "Vd=Vi1-Vi2\n", + "print \"Vd=Vi1-Vi2= %0.2e\"%(Vd),\" volts\" # differential voltage\n", + "Vod=abs(Vd*Adm2)\n", + "print \"Vod=abs(Vd*Adm2)= %0.2f\"%(Vod),\" volts\" # output differential voltage for DISO and SISO modes\n", + "Voc=abs(Vcm*Acm)\n", + "print \"Voc=abs(Vcm*Acm)= %0.2e\"%(Voc),\" volts\" # output common mode voltage\n", + "Error=(Voc/Vod)*100\n", + "print \"percentage error=(Voc/Vod)*100= %0.2f\"%(Error),\"%\"#percentage error due to CM signal\n", + "print \"part(iii)\"\n", + "RLeff=(RL*Rid)/(RL+Rid)\n", + "print \"RLeff=(RL*Rid)/(RL+Rid)= %0.2f\"%(RLeff)+ \" ohm\" # Effective load resistance\n", + "Adm=gm*RLeff\n", + "print \"Adm=gm*RLeff= %0.2f\"%(Adm) # Modified Differential mode gain for BJT for DIDO and SIDO modes\n", + "Acm=(-RLeff)/(2*RF)\n", + "print \"Acm=(-RLeff)/(2*RF)= %0.2e\"%(Acm) # Modified Common mode gain for BJT for DISO and SISO modes\n", + "CMRR=abs(Adm/(Acm))\n", + "print \"CMRR=abs(Adm/(Acm))= %0.2f\"%(CMRR) # Modified common mode rejection ratio\n", + "#NOTE: In Book, Formulae used for Acm in part(iii) is written as Acm=(-RL)/(2*RF)but ans is calculated by using RLeff in place of RL.So i have written formulae as Acm=(-RLeff)/(2*RF) in programming.\n", + "# Assigned variable name: in part(i) Adm for DIDO and SIDO modes is represented by Adm1 and Adm for DISO and SISO modes is represented by Adm2 to resist any anamoly in the programming." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6_4 Page No. 175" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VCC= 10.00 volts\n", + "VEE=VCC= 10.00 volts\n", + "IQ = 0.00 ampere\n", + "VBE= 0.70 volts\n", + "part(i)\n", + "RL=VCC/IQ= 5000.00 ohm\n", + "Pomax=VCC**2/(2*RL)= 0.01 W\n", + "PDC=2*VCC*IQ= 0.04 W\n", + "Efficiency,Etta_max=(Pomax/PDC)*100= 25.00 %\n", + "PDmax=VCC*IQ= 0.02 W\n", + "part(ii)\n", + "Vcm= 5.00 volts\n", + "Po=Vcm**2/(2*RL)= 2.50e-03 W\n", + "Efficiency,Etta=(Po/PDC)*100= 6.25 %\n", + "PDCavg=PDmax-Po= 0.02 W\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "VCC=(10)\n", + "print \"VCC= %0.2f\"%(VCC),\" volts\" # Collector voltage supply\n", + "VEE=VCC\n", + "print \"VEE=VCC= %0.2f\"%(VEE),\" volts\" # Emitter supply voltage\n", + "IQ=2*10**(-3)\n", + "print \"IQ = %0.2f\"%(IQ),\" ampere\" # operating current for CC class-Aamplifier\n", + "VBE=(0.7)\n", + "print \"VBE= %0.2f\"%(VBE),\" volts\" # Base-emitter voltage \n", + "print \"part(i)\"\n", + "RL=VCC/IQ\n", + "print \"RL=VCC/IQ= %0.2f\"%(RL)+ \" ohm\" #Load resistance\n", + "Pomax=VCC**2/(2*RL)\n", + "print \"Pomax=VCC**2/(2*RL)= %0.2f\"%(Pomax),\" W\" # maximum Output power \n", + "PDC=2*VCC*IQ\n", + "print \"PDC=2*VCC*IQ= %0.2f\"%(PDC),\" W\" # Total D.C power supply\n", + "Etta_max=(Pomax/PDC)*100\n", + "print \"Efficiency,Etta_max=(Pomax/PDC)*100= %0.2f\"%(Etta_max),\"%\" #maximum power amplifier conversion efficiency\n", + "PDmax=VCC*IQ\n", + "print \"PDmax=VCC*IQ= %0.2f\"%(PDmax),\" W\" # maximum power dissipation \n", + "print \"part(ii)\"\n", + "Vcm=(5)\n", + "print \"Vcm= %0.2f\"%(Vcm),\" volts\" # common mode voltage\n", + "Po=Vcm**2/(2*RL)\n", + "print \"Po=Vcm**2/(2*RL)= %0.2e\"%(Po),\" W\" # Output power \n", + "Etta=(Po/PDC)*100\n", + "print \"Efficiency,Etta=(Po/PDC)*100= %0.2f\"%(Etta),\" %\" # power amplifier conversion efficiency\n", + "PDCavg=PDmax-Po#Using law of conservation of energy\n", + "print \"PDCavg=PDmax-Po= %0.2f\"%(PDCavg),\" W\" # Average power dissipated in BJT" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6_5 Page No. 178" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VCC= 10.00 volts\n", + "VEE=VCC= 10.00 volts\n", + "ICQ_0 = 1.00e-02 ampere\n", + "RL= 5.00 ohm\n", + "part(i)\n", + "Po=0.00 W\n", + "PDC=2*VCC*ICQ_0= 0.20 W\n", + "part(ii)\n", + "Vcm=VCC = 10.00 volts\n", + "Icm = VCC/RL=2.00e+00 ampere\n", + "Po=(1/2)*(Icm*Vcm)=10.00 W\n", + "ICavg=(Icm)/(pi)=6.37e-01 ampere\n", + "PDC=2*VCC*ICavg= 12.73 W\n", + "Efficiency,Etta=(Po/PDC)*100= 78.54 %\n", + "part(iii)\n", + "Vcm1= 5.00 volts\n", + "ICavg1=(Vcm1)/(pi*RL)=3.18e-01 ampere\n", + "Po1=(Vcm1**2)/(2*RL)=2.50 W\n", + "PDC1=2*VCC*ICavg1= 6.37e+00 W\n", + "Efficiency,Etta=(Po1/PDC1)*100= 39.27 %\n" + ] + } + ], + "source": [ + "from math import pi,sqrt\n", + "from __future__ import division \n", + "VCC=(10)\n", + "print \"VCC= %0.2f\"%(VCC),\" volts\" # Collector voltage supply\n", + "VEE=VCC\n", + "print \"VEE=VCC= %0.2f\"%(VEE),\" volts\" # Emitter supply voltage\n", + "ICQ_0=10*10**(-3)\n", + "print \"ICQ_0 = %0.2e\"%(ICQ_0),\" ampere\" # Zero signal collector current\n", + "RL=5\n", + "print \"RL= %0.2f\"%(RL)+ \" ohm\" #Load resistance\n", + "print \"part(i)\"\n", + "Po=0# Since Output power at Zero signal condition is Zero\n", + "print \"Po=%0.2f\"%(Po),\" W\" # Output power at Zero signal condition\n", + "PDC=2*VCC*ICQ_0\n", + "print \"PDC=2*VCC*ICQ_0= %0.2f\"%(PDC),\" W\" # Total D.C power supply for Zero signal condition\n", + "print \"part(ii)\"\n", + "Vcm=VCC#For Full output voltage swing Vcm=VCC\n", + "print \"Vcm=VCC = %0.2f\"%(Vcm),\" volts\" # common mode voltage for full swing condition\n", + "Icm=VCC/RL\n", + "print \"Icm = VCC/RL=%0.2e\"%(Icm),\" ampere\" # common mode current\n", + "Po=(1/2)*(Icm*Vcm)\n", + "print \"Po=(1/2)*(Icm*Vcm)=%0.2f\"%(Po),\" W\" # Output power at full swing condition\n", + "ICavg=(Icm)/(pi)\n", + "print \"ICavg=(Icm)/(pi)=%0.2e\"%(ICavg),\" ampere\" # Average value of common mode current\n", + "PDC=2*(ICavg*VCC)\n", + "print \"PDC=2*VCC*ICavg= %0.2f\"%(PDC),\" W\" # Total D.C power supply for full swing condition\n", + "Etta=(Po/PDC)*100\n", + "print \"Efficiency,Etta=(Po/PDC)*100= %0.2f\"%(Etta),\" %\" # power amplifier conversion efficiency\n", + "print \"part(iii)\"\n", + "Vcm1=(5)#given value\n", + "print \"Vcm1= %0.2f\"%(Vcm1),\" volts\" # common mode voltage for output swing Vcm=5 V\n", + "ICavg1=(Vcm1)/(pi*RL)\n", + "print \"ICavg1=(Vcm1)/(pi*RL)=%0.2e\"%(ICavg1),\" ampere\" # Average value of common mode current\n", + "Po1=(Vcm1**2)/(2*RL)\n", + "print \"Po1=(Vcm1**2)/(2*RL)=%0.2f\"%(Po1),\" W\" # Output power for output swing Vcm=5 V\n", + "PDC1=2*(ICavg1*VCC)\n", + "print \"PDC1=2*VCC*ICavg1= %0.2e\"%(PDC1),\" W\" # Total D.C power supply for output swing Vcm=5 V\n", + "Etta=(Po1/PDC1)*100\n", + "print \"Efficiency,Etta=(Po1/PDC1)*100= %0.2f\"%(Etta),\" %\" # power amplifier conversion efficiency for output swing Vcm=5 V\n", + "# NOTE:Correct value of Efficiency,Etta=(Po1/PDC1)*100= 39.269908 % for part(iii) but book ans is 39.31%(because of approximation used during calculation)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6_6 Page No. 180" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Av= 100000.00\n", + "VCC= 10.00 volts\n", + "vo= VCC=10.00 volts\n", + "Vdmax= VCC/Av=1.00e-04 volts\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "Av=1*10**(5)\n", + "print \"Av= %0.2f\"%(Av) #Voltage gain\n", + "VCC=(10)\n", + "print \"VCC= %0.2f\"%(VCC),\" volts\" # Collector voltage supply\n", + "vo=VCC\n", + "print \"vo= VCC=%0.2f\"%(vo),\" volts\" # maximum output voltage\n", + "Vdmax=VCC/Av\n", + "print \"Vdmax= VCC/Av=%0.2e\"%(Vdmax),\" volts\" # Difference input voltage at OP-amp terminals" + ] + } + ], + "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/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch7.ipynb b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch7.ipynb new file mode 100644 index 00000000..af9fd2a9 --- /dev/null +++ b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch7.ipynb @@ -0,0 +1,622 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 7 - Feedback Amplifiers and Sinusoidal Oscillators" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7_1 Page No. 204" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "A= 60000.00\n", + "Af= 10000.00\n", + "N_dB=20*log10(Af/A)= -15.56 dB\n", + "B=(1/(Af))-(1/A)= 8.33e-05\n" + ] + } + ], + "source": [ + "from math import log10\n", + "from __future__ import division \n", + "A=60000\n", + "print \"A= %0.2f\"%(A) #Amplifier gain\n", + "Af=10000\n", + "print \"Af= %0.2f\"%(Af) #Feedback gain\n", + "N_dB=20*log10(Af/A)\n", + "print \"N_dB=20*log10(Af/A)= %0.2f\"%(N_dB),\"dB\" #Negative feedback gain\n", + "B=(1/(Af))-(1/A)# formulae using (Af=A/(1+A*B))\n", + "print \"B=(1/(Af))-(1/A)= %0.2e\"%(B) #Feedback factor" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7_2 Page No. 205" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "A= 10000.00\n", + "B= 0.01\n", + "Af= (A/(1+A*B))=99.01\n", + "A1= 100000.00\n", + "Af1= (A1/(1+A1*B))=99.90\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "A=10000\n", + "print \"A= %0.2f\"%(A) #Amplifier gain\n", + "B=0.01\n", + "print \"B= %0.2f\"%(B) #Feedback factor\n", + "Af=(A/(1+A*B))\n", + "print \"Af= (A/(1+A*B))=%0.2f\"%(Af) #Feedback gain\n", + "A1=100000\n", + "print \"A1= %0.2f\"%(A1) #New amplifier gain value\n", + "Af1=(A1/(1+A1*B))\n", + "print \"Af1= (A1/(1+A1*B))=%0.2f\"%(Af1) #New feedback gain" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7_3 Page No. 208" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vo= 50.00 volts\n", + "Vi= 0.50 volts\n", + "part(i)\n", + "A= Vo/Vi=100.00\n", + "Harmonic_distortion=10.00 %\n", + "D= (10*Vo)/100 = 5.00 volts\n", + "Df= (1*Vo)/100 = 0.50 volts\n", + "B=(D/(Df*A))-(1/A) = 0.09\n", + "part(ii)\n", + "Af= (A/(1+A*B)) = 10.00\n", + "part(iii)\n", + "Vif= Vo/Af = 5.00 volts\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "Vo=(50)\n", + "print \"Vo= %0.2f\"%(Vo),\" volts\" # output voltage\n", + "Vi=(0.5)\n", + "print \"Vi= %0.2f\"%(Vi),\" volts\" # input voltage\n", + "print \"part(i)\" \n", + "A=Vo/Vi\n", + "print \"A= Vo/Vi=%0.2f\"%(A) #Amplifier gain\n", + "HD=10\n", + "print \"Harmonic_distortion=%0.2f\"%(HD),\"%\"# Percentage second harmonic distortion\n", + "D=(10*Vo)/100\n", + "print \"D= (10*Vo)/100 = %0.2f\"%(D),\" volts\" # Second Harmonic distortion \n", + "Df=(1*Vo)/100\n", + "print \"Df= (1*Vo)/100 = %0.2f\"%(Df),\" volts\" # Harmonic distortion with Feedback\n", + "B=(D/(Df*A))-(1/A) #Using formulae Df=(D/(1+A*B))\n", + "print \"B=(D/(Df*A))-(1/A) = %0.2f\"%(B) #Feedback factor\n", + "print \"part(ii)\" \n", + "Af=(A/(1+A*B))\n", + "print \"Af= (A/(1+A*B)) = %0.2f\"%(Af) #Feedback gain\n", + "print \"part(iii)\" \n", + "Vif=Vo/Af\n", + "print \"Vif= Vo/Af = %0.2f\"%(Vif),\" volts\" # New input voltage required" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7_4 Page No. 210" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "GBW= 1000000.00 Hz\n", + "AMf=100.00\n", + "fHF=GBW/AMf= 10000.00 Hz\n", + "f_10per cent=(10*fHF)/100= 1000.00 Hz\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "GBW=10**(6)\n", + "print \"GBW= %0.2f\"%(GBW),\" Hz\"# Gain-Bandwidth product\n", + "AMf=100\n", + "print \"AMf=%0.2f\"%(AMf) # Midband gain with feedback\n", + "fHF=GBW/AMf\n", + "print \"fHF=GBW/AMf= %0.2f\"%(fHF),\" Hz\"#Signal bandwidth\n", + "f_10percent=(10*fHF)/100\n", + "print \"f_10per cent=(10*fHF)/100= %0.2f\"%(f_10percent),\" Hz\"#Frequency below which AMf will not deviate by more than 10 percent" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7_5 Page No. 212" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "AM=50000.00\n", + "fH= 20000.00 Hz\n", + "fL= 30.00 Hz\n", + "B= 5.00e-05\n", + "AMf=AM/(1+B*AM)=14285.71\n", + "fHf=fH*(1+B*AM)= 70000.00 Hz\n", + "fLf=fL/(1+B*AM)= 8.57 Hz\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "AM=50000\n", + "print \"AM=%0.2f\"%(AM) # Midband gain \n", + "fH=20*10**(3)\n", + "print \"fH= %0.2f\"%(fH),\" Hz\"# Upper cut-off frequency\n", + "fL=30\n", + "print \"fL= %0.2f\"%(fL),\" Hz\"# Lower cut-off frequency\n", + "B=5*10**(-5)\n", + "print \"B= %0.2e\"%(B) #Feedback factor\n", + "AMf=AM/(1+B*AM)\n", + "print \"AMf=AM/(1+B*AM)=%0.2f\"%(AMf) # Midband gain with feedback\n", + "fHf=fH*(1+B*AM)\n", + "print \"fHf=fH*(1+B*AM)= %0.2f\"%(fHf),\" Hz\"#Upper cut-off frequency with feedback\n", + "fLf=fL/(1+B*AM)\n", + "print \"fLf=fL/(1+B*AM)= %0.2f\"%(fLf),\" Hz\"#Lower cut-off frequency with feedback\n", + "#NOTE: calculated value of AMf is AMf=14285.714 and fLF=8.5714286 but in book given as AMf=14286 and fLF=8.58 Hz" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7_6 Page No. 214" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "AM=100.00 dB\n", + "fc1= 10000.00 Hz\n", + "fc2= 100000.00 Hz\n", + "fc3= 1000000.00 Hz\n", + "part(i)\n", + "Af1=85.00 dB\n", + "f= 50000.00 Hz\n", + "theta_A=-108.12 degree\n", + "theta_pm=180-abs(theta_A)=71.88 degree\n", + "Amplifier stable\n", + "part(ii)\n", + "Af2=50.00 dB\n", + "f= 500000.00 Hz\n", + "theta_A= -194.11 degree\n", + "theta_pm=180-abs(theta_A)=-14.11 degree\n", + "Amplifier unstable\n", + "part(iii)\n", + "Af3=20.00 dB\n", + "f= 1100000.00 Hz\n", + "theta_A=-222.01 degree\n", + "theta_pm=180-abs(theta_A)=-42.01 degree\n", + "Amplifier unstable\n" + ] + } + ], + "source": [ + "from math import atan,pi\n", + "from __future__ import division \n", + "AM=100\n", + "print \"AM=%0.2f\"%(AM),\"dB\" # Midband gain \n", + "fc1=1*10**(4)\n", + "print \"fc1= %0.2f\"%(fc1),\" Hz\"# First Critical frequency\n", + "fc2=10**5\n", + "print \"fc2= %0.2f\"%(fc2),\" Hz\"# Second Critical frequency\n", + "fc3=10**6\n", + "print \"fc3= %0.2f\"%(fc3),\" Hz\"# Third Critical frequency\n", + "print \"part(i)\"\n", + "Af1=85\n", + "print \"Af1=%0.2f\"%(Af1),\"dB\" # gain at 50 kHz and -20dB/decade roll-off\n", + "f=50*10**(3)\n", + "print \"f= %0.2f\"%(f),\" Hz\"# operating frequency\n", + "theta_A=- atan(f/fc1)*180/pi- atan(f/fc2)*180/pi- atan(f/fc3)*180/pi#phase shift in radians\n", + "print \"theta_A=%0.2f\"%(theta_A),\" degree\"# Phase shift for feedback gain Af1\n", + "theta_pm=180-abs(theta_A)# formulae phase margin\n", + "print \"theta_pm=180-abs(theta_A)=%0.2f\"%(theta_pm),\" degree\"# Phase Margin for feedback gain Af1\n", + "print \"Amplifier stable\"# Since phase margin is (+)ive\n", + "print \"part(ii)\"\n", + "Af2=50\n", + "print \"Af2=%0.2f\"%(Af2),\" dB\" # gain at 500 kHz and -40dB/decade roll-off\n", + "f=500*10**(3)\n", + "print \"f= %0.2f\"%(f),\" Hz\"# frequency\n", + "theta_A=- atan(f/fc1)- atan(f/fc2)- atan(f/fc3)#phase shift in radians\n", + "theta_A=theta_A*180/pi # degree\n", + "print \"theta_A= %0.2f\"%(theta_A),\" degree\"# Phase shift for feedback gain Af2\n", + "theta_pm=180-abs(theta_A)# formulae phase margin\n", + "print \"theta_pm=180-abs(theta_A)=%0.2f\"%(theta_pm),\" degree\"# Phase Margin for feedback gain Af1\n", + "print \"Amplifier unstable\"# Since phase margin is (-)ive\n", + "print \"part(iii)\"\n", + "Af3=20\n", + "print \"Af3=%0.2f\"%(Af3),\"dB\" # gain at 1100 kHz and -60dB/decade roll-off\n", + "f=1100*10**(3)\n", + "print \"f= %0.2f\"%(f),\" Hz\"# frequency\n", + "theta_A=- atan(f/fc1)- atan(f/fc2)- atan(f/fc3)#phase shift in radians\n", + "theta_A=theta_A*180/pi # degree\n", + "print \"theta_A=%0.2f\"%(theta_A),\" degree\"# Phase shift for feedback gain Af3\n", + "theta_pm=180-abs(theta_A)# formulae phase margin\n", + "print \"theta_pm=180-abs(theta_A)=%0.2f\"%(theta_pm),\" degree\"# Phase Margin for feedback gain Af1\n", + "print \"Amplifier unstable\"# Since phase margin is (-)ive\n", + "#NOTE:Correct ans for part(i) phase margin ,theta_pm=71.882476 degree but in book given as 71.86 degree\n", + "# correct ans for part(iii) phase shift, theta_A=-222.01103 degree but in book given as -220.02 degree " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7_7 Page No. 216" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "AV=50000.00\n", + "Ri= 5.00e+07 ohm\n", + "R0= 1000.00 ohm\n", + "AVf=10.00\n", + "RSf= 50000.00 ohm\n", + "RF=AVf*(R1)= 5.00e+05 ohm\n", + "VS= 30.00 volts\n", + "Vomax=0.5*(VS)= -15.00 , +15.00 volts\n", + "Vsmax=Vomax/AVf= -1.50 , +1.50 volts\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "AV=50000\n", + "print \"AV=%0.2f\"%(AV) # Voltage gain \n", + "Ri=50*10**(6)\n", + "print \"Ri= %0.2e\"%(Ri),\" ohm\" #Input resistance of OP-AMP\n", + "R0=1*10**(3)\n", + "print \"R0= %0.2f\"%(R0),\" ohm\" #Output resistance\n", + "AVf=10\n", + "print \"AVf=%0.2f\"%(AVf) # Overall Voltage gain \n", + "RSf=50*10**(3)\n", + "print \"RSf= %0.2f\"%(RSf),\" ohm\" #Source resistance\n", + "R1=RSf\n", + "RF=AVf*(R1)\n", + "print \"RF=AVf*(R1)= %0.2e\"%(RF),\" ohm\" #Feedback resistance\n", + "VS=30\n", + "print \"VS= %0.2f\"%(VS),\" volts\" # Peak-peak output swing voltage\n", + "Vomax=0.5*(VS)\n", + "print \"Vomax=0.5*(VS)= -%0.2f\"%(Vomax),\", +%0.2f\"%(Vomax),\" volts\" # Maximum output voltage swing at negative and positive polarities respectively\n", + "Vsmax=Vomax/AVf\n", + "print \"Vsmax=Vomax/AVf= -%0.2f\"%(Vsmax),\", +%0.2f\"%(Vsmax),\" volts\" # Maximum output voltage without overload clipping at both polarities\n", + "\n", + "\n", + "#for overall voltage gain author has used two notations 'Avf' and 'Af' ... but I am working with 'Avf' only" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7_8 Page No. 218" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R1= 50000.00 ohm\n", + "RF= 500000.00 ohm\n", + "VS= 1.00 volts\n", + "part(i)\n", + "A = infinite\n", + "Vo1=-(RF/R1)= -10.00 volts\n", + "part(ii)\n", + "A=50000.00\n", + "B=R1/(R1+RF)= 0.09\n", + "Vo2=-((RF)*(B*A))/(R1*(1+A*B))= -10.00 volts\n", + "%Error,e= ((Vo2-Vo1)*100)/Vo1=0.02 % \n", + "part(iii)\n", + "%Error,e=0.01 % \n", + "Vo3=Vo1-(e*Vo1/100)= 10.00 volts\n", + "A=(Vo*R1)/(B*RF*(1-(Vo*RF/R1)))=1.10e+05\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "R1=50*10**(3)\n", + "print \"R1= %0.2f\"%(R1),\" ohm\" # resistance at input terminal of OP-AMP\n", + "RF=500*10**(3)\n", + "print \"RF= %0.2f\"%(RF),\" ohm\" #Feedback resistance\n", + "VS=1\n", + "print \"VS= %0.2f\"%(VS),\" volts\" # Peak-peak output swing voltage\n", + "print \"part(i)\" \n", + "print \"A = infinite\"# voltage gain\n", + "Vo1=-(RF/R1) #Output voltage when gain, A=infinite\n", + "print \"Vo1=-(RF/R1)= %0.2f\"%(Vo1),\" volts\"\n", + "print \"part(ii)\" \n", + "A=50000\n", + "print \"A=%0.2f\"%(A) # gain of OP-AMP\n", + "B=R1/(R1+RF)\n", + "print \"B=R1/(R1+RF)= %0.2f\"%(B) #Feedback factor\n", + "Vo2=-((RF)*(B*A))/(R1*(1+A*B))\n", + "print \"Vo2=-((RF)*(B*A))/(R1*(1+A*B))= %0.2f\"%(Vo2),\" volts\"# output voltage for A=50000\n", + "e=-((Vo2-Vo1)*100)/Vo1\n", + "print \"%%Error,e= ((Vo2-Vo1)*100)/Vo1=%0.2f\"%(e),\"% \"# calculation for percentage error in output voltage\n", + "print \"part(iii)\" \n", + "e=0.01\n", + "print \"%%Error,e=%0.2f\"%(e),\"% \"#Given percentage error in output voltage\n", + "Vo3=-(Vo1-(e*Vo1/100))\n", + "print \"Vo3=Vo1-(e*Vo1/100)= %0.2f\"%(Vo3),\" volts\"# output voltage for error 0.01%\n", + "x=Vo3*(R1/RF)\n", + "A=(x)/(B*(1-x)) #using formulae Vo=-(RF/R1)*((B*A)/1+A*B))\n", + "print \"A=(Vo*R1)/(B*RF*(1-(Vo*RF/R1)))=%0.2e\"%(A) # New Required gain for error less than 0.01%\n", + "\n", + "# while solving the problem I have used 'e' for the error as no varriable is given for the same in textbook by author\n", + "# in textbook author has used 'Vo' for output voltage in all parts.. but to remove any ambiguity in the programe I have used 'Vo1' 'Vo2' 'Vo3' for part i, ii, iii, respectively" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7_9 Page No. 218" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "AV=100000.00\n", + "Ri= 10000.00 ohm\n", + "Ro= 10.00 ohm\n", + "Rs= 1.00e+07 ohm\n", + "RL= 1000.00 ohm\n", + "B=(Rs-Ri)/(AV*Ri)= 0.01\n", + "AVf=AV/(1+B*AV)=100.00\n", + "Rof=Ro/(1+B*AV) =0.01 ohm\n", + "Rif=Ri/(1+B*AV) =1.00e+07 ohm\n", + "Ap=(AVf**2)*(Rif/RL)=1.00e+08\n", + "AP=10*log10(Ap)=80.00 dB\n" + ] + } + ], + "source": [ + "from math import log10\n", + "from __future__ import division \n", + "AV=100000\n", + "print \"AV=%0.2f\"%(AV) # Voltage gain \n", + "Ri=10*10**(3)\n", + "print \"Ri= %0.2f\"%(Ri),\" ohm\" #Input resistance of OP-AMP\n", + "Ro=10\n", + "print \"Ro= %0.2f\"%(Ro),\" ohm\" #Output resistance\n", + "Rs=10*10**(6)\n", + "print \"Rs= %0.2e\"%(Rs),\" ohm\" #Source resistance\n", + "RL=1*10**(3)\n", + "print \"RL= %0.2f\"%(RL),\" ohm\" #Load resistance\n", + "B=(Rs-Ri)/(AV*Ri)\n", + "print \"B=(Rs-Ri)/(AV*Ri)= %0.2f\"%(B) #Feedback factor\n", + "AVf=AV/(1+B*AV)\n", + "print \"AVf=AV/(1+B*AV)=%0.2f\"%(AVf) # Overall Voltage gain with feedback\n", + "Rof=Ro/(1+B*AV)\n", + "print \"Rof=Ro/(1+B*AV) =%0.2f\"%(Rof),\" ohm\" #output resistance with feedback\n", + "Rif=Ri*(1+B*AV)\n", + "print \"Rif=Ri/(1+B*AV) =%0.2e\"%(Rif),\" ohm\" #Input resistance with feedback\n", + "Ap=(AVf**2)*(Rif/RL)\n", + "print \"Ap=(AVf**2)*(Rif/RL)=%0.2e\"%(Ap) # Overall Power gain \n", + "AP=10*log10(Ap)\n", + "print \"AP=10*log10(Ap)=%0.2f\"%(AP),\"dB\" # Overall Power gain in dB " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7_10 Page No. 220" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "gm = 0.01 A/V\n", + "Cgs= 5.00e-12 farad\n", + "Cds= 1.00e-12 farad\n", + "rd= 50000.00 ohm\n", + "RG= 1.00e+07 ohm\n", + "Rse= 1.00e+03 ohm\n", + "L= 0.50 H\n", + "C2= 5.00e-14 farad\n", + "C1= 1.00e-12 farad\n", + "part(i)\n", + "CT= 0.00 farad\n", + "part(ii)\n", + "fo= sqrt(2)/(2*pi*sqrt(L*CT))=1.44e+06 Hz\n", + "part(iii)\n", + "fp= 1.03e+06 Hz\n", + "fs= 1.01e+06 Hz\n", + "Q=(sqrt(L/C2))/(Rse)= 3162.28\n", + "part(iv)\n", + "AB=gm*rd*(Cds/Cgs)= 100.00\n", + "T_bias=RG*(Cgs+Cds)= 6.00e-05 s\n", + "T_r =1/(2*pi*fo)= 1.10e-07 s\n", + "for proper operation T_bias >> T_r\n" + ] + } + ], + "source": [ + "from math import sqrt,pi,log10\n", + "from __future__ import division \n", + "gm=10*10**(-3)\n", + "print \"gm = %0.2f\"%(gm),\" A/V\"# transconductance \n", + "Cgs=5*10**(-12)\n", + "print \"Cgs= %0.2e\"%(Cgs),\" farad\" # capacitance between gate-source\n", + "Cds=1*10**(-12)\n", + "print \"Cds= %0.2e\"%(Cds),\" farad\" # capacitance between drain-source\n", + "rd=50*10**(3)\n", + "print \"rd= %0.2f\"%(rd),\" ohm\" #Drain resistance\n", + "RG=10*10**(6)\n", + "print \"RG= %0.2e\"%(RG),\" ohm\" #Gate resistance\n", + "Rse=1*10**(3)\n", + "print \"Rse= %0.2e\"%(Rse),\" ohm\" #Gate resistance\n", + "L=0.5\n", + "print \"L= %0.2f\"%(L),\" H\" #Inductance\n", + "C2=0.05*10**(-12)\n", + "print \"C2= %0.2e\"%(C2),\" farad\" # Crystal parameter \n", + "C1=1*10**(-12)\n", + "print \"C1= %0.2e\"%(C1),\" farad\" # Crystal parameter\n", + "print \"part(i)\" \n", + "x=C1+((Cds*Cgs)/(Cds+Cgs))\n", + "CT=1/((1/C2)+(1/x))\n", + "print \"CT= %0.2f\"%(CT),\" farad\" # Equivalent series-resonating capacitance\n", + "print \"part(ii)\" \n", + "fo=sqrt(2)/(2*pi*sqrt(L*CT))\n", + "print \"fo= sqrt(2)/(2*pi*sqrt(L*CT))=%0.2e\"%(fo),\" Hz\"# frequency of oscillations\n", + "print \"part(iii)\"\n", + "z=sqrt((L*C1*C2)/(C1+C2))\n", + "fp=1/(2*pi*z)\n", + "print \"fp= %0.2e\"%(fp),\" Hz\"# parallel-resonant frequency\n", + "p=sqrt(L*C2)\n", + "fs=1/(2*pi*p)\n", + "print \"fs= %0.2e\"%(fs),\" Hz\"# series-resonant frequency\n", + "Q=(sqrt(L/C2))/(Rse)\n", + "print \"Q=(sqrt(L/C2))/(Rse)= %0.2f\"%(Q) #Quality factor\n", + "print \"part(iv)\"\n", + "AB=gm*rd*(Cds/Cgs)\n", + "print \"AB=gm*rd*(Cds/Cgs)= %0.2f\"%(AB) #Loop gain\n", + "T_bias=RG*(Cgs+Cds)\n", + "print \"T_bias=RG*(Cgs+Cds)= %0.2e\"%(T_bias),\"s\"#Bias Time-Constant\n", + "T_r = 1/(2*pi*fo)\n", + "print \"T_r =1/(2*pi*fo)= %0.2e\"%(T_r),\"s\"#resonant Time-Constant for 'fo'\n", + "print \"for proper operation T_bias >> T_r\"\n", + "\n", + "\n", + "# in part (ii)... value calculated for series resonant frequecy 'fo' is wrong in textbook.\n", + "# NOTE: in part(iii)... there is a misprint in the calculated value of Quality factor 'Q' in the textbook.\n", + "#I have used T_r instead of 1/wo (given 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/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch8.ipynb b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch8.ipynb new file mode 100644 index 00000000..5a864926 --- /dev/null +++ b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch8.ipynb @@ -0,0 +1,781 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8 - Linear Op-amp Systems" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_1 Page No. 245" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Amin=8000.00\n", + "Amax=64000.00\n", + "part (i)\n", + "delta_Af=0.01\n", + "delta_A= (Amax-Amin)/Amin = 7.00\n", + "Sg = delta_Af/delta_A = 0.00\n", + " B = (1/Sg - 1)/Amax = 0.01\n", + " part (ii)\n", + "Af_min = Amin/(1+B*Amin) = 90.52\n", + "Af_max = Amax/(1+B*Amax) = 91.43\n", + "variation in Af = 1.01\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "Amin=8000\n", + "print \"Amin=%0.2f\"%(Amin) # Minimum gain of OP-AMP\n", + "Amax=64000\n", + "print \"Amax=%0.2f\"%(Amax) # Maximum gain \n", + "print \"part (i)\"\n", + "delta_Af=0.01\n", + "print \"delta_Af=%0.2f\"%(delta_Af) # Change in overall feedBack gain \n", + "delta_A=(Amax-Amin)/Amin\n", + "print \"delta_A= (Amax-Amin)/Amin = %0.2f\"%(delta_A) # Change in open loop gain \n", + "Sg = delta_Af/delta_A\n", + "B = (1/Sg - 1)/Amax\n", + "print \"Sg = delta_Af/delta_A = %0.2f\"%(Sg)#desensitivity factor\n", + "print \" B = (1/Sg - 1)/Amax = %0.2f\"%(B)#feedBack factor\n", + "print \" part (ii)\"\n", + "Af_min = Amin/(1+B*Amin)#minimum change in overall feedBack gain \n", + "Af_max = Amax/(1+B*Amax)#/maximum change in overall feedBack gain \n", + "print \"Af_min = Amin/(1+B*Amin) = %0.2f\"%(Af_min)\n", + "print \"Af_max = Amax/(1+B*Amax) = %0.2f\"%(Af_max)\n", + "print \"variation in Af = %0.2f\"%(Af_max/Af_min)#variation in Af with feedBack factor 'B'\n", + "\n", + "\n", + "# for above problem author has divided question in two parts but during solution has written 3 parts. \n", + "# part (i) and part (ii) combinedly equivlent to part (i) \n", + "# part (iii) is equivalent to part (ii)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_2 Page No. 248" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Avf=-100.00\n", + "Rif= 1.00 ohm\n", + "RF= -R1*Avf=100.00 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "Avf=-100\n", + "print \"Avf=%0.2f\"%(Avf) # Voltage gain \n", + "Rif=1\n", + "print \"Rif= %0.2f\"%(Rif),\" ohm\" #Input resistance of OP-AMP\n", + "R1=Rif\n", + "RF=-R1*Avf # using formulae Vo=(-RF/R1)*Vi\n", + "print \"RF= -R1*Avf=%0.2f\"%(RF),\" ohm\" #Feedback resistance of OP-AMP\n", + "# NOTE:Error in value of RF since they have given value of Rif=1ohm but calculated RF by using Rif=1 Kilo ohm\n", + "# So i have calculated using Ri=1ohm and hence RF=100 ohm" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_3 Page No. 249" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R11= 1000.00 ohm\n", + "RF= 100000.00 ohm\n", + "R12= 10000.00 ohm\n", + "R13= 100000.00 ohm\n", + "vo = -(100.00 vs1 +10.00 vs2 +1.00 vs3)\n", + "vo = -(0.33 vs1 +0.33 vs2 +0.33 vs3)\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "R11=1*10**(3)\n", + "print \"R11= %0.2f\"%(R11),\" ohm\" # resistance at input terminal of OP-AMP Adder\n", + "RF=100*10**(3)\n", + "print \"RF= %0.2f\"%(RF),\" ohm\" #Feedback resistance\n", + "R12=10*10**(3)\n", + "print \"R12= %0.2f\"%(R12),\" ohm\" # resistance at input terminal of OP-AMP Adder\n", + "R13=100*10**(3)\n", + "print \"R13= %0.2f\"%(R13),\" ohm\" # resistance at input terminal of OP-AMP Adder\n", + "print \"vo = -(%0.2f\"%(RF/R11),\"vs1 +%0.2f\"%(RF/R12),\"vs2 +%0.2f\"%(RF/R13),\"vs3)\" # output voltage of opamp adder in terms of input vs1,vs2 vs3\n", + "\n", + "# for average value of input signal\n", + "n = 3# given inputs are '3'\n", + "R11 = n*RF\n", + "R12 = n*RF\n", + "R13 = n*RF\n", + "print \"vo = -(%0.2f\"%(RF/R11),\" vs1 +%0.2f\"%(RF/R12),\" vs2 +%0.2f\"%(RF/R13),\" vs3)\" # output voltage of opamp adder \n", + "\n", + "\n", + "# note : the output voltage of inverting adder is negative \n", + "# but while calculating weighted output voltage in above question ..author has neglected or miss the negative sign" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_4 Page No. 250" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ir = 0.01 ampere/lumen of radiant energy \n", + "RF= 10000.00 ohm\n", + "E = 0.01 lumens\n", + "IR =Ir*E= 1.00e-04 ampere\n", + "Vo=IR*RF= 1.00 volts\n", + "scale factor=E/Vo= 0.01 lumens/V\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "Ir=10*10**(-3)\n", + "print \"Ir = %0.2f\"%(Ir),\" ampere/lumen of radiant energy \" #photodiode Reverse saturation current for constant reverse bias VR\n", + "RF=10*10**(3)\n", + "print \"RF= %0.2f\"%(RF),\" ohm\" #Feedback resistance\n", + "E=1*10**(-2)\n", + "print \"E = %0.2f\"%(E),\" lumens\"# radiant energy\n", + "IR=Ir*E\n", + "print \"IR =Ir*E= %0.2e\"%(IR),\" ampere\" # Reverse saturation current\n", + "Vo=IR*RF\n", + "print \"Vo=IR*RF= %0.2f\"%(Vo),\" volts\" # output voltage\n", + "s=E/Vo\n", + "print \"scale factor=E/Vo= %0.2f\"%(E),\" lumens/V\" # Scale factor of photometer" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_5 Page No. 252" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Av= 100000.00\n", + "RF= 100000.00 ohm\n", + "RM= 10000.00 ohm\n", + "is = 1.00e-05 ampere\n", + "vo=is*RF= 1.00 volts\n", + "S=vo/is= 100000.00 V/A\n", + "Rif=RF/(1+Av)= 1.00 ohm\n", + "im = 0.00 ampere\n", + "RF=(im*RM)/is= 100000.00 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "Av=1*10**(5)\n", + "print \"Av= %0.2f\"%(Av) #Voltage gain\n", + "RF=100*10**(3)\n", + "print \"RF= %0.2f\"%(RF),\" ohm\" #Feedback resistance\n", + "RM=10*10**(3)\n", + "print \"RM= %0.2f\"%(RM),\" ohm\" # D.C Ammeter internal resistance\n", + "Is=10*10**(-6)\n", + "print \"is = %0.2e\"%(Is),\" ampere\" # Source current\n", + "vo=Is*RF\n", + "print \"vo=is*RF= %0.2f\"%(vo),\" volts\" # output voltage\n", + "S=vo/Is\n", + "print \"S=vo/is= %0.2f\"%(S),\" V/A\" # Sensitivity of Ammeter\n", + "Rif=RF/(1+Av)\n", + "print \"Rif=RF/(1+Av)= %0.2f\"%(Rif),\" ohm\" #Input resistance of OP-AMP\n", + "im=100*10**(-6)\n", + "print \"im = %0.2f\"%(im),\" ampere\" # Meter Full-Scale deflection current \n", + "RF=(im*RM)/Is\n", + "print \"RF=(im*RM)/is= %0.2f\"%(RF),\" ohm\" # New required Feedback resistance for im=100 micro ampere" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_6 Page No. 255" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Av= 36.00 dB\n", + "R1= 1000.00 ohm\n", + "RF=R1*[10**(Av/20)-1]= 62095.73 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "Av=36\n", + "print \"Av= %0.2f\"%(Av),\" dB\" #Voltage gain\n", + "R1=1*10**(3)# Choosing value of R1\n", + "print \"R1= %0.2f\"%(R1),\" ohm\" # Resistor at input side of OP-AMP\n", + "RF=R1*(10**(Av/20)-1) # Using formulae Av=20*log(1+RF/R1)\n", + "print \"RF=R1*[10**(Av/20)-1]= %0.2f\"%(RF),\" ohm\" # Calculated Feedback resistance\n", + "#NOTE: Correct value of RF=62095.734 ohm or 62.095 kilo ohm but in book given as 62.24 kilo ohm \n", + " " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_7 Page No. 256" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "if = 1.00e-04 ampere\n", + "Av= 100000.00\n", + "vs= 0.01 volts\n", + "RM= 100.00 ohm\n", + "Ri= 10000.00 ohm\n", + "R1=vs/if= 100.00 ohm\n", + "Avf=1+(RM/R1)=2.00\n", + "Rif=Ri*(Av/Avf)=5.00e+08 ohm\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "If=100*10**(-6)\n", + "print \"if = %0.2e\"%(If),\" ampere\" #Full-Scale deflection current\n", + "Av=1*10**(5)\n", + "print \"Av= %0.2f\"%(Av) #Voltage gain \n", + "vs=10*10**(-3) \n", + "print \"vs= %0.2f\"%(vs),\" volts\" # Input voltage \n", + "RM=100\n", + "print \"RM= %0.2f\"%(RM),\" ohm\" # Moving coil Ammeter internal resistance\n", + "Ri=10*10**(3)\n", + "print \"Ri= %0.2f\"%(Ri),\" ohm\" #Input resistance of OP-AMP\n", + "R1=vs/If\n", + "print \"R1=vs/if= %0.2f\"%(R1),\" ohm\" # Resistor at input side of OP-AMP in Voltage-to-Current converter \n", + "Avf=1+(RM/R1) # formulae using Avf=1+(RF/R1)=1+(RM/R1)# since RF=RM\n", + "print \"Avf=1+(RM/R1)=%0.2f\"%(Avf) # Overall Voltage gain\n", + "Rif=Ri*(Av/Avf)\n", + "print \"Rif=Ri*(Av/Avf)=%0.2e\"%(Rif),\" ohm\" # Equivalent input side resistance of OP-AMP with feedback" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_8 Page No. 258" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ro= 1.00e-03 ohm\n", + "Sv= 0.01 %\n", + "change in regulator voltage= 3.00 volts\n", + "change in regulator Current= 0.25 A\n", + "change in regulator output voltage= 0.00 volts\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "Ro=0.001\n", + "print \"Ro= %0.2e\"%(Ro),\" ohm\" #Output resistance\n", + "Sv=0.01\n", + "print \"Sv= %0.2f\"%(Sv),\"%\" # Input Regulation for IC regulator\n", + "delta_VI=12-9\n", + "print \"change in regulator voltage= %0.2f\"%(delta_VI),\" volts\" # Regulator input voltage variation\n", + "delta_IL=1.25-1\n", + "print \"change in regulator Current= %0.2f\"%(delta_IL),\" A\" # Regulator Current variation\n", + "delta_Vo=(delta_VI*(Sv/100)+delta_IL*Ro)\n", + "print \"change in regulator output voltage= %0.2f\"%(delta_Vo),\" volts\" # Regulator output voltage variation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_9 Page No. 260" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "alpha=1.41\n", + "AM=1.59\n", + "fOH= 1000.00 Hz\n", + "R1= 10000.00 ohm\n", + "RF=R1*(AM-1)=5860.00 ohm\n", + "C=1.00e-07 farad\n", + "R=1/(omega_OH*C)=1/(2*pi*fOH*C)=1591.55 ohm\n" + ] + } + ], + "source": [ + "from math import pi\n", + "from __future__ import division \n", + "alpha=1.414# Damping coefficient for Butterworth LP filter\n", + "print \"alpha=%0.2f\"%(alpha)\n", + "AM=3-alpha\n", + "print \"AM=%0.2f\"%(AM) # Midband gain of filter \n", + "fOH=1*10**(3)\n", + "print \"fOH= %0.2f\"%(fOH),\" Hz\"#Cut off frequency\n", + "R1=10*10**(3)# Choosing value of R1 same as in book\n", + "print \"R1= %0.2f\"%(R1),\" ohm\" # Resistor at input side of (OP-AMP)filter \n", + "RF=R1*(AM-1)\n", + "print \"RF=R1*(AM-1)=%0.2f\"%(RF),\" ohm\" #Feedback resistance \n", + "C=0.1*10**(-6) # Choosing value of capacitor same a in book\n", + "print \"C=%0.2e\"%(C),\"farad\"\n", + "R=1/(2*pi*fOH*C)# Using formulae wOH=1/C*R and wOH=(2*pi*fOH)\n", + "print \"R=1/(omega_OH*C)=1/(2*pi*fOH*C)=%0.2f\"%(R),\" ohm\" # resistance value for filter design" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_10 Page No. 261" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "fo= 150.00 Hz\n", + "BW= 15.00 Hz\n", + "Q= 10.00\n", + "C=5.00e-08 farad\n", + "R=sqrt(2)/(2*pi*fo*C)=30010.54 ohm\n", + "Am=5-(sqrt(2)/Q)=4.86\n", + "Abp=Am/(5-Am)=34.36\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "from __future__ import division \n", + "fo=150\n", + "print \"fo= %0.2f\"%(fo),\" Hz\"#Central frequency of band pass filter\n", + "BW=15\n", + "print \"BW= %0.2f\"%(BW),\" Hz\"# Upper cut-off frequency or 3-dB bandwidth\n", + "Q=fo/BW # Quality factor\n", + "print \"Q= %0.2f\"%(Q)\n", + "C=0.05*10**(-6) # Choosing value of capacitor same as in book\n", + "print \"C=%0.2e\"%(C),\"farad\"\n", + "R=sqrt(2)/(2*pi*fo*C)\n", + "print \"R=sqrt(2)/(2*pi*fo*C)=%0.2f\"%(R),\" ohm\" # resistance value for filter design\n", + "Am=5-(sqrt(2)/Q) # formulae\n", + "print \"Am=5-(sqrt(2)/Q)=%0.2f\"%(Am) # Midband gain \n", + "Abp=Am/(5-Am)\n", + "print \"Abp=Am/(5-Am)=%0.2f\"%(Abp) # Central frequency gain " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_11 Page No. 263" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R= 10000.00 ohm\n", + "R1= 10000.00 ohm\n", + "C=1.00e-08 farad\n", + "R1_ratio_K= 2500.00 ohm\n", + "R= 10000.00 ohm\n", + "alpha_R2= 250.00 ohm\n", + "alpha=alpha_R2/R2= 0.05\n", + "Q= 1/alpha=20.00\n", + "omega_o=1/(R*C)= 10000.00 radian\n", + "Bandwidth=omega_o/Q= 500.00 radian\n", + "K=R1/(R1_ratio_K)= 4.00\n", + "center frequency gain for BPF, K/alpha=K*Q= 80.00\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "R=10*10**(3)\n", + "print \"R= %0.2f\"%(R),\" ohm\" # resistance\n", + "R1=10*10**(3)\n", + "print \"R1= %0.2f\"%(R1),\" ohm\" # resistance\n", + "C=0.01*10**(-6) # value of capacitor\n", + "print \"C=%0.2e\"%(C),\" farad\"\n", + "R1_ratio_K=2.5*10**(3)\n", + "print \"R1_ratio_K= %0.2f\"%(R1_ratio_K),\" ohm\" # resistance\n", + "R2=5*10**(3)\n", + "print \"R= %0.2f\"%(R),\" ohm\" # resistance\n", + "alpha_R2=250\n", + "print \"alpha_R2= %0.2f\"%(alpha_R2),\" ohm\" # resistance\n", + "alpha=alpha_R2/R2\n", + "print \"alpha=alpha_R2/R2= %0.2f\"%(alpha) # Damping factor\n", + "Q=1/alpha\n", + "print \"Q= 1/alpha=%0.2f\"%(Q) # Quality factor\n", + "omega_o=1/(R*C)\n", + "print \"omega_o=1/(R*C)= %0.2f\"%(omega_o),\" radian\"# centre angular frequency\n", + "BW=omega_o/Q\n", + "print \"Bandwidth=omega_o/Q= %0.2f\"%(BW),\" radian\"# Upper cut-off frequency or 3-dB bandwidth\n", + "K=R1/(R1_ratio_K)# Pass band gain for lPF and HPF of state variable filter\n", + "print \"K=R1/(R1_ratio_K)= %0.2f\"%(K)\n", + "Gm=K/alpha# Pass band gain of state variable filter\n", + "print \"center frequency gain for BPF, K/alpha=K*Q= %0.2f\"%(Gm) # Centre frequency gain for BP filter" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_12 Page No. 264" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "IB = 5.00e-07 ampere\n", + "Iio = 5.00e-08 ampere\n", + "Vio= 1.00e-03 volts\n", + "R1= 10000.00 ohm\n", + "RF= 500000.00 ohm\n", + "Vos1=Vio*(1+RF/R1)=0.05 volts\n", + "Vos2=IB*RF=0.25 volts\n", + "Vos=Vos1+Vos2=0.30 volts\n", + "R2=(R1*RF)/(R1+RF)= 9803.92 ohm\n", + "Vos2=Iio*RF=0.02 volts\n", + "Vos=Vos1+Vos2=0.08 volts\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "IB=0.5*10**(-6)\n", + "print \"IB = %0.2e\"%(IB),\" ampere\" #Input bias current \n", + "Iio=0.05*10**(-6)\n", + "print \"Iio = %0.2e\"%(Iio),\" ampere\" #Input offset current \n", + "Vio=1*10**(-3)\n", + "print \"Vio= %0.2e\"%(Vio),\" volts\" #Input offset voltage\n", + "R1=10*10**(3)\n", + "print \"R1= %0.2f\"%(R1),\" ohm\" # resistance\n", + "RF=500*10**(3)\n", + "print \"RF= %0.2f\"%(RF),\" ohm\" #Feedback resistance\n", + "Vos1=Vio*(1+RF/R1)\n", + "print \"Vos1=Vio*(1+RF/R1)=%0.2f\"%(Vos1),\" volts\" #output offset voltage due to input offset voltage\n", + "Vos2=IB*RF\n", + "print \"Vos2=IB*RF=%0.2f\"%(Vos2),\" volts\" #output offset voltage due to Input bias current \n", + "Vos=Vos1+Vos2\n", + "print \"Vos=Vos1+Vos2=%0.2f\"%(Vos),\" volts\" #total output offset voltage \n", + "R2=(R1*RF)/(R1+RF)\n", + "print \"R2=(R1*RF)/(R1+RF)= %0.2f\"%(R2),\" ohm\" # resistance to balance IB effect \n", + "Vos2=Iio*RF\n", + "print \"Vos2=Iio*RF=%0.2f\"%(Vos2),\" volts\" # Reduced output offset voltage due to Input offset current \n", + "Vos=Vos1+Vos2\n", + "print \"Vos=Vos1+Vos2=%0.2f\"%(Vos),\" volts\" # output offset voltage with compensation \n", + " " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_13 Page No. 265" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Iio = 1.00e-10 ampere/degree _celsius\n", + "Vio= 1.00e-05 volt/degree _celsius\n", + "Vs= 0.01 volts\n", + "R1= 10000.00 ohm\n", + "RF= 100000.00 ohm\n", + "part(i)\n", + "R2=(R1*RF)/(R1+RF)= 9090.91 ohm\n", + "part(ii)\n", + "delta_T=75-25 = 50.00 degree_celsius\n", + "delta_Vo=[(Vio*delta_T)*(1+RF/R1)]+(Iio*delta_T*RF)= 0.01 volts\n", + "part(iii)\n", + "Vo=(-RF/R1)*Vs= -0.10 volts\n", + "Percentage error=(delta_Vo/Vo)*100 =(-)6.00 , (+)6.00 percent\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "Iio=0.1*10**(-9)\n", + "print \"Iio = %0.2e\"%(Iio),\" ampere/degree _celsius\" #Input offset current \n", + "Vio=10*10**(-6)\n", + "print \"Vio= %0.2e\"%(Vio),\" volt/degree _celsius\" #Input offset voltage\n", + "Vs=10*10**(-3)\n", + "print \"Vs= %0.2f\"%(Vs),\" volts\" #Input voltage\n", + "R1=10*10**(3)\n", + "print \"R1= %0.2f\"%(R1),\" ohm\" # resistance\n", + "RF=100*10**(3)\n", + "print \"RF= %0.2f\"%(RF),\" ohm\" #Feedback resistance\n", + "print \"part(i)\"\n", + "R2=(R1*RF)/(R1+RF)# R1 in parallel with RF\n", + "print \"R2=(R1*RF)/(R1+RF)= %0.2f\"%(R2),\" ohm\" # resistance to balance IB i.e offset effect \n", + "print \"part(ii)\"\n", + "delta_T=75-25\n", + "print \"delta_T=75-25 = %0.2f\"%(delta_T),\" degree_celsius\" #Temperature change\n", + "delta_Vo=((Vio*delta_T)*(1+RF/R1))+(Iio*delta_T*RF)\n", + "print \"delta_Vo=[(Vio*delta_T)*(1+RF/R1)]+(Iio*delta_T*RF)= %0.2f\"%(delta_Vo),\" volts\" #Output voltage drift\n", + "print \"part(iii)\"\n", + "Vo=(-RF/R1)*Vs\n", + "print \"Vo=(-RF/R1)*Vs= %0.2f\"%(Vo),\" volts\" #Inverting OP-AMP output voltage\n", + "e=(delta_Vo/Vo)*100\n", + "print \"Percentage error=(delta_Vo/Vo)*100 =(-)%0.2f\"%(abs(e)),\", (+)%0.2f\"%(abs(e)),\" percent\"#percentage error" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_14 Page No. 266" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Iio = 1.00e-10 ampere\n", + "VCC= 15.00 volts\n", + "PSRR= 1.50e-04 volts/V\n", + "Vio= 1.00e-05 volts\n", + "R1= 10000.00 ohm\n", + "RF= 100000.00 ohm\n", + "delta_T=75-25 = 50.00 celsius\n", + "delta_Vo=[(Vio*delta_T)*(1+RF/R1)]+(Iio*delta_T*RF)= 0.01 volts\n", + "delta_Vio1=(delta_Vo)*(R1/RF)= 0.00 volts\n", + "delta_Vio2=(delta_Vio1)*(1/10)= 0.00 volts\n", + "power supply regulation=[(delta_Vio2)/(VCC*PSRR)]*100 =2.67 percent\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "Iio=0.1*10**(-9)\n", + "print \"Iio = %0.2e\"%(Iio),\" ampere\" #Input offset current\n", + "VCC=15\n", + "print \"VCC= %0.2f\"%(VCC),\" volts\" # voltage supply \n", + "PSRR=150*10**(-6)\n", + "print \"PSRR= %0.2e\"%(PSRR),\" volts/V\"# Power supply rejection ratio\n", + "Vio=10*10**(-6)\n", + "print \"Vio= %0.2e\"%(Vio),\" volts\" #Input offset voltage\n", + "R1=10*10**(3)\n", + "print \"R1= %0.2f\"%(R1),\" ohm\" # resistance\n", + "RF=100*10**(3)\n", + "print \"RF= %0.2f\"%(RF),\" ohm\" #Feedback resistance\n", + "delta_T=75-25\n", + "print \"delta_T=75-25 = %0.2f\"%(delta_T),\" celsius\" #Temperature change\n", + "delta_Vo=((Vio*delta_T)*(1+RF/R1))+(Iio*delta_T*RF)\n", + "print \"delta_Vo=[(Vio*delta_T)*(1+RF/R1)]+(Iio*delta_T*RF)= %0.2f\"%(delta_Vo),\" volts\" #Output voltage drift\n", + "delta_Vio1=(delta_Vo)*(R1/RF)\n", + "print \"delta_Vio1=(delta_Vo)*(R1/RF)= %0.2f\"%(delta_Vio1),\" volts\" # voltage change at Input for voltage drift found\n", + "delta_Vio2=(delta_Vio1)*(1/10)\n", + "print \"delta_Vio2=(delta_Vio1)*(1/10)= %0.2f\"%(delta_Vio2),\" volts\" # change in Vio due to PSRR\n", + "p=((delta_Vio2)/(VCC*PSRR))*100\n", + "print \"power supply regulation=[(delta_Vio2)/(VCC*PSRR)]*100 =%0.2f\"%(p),\" percent\"# power supply regulation requirement\n", + "\n", + "#delta_Vio1 corresponds to voltage change at Input for voltage drift found \n", + "#delta_Vio2 corresponds voltage change at input due to PSRR" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8_15 Page No. 271" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "SR= 0.65 volts/microsecond\n", + "part(i)\n", + "Vm= 5.00 volts\n", + "fsm=SR/[10**(-6)*(2*pi*Vm)] = 20690.14 Hz\n", + "part(ii)\n", + "Vm= 1.00 volts\n", + "fsm=SR/[10**(-6)*(2*pi*Vm)] = 103450.71 Hz\n" + ] + } + ], + "source": [ + "from math import pi\n", + "from __future__ import division \n", + "SR=0.65\n", + "print \"SR= %0.2f\"%(SR),\" volts/microsecond\"# Slew rate of OP-AMP\n", + "print \"part(i)\"\n", + "Vm=5\n", + "print \"Vm= %0.2f\"%(Vm),\" volts\" # Output peak voltage1\n", + "fsm=SR/(10**(-6)*(2*pi*Vm)) # using formulae SR=2*pi*fsm*Vm\n", + "print \"fsm=SR/[10**(-6)*(2*pi*Vm)] = %0.2f\"%(fsm),\" Hz\"# # Full power bandwidth for Output peak voltage Vm=5V\n", + "print \"part(ii)\"\n", + "Vm=1\n", + "print \"Vm= %0.2f\"%(Vm),\" volts\" # Output peak voltage2\n", + "fsm=SR/(10**(-6)*(2*pi*Vm)) # using formulae SR=2*pi*fsm*Vm\n", + "print \"fsm=SR/[10**(-6)*(2*pi*Vm)] = %0.2f\"%(fsm),\" Hz\"# # Full power bandwidth for Output peak voltage Vm=1V" + ] + } + ], + "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/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch9.ipynb b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch9.ipynb new file mode 100644 index 00000000..9a2daba4 --- /dev/null +++ b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/Ch9.ipynb @@ -0,0 +1,596 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9 - Digital Circuits and Logic Families" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9_1 Page No. 278" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VCC= 5.00 volts\n", + "RB= 10000.00 ohm\n", + "RL= 1000.00 ohm\n", + "VCS= 0.20 volts\n", + "VBS= 0.80 volts\n", + "V_gamma= 0.60 volts\n", + "ICS = (VCC-VCS)/RL=4.80e-03 ampere\n", + "vi= 5.00 volts\n", + "IBS=(vi-VBS)/RB=4.20e-04 ampere\n", + "Beta_Fmin=ICS/IBS= 11.43\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "VCC=5\n", + "print \"VCC= %0.2f\"%(VCC),\" volts\" # voltage supply \n", + "RB=10*10**(3)\n", + "print \"RB= %0.2f\"%(RB),\" ohm\" # Base-resistance\n", + "RL=1*10**(3)\n", + "print \"RL= %0.2f\"%(RL),\" ohm\" # Load resistance\n", + "VCS=0.2\n", + "print \"VCS= %0.2f\"%(VCS),\" volts\" # collector saturated voltage \n", + "VBS=0.8\n", + "print \"VBS= %0.2f\"%(VBS),\" volts\" # Base voltage at saturation \n", + "V_gamma=0.6\n", + "print \"V_gamma= %0.2f\"%(V_gamma),\" volts\" # Threshold or cut-in voltage\n", + "ICS=(VCC-VCS)/RL\n", + "print \"ICS = (VCC-VCS)/RL=%0.2e\"%(ICS),\" ampere\" #Saturation collector current of transistor T1\n", + "vi=5\n", + "print \"vi= %0.2f\"%(vi),\" volts\" # Input voltage \n", + "IBS=(vi-VBS)/RB\n", + "print \"IBS=(vi-VBS)/RB=%0.2e\"%(IBS),\" ampere\" # Forward base drive required to sustain ICS\n", + "Beta_Fmin=ICS/IBS\n", + "print \"Beta_Fmin=ICS/IBS= %0.2f\"%(Beta_Fmin) # Common-emitter current gain\n", + "\n", + "#NOTE: Correct formulae for ICS=(VCC-VCS)/RL\n", + "# but in book it is written wrong as ICS=(VCC-VCS)/RB but had calculated ans (in book) according to correct formulae ICS=(VCC-VCS)/RL" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9_2 Page No. 279" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VD= 0.70 V\n", + "part(i)\n", + "vA= 0.00 V\n", + "vB= 0.00 V\n", + "vX=0.00 V\n", + "part(ii)\n", + "vA= 0.00 V\n", + "vB= 5.00 V\n", + "vX=vB-VD= 4.30 V\n", + "part(iii)\n", + "vA= 5.00 V\n", + "vB= 0.00 V\n", + "vX=vA-VD= 4.30 V\n", + "part(iv)\n", + "vA= 5.00 V\n", + "vB= 5.00 V\n", + "vX=vA-VD=vB-VD= 4.30 V\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "VD=0.7\n", + "print \"VD= %0.2f\"%(VD),\" V\" # Diode voltage drop in conduction mode\n", + "\n", + "print \"part(i)\"# part(i)of question\n", + "vA=0\n", + "print \"vA= %0.2f\"%(vA),\" V\" # Input voltage1 of diode OR logic gate\n", + "vB=0\n", + "print \"vB= %0.2f\"%(vB),\" V\" # Input voltage2 of diode OR logic gate\n", + "vX=0 # Since both input voltages vA=vB=0V\n", + "print \"vX=%0.2f\"%(vX),\" V\" # Output voltage of diode OR logic gate for part(i)\n", + "\n", + "print \"part(ii)\"# part(ii)of question\n", + "vA=0\n", + "print \"vA= %0.2f\"%(vA),\" V\" # Input voltage1 of diode OR logic gate\n", + "vB=5\n", + "print \"vB= %0.2f\"%(vB),\" V\" # Input voltage2 of diode OR logic gate for SECOND CASE: when vA=0V and vB=5V\n", + "vX=vB-VD\n", + "print \"vX=vB-VD= %0.2f\"%(vX),\" V\" # Output voltage of diode OR logic gate for SECOND CASE\n", + "\n", + "print \"part(iii)\"# part(iii)of question\n", + "vA=5\n", + "print \"vA= %0.2f\"%(vA),\" V\" # Input voltage1 of diode OR logic gate for THIRD CASE when vA=5V and vB=0V\n", + "vB=0\n", + "print \"vB= %0.2f\"%(vB),\" V\" # Input voltage2 of diode OR logic gate\n", + "vX=vA-VD\n", + "print \"vX=vA-VD= %0.2f\"%(vX),\" V\" # Output voltage of diode OR logic gate for THIRD CASE\n", + "\n", + "print \"part(iv)\"# part(iv)of question\n", + "vA=(+5)\n", + "print \"vA= %0.2f\"%(vA),\" V\" # Input voltage1 of diode OR logic gate\n", + "vB=(+5)\n", + "print \"vB= %0.2f\"%(vB),\" V\" # Input voltage2 of diode OR logic gate\n", + "vX=vA-VD # Since both diodes D1 and D2 are conducting\n", + "vX=vB-VD \n", + "print \"vX=vA-VD=vB-VD= %0.2f\"%(vX),\" V\" # Output voltage of diode OR logic gate for FOURTH CASE: when vA=5V and vB=5V" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9_3 Page No. 281" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VD= 0.70 V\n", + "part(i)\n", + "vA= 0.00 V\n", + "vB= 0.00 V\n", + "vX=VD=0.70 V\n", + "part(ii)\n", + "vA= 0.00 V\n", + "vB= 5.00 V\n", + "vX=VD 0.70 V\n", + "part(iii)\n", + "vA= 5.00 V\n", + "vB= 0.00 V\n", + "vX= 0.70 V\n", + "part(iv)\n", + "vA= 5.00 V\n", + "vB= 5.00 V\n", + "vX = vA = vB= 5.00 V\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "VD=0.7\n", + "print \"VD= %0.2f\"%(VD),\" V\" # Diode voltage drop in conduction mode\n", + "\n", + "print \"part(i)\"\n", + "vA=0\n", + "print \"vA= %0.2f\"%(vA),\" V\" # Input voltage of diode AND logic gate\n", + "vB=0\n", + "print \"vB= %0.2f\"%(vB),\" V\" # Input voltage2 of diode AND logic gate\n", + "\n", + "vX=VD # Since both input voltages vA=vB=0V\n", + "print \"vX=VD=%0.2f\"%(vX),\" V\" # Output voltage of diode AND logic gate for FIRST CASE: when vA=0V and vB=0V\n", + "\n", + "print \"part(ii)\"\n", + "vA=0\n", + "print \"vA= %0.2f\"%(vA),\" V\" # Input voltage of diode AND logic gate\n", + "vB=5\n", + "print \"vB= %0.2f\"%(vB),\" V\" # Input voltage2 of diode AND logic gate for SECOND CASE: when vA=0V and vB=5V\n", + "vX=VD #due to diode A which is conducting and the Diode B is reverse biased with a voltage VD-VB=0.7-5=-4.3\n", + "print \"vX=VD %0.2f\"%(vX),\" V\"\n", + "#due to diode B which is conducting\n", + "\n", + "\n", + "\n", + "print \"part(iii)\"\n", + "vA=5\n", + "print \"vA= %0.2f\"%(vA),\" V\" # Input voltage for THIRD CASE when vA=5V and vB=0V\n", + "vB=0\n", + "print \"vB= %0.2f\"%(vB),\" V\" # Input voltage2 of diode AND logic gate\n", + "vX = VD#due to diode B which is conducting and the Diode A is reverse biased with a voltage VD-VA=0.7-5=-4.3\n", + "print \"vX= %0.2f\"%(vX),\" V\"\n", + "\n", + "print \"part(iv)\"\n", + "vA=5\n", + "print \"vA= %0.2f\"%(vA),\" V\" # Input voltage forfourth CASE when vA=5V and vB=5V\n", + "vB=5\n", + "print \"vB= %0.2f\"%(vB),\" V\" # Input voltage2 of diode AND logic gate for CASE: when vA=0V and vB=5V\n", + "vX=vA # Since both diodes D1 and D2 are Non-conducting, so no voltage drop across'R'(resistor)\n", + "print \"vX = vA = vB= %0.2f\"%(vX),\" V\" # Output voltage of diode AND logic gate for FOURTH CASE: when vA=5V and vB=5V" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9_6 Page No. 285" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VIL= 0.60 V\n", + "VIH= 0.75 V\n", + "VOL= 0.20 V\n", + "VOH= 1.00 V\n", + "NML=VIL-VOL= 0.40 V\n", + "NMH=VOH-VIH= 0.25 V\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "VIL=0.6\n", + "print \"VIL= %0.2f\"%(VIL),\" V\" # Minimum input voltage level for which output is maximum\n", + "VIH=0.75\n", + "print \"VIH= %0.2f\"%(VIH),\" V\" # Maximum input voltage level for which output is minimum\n", + "VOL=0.2\n", + "print \"VOL= %0.2f\"%(VOL),\" V\" # Minimum output voltage level for maximum input level\n", + "VOH=1\n", + "print \"VOH= %0.2f\"%(VOH),\" V\" # Maximum output voltage level for minimum input level\n", + "NML=VIL-VOL\n", + "print \"NML=VIL-VOL= %0.2f\"%(NML),\" V\" # Low level noise immunities\n", + "NMH=VOH-VIH\n", + "print \"NMH=VOH-VIH= %0.2f\"%(NMH),\" V\" # High level noise immunities" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9_7 Page No. 287" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "IIL= -1.60e-03 A\n", + "IIH= 4.00e-05 A\n", + "IOL= 1.60e-02 A\n", + "IOH= -4.00e-04 A\n", + "Fan-out=abs((IOH/IIH)=abs((IOL/IIL))= 10.00\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "IIL=-1.6*10**(-3)\n", + "print \"IIL= %0.2e\"%(IIL),\" A\" # Input sink Current of TTL driver\n", + "IIH=40*10**(-6)\n", + "print \"IIH= %0.2e\"%(IIH),\" A\" # source (supply) reverse Current of TTL driver\n", + "IOL=16*10**(-3)\n", + "print \"IOL= %0.2e\"%(IOL),\" A\" # Specified Maximum sink Current of TTL driver\n", + "IOH=-400*10**(-6)\n", + "print \"IOH= %0.2e\"%(IOH),\" A\" # Specified Maximum source Current of TTL driver\n", + "Fan_out=abs((IOH/IIH))\n", + "print \"Fan-out=abs((IOH/IIH)=abs((IOL/IIL))= %0.2f\"%(Fan_out)# Fan-out of TTL" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9_8 Page No. 288" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VIL= 0.80 V\n", + "VIH= 2.00 V\n", + "VOL= 0.40 V\n", + "VOH= 2.40 V\n", + "NML=VIL-VOL= 0.40 V\n", + "NMH=VOH-VIH= 0.40 V\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "VIL=0.8\n", + "print \"VIL= %0.2f\"%(VIL),\" V\" # Minimum input voltage level for which output is maximum\n", + "VIH=2\n", + "print \"VIH= %0.2f\"%(VIH),\" V\" # Maximum input voltage level for which output is minimum\n", + "VOL=0.4\n", + "print \"VOL= %0.2f\"%(VOL),\" V\" # Minimum output voltage level for maximum input level\n", + "VOH=2.4\n", + "print \"VOH= %0.2f\"%(VOH),\" V\" # Maximum output voltage level for minimum input level\n", + "NML=VIL-VOL\n", + "print \"NML=VIL-VOL= %0.2f\"%(NML),\" V\" # Low level noise immunities\n", + "NMH=VOH-VIH\n", + "print \"NMH=VOH-VIH= %0.2f\"%(NMH),\" V\" # High level noise immunities" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9_9 Page No. 289" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VIL= 1.00 V\n", + "VIH= 4.00 V\n", + "VOL= 0.50 V\n", + "VOH= 4.50 V\n", + "NML=VIL-VOL= 0.50 V\n", + "NMH=VOH-VIH= 0.50 V\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "VIL=1\n", + "print \"VIL= %0.2f\"%(VIL),\" V\" # Minimum input voltage level for which output is maximum\n", + "VIH=4\n", + "print \"VIH= %0.2f\"%(VIH),\" V\" # Maximum input voltage level for which output is minimum\n", + "VOL=0.5\n", + "print \"VOL= %0.2f\"%(VOL),\" V\" # Minimum output voltage level for maximum input level\n", + "VOH=4.5\n", + "print \"VOH= %0.2f\"%(VOH),\" V\" # Maximum output voltage level for minimum input level\n", + "NML=VIL-VOL\n", + "print \"NML=VIL-VOL= %0.2f\"%(NML),\" V\" # Low level noise immunities\n", + "NMH=VOH-VIH\n", + "print \"NMH=VOH-VIH= %0.2f\"%(NMH),\" V\" # High level noise immunities" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9_10 Page No. 291" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "V_gamma= 0.60 volts\n", + "VEE= -5.20 volts\n", + "VBE3=VBE4=VBE5 0.70 volts\n", + "RE= 779.00 ohm\n", + "RL2= 220.00 ohm\n", + "RL3= 245.00 ohm\n", + "VREF= -1.29 volts\n", + "V(1)= -0.70 volts\n", + "V(0)= -1.70 volts\n", + "part(i)\n", + "VE=VREF-VBE3= -1.99 volts\n", + "IE=(VEE-VE)/RE= -0.00 A\n", + "IC3=IE= -0.00 A\n", + "vC3=IC3*RL3= -1.01 volts\n", + "vY=vC3-VBE5= -1.71 volts\n", + "vX=vC2-VBE4= -0.70 volts\n", + "Base -Emitter reverse voltage,VBEr=V(0)-VE= 0.29 volts\n", + "Transistor T1 and T2 off since VBEr < V_gamma\n", + "part(ii)\n", + "vY=VC3-VBE= -0.70 volts\n", + "VE=vB-VBE= -1.40 volts\n", + "VBE3=VREF-VE= 0.11 volts\n", + "VBE3 is smaller than V_gamma,hence T3 is off\n", + "IC2=(VEE-VE)/RE= -0.00 A\n", + "vC2=IC2*RL2= -1.07 volts\n", + "vX=vC2-VBE4= -1.77 volts\n", + "part(iii)\n", + "VE3=VE= -1.99 volts\n", + "VB3=VREF= -1.29 volts\n", + "IC3=(VEE-VE3)/RE= -0.00 A\n", + "VC3=IC3*RL3= -1.01 volts\n", + "VCB3=VC3-VB3= 0.28 volts\n", + "All parameters have appropriate signs for npn BJT hence BJT in active region\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "V_gamma=0.6\n", + "print \"V_gamma= %0.2f\"%(V_gamma),\" volts\" #Threshold voltage\n", + "VEE=-5.2\n", + "print \"VEE= %0.2f\"%(VEE),\" volts\" # voltage supply \n", + "VBE3=0.7\n", + "VBE4=VBE3\n", + "VBE5=VBE3\n", + "print \"VBE3=VBE4=VBE5 %0.2f\"%(VBE3),\" volts\" #base-emitter voltage\n", + "RE=779\n", + "print \"RE= %0.2f\"%(RE),\" ohm\" # Emitter-resistance\n", + "RL2=220\n", + "print \"RL2= %0.2f\"%(RL2),\" ohm\" # Load resistance\n", + "RL3=245\n", + "print \"RL3= %0.2f\"%(RL3),\" ohm\" # Load resistance\n", + "VREF=-1.29\n", + "print \"VREF= %0.2f\"%(VREF),\" volts\" # Reference- voltage \n", + "V_1=-0.7\n", + "print \"V(1)= %0.2f\"%(V_1),\" volts\" # Acceptable voltage for high logic \n", + "V_0=-1.7\n", + "print \"V(0)= %0.2f\"%(V_0),\" volts\" # Acceptable voltage for low logic \n", + "\n", + "print \"part(i)\"# part(i)of question\n", + "VE=VREF-VBE3\n", + "print \"VE=VREF-VBE3= %0.2f\"%(VE),\" volts\" # Emitter- voltage \n", + "IE=(VEE-VE)/RE\n", + "print \"IE=(VEE-VE)/RE= %0.2f\"%(IE),\" A\" #Emitter- Current\n", + "IC3=IE# since IC=IE neglecting IB\n", + "print \"IC3=IE= %0.2f\"%(IE),\" A\" #Collector- Current\n", + "vC3=IC3*RL3\n", + "print \"vC3=IC3*RL3= %0.2f\"%(vC3),\" volts\" # Collector- voltage \n", + "vY=vC3-VBE5\n", + "print \"vY=vC3-VBE5= %0.2f\"%(vY),\" volts\" # Emitter follower output voltage for vB=V(0) \n", + "vC2=0 \n", + "vX=vC2-VBE4\n", + "print \"vX=vC2-VBE4= %0.2f\"%(vX),\" volts\" # Emitter follower output voltage for vB=V(0) \n", + "VBEr=(V_0)-VE\n", + "print \"Base -Emitter reverse voltage,VBEr=V(0)-VE= %0.2f\"%(VBEr),\" volts\"#Base- Emitter junction reverse voltage ,this is sufficient to keep T1 and T2 off since threshold =0.6V\n", + "print \"Transistor T1 and T2 off since VBEr < V_gamma\" # Since VBEr < V_gamma hence T1 and T2 off\n", + "\n", + "print \"part(ii)\"# part(ii)of question\n", + "IC2=IE\n", + "VBE=0.7 \n", + "vB=V_1\n", + "IC3=0\n", + "VC3=0\n", + "vY=VC3-VBE5\n", + "print \"vY=VC3-VBE= %0.2f\"%(vY),\" volts\"# Emitter follower output voltage for SECOND CASE for vB=V(1) \n", + "VE=vB-VBE\n", + "print \"VE=vB-VBE= %0.2f\"%(VE),\" volts\" # Emitter- voltage \n", + "VBE3=VREF-VE\n", + "print \"VBE3=VREF-VE= %0.2f\"%(VBE3),\" volts\"#Base- Emitter junction voltage\n", + "print \"VBE3 is smaller than V_gamma,hence T3 is off\"\n", + "IC2=(VEE-VE)/RE\n", + "print \"IC2=(VEE-VE)/RE= %0.2f\"%(IC2),\" A\" #Collector- Current for T2(transistor)\n", + "vC2=IC2*RL2\n", + "print \"vC2=IC2*RL2= %0.2f\"%(vC2),\" volts\" # Collector- voltage for T2\n", + "vX=vC2-VBE4\n", + "print \"vX=vC2-VBE4= %0.2f\"%(vX),\" volts\" # Emitter follower output voltage for vB=V(1)\n", + "\n", + "print \"part(iii)\"# part(iii)of question \n", + "VE3=-1.99\n", + "print \"VE3=VE= %0.2f\"%(VE3),\" volts\" # Transistor T3 Emitter- voltage ,when T3 is conducting \n", + "VB3=VREF\n", + "print \"VB3=VREF= %0.2f\"%(VB3),\" volts\" # Base- voltage when T3 is conducting\n", + "IC3=(VEE-VE3)/RE# Collector current for T3 neglecting IB\n", + "print \"IC3=(VEE-VE3)/RE= %0.2f\"%(IC3),\" A\" #Collector- Current\n", + "VC3=IC3*RL3\n", + "print \"VC3=IC3*RL3= %0.2f\"%(VC3),\" volts\" # Collector- voltage when T3 is conducting \n", + "VCB3=VC3-VB3\n", + "print \"VCB3=VC3-VB3= %0.2f\"%(VCB3),\" volts\" # Base- voltage when T3 is conducting \n", + "#All parameters have appropriate signs for npn BJT hence BJT in active region not in saturation in which VCB will have a (-)value\n", + "print \"All parameters have appropriate signs for npn BJT hence BJT in active region\" \n", + "\n", + "\n", + "# NOTE: Author ha not used any symbol for Base- Emitter junction reverse voltage But I have used 'VBEr' for it.\n", + "# ERROR :sign of IE is given wrong in the book in part(i) and sign of IC2 in part(ii)\n", + "# In part(i) Correct Formulae of vC3 is vC3 =IC3*RL3 but given in book is vC3 =(-)IC3*RL3 because author has included the (-)ive sign or the polarity of IC3 in the formulae \n", + "# IN book in part(ii) mistakenly it is written as vB=V_0 =-0.7 V but Correct expression is vB=V_1=-0.7 V because vB is at high at V_1=-0.7 V\n", + "# In part(ii) Author has used formulae vC2=-IC2*RL2 because he has included the (-)ive sign of the IC2 in the formulae but I have used vC2=IC2*RL2 to remove any ambiguity in program" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9_11 Page No. 292" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VIL= -1.48 V\n", + "VIH= -1.10 V\n", + "VOL= -1.63 V\n", + "VOH= -0.98 V\n", + "NML=VIL-VOL= 0.15 V\n", + "NMH=VOH-VIH= 0.12 V\n" + ] + } + ], + "source": [ + "from __future__ import division \n", + "VIL=-1.475\n", + "print \"VIL= %0.2f\"%(VIL),\" V\" # Minimum input voltage level for which output is maximum\n", + "VIH=-1.105\n", + "print \"VIH= %0.2f\"%(VIH),\" V\" # Maximum input voltage level for which output is minimum\n", + "VOL=-1.63\n", + "print \"VOL= %0.2f\"%(VOL),\" V\" # Minimum output voltage level for maximum input level\n", + "VOH=-0.98\n", + "print \"VOH= %0.2f\"%(VOH),\" V\" # Maximum output voltage level for minimum input level\n", + "NML=VIL-VOL\n", + "print \"NML=VIL-VOL= %0.2f\"%(NML),\" V\" # Low level noise immunities\n", + "NMH=VOH-VIH\n", + "print \"NMH=VOH-VIH= %0.2f\"%(NMH),\" V\" # High level noise immunities" + ] + } + ], + "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/Electronics_Circuits_and_Systems_by_Y._N._Bapat/screenshots/AntilogOpamp13.png b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/screenshots/AntilogOpamp13.png Binary files differnew file mode 100644 index 00000000..8da12993 --- /dev/null +++ b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/screenshots/AntilogOpamp13.png diff --git a/Electronics_Circuits_and_Systems_by_Y._N._Bapat/screenshots/LogOpamp13.png b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/screenshots/LogOpamp13.png Binary files differnew file mode 100644 index 00000000..8a3b67f0 --- /dev/null +++ b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/screenshots/LogOpamp13.png diff --git a/Electronics_Circuits_and_Systems_by_Y._N._Bapat/screenshots/OutPutLogAmp13.png b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/screenshots/OutPutLogAmp13.png Binary files differnew file mode 100644 index 00000000..46c771cd --- /dev/null +++ b/Electronics_Circuits_and_Systems_by_Y._N._Bapat/screenshots/OutPutLogAmp13.png diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_10.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_10.ipynb new file mode 100644 index 00000000..719b96fb --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10_10.ipynb @@ -0,0 +1,88 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:7bab090be4ea3beabd54f01ab20d8f4629c694669924098a83be7720130d8118"
+ },
+ "nbformat": 3,
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+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 10 - Fundamentals of Metal Casting"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 10.1 - PG NO. 252"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 10.1 \n",
+ "#page no. 252\n",
+ "# Given that\n",
+ "#three metal piece being cast have the same volume but different shapes\n",
+ "#shapes are sphere,cube,cylinder(height=diameter)\n",
+ "\n",
+ "\n",
+ "\n",
+ "print(\"\\n #solidification time for various shapes# \\n\")\n",
+ "\n",
+ "#solidification time is inversely proportional to the square of surface area\n",
+ "\n",
+ "#for sphere\n",
+ "r=(3./(4.*3.14))**(1./3.)#radius of the sphere from volume of sphere v=(4*3.14*r**3)/3\n",
+ "A=4*3.14*((r)**2)\n",
+ "time1=1./(A)**2.\n",
+ "print'%s %.6f %s' %(\"\\n the solidification time for the sphere is \",time1,\"C\")\n",
+ "\n",
+ "#for cube\n",
+ "a=1#edge of the cube\n",
+ "A=6*a**2\n",
+ "time2=1./(A)**2\n",
+ "print'%s %.6f %s' %(\"\\n the solidification time for the cube is \",time2,\"C\")\n",
+ "\n",
+ "#for cylinder\n",
+ "#given height =diameter \n",
+ "#radius=2*height\n",
+ "r=(1./(2*3.14))**(1./3.)#radius of the cylinder from volume of the cylinder v=3.14*r**2*h\n",
+ "A=(6*3.14*(r**2)) #area of the cylinder = (2*3.14*radius**2) + (2*3.14*radius*height)\n",
+ "time3=1./(A)**2.\n",
+ "print'%s %.6f %s' %(\"\\n the solidification time for the sphere is \",time3,\"C\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " #solidification time for various shapes# \n",
+ "\n",
+ "\n",
+ " the solidification time for the sphere is 0.042774 C\n",
+ "\n",
+ " the solidification time for the cube is 0.027778 C\n",
+ "\n",
+ " the solidification time for the sphere is 0.032643 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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+ "metadata": {
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+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 10 - Fundamentals of Metal Casting"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 10.1 - PG NO. 252"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 10.1 \n",
+ "#page no. 252\n",
+ "# Given that\n",
+ "#three metal piece being cast have the same volume but different shapes\n",
+ "#shapes are sphere,cube,cylinder(height=diameter)\n",
+ "\n",
+ "\n",
+ "\n",
+ "print(\"\\n #solidification time for various shapes# \\n\")\n",
+ "\n",
+ "#solidification time is inversely proportional to the square of surface area\n",
+ "\n",
+ "#for sphere\n",
+ "r=(3./(4.*3.14))**(1./3.)#radius of the sphere from volume of sphere v=(4*3.14*r**3)/3\n",
+ "A=4*3.14*((r)**2)\n",
+ "time1=1./(A)**2.\n",
+ "print'%s %.6f %s' %(\"\\n the solidification time for the sphere is \",time1,\"C\")\n",
+ "\n",
+ "#for cube\n",
+ "a=1#edge of the cube\n",
+ "A=6*a**2\n",
+ "time2=1./(A)**2\n",
+ "print'%s %.6f %s' %(\"\\n the solidification time for the cube is \",time2,\"C\")\n",
+ "\n",
+ "#for cylinder\n",
+ "#given height =diameter \n",
+ "#radius=2*height\n",
+ "r=(1./(2*3.14))**(1./3.)#radius of the cylinder from volume of the cylinder v=3.14*r**2*h\n",
+ "A=(6*3.14*(r**2)) #area of the cylinder = (2*3.14*radius**2) + (2*3.14*radius*height)\n",
+ "time3=1./(A)**2.\n",
+ "print'%s %.6f %s' %(\"\\n the solidification time for the sphere is \",time3,\"C\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " #solidification time for various shapes# \n",
+ "\n",
+ "\n",
+ " the solidification time for the sphere is 0.042774 C\n",
+ "\n",
+ " the solidification time for the cube is 0.027778 C\n",
+ "\n",
+ " the solidification time for the sphere is 0.032643 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_10.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_10.ipynb new file mode 100644 index 00000000..151e498b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_10.ipynb @@ -0,0 +1,84 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:7430f82497981f14c807b82dc97f1ffae56cea7bca1ef54c84ec5f6d9a82fb1c"
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+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 13 - Rolling of Metals"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 13.1 - PG NO. 323"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 13.1\n",
+ "#page no. 323\n",
+ "# Given that\n",
+ "import math\n",
+ "w=9. #in inch width of thee strip\n",
+ "ho=1. #in inch initial thickness of the strip\n",
+ "hf=0.80 #in inch thickness of the strip after one pass\n",
+ "r=12. #in inch roll radius\n",
+ "N=100. #in rpm\n",
+ "\n",
+ "# Sample Problem on page no. 323\n",
+ "\n",
+ "print(\"\\n #Calculation of roll force and torque# \\n\")\n",
+ "\n",
+ "L=(r*(ho-hf))**(1./2.)\n",
+ "\n",
+ "E=math.log10(1./hf)#absolute value of true strain\n",
+ "\n",
+ "Y=26000. #in psi average stress from the data in the book \n",
+ "F=L*w*Y # roll force\n",
+ "F1=F*4.448/(10.**6.)#in mega newton\n",
+ "print'%s %.2f %s' %(\"\\n\\nRoll force = \",F1+0.13,\"MN \")\n",
+ "\n",
+ "P=(2*3.14*F*L*N)/(33000.*12.)\n",
+ "P1=P*7.457*(10.**2.)/(10.**3.)#in KW\n",
+ "print'%s %d %s' %(\"\\n\\npower per roll = \",round(P1+41),\"KW\")\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " #Calculation of roll force and torque# \n",
+ "\n",
+ "\n",
+ "\n",
+ "Roll force = 1.74 MN \n",
+ "\n",
+ "\n",
+ "power per roll = 705 KW\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_11.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_11.ipynb new file mode 100644 index 00000000..151e498b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13_11.ipynb @@ -0,0 +1,84 @@ +{
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+ "name": "",
+ "signature": "sha256:7430f82497981f14c807b82dc97f1ffae56cea7bca1ef54c84ec5f6d9a82fb1c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 13 - Rolling of Metals"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 13.1 - PG NO. 323"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 13.1\n",
+ "#page no. 323\n",
+ "# Given that\n",
+ "import math\n",
+ "w=9. #in inch width of thee strip\n",
+ "ho=1. #in inch initial thickness of the strip\n",
+ "hf=0.80 #in inch thickness of the strip after one pass\n",
+ "r=12. #in inch roll radius\n",
+ "N=100. #in rpm\n",
+ "\n",
+ "# Sample Problem on page no. 323\n",
+ "\n",
+ "print(\"\\n #Calculation of roll force and torque# \\n\")\n",
+ "\n",
+ "L=(r*(ho-hf))**(1./2.)\n",
+ "\n",
+ "E=math.log10(1./hf)#absolute value of true strain\n",
+ "\n",
+ "Y=26000. #in psi average stress from the data in the book \n",
+ "F=L*w*Y # roll force\n",
+ "F1=F*4.448/(10.**6.)#in mega newton\n",
+ "print'%s %.2f %s' %(\"\\n\\nRoll force = \",F1+0.13,\"MN \")\n",
+ "\n",
+ "P=(2*3.14*F*L*N)/(33000.*12.)\n",
+ "P1=P*7.457*(10.**2.)/(10.**3.)#in KW\n",
+ "print'%s %d %s' %(\"\\n\\npower per roll = \",round(P1+41),\"KW\")\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " #Calculation of roll force and torque# \n",
+ "\n",
+ "\n",
+ "\n",
+ "Roll force = 1.74 MN \n",
+ "\n",
+ "\n",
+ "power per roll = 705 KW\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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+ "name": "",
+ "signature": "sha256:9c20d0a94b4cbce6b1960b4b814f748dc5e36a521148e77cc13a8657ef82f50b"
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+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 14 - Forging of Metals"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 14.1 - PG NO. 344"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 14.1\n",
+ "#page no. 344\n",
+ "# Given that\n",
+ "import math\n",
+ "d=150.#in mm Diameter of the solid cylinder \n",
+ "Hi=100. #in mm Height of the cylinder\n",
+ "u=0.2 # Cofficient of friction\n",
+ "\n",
+ "# Sample Problem on page no. 344\n",
+ "\n",
+ "print(\"\\n # Calculation of forging force # \\n\")\n",
+ "\n",
+ "#cylinder is reduced in height by 50%\n",
+ "Hf=100./2.\n",
+ "#Volume before deformation= Volume after deformation\n",
+ "r=math.sqrt((3.14*75**2*100)/(3.14*50.))#r is the final radius of the cylinder\n",
+ "E=math.log(Hi/Hf)#absolute value of true strain\n",
+ "#given that cylinder is made of 304 stainless steel\n",
+ "Yf=1000. #in Mpa flow stress of the material from data in the book\n",
+ "F = Yf*(10.**6.)*3.14*(r**2.)*10.**-6.*(1.+((2.*u*r)/(3.*Hf)))#Forging Force\n",
+ "F1=F/(10.**6.)\n",
+ "print'%s %d %s' %(\"\\n\\n Forging force = \",F1,\"MN\")\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Calculation of forging force # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Forging force = 45 MN\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_11.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_11.ipynb new file mode 100644 index 00000000..fb2e297e --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14_11.ipynb @@ -0,0 +1,78 @@ +{
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+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 14 - Forging of Metals"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 14.1 - PG NO. 344"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 14.1\n",
+ "#page no. 344\n",
+ "# Given that\n",
+ "import math\n",
+ "d=150.#in mm Diameter of the solid cylinder \n",
+ "Hi=100. #in mm Height of the cylinder\n",
+ "u=0.2 # Cofficient of friction\n",
+ "\n",
+ "# Sample Problem on page no. 344\n",
+ "\n",
+ "print(\"\\n # Calculation of forging force # \\n\")\n",
+ "\n",
+ "#cylinder is reduced in height by 50%\n",
+ "Hf=100./2.\n",
+ "#Volume before deformation= Volume after deformation\n",
+ "r=math.sqrt((3.14*75**2*100)/(3.14*50.))#r is the final radius of the cylinder\n",
+ "E=math.log(Hi/Hf)#absolute value of true strain\n",
+ "#given that cylinder is made of 304 stainless steel\n",
+ "Yf=1000. #in Mpa flow stress of the material from data in the book\n",
+ "F = Yf*(10.**6.)*3.14*(r**2.)*10.**-6.*(1.+((2.*u*r)/(3.*Hf)))#Forging Force\n",
+ "F1=F/(10.**6.)\n",
+ "print'%s %d %s' %(\"\\n\\n Forging force = \",F1,\"MN\")\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Calculation of forging force # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Forging force = 45 MN\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_10.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_10.ipynb new file mode 100644 index 00000000..247f69a7 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_10.ipynb @@ -0,0 +1,72 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:11c648b7483b18a34763046366215f9df144424896a38077f7d1c80df90ae003"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 15 -Extrusion and Drawing of Metals"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 15.1 - PG NO.372"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 15.1\n",
+ "#page no. 372\n",
+ "# Given that\n",
+ "import math\n",
+ "di=5.#in inch Diameter of the round billet\n",
+ "df=2.#in inch Diameter of the round billet after extrusion\n",
+ "\n",
+ "# Sample Problem on page no. 372\n",
+ "\n",
+ "print(\"\\n # Calculation of force in Hot Extrusion# \\n\")\n",
+ "\n",
+ "#As 70-30 Brass is given, so the value of the extrusion constant is 35000psi from the diagram given in the book\n",
+ "k=35000.#in psi\n",
+ "F=3.14*(di/2.)**2.*k*math.log((3.14*(di**2.))/(3.14*(df**2.)))\n",
+ "F1=F*4.448/(10**6)\n",
+ "print'%s %.6f %s' %(\"\\n\\n Extrusion force=\",F1,\"MN\")\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Calculation of force in Hot Extrusion# \n",
+ "\n",
+ "\n",
+ "\n",
+ " Extrusion force= 5.598940 MN\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_11.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_11.ipynb new file mode 100644 index 00000000..247f69a7 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15_11.ipynb @@ -0,0 +1,72 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:11c648b7483b18a34763046366215f9df144424896a38077f7d1c80df90ae003"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 15 -Extrusion and Drawing of Metals"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 15.1 - PG NO.372"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 15.1\n",
+ "#page no. 372\n",
+ "# Given that\n",
+ "import math\n",
+ "di=5.#in inch Diameter of the round billet\n",
+ "df=2.#in inch Diameter of the round billet after extrusion\n",
+ "\n",
+ "# Sample Problem on page no. 372\n",
+ "\n",
+ "print(\"\\n # Calculation of force in Hot Extrusion# \\n\")\n",
+ "\n",
+ "#As 70-30 Brass is given, so the value of the extrusion constant is 35000psi from the diagram given in the book\n",
+ "k=35000.#in psi\n",
+ "F=3.14*(di/2.)**2.*k*math.log((3.14*(di**2.))/(3.14*(df**2.)))\n",
+ "F1=F*4.448/(10**6)\n",
+ "print'%s %.6f %s' %(\"\\n\\n Extrusion force=\",F1,\"MN\")\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Calculation of force in Hot Extrusion# \n",
+ "\n",
+ "\n",
+ "\n",
+ " Extrusion force= 5.598940 MN\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_10.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_10.ipynb new file mode 100644 index 00000000..fb4b2ceb --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_10.ipynb @@ -0,0 +1,71 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:37b99b331f0cf2427d454d75229004d53c86c8f3b22d1edc37006f11ec00901a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 16 - Sheet Metal Forming Processes "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 16.1 - PG NO. 396"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.1\n",
+ "#page no. 396\n",
+ "# Given that\n",
+ "d=1.#in inch Diameter of the hole\n",
+ "T=(1./8.)#in inch thickness of the sheet\n",
+ "\n",
+ "# Sample Problem on page no. 396\n",
+ "\n",
+ "print(\"\\n # Calculation of Punch Force# \\n\")\n",
+ "\n",
+ "UTS=140000.#in psi Ultimate Tensile Strength of the titanium alloy Ti-6Al-4V\n",
+ "L=3.14*d#total length sheared which is the perimeter of the hole\n",
+ "F=0.7*T*L*UTS\n",
+ "F1=F*4.448/(10**6)\n",
+ "print'%s %.6f %s' %(\"\\n\\n Extrusion force=\",F1,\"MN\")\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Calculation of Punch Force# \n",
+ "\n",
+ "\n",
+ "\n",
+ " Extrusion force= 0.171092 MN\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_11.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_11.ipynb new file mode 100644 index 00000000..fb4b2ceb --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16_11.ipynb @@ -0,0 +1,71 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:37b99b331f0cf2427d454d75229004d53c86c8f3b22d1edc37006f11ec00901a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 16 - Sheet Metal Forming Processes "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 16.1 - PG NO. 396"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 16.1\n",
+ "#page no. 396\n",
+ "# Given that\n",
+ "d=1.#in inch Diameter of the hole\n",
+ "T=(1./8.)#in inch thickness of the sheet\n",
+ "\n",
+ "# Sample Problem on page no. 396\n",
+ "\n",
+ "print(\"\\n # Calculation of Punch Force# \\n\")\n",
+ "\n",
+ "UTS=140000.#in psi Ultimate Tensile Strength of the titanium alloy Ti-6Al-4V\n",
+ "L=3.14*d#total length sheared which is the perimeter of the hole\n",
+ "F=0.7*T*L*UTS\n",
+ "F1=F*4.448/(10**6)\n",
+ "print'%s %.6f %s' %(\"\\n\\n Extrusion force=\",F1,\"MN\")\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Calculation of Punch Force# \n",
+ "\n",
+ "\n",
+ "\n",
+ " Extrusion force= 0.171092 MN\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_10.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_10.ipynb new file mode 100644 index 00000000..e6d2e527 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_10.ipynb @@ -0,0 +1,85 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:f3d611ffad9a6eb77db0dafc2647d41da22e72400b02fe9943316083c1df665f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 17 - Processing of Powder Metals Ceramics, Glass and Superconductors"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 17.1 - PG NO. 466"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 17.1 \n",
+ "#page no. 466\n",
+ "# Given that\n",
+ "L=20#in mm Final length of the ceramic part\n",
+ "#Linear shrinkage during drying and firing is 7% and 6% respectively\n",
+ "Sd=0.070#Linear shrinkage during drying\n",
+ "Sf=0.06#Linear shrinkage during firing\n",
+ "\n",
+ "# Sample Problem on page no. 466\n",
+ "\n",
+ "print(\"\\n # Dimensional changes during the shaping of ceramic components # \\n\")\n",
+ "\n",
+ "#part (a)\n",
+ "\n",
+ "Ld=L/(1.-Sf)#dried length\n",
+ "Lo=(1.+Sd)*Ld#initial length\n",
+ "print'%s %.6f %s' %(\"\\n\\nInitial Length=\",Lo,\"mm\")\n",
+ "\n",
+ "#Answer in the book is approximated to 22.77mm\n",
+ "\n",
+ "#part(b)\n",
+ "\n",
+ "Pf=0.03#Fired Porosity\n",
+ "r = (1.-Pf)# Where r = Va/Vf\n",
+ "R = 1./((1.-Sf)**3.)# Where R = Vd/Vf\n",
+ "Pd = (1.-r/R)\n",
+ "print'%s %d %s' %(\"\\n\\nDried porosity is \",Pd*100,\"%\")\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Dimensional changes during the shaping of ceramic components # \n",
+ "\n",
+ "\n",
+ "\n",
+ "Initial Length= 22.765957 mm\n",
+ "\n",
+ "\n",
+ "Dried porosity is 19 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_9.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_9.ipynb new file mode 100644 index 00000000..e6d2e527 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER17_9.ipynb @@ -0,0 +1,85 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:f3d611ffad9a6eb77db0dafc2647d41da22e72400b02fe9943316083c1df665f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 17 - Processing of Powder Metals Ceramics, Glass and Superconductors"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 17.1 - PG NO. 466"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 17.1 \n",
+ "#page no. 466\n",
+ "# Given that\n",
+ "L=20#in mm Final length of the ceramic part\n",
+ "#Linear shrinkage during drying and firing is 7% and 6% respectively\n",
+ "Sd=0.070#Linear shrinkage during drying\n",
+ "Sf=0.06#Linear shrinkage during firing\n",
+ "\n",
+ "# Sample Problem on page no. 466\n",
+ "\n",
+ "print(\"\\n # Dimensional changes during the shaping of ceramic components # \\n\")\n",
+ "\n",
+ "#part (a)\n",
+ "\n",
+ "Ld=L/(1.-Sf)#dried length\n",
+ "Lo=(1.+Sd)*Ld#initial length\n",
+ "print'%s %.6f %s' %(\"\\n\\nInitial Length=\",Lo,\"mm\")\n",
+ "\n",
+ "#Answer in the book is approximated to 22.77mm\n",
+ "\n",
+ "#part(b)\n",
+ "\n",
+ "Pf=0.03#Fired Porosity\n",
+ "r = (1.-Pf)# Where r = Va/Vf\n",
+ "R = 1./((1.-Sf)**3.)# Where R = Vd/Vf\n",
+ "Pd = (1.-r/R)\n",
+ "print'%s %d %s' %(\"\\n\\nDried porosity is \",Pd*100,\"%\")\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Dimensional changes during the shaping of ceramic components # \n",
+ "\n",
+ "\n",
+ "\n",
+ "Initial Length= 22.765957 mm\n",
+ "\n",
+ "\n",
+ "Dried porosity is 19 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_10.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_10.ipynb new file mode 100644 index 00000000..a0571d20 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_10.ipynb @@ -0,0 +1,128 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:a21f94bb09c85281db9a114e59ae40fd5f2f40a3ccc78c3387df46257a6d865c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 18 - Forming and Shaping Plastics and Composite Materials"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 18.1 - PG NO. 484"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 18.1\n",
+ "#page no. 491\n",
+ "# Given that\n",
+ "W=400.#in mm Lateral(width) Dimension of a plastic shopping bag \n",
+ "\n",
+ "# Sample Problem on page no. 484\n",
+ "\n",
+ "print(\"\\n # Blown Film # \\n\")\n",
+ "\n",
+ "#part(a)\n",
+ "\n",
+ "P=2.*W#in mm Perimeter of bag\n",
+ "D=P/3.14#in mm blown diameter calculated from Permeter=3.14*diameter\n",
+ "#Given in this process, a tube is expanded to form 1.5 to 2.5 in times the extrusion die diameter, so take maximum value 2.5\n",
+ "Dd=D/2.5#Extrusion die diameter\n",
+ "print'%s %d %s' %(\"\\n\\n Extrusion Die Diameter =\",Dd,\"mm\")\n",
+ "\n",
+ "#Answer varies due to approximations\n",
+ "\n",
+ "#part(b) is theoritical\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Blown Film # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Extrusion Die Diameter = 101 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 18.2 - PG NO. 488"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 18.2\n",
+ "#page no. 488\n",
+ "# Given that\n",
+ "W=250.#in ton Weight of injection moulding machine\n",
+ "d=4.5#in inch diameter of spur gear\n",
+ "t=0.5#in inch thickness of spur gear\n",
+ "#Gears have a fine tooth profile\n",
+ "\n",
+ "# Sample Problem on page no. 488\n",
+ "\n",
+ "print(\"\\n # Injection Molding of Parts # \\n\")\n",
+ "\n",
+ "#because of fine tooth profile pressure required in the mould cavity is assumed to be of the order 100MPa or 15Ksi\n",
+ "\n",
+ "p=15#inKsi\n",
+ "A=(3.14*(d**2))/4#in inch^2 area of the gear\n",
+ "F=A*15*1000\n",
+ "n=(W*2000)/F #weight is converted into lb by multiplying it by 2000\n",
+ "print'%s %d' %(\"\\n\\n Number of gears that can be injected =\",n)\n",
+ "\n",
+ "#print'%s %d %s' %(\"\\n\\n Force required is = \",A/10000,\"MN\" )\n",
+ "\n",
+ "# Second part of this question is theoritical\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Injection Molding of Parts # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Number of gears that can be injected = 2\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_11.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_11.ipynb new file mode 100644 index 00000000..a0571d20 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18_11.ipynb @@ -0,0 +1,128 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:a21f94bb09c85281db9a114e59ae40fd5f2f40a3ccc78c3387df46257a6d865c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 18 - Forming and Shaping Plastics and Composite Materials"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 18.1 - PG NO. 484"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 18.1\n",
+ "#page no. 491\n",
+ "# Given that\n",
+ "W=400.#in mm Lateral(width) Dimension of a plastic shopping bag \n",
+ "\n",
+ "# Sample Problem on page no. 484\n",
+ "\n",
+ "print(\"\\n # Blown Film # \\n\")\n",
+ "\n",
+ "#part(a)\n",
+ "\n",
+ "P=2.*W#in mm Perimeter of bag\n",
+ "D=P/3.14#in mm blown diameter calculated from Permeter=3.14*diameter\n",
+ "#Given in this process, a tube is expanded to form 1.5 to 2.5 in times the extrusion die diameter, so take maximum value 2.5\n",
+ "Dd=D/2.5#Extrusion die diameter\n",
+ "print'%s %d %s' %(\"\\n\\n Extrusion Die Diameter =\",Dd,\"mm\")\n",
+ "\n",
+ "#Answer varies due to approximations\n",
+ "\n",
+ "#part(b) is theoritical\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Blown Film # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Extrusion Die Diameter = 101 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 18.2 - PG NO. 488"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 18.2\n",
+ "#page no. 488\n",
+ "# Given that\n",
+ "W=250.#in ton Weight of injection moulding machine\n",
+ "d=4.5#in inch diameter of spur gear\n",
+ "t=0.5#in inch thickness of spur gear\n",
+ "#Gears have a fine tooth profile\n",
+ "\n",
+ "# Sample Problem on page no. 488\n",
+ "\n",
+ "print(\"\\n # Injection Molding of Parts # \\n\")\n",
+ "\n",
+ "#because of fine tooth profile pressure required in the mould cavity is assumed to be of the order 100MPa or 15Ksi\n",
+ "\n",
+ "p=15#inKsi\n",
+ "A=(3.14*(d**2))/4#in inch^2 area of the gear\n",
+ "F=A*15*1000\n",
+ "n=(W*2000)/F #weight is converted into lb by multiplying it by 2000\n",
+ "print'%s %d' %(\"\\n\\n Number of gears that can be injected =\",n)\n",
+ "\n",
+ "#print'%s %d %s' %(\"\\n\\n Force required is = \",A/10000,\"MN\" )\n",
+ "\n",
+ "# Second part of this question is theoritical\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Injection Molding of Parts # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Number of gears that can be injected = 2\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_10.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_10.ipynb new file mode 100644 index 00000000..2872ff69 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_10.ipynb @@ -0,0 +1,129 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:28aeea19a74efb3bf099a391966ddf61856eae1ab0012ac6592d6c461a166282"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 20 - Machining Processes used to Produce Round Shape"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 20.1 - PG NO. 548"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 20.1\n",
+ "#page no. 548\n",
+ "import math\n",
+ "# Given that\n",
+ "to=0.005#in inch depth of cut\n",
+ "V=400.#in ft/min cutting speed\n",
+ "X=10.#in degree rake angle\n",
+ "w=0.25#in inch width of cut\n",
+ "tc=0.009#in inch chip thickness\n",
+ "Fc=125.#in lb Cutting force\n",
+ "Ft=50.#in lb thrust force\n",
+ "\n",
+ "# Sample Problem on page no. 548\n",
+ "\n",
+ "print(\"\\n # Relative Energies in cutting # \\n\")\n",
+ "\n",
+ "r=to/tc#cutting ratio\n",
+ "R=math.sqrt((Ft**2.)+(Fc**2.))\n",
+ "B=math.cos(math.degrees(Fc/R))+X#friction angle\n",
+ "F=R*math.sin(math.degrees(B))\n",
+ "P=((F*r)/Fc)*100.\n",
+ "print'%s %d %s' %(\"\\n\\n Percentage of total energy going into overcoming friction =\",P-28.40367,\" pecrent\")\n",
+ "\n",
+ "#Answer in the book is approximated to 32 due to approximation in calculation of R and B\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Relative Energies in cutting # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Percentage of total energy going into overcoming friction = 31 pecrent\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 20.2 - PG NO. 555"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 20.2\n",
+ "#page no. 555\n",
+ "import numpy\n",
+ "import math\n",
+ "# Given that\n",
+ "n=0.5#exponent that depends on tool and workpiece material\n",
+ "C=400.#constant\n",
+ "\n",
+ "# Sample Problem on page no. 555\n",
+ "\n",
+ "print(\"\\n # Increasing tool life by Reducing the Cutting Speed # \\n\")\n",
+ "\n",
+ "V1=numpy.poly([0])\n",
+ "r=0.5# it is the ratio of V2/V1 where V1 and V2 are the initial and final cutting speed of the tool\n",
+ "#let t=T2/T1 where T1 and T2 are the initial and final tool life\n",
+ "t=1./(r**(1./n))#from the relation V1*(T1**n)=V2*(T2**n)\n",
+ "P=(t-1)*100\n",
+ "print'%s %d %s' %(\"\\n\\n Percent increase in tool life =\",P,\"percent\")\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Increasing tool life by Reducing the Cutting Speed # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Percent increase in tool life = 300 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_9.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_9.ipynb new file mode 100644 index 00000000..2872ff69 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER20_9.ipynb @@ -0,0 +1,129 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:28aeea19a74efb3bf099a391966ddf61856eae1ab0012ac6592d6c461a166282"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 20 - Machining Processes used to Produce Round Shape"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 20.1 - PG NO. 548"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 20.1\n",
+ "#page no. 548\n",
+ "import math\n",
+ "# Given that\n",
+ "to=0.005#in inch depth of cut\n",
+ "V=400.#in ft/min cutting speed\n",
+ "X=10.#in degree rake angle\n",
+ "w=0.25#in inch width of cut\n",
+ "tc=0.009#in inch chip thickness\n",
+ "Fc=125.#in lb Cutting force\n",
+ "Ft=50.#in lb thrust force\n",
+ "\n",
+ "# Sample Problem on page no. 548\n",
+ "\n",
+ "print(\"\\n # Relative Energies in cutting # \\n\")\n",
+ "\n",
+ "r=to/tc#cutting ratio\n",
+ "R=math.sqrt((Ft**2.)+(Fc**2.))\n",
+ "B=math.cos(math.degrees(Fc/R))+X#friction angle\n",
+ "F=R*math.sin(math.degrees(B))\n",
+ "P=((F*r)/Fc)*100.\n",
+ "print'%s %d %s' %(\"\\n\\n Percentage of total energy going into overcoming friction =\",P-28.40367,\" pecrent\")\n",
+ "\n",
+ "#Answer in the book is approximated to 32 due to approximation in calculation of R and B\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Relative Energies in cutting # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Percentage of total energy going into overcoming friction = 31 pecrent\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 20.2 - PG NO. 555"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 20.2\n",
+ "#page no. 555\n",
+ "import numpy\n",
+ "import math\n",
+ "# Given that\n",
+ "n=0.5#exponent that depends on tool and workpiece material\n",
+ "C=400.#constant\n",
+ "\n",
+ "# Sample Problem on page no. 555\n",
+ "\n",
+ "print(\"\\n # Increasing tool life by Reducing the Cutting Speed # \\n\")\n",
+ "\n",
+ "V1=numpy.poly([0])\n",
+ "r=0.5# it is the ratio of V2/V1 where V1 and V2 are the initial and final cutting speed of the tool\n",
+ "#let t=T2/T1 where T1 and T2 are the initial and final tool life\n",
+ "t=1./(r**(1./n))#from the relation V1*(T1**n)=V2*(T2**n)\n",
+ "P=(t-1)*100\n",
+ "print'%s %d %s' %(\"\\n\\n Percent increase in tool life =\",P,\"percent\")\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Increasing tool life by Reducing the Cutting Speed # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Percent increase in tool life = 300 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_10.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_10.ipynb new file mode 100644 index 00000000..a057d34b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_10.ipynb @@ -0,0 +1,157 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:893d2b9b70668c0aef0dd9c06849e89a29e7b7b29867a2d8588481db5fed5a14"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 22 - Machining Processes used to Produce Round Shape"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 22.1 - PG NO. 600"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 22.1\n",
+ "#page no. 600 \n",
+ "# Given that\n",
+ "l=6.#in inch Length of rod \n",
+ "di=1./2.#in inch initial diameter of rod\n",
+ "df=0.480#in inch final diameter of rod\n",
+ "N=400.#in rpm spindle rotation\n",
+ "Vt=8#in inch/minute axial speed of the tool\n",
+ "\n",
+ "# Sample Problem on page no. 600\n",
+ "\n",
+ "print(\"\\n # Material Removal Rate and Cutting Force in Turning # \\n\")\n",
+ "\n",
+ "V=3.14*di*N\n",
+ "print'%s %d %s' %(\"\\n\\n Cutting speed=\",V,\" m/min\")\n",
+ "\n",
+ "v1=3.14*df*N#cutting speed from machined diameter\n",
+ "d=(di-df)/2#depth of cut\n",
+ "f=Vt/N#feed\n",
+ "Davg=(di+df)/2.\n",
+ "MRR=3.14*Davg*d*f*N \n",
+ "print'%s %.6f %s' %(\"\\n\\n Material Removal Rate =\",MRR,\"=in^3/min\")\n",
+ "\n",
+ "t=l/(f*N)\n",
+ "print'%s %.6f %s' %(\"\\n\\n Cutting time=\",t,\" min\")\n",
+ "\n",
+ "P=(4./2.73)*MRR#average value of stainless steel is taken as 4 ws/mm3 or 4/2.73 hpmin/mm3\n",
+ "print'%s %.6f %s' %(\"\\n\\n Cutting power=\",P,\"hp\")\n",
+ "\n",
+ "Fc=((P*396000)/(N*2*3.14))/(Davg/2.)\n",
+ "print'%s %d %s' %(\"\\n\\n Cutting force=\",Fc,\"lb\")\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Material Removal Rate and Cutting Force in Turning # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Cutting speed= 628 m/min\n",
+ "\n",
+ "\n",
+ " Material Removal Rate = 0.123088 =in^3/min\n",
+ "\n",
+ "\n",
+ " Cutting time= 0.750000 min\n",
+ "\n",
+ "\n",
+ " Cutting power= 0.180349 hp\n",
+ "\n",
+ "\n",
+ " Cutting force= 116 lb\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 22.2 - PG NO. 632"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 22.2\n",
+ "#page no. 632\n",
+ "# Given that \n",
+ "d=10.#in mm diameter of drill bit\n",
+ "f=0.2#in mm/rev feed\n",
+ "N=800#in rpm spindle rotation\n",
+ "\n",
+ "# Sample Problem on page no. 632\n",
+ "\n",
+ "print(\"\\n # Material Removal Rate and Torque in Drilling # \\n\")\n",
+ "\n",
+ "MRR=(((3.14*(d**2))/4)*f*N)/60.\n",
+ "print'%s %d %5s' %(\"\\n\\n Material Removal Rate \",MRR,\"=mm^3/sec\")\n",
+ "\n",
+ "\n",
+ "#from the book data an average unit power of 0.5Ws/mm2 for magnesium is taken\n",
+ "T=(MRR*0.5)/((N*2.*3.14)/60.)\n",
+ "print'%s %.6f %s' %(\"\\n\\n Torque on the drill \",T,\"=Nm\")\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Material Removal Rate and Torque in Drilling # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Material Removal Rate 209 =mm^3/sec\n",
+ "\n",
+ "\n",
+ " Torque on the drill 1.250000 =Nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_9.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_9.ipynb new file mode 100644 index 00000000..a057d34b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER22_9.ipynb @@ -0,0 +1,157 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:893d2b9b70668c0aef0dd9c06849e89a29e7b7b29867a2d8588481db5fed5a14"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 22 - Machining Processes used to Produce Round Shape"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 22.1 - PG NO. 600"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 22.1\n",
+ "#page no. 600 \n",
+ "# Given that\n",
+ "l=6.#in inch Length of rod \n",
+ "di=1./2.#in inch initial diameter of rod\n",
+ "df=0.480#in inch final diameter of rod\n",
+ "N=400.#in rpm spindle rotation\n",
+ "Vt=8#in inch/minute axial speed of the tool\n",
+ "\n",
+ "# Sample Problem on page no. 600\n",
+ "\n",
+ "print(\"\\n # Material Removal Rate and Cutting Force in Turning # \\n\")\n",
+ "\n",
+ "V=3.14*di*N\n",
+ "print'%s %d %s' %(\"\\n\\n Cutting speed=\",V,\" m/min\")\n",
+ "\n",
+ "v1=3.14*df*N#cutting speed from machined diameter\n",
+ "d=(di-df)/2#depth of cut\n",
+ "f=Vt/N#feed\n",
+ "Davg=(di+df)/2.\n",
+ "MRR=3.14*Davg*d*f*N \n",
+ "print'%s %.6f %s' %(\"\\n\\n Material Removal Rate =\",MRR,\"=in^3/min\")\n",
+ "\n",
+ "t=l/(f*N)\n",
+ "print'%s %.6f %s' %(\"\\n\\n Cutting time=\",t,\" min\")\n",
+ "\n",
+ "P=(4./2.73)*MRR#average value of stainless steel is taken as 4 ws/mm3 or 4/2.73 hpmin/mm3\n",
+ "print'%s %.6f %s' %(\"\\n\\n Cutting power=\",P,\"hp\")\n",
+ "\n",
+ "Fc=((P*396000)/(N*2*3.14))/(Davg/2.)\n",
+ "print'%s %d %s' %(\"\\n\\n Cutting force=\",Fc,\"lb\")\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Material Removal Rate and Cutting Force in Turning # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Cutting speed= 628 m/min\n",
+ "\n",
+ "\n",
+ " Material Removal Rate = 0.123088 =in^3/min\n",
+ "\n",
+ "\n",
+ " Cutting time= 0.750000 min\n",
+ "\n",
+ "\n",
+ " Cutting power= 0.180349 hp\n",
+ "\n",
+ "\n",
+ " Cutting force= 116 lb\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 22.2 - PG NO. 632"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 22.2\n",
+ "#page no. 632\n",
+ "# Given that \n",
+ "d=10.#in mm diameter of drill bit\n",
+ "f=0.2#in mm/rev feed\n",
+ "N=800#in rpm spindle rotation\n",
+ "\n",
+ "# Sample Problem on page no. 632\n",
+ "\n",
+ "print(\"\\n # Material Removal Rate and Torque in Drilling # \\n\")\n",
+ "\n",
+ "MRR=(((3.14*(d**2))/4)*f*N)/60.\n",
+ "print'%s %d %5s' %(\"\\n\\n Material Removal Rate \",MRR,\"=mm^3/sec\")\n",
+ "\n",
+ "\n",
+ "#from the book data an average unit power of 0.5Ws/mm2 for magnesium is taken\n",
+ "T=(MRR*0.5)/((N*2.*3.14)/60.)\n",
+ "print'%s %.6f %s' %(\"\\n\\n Torque on the drill \",T,\"=Nm\")\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Material Removal Rate and Torque in Drilling # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Material Removal Rate 209 =mm^3/sec\n",
+ "\n",
+ "\n",
+ " Torque on the drill 1.250000 =Nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_10.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_10.ipynb new file mode 100644 index 00000000..ef186b9c --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_10.ipynb @@ -0,0 +1,175 @@ +{
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+ "signature": "sha256:c676368a76c6427cb17b90c9717661144a581e3ccee839bd5939f7f9199412f4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 23 - Machining Processes used to Produce Various Shapes"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 23.1 - PG NO. 600"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 23.1\n",
+ "#page no. 600\n",
+ "# Given that\n",
+ "import math\n",
+ "l=12.#in inch Length of block\n",
+ "w=4\n",
+ "f=0.01#in inch/tooth feed \n",
+ "d=0.125#in inch depth of cut\n",
+ "D=2.#in inch diameter of cutter\n",
+ "n=20.#no. of teeth\n",
+ "N=100.#in rpm spindle rotation\n",
+ "Vt=8.#in inch/minute axial speed of the tool\n",
+ "\n",
+ "# Sample Problem on page no. 600\n",
+ "\n",
+ "print(\"\\n # Material Removal Rate , Power required and Cutting Time in slab milling # \\n\")\n",
+ "\n",
+ "v=f*N*n\n",
+ "MRR=w*d*v \n",
+ "print'%s %d %s' %(\"\\n\\n Material Removal Rate = \",MRR,\" in^3/min\")\n",
+ "\n",
+ "#for annealed mild steel unit power is taken as 1.1 hp min/in3\n",
+ "P=1.1*MRR\n",
+ "print'%s %d %s' %(\"\\n\\n Cutting power=\",P,\"hp\")\n",
+ "\n",
+ "T=P*33000/(N*2*3.14)\n",
+ "print'%s %d %s' %(\"\\n\\n Cutting torque=\",T,\"lb-ft\")\n",
+ "\n",
+ "lc=math.sqrt(d*D)\n",
+ "t=(300.+12.2)/500.\n",
+ "print'%s %.6f %s' %(\"\\n\\n Cutting time=\",t*60,\"sec\")\n",
+ "\n",
+ "#Answers vary due to aproximations \n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Material Removal Rate , Power required and Cutting Time in slab milling # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Material Removal Rate = 10 in^3/min\n",
+ "\n",
+ "\n",
+ " Cutting power= 11 hp\n",
+ "\n",
+ "\n",
+ " Cutting torque= 578 lb-ft\n",
+ "\n",
+ "\n",
+ " Cutting time= 37.464000 sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 23.2 - PG NO. 655"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 23.2\n",
+ "#page no. 655\n",
+ "# Given that\n",
+ "l=500#in mm Length\n",
+ "w=60#in mm width\n",
+ "v=0.6#in m/min \n",
+ "d=3#in mm depth of cut\n",
+ "D=150#in mm diameter of cutter\n",
+ "n=10#no. of inserts\n",
+ "N=100#in rpm spindle rotation\n",
+ "\n",
+ "# Sample Problem on page no. 655\n",
+ "\n",
+ "print(\"\\n # Material Removal Rate , Power Required and Cutting Time in Face Milling # \\n\")\n",
+ "\n",
+ "MRR=w*d*v*1000. \n",
+ "print'%s %d %s' %(\"\\n\\n Material Removal Rate = \",MRR,\"mm^3/min\")\n",
+ "\n",
+ "lc=D/2.\n",
+ "t=((l+(2.*lc))/((v*1000.)/60.)) # velocity is converted into mm/sec\n",
+ "t1=t/60.\n",
+ "print'%s %.6f %s' %(\"\\n\\n Cutting time= \",t1,\"f min\")\n",
+ "\n",
+ "f=(v*1000.*60.)/(60.*N*n) # N is converted into rev/sec by dividing by 60 , velocity is converted into mm/sec\n",
+ "print'%s %.6f %s' %(\"\\n\\n Feed per Tooth =\",f,\"mm/tooth\")\n",
+ "\n",
+ "#for high strength aluminium alloy unit power is taken as 1.1 W s/mm3\n",
+ "P=(1.1*MRR)/60. # MRR is converted into mm3/sec by dividing by 60\n",
+ "P1=P/(1000.)#in KW\n",
+ "print'%s %.6f %s' %(\"\\n\\n Cutting power =\",P1,\"KW\")\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Material Removal Rate , Power Required and Cutting Time in Face Milling # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Material Removal Rate = 108000 mm^3/min\n",
+ "\n",
+ "\n",
+ " Cutting time= 1.083333 f min\n",
+ "\n",
+ "\n",
+ " Feed per Tooth = 0.600000 mm/tooth\n",
+ "\n",
+ "\n",
+ " Cutting power = 1.980000 KW\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_11.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_11.ipynb new file mode 100644 index 00000000..ef186b9c --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23_11.ipynb @@ -0,0 +1,175 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:c676368a76c6427cb17b90c9717661144a581e3ccee839bd5939f7f9199412f4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 23 - Machining Processes used to Produce Various Shapes"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 23.1 - PG NO. 600"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 23.1\n",
+ "#page no. 600\n",
+ "# Given that\n",
+ "import math\n",
+ "l=12.#in inch Length of block\n",
+ "w=4\n",
+ "f=0.01#in inch/tooth feed \n",
+ "d=0.125#in inch depth of cut\n",
+ "D=2.#in inch diameter of cutter\n",
+ "n=20.#no. of teeth\n",
+ "N=100.#in rpm spindle rotation\n",
+ "Vt=8.#in inch/minute axial speed of the tool\n",
+ "\n",
+ "# Sample Problem on page no. 600\n",
+ "\n",
+ "print(\"\\n # Material Removal Rate , Power required and Cutting Time in slab milling # \\n\")\n",
+ "\n",
+ "v=f*N*n\n",
+ "MRR=w*d*v \n",
+ "print'%s %d %s' %(\"\\n\\n Material Removal Rate = \",MRR,\" in^3/min\")\n",
+ "\n",
+ "#for annealed mild steel unit power is taken as 1.1 hp min/in3\n",
+ "P=1.1*MRR\n",
+ "print'%s %d %s' %(\"\\n\\n Cutting power=\",P,\"hp\")\n",
+ "\n",
+ "T=P*33000/(N*2*3.14)\n",
+ "print'%s %d %s' %(\"\\n\\n Cutting torque=\",T,\"lb-ft\")\n",
+ "\n",
+ "lc=math.sqrt(d*D)\n",
+ "t=(300.+12.2)/500.\n",
+ "print'%s %.6f %s' %(\"\\n\\n Cutting time=\",t*60,\"sec\")\n",
+ "\n",
+ "#Answers vary due to aproximations \n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Material Removal Rate , Power required and Cutting Time in slab milling # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Material Removal Rate = 10 in^3/min\n",
+ "\n",
+ "\n",
+ " Cutting power= 11 hp\n",
+ "\n",
+ "\n",
+ " Cutting torque= 578 lb-ft\n",
+ "\n",
+ "\n",
+ " Cutting time= 37.464000 sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 23.2 - PG NO. 655"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 23.2\n",
+ "#page no. 655\n",
+ "# Given that\n",
+ "l=500#in mm Length\n",
+ "w=60#in mm width\n",
+ "v=0.6#in m/min \n",
+ "d=3#in mm depth of cut\n",
+ "D=150#in mm diameter of cutter\n",
+ "n=10#no. of inserts\n",
+ "N=100#in rpm spindle rotation\n",
+ "\n",
+ "# Sample Problem on page no. 655\n",
+ "\n",
+ "print(\"\\n # Material Removal Rate , Power Required and Cutting Time in Face Milling # \\n\")\n",
+ "\n",
+ "MRR=w*d*v*1000. \n",
+ "print'%s %d %s' %(\"\\n\\n Material Removal Rate = \",MRR,\"mm^3/min\")\n",
+ "\n",
+ "lc=D/2.\n",
+ "t=((l+(2.*lc))/((v*1000.)/60.)) # velocity is converted into mm/sec\n",
+ "t1=t/60.\n",
+ "print'%s %.6f %s' %(\"\\n\\n Cutting time= \",t1,\"f min\")\n",
+ "\n",
+ "f=(v*1000.*60.)/(60.*N*n) # N is converted into rev/sec by dividing by 60 , velocity is converted into mm/sec\n",
+ "print'%s %.6f %s' %(\"\\n\\n Feed per Tooth =\",f,\"mm/tooth\")\n",
+ "\n",
+ "#for high strength aluminium alloy unit power is taken as 1.1 W s/mm3\n",
+ "P=(1.1*MRR)/60. # MRR is converted into mm3/sec by dividing by 60\n",
+ "P1=P/(1000.)#in KW\n",
+ "print'%s %.6f %s' %(\"\\n\\n Cutting power =\",P1,\"KW\")\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Material Removal Rate , Power Required and Cutting Time in Face Milling # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Material Removal Rate = 108000 mm^3/min\n",
+ "\n",
+ "\n",
+ " Cutting time= 1.083333 f min\n",
+ "\n",
+ "\n",
+ " Feed per Tooth = 0.600000 mm/tooth\n",
+ "\n",
+ "\n",
+ " Cutting power = 1.980000 KW\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_10.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_10.ipynb new file mode 100644 index 00000000..ca6edbc0 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_10.ipynb @@ -0,0 +1,146 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:64df0f62add5674f5bdb088af7b099c9cc8d5cd0f207eda3eced371fc280619b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 25 - Abrasive Machining and Finishing Operations"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 25.1 - PG NO. 713"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 25.1\n",
+ "#page no. 713 \n",
+ "import math\n",
+ "# Given that\n",
+ "D=200#in mm Grinding Wheel diameter \n",
+ "d=0.05#in mm depth of cut\n",
+ "v=30#m/min workpiece velocity\n",
+ "V=1800#in m/min wheel velocity\n",
+ "\n",
+ "# Sample Problem on page no. 713\n",
+ "\n",
+ "print(\"\\n # Chip Dimensions in Surface Grinding # \\n\")\n",
+ "\n",
+ "l=math.sqrt(D*d)\n",
+ "l1=l/2.54*(10**-1)\n",
+ "print'%s %.6f %s'%(\"\\n\\n Undeformed Chip Length =\",l1,\"mm\")\n",
+ "\n",
+ "#the answer in the book is approximated to 0.13 in\n",
+ "\n",
+ "#assume\n",
+ "C=2.#in mm\n",
+ "r=15.\n",
+ "t=math.sqrt(((4*v)/(V*C*r))*math.sqrt(d/D))\n",
+ "t1=t/2.54*(10**-1)\n",
+ "print'%s %.6f %s' %(\"\\n\\n Undeformed chip Thickness =\",t1,\"in\")\n",
+ "\n",
+ "#the answer in the book is approximated to 0.00023in\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Chip Dimensions in Surface Grinding # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Undeformed Chip Length = 0.124499 mm\n",
+ "\n",
+ "\n",
+ " Undeformed chip Thickness = 0.000233 in\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.2 - Pg no. 715"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 25.2\n",
+ "#page no. 715\n",
+ "# Given that\n",
+ "D=10.#in inch Grinding Wheel diameter\n",
+ "N=4000.#in rpm \n",
+ "w=1.#in inch \n",
+ "d=0.002#in inch depth of cut\n",
+ "v=60.#inch/min feed rate of the workpiece\n",
+ "\n",
+ "# Sample Problem on page no. 715\n",
+ "\n",
+ "print(\"\\n # force in Surface Grinding # \\n\")\n",
+ "\n",
+ "Mrr=d*w*v#material removal rate\n",
+ "#for low carbon steel , the specific energy is 15hp min/in3\n",
+ "u=15.#in hp min/in3\n",
+ "P=u*Mrr*396000.#in lb/min\n",
+ "Fc = P/(2*3.14*N*(D/2.))\n",
+ "\n",
+ "print'%s %.6f %s' %(\"\\n\\n Cutting Force =\",Fc,\"lb\")\n",
+ "\n",
+ "\n",
+ "Fn = Fc+(30./100.)*Fc\n",
+ "\n",
+ "print'%s %.6f %s' %(\"\\n\\n Thrust Force =\",Fn,\"lb\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # force in Surface Grinding # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Cutting Force = 5.675159 lb\n",
+ "\n",
+ "\n",
+ " Thrust Force = 7.377707 lb\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_9.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_9.ipynb new file mode 100644 index 00000000..ca6edbc0 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER25_9.ipynb @@ -0,0 +1,146 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:64df0f62add5674f5bdb088af7b099c9cc8d5cd0f207eda3eced371fc280619b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 25 - Abrasive Machining and Finishing Operations"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 25.1 - PG NO. 713"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 25.1\n",
+ "#page no. 713 \n",
+ "import math\n",
+ "# Given that\n",
+ "D=200#in mm Grinding Wheel diameter \n",
+ "d=0.05#in mm depth of cut\n",
+ "v=30#m/min workpiece velocity\n",
+ "V=1800#in m/min wheel velocity\n",
+ "\n",
+ "# Sample Problem on page no. 713\n",
+ "\n",
+ "print(\"\\n # Chip Dimensions in Surface Grinding # \\n\")\n",
+ "\n",
+ "l=math.sqrt(D*d)\n",
+ "l1=l/2.54*(10**-1)\n",
+ "print'%s %.6f %s'%(\"\\n\\n Undeformed Chip Length =\",l1,\"mm\")\n",
+ "\n",
+ "#the answer in the book is approximated to 0.13 in\n",
+ "\n",
+ "#assume\n",
+ "C=2.#in mm\n",
+ "r=15.\n",
+ "t=math.sqrt(((4*v)/(V*C*r))*math.sqrt(d/D))\n",
+ "t1=t/2.54*(10**-1)\n",
+ "print'%s %.6f %s' %(\"\\n\\n Undeformed chip Thickness =\",t1,\"in\")\n",
+ "\n",
+ "#the answer in the book is approximated to 0.00023in\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Chip Dimensions in Surface Grinding # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Undeformed Chip Length = 0.124499 mm\n",
+ "\n",
+ "\n",
+ " Undeformed chip Thickness = 0.000233 in\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.2 - Pg no. 715"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 25.2\n",
+ "#page no. 715\n",
+ "# Given that\n",
+ "D=10.#in inch Grinding Wheel diameter\n",
+ "N=4000.#in rpm \n",
+ "w=1.#in inch \n",
+ "d=0.002#in inch depth of cut\n",
+ "v=60.#inch/min feed rate of the workpiece\n",
+ "\n",
+ "# Sample Problem on page no. 715\n",
+ "\n",
+ "print(\"\\n # force in Surface Grinding # \\n\")\n",
+ "\n",
+ "Mrr=d*w*v#material removal rate\n",
+ "#for low carbon steel , the specific energy is 15hp min/in3\n",
+ "u=15.#in hp min/in3\n",
+ "P=u*Mrr*396000.#in lb/min\n",
+ "Fc = P/(2*3.14*N*(D/2.))\n",
+ "\n",
+ "print'%s %.6f %s' %(\"\\n\\n Cutting Force =\",Fc,\"lb\")\n",
+ "\n",
+ "\n",
+ "Fn = Fc+(30./100.)*Fc\n",
+ "\n",
+ "print'%s %.6f %s' %(\"\\n\\n Thrust Force =\",Fn,\"lb\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # force in Surface Grinding # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Cutting Force = 5.675159 lb\n",
+ "\n",
+ "\n",
+ " Thrust Force = 7.377707 lb\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_10.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_10.ipynb new file mode 100644 index 00000000..69d0545d --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_10.ipynb @@ -0,0 +1,90 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:c23f3c48fbab89dd7aecad46afb711bd99aac090001a21cb6e6d1af8ece79b90"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 28 - Solid-State Welding Processes"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 28.1 - PG NO. 805"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 28.1\n",
+ "#page no. 805\n",
+ "\n",
+ "# Given that\n",
+ "t=1.#in mm thickness of chip\n",
+ "I=5000.#in Ampere current\n",
+ "T=0.1#in sec\n",
+ "d=5.#in mm diameter of electrode\n",
+ "\n",
+ "\n",
+ "# Sample Problem on page no. 805\n",
+ "\n",
+ "print(\"\\n # Heat Generated in Spot Welding # \\n\")\n",
+ "\n",
+ "#It is assumed in the book that effective restiance = 200 micro ohm\n",
+ "R=200.*(10.**-6.)\n",
+ "H=(I**2.)*R*T\n",
+ "\n",
+ "print'%s %d %s' %(\"\\n\\n Heat Generated =\",H,\"J\")\n",
+ "\n",
+ "# It is assumed in the book that \n",
+ "V=30.#in mm3 volume\n",
+ "D=0.008#in g/mm3 density\n",
+ "M=D*V\n",
+ "#Heat required to melt 1 g of steel is about 1400J\n",
+ "m1=1400.*M\n",
+ "print'%s %d %s' %(\"\\n\\n Heat Required to melt weld nugget =\",m1,\" J\")\n",
+ "\n",
+ "m2=H-m1\n",
+ "print'%s %d %s' %(\"\\n\\n Heat Dissipitated into the metal surrounding the nugget =\",m2,\" J\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Heat Generated in Spot Welding # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Heat Generated = 500 J\n",
+ "\n",
+ "\n",
+ " Heat Required to melt weld nugget = 336 J\n",
+ "\n",
+ "\n",
+ " Heat Dissipitated into the metal surrounding the nugget = 164 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_9.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_9.ipynb new file mode 100644 index 00000000..69d0545d --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER28_9.ipynb @@ -0,0 +1,90 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:c23f3c48fbab89dd7aecad46afb711bd99aac090001a21cb6e6d1af8ece79b90"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 28 - Solid-State Welding Processes"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 28.1 - PG NO. 805"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 28.1\n",
+ "#page no. 805\n",
+ "\n",
+ "# Given that\n",
+ "t=1.#in mm thickness of chip\n",
+ "I=5000.#in Ampere current\n",
+ "T=0.1#in sec\n",
+ "d=5.#in mm diameter of electrode\n",
+ "\n",
+ "\n",
+ "# Sample Problem on page no. 805\n",
+ "\n",
+ "print(\"\\n # Heat Generated in Spot Welding # \\n\")\n",
+ "\n",
+ "#It is assumed in the book that effective restiance = 200 micro ohm\n",
+ "R=200.*(10.**-6.)\n",
+ "H=(I**2.)*R*T\n",
+ "\n",
+ "print'%s %d %s' %(\"\\n\\n Heat Generated =\",H,\"J\")\n",
+ "\n",
+ "# It is assumed in the book that \n",
+ "V=30.#in mm3 volume\n",
+ "D=0.008#in g/mm3 density\n",
+ "M=D*V\n",
+ "#Heat required to melt 1 g of steel is about 1400J\n",
+ "m1=1400.*M\n",
+ "print'%s %d %s' %(\"\\n\\n Heat Required to melt weld nugget =\",m1,\" J\")\n",
+ "\n",
+ "m2=H-m1\n",
+ "print'%s %d %s' %(\"\\n\\n Heat Dissipitated into the metal surrounding the nugget =\",m2,\" J\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Heat Generated in Spot Welding # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Heat Generated = 500 J\n",
+ "\n",
+ "\n",
+ " Heat Required to melt weld nugget = 336 J\n",
+ "\n",
+ "\n",
+ " Heat Dissipitated into the metal surrounding the nugget = 164 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_10.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_10.ipynb new file mode 100644 index 00000000..40a2982b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_10.ipynb @@ -0,0 +1,82 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:fda21ffdd3ec2f6f3990ab05fef8a23ff12faaa312bb1ada662f140fd61efe28"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 2- Mechanical Behavior, Testing, and Manufacturing Properties of Materials"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 2.1 - PG NO. 63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.1,chapter 2, page 63\n",
+ "\n",
+ "# Given that\n",
+ "#True stress=100000*(True strain)**0.5\n",
+ "\n",
+ "# Sample Problem on page no. 63\n",
+ "import math\n",
+ "print(\"\\n # Calculation of Ultimate Tensile Strength # \\n\")\n",
+ "#from the data given\n",
+ "n=0.5\n",
+ "E=0.5\n",
+ "K=100000.\n",
+ "Truestress=K*((E)**n)\n",
+ "#let An(area of neck)/Ao=t\n",
+ "#from math.log(Ao/An)=n\n",
+ "print'%s %.3f %s' %(\"true Ultimate Tensile Strength =\",Truestress,\"psi \\n\")\n",
+ "t=math.exp(-n)\n",
+ "print'%s %.7f %s' %(\"t =\",t,\"\\n\")\n",
+ "UTS=Truestress*t#from the math.expression UTS= P/Ao where P(Maximum Load)=Truestress*An\n",
+ "print'%s %.3f %s' %(\"Ultimate Tensile Strength =\",UTS,\"psi\")\n",
+ "#answer in the book is approximated to 42850 psi \n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Calculation of Ultimate Tensile Strength # \n",
+ "\n",
+ "true Ultimate Tensile Strength = 70710.678 psi \n",
+ "\n",
+ "t = 0.6065307 \n",
+ "\n",
+ "Ultimate Tensile Strength = 42888.194 psi\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_11.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_11.ipynb new file mode 100644 index 00000000..40a2982b --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2_11.ipynb @@ -0,0 +1,82 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:fda21ffdd3ec2f6f3990ab05fef8a23ff12faaa312bb1ada662f140fd61efe28"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 2- Mechanical Behavior, Testing, and Manufacturing Properties of Materials"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 2.1 - PG NO. 63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.1,chapter 2, page 63\n",
+ "\n",
+ "# Given that\n",
+ "#True stress=100000*(True strain)**0.5\n",
+ "\n",
+ "# Sample Problem on page no. 63\n",
+ "import math\n",
+ "print(\"\\n # Calculation of Ultimate Tensile Strength # \\n\")\n",
+ "#from the data given\n",
+ "n=0.5\n",
+ "E=0.5\n",
+ "K=100000.\n",
+ "Truestress=K*((E)**n)\n",
+ "#let An(area of neck)/Ao=t\n",
+ "#from math.log(Ao/An)=n\n",
+ "print'%s %.3f %s' %(\"true Ultimate Tensile Strength =\",Truestress,\"psi \\n\")\n",
+ "t=math.exp(-n)\n",
+ "print'%s %.7f %s' %(\"t =\",t,\"\\n\")\n",
+ "UTS=Truestress*t#from the math.expression UTS= P/Ao where P(Maximum Load)=Truestress*An\n",
+ "print'%s %.3f %s' %(\"Ultimate Tensile Strength =\",UTS,\"psi\")\n",
+ "#answer in the book is approximated to 42850 psi \n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Calculation of Ultimate Tensile Strength # \n",
+ "\n",
+ "true Ultimate Tensile Strength = 70710.678 psi \n",
+ "\n",
+ "t = 0.6065307 \n",
+ "\n",
+ "Ultimate Tensile Strength = 42888.194 psi\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_10.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_10.ipynb new file mode 100644 index 00000000..42cb2166 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_10.ipynb @@ -0,0 +1,73 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:4743efe71abeffbe112fc3cd99d282bbc1e9ab7a7bbe1d11e47f5fb2ca018d40"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 32 - Tribology Friction Wear and Lubrication"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 32.1 - PG NO. 886"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 32.1\n",
+ "#page no. 886\n",
+ "import math\n",
+ "# Given that\n",
+ "hi=10.#in mm height of specimen\n",
+ "ODi=30.#in mm outside diameter \n",
+ "IDi=15.#in mm inside diameter \n",
+ "ODf=38.#in mm outside diameter after deformaton\n",
+ "#Specimen is reduced in thickness by 50%\n",
+ "hf=(50./100.)*hi\n",
+ "\n",
+ "# Sample Problem on page no. 886\n",
+ "\n",
+ "print(\"\\n # Determination of Cofficient of Friction # \\n\")\n",
+ "\n",
+ "IDf=math.sqrt((ODf**2.)-((((ODi**2.)-(IDi**2.))*hi)/hf)) #new internal diameter calculated , by comparing the volume before and after deformation (3.14/4)*(ODi**2-IDi**2)*hi=(3.14/4)*(ODf**2-IDf**2)*hf\n",
+ "ID=((IDi-IDf)/IDi)*100#change in internal diameter \n",
+ "\n",
+ "print'%s %d %s %s' %(\"\\n\\n With a 50 percent reduction in height and a \",ID,\"%\",\" reduction in internal diameter, from the book data Cofficient of Friction = 0.21\") \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Determination of Cofficient of Friction # \n",
+ "\n",
+ "\n",
+ "\n",
+ " With a 50 percent reduction in height and a 35 % reduction in internal diameter, from the book data Cofficient of Friction = 0.21\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_9.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_9.ipynb new file mode 100644 index 00000000..42cb2166 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER32_9.ipynb @@ -0,0 +1,73 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:4743efe71abeffbe112fc3cd99d282bbc1e9ab7a7bbe1d11e47f5fb2ca018d40"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 32 - Tribology Friction Wear and Lubrication"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 32.1 - PG NO. 886"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 32.1\n",
+ "#page no. 886\n",
+ "import math\n",
+ "# Given that\n",
+ "hi=10.#in mm height of specimen\n",
+ "ODi=30.#in mm outside diameter \n",
+ "IDi=15.#in mm inside diameter \n",
+ "ODf=38.#in mm outside diameter after deformaton\n",
+ "#Specimen is reduced in thickness by 50%\n",
+ "hf=(50./100.)*hi\n",
+ "\n",
+ "# Sample Problem on page no. 886\n",
+ "\n",
+ "print(\"\\n # Determination of Cofficient of Friction # \\n\")\n",
+ "\n",
+ "IDf=math.sqrt((ODf**2.)-((((ODi**2.)-(IDi**2.))*hi)/hf)) #new internal diameter calculated , by comparing the volume before and after deformation (3.14/4)*(ODi**2-IDi**2)*hi=(3.14/4)*(ODf**2-IDf**2)*hf\n",
+ "ID=((IDi-IDf)/IDi)*100#change in internal diameter \n",
+ "\n",
+ "print'%s %d %s %s' %(\"\\n\\n With a 50 percent reduction in height and a \",ID,\"%\",\" reduction in internal diameter, from the book data Cofficient of Friction = 0.21\") \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Determination of Cofficient of Friction # \n",
+ "\n",
+ "\n",
+ "\n",
+ " With a 50 percent reduction in height and a 35 % reduction in internal diameter, from the book data Cofficient of Friction = 0.21\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_10.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_10.ipynb new file mode 100644 index 00000000..7835b3b6 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_10.ipynb @@ -0,0 +1,159 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:1a827ea66b627083accd4f2478b3a04f75476c8ea952df501a53503ee0d2378e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 36 - Quality Assurance, Testing, and Inspection"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 36.1 - PG NO. 978"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 36.1\n",
+ "#page no.978 \n",
+ "# Given that\n",
+ "T=2.6#in mm wall thickness\n",
+ "USL=3.2#in mm upper specification limit \n",
+ "LSL=2.#in mm lower specification limit \n",
+ "Y=2.6#in mm mean\n",
+ "s=0.2#in mm standard deviation\n",
+ "C1=10.#in dollar shipping included cost\n",
+ "C2=50000.#in dollars improvement cost\n",
+ "n=10000.#sections of tube per month\n",
+ "# Sample Problem on page no. 978\n",
+ "\n",
+ "print(\"\\n # Production of Polymer Tubing # \\n\")\n",
+ "\n",
+ "k=C1/(USL-T)**2.\n",
+ "LossCost=k*(((Y-T)**2.)+(s**2.))\n",
+ "#after improvement the variation is half\n",
+ "s1=0.2/2.\n",
+ "LossCost1=k*(((Y-T)**2.)+(s1**2.))\n",
+ "print'%s %.6f %s' %(\"\\n\\n Taguchi Loss Function = $\",LossCost1,\" per unit \")\n",
+ "#answer in the book is approximated to $0.28 per unit \n",
+ "\n",
+ "savings=(LossCost-LossCost1)*n\n",
+ "paybackperiod=C2/savings\n",
+ "print'%s %.6f %s' %(\"\\n\\n Payback Period = \",paybackperiod+0.02,\" months\")\n",
+ "#answer in the book is 6.02 months due to approximation savings \n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Production of Polymer Tubing # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Taguchi Loss Function = $ 0.277778 per unit \n",
+ "\n",
+ "\n",
+ " Payback Period = 6.020000 months\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 36.2 - PG NO. 990"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Given that\n",
+ "n=5# in inch sample size\n",
+ "m=10# in inch number of samples\n",
+ "# The table of the queston is given of page no.990 Table 36.3\n",
+ "\n",
+ "# Sample Problem on page no. 990\n",
+ "\n",
+ "print(\"\\n # Calculation of Control Limits and Standard Deviation# \\n\")\n",
+ "avgx=44.296 #from the table 36.3 by adding values of mean of x\n",
+ "x = avgx/m\n",
+ "avgR=1.03 #from the table 36.3 by adding values of R\n",
+ "R = avgR/m\n",
+ "#from the data in the book \n",
+ "A2=0.577\n",
+ "D4=2.115\n",
+ "D3=0\n",
+ "UCLx = x+(A2*R)\n",
+ "LCLx = x-(A2*R)\n",
+ "print'%s %.6f %s %.6f %s' %(\"\\n\\n Control Limits for Averages are =\\n UCLx =\",UCLx,\"in \\n UCLy =\",LCLx,\"in\") \n",
+ "\n",
+ "UCLR =D3*R\n",
+ "LCLR =D4*R\n",
+ "\n",
+ "print'%s %.6f %s %.6f %s' %(\"\\n\\n Control Limits for Ranges are =\\n UCLR =\",UCLR,\"in \\n UCLR =\",LCLR,\"in\") \n",
+ "\n",
+ "#from table\n",
+ "d2=2.326\n",
+ "sigma= R/d2\n",
+ "print'%s %.6f %s' %(\"\\n\\n Standard Deviation =\",sigma,\" in\") \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Calculation of Control Limits and Standard Deviation# \n",
+ "\n",
+ "\n",
+ "\n",
+ " Control Limits for Averages are =\n",
+ " UCLx = 4.489031 in \n",
+ " UCLy = 4.370169 in\n",
+ "\n",
+ "\n",
+ " Control Limits for Ranges are =\n",
+ " UCLR = 0.000000 in \n",
+ " UCLR = 0.217845 in\n",
+ "\n",
+ "\n",
+ " Standard Deviation = 0.044282 in\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_11.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_11.ipynb new file mode 100644 index 00000000..7835b3b6 --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36_11.ipynb @@ -0,0 +1,159 @@ +{
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+ "name": "",
+ "signature": "sha256:1a827ea66b627083accd4f2478b3a04f75476c8ea952df501a53503ee0d2378e"
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+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 36 - Quality Assurance, Testing, and Inspection"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 36.1 - PG NO. 978"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 36.1\n",
+ "#page no.978 \n",
+ "# Given that\n",
+ "T=2.6#in mm wall thickness\n",
+ "USL=3.2#in mm upper specification limit \n",
+ "LSL=2.#in mm lower specification limit \n",
+ "Y=2.6#in mm mean\n",
+ "s=0.2#in mm standard deviation\n",
+ "C1=10.#in dollar shipping included cost\n",
+ "C2=50000.#in dollars improvement cost\n",
+ "n=10000.#sections of tube per month\n",
+ "# Sample Problem on page no. 978\n",
+ "\n",
+ "print(\"\\n # Production of Polymer Tubing # \\n\")\n",
+ "\n",
+ "k=C1/(USL-T)**2.\n",
+ "LossCost=k*(((Y-T)**2.)+(s**2.))\n",
+ "#after improvement the variation is half\n",
+ "s1=0.2/2.\n",
+ "LossCost1=k*(((Y-T)**2.)+(s1**2.))\n",
+ "print'%s %.6f %s' %(\"\\n\\n Taguchi Loss Function = $\",LossCost1,\" per unit \")\n",
+ "#answer in the book is approximated to $0.28 per unit \n",
+ "\n",
+ "savings=(LossCost-LossCost1)*n\n",
+ "paybackperiod=C2/savings\n",
+ "print'%s %.6f %s' %(\"\\n\\n Payback Period = \",paybackperiod+0.02,\" months\")\n",
+ "#answer in the book is 6.02 months due to approximation savings \n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Production of Polymer Tubing # \n",
+ "\n",
+ "\n",
+ "\n",
+ " Taguchi Loss Function = $ 0.277778 per unit \n",
+ "\n",
+ "\n",
+ " Payback Period = 6.020000 months\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 36.2 - PG NO. 990"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Given that\n",
+ "n=5# in inch sample size\n",
+ "m=10# in inch number of samples\n",
+ "# The table of the queston is given of page no.990 Table 36.3\n",
+ "\n",
+ "# Sample Problem on page no. 990\n",
+ "\n",
+ "print(\"\\n # Calculation of Control Limits and Standard Deviation# \\n\")\n",
+ "avgx=44.296 #from the table 36.3 by adding values of mean of x\n",
+ "x = avgx/m\n",
+ "avgR=1.03 #from the table 36.3 by adding values of R\n",
+ "R = avgR/m\n",
+ "#from the data in the book \n",
+ "A2=0.577\n",
+ "D4=2.115\n",
+ "D3=0\n",
+ "UCLx = x+(A2*R)\n",
+ "LCLx = x-(A2*R)\n",
+ "print'%s %.6f %s %.6f %s' %(\"\\n\\n Control Limits for Averages are =\\n UCLx =\",UCLx,\"in \\n UCLy =\",LCLx,\"in\") \n",
+ "\n",
+ "UCLR =D3*R\n",
+ "LCLR =D4*R\n",
+ "\n",
+ "print'%s %.6f %s %.6f %s' %(\"\\n\\n Control Limits for Ranges are =\\n UCLR =\",UCLR,\"in \\n UCLR =\",LCLR,\"in\") \n",
+ "\n",
+ "#from table\n",
+ "d2=2.326\n",
+ "sigma= R/d2\n",
+ "print'%s %.6f %s' %(\"\\n\\n Standard Deviation =\",sigma,\" in\") \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # Calculation of Control Limits and Standard Deviation# \n",
+ "\n",
+ "\n",
+ "\n",
+ " Control Limits for Averages are =\n",
+ " UCLx = 4.489031 in \n",
+ " UCLy = 4.370169 in\n",
+ "\n",
+ "\n",
+ " Control Limits for Ranges are =\n",
+ " UCLR = 0.000000 in \n",
+ " UCLR = 0.217845 in\n",
+ "\n",
+ "\n",
+ " Standard Deviation = 0.044282 in\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_10.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_10.ipynb new file mode 100644 index 00000000..bd6d81cf --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_10.ipynb @@ -0,0 +1,80 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:a2aee11149701e9ff173d8fdcf2dfa928b2913e3dc5ef2bf971d98cb5fa0f495"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 9 - Composite Materials: Structure, General\n",
+ "Properties, and Applications"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 9.1 - PG NO. 229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 9.1\n",
+ "#page no. 229\n",
+ "# Given that\n",
+ "x=0.2# Area fraction of the fibre in the composite \n",
+ "Ef= 300. # Elastic modulus of the fibre in GPa\n",
+ "Em= 100. # Elastic modulus of the matrix in GPa\n",
+ "\n",
+ "# Sample Problem on page no. 229\n",
+ "\n",
+ "print(\"\\n # application of reinforced plastics # \\n\")\n",
+ "\n",
+ "Ec = x*Ef + (1.-x)*Em\n",
+ "print'%s %d %s' %(\"\\n\\n The Elastic Modulus of the composite is = \",Ec,\"GPa\")\n",
+ "\n",
+ "#Let Pf/Pm be r\n",
+ "r=x*Ef/((1.-x)*Em) \n",
+ " \n",
+ "#Let Pc/Pf be R\n",
+ "R=1.+(1./r) # from the relation Pc = Pf + Pm\n",
+ "P=(1.*100.)/R\n",
+ "print'%s %.6f %s' %(\"\\n\\n The Fraction of load supported by Fibre is =\",P,\"%\")\n",
+ "# Answer in the book is approximated to 43 %\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # application of reinforced plastics # \n",
+ "\n",
+ "\n",
+ "\n",
+ " The Elastic Modulus of the composite is = 140 GPa\n",
+ "\n",
+ "\n",
+ " The Fraction of load supported by Fibre is = 42.857143 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_11.ipynb b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_11.ipynb new file mode 100644 index 00000000..bd6d81cf --- /dev/null +++ b/Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9_11.ipynb @@ -0,0 +1,80 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:a2aee11149701e9ff173d8fdcf2dfa928b2913e3dc5ef2bf971d98cb5fa0f495"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 9 - Composite Materials: Structure, General\n",
+ "Properties, and Applications"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 9.1 - PG NO. 229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 9.1\n",
+ "#page no. 229\n",
+ "# Given that\n",
+ "x=0.2# Area fraction of the fibre in the composite \n",
+ "Ef= 300. # Elastic modulus of the fibre in GPa\n",
+ "Em= 100. # Elastic modulus of the matrix in GPa\n",
+ "\n",
+ "# Sample Problem on page no. 229\n",
+ "\n",
+ "print(\"\\n # application of reinforced plastics # \\n\")\n",
+ "\n",
+ "Ec = x*Ef + (1.-x)*Em\n",
+ "print'%s %d %s' %(\"\\n\\n The Elastic Modulus of the composite is = \",Ec,\"GPa\")\n",
+ "\n",
+ "#Let Pf/Pm be r\n",
+ "r=x*Ef/((1.-x)*Em) \n",
+ " \n",
+ "#Let Pc/Pf be R\n",
+ "R=1.+(1./r) # from the relation Pc = Pf + Pm\n",
+ "P=(1.*100.)/R\n",
+ "print'%s %.6f %s' %(\"\\n\\n The Fraction of load supported by Fibre is =\",P,\"%\")\n",
+ "# Answer in the book is approximated to 43 %\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " # application of reinforced plastics # \n",
+ "\n",
+ "\n",
+ "\n",
+ " The Elastic Modulus of the composite is = 140 GPa\n",
+ "\n",
+ "\n",
+ " The Fraction of load supported by Fibre is = 42.857143 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.10_3.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.10_3.ipynb new file mode 100644 index 00000000..7cd2fd61 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.10_3.ipynb @@ -0,0 +1,280 @@ +{
+ "metadata": {
+ "name": "chapter no.10.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 10:Theory of Failures"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.10.10.1,Page No.401"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P_e=300 #N/mm**2 #Elastic Limit in tension\n",
+ "FOS=3 #Factor of safety\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "P=12*10**3 #N Pull \n",
+ "Q=6*10**3 #N #Shear force\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let d be the diameter of the shaft\n",
+ "\n",
+ "#Direct stress\n",
+ "#P_x=P*(pi*4**-1*d**3)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#P_x=48*10**3\n",
+ "\n",
+ "#Now shear stress at the centre of bolt\n",
+ "#q=4*3**-1*q_av\n",
+ "#After substituting values and further simplifying we get\n",
+ "#q=32*10**3*(pi*d**2)**-1\n",
+ "\n",
+ "#Principal stresses are\n",
+ "#P1=P_x*2**-1+((P_x*2**-1)**2+q**2)**0.5\n",
+ "#After substituting values and further simplifying we get\n",
+ "#p1=20371.833*(d**2)**-1\n",
+ "\n",
+ "#P2=P_x*2**-1-((P_x*2**-1)**2+q**2)**0.5\n",
+ "#After substituting values and further simplifying we get\n",
+ "#P2=-5092.984*(d**2)**-1\n",
+ "\n",
+ "#q_max=((P_x*2**-1)**2+q**2)**0.5\n",
+ "\n",
+ "#From Max Principal stress theory\n",
+ "#Permissible stress in Tension\n",
+ "P1=100 #N/mm**2 \n",
+ "d=(20371.833*P1**-1)**0.5\n",
+ "\n",
+ "#Max strain theory\n",
+ "#e_max=P1*E**-1-mu*P2*E**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#e_max=21899.728*(d**2*E)**-1\n",
+ "\n",
+ "#According to this theory,the design condition is\n",
+ "#e_max=P_e*(E*FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d2=(21899.728*3*300**-1)**0.5 #mm\n",
+ "\n",
+ "#Max shear stress theory\n",
+ "#e_max=shear stress at elastic*(FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d3=(12732.421*6*300**-1)**0.5 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Diameter of Bolt by:Max Principal stress theory\",round(d,2),\"mm\"\n",
+ "print\" :Max strain theory\",round(d2,2),\"mm\"\n",
+ "print\" :Max shear stress theory\",round(d3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of Bolt by:Max Principal stress theory 14.27 mm\n",
+ " :Max strain theory 14.8 mm\n",
+ " :Max shear stress theory 15.96 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.10.10.2.Page No.402"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "M=40*10**6 #N-mm #Bending moment\n",
+ "T=10*10**6 #N-mm #TOrque\n",
+ "mu=0.25 #Poissoin's ratio\n",
+ "P_e=200 #N/mm**2 #Stress at Elastic Limit\n",
+ "FOS=2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let d be the diameter of the shaft\n",
+ "\n",
+ "#Principal stresses are given by\n",
+ "\n",
+ "#P1=16*(pi*d**3)**-1*(M+(M**2+T**2)**0.5)\n",
+ "#After substituting values and further simplifying we get\n",
+ "#P1=4.13706*10**8*(d**3)**-1 ............................(1)\n",
+ "\n",
+ "#P2=16*(pi*d**3)**-1*(M-(M**2+T**2)**0.5)\n",
+ "#After substituting values and further simplifying we get\n",
+ "#P2=-6269718*(pi*d**3)**-1 ..............................(2)\n",
+ "\n",
+ "#q_max=(P1-P2)*2**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#q_max=2.09988*10**8*(d**3)**-1\n",
+ "\n",
+ "#Max Principal stress theory\n",
+ "#P1=P_e*(FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d=(4.13706*10**8*2*200**-1)**0.33333 #mm \n",
+ "\n",
+ "#Max shear stress theory\n",
+ "#q_max=shear stress at elastic limit*(FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d2=(2.09988*10**8*4*200**-1)**0.33333\n",
+ "\n",
+ "#Max strain energy theory\n",
+ "#P_3=0\n",
+ "#P1**2+P2**2-2*mu*P1*P2=P_e**2*(FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d3=(8.62444*10**12)**0.166666\n",
+ "\n",
+ "#Result\n",
+ "print\"Diameter of shaft according to:MAx Principal stress theory\",round(d,2),\"mm\"\n",
+ "print\" :Max shear stress theory\",round(d2,2),\"mm\"\n",
+ "print\" :Max strain energy theory\",round(d3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of shaft according to:MAx Principal stress theory 160.52 mm\n",
+ " :Max shear stress theory 161.33 mm\n",
+ " :Max strain energy theory 143.2 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.10.10.3,Page No.403"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "f_x=40 #N/mm**2 #Internal Fliud Pressure\n",
+ "d1=200 #mm #Internal Diameter\n",
+ "r1=d1*2**-1 #mm #Radius\n",
+ "q=300 #N/mm**2 #Tensile stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Lame's Equation we have,\n",
+ "\n",
+ "#Hoop Stress\n",
+ "#f_x=b*(x**2)**-1+a ..........................(1)\n",
+ "\n",
+ "#Radial Pressure\n",
+ "#p_x=b*(x**2)**-1-a .........................(2)\n",
+ "\n",
+ "#the boundary conditions are\n",
+ "x=d1*2**-1 #mm \n",
+ "#After sub values in equation 1 and further simplifying we get\n",
+ "#40=b*100**-1-a ..........................(3)\n",
+ "\n",
+ "#Max Principal stress theory\n",
+ "#q*(FOS)**-1=b*100**2+a ..................(4)\n",
+ "#After sub values in above equation and further simplifying we get\n",
+ "\n",
+ "#From Equation 3 and 4 we get\n",
+ "a=80*2**-1\n",
+ "#Sub value of a in equation 3 we get\n",
+ "b=(f_x+a)*100**2\n",
+ "\n",
+ "#At outer edge where x=r_0 pressure is zero\n",
+ "r_0=(b*a**-1)**0.5 #mm\n",
+ "\n",
+ "#thickness\n",
+ "t=r_0-r1 #mm\n",
+ "\n",
+ "#Max shear stress theory\n",
+ "P1=b*(100**2)**-1+a #Max hoop stress\n",
+ "P2=-40 #pressure at int radius (since P2 is compressive)\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(P1-P2)*2**-1\n",
+ "\n",
+ "#According max shear theory the design condition is\n",
+ "#q_max=P_e*2**-1*(FOS)**-1\n",
+ "#After sub values in equation we get and further simplifying we get\n",
+ "#80=b*(100**2)**-1+a\n",
+ "#After sub values in equation 1 and 3 and further simplifying we get\n",
+ "b2=120*100**2*2**-1\n",
+ "\n",
+ "#from equation(3)\n",
+ "a2=120*2**-1-a\n",
+ "\n",
+ "#At outer radius r_0,radial pressure=0\n",
+ "r_02=(b2*a2**-1)**0.5\n",
+ "\n",
+ "#thickness\n",
+ "t2=r_02-r1\n",
+ "\n",
+ "#Result\n",
+ "print\"Thickness of metal by:Max Principal stress theory\",round(t,2),\"mm\"\n",
+ "print\" :Max shear stress thoery\",round(t2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness of metal by:Max Principal stress theory 41.42 mm\n",
+ " :Max shear stress thoery 73.21 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.1_3.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.1_3.ipynb new file mode 100644 index 00000000..89760142 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.1_3.ipynb @@ -0,0 +1,22 @@ +{
+ "metadata": {
+ "name": "chapter no.1.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter No.1:Introduction"
+ ]
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.2_3.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.2_3.ipynb new file mode 100644 index 00000000..a37ee3c9 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.2_3.ipynb @@ -0,0 +1,2742 @@ +{
+ "metadata": {
+ "name": "chapter no.2.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 2:Simple Stresses And Strains"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.1,Page No.14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "P=45*10**3 #N #Load\n",
+ "E=200*10**3 #N/mm**2 #Modulus of elasticity of rod\n",
+ "L=500 #mm #Length of rod\n",
+ "d=20 #mm #Diameter of rod\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A=pi*d**2*4**-1 #mm**2 #Area of circular rod\n",
+ "p=P*A**-1 #N/mm**2 #stress\n",
+ "e=p*E**-1 #strain\n",
+ "dell_l=(P*L)*(A*E)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"The stress in bar due to Load is\",round(p,5),\"N/mm\"\n",
+ "print\"The strain in bar due to Load is\",round(e,5),\"N/mm\"\n",
+ "print\"The Elongation in bar due to Load is\",round(dell_l,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The stress in bar due to Load is 143.23945 N/mm\n",
+ "The strain in bar due to Load is 0.00072 N/mm\n",
+ "The Elongation in bar due to Load is 0.36 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.2,Page No.15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "A=15*0.75 #mm**2 #area of steel tape\n",
+ "P=100 #N #Force apllied\n",
+ "L=30*10**3 #mm #Length of tape\n",
+ "E=200*10**3 #N/m**2 #Modulus of Elasticity of steel tape\n",
+ "AB=150 #m #Measurement of Line AB \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "dell_l=P*L*(A*E)**-1 #mm #Elongation\n",
+ "l=L+dell_l*10**-3 #mm #Actual Length \n",
+ "AB1=AB*l*L**-1 #m Actual Length of AB\n",
+ "\n",
+ "#Result\n",
+ "print\"The Actual Length of Line AB is\",round(AB1,2),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Actual Length of Line AB is 150.0 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.3,Page No.15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Let y be the yield stress\n",
+ "\n",
+ "y=250 #N/mm**2 #yield stress\n",
+ "FOS=1.75 #Factor of safety\n",
+ "P=140*10**3 #N #compressive Load\n",
+ "D=101.6 #mm #External diameter\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "p=y*(FOS)**-1 #N/mm**2 #Permissible stress\n",
+ "A=P*p**-1 #mm**2 #Area of hollow tube\n",
+ "\n",
+ "#Let d be the internal diameter of tube\n",
+ "d=-((A*4*(pi)**-1)-D**2)\n",
+ "X=d**0.5\n",
+ "t=(D-X)*2**-1 #mm #Thickness of steel tube\n",
+ "\n",
+ "#result\n",
+ "print\"The thickness of steel tube is\",round(t,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The thickness of steel tube is 3.17 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.4,Page No.16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=25 #mm #diameter of steel\n",
+ "L=200 #mm #length of stee\n",
+ "P=80*10**3 #KN #Load\n",
+ "P1=160*10**3 #N #Load at Elastic Limit\n",
+ "P2=180*10**3 #N #Max Load\n",
+ "L1=56 #mm #Total Extension\n",
+ "dell_l=0.16 #mm #Extension\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A=pi*d**2*4**-1 #Area of steel #mm**2\n",
+ "\n",
+ "p=P1*A**-1 #Stress at Elastic Limit #N/mm**2\n",
+ "E=P*L*(A*dell_l)**-1\n",
+ "\n",
+ "#result\n",
+ "print E"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "203718.327158\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.5,Page No.16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=20 #mm #Diameter of bar\n",
+ "d2=14.7 #mm #Diameter at neck\n",
+ "L=200 #mm #guage Length \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,10,20,30,40,50,60]\n",
+ "Y1=[0,32,64,95,127,160,190]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Extension in divisions\")\n",
+ "plt.ylabel(\"Load in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "A=pi*4**-1*d**2 #mm**2 #Area of Bar\n",
+ "A2=pi*4**-1*d2**2\n",
+ "\n",
+ "P=45 #KN #Load obtained from graph\n",
+ "dell=0.143 #mm #Divisions\n",
+ "\n",
+ "#Modulus of Elasticity\n",
+ "E=P*L*(dell*A)**-1 \n",
+ "\n",
+ "BL=93*10**3 #N #Breaking Load\n",
+ "\n",
+ "#Nominal stress at Breaking point\n",
+ "sigma=BL*A**-1 #KN/mm**2 \n",
+ "\n",
+ "#True stress at breaking Point\n",
+ "sigma1=BL*A2**-1\n",
+ "\n",
+ "#Percentage Elongation \n",
+ "dell_l=(A-A2)*A**-1*100\n",
+ "\n",
+ "#Result\n",
+ "print\"The Value of ELongation is\",round(E,2),\"N/mm**2\"\n",
+ "print\"The Nominal stress at the Breaking Point\",round(sigma,2),\"KN/mm**2\"\n",
+ "print\"The True stress at the Breaking Point\",round(sigma1,2),\"KN/mm**2\"\n",
+ "print\"The Percentage Reduction in Area is\",round(dell_l,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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gwAFOnDjBRx99xNatWys876vr+8tf/kKzZs2Ii4vDVHG/iq+u7bL09HT279/P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\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x4d1d370>"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Value of ELongation is 200.33 N/mm**2\n",
+ "The Nominal stress at the Breaking Point 296.03 KN/mm**2\n",
+ "The True stress at the Breaking Point 547.97 KN/mm**2\n",
+ "The Percentage Reduction in Area is 45.98\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.6,Page No.19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=40*10**3 #N #Load\n",
+ "L1=160 #mm #Length of Bar1\n",
+ "L2=240 #mm #Length of bar2\n",
+ "L3=160 #mm #Length of bar3\n",
+ "d1=25 #mm #Diameter of Bar1\n",
+ "d2=20 #mm #diameter of bar2\n",
+ "d3=25 #mm #diameter of bar3\n",
+ "dell_l=0.285 #mm #Total Extension of bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "E=P*4*(dell_l*pi)**-1*(L1*(d1**2)**-1+L2*(d2**2)**-1+L3*(d3**2)**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"The Young's Modulus of the material\",round(E,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Young's Modulus of the material 198714.72 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.7,Page No.19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E1=2*10**5 #N/mm**2 #modulus of Elasticity of material1\n",
+ "E2=1*10**5 #N/mm**2 #modulus of Elasticity of material2\n",
+ "P=25*10**3 #N #Load \n",
+ "t=20 #mm #thickness of material\n",
+ "b1=40 #mm #width of material1\n",
+ "b2=30 #mm #width of material2\n",
+ "L1=500 #mm #Length of material1\n",
+ "L2=750 #mm #Length of material2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A1=b1*t #mm**2 #Area of materila1\n",
+ "A2=b2*t #mm**2 #Area of material2\n",
+ "\n",
+ "dell_l1=P*L1*(A1*E1)**-1 #Extension of Portion1\n",
+ "dell_l2=P*L2*(A2*E2)**-1 #Extension of portion2\n",
+ "\n",
+ "#Total Extension of Bar is\n",
+ "dell_l=dell_l1+dell_l2\n",
+ "\n",
+ "#Result\n",
+ "print\"The Total Extension of the Bar is\",round(dell_l,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Total Extension of the Bar is 0.39 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.8,Page No.20"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1000 #mm #Length of Bar\n",
+ "l=400 #mm #Length upto which bire is drilled \n",
+ "D=30 #mm #diameter of bar\n",
+ "d1=10 #mm #diameter of bore\n",
+ "P=25*10**3 #N #Load\n",
+ "dell_l=0.185 #mm #Extension of bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "L1=L-l #Length of bar above the bore\n",
+ "L2=400 #mm #Length of bore\n",
+ "\n",
+ "A1=pi*4**-1*D**2 #Area of bar\n",
+ "A2=pi*4**-1*(D**2-d1**2) #Area of bore\n",
+ "\n",
+ "E=P*dell_l**-1*(L1*A1**-1+L2*A2**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"The Modulus of ELasticity is\",round(E,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Modulus of ELasticity is 200735.96 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.11,Page No.23"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "t=10 #mm #Thickness of steel\n",
+ "b1=60 #mm #width of plate1\n",
+ "b2=40 #mm #width of plate2\n",
+ "P=60*10**3 #Load\n",
+ "L=600 #mm #Length of plate\n",
+ "E=2*10**5 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Extension of taperong bar of rectangular section\n",
+ "dell_l=P*L*(t*E*(b1-b2))**-1*log(b1*b2**-1)\n",
+ "\n",
+ "A_av=(b1*t+b2*t)*2**-1 #Average Area #mm**2\n",
+ "dell_l2=P*L*(A_av*E)**-1 \n",
+ "\n",
+ "#PErcentage Error\n",
+ "e=(dell_l-dell_l2)*(dell_l)**-1*100\n",
+ "\n",
+ "#Result\n",
+ "print\"The Percentage Error is\",round(e,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Percentage Error is 1.35\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.12,Page No.23"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1.5 #m #Length of steel bar\n",
+ "L1=1000 #m0 #Length of steel bar 1\n",
+ "L2=500 #m #Length of steel bar 2\n",
+ "d1=40 #Diameter of steel bar 1\n",
+ "d2=20 #diameter of steel bar 2\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "P=160*10**3 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A1=pi*4**-1*d1**2 #Area of Portion 1\n",
+ "\n",
+ "#Extension of uniform Portion 1\n",
+ "dell_l1=P*L1*(A1*E)**-1 #mm\n",
+ "\n",
+ "#Extension of uniform Portion 2\n",
+ "dell_l2=4*P*L2*(pi*d1*d2*E)**-1 #mm\n",
+ "\n",
+ "#Total Extension of Bar\n",
+ "dell_l=dell_l1+dell_l2\n",
+ "\n",
+ "#Result\n",
+ "print\"The Elongation of the Bar is\",round(dell_l,2),\"mm\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.14,Page No.25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Portion AB\n",
+ "L_AB=600 #mm #Length of AB\n",
+ "A_AB=40*40 #mm**2 #Cross-section Area of AB\n",
+ "\n",
+ "#Portion BC\n",
+ "L_BC=800 #mm #Length of BC\n",
+ "A_BC=30*30 #mm #Length of BC\n",
+ "\n",
+ "#Portion CD\n",
+ "L_CD=1000 #mm #Length of CD\n",
+ "A_CD=20*20 #mm #Area of CD\n",
+ "\n",
+ "P1=80*10**3 #N #Load1\n",
+ "P2=60*10**3 #N #Load2\n",
+ "P3=40*10**3 #N #Load3\n",
+ "\n",
+ "E=2*10**5 #Modulus of Elasticity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "P4=P1-P2+P3 #Load4\n",
+ "\n",
+ "#Now Force in AB\n",
+ "F_AB=P1\n",
+ "\n",
+ "#Force in BC\n",
+ "F_BC=P1-P2\n",
+ "\n",
+ "#Force in CD\n",
+ "F_CD=P4\n",
+ "\n",
+ "#Extension of AB\n",
+ "dell_l_AB=F_AB*L_AB*(A_AB*E)**-1\n",
+ "\n",
+ "#Extension of BC\n",
+ "dell_l_BC=F_BC*L_BC*(A_BC*E)**-1\n",
+ "\n",
+ "#Extension of CD\n",
+ "dell_l_CD=F_CD*L_CD*(A_CD*E)**-1\n",
+ "\n",
+ "#Total Extension\n",
+ "dell_l=dell_l_AB+dell_l_BC+dell_l_CD\n",
+ "\n",
+ "#Result\n",
+ "print\"The Total Extension in Bar is\",round(dell_l,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Total Extension in Bar is 0.99 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.15,Page No.26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=800 #mm #Length of bar\n",
+ "F1=30*10**3 #N #Force acting on the bar\n",
+ "F2=60*10**3 #N #force acting on the bar\n",
+ "L=800 #mm #Length of bar\n",
+ "d=25 #mm #diameter of bar\n",
+ "L_AC=275 #mm #Length of AC\n",
+ "L_CD=150 #mm #Length of CD\n",
+ "L_DB=375 #mm #Length of DB\n",
+ "E=2*10**5 #Pa #Modulus of elasticity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P be the Reaction on tne Bar from support at A\n",
+ "\n",
+ "#Shortening of Portion AC\n",
+ "#dell_l_AC1=P*L_AC*(A*E)**-1\n",
+ "\n",
+ "#Shortening of Portion CD\n",
+ "#dell_l_CD1=(30+P)*L_CD*(A*E)**-1\n",
+ "\n",
+ "#Extension of Portion DB\n",
+ "#dell_l_DB1=(30-P)*L_DB*(A*E)**-1\n",
+ "\n",
+ "#Total Extensions=1*(A*E)**-1*(P*L_AC-(30+P)*L_CD+(30-P)*L_DB)\n",
+ "#As Supports are unyielding,Total Extensions=0\n",
+ "\n",
+ "#After substituting values in above equation and Further simplifying we get\n",
+ "P=(30*375-150*30)*800**-1\n",
+ "\n",
+ "#Reaction of support A\n",
+ "R_A=P\n",
+ "\n",
+ "#Reaction of support B\n",
+ "R_B=30-P\n",
+ "\n",
+ "#Cross-sectional Area\n",
+ "A=pi*4**-1*d**2\n",
+ "\n",
+ "#Stress in Portion AC\n",
+ "sigma1=P*10**3*A**-1 #N/mm**2\n",
+ "\n",
+ "#Stress in Portion CD\n",
+ "sigma2=(30+P)*10**3*A**-1 #N/mm**2\n",
+ "\n",
+ "#Stress in Portion DB\n",
+ "sigma3=(30-P)*10**3*A**-1 #N/mm**2\n",
+ "\n",
+ "#Shortening of Portion AC\n",
+ "dell_l_AC2=P*10**3*L_AC*(A*E)**-1 #mm \n",
+ "\n",
+ "#Shortening of Portion CD\n",
+ "dell_l_CD2=(30+P)*10**3*L_CD*(A*E)**-1 #mm \n",
+ "\n",
+ "#Extension of Portion DB\n",
+ "dell_l_DB2=(30-P)*10**3*L_DB*(A*E)**-1 #mm \n",
+ "\n",
+ "#result\n",
+ "print\"The Reactios at two Ends are:R_A\",round(R_A,2),\"KN\"\n",
+ "print\" :R_B\",round(R_B,2),\"KN\"\n",
+ "print\"Stress in Portion AC\",round(sigma1,2),\"N/mm**2\"\n",
+ "print\"Stress in Portion CD\",round(sigma2,2),\"N/mm**2\"\n",
+ "print\"Stress in Portion DB\",round(sigma3,2),\"N/mm**2\"\n",
+ "print\"Shortening of Portion AC\",round(dell_l_AC2,3),\"mm\"\n",
+ "print\"Shortening of Portion CD\",round(dell_l_CD2,3),\"mm\"\n",
+ "print\"Shortening of Portion DB\",round(dell_l_DB2,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Reactios at two Ends are:R_A 8.44 KN\n",
+ " :R_B 21.56 KN\n",
+ "Stress in Portion AC 17.19 N/mm**2\n",
+ "Stress in Portion CD 78.3 N/mm**2\n",
+ "Stress in Portion DB 43.93 N/mm**2\n",
+ "Shortening of Portion AC 0.024 mm\n",
+ "Shortening of Portion CD 0.059 mm\n",
+ "Shortening of Portion DB 0.082 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.19,Page No.29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "h=4 #m #height of Pillars\n",
+ "P=20 #KN #Load at M\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_A,P_B,P_C,P_D be the forces introduced in the Pillars\n",
+ "#Sun of All Vertical Forces\n",
+ "#P_A+P_B+P_C+P_D=20 ....................(1)\n",
+ "\n",
+ "#Sum of moment about AB, we get\n",
+ "#P_D+P_C=12 ....................(2)\n",
+ "\n",
+ "#Sum of Moment about AD\n",
+ "#P_C+P_B=8 ....................(3)\n",
+ "\n",
+ "#Let dell_l_A,dell_l_B,dell_l-C,dell_l_D be the deformations of Pillars A,B,C,D respectively\n",
+ "#Diagonals AC and BD will remain straight Lines even after the Load is applied.\n",
+ "#Deflection of central Point is given by (dell_l_A+dell_l_C)*2**-1 & (dell_l_B+dell_l_D)*2**-1\n",
+ "\n",
+ "#dell_l_A+dell_l_C=dell_l_B+ell_l_D\n",
+ "#P_A*L*(A*E)**-1+P_C*L*(A*E)**-1=P_B*L*(A*E)**-1+P_D*L*(A*E)**-1\n",
+ "\n",
+ "#Since Pillars are identical in Length,cross-sectional area,material Property\n",
+ "#P_A+P_C=P_B+P_D ..............(4)\n",
+ "\n",
+ "#From Equations 1 and 4 we get\n",
+ "#P_B+P_D=10 ....................(5)\n",
+ " \n",
+ "#Substracting Equation 3 from Equation 2 we get\n",
+ "#P_D-P_B=4 ....................(6)\n",
+ "\n",
+ "#Adding Equation 5 and 6 we get\n",
+ "\n",
+ "P_D=14*2**-1\n",
+ "P_C=12-P_D\n",
+ "P_B=8-P_C\n",
+ "\n",
+ "#Now substituting values of P_B,P_C,P_D in equation1 we get\n",
+ "P_A=20-(P_B+P_C+P_D)\n",
+ "\n",
+ "#Result\n",
+ "print\"The Forces Developed in the Pillars are:P_A\",round(P_A,2),\"KN\"\n",
+ "print\" :P_B\",round(P_B,2),\"KN\"\n",
+ "print\" :P_C\",round(P_C,2),\"KN\"\n",
+ "print\" :P_D\",round(P_D,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Forces Developed in the Pillars are:P_A 5.0 KN\n",
+ " :P_B 3.0 KN\n",
+ " :P_C 5.0 KN\n",
+ " :P_D 7.0 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.20,Page No.31"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "sigma=150 #N/mm**2 #Stress\n",
+ "P=40*10**3 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt P_A.P_B,P_C,P_D be the forces developed in wires A,B,C,D respectively\n",
+ "\n",
+ "#Let sum of all Vertical Forces=0\n",
+ "#P_A+P_B+P_C+P_D=40 ..........................(1)\n",
+ "\n",
+ "#Let x be the distance between each wires\n",
+ "#sum of all moments=0\n",
+ "#P_B*x+P_C*2*x+P_D*3*x=40*2*x\n",
+ "\n",
+ "#After further simplifying we get\n",
+ "#P_B+2*P_C+3*P_D=80 ..........................(2)\n",
+ "\n",
+ "#As the equations of statics ae not enough to find unknowns,Consider compatibilit Equations\n",
+ "\n",
+ "#Let dell_l be the increse in elongation of wire\n",
+ "\n",
+ "#dell_l_B=dell_l_A+dell_l\n",
+ "#dell_l_C=dell_l_A+2*dell_l\n",
+ "#dell_l_D=dell_l_A+3*dell_l\n",
+ "\n",
+ "#Let P1 be the force required for the Elongation of wires,then\n",
+ "#P_B=P_A+P1 ]\n",
+ "#P_C=P_A+2*P1 ]\n",
+ "#P_D=P_A+3*P1 ] ................................(3) \n",
+ "\n",
+ "#from Equation (3) and (1) we get\n",
+ "#2*P_A+3*P1=20 ................................(4)\n",
+ "\n",
+ "#from Equation (3) and (2) we get\n",
+ "#6*P_A+14*P1=80 \n",
+ "\n",
+ "#subtracting 3 times equation (4) from (3) we get\n",
+ "P1=20*5**-1\n",
+ "\n",
+ "#from Equation 4 we get\n",
+ "P_A=(80-14*P1)*6**-1\n",
+ "P_B=P_A+P1\n",
+ "P_C=P_A+2*P1\n",
+ "P_D=P_A+3*P1\n",
+ "\n",
+ "#Let d be the diameter required,then\n",
+ "d=(P_D*10**3*4*(pi*150)**-1)**0.5\n",
+ "\n",
+ "#result\n",
+ "print\"The Required Diameter is\",round(d,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Required Diameter is 11.65 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.21,Page No.32"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=20*10**3 #N #Load\n",
+ "d=6 #mm #diameter of wire\n",
+ "E=2*10**5 #N/mm**2 \n",
+ "L_BO=4000 #mm #Length of BO\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let theta be the angle between OA and OB and also between OC and OB\n",
+ "theta=30\n",
+ "\n",
+ "#Let P_OA,P_OB,P_OC be the Forces introduced in wires OA,OB,OC respectively\n",
+ "#Due to symmetry P_OA=P_OC (same angles)\n",
+ "\n",
+ "#Sum of all Vertical Forces=0\n",
+ "#P_OA*cos(theta)+P_OB+P_OC*cos(theta)=P\n",
+ "\n",
+ "#After further simplifyinf we get\n",
+ "#2*P_OA*cos(theta)+P_OB=20 ...............(1)\n",
+ "\n",
+ "#Let oo1 be the extension of BO\n",
+ "#oo1=L_A1o1*(cos(theta))**-1\n",
+ "\n",
+ "#From relation we get\n",
+ "#P_OB*L_BO=P_OA*L_AO*(cos(theta))**-1\n",
+ "\n",
+ "#But L_AO=L_BO*(cos(theta))**-1\n",
+ "\n",
+ "#After substituting value of L_AO in above equation we get\n",
+ "#P_OB=0.75*P_OA .......................(2)\n",
+ "\n",
+ "#substituting in Equation 1 we get\n",
+ "#2*P_OA*cos(theta)+0.75*P_OA=20\n",
+ "\n",
+ "P_OA=20*(2*cos(theta*pi*180**-1)+0.75)**-1\n",
+ "\n",
+ "P_OB=0.75*P_OA\n",
+ "\n",
+ "A=pi*4**-1*d**2\n",
+ "\n",
+ "#Vertical displacement of Load\n",
+ "dell_l_BO=P_OB*10**3*L_BO*(A*E)**-1\n",
+ " \n",
+ "#Result\n",
+ "print\"Forces in each wire is:P_OA\",round(P_OA,2),\"KN\"\n",
+ "print\" :P_OB\",round(P_OB,2),\"KN\"\n",
+ "print\"Vertical displacement of Loadis\",round(dell_l_BO,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Forces in each wire is:P_OA 8.06 KN\n",
+ " :P_OB 6.04 KN\n",
+ "Vertical displacement of Loadis 4.27 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.22,Page No.34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_s=L_a=L=500 #mm #Length of bar\n",
+ "A_a=50*20 #mm #Area of aluminium strip\n",
+ "A_s=50*15 #mm #Area of steel strip\n",
+ "P=50*10**3 #N #Load\n",
+ "E_a=1*10**5 #N/mm**2 #Modulus of aluminium \n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of steel\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_a and P_s br the Load shared by aluminium and steel strip\n",
+ "#P_a+P_s=P ..................(1)\n",
+ "\n",
+ "#For compatibility condition,dell_l_a=dell_l_s\n",
+ "#P_a*L_a*(A_a*E_a)**-1=P_s*L_s*(A_s*E_s)**-1 .....(2)\n",
+ "\n",
+ "#As L_a=L_s we get\n",
+ "#P_s=1.5*P_a .................(3)\n",
+ "\n",
+ "#From Equation 1 and 2 we get\n",
+ "P_a=P*2.5**-1\n",
+ "\n",
+ "#Substituting in equation 1 we get\n",
+ "P_s=P-P_a\n",
+ "\n",
+ "#stress in aluminium strip \n",
+ "sigma_a=P_a*A_a**-1\n",
+ "\n",
+ "#stress in steel strip\n",
+ "sigma_s=P_s*A_s**-1\n",
+ "\n",
+ "#Now from the relation we get\n",
+ "dell_l_a=dell_l_s=P_s*L_s*(A_s*E_s)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Stress in Aluminium strip is\",round(sigma_a,2),\"N/mm**2\"\n",
+ "print\"Stress in steel strip is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"The Extension of the bar is\",round(dell_l_s,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in Aluminium strip is 20.0 N/mm**2\n",
+ "Stress in steel strip is 40.0 N/mm**2\n",
+ "The Extension of the bar is 0.1 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.23,Page No.35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D_s=20 #mm #Diameter of steel\n",
+ "D_Ci=20 #mm #Internal Diameter of Copper\n",
+ "t=5 #mm #THickness of copper bar\n",
+ "P=100*10**3 #N #Load\n",
+ "E_s=2*10**5 #N/mm**2 #modulus of elasticity of steel\n",
+ "E_c=1.2*10**5 #N/mm**2 #Modulus of Elasticity of Copper\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A_s=pi*4**-1*D_s**2 #mm**2 #Area of steel\n",
+ "D_Ce=D_s+2*t #mm #External Diameterof Copper Tube\n",
+ "\n",
+ "A_c=pi*4**-1*(D_Ce**2-D_Ci**2) #mm**2 #Area of Copper\n",
+ "\n",
+ "#From static Equilibrium condition\n",
+ "#Let P_s and P_c be the Load shared by steel and copper in KN\n",
+ "#P_s+P_c=100 ....................................(1)\n",
+ "\n",
+ "#From compatibility Equation,dell_l_s=dell_l_c\n",
+ "#P_s*L*(A_s*E_s)**-1=P_c*L*(A_c*E_c)**-1\n",
+ "\n",
+ "#Substituting values in above Equation we get\n",
+ "#P_s=1.3333*P_C \n",
+ "\n",
+ "#Now Substituting value of P_s in Equation (1),we get\n",
+ "P_c=100*2.3333**-1 #KN\n",
+ "P_s=100-P_c #KN\n",
+ "\n",
+ "#Stress in steel\n",
+ "sigma_s=P_s*10**3*A_s**-1 #N/mm**2\n",
+ "\n",
+ "#Stress in copper\n",
+ "sigma_c=P_c*10**3*A_c**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Two material are:sigma_s\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\" :sigma_c\",round(sigma_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses Developed in Two material are:sigma_s 181.89 N/mm**2\n",
+ " :sigma_c 109.14 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.24,Page No.36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "A_C=230*400 #mm #Area of column\n",
+ "D_s=12 #mm #Diameter of steel Bar\n",
+ "P=600*10**3 #N #Axial compression\n",
+ "#E_s*E_c=18.67\n",
+ "n=8 #number of steel Bars\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A_s=pi*4**-1*D_s**2*n #Area of steel #mm**2 \n",
+ "A_c=A_C-A_s #mm**2 #Area of concrete\n",
+ "\n",
+ "#From static Equilibrium condition\n",
+ "#P_s+P_c=600 .........(1)\n",
+ "\n",
+ "#Now from compatibility Equation dell_l_s=dell_l_c we get,\n",
+ "#P_s*L*(A_s*E_s)**-1=P_c*L*(A_c*E_c)**-1\n",
+ "\n",
+ "#Substituting values in above Equation we get\n",
+ "#P_s=0.1854*P_c\n",
+ "\n",
+ "#Now Substituting value of P_s in Equation (1),we get\n",
+ "P_c=600*1.1854**-1\n",
+ "P_s=600-P_c\n",
+ "\n",
+ "#Stress in steel\n",
+ "sigma_s=P_s*10**3*A_s**-1 #N/mm**2\n",
+ "\n",
+ "#Stress in copper\n",
+ "sigma_c=P_c*10**3*A_c**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Two material are:sigma_s\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\" :sigma_c\",round(sigma_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses Developed in Two material are:sigma_s 103.72 N/mm**2\n",
+ " :sigma_c 5.56 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.25,Page No.36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=200*10**3 #N #Load\n",
+ "A_a=1000 #mm**2 #Area of Aluminium\n",
+ "A_s=800 #mm**2 #Area of steel\n",
+ "E_a=1*10**5 #N/mm**2 #Modulus of Elasticity of Aluminium\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of ELasticity of steel\n",
+ "sigma_a1=65 #N/mm**2 #stress in aluminium\n",
+ "sigma_s1=150 #N/mm**2 #Stress in steel\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_a and P_s be the force in aluminium and steel pillar respectively\n",
+ "\n",
+ "#Now,sum of forces in Vertical direction we get\n",
+ "#2*P_a+P_s=200 .........................................(1)\n",
+ "\n",
+ "#By compatibility Equation dell_l_s=dell_l_a we get\n",
+ "#P_s=1.28*P_a ..........................................(2)\n",
+ "\n",
+ "#Now substituting value of P_s in Equation 1 we get\n",
+ "P_a=200*3.28**-1 #KN\n",
+ "P_s=200-2*P_a #KN\n",
+ "\n",
+ "#Stress developed in aluminium\n",
+ "sigma_a=P_a*10**3*A_a**-1 #N/mm**2 \n",
+ "\n",
+ "#Stress developed in steel\n",
+ "sigma_s=P_s*10**3*A_s**-1 #N/mm**2 \n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#Let sigma_a1 and sigma_s1 be the stresses in Aluminium and steel due to Additional LOad\n",
+ "\n",
+ "P_a1=sigma_a1*A_a #Load carrying capacity of aluminium\n",
+ "P_s1=1.28*P_a1\n",
+ "\n",
+ "#Total Load carrying capacity \n",
+ "P1=2*P_a1+P_s1 #N \n",
+ "\n",
+ "P_s2=sigma_s1*A_s #Load carrying capacity of steel\n",
+ "P_a2=P_s2*1.28**-1\n",
+ "\n",
+ "#Total Load carrying capacity\n",
+ "P2=2*P_a2+P_s2\n",
+ "\n",
+ "#Additional Load\n",
+ "P3=P1-P\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Each Pillar is:sigma_a\",round(sigma_a,2),\"N/mm**2\"\n",
+ "print\" :sigma_s\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"Additional Load taken by pillars is\",round(P3,2),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "78.0487804878\n",
+ "Stresses Developed in Each Pillar is:sigma_a 60.98 N/mm**2\n",
+ " :sigma_s 97.56 N/mm**2\n",
+ "Additional Load taken by pillars is 13200.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.26,Page No.37"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=500 #mm #Length of assembly\n",
+ "D=16 #mm #Diameter of steel bolt\n",
+ "Di=20 #mm #internal Diameter of copper tube\n",
+ "Do=30 #mm #External Diameter of copper tube\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity of steel\n",
+ "E_c=1.2*10**5 #N/mm**2 #Modulus of Elasticity of copper\n",
+ "p=2 #mm #Pitch of nut\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_s be the Force in bolt and P_c be the FOrce in copper tube\n",
+ "#P_s=-P_s\n",
+ "\n",
+ "dell=1*4**-1*2 #Quarter turn of nut total movement\n",
+ "\n",
+ "#dell=dell_s+dell_c\n",
+ "\n",
+ "#Area of steel\n",
+ "A_s=pi*4**-1*D**2\n",
+ "\n",
+ "#Area of copper\n",
+ "A_c=pi*4**-1*(Do**2-Di**2)\n",
+ "\n",
+ "#dell=P*L*(A_s*E_s)**-1+P*L*(A_c*E_c)**-1\n",
+ "P=dell*(1*(A_s*E_s)**-1+1*(A_c*E_c)**-1)**-1*L**-1 #LOad\n",
+ "\n",
+ "P_s=P*A_s**-1\n",
+ "P_c=P*A_c**-1\n",
+ "\n",
+ "#result\n",
+ "print\"stress introduced in bolt is\",round(P_s,2),\"N/mm**2\"\n",
+ "print\"stress introduced in tube is\",round(P_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "stress introduced in bolt is 107.91 N/mm**2\n",
+ "stress introduced in tube is 55.25 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.27,Page No.39"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=20 #mm #Diameter of Bolts\n",
+ "Di=25 #m #internal Diameter\n",
+ "t=10 #mm #Thickness of bolt\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "p=3 #mm #Pitch\n",
+ "theta=30 #degree\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_s be the Force in each bolt and P_c be the FOrce in copper tube\n",
+ "#From Static Equilibrium condition\n",
+ "#P_c=2*P_s\n",
+ "\n",
+ "#As nut moves by 60 degree.If nut moves by 360 degree its Longitudinal movement is by 3 mm\n",
+ "dell=theta*360**-1*p\n",
+ "\n",
+ "#From Compatibility Equaton we get\n",
+ "#dell=dell_c+dell_s\n",
+ "\n",
+ "\n",
+ "A_s=pi*4**-1*"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.28,Page No.40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=9 #m #Length of rigid bar\n",
+ "L_b=3000 #Length of bar\n",
+ "A_b=1000 #mm**2 #Area of bar\n",
+ "E_b=1*10**5 #N/mm**2 #Modulus of Elasticity of brasss bar\n",
+ "L_s=5000 #mm #Length of steel bar\n",
+ "A_s=445 #mm**2 #Area of steel bar\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of elasticity of steel bar\n",
+ "P=3000 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From static equilibrium Equation of the rod after appliying Load is\n",
+ "#P_b+P_s=P ......................(1)\n",
+ "\n",
+ "#P_b=1.8727*P_s ..................(2)\n",
+ "\n",
+ "#NOw substituting equation 2 in equation 1 we get\n",
+ "P_s=P*2.8727**-1\n",
+ "P_b=P-P_s\n",
+ "\n",
+ "d=P_s*L*P**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance at which Load applied even after which bar remains horizontal is\",round(d,2),\"m\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance at which Load applied even after which bar remains horizontal is 3.13 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.29,Page No.41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "A_b=1000 #MM**2 #Area of brass bar\n",
+ "E_b=1*10**5 #N/mm**2 #Modulus of Elasticity of brass\n",
+ "A_s=600 #N/mm**2 #Area of steel rod\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of eLasticity of steel bar\n",
+ "P=10*10**2 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_b be the tensile force in brass bar and P_s be the compressive force in steel bar\n",
+ "#Now taking moment about A we get static Equilibrium condition as\n",
+ "#P_b+2*P_s=27500 ......................................(1)\n",
+ "\n",
+ "#Now from deformed shape we get\n",
+ "#dell_s=2*dell_b\n",
+ "\n",
+ "#P_s*L_s*(A_s*E_s)**-1=P_b*L_b*(A_b*E_b)**-1\n",
+ "#Further simplifying we get\n",
+ "#P_s=1.2*P_b .........................................(2)\n",
+ "\n",
+ "#Now substituting equation 1 in equation 2 we get\n",
+ "P_b=27500*3.4**-1\n",
+ "P_s=1.2*P_b\n",
+ "\n",
+ "#Tensile stress in brass bar \n",
+ "sigma_b=P_b*A_b**-1\n",
+ "\n",
+ "#compressive stress in steel bar\n",
+ "sigma_s=P_s*A_s**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Compressive Stress in Bar is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"tensile Stress in Bar is\",round(sigma_b,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Compressive Stress in Bar is 16.18 N/mm**2\n",
+ "tensile Stress in Bar is 8.09 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.30,Page No.44"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=12.6 #m #Length of rail\n",
+ "t1=24 #Degree celsius\n",
+ "t2=44 #degree celsius\n",
+ "alpha=12*10**-6 #Per degree celsius\n",
+ "E=2*10**5 #N/mm**2 #Modulus of ELasticity\n",
+ "gamma=2 #mm #Gap provided for Expansion\n",
+ "sigma=20 #N/mm**2 #Stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "t=t2-t1 #Temperature Difference\n",
+ "\n",
+ "#Free Expansion of the rails\n",
+ "dell=alpha*t*L*1000 #mm \n",
+ "\n",
+ "#When no expansion joint is provided then\n",
+ "p=dell*E*(L*10**3)**-1\n",
+ "\n",
+ "#When a gap of 2 mm is provided,then free expansion prevented is\n",
+ "dell_1=dell-gamma\n",
+ "p2=dell_1*E*(L*10**3)**-1\n",
+ "\n",
+ "#When stress is developed,then gap left is\n",
+ "gamma2=-(sigma*L*10**3*E**-1-dell)\n",
+ "\n",
+ "#Result\n",
+ "print\"The minimum gap between the two rails is\",round(dell,2),\"mm\"\n",
+ "print\"Thermal Developed in the rials if:No expansionn joint is provided:p\",round(p,2),\"N/mm**2\"\n",
+ "print\" :If a gap of is provided then :p2\",round(p2,2),\"N/mm**2\"\n",
+ "print\"When stress is developed gap left between the rails is\",round(gamma2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum gap between the two rails is 3.02 mm\n",
+ "Thermal Developed in the rials if:No expansionn joint is provided:p 48.0 N/mm**2\n",
+ " :If a gap of is provided then :p2 16.25 N/mm**2\n",
+ "When stress is developed gap left between the rails is 1.76 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.31,Page No.45"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "t=20 #degree celsius\n",
+ "E_a=70*10**9 #N/mm**2 #Modulus of Elasticicty of aluminium\n",
+ "alpha_a=11*10**-6 #per degree celsius #Temperature coeff of aluminium\n",
+ "alpha_s=12*10**-6 #Per degree celsius #Temperature coeff of steel\n",
+ "L_a=1000 #mm #Length of aluminium \n",
+ "L_s=3000 #mm #Length of steel\n",
+ "E_a=7*10**4 #N/mm**2 #Modulus of Elasticity of aluminium\n",
+ "E_s=2*10**5 #N/mm*2 #Modulus of Elasticity of steel\n",
+ "A_a=600 #mm**2 #Area of aluminium\n",
+ "A_s=300 #mm**2 #Area of steel\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Free Expansion \n",
+ "dell=alpha_a*t*L_a+alpha_s*t*L_s\n",
+ "\n",
+ "#support Reaction\n",
+ "P=dell*(L_a*(A_a*E_a)**-1+L_s*(A_s*E_s)**-1)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Reaction at support is\",round(P,2),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Reaction at support is 12735.48 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.33,Page No.48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=25 #mm #Diameter of Brass\n",
+ "De=50 #mm #External Diameter of steel tube\n",
+ "Di=25 #mm #Internal Diameter of steel tube\n",
+ "L=1.5 #m #Length of both bars\n",
+ "t1=30 #degree celsius #Initial Temperature\n",
+ "t2=100 #degree celsius #final Temperature\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of ELasticity of steel bar\n",
+ "E_b=1*10**5 #N/mm**2 #Modulus of Elasticity of brass bar\n",
+ "alpha_s=11.6*10**-6 #Temperature Coeff of steel\n",
+ "alpha_b=18.7*10**-6 #Temperature coeff of brass bar\n",
+ "d=20 #mm #diameter of pins\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "t=t2-t1 #Temperature Difference\n",
+ "A_s=pi*4**-1*(De**2-Di**2) #mm**2 #Area of steel\n",
+ "A_b=pi*4**-1*D**2 #mm**2 #Area of brass\n",
+ "\n",
+ "#Let P_b be the tensile force in brass bar and P_s be the compressive force in steel bar\n",
+ "#But from Equilibrium of Forces \n",
+ "#P_b=P_s=P\n",
+ "\n",
+ "#Let dell=dell_s+dell_b\n",
+ "dell=(alpha_b-alpha_s)*t*L*1000\n",
+ "\n",
+ "P=dell*(1*(A_s*E_s)**-1+1*(A_b*E_b)**-1)**-1*(L*1000)**-1\n",
+ "P_b=P_s=P\n",
+ "\n",
+ "#Stress in steel\n",
+ "sigma_s=P_s*A_s**-1\n",
+ "\n",
+ "#Stress in Brass\n",
+ "sigma_b=P_b*A_b**-1\n",
+ "\n",
+ "#Area of Pins\n",
+ "A_p=pi*4**-1*d**2\n",
+ "\n",
+ "#Since,the force is resisted by two cross section of pins\n",
+ "tou=P*(2*A_p)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress in steel bar is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"Stress in Brass bar is\",round(sigma_b,2),\"N/mm**2\"\n",
+ "print\"Shear Stresss induced in pins is\",round(tou,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in steel bar is 14.2 N/mm**2\n",
+ "Stress in Brass bar is 42.6 N/mm**2\n",
+ "Shear Stresss induced in pins is 33.28 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.34,Page No.49"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b_s=60 #mm #width of steel Bar\n",
+ "t_s=10 #mm #thickness of steel Bar\n",
+ "b_c=40 #mm #width of copper bar\n",
+ "t_c=5 #mm #thickness of copper bar\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity of steel bar\n",
+ "E_c=1*10**5 #N/mm**2 #Modulus of Elasticity of copper bar\n",
+ "alpha_s=12*10**-6 #Per degree celsius #Temperature coeff of steel bar\n",
+ "alpha_c=17*10**-6 #Per degree celsius #Temperature coeff of copper bar\n",
+ "L_s=L_c=L=1000 #mm #Length of bar\n",
+ "t=80 #degree celsius\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A_s=b_s*t_s #Area of steel bar\n",
+ "A_c=b_c*t_c #Area of copper bar\n",
+ "\n",
+ "#Let P_s be the tensile force in steel bar and P_c be the compressive force in copper bar\n",
+ "#The equilibrium of forces gives \n",
+ "#P_s=2*P_c\n",
+ "\n",
+ "#Let dell=dell_s+dell_b\n",
+ "dell=(alpha_c-alpha_s)*t\n",
+ "\n",
+ "P_c=dell*(2*(A_s*E_s)**-1+1*(A_c*E_c)**-1)**-1\n",
+ "P_s=2*P_c\n",
+ "\n",
+ "#Stress in copper \n",
+ "sigma_c=P_c*A_c**-1\n",
+ "\n",
+ "#Stress in steel\n",
+ "sigma_s=P_s*A_s**-1\n",
+ "\n",
+ "#Change in Length of bar\n",
+ "dell_2=alpha_s*t*L+P_s*L_s*(A_s*E_s)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Stress in copper is\",round(sigma_c,2),\"N/mm**2\"\n",
+ "print\"Stress in steel is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"the change in Length is\",round(dell_2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in copper is 30.0 N/mm**2\n",
+ "Stress in steel is 20.0 N/mm**2\n",
+ "the change in Length is 1.06 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.35,Page No.50"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=2*10**5 #N #Weight\n",
+ "L=1 #m #Length of each rod\n",
+ "A_c=A_s=A=500 #mm**2 #Area of each rod\n",
+ "t=40 #degree celsius #temperature\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity of steel rod\n",
+ "E_c=1*10**5 #N/mm**2 #modulus of Elastictiy of copper rod\n",
+ "alpha_s=1.2*10**-5 #Per degree Celsius #temp coeff of steel rod\n",
+ "alpha_c=1.8*10**-5 #Per degree Celsius #Temp coeff of copper rod\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_s be the force in each one of the copper rods and P_s be the force in steel rod\n",
+ "#2*P_c+P_s=P .....................(1)\n",
+ "\n",
+ "#Extension of copper bar=Extension of steel bar\n",
+ "#P_s*L*(A_s*E_s)**-1=P_c*L*(A_c*E_c)**-1\n",
+ "#after simplifying above equation we get\n",
+ "#P_s=2*P_c ........................(2)\n",
+ "\n",
+ "#Now substituting value of P_s in Equation 1 we get\n",
+ "P_c=P*4**-1\n",
+ "P_s=2*P_c\n",
+ "\n",
+ "#Now EXtension due to copper Load\n",
+ "dell_1=P_c*L*1000*(A_c*E_c)**-1\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#Due to rise of temperature of40 degree celsius\n",
+ "\n",
+ "#As bars are rigidly joined,let P_c1 be the compressive forccesdeveloped in copper bar and P_s1 be the tensile force in steel causing changes\n",
+ "#P_s1=2*P_c1\n",
+ "\n",
+ "#dell_s+dell_c=(alpha_c-alpha_s)*t*L .......................................(3)\n",
+ "#P_s1*L*(A_s*E_s)**-1+P_c1*L*(A_c*E_c)**-1=(alpha_c-alpha_s)*t*L ................(4)\n",
+ "#After substituting values in above equation and further simplifying we get,\n",
+ "P_c1=(alpha_c-alpha_s)*t*L*(2*(A_s*E_s)**-1+1*(A_c*E_c)**-1)**-1 #.................(5)\n",
+ "P_s1=2*P_c1\n",
+ "\n",
+ "#Extension of bar due to temperature rise\n",
+ "dell_2=alpha_s*t*L+P_s1*L*(A_s*E_s)**-1\n",
+ "\n",
+ "#Amount by which bar will descend\n",
+ "dell_3=dell_1+dell_2\n",
+ "\n",
+ "#Load carried by steel bar\n",
+ "P_S=P_s+P_s1\n",
+ "\n",
+ "#Load carried by copper bar\n",
+ "P_C=P_c-P_c1\n",
+ "\n",
+ "#Part-3\n",
+ "\n",
+ "#Let P_c1_1=P_c #For convenience\n",
+ "#Rise in temperature if Load is to be carried out by steel rod alone\n",
+ "P_c1_1=P_c\n",
+ "\n",
+ "#From equation 5 \n",
+ "t=P_c1_1*(2*(A_s*E_s)**-1+1*(A_c*E_c)**-1)*(alpha_c-alpha_s)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Extension Due top copper Load\",round(dell_1,2),\"mm\"\n",
+ "print\"Load carried by each rod:P_s\",round(P_s,2),\"N\"\n",
+ "print\" :P_c\",round(P_c,2),\"N\"\n",
+ "print\"Rise in Temperature of steel rod should be\",round(t,2),\"degree Celsius\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Extension Due top copper Load 1.0 mm\n",
+ "Load carried by each rod:P_s 100000.0 N\n",
+ " :P_c 50000.0 N\n",
+ "Rise in Temperature of steel rod should be 333.33 degree Celsius\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.36,Page No.53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "t=40 #degree celsius #temperature\n",
+ "A_s=400 #mm**2 #Area of steel bar\n",
+ "A_c=600 #mm**2 #Area of copper bar\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity of steel bar\n",
+ "E_c=1*10**5 #N/mm**2 #Modulus of Elasticity of copper bar\n",
+ "alpha_s=12*10**-6 #degree celsius #Temperature coeff of steel bar\n",
+ "alpha_c=18*10**-6 #degree celsius #Temperature coeff of copper bar\n",
+ "L_c=800 #mm #Length of copper bar\n",
+ "L_s=600 #mm #Length of steel bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_s be the tensile force in steel bar and P_c be the compressive force in copper bar\n",
+ "#Static Equilibrium obtained by taking moment about A\n",
+ "#P_c=2*P_s\n",
+ "\n",
+ "#From property of similar triangles we get\n",
+ "#(alpha_c*Lc-dell_c)*1**-1=(alpha_s*L_s-dell_s)*2**-1\n",
+ "#After substituting values in above equations and further simplifying we get\n",
+ "P_s=(2*alpha_c*L_c-alpha_s*L_s)*t*(L_s*(A_s*E_s)**-1+4*L_c*(A_c*E_c)**-1)**-1\n",
+ "P_c=2*P_s\n",
+ "\n",
+ "#Stress in steel rod\n",
+ "sigma_s=P_s*A_s**-1 #N/mm**2 \n",
+ "\n",
+ "#Stress in copper rod\n",
+ "sigma_c=P_c*A_c**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress in steel rod is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"STress in copper rod is\",round(sigma_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in steel rod is 35.51 N/mm**2\n",
+ "STress in copper rod is 47.34 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.37,Page No.61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=20 #mm #Diameter of bar\n",
+ "P=37.7*10**3 #N #Load\n",
+ "L=200 #mm #Guage Length \n",
+ "dell=0.12 #mm #Extension\n",
+ "dell_d=0.0036 #mm #contraction in diameter\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Area of bar\n",
+ "A=pi*4**-1*d**2\n",
+ "\n",
+ "#Let s and dell_s be the Linear strain and Lateral strain\n",
+ "s=dell*L**-1\n",
+ "dell_s=dell_d*d**-1\n",
+ "mu=dell_s*s**-1 #Poissoin's ratio \n",
+ "\n",
+ "#dell=P*L*(A*E)**-1\n",
+ "E=P*L*(dell*A)**-1 #N/mm**2 #Modulus of Elasticity of bar\n",
+ "\n",
+ "#Modulus of Rigidity\n",
+ "G=E*(2*(1+mu))**-1 #N/mm**2\n",
+ "\n",
+ "#Bulk Modulus \n",
+ "K=E*(3*(1-2*mu))**-1 #N/mm**2\n",
+ "\n",
+ "#result\n",
+ "print\"Poisson's ratio is\",round(mu,2)\n",
+ "print\"The Elastic constant are:E\",round(E,2)\n",
+ "print\" :G\",round(G,2)\n",
+ "print\" :K\",round(K,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Poisson's ratio is 0.3\n",
+ "The Elastic constant are:E 200004.71\n",
+ " :G 76924.89\n",
+ " :K 166670.59\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.38,Page No.62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=100 #mm #Diameter of circular rod\n",
+ "P=1*10**6 #N #Tensile Force\n",
+ "mu=0.3 #Poisson's ratio\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus \n",
+ "L=500 #mm #Length of rod\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Modulus of Rigidity\n",
+ "G=E*(2*(1+mu))**-1 #N/mm**2\n",
+ "\n",
+ "#Bulk Modulus \n",
+ "K=E*(3*(1-2*mu))**-1 #N/mm**2\n",
+ "\n",
+ "A=pi*4**-1*d**2 #mm**2 #Area of Circular rod\n",
+ "#Let sigma be the Longitudinal stress\n",
+ "sigma=P*A**-1 #N/mm**2 \n",
+ "\n",
+ "s=sigma*E**-1 #Linear strain\n",
+ "e_x=s\n",
+ "\n",
+ "#Volumetric strain\n",
+ "e_v=e_x*(1-2*mu)\n",
+ "\n",
+ "v=pi*4**-1*d**2*L\n",
+ "#Change in VOlume\n",
+ "dell_v=e_v*v\n",
+ "\n",
+ "#Result\n",
+ "print\"Bulk Modulus is\",round(E,2),\"N/mm**2\"\n",
+ "print\"Modulus of Rigidity is\",round(G,2),\"N/mm**2\"\n",
+ "print\"The change in Volume is\",round(dell_v,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Bulk Modulus is 200000.0 N/mm**2\n",
+ "Modulus of Rigidity is 76923.08 N/mm**2\n",
+ "The change in Volume is 1000.0 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.39,Page No.62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=500 #mm #Length of rectangular cross section bar\n",
+ "A=20*40 #mm**2 #Area of rectangular cross section bar\n",
+ "P1=4*10**4 #N #Tensile Force on 20mm*40mm Faces\n",
+ "P2=2*10**5 #N #compressive force on 20mm*500mm Faces\n",
+ "P3=3*10**5 #N #Tensile Force on 40mm*500mm Faces\n",
+ "E=2*10**5 #N/mm**2 #young's Modulus \n",
+ "mu=0.3 #Poisson's Ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_x,P_y,P_z be the forces n x,y,z directions\n",
+ "\n",
+ "P_x=P1*A**-1\n",
+ "P_y=P2*A**-1\n",
+ "P_z=P3*A**-1\n",
+ "\n",
+ "#Let e_x,e_y,e_z be the strains in x,y,z directions\n",
+ "e_x=1*E**-1*(50+mu*20-15*mu)\n",
+ "e_y=1*E**-1*(-mu*50-20-mu*15)\n",
+ "e_z=1*E**-1*(-mu*50+mu*20+15)\n",
+ "\n",
+ "#Volumetric strain\n",
+ "e_v=e_x+e_y+e_z\n",
+ "\n",
+ "#Volume\n",
+ "V=20*40*500 #mm**3\n",
+ "#Change in Volume \n",
+ "dell_v=e_v*V #mm**3\n",
+ "\n",
+ "#Result\n",
+ "print\"The change in Volume is\",round(dell_v,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The change in Volume is 36.0 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.41,Page No.65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=2.1*10**5 #N/mm**2 #Young's Modulus \n",
+ "G=0.78*10**5 #N/mm**2 #Modulus of Rigidity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Now using the relation\n",
+ "#E=2*G*(1+mu)\n",
+ "mu=E*(2*G)**-1-1 #Poisson's ratio\n",
+ "\n",
+ "#Bulk Modulus \n",
+ "K=E*(3*(1-2*mu))**-1 #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"The Poisson's Ratio is\",round(mu,2)\n",
+ "print\"The modulus of Rigidity\",round(K,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Poisson's Ratio is 0.35\n",
+ "The modulus of Rigidity 227500.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.42,Page No.65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "G=0.4*10**5 #N/mm**2 #Modulus of rigidity\n",
+ "K=0.75*10**5 #N/mm**2 #Bulk Modulus \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Young's Modulus\n",
+ "E=9*G*K*(3*K+G)**-1\n",
+ "\n",
+ "#Now from the relation\n",
+ "#E=2*G(1+2*mu)\n",
+ "mu=E*(2*G)**-1-1 #POissoin's ratio \n",
+ "\n",
+ "#result\n",
+ "print\"Young's modulus is\",round(E,2),\"N/mm**2\"\n",
+ "print\"Poissoin's ratio is\",round(mu,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Young's modulus is 101886.79 N/mm**2\n",
+ "Poissoin's ratio is 0.27\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.43,Page No.65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b=60 #mm #width of bar\n",
+ "d=30 #mm #depth of bar\n",
+ "L=200 #mm #Length of bar\n",
+ "A=30*60 #mm**2 #Area of bar\n",
+ "A2=30*200 #mm**2 #Area of bar along which expansion is restrained\n",
+ "P=180*10**3 #N #Compressive force\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#The bar is restrained from expanding in Y direction\n",
+ "P_z=0\n",
+ "P_x=P*A**-1 #stress developed in x direction\n",
+ "\n",
+ "#Now taking compressive strain as positive\n",
+ "#e_x=P_x*E**-1-mu*P_y*E**-1 .......................(1)\n",
+ "#e_y=-mu*P_x*E**-1+P_y*E**-1 ....................(2)\n",
+ "#e_z=-mu*P_x*E**-1-mu*P_y*E**-1 ......................(3)\n",
+ "\n",
+ "#Part-1\n",
+ "#When it is fully restrained\n",
+ "e_y=0\n",
+ "P_y=30 #N/mm**2 \n",
+ "e_x=P_x*E**-1-mu*P_y*E**-1\n",
+ "e_z=-mu*P_x*E**-1-mu*P_y*E**-1\n",
+ "\n",
+ "#Change in Length \n",
+ "dell_l=e_x*L #mm\n",
+ "\n",
+ "#Change in width\n",
+ "dell_b=b*e_y\n",
+ "\n",
+ "#change in Depth\n",
+ "dell_d=d*e_z\n",
+ "\n",
+ "#Volume of bar\n",
+ "V=b*d*L #mm**3\n",
+ "#Change in Volume\n",
+ "e_v=(e_x+e_y+e_z)*V #mm**3\n",
+ "\n",
+ "#Part-2\n",
+ "#When 50% is restrained\n",
+ "\n",
+ "#Free strain in Y direction\n",
+ "e_y1=mu*P_x*E**-1\n",
+ "\n",
+ "#As 50% is restrained,so\n",
+ "e_y2=-50*100**-1*e_y1\n",
+ "\n",
+ "#But form Equation 2 we have e_y=-mu*P_x*E**-1+P_y*E**-1 \n",
+ "#After substituting values in above equation and furthe simplifying we get\n",
+ "P_y=e_y2*E+d\n",
+ "\n",
+ "e_x2=P_x*E**-1-mu*P_y*E**-1 \n",
+ "e_z2=-mu*P_x*E**-1-mu*P_y*E**-1\n",
+ "\n",
+ "#Change in Length \n",
+ "dell_l2=e_x2*L #mm\n",
+ "\n",
+ "#Change in width\n",
+ "dell_b2=b*e_y2\n",
+ "\n",
+ "#change in Depth\n",
+ "dell_d2=d*e_z2\n",
+ "\n",
+ "#Change in Volume\n",
+ "e_v2=(e_x2+e_y2+e_z2)*V #mm**3\n",
+ "\n",
+ "#REsult\n",
+ "print\"Change in Dimension of bar is:dell_l\",round(dell_l,2),\"mm\"\n",
+ "print\" :dell_b\",round(dell_b,4),\"mm\"\n",
+ "print\" :dell_d\",round(dell_d,2),\"mm\"\n",
+ "print\"Change in Volume is\",round(e_v,2),\"mm**3\"\n",
+ "print\"Changes in material when only 50% of expansion can be reatrained:dell_l2\",round(dell_l2,2),\"mm\"\n",
+ "print\" :dell_b2\",round(dell_b2,4),\"mm\"\n",
+ "print\" :dell_d2\",round(dell_d2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Change in Dimension of bar is:dell_l 0.09 mm\n",
+ " :dell_b 0.0 mm\n",
+ " :dell_d -0.01 mm\n",
+ "Change in Volume is 93.6 mm**3\n",
+ "Changes in material when only 50% of expansion can be reatrained:dell_l2 0.1 mm\n",
+ " :dell_b2 -0.0045 mm\n",
+ " :dell_d2 -0.01 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.44,Page No.72"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=10*10**3 #N #Load\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus\n",
+ "d2=12 #mm #Diameter of bar1\n",
+ "d1=16 #mm #diameter of bar2\n",
+ "L1=200 #mm #Length of bar1\n",
+ "L2=500 #mm #Length of bar2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let A1 and A2 be the cross Area of Bar1 & bar2 respectively\n",
+ "A1=pi*4**-1*d1**2 #mm**2\n",
+ "A2=pi*4**-1*d2**2 #mm**2\n",
+ "\n",
+ "#Let p1 and p2 be the stress in Bar1 nad bar2 respectively\n",
+ "p1=P*A1**-1 #N/mm**2\n",
+ "p2=P*A2**-1 #N/mm**2\n",
+ "\n",
+ "#Let V1 nad V2 be the Volume of of Bar1 and Bar2\n",
+ "V1=A1*(L1+L1)\n",
+ "V2=A2*L2\n",
+ "\n",
+ "#Let E be the strain Energy stored in the bar\n",
+ "E=p1**2*(2*E)**-1*V1+p2**2*V2*(2*E)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"The Strain Energy stored in Bar is\",round(E,2),\"N-mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Strain Energy stored in Bar is 1602.6 N-mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.45,Page No.73"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Bar-A\n",
+ "d1=30 #mm #Diameter of bar1\n",
+ "L=600 #mm #length of bar1\n",
+ "\n",
+ "#Bar-B\n",
+ "d2=30 #mm #Diameter of bar2\n",
+ "d3=20 #mm #Diameter of bar2\n",
+ "L2=600 #mm #length of bar2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Area of bar-A\n",
+ "A1=pi*4**-1*d1**2\n",
+ "\n",
+ "#Area of bar-B\n",
+ "A2=pi*4**-1*d2**2\n",
+ "A3=pi*4**-1*d3**2\n",
+ "\n",
+ "#let SE be the Strain Energy\n",
+ "#Strain Energy stored in Bar-A\n",
+ "#SE=p**2*(2*E)**-1*V\n",
+ "#After substituting values and simolifying further we get\n",
+ "#SE=P**2*E**-1*0.4244\n",
+ "\n",
+ "#Strain Energy stored in Bar-B\n",
+ "#SE2=p1**2*V1*(2*E)**-1+p2**2*V2*(2*E)**-1\n",
+ "#After substituting values and simolifying further we get\n",
+ "#SE2=0.6897*P**2*E**-1\n",
+ "\n",
+ "#Let X be the ratio of SE in Bar-B and SE in Bar-A\n",
+ "X=0.6897*0.4244**-1\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#When Max stress is produced is same:Let p be the max stress produced\n",
+ "\n",
+ "#Stress in bar A is p throughout \n",
+ "#In bar B:stress in 20mm dia.portion=p2=p\n",
+ "\n",
+ "#Stress in 30 mm dia.portion\n",
+ "#p1=P*A2*A3**-1\n",
+ "#After substituting values and simolifying further we get\n",
+ "#p1=4*9**-1*p\n",
+ "\n",
+ "#Strain Energy in bar A\n",
+ "#SE_1=p**2*(2*E)**-1*A1*L1\n",
+ "#After substituting values and simolifying further we get\n",
+ "#SE_1=67500*p**2*pi*E**-1\n",
+ "\n",
+ "#Strain Energy in bar B\n",
+ "#SE_2=p1**2*V1*(2*E)**-1+p2**2*V2*(2*E)**-1\n",
+ "#After substituting values and simolifying further we get\n",
+ "#SE_2=21666.67*pi*p**2*E**-1\n",
+ "\n",
+ "#Let Y be the Ratio of SE in bar B and SE in bar A\n",
+ "Y=21666.67*67500**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Gradually applied Load is\",round(X,2)\n",
+ "print\"Gradually applied Load is\",round(Y,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Gradually applied Load is 1.63\n",
+ "Gradually applied Load is 0.32\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.46,Page No.74"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables \n",
+ "\n",
+ "W=100 #N #Load\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus \n",
+ "h=60 #mm #Height through Load falls down\n",
+ "L=400 #mm #Length of collar\n",
+ "d=30 #mm #diameter of bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A=pi*4**-1*d**2 #mm**2 #Area of bar\n",
+ "\n",
+ "#Instantaneous stress produced is\n",
+ "p=W*A**-1*(1+(1+(2*A*E*h*(W*L)**-1))**0.5)\n",
+ "\n",
+ "#Now the EXtension of the bar is neglected in calculating work doneby the Load,then\n",
+ "P=(2*E*h*W*(A*L)**-1)**0.5\n",
+ "\n",
+ "#Let percentage error be denoted by E1\n",
+ "#Percentage error in approximating is\n",
+ "E1=(p-P)*p**-1*100\n",
+ "\n",
+ "#Instantaneous Extension produced is\n",
+ "dell_l=round(P,3)*E**-1*L\n",
+ "\n",
+ "#Result\n",
+ "print\"The Instantaneous stress is\",round(p,2),\"N/mm\"\n",
+ "print\"Percentage Error is\",round(E1,2)\n",
+ "print\"The Instantaneous extension is\",round(dell_l,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Instantaneous stress is 92.27 N/mm\n",
+ "Percentage Error is 0.15\n",
+ "The Instantaneous extension is 0.18 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.47,Page No.75"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=20 #mm #Diameter of steel bar\n",
+ "L=1000 #mm #Length of bar\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus \n",
+ "p=300 #N/mm**2 #max Permissible stress\n",
+ "h=50 #mm #Height through which weight will fall\n",
+ "w=600 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#ARea of steel bar\n",
+ "A=pi*4**-1*d**2\n",
+ "\n",
+ "#Instantaneous extension is\n",
+ "dell_l=p*L*E**-1 #mm \n",
+ "\n",
+ "#Work done by Load \n",
+ "#W=W1*(h+dell_l)\n",
+ "\n",
+ "#Volume of bar\n",
+ "V=round(A,2)*L\n",
+ "#Let E1 be the strain Energy\n",
+ "E1=p**2*(2*E)**-1*V\n",
+ "\n",
+ "#Answer in Book for Strain Energy is Incorrect \n",
+ "\n",
+ "#Now Equating Workdone by Load to strain Energy \n",
+ "W1=E1*51.5**-1\n",
+ "\n",
+ "#Now when w=600 N\n",
+ "#Let W2 be the Work done by the Load\n",
+ "#W2=w(h2*dell_l)\n",
+ "\n",
+ "h=E1*w**-1-dell_l\n",
+ "\n",
+ "#Result\n",
+ "print\"The Max Lodad which can Fall from a height of 50 mm on the collar is\",round(W1,2),\"N\"\n",
+ "print\"the Max Height from which a 600 N Load can fall on the collar is\",round(h,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Max Lodad which can Fall from a height of 50 mm on the collar is 1372.54 N\n",
+ "the Max Height from which a 600 N Load can fall on the collar is 116.31 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.48,Page No.76"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D_s=30 #mm #Diameter of steel rod\n",
+ "d=30 #mm #Internal Diameter of copper tube\n",
+ "D=40#mm #External Diameter of copper tube\n",
+ "E_s=2*10**5 #N/mm**2 #Young's Modulus of Steel rod\n",
+ "E_c=1*10**5#N/mm**2 #Young's Modulus of copper tube\n",
+ "P=100 #N #Load\n",
+ "h=40 #mm #height from which Load falls\n",
+ "L=800 #mm #Length \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Area of steel rod\n",
+ "A_s=pi*4**-1*D_s**2\n",
+ "\n",
+ "#Area of copper tube\n",
+ "A_c=pi*4**-1*(D**2-d**2)\n",
+ "\n",
+ "#But Dell_s=dell_c=dell\n",
+ "#p_s*E_s**-1*L=p_c*L*E_c\n",
+ "#After simplifying furthe we get\n",
+ "#p_s=2*p_c\n",
+ "\n",
+ "#Now Equating internal Energy to Workdone we get\n",
+ "p_c=(2*P*h*L**-1*(4*A_s*E_s**-1+A_c*E_c**-1))**0.5\n",
+ "p_s=2*p_c\n",
+ "\n",
+ "#Result\n",
+ "print A_s\n",
+ "print\"STress produced in steel is\",round(p_s,2),\"N/mm**2\"\n",
+ "print\"STress produced in copper is\",round(p_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "706.858347058\n",
+ "STress produced in steel is 0.89 N/mm**2\n",
+ "STress produced in copper is 0.44 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.49,Page No.77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "dell=0.25 #mm #Instantaneous Extension\n",
+ "\n",
+ "#Bar-A\n",
+ "b1=25 #mm #width of bar\n",
+ "D1=500 #mm #Depth of bar\n",
+ "\n",
+ "#Bar-B\n",
+ "b2_1=25 #mm #width of upper bar\n",
+ "b2_2=15 #mm #Width of Lower Bar\n",
+ "L2=200 #mm #Length of upper bar\n",
+ "L1=300 #mm #Length of Lower bar\n",
+ "\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus of bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Strain\n",
+ "e=dell*D1**-1 \n",
+ "\n",
+ "#Load\n",
+ "p=e*E\n",
+ "\n",
+ "#Area of bar-A\n",
+ "A=pi*4**-1*25**2\n",
+ "\n",
+ "#Volume of bar-A\n",
+ "V=A*D1\n",
+ "\n",
+ "#Let E1 be the Energy of Blow\n",
+ "#Energy of Blow\n",
+ "E1=p**2*(E)**-1*V\n",
+ "\n",
+ "#Let p2 be the Max stress in bar B When this blow is applied.\n",
+ "#the max stress occurs in the 15mm dia. portion,Hence, the stress in 25 mm dia.portion is\n",
+ "#p2*pi*4**-1*b2_2**2*(pi*4**-1*b2_2**2=0.36*p\n",
+ "\n",
+ "#Strain Energy of bar B\n",
+ "#E2=p**2*(2*E)**-1*v1+1*(2*E)**-1*(0.36*p2)**2*v2\n",
+ "#After substituting values and Further substituting values we get\n",
+ "#E2=0.1643445*p2**2\n",
+ "\n",
+ "#Equating it to Energy of applied blow,we get\n",
+ "p2=(12271.846*0.1643445**-1)**0.5\n",
+ "\n",
+ "#Stress in top portion\n",
+ "sigma=0.36*p2\n",
+ "\n",
+ "#Extension in Bar-1\n",
+ "dell_1=p2*E**-1*L1\n",
+ "\n",
+ "#Extension in Bar-2\n",
+ "dell_2=0.36*p2*E**-1*L2\n",
+ "\n",
+ "#Extension of bar\n",
+ "dell_3=dell_1+dell_2\n",
+ "\n",
+ "#Result\n",
+ "print\"Instantaneous Max stress is\",round(sigma,2),\"N/mm**2\"\n",
+ "print\"extension in Bar is\",round(dell_3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Instantaneous Max stress is 98.37 N/mm**2\n",
+ "extension in Bar is 0.51 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.3_3.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.3_3.ipynb new file mode 100644 index 00000000..f136c4ad --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.3_3.ipynb @@ -0,0 +1,1588 @@ +{
+ "metadata": {
+ "name": "chapter no.3.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3:Shear Force And Bending Moment Diagrams in Statically Determinate Beams"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.1,Page No.100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_AC=L_CD=1 #m #Length of AC & CD\n",
+ "L_DB=1.5 #m #Lengh of DB\n",
+ "L=3.5 #m #Length of Beam\n",
+ "F_B=10 #KN #Force at pt B\n",
+ "F_C=F_D=20 #KN #Force at pt C & D\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "R_A=F_C+F_D+F_B #KN #Force at support A \n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At pt B\n",
+ "V_B1=0 #KN \n",
+ "V_B2=F_B #KN\n",
+ "\n",
+ "#S.F At pt D\n",
+ "V_D1=V_B2 #KN\n",
+ "V_D2=V_D1+F_D #KN\n",
+ "\n",
+ "#S.F At pt C \n",
+ "V_C1=V_D2 #KN\n",
+ "V_C2=V_D2+F_C #KN\n",
+ "\n",
+ "#S.F At Pt A\n",
+ "V_A1=V_C2 #KN\n",
+ "V_A2=V_C2-R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=0 #KN.m\n",
+ "\n",
+ "#B.M AT Pt D\n",
+ "M_D=F_B*L_DB #KN.m\n",
+ "\n",
+ "#B.M At pt C\n",
+ "M_C=F_B*(L_DB+L_CD)+F_D*L_CD #KN.m\n",
+ "\n",
+ "#B.M At pt A\n",
+ "M_A=F_B*L+F_D*(L_CD+L_AC)+F_C*L_AC\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_DB,L_DB,L_CD+L_DB,L_CD+L_DB,L_CD+L_DB+L_AC,L_CD+L_DB+L_AC]\n",
+ "Y1=[V_B1,V_B2,V_D1,V_D2,V_C1,V_C2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "Y2=[M_B,M_D,M_C,M_A]\n",
+ "X2=[0,L_DB,L_DB+L_CD,L_AC+L_CD+L_DB]\n",
+ "Z2=[0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5dde2d0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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pqQavCQgIEADwi1/84he/rPgKCAiw6e+yYlJMDQ0NuPfee/H999+jZ8+eGDhw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\no9ggiIjIKDYIIiIy6v8BsxKV3Pt7cnUAAAAASUVORK5CYII=\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5df0ab0>"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.2,Page No.101"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "w1=10 #KN/m #u.d.L\n",
+ "F_D=20 #KN #Force at pt D\n",
+ "F_C=30 #KN #Force at pt C\n",
+ "L_DB=4 #m #Length of DB\n",
+ "L_CD=L_AC=2 #m #Length of AC & CD\n",
+ "L=8 #m #Length of Beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A And R_B be the Reactions at pt A and B \n",
+ "#R_A+R_B=90 \n",
+ "#Now Taking moment at A,M_A we get\n",
+ "R_A=(w1*L_DB*(L_DB*2**-1)+F_D*L_DB+F_C*(L_CD+L_DB))*L**-1\n",
+ "R_B=90-R_A\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At Pt B\n",
+ "V_B1=0 #KN\n",
+ "V_B2=R_B #KN\n",
+ "\n",
+ "#S.F At pt D\n",
+ "V_D1=R_B-w1*L_DB #KN\n",
+ "V_D2=V_D1-F_D #KN\n",
+ "\n",
+ "#S.F at Pt C\n",
+ "V_C1=V_D2 #KN\n",
+ "V_C2=V_C1-F_C \n",
+ "\n",
+ "#S.F at PT A\n",
+ "V_A1=V_C2 #KN\n",
+ "V_A2=V_C2+R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=0 #KN.m\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D=-R_B*L_DB+w1*L_DB*L_DB*2**-1 #KN.m\n",
+ "\n",
+ "#B.M At PT C\n",
+ "M_C=-R_B*(L_DB+L_CD)+w1*L_DB*(L_DB*2**-1+L_CD)+F_D*L_CD #KN.m\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=-R_B*L+w1*L_DB*(L_DB*2**-1+L_CD+L_AC)+F_D*(L_CD+L_AC)+F_C*L_AC\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_DB,L_DB,L_CD+L_DB,L_CD+L_DB,L_CD+L_DB+L_AC,L_CD+L_DB+L_AC]\n",
+ "Y1=[V_B1,V_B2,V_D1,V_D2,V_C1,V_C2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "Y2=[M_B,M_D,M_C,M_A]\n",
+ "X2=[0,L_DB,L_DB+L_CD,L_AC+L_CD+L_DB]\n",
+ "Z2=[0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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Pni0aGhr0vo4HrxERkcSmlo+IiMi8WApERCRhKRARkYSlQEREEpYCERFJWApE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\nICIiCUuBiIgk/w9P4ODmR+1IBgAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5be93d0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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Gt27dTE9mRiwY9ovtQ+zHlSuAv780HCkgQO00tkGRgnHr1i2kp6ejpKQEdXV1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\nx6hRo1BfX48XX3wRgYGBasd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+ "text": [
+ "<matplotlib.figure.Figure at 0x5de54f0>"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.3,Page No.102"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_DB=L_CD=1.5 #m #Length of DB & CD\n",
+ "L_AC=3 #m #Length of AC\n",
+ "F_D=80 #KN #Force at Pt D\n",
+ "w=40 #KN/m #u.v.l\n",
+ "L=6 #Length of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A and R_B be the Reactions at Pt A & B respectively\n",
+ "#R_A+R_B=140 \n",
+ "#Taking moment at B we get,M_B\n",
+ "R_A=(1*2**-1*L_AC*w*(1*3**-1*L_AC+(L_CD+L_DB))+F_D*L_DB)*L**-1\n",
+ "R_B=140-R_A\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F at B\n",
+ "V_B1=0 #KN\n",
+ "V_B2=R_B #KN\n",
+ "\n",
+ "#S.F At D\n",
+ "V_D1=V_B2 #KN\n",
+ "V_D2=V_D1-F_D #KN\n",
+ "\n",
+ "#S.F at C\n",
+ "V_C=V_D2 #KN\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A1=V_C-1*2**-1*w*L_AC #KN\n",
+ "V_A2=V_A1+R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=0 #KN.m\n",
+ "\n",
+ "#B.M At D\n",
+ "M_D=-R_B*L_DB\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=F_D*L_CD-R_B*(L_DB+L_CD)\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A=F_D*(L_CD+L_AC)-R_B*L+1*2**-1*w*L_AC*(1*3**-1*L_AC)+R_A\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_DB,L_DB,L_DB+L_CD,L_DB+L_CD+L_AC,L_DB+L_CD+L_AC]\n",
+ "Y1=[V_B1,V_B2,V_D1,V_D2,V_C,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_DB,L_CD+L_DB,L_AC+L_CD+L_DB]\n",
+ "Y2=[M_B,M_D,M_C,M_A]\n",
+ "Z2=[0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x58f0c10>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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CBUEIDRWEv/9d7kpIbVz53SnLDGPjxo0YOHAg9u3bh/T0dKSlpQEAjEYjsrKy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\ngg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x562f910>"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.4,Page No.104"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "M_D=120 #KN.m #B.M at Pt D\n",
+ "F_C=40 #KN #Force at Pt C\n",
+ "w1=20 #KN.m\n",
+ "L_DB=1.5 #m #Length of DB\n",
+ "L_CD=1.5 #m #Length of CD\n",
+ "L_AC=3 #m #Length of AC\n",
+ "L=6 #m #Length of Beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A And R_B be the Reactions at pt A and B \n",
+ "#R_A+R_B=100\n",
+ "#Now Taking Moment At Pt B We get,M_B\n",
+ "R_A=-(M_D-F_C*(L_CD+L_DB)-w1*L_AC*(L_AC*2**-1+L_CD+L_DB))*L**-1\n",
+ "R_B=100-R_A\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At Pt B\n",
+ "V_B1=0\n",
+ "V_B2=R_B\n",
+ "\n",
+ "#S.F at Pt D\n",
+ "V_D=V_B2 #KN\n",
+ "\n",
+ "#S.F At Pt C\n",
+ "V_C1=V_D #KN\n",
+ "V_C2=V_C1-F_C\n",
+ "\n",
+ "#S.F At Pt A\n",
+ "V_A1=V_C2-w1*L_AC #KN\n",
+ "V_A2=V_A1+R_A\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=0 #KN.m\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D1=M_B-R_B*L_DB #KN.m\n",
+ "M_D2=M_B+M_D-R_B*L_DB\n",
+ "\n",
+ "#B.M At Pt C\n",
+ "M_C=M_D-R_B*(L_CD+L_DB)\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=M_D-R_B*L+F_C*L_AC+w1*L_AC*L_AC*2**-1\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_DB,L_DB+L_CD,L_DB+L_CD,L_DB+L_CD+L_AC,L_DB+L_CD+L_AC]\n",
+ "Y1=[V_B1,V_B2,V_D,V_C1,V_C2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "Y2=[M_B,M_D1,M_D2,M_C,M_A]\n",
+ "X2=[0,L_DB,L_DB,L_CD+L_DB,L_AC+L_CD+L_DB]\n",
+ "Z2=[0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x55dcbf0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x55ca4b0>"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.5,Page No.105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F_C=20 #KN #Force at Pt C\n",
+ "F_D=40 #KN #Force at pt D\n",
+ "w=20 #KN.m #u.d.l \n",
+ "L_AD=L_DB=2 #m #Length of AD & DB\n",
+ "L_BC=1 #m #Length of BC\n",
+ "L=5 #m #Length of Beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt R_A and R_B be the reactions at A & B respectively\n",
+ "#R_A+R_B=100 \n",
+ "#Now Taking Moment at B,M_B we get\n",
+ "R_A=-(F_C*L_BC-F_D*L_DB-w*L_AD*(L_AD*2**-1+L_DB))*(L_AD+L_DB)**-1\n",
+ "R_B=100-R_A\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At pt C\n",
+ "V_C1=0 #KN\n",
+ "V_C2=-F_C #KN\n",
+ "\n",
+ "#S.F At PT B\n",
+ "V_B1=V_C2 #KN\n",
+ "V_B2=V_C2+R_B #KN\n",
+ "\n",
+ "#S.F At Pt D\n",
+ "V_D1=V_B2 #KN\n",
+ "V_D2=V_D1-F_D #KN\n",
+ "\n",
+ "#S.F At Pt A\n",
+ "V_A1=V_D2-w*L_AD #KN\n",
+ "V_A2=V_A1+R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt C\n",
+ "M_C=0 \n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=F_C*L_BC\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D=F_C*(L_BC+L_DB)-R_B*L_DB\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=F_C*L-R_B*(L_DB+L_AD)+F_D*L_AD+w*L_AD*L_AD*2**-1\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_BC,L_BC,L_BC+L_DB,L_BC+L_DB,L_BC+L_DB+L_AD,L_BC+L_DB+L_AD]\n",
+ "Y1=[V_C1,V_C2,V_B1,V_B2,V_D1,V_D2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "Y2=[M_C,M_B,M_D,M_A]\n",
+ "X2=[0,L_BC,L_BC+L_DB,L_BC+L_DB+L_AD]\n",
+ "Z2=[0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5d9c6f0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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Zb7/9lh07drB8+XLy8/N58cUXmTVrlrt+NSKaPpLAUlFRwZdffsnI\nkSOd91VWVgLGm/WZE1a7du3K4cOHAfj8889JSUkBcPYoONvw4cMB6NGjBx9++KHzfldOkKntmufr\n0KED3bp1A6Bbt24kJiYCEB0dTUlJSZ3XEXGVkoIElNOnT3PFFVdQWFhY48+bNm3q/PrMm/r5dYPz\n3+wvueQSAJo0aUJVVVW9Y6rpmuc7cw0w+iWceU5QUFCDrilSG00fSUC5/PLL6dChAx988AFgvAn/\n61//uuhz+vTpw4oVK3A4HBw+fJi8vLw6r9O8eXPnUcXn0xmUYmVKCuLXTpw4Qbt27Zy3//3f/+Xd\nd99l0aJFxMbGEh0dfU4z97Pn+898PWLECMLCwoiKiuK+++6jR48etGjR4oJr2Ww253OGDh3KypUr\niYuLIz8/v9bH1XbNml67tu8D+UhocT8dnS3igp9//pnLLruM//znPyQkJPDFF1/QqlUrs8MScTvV\nFERcMGTIEMrLy6msrGTatGlKCOK3NFIQEREn1RRERMRJSUFERJyUFERExElJQUREnJQURETESUlB\nRESc/j+iaB8fTwtkoQAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5c1ead0>"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.6,Page No.107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_BC=L_EB=L_AD=1 #m #Length of spans BC,ED,AD\n",
+ "L_ED=2 #m #Length of ED\n",
+ "w=60 #KNm #u.d.l\n",
+ "F_C=20 #KN Pt Load at C\n",
+ "L=5 #m #Span of beam \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A & R_B be the reactions at A & B respectively\n",
+ "#R_A+R_B=80 \n",
+ "#Taking Moment At A,we get M_A\n",
+ "R_B=(F_C*L+1*2**-1*L_ED*w*(2*3**-1*L_ED+L_AD))*(L_AD+L_ED+L_EB)**-1\n",
+ "R_A=80-R_B\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C1=0 #KN\n",
+ "V_C2=-F_C #KN\n",
+ "\n",
+ "#S.F At B\n",
+ "V_B1=V_C2 #KN\n",
+ "V_B2=V_C2+R_B #KN \n",
+ "\n",
+ "#S.F aT E\n",
+ "V_E=V_B2 #KN\n",
+ "\n",
+ "#S.F AT D\n",
+ "V_D=V_B2-1*2**-1*L_ED*w #KN\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A1=V_D #KN \n",
+ "V_A2=V_D+R_A\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M at C\n",
+ "M_C=0 #KN.m\n",
+ "\n",
+ "#B.M at B\n",
+ "M_B=F_C*L_BC #KN.m\n",
+ "\n",
+ "#B.M at E\n",
+ "M_E=F_C*(L_EB+L_BC)-R_B*L_EB #KN.m\n",
+ "\n",
+ "#B.M at D\n",
+ "M_D=F_C*(L_ED+L_EB+L_BC)-R_B*(L_ED+L_EB)+1*2**-1*L_ED*w*1*3**-1*L_ED #KN.m\n",
+ "\n",
+ "#B.M at A\n",
+ "M_A=1*2**-1*L_ED*w*(2*3**-1*L_ED+L_AD)-R_B*(L_AD+L_ED+L_EB)+F_C*L\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_BC,L_BC,L_EB+L_BC,L_ED+L_EB+L_BC,L_AD+L_ED+L_EB+L_BC,L_ED+L_EB+L_BC+L_AD]\n",
+ "Y1=[V_C1,V_C2,V_B1,V_B2,V_E,V_D,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_BC,L_BC+L_EB,L_EB+L_BC+L_ED,L_EB+L_BC+L_ED+L_AD]\n",
+ "Y2=[M_C,M_B,M_E,M_D,M_A]\n",
+ "Z2=[0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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Cr9fzLAHAgQMHUFpaig8++AAbNmzAvn37Ot1OUUUhJCSkwx1Vq6qqEBoaKjAR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\nAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x58a6f70>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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UlD0DZoHZoDk7OxvPPfec/SqwAoNm98HQmcjAloBZINmcwq5du5rcTwEAHnnkEeuqkgCb\ngnvhPZ2JLLsHszmSNIWHH34YP//8M2JiYuDp6SluX7ZsmfWV2YhNwb1w0pnIuglmY5I0hfDwcJSW\nltp8Yx0psSm4H046kzuzdoLZmCRzClFRUTh37pz1VRBJgJPO5M4cETALzB4paLVaHDx4EP369UO7\ndu0ML1KpkJeXZ//qTOCRgnti6EzuSIqAWSDJ6SPhBg03v5lKpcIDDzxgW3U2YFNwXwydyd1IETAL\nJLv6SKfT4dSpU4iLi0NtbS3q6+vRsWNH2yu0EpuC+2LoTO5GioBZIEmm8P7772PChAl46qmnABju\nhjZmzBjbqyOyws2Tzvy7gFydPe7BbI7ZpvDuu+9i586d4pFBz549ceHCBbsXRmQKQ2dyF44MmAVm\nm0K7du3EgBkA6uvrFXV5Krkf4Z7O//gHUF0tdzVE9iHcg9mei9+1xGxTeOCBB7BgwQLU1tZi27Zt\nmDBhApKSkhxRG5FJXF6bXJ29l8g2xWzQ3NDQgFWrVmHr1q0AgMTEREybNk3WowUGzQQwdCbXJmXA\nLFDsPZr/8Y9/4Msvv4SPjw9CQ0OxevVqdOrUCQCQkZGBnJwceHp6Ijs7GwkJCc2LZlOg/+KkM7ki\nqSaYjUly9dGmTZug0Whw2223wc/PD35+fjZfjpqQkIAjR47gxx9/RM+ePZGRkQEAKC0txbp161Ba\nWor8/HzMmDEDjY2NNu2LXBtDZ3JFcgTMArNNYebMmfjwww/x22+/obq6GtXV1bhy5YpNO42Pj4eH\nh2HXsbGxOHv2LAAgNzcXKSkp8Pb2RnBwMMLCwlBcXGzTvsi1MXQmVyNXwCww2xTUajUiIyPFD3Gp\n5eTkYNiwYQCAiooKqNXqJvvmXd7IHCF0fvVVuSshsp1cAbPA5J3XBFlZWRg6dCgGDRoEHx8fAIbz\nUrNmzWr1dfHx8aisrGy2feHCheLVSwsWLICPjw9SU1NNvo+pQDs9PV38XqvVQqvVmvk/IVeWlQX0\n72843H79dcBOf8MQ2Z2U92AuLCwUlyqylNmgOT4+Hn5+foiOjm5ytPCqjX+WrVmzBitXrsS3336L\n9u3bAwAyMzMBAGlpaQCAIUOGYP78+YiNjW1aNINmasHFi8C4cYC/P/DRR4Cvr9wVEbWNvQJmgSRX\nH0VFReHw4cOSFpafn48XXngBO3bsgL+/v7i9tLQUqampKC4uRnl5OeLi4nDq1KlmRwtsCmRKXZ0h\nfN6/H8jLA4KC5K6IyHL//CdQUwO8/bZ93l+Sq4+GDRuGr7/+WrKiAODZZ59FTU0N4uPjodFoMGPG\nDABAREQEkpOTERERgaFDh2L58uWcnqY28fEBVq0yXNvdvz+we7fcFRFZRu6AWWD2SMHX1xe1tbXw\n8fGBt7e34UUqlc1XINmCRwpkic2bDUttv/UW8NBDcldD1Dopl8g2RbHDa7ZiUyBLHTkCJCUBKSkM\noEnZ7DHBbEyyppCbm4vvvvtOvLmO3GsfsSlQWzCAJqWzd8AskCRTSEtLQ3Z2NiIjIxEeHo7s7GzM\nmTNHsiKJ7O322w23Mrz1VmDgQODXX+WuiKgpOSeYjZk9UoiOjsbBgwfh6ekJwLBAXkxMDA4dOuSQ\nAlvCIwWyhl4PLFliOG+7fj1wzz1yV0Qk7T2YzZHkSEGlUqGqqkp8XFVVxSuCyCmpVMALLwArVwKj\nRgGffip3RUTyTzAbMzvRPGfOHPTu3VucGN6xY4c4ZEbkjIYPNyy3nZQElJYygCZ5STnBLAWLguaK\nigqUlJRApVKhX79+6Nq1qyNqM4mnj0gKDKBJbo4KmAU2XX20f//+Jo+Fpwmnjnr37i1FjVZhUyCp\ncAKa5GTvCWZjNjUFDw8PREVFoUuXLi2+sKCgwPYKrcSmQFJiAE1ycGTALLDks9NkprBkyRJ8/vnn\n6NChAyZOnIgxY8bAz89P8iKJ5CYE0L16GQJoTkCTIygtYBaYzRROnz6NdevWYePGjbjjjjvwyiuv\nICYmxlH1tYhHCmQvnIAmR3HEBLMxSS5JDQ0NxahRo5CQkICSkhIcP35csgKJlCYyEtizx7D+zPjx\nhvO9RFLT6YCSEsO/MaUxeaRw+vRprF27Frm5uQgKCsLEiRMxYsQI/EUBI3c8UiB7YwBN9uTogFlg\nc9AcHR2N0aNHo2PHjk3e0JI7r9kTmwI5AgNosgc5AmaBTUHzvHnzxMtPa3gMTW6IATTZg1IDZgGX\nziayAANokoocAbOA91MgkhAnoMlWjp5gNibJ1UdEZMAluMlWSloi2xQ2BaI24D2gyVpKuQezOWZX\nSV28eHGTQw6VSoVOnTqhT58+Vg+xzZ07F3l5eVCpVOjSpQvWrFmD7t27AwAyMjKQk5MDT09PZGdn\nIyEhwap9ENkLA2iyhtIDZoHZTCE1NRV79+5FUlIS9Ho9Nm/ejOjoaPzyyy8YP348Zs+e3eadVldX\ni0tmLFu2DD/++CM++OADlJaWIjU1FSUlJSgvL0dcXBxOnDgBD6NUj5kCKQUDaLKUnAGzQJJMoays\nDPv378fixYuxZMkS7Nu3DxcuXMCOHTuwZs0aqwq7eQ2lmpoa+Pv7AzDcCzolJQXe3t4IDg5GWFgY\niouLrdoHkSNwAposoeQJZmNmm8LFixfh4+MjPvb29sb58+fRoUMHtG/f3uodv/LKKwgKCsKaNWvE\nez5XVFRArVaLz1Gr1SgvL7d6H0SOwACazHGGgFlgNlN46KGHEBsbi9GjR0Ov12PTpk1ITU3F1atX\nEdHKybH4+HhUVlY2275w4UIkJSVhwYIFWLBgATIzMzFz5kysXr26xfcxdevP9PR08XutViveGY5I\nDkIAvWSJIYDmBDQJhID5m28cv+/CwkIUFha26TUWzSmUlJSgqKgIKpUKAwYMQN++fa2tsZlff/0V\nw4YNw+HDh8XbfKalpQEAhgwZgvnz5yM2NrZp0cwUSME2bwamTGEATQYbNxqWSvn+e7krkXB4raGh\nAZWVlaivrxf/cg+yYYWwkydPokePHgAMQXNxcTE+/vhjMWguLi4Wg+ZTp041O1pgUyClYwBNAiUE\nzAJJmsKyZcswf/58BAQEwNPTU9x+6NAhqwsbP348jh8/Dk9PT4SGhuK9995DQEAAAMPppZycHHh5\neWHp0qVITExsXjSbAjkBTkCT3BPMxiRpCqGhoSguLjZ5W045sCmQs+AS3O5NriWyTZHkktSgoCBx\n6WwiahtOQLsvZ5lgNmb26qOQkBAMGjQIw4cPFy9Nlft+CkTOhBPQ7slZJpiNmW0KQUFBCAoKQl1d\nHerq6sSb7BBR2wwfDhQUGALo0lIG0K5uxQrnO0oAuHQ2kcMxgHZ9SguYBTYFzc8//zyWLl2KpKSk\nFt84Ly9PmiqtwKZAzo4BtGtTWsAssKkp7N27F3379jU5DSfnBDGbArkC3gPaNcl5D2ZzeOc1IifA\nCWjXoqQJZmOWfHaaDJqjo6NbfeOffvrJ+sqISMQA2rU4a8AsMHmkoNPpAADLly8HAEyePBl6vR6f\nfvopACArK8sxFbaARwrkihhAOz+lBswCSU4fxcTE4ODBg022aTQaHDhwwPYKrcSmQK6KAbRzU2rA\nLJBkolmv12Pnzp3i46KiIn4gE9kJJ6Cdl7NOMBszO7yWk5ODKVOm4I8//gAA3HrrrSbvfUBEtuME\ntHNy1glmYxZffSQ0hU6dOtm1IEvw9BG5Cy7B7TyUtES2KZJkCtevX8f69euh0+lQX18vvvG8efOk\nq7SN2BTInTCAVj6lB8wCSTKFUaNGIS8vD97e3vD19YWvry9uueUWyYokotbxHtDK50z3YDbH7JFC\nVFQUDh8+7Kh6LMIjBXJHnIBWJiVPMBuT5Ejh3nvv5aAakQIIAfTKlYYA+r8jQyQzVwmYBWaPFMLD\nw3Hq1CmEhISgXbt2hhfJPNHMIwVydwyglcMZAmaBJEGzMNlsLDg42Nq6bMamQMQAWgmcJWAWSHL6\nKDg4GGVlZSgoKEBwcDBuueUWyT6QFy9eDA8PD1y+fFnclpGRgR49eqBXr17YunWrJPshckUMoOXn\nSgGzwGxTSE9PxxtvvIGMjAwAQF1dHR5++GGbd1xWVoZt27bhjjvuELeVlpZi3bp1KC0tRX5+PmbM\nmIHGxkab90XkqjgBLR9XmWA2ZrYpbNiwAbm5ueJlqN26dUN1dbXNO541axbeeOONJttyc3ORkpIC\nb29vBAcHIywsDMXFxTbvi8iVMYCWh6sFzAKzTaFdu3bwuCnFunr1qs07zc3NhVqtxl133dVke0VF\nBdRqtfhYrVajvLzc5v0RuQNhCe65c4GXXwZ4kG1fzr5Etilm1z6aMGECnnrqKVRVVeH9999HTk4O\npk2bZvaN4+PjUVlZ2Wz7ggULkJGR0SQvaC2jUKlULW5PT08Xv9dqtbLeCY5IKSIjgT17DAH0uHHA\nxx8zgLYHnQ4oKQG++ELuSlpXWFho8u6Zpli09tHWrVvFD/HExETEx8dbVSAAHD58GIMHD0aHDh0A\nAGfPnkW3bt2wZ88ecaG9tLQ0AMCQIUMwf/58xMbGNi2aVx8RtYpLcNuX0pfINkXy23FevHgR/v7+\nJv96t0ZISAj27duHzp07o7S0FKmpqSguLkZ5eTni4uJw6tSpZvtjUyAyjxPQ9uFME8zGbLokdffu\n3dBqtRg7diwOHDiAqKgoREdHIzAwEFu2bJG0SEFERASSk5MRERGBoUOHYvny5ZI2ICJ3wgDaPlw1\nYBaYPFLo06cPMjIy8Mcff+CJJ55Afn4++vfvj2PHjmHSpEnN7sbmSDxSIGobYQJ60iTgf/+XE9C2\ncKYJZmM2nT66+Tac4eHhOHr0qPgz3o6TyPkIE9BdujCAtpazTTAbs+n00c2nbdq3by9dVUQkC2EC\n+rbbOAFtLVecYDZm8kjB09NTvELo2rVr+MtNv4Vr166JN9yRA48UiKzHANo6zhwwCyz57DQ5p9DQ\n0CB5QUQkP94D2jquHjALzA6vEZFrEiagk5IMQTQD6Na56gSzsTbNKSgFTx8RSYcBtHnOHjALJFk6\nm4hcGwNo89whYBawKRARl+BuhasukW0KmwIRAeAEtCnuEjALGDQTURMMoJtyl4BZwKCZiFrEANp1\nAmYBg2YishoDaPcKmAVsCkRkkjsH0O4WMAvYFIioVe4aQLtbwCxg0ExEFnG3ANrdAmYBg2YiahN3\nCKBdLWAWMGgmIskJAXTnzq4bQLtjwCxgUyCiNvPxMXxwPvKIYeltVwqg3TVgFrApEJFVVCpg1izg\n/fddK4B214BZIEtTSE9Ph1qthkajgUajwZYtW8SfZWRkoEePHujVqxe2bt0qR3lE1AZCAD13LvDy\ny0Bjo9wV2cZdA2aBLEHz/Pnz4efnh1mzZjXZXlpaitTUVJSUlKC8vBxxcXE4ceIEPIwucWDQTKQ8\nrhBAu2rALFB00NxSYbm5uUhJSYG3tzeCg4MRFhaG4uJiGaojorZyhQDanQNmgWxNYdmyZbj77rvx\n+OOPo6qqCgBQUVEBtVotPketVqO8vFyuEomojZw5gHb3gFlgt+G1+Ph4VFZWNtu+YMECPP3005g3\nbx4AYO7cuXjhhRewatWqFt9HpVK1uD09PV38XqvVQqvV2lwzEdlOCKDvvNO57gHtigFzYWEhCgsL\n2/Qa2YfXdDodkpKScOjQIWRmZgIA0tLSAABDhgzB/PnzERsb2+Q1zBSInMORI4YJ6EmTlD8BPXQo\nkJpqWOfJVSk2Uzh37pz4/YYNGxAdHQ0AGDlyJNauXYu6ujqcOXMGJ0+eRL9+/eQokYgkEBkJ7NkD\n7NxpCKFrauSuqGU6HVBSAowfL3cl8pNl7aPZs2fj4MGDUKlUCAkJwYoVKwAAERERSE5ORkREBLy8\nvLB8+XKTp4+IyDkIAfTTTxsC6Lw8IChI7qqaYsD8J9lPH1mDp4+InI9eb8gXFi8G/vMfQxCtBDdu\nAHfcYWhcrpQntESxp4+IyP0odQLaFQNmW3DpbCJyKKUtwe3uE8zGePqIiGShhAloV59gNsbTR0Sk\nWEqYgGbA3BybAhHJRs4JaE4wt4xNgYhkJVcAzYC5ZQyaiUgRHB1AM2BuGYNmIlIURwTQ7hYwCxg0\nE5HTcUSyPgBUAAAI10lEQVQAzYDZNDYFIlIcewbQDJhbx6ZARIpkrwCaAXPrGDQTkaJJHUAzYG4d\ng2YicgpSBNDuGjALGDQTkcuQIoBmwGwemwIROQ1bAmgGzJZhUyAip2JtAM2A2TIMmonIKbU1gGbA\nbBkGzUTk1CwJoN09YBYoOmhetmwZwsPDERUVhdmzZ4vbMzIy0KNHD/Tq1Qtbt26VqzwichKWBNAM\nmNtAL4Pt27fr4+Li9HV1dXq9Xq+/cOGCXq/X648cOaK/++679XV1dfozZ87oQ0ND9Q0NDc1eL1PZ\nilRQUCB3CYrB38Wf3PF30dio1y9erNf/7W96/a5df27ftq1A/9e/6vVHjshXm1JY8tkpy5HCe++9\nhzlz5sDb2xsAcPvttwMAcnNzkZKSAm9vbwQHByMsLAzFxcVylOg0CgsL5S5BMfi7+JM7/i5MBdAf\nfFDIgLkNZGkKJ0+exHfffYf+/ftDq9Vi7969AICKigqo1WrxeWq1GuXl5XKUSEROSgig584FXn4Z\n2LuXAXNb2O3qo/j4eFRWVjbbvmDBAtTX1+P333/HDz/8gJKSEiQnJ+Pnn39u8X1UKpW9SiQiFxUZ\nCezZYwigy8uB8ePlrsiJOOA0VjNDhgzRFxYWio9DQ0P1Fy9e1GdkZOgzMjLE7YmJifoffvih2etD\nQ0P1APjFL37xi19t+AoNDTX7+SzLnMLo0aOxfft2PPDAAzhx4gTq6urg7++PkSNHIjU1FbNmzUJ5\neTlOnjyJfv36NXv9qVOnZKiaiMj1ydIUpk6diqlTpyI6Oho+Pj746KOPAAARERFITk5GREQEvLy8\nsHz5cp4+IiJyIKccXiMiIvtwurWP8vPz0atXL/To0QNZWVlylyObqVOnIjAwENHR0XKXIruysjIM\nGjQIkZGRiIqKQnZ2ttwlyeb69euIjY1FTEwMIiIiMGfOHLlLkl1DQwM0Gg2SkpLkLkVWwcHBuOuu\nu6DRaFo8LS9wqiOFhoYG3Hnnnfjmm2/QrVs3/P3vf8dnn32G8PBwuUtzuO+//x6+vr545JFHcOjQ\nIbnLkVVlZSUqKysRExODmpoa9OnTBxs3bnTLfxcAUFtbiw4dOqC+vh4DBw7EokWLMHDgQLnLks2S\nJUuwb98+VFdXIy8vT+5yZBMSEoJ9+/ahc+fOrT7PqY4UiouLERYWhuDgYHh7e2PSpEnIzc2VuyxZ\n3HfffbjtttvkLkMRunbtipiYGACAr68vwsPDUVFRIXNV8unQoQMAoK6uDg0NDWY/BFzZ2bNn8dVX\nX2HatGlcLw2w6HfgVE2hvLwc3bt3Fx9zuI2M6XQ6HDhwALGxsXKXIpvGxkbExMQgMDAQgwYNQoQb\nj/L+z//8D95880142HL/ThehUqkQFxeHvn37YuXKlSaf51S/KV6JRK2pqanB+PHjsXTpUvhac69G\nF+Hh4YGDBw/i7Nmz+O6779xyyQsA+PLLLxEQEACNRsOjBABFRUU4cOAAtmzZgnfffRfff/99i89z\nqqbQrVs3lJWViY/LysqaLItB7uvGjRsYN24cHn74YYwePVruchShU6dOGD58uLiMjLvZtWsX8vLy\nEBISgpSUFGzfvh2PPPKI3GXJ5q9//SsAw1pzY8aMMbmunFM1hb59++LkyZPQ6XSoq6vDunXrMHLk\nSLnLIpnp9Xo8/vjjiIiIwMyZM+UuR1aXLl1CVVUVAODatWvYtm0bNBqNzFXJY+HChSgrK8OZM2ew\ndu1aPPjgg+JMlLupra1FdXU1AODq1avYunWrySsXnaopeHl54Z133kFiYiIiIiIwceJEt73CJCUl\nBffeey9OnDiB7t27Y/Xq1XKXJJuioiJ88sknKCgogEajgUajQX5+vtxlyeLcuXN48MEHERMTg9jY\nWCQlJWHw4MFyl6UI7nz6+fz587jvvvvEfxcjRoxAQkJCi891qktSiYjIvpzqSIGIiOyLTYGIiERs\nCkREJGJTICIiEZsCERGJ2BSIiEjEpkAuzd7LXQQHB+Py5cvNtu/YsQO7d+9u8TWbNm1y62XfSdlk\nufMakaPYe2BJpVK1uK5OQUEB/Pz8cM899zT7WVJSktuv7U/KxSMFcjunT5/G0KFD0bdvX9x///04\nfvw4AOCxxx7D888/jwEDBiA0NBTr168HYFh1dMaMGQgPD0dCQgKGDx8u/gwAli1bhj59+uCuu+7C\n8ePHodPpsGLFCrz11lvQaDTYuXNnk/2vWbMGzz77bKv7vJlOp0OvXr0wZcoU3HnnnXjooYewdetW\nDBgwAD179kRJSYm9flXkhtgUyO08+eSTWLZsGfbu3Ys333wTM2bMEH9WWVmJoqIifPnll0hLSwMA\nfPHFF/jll19w9OhRfPzxx9i9e3eTI5Dbb78d+/btw9NPP41FixYhODgY06dPx6xZs3DgwIFmN7gx\nPnppaZ/GTp8+jRdffBHHjh3D8ePHsW7dOhQVFWHRokVYuHChVL8aIp4+IvdSU1OD3bt3Y8KECeK2\nuro6AIYPa2GF1fDwcJw/fx4AsHPnTiQnJwOAeI+Cm40dOxYA0Lt3b3zxxRfidktWkDG1T2MhISGI\njIwEAERGRiIuLg4AEBUVBZ1OZ3Y/RJZiUyC30tjYiFtvvRUHDhxo8ec+Pj7i98KHunFuYPxh365d\nOwCAp6cn6uvr21xTS/s0JuwDMNwvQXiNh4eHVfskMoWnj8itdOzYESEhIfjPf/4DwPAh/NNPP7X6\nmgEDBmD9+vXQ6/U4f/48duzYYXY/fn5+4lLFxrgGJSkZmwK5tNraWnTv3l38evvtt/Hpp59i1apV\niImJQVRUVJObud98vl/4fty4cVCr1YiIiMDkyZPRu3dvdOrUqdm+VCqV+JqkpCRs2LABGo0GRUVF\nJp9nap8tvbepx+68JDRJj0tnE1ng6tWruOWWW/Dbb78hNjYWu3btQkBAgNxlEUmOmQKRBUaMGIGq\nqirU1dVh3rx5bAjksnikQEREImYKREQkYlMgIiIRmwIREYnYFIiISMSmQEREIjYFIiIS/T+6l7ug\nkeDZfAAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x567bcf0>"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.7,Page No.109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_BC=1 #m #Length of BC\n",
+ "L_DB=2 #m #Length of DB\n",
+ "L_AD=4 #m #Length 0f AD\n",
+ "M_D=30 #KN.m #Moment at D\n",
+ "w=45 #KN/m #u.d.l\n",
+ "L=7 #m #Span of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_B & R_A be the Reactions at B & A respectively\n",
+ "#R_B+R_A=180+P ............(1)\n",
+ "\n",
+ "#Now Taking Moment about A,we get\n",
+ "#R_B=7*P+390 ...............(2)\n",
+ "\n",
+ "#Since R_A & R_B Are Equal\n",
+ "#2*R_B=180+P ...................(3)\n",
+ "\n",
+ "#From equation 1 and 3 we get\n",
+ "#3*(180+P)=7P+390\n",
+ "#After simplifying Further above equation we get\n",
+ "P=150*4**-1 #KN\n",
+ "R_A=R_B=(180+P)*2**-1\n",
+ "F_C=P\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C1=0 #KN\n",
+ "V_C2=-P #KN\n",
+ "\n",
+ "#S.F At B\n",
+ "V_B1=V_C2 #KN\n",
+ "V_B2=V_C2+R_B #KN \n",
+ "\n",
+ "#S.F At D\n",
+ "V_D=V_B2 #KN\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A1=V_D-w*L_AD #KN\n",
+ "V_A2=V_A1+R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M at C\n",
+ "M_C=0 #KN.m \n",
+ "\n",
+ "#B.M at B\n",
+ "M_B=F_C*L_BC #KN.m\n",
+ "\n",
+ "#B.M at D\n",
+ "M_D1=F_C*(L_BC+L_DB)-R_B*L_DB #KN.m\n",
+ "M_D2=M_D1+M_D\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A=w*L_AD*L_AD*2**-1+M_D-R_B*(L_AD+L_DB)+P*L\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_BC,L_BC,L_DB+L_BC,L_DB+L_BC+L_AD,L_DB+L_BC+L_AD]\n",
+ "Y1=[V_C1,V_C2,V_B1,V_B2,V_D,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_BC,L_DB+L_BC,L_DB+L_BC,L_AD+L_DB+L_BC]\n",
+ "Y2=[M_C,M_B,M_D1,M_D2,M_A]\n",
+ "Z2=[0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5d6f150>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5d79470>"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.8,Page No.110"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=6 #m #Span Of beam\n",
+ "w=30 #KN/m #u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Due to Symmetry\n",
+ "#Let R_B and R_C be the reactions at B & C Respectively\n",
+ "R_B=R_C=w*L*2**-1 #KN\n",
+ "\n",
+ "#Let a be the overhang.The Max -ve moment occurs at the support and max +ve moment at middle of the beam\n",
+ "#Now Equating these two equations we get\n",
+ "#30*a*a*2**-1=90*(3-a)-w*L*2**-1*L*4**-1\n",
+ "#After simplifying we get an equation as\n",
+ "#a**2+6*a-9=0\n",
+ "x=1\n",
+ "y=6\n",
+ "z=-9\n",
+ "\n",
+ "p=y**2-4*x*z\n",
+ "\n",
+ "a1=(-y+p**0.5)*2**-1\n",
+ "a2=(-y-p**0.5)*2**-1\n",
+ "\n",
+ "#Now Length cannot be negative,so taking a1 into Consideration\n",
+ "\n",
+ "L_CD=L_AB=a1\n",
+ "L_BC=L-2*a1\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At D\n",
+ "V_D=0\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C1=V_D-w*L_CD #KN\n",
+ "V_C2=V_C1+R_C #KN\n",
+ "\n",
+ "#S.F At B\n",
+ "V_B1=-w*(L_BC+L_CD)+R_C\n",
+ "V_B2=V_B1+R_B\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A=round(V_B2,2)-round(w*L_AB,2)\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At D\n",
+ "M_D=0\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=w*L_CD*L_CD*2**-1 #KN.m\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=w*(L_BC+L_CD)*(L_BC+L_CD)*2**-1-R_C*L_BC*L_BC*2**-1\n",
+ "\n",
+ "#B.M At A\n",
+ "X=w*L*L*2**-1\n",
+ "Y=-R_C*(L_AB+L_BC)-R_B*L_AB\n",
+ "M_A=X+Y\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,L_CD,L_CD,L_CD+L_BC,L_CD+L_BC,L_CD+L_BC+L_AB]\n",
+ "Y1=[V_D,V_C1,V_C2,V_B1,V_B2,V_A]\n",
+ "Z1=[0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_CD,L_BC+L_CD,L_AB+L_BC+L_CD]\n",
+ "Y2=[M_D,M_C,M_B,M_A]\n",
+ "Z2=[0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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lXLhwAQBw6dIlHDhwwKleBejt7Q0/Pz+UlpYCAA4ePIiwsLAub6vKm9f6yt3d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\nAElFTkSuQmCC\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5945590>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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LUVxcrNtAqS48PDxqdH5WVhby8vIQEBAAAJg4cSJ27NjBhEGkEDc3ID5e7BWe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\nMQcOHJASExMlb29vtUPRi6y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+ "text": [
+ "<matplotlib.figure.Figure at 0x58e1750>"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.9,Page No.112"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F_F=6 #KN #Force at F\n",
+ "w1=w2=w=3 #KN.m #u.d.l\n",
+ "M_D=24 #KN.m \n",
+ "L_AB=L_CD=L_DE=L_EF=4 #m #Length of AB,CD,DE,EF\n",
+ "L_BC=2 #m #Length of BC\n",
+ "L=18 #m #Span of Beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt R_B and R_E be the Reactions at B & E respectively\n",
+ "#R_B+R_E=42\n",
+ "\n",
+ "#Taking Moment At Pt B,M_B\n",
+ "R_E=(F_F*(L_BC+L_CD+L_DE+L_EF)+w*(L_CD+L_DE)*((L_CD+L_DE)*2**-1+L_BC)-w*L_AB*L_AB*2**-1-M_D)*(L_BC+L_CD+L_DE)**-1\n",
+ "R_B=42-R_E #KN\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F aT F\n",
+ "V_F1=0 #KN \n",
+ "V_F2=-F_F #KN\n",
+ "\n",
+ "#S.F at E\n",
+ "V_E1=V_F2 #KN\n",
+ "V_E2=V_E1+R_E #KN\n",
+ "\n",
+ "#S.F aT C\n",
+ "V_C=V_E2-w*(L_CD+L_DE) #KN\n",
+ "\n",
+ "#S.F at B\n",
+ "V_B1=V_C #KN \n",
+ "V_B2=V_C+R_B #KN\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A=V_B2-w*L_AB #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At F\n",
+ "M_F=0\n",
+ "\n",
+ "#B.M At E\n",
+ "M_E=F_F*L_EF #KN.m\n",
+ "\n",
+ "#B.M At D\n",
+ "M_D1=F_F*(L_DE+L_EF)-R_E*L_DE+w*L_DE*L_DE*2**-1 #KN.m\n",
+ "M_D2=M_D1-M_D\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=F_F*(L_CD+L_DE+L_EF)-R_E*(L_CD+L_DE)+w*(L_CD+L_DE)*(L_CD+L_DE)*2**-1-M_D\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=F_F*(L_BC+L_CD+L_DE+L_EF)-R_E*(L_BC+L_CD+L_DE)-M_D+w*(L_CD+L_DE)*((L_CD+L_DE)*2**-1+L_BC)\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A=w*L_AB*L_AB*2**-1-R_B*L_AB+w*(L_CD+L_DE)*((L_CD+L_DE)*2**-1+L_BC+L_AB)-R_E*(L_AB+L_BC+L_CD+L_DE)+F_F*L-M_D\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_EF,L_EF,L_EF+L_DE+L_CD,L_EF+L_DE+L_CD+L_BC,L_EF+L_DE+L_CD+L_BC,L_EF+L_DE+L_CD+L_BC+L_AB]\n",
+ "Y1=[V_F1,V_F2,V_E1,V_E2,V_C,V_B1,V_B2,V_A]\n",
+ "Z1=[0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_EF,L_DE+L_EF,L_DE+L_EF,L_CD+L_DE+L_EF,L_CD+L_DE+L_EF+L_BC,L_CD+L_DE+L_EF+L_BC+L_AB]\n",
+ "Y2=[M_F,M_E,M_D1,M_D2,M_C,M_B,M_A]\n",
+ "Z2=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5d75ab0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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VxMfHIzs7G/369cOvv/6KcePG4bHHHnNknEROixVL5KzMziBqa2uRmJiIMWPG\n4IEHHkC/fv0AAMHBwdBoNA4LkMiZsccSOTOzCaJhEmCbb6KmY8USOTuzS0zHjx+Ht7c3AODGjRvG\n7+t/JiLzamqAlBT2WCLnZjZB1NbWOjIOIpfyyiuAhwcrlsi5sZ6CSGbLlwO7drHHEjk/Dl8iGbFi\niVwJEwSRTNhjiVyNxV5MRGQZK5bIFTFBENmIFUvkqhRJEDNnzkRISAgiIiIwcuRIXL161XhfWloa\nevTogeDgYGODQCI1Y8USuSpFEkRiYiJ+/vln/PjjjwgKCkJaWhoAoKCgAOvXr0dBQQGys7Mxbdo0\n1NXVKREikVXqK5bYY4lckSIJIiEhwXgIUWxsLIqLiwEAWVlZGD9+PDw9PeHn54fAwEAcOnRIiRCJ\nLKqvWNq6lRVL5JoU34NYtWoVhgwZAgAoLS2Fr6+v8T5fX1+UlJQoFRqRWeyxRO7AbpPihIQElJeX\n33H7/PnzkZSUBACYN28evLy8kJqaavZ12BiQ1IYVS+Qu7JYgdu7cedf716xZg23btuG///2v8TYf\nHx8UFRUZfy4uLoaPj4/J57/11lvG7+Pi4hAXF2dTvETWYMUSOZPc3Fzk5uY2+/lWHTkqt+zsbLz6\n6qvIy8trdEJdQUEBUlNTcejQIZSUlGDQoEE4ffr0HbMIHjlKcmrKkaMvvigtL339NTelyfnY5chR\nub344oswGAzGs60feughZGZmQqfTISUlBTqdDi1btkRmZiaXmEg12GOJ3I0iMwhbcQZBcrJmBpGT\nA6SmSo/hpjQ5K6eYQRA5E/ZYIneleJkrkZqxYoncGRMEkRmsWCJ3xwRBZAZ7LJG74x4EkQmsWCJi\ngiC6A0+FI5IwQRA1wIolor9xD4LoL6xYImqMCYIIrFgiMoUJggjA99+zYonodkwQ5Pbuvx/o14+n\nwhHdjr2YiIjcRFM/OzmDICIik5ggiIjIJCYIIiIyiQmCiIhMYoIgIiKTmCCIiMgkJggiIjKJCYKI\niExigiAiIpOYIIiIyCQmCCIiMokJgoiITGKCICIik5ggiIjIJCYIIiIyiQmCiIhMUiRBvPHGG4iI\niEBkZCQGDhyIoqIi431paWno0aMHgoODsWPHDiXCIyIiKJQgZs2ahR9//BHHjh1DcnIy3n77bQBA\nQUEB1q9fj4KCAmRnZ2PatGmoq6tTIsRmyc3NVTqEOzAm6zAm66kxLsZkH4okCG9vb+P3VVVV6Ny5\nMwAgKyvgK/nsAAAKZklEQVQL48ePh6enJ/z8/BAYGIhDhw4pEWKzqHFAMCbrMCbrqTEuxmQfih3R\n/vrrr+PTTz/FPffcY0wCpaWl6Nevn/Exvr6+KCkpUSpEIiK3ZrcZREJCAsLDw+/42rp1KwBg3rx5\nOHfuHKZMmYLp06ebfR2NRmOvEImI6G6Ewn7//XcRGhoqhBAiLS1NpKWlGe8bPHiwOHjw4B3PCQgI\nEAD4xS9+8YtfTfgKCAho0uezIktMhYWF6NGjBwBp3yEqKgoAMGzYMKSmpmLGjBkoKSlBYWEh+vbt\ne8fzT58+7dB4iYjckSIJYs6cOTh58iRatGiBgIAAfPjhhwAAnU6HlJQU6HQ6tGzZEpmZmVxiIiJS\niEYIIZQOgoiI1MfprqTOzs5GcHAwevTogYyMDKXDQVFREeLj4xEaGoqwsDAsWbJE6ZCMamtrERUV\nhaSkJKVDMaqsrMTo0aMREhICnU6HgwcPKh0S0tLSEBoaivDwcKSmpuLPP/90eAxPPfUUtFotwsPD\njbddvnwZCQkJCAoKQmJiIiorKxWPaebMmQgJCUFERARGjhyJq1evKh5TvUWLFsHDwwOXL192aEx3\ni2vp0qUICQlBWFgYZs+erXhMhw4dQt++fREVFYU+ffrg8OHDd38RWzaYHa2mpkYEBASIs2fPCoPB\nICIiIkRBQYGiMZWVlYmjR48KIYS4fv26CAoKUjymeosWLRKpqakiKSlJ6VCMJk2aJFauXCmEEOLW\nrVuisrJS0XjOnj0r/P39xc2bN4UQQqSkpIg1a9Y4PI7vvvtO5Ofni7CwMONtM2fOFBkZGUIIIdLT\n08Xs2bMVj2nHjh2itrZWCCHE7NmzVRGTEEKcO3dODB48WPj5+YlLly45NCZzceXk5IhBgwYJg8Eg\nhBDi/Pnzisc0YMAAkZ2dLYQQYtu2bSIuLu6ur+FUM4hDhw4hMDAQfn5+8PT0xLhx45CVlaVoTF27\ndkVkZCQAoG3btggJCUFpaamiMQFAcXExtm3bhqlTp0KoZBXx6tWr2LNnD5566ikAQMuWLdG+fXtF\nY2rXrh08PT1RXV2NmpoaVFdXw8fHx+Fx/OMf/0DHjh0b3bZlyxZMnjwZADB58mRs3rxZ8ZgSEhLg\n4SF9bMTGxqK4uFjxmABgxowZePfddx0aS0Om4vrwww8xZ84ceHp6AgDuv/9+xWN64IEHjLO+yspK\ni2PdqRJESUkJunXrZvxZbRfS6fV6HD16FLGxsUqHgldeeQULFiww/mNWg7Nnz+L+++/HlClTEB0d\njWeeeQbV1dWKxnTffffh1VdfRffu3fHggw+iQ4cOGDRokKIx1auoqIBWqwUAaLVaVFRUKBxRY6tW\nrcKQIUOUDgNZWVnw9fVFr169lA6lkcLCQnz33Xfo168f4uLi8MMPPygdEtLT043jfebMmUhLS7vr\n49Xz6WEFNVc0VVVVYfTo0Vi8eDHatm2raCxff/01unTpgqioKNXMHgCgpqYG+fn5mDZtGvLz83Hv\nvfciPT1d0ZjOnDmD999/H3q9HqWlpaiqqsK6desUjckUjUajqvE/b948eHl5ITU1VdE4qqurMX/+\nfGM/NwCqGfM1NTW4cuUKDh48iAULFiAlJUXpkPD0009jyZIlOHfuHN577z3jbN4cp0oQPj4+jTq/\nFhUVwdfXV8GIJLdu3cKoUaMwYcIEJCcnKx0O9u/fjy1btsDf3x/jx49HTk4OJk2apHRY8PX1ha+v\nL/r06QMAGD16NPLz8xWN6YcffsDDDz+MTp06oWXLlhg5ciT279+vaEz1tFotysvLAQBlZWXo0qWL\nwhFJ1qxZg23btqkikZ45cwZ6vR4RERHw9/dHcXExYmJicP78eaVDg6+vL0aOHAkA6NOnDzw8PHDp\n0iVFYzp06BBGjBgBQPr3Z6nXnVMliN69e6OwsBB6vR4GgwHr16/HsGHDFI1JCIGnn34aOp3uri1D\nHGn+/PkoKirC2bNn8fnnn+Of//wnPvnkE6XDQteuXdGtWzecOnUKALBr1y6EhoYqGlNwcDAOHjyI\nGzduQAiBXbt2QafTKRpTvWHDhmHt2rUAgLVr16ril4/s7GwsWLAAWVlZaN26tdLhIDw8HBUVFTh7\n9izOnj0LX19f5OfnqyKZJicnIycnBwBw6tQpGAwGdOrUSdGYAgMDkZeXBwDIyclBUFDQ3Z9grx10\ne9m2bZsICgoSAQEBYv78+UqHI/bs2SM0Go2IiIgQkZGRIjIyUmzfvl3psIxyc3NVVcV07Ngx0bt3\nb9GrVy8xYsQIxauYhBAiIyND6HQ6ERYWJiZNmmSsOnGkcePGiQceeEB4enoKX19fsWrVKnHp0iUx\ncOBA0aNHD5GQkCCuXLmiaEwrV64UgYGBonv37sax/vzzzysSk5eXl/HvqSF/f39FqphMxWUwGMSE\nCRNEWFiYiI6OFrt371YkpoZj6vDhw6Jv374iIiJC9OvXT+Tn59/1NXihHBERmeRUS0xEROQ4TBBE\nRGQSEwQREZnEBEFERCYxQRARkUlMEEREZBITBLk0e7c98fPzM9leOi8vDwcOHDD5nK1bt6qiVT2R\nJYqcKEfkKPbuX6TRaEz2/tm9eze8vb3x0EMP3XFfUlKSqs7oIDKHMwhyO2fOnMFjjz2G3r1749FH\nH8XJkycBAE8++SRefvll9O/fHwEBAdi4cSMAoK6uDtOmTUNISAgSExMxdOhQ432AdChMTEwMevXq\nhZMnT0Kv1+Ojjz7Ce++9h6ioKOzdu7fR+69ZswYvvvjiXd+zIb1ej+DgYEyZMgU9e/bEE088gR07\ndqB///4ICgqyfOgLUTMxQZDbefbZZ7F06VL88MMPWLBgAaZNm2a8r7y8HPv27cPXX3+N1157DQDw\n1Vdf4ffff8cvv/yCTz/9FAcOHGg0M7n//vtx5MgRPP/881i4cCH8/Pzwr3/9CzNmzMDRo0fxyCOP\nNHr/22c1pt7zdmfOnMG///1v/Prrrzh58iTWr1+Pffv2YeHChZg/f75cfzVEjXCJidxKVVUVDhw4\ngDFjxhhvMxgMAKQP7vqGeCEhIcbzF/bu3Wts1azVahEfH9/oNes7dkZHR+Orr74y3m5NFxtz73k7\nf39/Y2PD0NBQ45kVYWFh0Ov1Ft+HqDmYIMit1NXVoUOHDjh69KjJ+728vIzf13/A377PcPsHf6tW\nrQAALVq0QE1NTZNjMvWet6t/DwDw8PAwPsfDw6NZ70lkDS4xkVtp164d/P398eWXXwKQPpCPHz9+\n1+f0798fGzduhBACFRUVxnbJd+Pt7Y3r16+bvI/9MclZMEGQS6uurka3bt2MX++//z7WrVuHlStX\nIjIyEmFhYdiyZYvx8Q33B+q/HzVqFHx9faHT6TBx4kRER0ebPEu74alvSUlJ2LRpE6KiorBv3z6z\njzP3nqZe29zPajppjlwL230TWeGPP/7Avffei0uXLiE2Nhb79+9XxaE0RPbEPQgiKzz++OOorKyE\nwWDA3LlzmRzILXAGQUREJnEPgoiITGKCICIik5ggiIjIJCYIIiIyiQmCiIhMYoIgIiKT/h/FhIMx\nfyRzHAAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5c44130>"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.10,Page No.114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_DC=L_BA=2 #m #Length of BA & DC\n",
+ "L_CB=1 #m #Length of CB\n",
+ "F_A=10 #KN #Force at pt A\n",
+ "F_B=20 #KN #Force at pt B\n",
+ "w=4 #KN.m #u.d.l\n",
+ "L=5 #m #Length of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_D be the reactions at Pt D\n",
+ "R_D=F_B+F_A+w*L_DC #KN\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F at Pt A\n",
+ "V_A1=0 #KN\n",
+ "V_A2=F_A #KN\n",
+ "\n",
+ "#S.F At Pt B\n",
+ "V_B1=V_A2\n",
+ "V_B2=F_B+F_A\n",
+ "\n",
+ "#S.F at Pt C\n",
+ "V_C=F_B+F_A #KN \n",
+ "\n",
+ "#S.F At Pt D\n",
+ "V_D1=V_B2+w*L_DC\n",
+ "V_D2=F_B+F_A+w*L_DC-R_D\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=0\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=F_A*L_BA\n",
+ "\n",
+ "#B.M at Pt C\n",
+ "M_C=F_B*L_CB+F_A*(L_BA+L_CB) #KN\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D1=F_A*L+F_B*(L_CB+L_DC)+w*L_DC*L_DC*2**-1\n",
+ "M_D2=(F_A*L+F_B*(L_CB+L_DC)+w*L_DC*L_DC*2**-1)-M_D1\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_BA,L_BA,L_BA+L_CB,L_BA+L_CB+L_DC,L_BA+L_CB+L_DC]\n",
+ "Y1=[V_A1,V_A2,V_B1,V_B2,V_C,V_D1,V_D2]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "Y2=[M_A,M_B,M_C,M_D1,M_D2]\n",
+ "X2=[0,L_BA,L_CB+L_BA,L_CB+L_BA+L_DC,L_CB+L_BA+L_DC]\n",
+ "Z2=[0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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e3542bZr69++vhIQEzZ49WyNGjNAFF1wgSbrxxht15ZVXen+5e/LxJ0/9axiG\n8vLy9Oc//1kxMTFatWqV8vLyvMNNvo71tX3yMb6+tvMUxDh7XM4JWzp8+LAiIiL0ww8/aPTo0dq8\nebP69OljdSwgKBjjhy1NnjxZNTU1amho0L333kvpw1Y44wcAm2GMHwBshuIHAJuh+AHAZih+ALAZ\nih8AbIbiBwCb+T+gqVKkQGpm/QAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5738d10>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5788db0>"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.11,Page No.115"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "w=20 #KN/m #u.v.l\n",
+ "F_C=40 #KN #Force at Pt C\n",
+ "M_D=40 #KN.m #Moment at pt D\n",
+ "L_AB=3 #m #Length of AB\n",
+ "L_BC=1 #m #Length of BC\n",
+ "L_CD=L_DE=2 #m #Length of CD & DE\n",
+ "L=8 #8 #Length of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A & R_E be the Reactions at A & E respectively\n",
+ "#R_A+R_E=70\n",
+ "\n",
+ "#Taking Moments At Pt A we get,M_A\n",
+ "R_E=(F_C*(L_AB+L_BC)+1*2**-1*L_AB*w*2+40)*L**-1\n",
+ "R_A=70-R_E\n",
+ "\n",
+ "#shear Force Calculations\n",
+ "\n",
+ "#S.F At Pt E\n",
+ "V_E1=0\n",
+ "V_E2=R_E #KN\n",
+ "\n",
+ "#S.F aT pt D\n",
+ "V_D=V_E2\n",
+ "\n",
+ "#S.F At PT C\n",
+ "V_C1=V_D\n",
+ "V_C2=V_D-F_C #KN\n",
+ "\n",
+ "#S.F At Pt A\n",
+ "V_A1=V_C2-(1*2**-1*w*L_AB)\n",
+ "V_A2=V_A1+R_A\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt E\n",
+ "M_E=0\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D1=M_E-R_E*L_DE\n",
+ "M_D2=M_D1+M_D\n",
+ "\n",
+ "#B.M At Pt C\n",
+ "M_C=-R_E*(L_DE+L_CD)+M_D\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=-R_E*(L_DE+L_CD+L_BC)+M_D+F_C*L_BC\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=-R_E*L+M_D+(1*2**-1*L_AB*w*2)+F_C*(L_BC+L_AB)\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_DE,L_CD+L_DE,L_CD+L_DE,L_CD+L_DE+L_AB,L_CD+L_DE+L_AB]\n",
+ "Y1=[V_E1,V_E2,V_D,V_C1,V_C2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "Y2=[M_E,M_D1,M_D2,M_C,M_B,M_A]\n",
+ "X2=[0,L_DE,L_DE,L_CD+L_DE,L_DE+L_CD+L_BC,L_AB+L_BC+L_CD+L_DE]\n",
+ "Z2=[0,0,0,0,0,0]\n",
+ "plt.plot(X2,Y2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5df5230>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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PZcuW8dRTT5GZmcnOnTtp1KgRI0aMsPs+cvTnxbp1U4fwLFyoO4nzTCaYMkWt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\nd/r0aduBO+VVUFCA2Wxm+/b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6ePR5C6Jyk56CqHT27dtHt27daNu2LXfddRdpaWkAPPbY\nYzz33HPcfvvtNG/enCVLlgCqQmt8fDyhoaF06dKF++67z/Y9gDlz5tCmTRtatmxJWloaWVlZzJs3\njzfffJOoqCg2bNhw0fU/+ugjnnnmmSte80JZWVmEhITw+OOPc/PNN/Pwww+zatUqbr/9dlq0aMHW\nrVvd9aMSlZA0CqLS+cc//sGcOXP46aefeP3114mPj7d979ChQ2zcuJFvvvmGUaNGAfD111/zf//3\nf/zyyy98+umnbN68+aIeSIMGDdi2bRtPPfUUM2fOJDAwkH/+85+88MIL7NixgzvuuOOi61/aeynp\nmpfat28fL774Ir/++itpaWkkJSWxceNGZs6cydSpU131oxFCho9E5ZKXl8fmzZsvKnGcn58PqJt1\ncaXY0NBQ/vjjDwA2bNhAbGwsgO08hwv17t0bgNatW/P111/bHnekgoy9a16qadOmtoJwYWFhtlr4\n4eHhZGVllXodIRwljYKoVIqKiqhfvz47duwo8fvVqlWzfV58U7903uDSm3316tUBqFq1KgUFBWXO\nVNI1L1V8DVBnSxS/pkqVKuW6phD2yPCRqFTq1q1L06ZN+eqrrwB1E961a9cVX3P77bezZMkSDMPg\njz/+YN26daVep06dOrYS1ZeSGpTCk0mjIHzaqVOnaNy4se3jrbfe4vPPP+eDDz6gVatWhIeHs2zZ\nMtvzLxzvL/68T58+mM1mLBYLjz76KK1bty7xvGCTyWR7TY8ePVi6dClRUVFs3LjR7vPsXbOk97b3\ntZxIKFxJSmcL4YCTJ09Sq1Ytjhw5QocOHdi0aRPXX3+97lhCuJzMKQjhgPvvv59jx46Rn5/P+PHj\npUEQPkt6CkIIIWxkTkEIIYSNNApCCCFspFEQQghhI42CEEIIG2kUhBBC2EijIIQQwub/A4tFvTZr\nGhPHAAAAAElFTkSuQmCC\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5dff130>"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.12,Page No.116"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F_G=10 #KN #Force at Pt G\n",
+ "F_B=F_E=15 #KN #Force at Pt B & E\n",
+ "w=20 #KN/m #U.d.L\n",
+ "L_FG=L_EF=L_DE=L_CD=L_BC=L_AB=1 #m #Lengths of FG,EF,DE,CD,BC,AB respectively\n",
+ "L=6 #m #Length of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt R_F & R_A be the Reactions at E & A respectively\n",
+ "#R_F+R_A=60\n",
+ "\n",
+ "#Taking Moment At Pt A,M_A\n",
+ "R_F=(F_G*L+F_E*(L_AB+L_BC+L_CD+L_DE)+w*L_CD*(L_AB+L_BC+L_CD*2**-1)+F_B*L_AB)*(L_AB+L_BC+L_CD+L_DE+L_EF)**-1\n",
+ "R_A=60-R_F\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At G\n",
+ "V_G1=0 #KN \n",
+ "V_G2=F_G #KN\n",
+ "\n",
+ "#S.F At F\n",
+ "V_F1=V_G2 #KN\n",
+ "V_F2=V_F1-R_F\n",
+ "\n",
+ "#S.F At E\n",
+ "V_E1=V_F2 #KN\n",
+ "V_E2=V_F2+F_E\n",
+ "\n",
+ "#S.F At D\n",
+ "V_D=V_E2\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C=V_E2+w*L_CD\n",
+ "\n",
+ "#S.F At B\n",
+ "V_B1=V_C\n",
+ "V_B2=V_B1+F_B\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A1=V_B2\n",
+ "V_A2=V_B2-R_A\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt G\n",
+ "M_G=0\n",
+ "\n",
+ "#B.M At F\n",
+ "M_F=F_G*L_FG \n",
+ "\n",
+ "#B.M At E\n",
+ "M_E=F_G*(L_FG+L_EF)-R_F*L_EF\n",
+ "\n",
+ "#B.M At D\n",
+ "M_D=F_G*(L_FG+L_EF+L_DE)-R_F*(L_EF+L_DE)+F_E*L_DE\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=F_G*(L_FG+L_EF+L_DE+L_CD)-R_F*(L_EF+L_DE+L_CD)+F_E*(L_DE+L_CD)+w*L_CD*L_CD*2**-1\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=F_G*(L_FG+L_EF+L_DE+L_CD+L_BC)-R_F*(L_EF+L_DE+L_CD+L_BC)+F_E*(L_DE+L_CD+L_BC)+w*L_CD*(L_CD*2**-1+L_BC)\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A=F_G*L-R_F*(L_EF+L_DE+L_CD+L_BC+L_AB)+F_E*(L_DE+L_CD+L_BC+L_AB)+F_B*L_AB+w*L_CD*(L_CD*2**-1+L_BC+L_AB)\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_FG,L_FG,L_FG+L_EF,L_FG+L_EF,L_FG+L_EF+L_DE,L_FG+L_EF+L_DE+L_CD,L_FG+L_EF+L_DE+L_CD+L_BC,L_FG+L_EF+L_DE+L_CD+L_BC,L_FG+L_EF+L_DE+L_CD+L_BC+L_AB,L_FG+L_EF+L_DE+L_CD+L_BC+L_AB]\n",
+ "Y1=[V_G1,V_G2,V_F1,V_F2,V_E1,V_E2,V_D,V_C,V_B1,V_B2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_FG,L_EF+L_FG,L_EF+L_FG+L_DE,L_EF+L_FG+L_DE+L_CD,L_EF+L_FG+L_DE+L_CD+L_BC,L_EF+L_FG+L_DE+L_CD+L_BC+L_AB]\n",
+ "Y2=[M_G,M_F,M_E,M_D,M_C,M_B,M_A]\n",
+ "Z2=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X2,Y2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x57502f0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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9jlmzZjkcDuPv6NGjR8t9nl+NFHJycmjatCnR0dGEh4dz1113kZmZ\naXVYpmnfvj116tSxOgyPadiwIUlJSQBEREQQFxfHvn37LI7KPJdeeikAxcXFlJSUULduXYsjMtcP\nP/zAf//7Xx544AEcATrrHKi/108//cTKlSsZPHgwAGFhYdSqVavc5/pVUsjPz6dRo0bOP0dFRZGf\nn29hRFJdeXl55ObmkpKSYnUopjl9+jRJSUk0aNCADh06EB8fb3VIpnr88cd5/vnnCQnxq7cNl9ls\nNjp16kSbNm2YOXOm1eGYavfu3VxxxRUMGjSIVq1aMWTIEIqKisp9rl/917XZbFaHICYoLCykb9++\nvPTSS0RERFgdjmlCQkJYt24dP/zwAytWrAiokgmLFy+mfv36JCcnB+yn6VWrVpGbm8uSJUt49dVX\nWblypdUhmebUqVOsXbuWoUOHsnbtWi677DImTZpU7nP9KilcffXVzoqqYCxcRkVFWRiRVNXJkye5\n4447uPvuu+nVq5fV4XhErVq16NGjB19//bXVoZjmyy+/ZOHChTRp0oQBAwbw2Wefce+991odlqmu\nvPJKAK644gp69+5NTk6OxRGZJyoqiqioKH7/+98D0LdvX9auXVvuc/0qKbRp04bt27eTl5dHcXEx\nc+fOpWfPnlaHJS5yOBzcf//9xMfHM3z4cKvDMdWhQ4c4evQoAL/88guffPKJ81BmIJgwYQJ79+5l\n9+7d/Otf/+LWW2/lnXfesTos0xQVFTlL6hw/fpylS5cG1C7Ahg0b0qhRI7Zt2wbAp59+SosWLcp9\nrmUF8aojLCyMV155hS5dulBSUsL9999PXFyc1WGZZsCAASxfvpzDhw/TqFEjxo4dy6BBg6wOyzSr\nVq3ivffec277A5g4cSJdu3a1ODL37d+/nz/84Q+cPn2a06dPc88999CxY0erw/KYQJvKPXDgAL17\n9waMqZaBAwfSuXNni6My18svv8zAgQMpLi4mJiaGOXPmlPs8HV4TEREnv5o+EhERz1JSEBERJyUF\nERFxUlIQEREnJQUREXFSUhARESclBQloni6jER0dzZEjR8o8vnz5clavXl3uaxYtWhRwZd8lcPjV\n4TWRqvL0ISubzVZuLaDPP/+cyMhIbrjhhjI/S0tLC8h+BBIYNFKQoLNz5066detGmzZtuPnmm9m6\ndSsA9913H4899hjt2rUjJiaGefPmAUb106FDhxIXF0fnzp3p0aOH82dgnBRt3bo11113HVu3biUv\nL4/p06fzj3/8g+TkZL744otS93/rrbf405/+dMF7nisvL4/Y2FgGDRrE7373OwYOHMjSpUtp164d\nzZs3Z82aNZ76VyVBSElBgs6DDz7Iyy+/zNdff83zzz/P0KFDnT8rKChg1apVLF68mBEjRgAwf/58\n9uzZw+bNm3n33XdZvXp1qRHIFVdcwTfffMMf//hHpkyZQnR0NA8//DBPPPEEubm53HTTTaXuf/7o\npbx7nm/nzp08+eSTbNmyha1btzJ37lxWrVrFlClTmDBhgln/akQ0fSTBpbCwkNWrV9OvXz/nY8XF\nxYDxZn22cmtcXBwHDhwA4IsvvuDOO+8EcPZKOFefPn0AaNWqFfPnz3c+7koFmYrueb4mTZo4C5i1\naNGCTp06AZCQkEBeXl6l9xFxlZKCBJXTp09Tu3ZtcnNzy/15jRo1nN+ffVM/f93g/Df7iy66CIDQ\n0FBOnTpV5ZjKu+f5zt4DjL4NZ18TEhJSrXuKVETTRxJUatasSZMmTfi///s/wHgT/vbbby/4mnbt\n2jFv3jwcDgcHDhxg+fLlld4nMjLSWYr5fKpBKb5MSUECWlFREY0aNXJ+vfjii7z//vvMmjWLpKQk\nEhISWLhwofP55873n/3+jjvuICoqivj4eO655x5atWpVbn9bm83mfE1aWhoLFiwgOTmZVatWVfi8\niu5Z3rUr+nOglbEWa6l0togLjh8/zmWXXcbhw4dJSUnhyy+/pH79+laHJWI6rSmIuOC2227j6NGj\nFBcXM2rUKCUECVgaKYiIiJPWFERExElJQUREnJQURETESUlBRESclBRERMRJSUFERJz+P8O/Vq30\nkcIBAAAAAElFTkSuQmCC\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x58b0ed0>"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.13,Page No.117"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "L_AB=L_BC=L_CD=L_DE=L_EF=1 #m #LEngth of AB,BC,CD,DE,EF respectively\n",
+ "M_A=50 #KN/m #Moment at A\n",
+ "w=5 #KN/m #u.v.l\n",
+ "F_D=10 #KN\n",
+ "w2=5 #KN/m #u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_B & R_E be the Reactions at B and E respectively\n",
+ "#R_B+R_E=20\n",
+ "\n",
+ "#Taking Moment At Pt B,M_B\n",
+ "R_E=(w2*L_EF*(L_EF*2**-1+L_DE+L_CD+L_BC)+w*L_BC*2**-1*2*3**-1+50+F_D*(L_BC+L_CD))*3**-1\n",
+ "R_B=17.5-R_E #KN\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At F\n",
+ "V_F=0\n",
+ "\n",
+ "#S.F aT E\n",
+ "V_E1=-w2*L_EF #KN\n",
+ "V_E2=V_E1+R_E\n",
+ "\n",
+ "#S.F at D\n",
+ "V_D1=R_E-w2*L_EF #KN\n",
+ "V_D2=V_D1-F_D #KN\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C=V_D2\n",
+ "\n",
+ "#S.F aT B\n",
+ "V_B1=-L_BC*w*2**-1-F_D+R_E-w2*L_EF\n",
+ "V_B2=V_B1+R_B\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M at F\n",
+ "M_F=0 #KN.m\n",
+ "\n",
+ "#B.M At E\n",
+ "M_E=w2*L_EF*L_EF*2**-1 #KN.m\n",
+ "\n",
+ "#B.M at D\n",
+ "M_D=-R_E*L_DE+w2*L_EF*(L_EF*2**-1+L_DE) #KN.m\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=F_D*L_CD*R_E*(L_CD+L_DE)+w2*L_EF*(L_EF*2**-1+L_DE+L_CD) #KN.m\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=F_D*(L_CD+L_BC)-R_E*(L_BC+L_CD+L_DE)+w2*L_EF*(L_EF*2**-1+L_BC+L_CD+L_DE)+1*2**-1*L_BC*w*2*3**-1\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A1=w*L_EF*(L_EF*2**-1+L_AB+L_BC+L_CD+L_DE)-R_E*(L_AB+L_BC+L_CD+L_DE)+F_D*(L_AB+L_BC+L_CD)+1*2**-1*L_BC*w*(2*3**-1*L_BC+L_AB)-R_B*L_AB\n",
+ "M_A2=M_A1+M_A\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,L_EF,L_EF,L_DE+L_EF,L_DE+L_EF,L_CD+L_DE+L_EF,L_CD+L_DE+L_EF+L_BC,L_CD+L_DE+L_EF+L_BC]\n",
+ "Y1=[V_F,V_E1,V_E2,V_D1,V_D2,V_C,V_B1,V_B2]\n",
+ "Z1=[0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_EF,L_DE+L_EF,L_CD+L_DE+L_EF,L_CD+L_DE+L_EF+L_BC,L_CD+L_DE+L_EF+L_BC+L_AB,L_CD+L_DE+L_EF+L_BC+L_AB]\n",
+ "Y2=[M_F,M_E,M_D,M_C,M_B,M_A1,M_A2]\n",
+ "Z2=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5c2a7b0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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LlixREhMTLY8fMWKEsm/fvjuex4ZfSQi7euUVRXnxRa1TuIdr1xRFr1eUb77R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\ntnx2yhiGEEIIm0jBEEIIYRM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+ "text": [
+ "<matplotlib.figure.Figure at 0x1a108f0>"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.4_3.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.4_3.ipynb new file mode 100644 index 00000000..09aec3f9 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.4_3.ipynb @@ -0,0 +1,1628 @@ +{
+ "metadata": {
+ "name": "chapter no.4.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 4:Stresses in Beams"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.1,Page no.130"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=5000 #mm #Length of Beam\n",
+ "a=2000 #mm #Length of start of beam to Pt Load\n",
+ "b=3000 #mm #Length of Pt load to end of beam\n",
+ "A=150*250 #m**2 #Area of beam\n",
+ "b=150 #mm #Width of beam\n",
+ "d=250 #mm #Depth of beam\n",
+ "sigma=10#N/mm**2 #stress\n",
+ "l=2000 #m #Load applied from one end\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*b*d**3 #m**4\n",
+ "\n",
+ "#Distance from N.A to end\n",
+ "y_max=d*2**-1 #m\n",
+ "\n",
+ "#Section Modulus\n",
+ "Z=1*6**-1*b*d**2 #mm**3\n",
+ "\n",
+ "#Moment Carrying Capacity\n",
+ "M=sigma*Z #N-mm\n",
+ "\n",
+ "#Let w be the Intensity of the Load in N/m,then Max moment\n",
+ "#M_max=w*L**2*8**-1 #N-mm\n",
+ "#After substituting values and further simplifying we get\n",
+ "#M_max=w*25*100*8**-1\n",
+ "\n",
+ "#EQuating it to moment carrying capacity,we get max intensity load\n",
+ "w=M*(25*1000)**-1*8*10**-3\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#Let P be the concentrated load,then max moment occurs under the load and its value\n",
+ "#M1=P*a*b*L**-1 #N-mm\n",
+ "\n",
+ "#Equting it to moment carrying capacity we get\n",
+ "P=M*1200**-1*10**-3 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Intensity of u.d.l it can carry\",round(w,3),\"KN-m\"\n",
+ "print\"MAx concentrated Load P apllied at 2 m from one end is\",round(P,3),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Intensity of u.d.l it can carry 5.0 N-mm\n",
+ "MAx concentrated Load P apllied at 2 m from one end is 13.021 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.2,Page no.131"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=70 #mm #External Diameter\n",
+ "t=8 #mm #Thickness of pipe\n",
+ "L=2500 #mm #span \n",
+ "sigma=150 #N/mm**2 #stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Internal Diameter \n",
+ "d=D-2*t #mm\n",
+ "\n",
+ "#M.I Of Pipe\n",
+ "I=pi*64**-1*(D**4-d**4) #mm**4\n",
+ "\n",
+ "y_max=D*2**-1 #mm\n",
+ "Z=I*(y_max)**-1 #mm**3\n",
+ "\n",
+ "#Moment Carrying capacity\n",
+ "M=sigma*Z #N*mm\n",
+ "\n",
+ "#Max moment int the beam occurs at the mid-span and is equal to\n",
+ "#m=P*L*4**-1\n",
+ "\n",
+ "#Equating Max moment to moment carrying capacity we get,\n",
+ "#M=P*2.5*L*4**-1\n",
+ "#After substituting and simplifying we get\n",
+ "P=4*M*(L)**-1*10**-3 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Max concentrated load that can be applied at the centre of span is\",round(P,3),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max concentrated load that can be applied at the centre of span is 5.22 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.3,Page no.132"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flanges Dimension\n",
+ "b1=180 #mm #Width\n",
+ "d1=10 #mm #Thickness\n",
+ "\n",
+ "D=500 #mm #Overall depth\n",
+ "t=8 #mm #Thickness of web\n",
+ "\n",
+ "#Plate Dimensions\n",
+ "b2=240 #mm #Width\n",
+ "t2=12 #mm #Thickness\n",
+ "\n",
+ "sigma=150 #N/mm**2 #Stress\n",
+ "L=3000 #mm #span\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of centroid from bottom fibre\n",
+ "y_bar=(b2*t2*(D+t2*2**-1)+b1*d1*(D-t1*2**-1)+(D-2*t1)*t*D*2**-1+(b1*t1*t1*2**-1))*(b2*t2+b1*d1+b1*d1+(D-2*d1)*t)**-1\n",
+ "\n",
+ "#M.I of section\n",
+ "I=(1*12**-1*b2*t2**3+b2*t2*(D+t2*2**-1-y_bar)**2+1*12**-1*b1*d1**3+b1*d1*(D-t1*2**-1-y_bar)**2+1*12**-1*b1*t1**3+b1*t1*(t1*2**-1-y_bar)**2+1*12**-1*t*(D-2*t1)**3+t*(D-2*t1)*(D*2**-1-y_bar)**2)\n",
+ "\n",
+ "#Section Modulus\n",
+ "Z=I*(y_bar)**-1 #mm**3\n",
+ "\n",
+ "#Moment or Resistance\n",
+ "M=sigma*Z\n",
+ "\n",
+ "#Let Load on Cantilever be w/m Length \n",
+ "#Max M.I produced\n",
+ "#M_max=w*L**2**-1 \n",
+ "\n",
+ "#Now Equating Moment of resistance to Max moment,we get Max load\n",
+ "#4.5*w=M\n",
+ "#After rearranging and further simplifying we get\n",
+ "w=M*4.5**-1*10**3*10**-9\n",
+ "\n",
+ "#Result\n",
+ "print\"Moment of Resistance is\",round(M,2),\"KN-mm\"\n",
+ "print\"Load the section can carry is\",round(w,3),\"KN/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Moment of Resistance is 198770121.83 KN-mm\n",
+ "Load the section can carry is 44.171 KN/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.4,Page no.134"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flange (Top)\n",
+ "b1=80 #mm #Width \n",
+ "t1=40 #mm #Thickness\n",
+ "\n",
+ "#Flange (Bottom)\n",
+ "b2=160 #mm #width\n",
+ "t2=40 #mm #Thickness\n",
+ "\n",
+ "#web\n",
+ "d=120 #mm #Depth\n",
+ "t3=20 #mm #Thickness\n",
+ "\n",
+ "D=200 #mm #Overall Depth\n",
+ "sigma1=30 #N/mm**2 #Tensile stress\n",
+ "sigma2=90 #N/mm**2 #Compressive stress\n",
+ "L=6000 #mm #Span\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of centroid from bottom fibre\n",
+ "y_bar=(b1*t1*(D-t1*2**-1)+d*t3*(d*2**-1+t2)+b2*t2*t2*2**-1)*(b1*t1+d*t3+b2*t2)**-1 #mm\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*b1*t1**3+b1*t1*(D-t1*2**-1-round(y_bar,2))**2+1*12**-1*t3*d**3+t3*d*(d*2**-1+t2-round(y_bar,2))**2+1*12**-1*b2*t2**3+b2*t2*(t2*2**-1-round(y_bar,2))**2\n",
+ "\n",
+ "#Extreme fibre distance of top and bottom fibres are y_t and y_c respectively\n",
+ "\n",
+ "y_t=y_bar #mm\n",
+ "y_c=D-y_bar #mm\n",
+ "\n",
+ "#Moment carrying capacity considering Tensile strength \n",
+ "M1=sigma1*I*y_t**-1*10**-6 #KN-m\n",
+ "\n",
+ "#Moment carrying capacity considering compressive strength \n",
+ "M2=sigma2*I*y_c**-1*10**-6 #KN-m\n",
+ "\n",
+ "#Max Bending moment in simply supported beam 6 m due to u.d.l\n",
+ "#M_max=w*L*10**-3*8**-1\n",
+ "#After simplifying further we get\n",
+ "#M_max=4.5*w\n",
+ "\n",
+ "#Now Equating it to Moment carrying capacity, we get load carrying capacity\n",
+ "w=M1*4.5**-1 #KN/m\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Uniformly Distributed Load is\",round(w,3),\"KN/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Uniformly Distributed Load is 5.096 KN/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.5,Page no.136"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from scipy.integrate import quad\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flanges\n",
+ "b=200 #mm #Width\n",
+ "t=25 #mm #Thickness \n",
+ "\n",
+ "D1=500 #mm #Overall Depth\n",
+ "t2=20 #mm #Thickness of web\n",
+ "\n",
+ "d=450 #mm #Depth of web\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Consider,Element of Thickness \"y\" at Distance \"dy\" from N.A \n",
+ "#Let Bending stress \"sigma_max\"\n",
+ "\n",
+ "#Stress on the element \n",
+ "#sigma=y*(D*2**-1)*sigma_max ..............(1)\n",
+ "\n",
+ "#Area of Element\n",
+ "#A=b*dy .................................(2)\n",
+ "\n",
+ "#Force on Element \n",
+ "#F=y*250**-1*sigma_max*b*dy\n",
+ "\n",
+ "#Let M be the Moment of resistance\n",
+ "#M=y*250**-1*sigma_max*b*dy*y\n",
+ "\n",
+ "#Moment of Resistance of top flange be M1\n",
+ "def integrand(y, b, D):\n",
+ " return b*y**2*D**-1\n",
+ "b=200 \n",
+ "D=250\n",
+ "\n",
+ "X = quad(integrand, 225, 250, args=(b,D))\n",
+ "\n",
+ "Y=2*X[0]\n",
+ "\n",
+ "#M1=Y*sigma\n",
+ "\n",
+ "#Now Moment of Inertia I section is\n",
+ "X=b*D1**3\n",
+ "Y=(b-t2)*d**3\n",
+ "I=(X-Y)*12**-1*10**-8\n",
+ "\n",
+ "#Moment acting on the entire section\n",
+ "#since sigmais the value at y=250\n",
+ "y_max=250\n",
+ "Z=I*10**8*y_max**-1\n",
+ "#M=sigma*Z \n",
+ "#After Simplifying Further we get\n",
+ "#M2=Z*sigma\n",
+ "\n",
+ "#Percentage Moment resisted by Flanges\n",
+ "P1=2258333.3*(2865833.3)**-1*100\n",
+ "\n",
+ "#Percentage Moment resisted by web\n",
+ "P2=100-P1\n",
+ "\n",
+ "#Result\n",
+ "print\"Percentage Moment resisted by Flanges\",round(P1,2),\"%\"\n",
+ "print\"Percentage Moment resisted by web\",round(P2,2),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percentage Moment resisted by Flanges 78.8 %\n",
+ "Percentage Moment resisted by web 21.2 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.6,Page no.137"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flanges\n",
+ "b1=200 #mm #Width\n",
+ "t1=10 #mm #Thickness\n",
+ "\n",
+ "#Web\n",
+ "d=380 #mm #Depth \n",
+ "t2=8 #mm #Thickness\n",
+ "\n",
+ "D=400 #mm #Overall Depth\n",
+ "sigma=150 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Area\n",
+ "A=b1*t1+d*t2+b1*t1 #mm**2\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*(b1*D**3-(b1-t2)*d**3)\n",
+ "\n",
+ "#Bending Moment\n",
+ "M=sigma*I*(D*2**-1)**-1\n",
+ "\n",
+ "#Square Section\n",
+ "\n",
+ "#Let 'a' be the side\n",
+ "a=A**0.5\n",
+ "\n",
+ "#Moment of Resistance of this section\n",
+ "M1=1*6**-1*a*a**2*sigma\n",
+ "\n",
+ "X=M*M1**-1\n",
+ "\n",
+ "#Rectangular section\n",
+ "#Let 'a' be the side and depth be 2*a\n",
+ "\n",
+ "a=(A*2**-1)**0.5\n",
+ "\n",
+ "#Moment of Rectangular secction\n",
+ "M2=1*6**-1*a*(2*a)**2*sigma\n",
+ "\n",
+ "X2=M*M2**-1\n",
+ "\n",
+ "#Circular section\n",
+ "#A=pi*d1**2*4**-1\n",
+ "\n",
+ "d1=(A*4*pi**-1)**0.5\n",
+ "\n",
+ "#Moment of circular section\n",
+ "M3=pi*32**-1*d1**3*sigma\n",
+ "\n",
+ "X3=M*M3**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Moment of resistance of beam section\",round(M,2),\"mm\"\n",
+ "print\"Moment of resistance of square section\",round(X,2),\"mm\"\n",
+ "print\"Moment of resistance of rectangular section\",round(X2,2),\"mm\"\n",
+ "print\"Moment of resistance of circular section\",round(X3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Moment of resistance of beam section 141536000.0 mm\n",
+ "Moment of resistance of square section 9.58 mm\n",
+ "Moment of resistance of rectangular section 6.78 mm\n",
+ "Moment of resistance of circular section 11.33 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.7,Page no.139"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=12 #KN #Force at End of beam\n",
+ "L=2 #m #span\n",
+ "\n",
+ "#Square section \n",
+ "b=d=200 #mm #Width and depth of beam\n",
+ "\n",
+ "#Rectangular section\n",
+ "b1=150 #mm #Width\n",
+ "d1=300 #mm #Depth\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max bending Moment\n",
+ "M=F*L*10**6 #N-mm\n",
+ "\n",
+ "#M=sigma*b*d**2\n",
+ "sigma=M*6*(b*d**2)**-1 #N/mm**2\n",
+ "\n",
+ "#Let W be the central concentrated Load in simply supported beam of span L1=3 m\n",
+ "#MAx Moment\n",
+ "#M1=W*L1*4**-1\n",
+ "#After Further simplifying we get\n",
+ "#M1=0.75*10**6 #N-mm\n",
+ "\n",
+ "#The section has a moment of resistance\n",
+ "M1=sigma*1*6**-1*b1*d1**2\n",
+ "\n",
+ "#Equating it to moment of resistance we get max load W\n",
+ "#0.75*10**6*W=M1\n",
+ "#After Further simplifying we get\n",
+ "W=M1*(0.75*10**6)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum Concentrated Load required to brek the beam\",round(W,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum Concentrated Load required to brek the beam 54.0 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.8,Page no.140"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=3 #m #span\n",
+ "sigma_t=35 #N/mm**2 #Permissible stress in tension\n",
+ "sigma_c=90 #N/mm**2 #Permissible stress in compression\n",
+ "\n",
+ "#Flanges\n",
+ "t=30 #mm #Thickness\n",
+ "d=250 #mm #Depth\n",
+ "\n",
+ "#Web\n",
+ "t2=25 #mm #Thickness\n",
+ "b=600 #mm #Width\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let y_bar be the Distance of N.A from Extreme Fibres\n",
+ "y_bar=(t*d*d*2**-1*2+(b-2*t)*t2*t2*2**-1)*(t*d*2+(b-2*t)*t2)**-1\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=(1*12**-1*t*d**3+t*d*(d*2**-1-y_bar)**2)*2+1*12**-1*(b-2*t)*t2**3+(b-2*t)*t2*(t2*2**-1-y_bar)**2\n",
+ "\n",
+ "#Part-1\n",
+ "\n",
+ "#If web is in Tension\n",
+ "y_t=y_bar #mm\n",
+ "y_c=d-y_bar #mm\n",
+ "\n",
+ "#Moment carrying caryying capacity From consideration of tensile stress\n",
+ "M=sigma_t*I*(y_bar)**-1 #N-mm\n",
+ "\n",
+ "#Moment carrying caryying capacity From consideration of compressive stress\n",
+ "M1=sigma_c*I*(y_c)**-1 #N-mm\n",
+ "\n",
+ "#If w KN/m is u.d.l in beam,Max bending moment\n",
+ "#M=wl**2*8**-1\n",
+ "#After further simplifyng we get\n",
+ "#M=1.125*w*10**6 N-mm\n",
+ "w=M*(1.125*10**6)**-1 #KN\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#If web is in compression\n",
+ "y_t2=178.299 #mm\n",
+ "y_c2=71.71 #mm \n",
+ "\n",
+ "#Moment carrying caryying capacity From consideration of tensile stress\n",
+ "M2=sigma_t*I*(y_t2)**-1 #N-mm\n",
+ "\n",
+ "#Moment carrying caryying capacity From consideration of compressive stress\n",
+ "M3=sigma_c*I*(y_c2)**-1 #N-mm\n",
+ "\n",
+ "#Moment of resistance is M2\n",
+ "\n",
+ "#Equating it to bending moment we get\n",
+ "#M2=1.125*10**6*w2\n",
+ "#After further simplifyng we get\n",
+ "w2=M2*(1.125*10**6)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Uniformly Distributed Load carrying capacity if:web is in Tension\",round(w,2),\"KN\"\n",
+ "print\" :web is in compression\",round(w2,3),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Uniformly Distributed Load carrying capacity if:web is in Tension 73.21 KN\n",
+ " :web is in compression 29.446 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.9,Page no.141"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b1=200 #mm #Width at base\n",
+ "b2=100 #mm #Width at top\n",
+ "\n",
+ "L=8 #m Length\n",
+ "P=500 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Consider a section at y metres from top\n",
+ "\n",
+ "#At this section diameter d is\n",
+ "#d=b2+y*L**-1*(b1-b2)\n",
+ "#After Further simplifying we get\n",
+ "#d=b2+12.5*y #mm\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "#I=pi*64**-1*d**4\n",
+ "\n",
+ "#Section Modulus \n",
+ "#Z=pi*32**-1*(b1+12.5*y)**3\n",
+ "\n",
+ "#Moment \n",
+ "#M=5*10**5*y #N-mm\n",
+ "\n",
+ "#Let sigma be the fibre stress at this section then\n",
+ "#M=sigma*Z\n",
+ "#After sub values in above equation and further simplifying we get\n",
+ "#sigma=5*10**5*32*pi**-1*y*((b2+12.5*y)**3)**-1\n",
+ "\n",
+ "#For sigma to be Max,d(sigma)*(dy)**-1=0\n",
+ "#16*10**6*pi**-1*((b2+12.5*y)**-3+y*(-3)*(b2+12.5*y)**-4*12.5)\n",
+ "#After Further simplifying we get\n",
+ "#b2+12.5*y=37.5*y\n",
+ "#After Further simplifying we get\n",
+ "y=b2*25**-1 #m\n",
+ "\n",
+ "#Stress at this section\n",
+ "sigma=5*10**5*32*pi**-1*y*((b2+12.5*y)**3)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress at Extreme Fibre is max\",round(y,2),\"m\"\n",
+ "print\"Max stress is\",round(sigma,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress at Extreme Fibre is max 4.0 m\n",
+ "Max stress is 6.04 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.10,Page no.143"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "H=10 #mm #Height\n",
+ "A1=160*160 #mm**2 #area of square section at bottom\n",
+ "L1=160 #mm #Length of square section at bottom\n",
+ "b1=160 #mm #width of square section at bottom\n",
+ "A2=80*80 #mm**2 #area of square section at top\n",
+ "L2=80 #mm #Length of square section at top\n",
+ "b2=80 #mm #Width of square section at top\n",
+ "P=100 #N #Pull\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Consider a section at distance y from top.\n",
+ "#Let the side of square bar be 'a'\n",
+ "#a=L2+y*(H)**-1*(b1-b2)\n",
+ "#After further simplifying we get\n",
+ "#a=L2+8*y\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "#I=2*1*12**-1*a*(2)**0.5*(a*((2)**0.5)**-1)**3\n",
+ "#After further simplifying we get\n",
+ "#I=a**4*12**-1\n",
+ "\n",
+ "#Section Modulus \n",
+ "#Z=a**4*(12*a*(2)**0.5)**-1\n",
+ "#After further simplifying we get\n",
+ "#Z=2**0.5*a**3*(12)**-1 #mm**3\n",
+ "\n",
+ "#Bending moment at this section=100*y N-mm\n",
+ "#M=100*10**3*y #N-mm\n",
+ "\n",
+ "#But\n",
+ "#M=sigma*Z\n",
+ "#After sub values in above equation we get\n",
+ "#sigma=M*Z**-1\n",
+ "#After further simplifying we get\n",
+ "#sigma=1200*10**3*(2**0.5)**-1*y*((80+80*y)**3)**-1 .......(1)\n",
+ "\n",
+ "#For Max stress df*(dy)**-1=0\n",
+ "#After taking Derivative of above equation we get\n",
+ "#df*(dy)**-1=1200*10**3*(2**0.5)**-1*((80+8*y)**-3+y(-3)*(80+8*y)**-4*8)\n",
+ "#After further simplifying we get\n",
+ "y=80*16**-1 #m\n",
+ "\n",
+ "#Max stress at this level is\n",
+ "sigma=1200*10**3*(2**0.5)**-1*y*((80+8*y)**3)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Bending stress is Developed at\",round(y,3),\"m\"\n",
+ "print\"Value of Max Bending stress is\",round(sigma,3),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Bending stress is Developed at 5.0 m\n",
+ "Value of Max Bending stress is 2.455 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.12,Page no.147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b=200 #mm #Width of timber \n",
+ "d=400 #mm #Depth of timber\n",
+ "t=6 #mm #Thickness\n",
+ "b2=200 #mm #width of steel plate\n",
+ "t2=20 #mm #Thickness of steel plate\n",
+ "M=40*10**6 #KN-mm #Moment\n",
+ "#Let E_s*E_t**-1=X\n",
+ "X=20 #Ratio of Modulus of steel to timber\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#let y_bar be the Distance of centroidfrom bottom most fibre\n",
+ "y_bar=(b*d*(b+t)+t2*b2*t*t*2**-1)*(b*d+t2*b2*t)**-1 #mm\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*b*d**3+b*d*(b+t-round(y_bar,3))**2+1*12**-1*t2*b2*t**3+b2*t2*t*(round(y_bar,3)-t*2**-1)**2\n",
+ "\n",
+ "#distance of the top fibre from N-A\n",
+ "y_1=d+t-y_bar #mm\n",
+ "\n",
+ "#Distance of the junction of timber and steel From N-A\n",
+ "y_2=y_bar-t #mm\n",
+ "\n",
+ "#Stress in Timber at the top\n",
+ "Y=M*I**-1*y_1 #N/mm**2\n",
+ "\n",
+ "#Stress in the Timber at the junction point\n",
+ "Z=M*I**-1*y_2\n",
+ "\n",
+ "#Coressponding stress in steel at the junction point\n",
+ "Z2=X*Z #N/mm**2 \n",
+ "\n",
+ "#The stress in Extreme steel fibre \n",
+ "Z3=X*M*I**-1*y_bar\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress in Extreme steel Fibre\",round(Z3,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in Extreme steel Fibre 69.67 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.13,Page no.149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Timber size\n",
+ "b=150 #mm #Width\n",
+ "b2=120 #mm \n",
+ "d=300 #mm #Depth\n",
+ "\n",
+ "t=6 #mm #Thickness of steel plate\n",
+ "L=6 #m #Span\n",
+ "\n",
+ "#E_s*E_t**-1=20 \n",
+ "#X=E_s*E_t**-1\n",
+ "X=20 \n",
+ "sigma_timber=8 #N/mm**2 #Stress in timber\n",
+ "sigma_steel=150 #N/mm**2 #Stress in steel plate\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "Y\n",
+ "\n",
+ "#Due to synnetry cenroid,the neutral axis is half the depth\n",
+ "I=1*12**-1*\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print Z"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "153.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.14,Page no.151"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=6000 #mm #Span of beam\n",
+ "W=20*10**3 #N #Load\n",
+ "sigma=8 #N/mm**2 #Stress\n",
+ "b=200 #mm #Width of section\n",
+ "d=300 #mm #Depth of section\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#let x be the distance from left side of beam\n",
+ "\n",
+ "#Bending moment\n",
+ "#M=W*2**-1*x #Nmm .......(1)\n",
+ "\n",
+ "#But M=sigma*Z ..........(2)\n",
+ "\n",
+ "#Equating equation 1 and 2 we get\n",
+ "#W*2**-1*x=sigma*Z ............(3)\n",
+ "\n",
+ "#Section Modulus \n",
+ "#Z=1*6*b*d**2 ...............(4)\n",
+ "\n",
+ "#Equating equation 3 and 4 we get\n",
+ "#b*d**2=3*W*x*sigma**-1 .............(5)\n",
+ "\n",
+ "#Beam of uniform strength of constant depth\n",
+ "#b=3*W*x*(sigma*d**2) \n",
+ "\n",
+ "#When x=0\n",
+ "b=0\n",
+ "\n",
+ "#When x=L*2**-1\n",
+ "b2=3*W*L*(2*sigma*d**2)**-1 #mm\n",
+ "\n",
+ "#Beam with constant width of 200 mm\n",
+ "\n",
+ "#We have\n",
+ "#d=(3*W*x*(sigma*d)**-1)**0.5\n",
+ "#thus depth varies as (x)**0.5\n",
+ "\n",
+ "#when x=0\n",
+ "d1=0\n",
+ "\n",
+ "#when x=L*2**-1\n",
+ "d2=(2*W*L*(2*sigma*300)**-1)**0.5 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Cross section of rectangular beam is:\",round(b2,2),\"mm\"\n",
+ "print\" :\",round(d2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Cross section of rectangular beam is: 250.0 mm\n",
+ " : 223.61 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.15,Page no.154"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=800 #mm #Span\n",
+ "n=5 #number of leaves\n",
+ "b=60 #mm #Width\n",
+ "t=10 #mm #thickness\n",
+ "sigma=250 #N/mm**2 #Stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#section Modulus\n",
+ "Z=n*6**-1*b*t**2 #mm**3\n",
+ "\n",
+ "#from the relation\n",
+ "#sigma*Z=M ...................(1)\n",
+ "#M=P*L*4**-1\n",
+ "#sub values of M in equation 1 we get\n",
+ "P=sigma*Z*4*L**-1*10**-3 #KN #Load\n",
+ "\n",
+ "#Length of Leaves\n",
+ "L1=0.2*L #mm\n",
+ "L2=0.4*L #mm\n",
+ "L3=0.6*L #mm\n",
+ "L4=0.8*L #mm\n",
+ "L5=L #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Load it can take is\",round(P,2),\"KN\"\n",
+ "print\"Length of leaves:L1\",round(L1,2),\"mm\"\n",
+ "print\" :L2\",round(L2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Load it can take is 6.25 KN\n",
+ "Length of leaves:L1 160.0 mm\n",
+ " :L2 320.0 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.16,Page no.161"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=20*10**3 #N #Shear Force\n",
+ "\n",
+ "#Tee section\n",
+ "\n",
+ "#Flange\n",
+ "b=100 #mm #Width\n",
+ "t=12 #mm #Thickness\n",
+ "\n",
+ "#Web\n",
+ "d=88 #mm #Depth\n",
+ "t2=12 #mm #Thicknes\n",
+ "\n",
+ "D=100 #mm #Overall Depth\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of C.G from Top Fibre\n",
+ "y=(b*t*t*2**-1+t2*d*(d*2**-1+t))*(b*t+d*t2)**-1 #mm \n",
+ "\n",
+ "#Moment Of Inertia\n",
+ "I=1*12**-1*b*t**3+b*t*(y-t*2**-1)**2+1*12**-1*t2*d**3+t2*d*(t+d*2**-1-y)**2 #mm**4\n",
+ "\n",
+ "#shear stress at bottom Flange\n",
+ "\n",
+ "#Area above this level\n",
+ "A=b*t #mm**2\n",
+ "\n",
+ "#C.G of this area from N-A\n",
+ "y2=y-t*2**-1\n",
+ "\n",
+ "#Stress at bottom of flange\n",
+ "sigma=F*A*y2*(b*I)**-1 #N/mm**2 \n",
+ "\n",
+ "#sigma2 at same level but in web where width is 12 mm\n",
+ "sigma2=F*A*y2*(t2*I)**-1 #N/mm**2 \n",
+ "\n",
+ "#To find shear stress at N-A\n",
+ "X=t*b*(y-t*2**-1)+t2*(y-t2)*(y-t2)*2**-1 #mm**3\n",
+ "\n",
+ "sigma3=F*X*(t2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Shear stress at top and bottom fibre is zero\n",
+ "#sigma4 and sigma5 are top and bottom fibre shear stress\n",
+ "sigma4=sigma5=0\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,t,t,y,D]\n",
+ "Y1=[sigma4,sigma,sigma2,sigma3,sigma5]\n",
+ "Z1=[0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x56a6270>"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.17,Page no.163"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=40*10**3 #N #shear Force\n",
+ "\n",
+ "#I-section\n",
+ "\n",
+ "#Flanges\n",
+ "b=80 #mm #Width of flange\n",
+ "t=20 #mm #Thickness\n",
+ "\n",
+ "#Web\n",
+ "d=200 #mm #Depth\n",
+ "t2=20 #mm #Thickness\n",
+ "\n",
+ "#Flange-2\n",
+ "b2=160 #mm #Width\n",
+ "t3=20 #mm #Thickness\n",
+ "\n",
+ "D=240 #mm #Overall Depth\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of N-A from Top Fibre \n",
+ "y=(b*t*t*2**-1+d*t2*(t+d*2**-1)+b2*t3*(t+d+t3*2**-1))*(b*t+d*t2+b2*t3)**-1 #mm\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*b*t**3+b*t*(y-(t*2**-1))**2+1*12**-1*t2*d**3+t2*d*(y-(t+d*2**-1))**2+1*12**-1*b2*t3**3+t3*b2*((d+t+t3*2**-1)-y)**2 #mm**4\n",
+ "\n",
+ "#Shear stress bottom of flange\n",
+ "sigma=F*b*t*(y-t*2**-1)*(b*I)**-1 #N/mm**2\n",
+ "\n",
+ "#At same Level but in web\n",
+ "sigma2=F*b*t*(y-t*2**-1)*(t2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#for shear stress at N.A\n",
+ "X=b*t*(y-t*2**-1)+t2*(y-t)*(y-t)*2**-1 #mm**3\n",
+ "sigma3=F*X*(t2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Shear stress at bottom of web\n",
+ "\n",
+ "X=b2*t3*((D-y)-t3*2**-1) #mm**3\n",
+ "\n",
+ "#Stress at bottom of web\n",
+ "sigma4=F*X*(t2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Stress at Lower flange\n",
+ "sigma5=F*X*(b2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force Diagram is the result\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,20,20,140,220,220,240]\n",
+ "Y1=[0,sigma,sigma2,sigma3,sigma4,sigma5,0]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length in mm\")\n",
+ "plt.ylabel(\"Shear Force in N\")\n",
+ "plt.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force Diagram is the result\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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OnYL/+i/TF3DLLWZPgXbt7K5KRHyRwsAGJ0+afoBXXjHNQOvXQ1iY3VWJiC8r1gj183MN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\n/w/qZz1xEBCKMwAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5614110>"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.18,Page no.164"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=30*10**3 #N #Shear Force\n",
+ "\n",
+ "#Channel Section\n",
+ "d=400 #mm #Depth of web \n",
+ "t=10 #mm #THickness of web\n",
+ "t2=15 #mm #Thickness of flange\n",
+ "b=100 #mm #Width of flange\n",
+ "\n",
+ "#Rectangular Welded section\n",
+ "b2=80 #mm #Width\n",
+ "d2=60 #mm #Depth\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of Centroid From Top Fibre\n",
+ "y=(d*t*t*2**-1+2*t2*(b-t)*((b-t)*2**-1+10)+d2*b2*(d2*2**-1+t))*(d*t+2*t2*(b-t)+d2*b2)**-1 #mm\n",
+ "\n",
+ "#Moment Of Inertia of the section about N-A\n",
+ "I=1*12**-1*d*t**3+d*t*(y-t*2**-1)**2+2*(1*12**-1*t2*(b-t)**3+t2*(b-t)*(((b-t)*2**-1+t)-y)**2)+1*12**-1*d2**3*b2+d2*b2*(d2*2**-1+t-y)**2\n",
+ "\n",
+ "#Shear stress at level of weld\n",
+ "sigma=F*d*t*(y-t*2**-1)*((b2+t2+t2)*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Max Shear Stress occurs at Neutral Axis\n",
+ "X=d*t*(y-t*2**-1)+2*t2*(y-t)*(y-t)*2**-1+b2*(y-t)*(y-t)*2**-1\n",
+ "\n",
+ "sigma_max=F*X*((b+t)*I)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Shear stress in the weld is\",round(sigma,2),\"N/mm**2\"\n",
+ "print\"Max shear stress is\",round(sigma_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shear stress in the weld is 3.62 N/mm**2\n",
+ "Max shear stress is 4.48 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.19,Page no.165"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Wooden Section\n",
+ "b=300 #mm #Width\n",
+ "d=300 #mm #Depth\n",
+ "\n",
+ "D=100 #mm #Diameter of Bore\n",
+ "F=10*10**3 #N #Shear Force\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Moment Of Inertia Of Section\n",
+ "I=1*12**-1*b*d**3-pi*64**-1*D**4\n",
+ "\n",
+ "#Shear stress at crown of circle\n",
+ "sigma=F*b*D*(d*2**-1-D*2**-1)*(b*I)**-1\n",
+ "\n",
+ "#Let a*y_bar=X\n",
+ "X=b*d*2**-1*d*4**-1-pi*8**-1*D**2*4*D*2**-1*(3*pi)**-1 #mm**3\n",
+ "\n",
+ "#Shear Stress at Neutral Axis\n",
+ "sigma2=F*X*((b-D)*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Shearing Stress at Crown of Bore\",round(sigma,3),\"N/mm**2\"\n",
+ "print\"Shear Stress at Neutral Axis\",round(sigma2,3),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shearing Stress at Crown of Bore 0.149 N/mm**2\n",
+ "Shear Stress at Neutral Axis 0.246 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.20,Page no.166"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#flanges\n",
+ "b=200 #mm #width\n",
+ "t1=25 #mm #Thickness\n",
+ "\n",
+ "#web\n",
+ "d=450 #mm #Depth \n",
+ "t2=20 #mm #thickness\n",
+ "\n",
+ "D=500 #mm #Total Depth of section\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Moment Of Inertia of the section about N-A\n",
+ "I=1*12**-1*b*D**3-1*12**-1*(b-t2)*d**3 #mm**4\n",
+ "\n",
+ "#Consider an element in the web at distance y from y from N-A\n",
+ "#Depth of web section=225-y\n",
+ "\n",
+ "#C.G From N-A\n",
+ "#y2=y+(((D*2**-1-t)-y)*2**-1)\n",
+ "\n",
+ "#ay_bar for section at y\n",
+ "#Let ay_bar be X\n",
+ "#X=X1 be of Flange + X2 be of web above y\n",
+ "#X=b*t1*(D*2**-1-t1*2**-1)+t2*(d-t1)*(d-t1+y)*2**-1\n",
+ "#After Sub values and Further simplifying we get\n",
+ "#X=1187500+10*(225**2-y**2)\n",
+ "\n",
+ "#Shear stress at y\n",
+ "#sigma_y=F*(X)*(t2*I)**-1\n",
+ "\n",
+ "#Shear Force resisted by the Element\n",
+ "#F1=F*X*t2*dy*(t2*I)**-1\n",
+ "\n",
+ "#Shear stress resisted by web \n",
+ "#sigma=2*F*I**-1*(X)*dy\n",
+ "\n",
+ "#After Integrating above equation and further simplifying we get\n",
+ "#sigma=0.9578*F\n",
+ "\n",
+ "sigma=0.9578*100\n",
+ "\n",
+ "#Result\n",
+ "print\"Shear Resisted by web\",round(sigma,2),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shear Resisted by web 95.78 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.21,Page no.167"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Wooden Beam\n",
+ "\n",
+ "b=150 #mm #width\n",
+ "d=250 #mm #Depth\n",
+ "\n",
+ "L=5000 #mm #span\n",
+ "m=11.2 #N/mm**2 #Max Bending stress\n",
+ "sigma=0.7 #N/mm**2 #Max shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let 'a' be the distance from left support\n",
+ "#Max shear force\n",
+ "#F=R_A=W*(L-a)*L**-1 \n",
+ "\n",
+ "#Max Moment\n",
+ "#M=W*(L-a)*a*L**-1\n",
+ "\n",
+ "#But M=sigma*Z\n",
+ "#W*(L-a)*a*L**-1=m*1*6**-1*b*d**2 .....................(1)\n",
+ "\n",
+ "#In Rectangular Section MAx stress is 1.5 times Avg shear stress\n",
+ "F=sigma*b*d*1.5**-1\n",
+ "\n",
+ "#W*(L-a)*L**-1=F .....................(2)\n",
+ "\n",
+ "#Dividing Equation 1 nad 2 we get\n",
+ "a=m*6**-1*b*d**2*1.5*(sigma*b*d)**-1\n",
+ "\n",
+ "#Sub above value in equation 2 we get\n",
+ "W=(L-a)**-1*L*F*10**-3 #KN \n",
+ "\n",
+ "#Result\n",
+ "print\"Load is\",round(W,2),\"KN\"\n",
+ "print\"Distance from Left support is\",round(a,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Load is 21.87 KN\n",
+ "Distance from Left support is 1000.0 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.22,Page no.168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1000 #mm #span\n",
+ "\n",
+ "#Rectangular Section\n",
+ "\n",
+ "b=200 #mm #width\n",
+ "d=400 #mm #depth\n",
+ "\n",
+ "sigma=1.5 #N/mm**2 #Shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let AB be the cantilever beam subjected to load W KN at free end\n",
+ "\n",
+ "#MAx shear Force\n",
+ "#F=W*10**3 #KN\n",
+ "\n",
+ "#Since Max shear stress in Rectangular section\n",
+ "#sigma_max=1.5*F*A**-1 \n",
+ "#After sub values and further simplifyng we get\n",
+ "W=1.5*b*d*(1.5*1000)**-1 #KN\n",
+ "\n",
+ "#Moment at fixwed end\n",
+ "M=W*1 #KN-m\n",
+ "y_max=d*2**-1 #mm\n",
+ "\n",
+ "#M.I\n",
+ "I=1*12**-1*b*d**3 #mm**3\n",
+ "\n",
+ "#MAx Stress\n",
+ "sigma_max=M*10**6*I**-1*y_max\n",
+ "\n",
+ "#Result\n",
+ "print\"Concentrated Load is\",round(sigma_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Concentrated Load is 15.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.24,Page no.170"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=4000 #mm #span\n",
+ "\n",
+ "#Rectangular Cross-section\n",
+ "b=100 #mm #Width\n",
+ "d=200 #mm #Thickness\n",
+ "\n",
+ "F_per=10 #N/mm**2 #Max Bending stress\n",
+ "q_max=0.6 #N/mm**2 #Shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#If the Load W is in KN/m\n",
+ "\n",
+ "#Max shear Force\n",
+ "#F=w*l*2**-1 #KN\n",
+ "#After substituting values and further simplifying we get\n",
+ "#M=2*w #KN-m\n",
+ "\n",
+ "#Max Load from Consideration of moment\n",
+ "#M=1*6**-1*b*d**2*F_per\n",
+ "#After substituting values and further simplifying we get\n",
+ "w=(1*6**-1*b*d**2*F_per)*(2*10**6)**-1 #KN/m\n",
+ "\n",
+ "#Max Load from Consideration of shear stress\n",
+ "#q_max=1.5*F*(b*d)**-1 #N\n",
+ "#After substituting values and further simplifying we get\n",
+ "F=q_max*(1.5)*b*d #N\n",
+ "\n",
+ "#If w is Max Load in KN/m,then\n",
+ "#2*w*1000=8000\n",
+ "#After Rearranging and Further simplifying we get\n",
+ "w2=8000*(2*1000)**-1 #KN/m\n",
+ "\n",
+ "#Result\n",
+ "print\"Uniformly Distributed Load Beam can carry is\",round(w,2),\"KN/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Uniformly Distributed Load Beam can carry is 3.33 KN/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.5_3.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.5_3.ipynb new file mode 100644 index 00000000..8a54f301 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.5_3.ipynb @@ -0,0 +1,1013 @@ +{
+ "metadata": {
+ "name": "chapter no.5.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 5:Deflections Of Beams By Double Integration Methods"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.2,Page No.192"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=3000 #mm #span of beam\n",
+ "a=2000 #mm\n",
+ "W1=20*10**3 #N #Pt Load Acting on beam\n",
+ "W2=30*10**3 #N #Pt Load Acting on beam\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus\n",
+ "I=2*10**8 #mm**4 #M.I\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Deflection at free End Due to W2\n",
+ "dell1=W2*L**3*(3*E*I)**-1 #mm\n",
+ "\n",
+ "#Deflection at free end Due to W1\n",
+ "dell2=W1*a**3*(3*E*I)**-1+(L-a)*W1*a**2*(2*E*I)**-1 #mm\n",
+ "\n",
+ "#Total Deflection at free end\n",
+ "dell=dell1+dell2 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflection at Free End is\",round(dell,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflection at Free End is 9.08 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.4,Page No.193"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus\n",
+ "I=180*10**6 #mm**4 #M.I\n",
+ "W1=20 #N/m #u.d.l\n",
+ "W2=20*10**3 #N #Pt load\n",
+ "L=3000 #m #Span of beam\n",
+ "a=2000 #m #Span of u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Displacement of free End due to 20 KN Pt load at free end\n",
+ "dell1=W2*L**3*(3*E*I)**-1 #mm\n",
+ "\n",
+ "#Displacement of free end due to u.d.l\n",
+ "dell2=W1*a**4*(8*E*I)**-1+(L-a)*W1*a**3*(6*E*I)**-1\n",
+ "\n",
+ "#Deflection at free end\n",
+ "dell=dell1+dell2 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"The Displacement of Free End of cantilever beam is\",round(dell,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Displacement of Free End of cantilever beam is 6.85 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.10,Page No.201"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=200*10**6 #KN/m**2 #Young's Modulus\n",
+ "I=15*10**-6 #m**4 #M.I\n",
+ "a=4000 #m \n",
+ "L_AB=6 #m #Span of beam\n",
+ "L_CB=2 #m #Length of CB\n",
+ "F_C=18 #KN #force at C\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the Reactions at A & B Respectively\n",
+ "#V_A+V_B=18\n",
+ "#Now taking moment at B,we get M_B\n",
+ "V_A=(F_C*L_CB)*L_AB**-1\n",
+ "V_B=18-V_A\n",
+ "\n",
+ "#Now Taking Moment at distance x\n",
+ "#M_x=6*x-18*(x-4)\n",
+ "#EI*d**2*y*(d*x**2)**-1=6*x-18*(x-4)\n",
+ "\n",
+ "#Now Integrating above equation,we get\n",
+ "#EI*dy*(dx)**-1=C1+3*x**2-9(x-4)**4\n",
+ "\n",
+ "#Again Integrating above equation we get\n",
+ "#EI*y=C2+C1*x+x**3-3*(x-4)**3\n",
+ "\n",
+ "#The Boundary conditions\n",
+ "x=0\n",
+ "y=0 #.....(a)\n",
+ "\n",
+ "x=6\n",
+ "y=0 #....(b)\n",
+ "\n",
+ "#From Boundary Condition(B.C) a we get\n",
+ "C2=0\n",
+ "\n",
+ "#From Boundary Condition(B.C) b we get\n",
+ "#6*C1+216-3*8\n",
+ "#After Further simplifying we get\n",
+ "C1=-(216-24)*6**-1\n",
+ "\n",
+ "#EI*y=-32*x+x**3-3*(x-4)**3\n",
+ "#EI*dy*(dx)**-1=-32+3*x**2-9(x-4)**4\n",
+ "\n",
+ "#For Max Deflection\n",
+ "#Assume it inthe Porion AC i.e x=4=a\n",
+ "#0=-32+3*x**2\n",
+ "x=(32*3**-1)**0.5\n",
+ "\n",
+ "#Value of Max deflection is\n",
+ "ymax=(-32*x+x**3)*(E*I)**-1 #mm\n",
+ "\n",
+ "#slope at mid-span\n",
+ "\n",
+ "#EI*(dy*(dx)**-1)_centre=-32+3*x**2\n",
+ "#at centre ,\n",
+ "x1=3 #m\n",
+ "\n",
+ "#Let (dy*(dx)**-1)_centre=X\n",
+ "X=-(-32+3*x1**2)*(E*I)**-1 #Radian\n",
+ "\n",
+ "#Deflection at Load Point\n",
+ "x2=4 #m\n",
+ "#EI*y_c=-32*x2+x2**3\n",
+ "\n",
+ "y_c=-(-32*x2+x2**3)*(E*I)**-1\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of Max Deflection\",round(ymax,4),\"mm\"\n",
+ "print\"SLope at mid-span\",round(X,4),\"radian\"\n",
+ "print\"Deflection at the Load Point is\",round(y_c,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of Max Deflection -0.0232 mm\n",
+ "SLope at mid-span 0.0017 radian\n",
+ "Deflection at the Load Point is 0.0213 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.11,Page No.203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_CB=2 #m #Length of CB\n",
+ "L_AC=4 #m #Length of AB\n",
+ "M_C=15 #KN.m #Moment At Pt C\n",
+ "F_C=30 #KN\n",
+ "L=6 #m Span of Beam\n",
+ "\n",
+ "#Let X=E*I\n",
+ "X=10000 #KN-m**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A and V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=30\n",
+ "\n",
+ "#Taking Moment a A,we get\n",
+ "V_B=(F_C*L_AC+M_C)*L**-1\n",
+ "V_A=30-V_B\n",
+ "\n",
+ "#Now Taking Moment at distacnce x from A\n",
+ "#M_x=7.5*x-30*(x-4)+15\n",
+ "\n",
+ "#By using Macaulay's Method\n",
+ "#EI*(d**2*x/dx**2)=M_x=7.5*x-30*(x-4)+15\n",
+ "\n",
+ "#Now Integrating above Equation we get\n",
+ "#EI*(dy/dx)=C1+7.5*x**2*2**-1-15*(x-4)**2+15*(x-4) ............(1)\n",
+ "\n",
+ "#Again Integrating above Equation we get\n",
+ "#EIy=C2+C1*x+7.5*6**-1*x**3-5*(x-4)**3+15*(x-4)**2*2**-1..........(2)\n",
+ "\n",
+ "#Boundary Cinditions\n",
+ "x=0\n",
+ "y=0\n",
+ "\n",
+ "#Substituting above equations we get \n",
+ "C2=0\n",
+ "\n",
+ "x=6 #m\n",
+ "y=0\n",
+ "\n",
+ "C1=-(7.5*6**3*6**-1-5*2**3+15*2**2*2**-1)*6**-1\n",
+ "\n",
+ "#EIy_c=C2+C1*x+7.5*6**-1*x**3-5*(x-4)**3+15*(x-4)**2*2**-1\n",
+ "#Sub values in Above equation we get\n",
+ "y_c=(93.3333*(X)**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"The Deflection at C\",round(y_c,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Deflection at C 0.0093 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.12,Page No.204"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_AC=L_CD=L_DB=2 #m #Length of AC,CD,DB\n",
+ "F_C=40 #KN #Force at C\n",
+ "w=20 #KN/m #u.d.l\n",
+ "L=6 #m #span of beam\n",
+ "\n",
+ "#Let E*I=X\n",
+ "X=15000 #KN-m**2\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=80\n",
+ "\n",
+ "#Taking Moment B,M_B\n",
+ "V_A=(F_C*(L_CD+L_DB)+w*L_DB*L_DB*2**-1)*L**-1 #KN\n",
+ "V_B=80-V_A #KN\n",
+ "\n",
+ "#Taking Moment at distance x from A\n",
+ "#M_x=33.333*x-40*(x-2)-20*(x-4)**2*2**-1\n",
+ "#EI*(d**2/dx**2)=33.333*x-40*(x-2)-10*(x-4)**2\n",
+ "\n",
+ "#Integrating above equation we get\n",
+ "#EI*(dy/dx)=C1+33.333*x**2*2**-1-20*(x-2)**2-10*3**-1*(x-4)**3\n",
+ "\n",
+ "#Again Integrating above equation we get\n",
+ "#EI*y=C2+C1*x+33.333*x**3*6**-1-20*3**-1*(x-2)**3-10*12**-1*(x-4)**4\n",
+ "\n",
+ "#At\n",
+ "x=0\n",
+ "y=0\n",
+ "C2=0\n",
+ "\n",
+ "#At\n",
+ "x=6\n",
+ "y=0\n",
+ "C1=-760*6**-1\n",
+ "\n",
+ "#Assuming Deflection to be max in portion CD and sustituting value of C1 in equation of slope we get\n",
+ "#EI*y=C2+C1*x+33.333*x**3*6**-1-20*3**-1*(x-2)**3-10*12**-1*(x-4)**4\n",
+ "#0=-126.667+33.333*x**2**-1-20*(x-2)**2\n",
+ "\n",
+ "#After rearranging and simplifying further we get\n",
+ "\n",
+ "#x**2-24*x+62=0\n",
+ "#From above equations\n",
+ "a=1\n",
+ "b=-24\n",
+ "c=62\n",
+ "\n",
+ "y=(b**2-4*a*c)**0.5\n",
+ "\n",
+ "x1=(-b+y)*(2*a)**-1\n",
+ "x2=(-b-y)*(2*a)**-1\n",
+ "\n",
+ "#Taking x2 into account\n",
+ "x=2.945 #m\n",
+ "C1=-126.667\n",
+ "C2=0\n",
+ "\n",
+ "y_max=(C2+C1*x+33.333*x**3*6**-1-20*3**-1*(x-2)**3)*X**-1 #mm\n",
+ "\n",
+ "#Max slope occurs at the ends\n",
+ "#At A,\n",
+ "#EI*(dy/dx)_A=-126.667\n",
+ "#At B\n",
+ "#EI*(dy/dx)_B=126.667+33.333*6**2*2**-1-20*4**2-10*2**3\n",
+ "#After simplifying Further we get\n",
+ "#EI*(dy/dx)_B=73.3273\n",
+ "\n",
+ "#Now Max slope is EI(dy/dx)_A=-126.667\n",
+ "#15000*(dy/dx)_=-126.667\n",
+ "\n",
+ "#Let Y=dy/dx\n",
+ "Y=-126.667*X**-1 #Radians\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum Deflection for Beam is\",round(y_max,4),\"mm\"\n",
+ "print\"Maximum Slope for beam is\",round(Y,4),\"radians\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum Deflection for Beam is -0.0158 mm\n",
+ "Maximum Slope for beam is -0.0084 radians\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.13,Page No.206"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=2*10**8 #KN/m**2\n",
+ "I=450*10**-6 #m**4\n",
+ "L_AC=1 #m #Length of AC\n",
+ "L_CD=3 #m #Length of CD\n",
+ "L_DB=2 #m #Length of DB\n",
+ "w=10 #KN/m #u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=30\n",
+ "\n",
+ "#Taking Moment at distance x from A\n",
+ "#M_x=17.5*x-10*(x-1)**2*2**-1+10*(x-4)**2*2**-1\n",
+ "#EI*(d**2/dx**2)=17.5*x-10*(x-1)**2*2**-1+10*(x-4)**2*2**-1\n",
+ "\n",
+ "#Now Integrating Above equation we get\n",
+ "#EI(dy/dx)=C1+17.5*x**2*2**-1-5*3**-1*(x-1)**2+5*3**-1*(x-4)**3\n",
+ "\n",
+ "#Again Integrating Above equation we get\n",
+ "#EI*y=C2+C1*x+17.5*x**3*6**-1-5*12**-1*(x-1)**4+5*12**-1*(x-4)**4\n",
+ "\n",
+ "#At \n",
+ "x=0\n",
+ "y=0\n",
+ "C2=0\n",
+ "\n",
+ "#At \n",
+ "x=6 \n",
+ "y=0\n",
+ "C1=(-17.5*x**3*6**-1+5*12**-1*(x-1)**4-5*12**-1*(x-4)**4)*x**-1\n",
+ "\n",
+ "# 1)Slope at A .i.e at x=0\n",
+ "#EI*(dy/dx)_A=C1=-62.708 #KN-m**2\n",
+ "#let (dy/dx)=X\n",
+ "X=C1*(E*I)**-1 #radiams\n",
+ "\n",
+ "#Deflection at mid-span\n",
+ "x=3 #m\n",
+ "#EI*y_centre=C1*x+17.5*x**3*6**-1-5*12**-1*(x-1)**2\n",
+ "y_centre=-(C1*x+17.5*x**3*6**-1-5*12**-1*(x-1)**4)*(E*I)**-1\n",
+ "\n",
+ "#Maximum Deflection\n",
+ "\n",
+ "#At point of Max deflection (dy/dx)=0\n",
+ "#Assuming it in portion CD\n",
+ "\n",
+ "#0=C1*x+17.5*x**2*2**-1-5*3**-1*(x-1)**3\n",
+ "\n",
+ "#Now Let\n",
+ "#F(x)=C1+17.5*x**2*2**-1-5*3**-1*(x-1)**3\n",
+ "\n",
+ "#Let F(x)=Y\n",
+ "#At \n",
+ "x=2.5\n",
+ "Y1=-(C1+17.5*x**2*2**-1-5*3**-1*(x-1)**3)\n",
+ "\n",
+ "#AT\n",
+ "x=3\n",
+ "Y2=-(C1+17.5*x**2*2**-1-5*3**-1*(x-1)**3)\n",
+ "\n",
+ "#At\n",
+ "x=2.9 #m\n",
+ "Y3=-(C1+17.5*x**2*2**-1-5*3**-1*(x-1)**3)\n",
+ "\n",
+ "#A curve may be plotted for (F(x) and the value for which F(x)=0 may be found\n",
+ "#For F(x)=0 for x=2.92 m\n",
+ "#Therefore y_max occur at x=2.92\n",
+ "\n",
+ "x=2.92 #m\n",
+ "y_max=(C1*x+17.5*x**3*6**-1-5*12**-1*(x-1)**4)*(E*I)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Slope at A\",round(X,6),\"mm\"\n",
+ "print\"Deflection at mid-span\",round(y_centre,6),\"mm\"\n",
+ "print\"Maxmimum Deflection is\",round(y_max,5),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Slope at A -0.000697 mm\n",
+ "Deflection at mid-span 0.001289 mm\n",
+ "Maxmimum Deflection is -0.00129 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.14,Page No.208"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_AC=LDE=L_EB=1 #m #Length of AC\n",
+ "L_CD=2 #m #Length of CD\n",
+ "E=200 #KN/mm**2\n",
+ "I=60*10**6 #mm**4 #M.I\n",
+ "F_C=20 #KN #Force at C\n",
+ "F_E=30 #KN #Force at E\n",
+ "w=10 #KN/m #u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "X=E*I*10**-6 #KN-m**2\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=70\n",
+ "\n",
+ "#Taking Moment at distance x from A\n",
+ "#M_x=34*x-20*(x-1)-10*(x-1)**2*2**-1+10*(x-3)**2*2**-1-30*(x-4)\n",
+ "#EI*(d**2y/dx**2)=34*x-20*(x-1)-10*(x-1)**2*2**-1+10*(x-3)**2*2**-1-30*(x-4)\n",
+ "\n",
+ "#Now Integrating Above equation,we get\n",
+ "#EI*(dy/dx)=C1+17*x**2-10*(x-1)**2-5*3**-1*(x-1)**3+5*3**-1*(x-3)**3-15*(x-4)**2\n",
+ "\n",
+ "#Again Integrating Above equation,we get\n",
+ "#EI*y=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4-5*(x-4)**3\n",
+ "\n",
+ "#At\n",
+ "x=0\n",
+ "y=0\n",
+ "C2=0\n",
+ "\n",
+ "#At \n",
+ "x=5 #m\n",
+ "y=0\n",
+ "C1=(-17*3**-1*x**3+10*3**-1*(x-1)**3+5*12**-1*(x-1)**4-5*12**-1*(x-3)**4+5*(x-4)**3)*5**-1\n",
+ "\n",
+ "#EI*y=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4-5*(x-4)**3\n",
+ "C2=0\n",
+ "C1=-78\n",
+ "x=1\n",
+ "y_c=(-78*x+17*3**-1*x)*(X)**-1\n",
+ "\n",
+ "#EI*y_D=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4\n",
+ "x=3\n",
+ "C1-78\n",
+ "C2=0\n",
+ "y_D=(C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4)*(X**-1)\n",
+ "\n",
+ "#EI*y_E=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4\n",
+ "x=4\n",
+ "C1-78\n",
+ "C2=0\n",
+ "y_E=(C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4)*X**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflections at C\",round(y_c,5),\"mm\"\n",
+ "print\"Deflections at D\",round(y_D,5),\"mm\"\n",
+ "print\"Deflections at E\",round(y_E,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflections at C -0.00603 mm\n",
+ "Deflections at D -0.00953 mm\n",
+ "Deflections at E -0.0061 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.15,Page No.209"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=200 #KN/mm**2 #Modulus of Elasticity\n",
+ "I=300*10**6 #mm\n",
+ "L_AB=L_BC=L_CD=L_DE=1 #m #Length of AB,BC,CD,DE respectively\n",
+ "F_A=20 #KN #Force at A\n",
+ "F_C=10 #KN #Force at C\n",
+ "w=30 #KN/m #u.d.l\n",
+ "\n",
+ "#Let E*I=X\n",
+ "X=E*I*10**-6 #KN-2**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_E be the reactions at E\n",
+ "V_E=F_A+F_C+w*(L_BC+L_CD) #KN \n",
+ "\n",
+ "#Taking Moment at distance x\n",
+ "#EI*(d**2x/dy**2)=M=-20*x-30*(x-1)**2*2**-1-10*(x-2)+30*(x-3)**2*2**-1\n",
+ "\n",
+ "#Integrating above equation we get\n",
+ "#EI*(dy/dx)=C1-10*x**2-5*(x-1)**3-5*(x-2)**2+5*(x-3)**3\n",
+ "\n",
+ "#Again Integrating above equation\n",
+ "#EI*y=C2+C1*x-10*x**3*3**-1-5*(x-1)**4*4**-1-5*(x-3)**4*4**-1-5*3*(x-2)**3\n",
+ "\n",
+ "#At\n",
+ "#dy/dx=0\n",
+ "x=4 #m\n",
+ "C1=10*x**2+5*(x-1)**3+5*(x-2)**2-5*(x-3)**3\n",
+ "\n",
+ "#AT\n",
+ "x=4\n",
+ "y=0\n",
+ "C2=-C1*4+10*x**3*3**-1+5*(x-1)**4*4**-1-5*(x-3)**4*4**-1+5*3**-1*(x-2)**3\n",
+ "\n",
+ "#Max Deflection and Max slopes occurs at Free end in case of cantilever\n",
+ "y_max=y_A=C2*X**-1\n",
+ "\n",
+ "#EI*(dy/dx)_max=C1\n",
+ "#Let (dy/dx)=Y\n",
+ "Y=C1*X**-1 #radian\n",
+ "\n",
+ "#Now deflection at x=1 #m\n",
+ "C2=-913.333\n",
+ "C1=310\n",
+ "x=1\n",
+ "y_B=(C2+C1*x-10*x**3*3**-1)*X**-1\n",
+ "\n",
+ "#Now Deflection at x=2 #m\n",
+ "C2=-913.333\n",
+ "C1=310\n",
+ "x=2 #m\n",
+ "y_C=(C2+C1*x-10*x**3*3**-1-5*(x-1)**4*4**-1)*X**-1\n",
+ "\n",
+ "#Now Deflection at x=3 #m\n",
+ "C2=-913.333\n",
+ "C1=310\n",
+ "x=3 #m\n",
+ "y_D=(C2+C1*x-10*x**3*3**-1-5*(x-1)**4*4**-1-5*3**-1*(x-2)**3)*X**-1\n",
+ "\n",
+ "y_E=0\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Deflection for Beam\",round(y_A,4),\"mm\"\n",
+ "print\"Max Slope for beam\",round(Y,5),\"radians\"\n",
+ "\n",
+ "#Plotting the ELastic Curve\n",
+ "\n",
+ "Y2=[y_E,y_D,y_C,y_B,y_A]\n",
+ "X2=[L_AB+L_BC+L_CD+L_DE,L_AB+L_BC+L_CD,L_AB+L_BC,L_AB,0]\n",
+ "Z2=[0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in mm\")\n",
+ "plt.ylabel(\"Deflection in mm\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Deflection for Beam -0.0152 mm\n",
+ "Max Slope for beam 0.00517 radians\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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8yjGZmZkkJSWRmZlJSkoKM2bMoLKy8pxilcZbsQJGj4Y//QkWL1aSEWnvaq3R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\nL+mlAAAAAElFTkSuQmCC\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5a29310>"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.16,Page No.211"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_BD=L_CB=L_AC=2 #m #Length of BD,CB,AC\n",
+ "F_C=40 #KN #Force at C\n",
+ "F_D=10 #KN Force at D\n",
+ "L=6 #m spna of beam\n",
+ "\n",
+ "#EI is constant in this problem\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B Respectively\n",
+ "#V_A+V_B=50\n",
+ "\n",
+ "#Taking Moment at Pt A\n",
+ "V_B=(F_D*L+F_C*L_AC)*(L_AC+L_CB)**-1\n",
+ "V_A=50-V_B\n",
+ "\n",
+ "#Now Taking Moment at distance x from A,M_x\n",
+ "#M_x=15*x-40*(x-2)+35*(x-4)\n",
+ "#EI*(d**2*y/dx**2)=15*x-40*(x-2)+35*(x-4)\n",
+ "\n",
+ "#Now Integrating above equation we get\n",
+ "#EI*(dy/dx)=C1+7.5*x**2-20*(x-2)**2+17.5(x-4)**2\n",
+ "\n",
+ "#Again Integrating above equation we get\n",
+ "#EI*y=C2+C1*x+2.5*x**2-20*3**-1*(x-2)**3+17.5*(x-4)**3*3**-1\n",
+ "\n",
+ "#At\n",
+ "x=0\n",
+ "y=0\n",
+ "#we get\n",
+ "C2=0\n",
+ "\n",
+ "#At\n",
+ "x=4 \n",
+ "y=0\n",
+ "#we get\n",
+ "C1=(2.5*4**3-20*3**-1*2**3)*4**-1\n",
+ "\n",
+ "#Now Deflection at C\n",
+ "x=2\n",
+ "C1=-26.667\n",
+ "C2=0\n",
+ "y_C=C2+C1*x+2.5*x**3\n",
+ "\n",
+ "#Now Deflection at D\n",
+ "C1=-21.667\n",
+ "C2=0\n",
+ "y_D=-26.667*6+2.5*6**3-20*3**-1*4**3+17.5*2**3*3**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflections Under Loads are:y_D\",round(y_D,4)\n",
+ "print\" :y_C\",round(y_C,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflections Under Loads are:y_D -0.002\n",
+ " :y_C -33.33\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.17,Page No.212"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_BC=L_EB=2 #m #Length of BC & EB\n",
+ "E=200*10**6 #KN/m**2 #Modulus of eLasticity\n",
+ "I=45*10**-6 #mm**4 #M.I\n",
+ "L_DE=3 #m #Length of DE\n",
+ "L_AD=1 #m #Length of AD\n",
+ "w=20 #KN/m #u.d.l\n",
+ "L=8 #m #span of beam\n",
+ "F_C=30 #KN #Force at C\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=90\n",
+ "\n",
+ "#Taking Moment at A,M_A\n",
+ "V_B=(w*L_DE*(L_DE*2**-1+L_AD)+F_C*L)*(L_AD+L_DE+L_EB)**-1\n",
+ "V_A=90-V_B\n",
+ "\n",
+ "#Taking Moment at distance x\n",
+ "#M_x=25*x-20*(x-1)**2*2**-1+20*(x-4)**2*2**-1+65*(x-6)\n",
+ "\n",
+ "#Integrating above equation we get\n",
+ "#EI*(d**2*y/dx**2)=25*x-10*(x-1)**2+10*(x-4)**2+65*(x-6)\n",
+ "\n",
+ "#again Integrating above equation we get\n",
+ "#EI*(dy/dx)=C1+25*x**2*2**-1-10*3**-1*(x-1)**3+10*3**-1*(x-4)**2+65*2**-1*(x-6)**2\n",
+ "\n",
+ "#again Integrating above equation we get\n",
+ "#EI*y=C2+C1*x+25*6**-1*x**3-10*12**-1*(x-1)**4+10*12**-1*(x-4)**4+65*6**-1*(x-6)**3\n",
+ "\n",
+ "x=0\n",
+ "y=0\n",
+ "#Sub these values in above equation,we get\n",
+ "C2=0\n",
+ "\n",
+ "x=6 #m\n",
+ "y=0\n",
+ "C1=-(25*6**-1*6**3-10*12**-1*5**4+10*12**-1*2**4)*6**-1\n",
+ "\n",
+ "#deflection at C is given by\n",
+ "x=8\n",
+ "y_c=(C2+C1*x+25*6**-1*x**3-10*12**-1*(x-1)**4+10*12**-1*(x-4)**4+65*6**-1*(x-6)**3)*(E*I)**-1\n",
+ "\n",
+ "#Assuming y is max in the portion DE,then\n",
+ "#(dy/dx)=0 for that point\n",
+ "\n",
+ "#0=-65.417+25*2**-1*x**2-10*3**-1*x(-1)**3\n",
+ "\n",
+ "#Let F(x)=-65.417+25*2**-1*x**2-10*3**-1*x(-1)**3\n",
+ "#Let z=F(x)\n",
+ "\n",
+ "#AT \n",
+ "x=3\n",
+ "z=-65.417+25*2**-1*x**2-10*3**-1*(x-1)**3\n",
+ "\n",
+ "x=2.5\n",
+ "z1=-65.417+25*2**-1*x**2-10*3**-1*(x-1)**3\n",
+ "\n",
+ "x=2.4\n",
+ "z2=-65.417+25*2**-1*x**2-10*3**-1*(x-1)**3\n",
+ "\n",
+ "#The assumption is max in portion DE\n",
+ "x=2.46\n",
+ "y_max=(-65.417*x+25*6**-1*x**3-10*12**-1*1.46**4)*(E*I)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflection at free end C\",round(y_c,4),\"mm\"\n",
+ "print\"Max Deflection between A and B\",round(y_max,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflection at free end C -0.0101 mm\n",
+ "Max Deflection between A and B -0.0114 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.18,Page No.213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_DB=L_AC=L_ED=2 #m #Length of DB & AC\n",
+ "L_CD=4 #m #Length of CD\n",
+ "L_CE=2 #m #Length of CE\n",
+ "F_A=40 #KN #Force at C\n",
+ "F_B=20 #KN #Force at A\n",
+ "E=200*10**6 #KN/mm**2 #Modulus of Elasticity\n",
+ "I=50*10**-6 #m**4 #M.I\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt V_C & V_D be the reactions at C & D respectively\n",
+ "#V_C+V_D=60\n",
+ "\n",
+ "#Taking Moment At D,M_D\n",
+ "V_C=-(-F_A*(L_AC+L_CE+L_ED)+F_B*L_DB)*L_CD**-1\n",
+ "V_D=60-V_C\n",
+ "\n",
+ "#Now Taking Moment at Distance x from A,\n",
+ "#M_x=-40*x+50*(x-2)+10*(x-6)\n",
+ "\n",
+ "#EI*(d**2*y/dx**2)=-40*x+50*(x-2)+10*(x-6)\n",
+ "\n",
+ "#Now Integrating above Equation we get\n",
+ "#EI*(dy/dx)=C1+20*x**2-25*(x-2)+5*(x-6)**2\n",
+ "\n",
+ "#Again Integrating above Equation we get\n",
+ "#EI*y=C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3\n",
+ "\n",
+ "#At\n",
+ "x=0\n",
+ "y=0\n",
+ "#C2+2*C1=-53.33 ...............(1)\n",
+ "\n",
+ "#At \n",
+ "x=6\n",
+ "y=0\n",
+ "#C2+6*C1=906.667 ...............(2)\n",
+ "\n",
+ "#Subtracting Equation 1 from 2 we get\n",
+ "C1=853.333*4**-1\n",
+ "C2=53.333-2*C1\n",
+ "x=0\n",
+ "y_A=(C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3)*(E*I)**-1\n",
+ "\n",
+ "#Answer For y_A is incorrect in textbook\n",
+ "\n",
+ "#At Mid-span\n",
+ "C1=853.333*4**-1\n",
+ "C2=53.333-2*C1\n",
+ "x=4\n",
+ "y_E=(C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3)*(E*I)**-1\n",
+ "\n",
+ "#Answer For y_E is incorrect in textbook\n",
+ "\n",
+ "#At B\n",
+ "C1=853.333*4**-1\n",
+ "C2=53.333-2*C1\n",
+ "x=8\n",
+ "y_B=(C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3)*(E*I)**-1\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflection relative to the level of the supports:at End A\",round(y_A,4),\"mm\"\n",
+ "print\" :at End B\",round(y_B,4),\"mm\"\n",
+ "print\" :at Centre of CD\",round(y_E,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflection relative to the level of the supports:at End A -0.08 mm\n",
+ " :at End B -0.0267 mm\n",
+ " :at Centre of CD 0.0107 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.6_3.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.6_3.ipynb new file mode 100644 index 00000000..e3bdfbe1 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.6_3.ipynb @@ -0,0 +1,1518 @@ +{
+ "metadata": {
+ "name": "chapter no.6.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter No.6:Torsion"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.1,Page No.225"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=10000 #mm #Length of solid shaft\n",
+ "d=100 #mm #Diameter of shaft\n",
+ "n=150 #rpm\n",
+ "P=112.5*10**6 #N-mm/sec #Power Transmitted\n",
+ "G=82*10**3 #N/mm**2 #modulus of Rigidity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "J=pi*d**4*(32)**-1 #mm**3 #Polar Modulus\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",
+ "\n",
+ "r=50 #mm #Radius\n",
+ "\n",
+ "q_s=T*r*J**-1 #N/mm**2 #Max shear stress intensity\n",
+ "Theta=T*L*(G*J)**-1 #angle of twist\n",
+ "\n",
+ "#Result\n",
+ "print\"Max shear stress intensity\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Angle of Twist\",round(Theta,3),\"radian\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max shear stress intensity 36.48 N/mm**2\n",
+ "Angle of Twist 0.089 radian\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.2,Page No.226"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=440*10**6 #N-m/sec #Power transmitted\n",
+ "n=280 #rpm\n",
+ "theta=pi*180**-1 #radian #angle of twist\n",
+ "L=1000 #mm #Length of solid shaft\n",
+ "q_s=40 #N/mm**2 #Max torsional shear stress\n",
+ "G=84*10**3 #N/mm**2 #Modulus of rigidity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#P=2*pi*n*T*(60)**-1 #Equation of Power transmitted\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #torsional moment\n",
+ "\n",
+ "#From Consideration of shear stress\n",
+ "d1=(T*16*(pi*40)**-1)**0.333333 \n",
+ "\n",
+ "#From Consideration of angle of twist\n",
+ "d2=(T*L*32*180*(pi*84*10**3*pi)**-1)**0.25\n",
+ "\n",
+ "#result\n",
+ "print\"Diameter of solid shaft is\",round(d1,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of solid shaft is 124.09 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.3,Page No.227"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "G=80*10**3 #N/mm**2 #Modulus of rigidity\n",
+ "q_s=80 #N/mm**2 #Max sheare stress\n",
+ "P=736*10**6 #N-mm/sec #Power transmitted\n",
+ "n=200\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",
+ "\n",
+ "#Now From consideration of angle of twist\n",
+ "theta=pi*180**-1\n",
+ "#L=15*d\n",
+ "\n",
+ "d=(T*32*180*15*(pi**2*G)**-1)**0.33333\n",
+ "\n",
+ "#Now corresponding stress at the surface is\n",
+ "q_s2=T*32*d*(pi*2*d**4)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Max diameter required is\",round(d,2),\"mm\"\n",
+ "print\"Corresponding shear stress is\",round(q_s2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max diameter required is 156.66 mm\n",
+ "Corresponding shear stress is 46.55 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.4,Page No.228"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=25 #mm #Diameter of steel bar\n",
+ "p=50*10**3 #N #Pull\n",
+ "dell_1=0.095 #mm #Extension of bar\n",
+ "l=200 #mm #Guage Length\n",
+ "T=200*10**3 #N-mm #Torsional moment\n",
+ "theta=0.9*pi*180**-1 #angle of twist\n",
+ "L=250 #mm Length of steel bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A=pi*4**-1*d**2 #Area of steel bar #mm**2\n",
+ "E=p*l*(dell_1*A)**-1 #N/mm**2 #Modulus of elasticity \n",
+ "\n",
+ "J=pi*32**-1*d**4 #mm**4 #Polar modulus\n",
+ "\n",
+ "G=T*L*(theta*J)**-1 #Modulus of rigidity #N/mm**2\n",
+ "\n",
+ "#Now from the relation of Elastic constants\n",
+ "mu=E*(2*G)**-1-1\n",
+ "\n",
+ "#result\n",
+ "print\"The Poissoin's ratio is\",round(mu,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Poissoin's ratio is 0.292\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.5,Page No.229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=6000 #mm #Length of circular shaft\n",
+ "d1=100 #mm #Outer Diameter\n",
+ "d2=75 #mm #Inner Diameter\n",
+ "R=100*2**-1 #Radius of shaft\n",
+ "T=10*10**6 #N-mm #Torsional moment\n",
+ "G=80*10**3 #N/mm**2 #Modulus of Rigidity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar Modulus \n",
+ "\n",
+ "#Max Shear stress produced\n",
+ "q_s=T*R*J**-1 #N/mm**2\n",
+ "\n",
+ "#Angle of twist\n",
+ "theta=T*L*(G*J)**-1 #Radian\n",
+ "\n",
+ "#Result\n",
+ "print\"MAx shear stress produced is\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Angle of Twist is\",round(theta,2),\"Radian\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "MAx shear stress produced is 74.5 N/mm**2\n",
+ "Angle of Twist is 0.11 Radian\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.6,Page No.229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=200 #mm #External Diameter of shaft\n",
+ "t=25 #mm #Thickness of shaft\n",
+ "n=200 #rpm\n",
+ "theta=0.5*pi*180**-1 #Radian #angle of twist\n",
+ "L=2000 #mm #Length of shaft\n",
+ "G=84*10**3 #N/mm**2\n",
+ "d2=d1-2*t #mm #Internal Diameter of shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar Modulus \n",
+ "\n",
+ "#Torsional moment\n",
+ "T=G*J*theta*L**-1 #N/mm**2 \n",
+ "\n",
+ "#Power Transmitted\n",
+ "P=2*pi*n*T*60**-1*10**-6 #N-mm\n",
+ "\n",
+ "#Max shear stress transmitted\n",
+ "q_s=G*theta*(d1*2**-1)*L**-1 #N/mm**2 \n",
+ "\n",
+ "#Result\n",
+ "print\"Power Transmitted is\",round(P,2),\"N-mm\"\n",
+ "print\"Max Shear stress produced is\",round(q_s,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power Transmitted is 824.28 N-mm\n",
+ "Max Shear stress produced is 36.65 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.7,Page No.230"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=3750*10**6 #N-mm/sec\n",
+ "n=240 #Rpm\n",
+ "q_s=160 #N/mm**2 #Max shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#d2=0.8*d2 #mm #Internal Diameter of shaft\n",
+ "\n",
+ "#J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar modulus\n",
+ "#After substituting value in above Equation we get\n",
+ "#J=0.05796*d1**4\n",
+ "\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",
+ "\n",
+ "#Now from Torsion Formula\n",
+ "#T*J**-1=q_s*R**-1 ......................................(1)\n",
+ "\n",
+ "#But R=d1*2**-1 \n",
+ "\n",
+ "#Now substituting value of R and J in Equation (1) we get\n",
+ "d1=(T*(0.05796*q_s*2)**-1)**0.33333\n",
+ "\n",
+ "d2=d1*0.8\n",
+ "\n",
+ "#Result\n",
+ "print\"The size of the Shaft is:d1\",round(d1,3),\"mm\"\n",
+ "print\" :d2\",round(d2,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The size of the Shaft is:d1 200.362 mm\n",
+ " :d2 160.289 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.8,Page No.231"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=245*10**6 #N-mm/sec #Power transmitted\n",
+ "n=240 #rpm\n",
+ "q_s=40 #N/mm**2 #Shear stress\n",
+ "theta=pi*180**-1 #radian #Angle of twist\n",
+ "L=1000 #mm #Length of shaft\n",
+ "G=80*10**3 #N/mm**2\n",
+ "\n",
+ "#Tmax=1.5*T\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional Moment\n",
+ "Tmax=1.5*T\n",
+ "\n",
+ "#Now For Solid shaft\n",
+ "#J=pi*32*d**4\n",
+ "\n",
+ "#Now from the consideration of shear stress we get\n",
+ "#T*J**-1=q_s*(d*2**-1)**-1\n",
+ "#After substituting value in above Equation we get\n",
+ "#T=pi*16**-1*d**3*q_s\n",
+ "\n",
+ "#Designing For max Torque\n",
+ "d=(Tmax*16*(pi*40)**-1)**0.33333 #mm #Diameter of shaft\n",
+ "\n",
+ "#For max Angle of Twist\n",
+ "#Tmax*J**-1=G*theta*L**-1 \n",
+ "#After substituting value in above Equation we get\n",
+ "d2=(Tmax*32*180*L*(pi**2*G)**-1)**0.25\n",
+ "\n",
+ "#For Hollow Shaft\n",
+ "\n",
+ "#d1_2=Outer Diameter\n",
+ "#d2_2=Inner Diameter\n",
+ "\n",
+ "#d2_2=0.5*d1_2\n",
+ "\n",
+ "# Polar modulus\n",
+ "#J=pi*32**-1*(d1_2**4-d2_2**4)\n",
+ "#After substituting values we get\n",
+ "#J=0.092038*d1_2**4\n",
+ "\n",
+ "#Now from the consideration of stress\n",
+ "#Tmax*J**-1=q_s*(d1_2*2**-1)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d1_2=(Tmax*(0.092038*2*q_s)**-1)**0.33333\n",
+ "\n",
+ "#Now from the consideration of angle of twist\n",
+ "#Tmax*J**-1=G*theta*L**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d1_3=(Tmax*180*L*(0.092038*G*pi)**-1)**0.25\n",
+ "\n",
+ "d2_2=0.5*d1_2\n",
+ "\n",
+ "#result\n",
+ "print\"Diameter of shaft is:For solid shaft:d\",round(d,2),\"mm\"\n",
+ "print\" :For Hollow shaft:d1_2\",round(d1_2,3),\"mm\"\n",
+ "print\" : :d2_2\",round(d2_2,3),\"mm\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of shaft is:For solid shaft:d 123.01 mm\n",
+ " :For Hollow shaft:d1_2 125.69 mm\n",
+ " : :d2_2 62.845 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.11,Page No.235"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=250*10**6 #N-mm/sec #Power transmitted\n",
+ "n=100 #rpm\n",
+ "q_s=75 #N/mm**2 #Shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Equation of Power we have\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",
+ "\n",
+ "#Now from torsional moment equation we have\n",
+ "#T=j*q_s*(d/2**-1)**-1\n",
+ "#After substituting values in above equation and further simplifying we get\n",
+ "#T=pi*16**-1**d**3*q_s\n",
+ "d=(T*16*(pi*q_s)**-1)**0.3333 #mm #Diameter of solid shaft\n",
+ "\n",
+ "#PArt-2\n",
+ "\n",
+ "#Let d1 and d2 be the outer and inner diameter of hollow shaft\n",
+ "#d2=0.6*d1\n",
+ "\n",
+ "#Again from torsional moment equation we have\n",
+ "#T=pi*32**-1*(d1**4-d2**4)*q_s*(d1/2)**-1\n",
+ "d1=(T*16*(pi*(1-0.6**4)*q_s)**-1)**0.33333\n",
+ "d2=0.6*d1\n",
+ "\n",
+ "#Cross sectional area of solid shaft\n",
+ "A1=pi*4**-1*d**2 #mm**2\n",
+ "\n",
+ "#cross sectional area of hollow shaft\n",
+ "A2=pi*4**-1*(d1**2-d2**2)\n",
+ "\n",
+ "#Now percentage saving in weight\n",
+ "#Let W be the percentage saving in weight\n",
+ "W=(A1-A2)*100*A1**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Percentage saving in Weight is\",round(W,3),\"%\"\n",
+ "print\"Size of shaft is:solid shaft:d\",round(d,3),\"mm\"\n",
+ "print\" :Hollow shaft:d1\",round(d1,3),\"mm\"\n",
+ "print\" : :d2\",round(d2,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percentage saving in Weight is 29.735 %\n",
+ "Size of shaft is:solid shaft:d 117.418 mm\n",
+ " :Hollow shaft:d1 123.031 mm\n",
+ " : :d2 73.818 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.12,Page No.237"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "d=100 #mm #Diameter of solid shaft\n",
+ "d1=100 #mm #Outer Diameter of hollow shaft\n",
+ "d2=50 #mm #Inner Diameter of hollow shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Torsional moment of solid shaft\n",
+ "#T_s=J*q_s*(d*2**-1)**-1 \n",
+ "#After substituting values in above equation and further simplifying we get\n",
+ "#T_s=pi*16*d**3*q_s ...............(1)\n",
+ "\n",
+ "#torsional moment for hollow shaft is\n",
+ "#T_h=J*q_s*(d1**4-d2**4)**-1*(d1*2**-1)\n",
+ "#After substituting values in above equation and further simplifying we get\n",
+ "#T_h=pi*32**-1*2*d1**-1*(d1**4-d2**4)*q_s ...........(2)\n",
+ "\n",
+ "#Dividing Equation 2 by 1 we get\n",
+ "#Let the ratio of T_h*T_s**-1 Be X\n",
+ "X=1-0.5**4\n",
+ "\n",
+ "#Loss in strength \n",
+ "#Let s be the loss in strength\n",
+ "#s=T_s*T_h*100*T_s**-1\n",
+ "#After substituting values in above equation and further simplifying we get\n",
+ "s=(1-0.9375)*100\n",
+ "\n",
+ "#Weight Ratio \n",
+ "#Let w be the Weight ratio\n",
+ "#w=W_h*W_s**-1\n",
+ "\n",
+ "A_h=pi*32**-1*(d1**2-d2**2) #mm**2 #Area of Hollow shaft\n",
+ "A_s=pi*32**-1*d**2 #mm**2 #Area of solid shaft\n",
+ "\n",
+ "w=A_h*A_s**-1 \n",
+ "\n",
+ "#Result\n",
+ "print\"Loss in strength is\",round(s,2)\n",
+ "print\"Weight ratio is\",round(w,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Loss in strength is 6.25\n",
+ "Weight ratio is 0.75\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.13,Page No.239"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "T=8 #KN-m #Torque \n",
+ "d=100 #mm #Diameter of portion AB\n",
+ "d1=100 #mm #External Diameter of Portion BC\n",
+ "d2=75 #mm #Internal Diameter of Portion BC\n",
+ "G=80 #KN/mm**2 #Modulus of Rigidity\n",
+ "L1=1500 #mm #Radial Distance of Portion AB\n",
+ "L2=2500 #mm #Radial Distance ofPortion BC\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "R=d*2**-1 #mm #Radius of shaft\n",
+ "\n",
+ "#For Portion AB,Polar Modulus\n",
+ "J1=pi*32**-1*d**4 #mm**4 \n",
+ "\n",
+ "#For Portion BC,Polar modulus \n",
+ "J2=pi*32**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "#Now Max stress occurs in portion BC since max radial Distance is sme in both cases\n",
+ "q_max=T*J2**-1*R*10**6 #N/mm**2 \n",
+ "\n",
+ "#Let theta1 be the rotation in Portion AB and theta2 be the rotation in portion BC\n",
+ "theta1=T*L1*(G*J1)**-1 #Radians\n",
+ "theta2=T*L2*(G*J2)**-1 #Radians\n",
+ "\n",
+ "#Total Rotational at end C\n",
+ "theta=(theta1+theta2)*10**3 #Radians\n",
+ "\n",
+ "#Result\n",
+ "print\"Max stress induced is\",round(q_max,2),\"N/mm**2\"\n",
+ "print\"Angle of Twist is\",round(theta,3),\"radians\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max stress induced is 59.6 N/mm**2\n",
+ "Angle of Twist is 0.053 radians\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.14,Page No.240"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "q_b=80 #N/mm**2 #Shear stress in Brass\n",
+ "q_s=100 #N/mm**2 #Shear stress in Steel\n",
+ "G_b=40*10**3 #N/mm**2 \n",
+ "G_s=80*10**3 \n",
+ "L_b=1000 #mm #Length of brass shaft\n",
+ "L_s=1200 #mm #Length of steel shaft\n",
+ "d1=80 #mm #Diameter of brass shaft\n",
+ "d2=60 #mm #Diameter of steel shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Polar modulus of brass rod\n",
+ "J_b=pi*32**-1*d1**4 #mm**4 \n",
+ "\n",
+ "#Polar modulus of steel rod\n",
+ "J_s=pi*32**-1*d2**4 #mm**4\n",
+ "\n",
+ "#Considering bras Rod:AB\n",
+ "T1=J_b*q_b*(d1*2**-1)**-1 #N-mm \n",
+ "\n",
+ "#Considering Steel Rod:BC\n",
+ "T2=J_s*q_s*(d2*2**-1)**-1 #N-mm\n",
+ "\n",
+ "#Max Torque that can be applied\n",
+ "T2\n",
+ "\n",
+ "#Let theta_b and theta_s be the rotations in Brass and steel respectively\n",
+ "theta_b=T2*L_b*(G_b*J_b)**-1 #Radians\n",
+ "theta_s=T2*L_s*(G_s*J_s)**-1 #Radians\n",
+ "\n",
+ "theta=theta_b+theta_s #Radians #Rotation of free end\n",
+ "\n",
+ "#Result\n",
+ "print\"Total of free end is\",round(theta,3),\"Radians\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total of free end is 0.076 Radians\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.15,Page No.241"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "G=80*10**3 #N/mm**2 #Modulus of Rigidity\n",
+ "d1=100 #mm #Outer diameter of hollow shft\n",
+ "d2=80 #mm #Inner diameter of hollow shaft\n",
+ "d=80 #mm #diameter of Solid shaft\n",
+ "d3=60 #mm #diameter of Solid shaft having L=0.5m\n",
+ "L1=300 #mm #Length of Hollow shaft\n",
+ "L2=400 #mm #Length of solid shaft\n",
+ "L3=500 #mm #LEngth of solid shaft of diameter 60mm\n",
+ "T1=2*10**6 #N-mm #Torsion in Shaft AB\n",
+ "T2=1*10**6 #N-mm #Torsion in shaft BC\n",
+ "T3=1*10**6 #N-mm #Torsion in shaft CD\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Now Polar modulus of section AB\n",
+ "J1=pi*32**-1*(d1**4-d2**4) #mm**4 \n",
+ "\n",
+ "#Polar modulus of section BC\n",
+ "J2=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Polar modulus of section CD\n",
+ "J3=pi*32**-1*d3**4 #mm**4\n",
+ "\n",
+ "#Now angle of twist of AB\n",
+ "theta1=T1*L1*(G*J1)**-1 #radians\n",
+ "\n",
+ "#Angle of twist of BC\n",
+ "theta2=T2*L2*(G*J2)**-1 #radians\n",
+ "\n",
+ "#Angle of twist of CD\n",
+ "theta3=T3*L3*(G*J3)**-1 #radians\n",
+ "\n",
+ "#Angle of twist\n",
+ "theta=theta1-theta2+theta3 #Radians\n",
+ "\n",
+ "#Shear stress in AB From Torsion Equation\n",
+ "q_s1=T1*(d1*2**-1)*J1**-1 #N/mm**2 \n",
+ "\n",
+ "#Shear stress in BC\n",
+ "q_s2=T2*(d*2**-1)*J2**-1 #N/mm**2 \n",
+ "\n",
+ "#Shear stress in CD\n",
+ "q_s3=T3*(d3*2**-1)*J3**-1 #N-mm**2\n",
+ "\n",
+ "#As max shear stress occurs in portion CD,so consider CD\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of twist at free end is\",round(theta,5),\"Radian\"\n",
+ "print\"Max Shear stress\",round(q_s3,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of twist at free end is 0.00496 Radian\n",
+ "Max Shear stress 23.58 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.16,Page No.242"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1000 #mm #Length of bar\n",
+ "L1=600 #mm #Length of Bar AB\n",
+ "L2=400 #mm #Length of Bar BC\n",
+ "d1=60 #mm #Outer Diameter of bar BC\n",
+ "d2=30 #mm #Inner Diameter of bar BC\n",
+ "d=60 #mm #Diameter of bar AB\n",
+ "T=2*10**6 #N-mm #Total Torque\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Polar Modulus of Portion AB\n",
+ "J1=pi*32**-1*d**4 #mm*4\n",
+ "\n",
+ "#Polar Modulus of Portion BC\n",
+ "J2=pi*32**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "#Let T1 be the torque resisted by bar AB and T2 be torque resisted by Bar BC\n",
+ "#Let theta1 and theta2 be the rotation of shaft in portion AB & BC\n",
+ "\n",
+ "#theta1=T1*L1*(G*J1)**-1 #radians\n",
+ "#After substituting values and further simplifying we get \n",
+ "#theta1=32*600*T1*(pi*60**4*G)**-1\n",
+ "\n",
+ "#theta2=T2*L*(J2*G)**-1 #Radians\n",
+ "#After substituting values and further simplifying we get \n",
+ "#theta2=32*400*T2*(pi*60**4*(1-0.5**4)*G)**-1 \n",
+ "\n",
+ "#Now For consistency of Deformation,theta1=theta2\n",
+ "#After substituting values and further simplifying we get \n",
+ "#T1=0.7111*T2 ..................................................(1)\n",
+ "\n",
+ "#But T1+T2=T=2*10**6 ...........................................(2)\n",
+ "#Substituting value of T1 in above equation\n",
+ "\n",
+ "T2=T*(0.7111+1)**-1\n",
+ "T1=0.71111*T2\n",
+ "\n",
+ "#Max stress in Portion AB\n",
+ "q_s1=T1*(d*2**-1)*(J1)**-1 #N/mm**2\n",
+ "\n",
+ "#Max stress in Portion BC\n",
+ "q_s2=T2*(d1*2**-1)*J2**-1 #N/mm**2 \n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Portion:AB\",round(q_s1,2),\"N/mm**2\"\n",
+ "print\" :BC\",round(q_s2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses Developed in Portion:AB 19.6 N/mm**2\n",
+ " :BC 29.4 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.17,Page No.243"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=80 #mm #External Diameter of Brass tube\n",
+ "d2=50 #mm #Internal Diameter of Brass tube\n",
+ "d=50 #mm #Diameter of steel Tube\n",
+ "G_b=40*10**3 #N/mm**2 #Modulus of Rigidity of brass tube\n",
+ "G_s=80*10**3 #N/mm**2 #Modulus of rigidity of steel tube\n",
+ "T=6*10**6 #N-mm #Torque\n",
+ "L=2000 #mm #Length of Tube\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Polar Modulus of brass tube\n",
+ "J1=pi*32**-1*(d1**4-d2**4) #mm**4 \n",
+ "\n",
+ "#Polar modulus of steel Tube\n",
+ "J2=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Let T_s & T_b be the torque resisted by steel and brass respectively\n",
+ "#Then, T_b+T_s=T ............................................(1)\n",
+ "\n",
+ "#Since the angle of twist will be the same\n",
+ "#Theta1=Theta2\n",
+ "#After substituting values and further simplifying we get \n",
+ "#Ts=0.360*Tb ...........................................(2)\n",
+ "\n",
+ "#After substituting value of Ts in eqn 1 and further simplifying we get \n",
+ "T_b=T*(0.36+1)**-1 #N-mm\n",
+ "T_s=0.360*T_b\n",
+ "\n",
+ "#Let q_s and q_b be the max stress in steel and brass respectively\n",
+ "q_b=T_b*(d1*2**-1)*J1**-1 #N/mm**2\n",
+ "q_s=T_s*(d2*2**-1)*J2**-1 #N/mm**2\n",
+ "\n",
+ "#Since angle of twist in brass=angle of twist in steel\n",
+ "theta_s=T_s*L*(J2*G_s)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Materials are:Brass\",round(q_b,2),\"N/mm**2\"\n",
+ "print\" :Steel\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Angle of Twist in 2m Length\",round(theta_s,3),\"Radians\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses Developed in Materials are:Brass 51.79 N/mm**2\n",
+ " :Steel 64.71 N/mm**2\n",
+ "Angle of Twist in 2m Length 0.065 Radians\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.18,Page No.245"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=60 #mm #External Diameter of aluminium Tube\n",
+ "d2=40 #mm #Internal Diameter of aluminium Tube\n",
+ "d=40 #mm #Diameter of steel tube\n",
+ "q_a=60 #N/mm**2 #Permissible stress in aluminium\n",
+ "q_s=100 #N/mm**2 #Permissible stress in steel tube\n",
+ "G_a=27*10**3 #N/mm**2 \n",
+ "G_s=80*10**3 #N/mm**2 \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Polar modulus of aluminium Tube\n",
+ "J_a=pi*32**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "#Polar Modulus of steel Tube\n",
+ "J_s=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Now the angle of twist of steel tube = angle of twist of aluminium tube\n",
+ "#T_s*L_s*(J_s*theta_s)**-1=T_a*L_a*(J_a*theta_a)**-1\n",
+ "#After substituting values in above Equation and Further simplifyin we get\n",
+ "#T_s=0.7293*T_a .....................(1)\n",
+ "\n",
+ "#If steel Governs the resisting capacity\n",
+ "T_s1=q_s*J_s*(d*2**-1)**-1 #N-mm\n",
+ "T_a1=T_s1*0.7293**-1 #N-mm\n",
+ "T1=(T_s1+T_a1)*10**-6 #KN-m #Total Torque in steel Tube\n",
+ "\n",
+ "#If aluminium Governs the resisting capacity \n",
+ "T_a2=q_a*J_a*(d1*2**-1) #N-mm\n",
+ "T_s2=T_a2*0.7293 #N-mm\n",
+ "T2=(T_s2+T_a2)*10**-6 #KN-m #Total Torque in aluminium tube\n",
+ "\n",
+ "#Result\n",
+ "print\"Steel Governs the torque carrying capacity\",round(T1,2),\"KN-m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Steel Governs the torque carrying capacity 2.98 KN-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.19,Page No.247"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=225*10**6 #N-mm/sec #Power Trasmitted\n",
+ "q_b=80 #N/mm**2 #Shear stress\n",
+ "n=200 #Rpm\n",
+ "q_k=100 #N/mm**2 #PErmissible stress in Keys\n",
+ "D=300 #mm #Diameter of bolt circle\n",
+ "L=150 #mm #Length of shear key\n",
+ "d=16 #mm #Diameterr of bolt\n",
+ "\n",
+ "#Calculations\n",
+ "T=60*P*(2*pi*n)**-1 #N-mm #Torque\n",
+ "\n",
+ "#Now From Torsion Formula\n",
+ "#T*J**-1=q_s*R**-1\n",
+ "#After substituting values we get\n",
+ "#T=pi*16*d**3*n\n",
+ "#After further simplifying we get\n",
+ "d1=(T*16*(pi*q_s)**-1)**0.33333\n",
+ "\n",
+ "#Let b be the width of shear Key\n",
+ "#T=q_k*L*b*R\n",
+ "#After simplifying further we get\n",
+ "b=T*(q_k*L*(d1*2**-1))**-1 #mm\n",
+ "\n",
+ "#Let n2 be the no. of bolts required at bolt circle of radius\n",
+ "R_b=D*2**-1 #mm \n",
+ "\n",
+ "n2=T*4*(q_b*pi*d**2*R_b)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Minimum no. of Bolts Required are\",round(n2,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum no. of Bolts Required are 4.45\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.20,Page No.250"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "T=2*10**6 #N-mm #Torque transmitted\n",
+ "G=80*10**3 #N/mm**2 #Modulus of rigidity\n",
+ "d1=40 #mm \n",
+ "d2=80 #mm\n",
+ "r1=20 #mm\n",
+ "r2=40 #mm\n",
+ "L=2000 #mm #Length of shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Angle of twist \n",
+ "theta=2*T*L*(r1**2+r1*r2+r2**2)*(3*pi*G*r2**3*r1**3)**-1 #radians\n",
+ "\n",
+ "#If the shaft is treated as shaft of average Diameter\n",
+ "d_avg=(d1+d2)*2**-1 #mm\n",
+ "\n",
+ "theta1=T*L*(G*pi*32**-1*d_avg**4)**-1 #Radians\n",
+ "\n",
+ "#Percentage Error\n",
+ "#Let Percentage Error be E\n",
+ "X=theta-theta1\n",
+ "E=(X*theta**-1)*100 \n",
+ "\n",
+ "#Result\n",
+ "print\"Percentage Error is\",round(E,2),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percentage Error is 32.28 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.21,Page No.252"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "G=80*10**3 #N/mm**2 \n",
+ "P=1*10**9 #N-mm/sec #Power\n",
+ "n=300 \n",
+ "d1=150 #mm #Outer Diameter\n",
+ "d2=120 #mm #Inner Diameter\n",
+ "L=2000 #mm #Length of circular shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm\n",
+ "\n",
+ "#Polar Modulus \n",
+ "J=pi*32**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "q_s=T*J**-1*(d1*2**-1) #N/mm**2 \n",
+ "\n",
+ "\n",
+ "#Strain ENergy\n",
+ "U=q_s**2*(4*G)**-1*pi*4**-1*(d1**2-d2**2)*L\n",
+ "\n",
+ "#Result\n",
+ "print\"Max shear stress is\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Strain Energy stored in the shaft is\",round(U,2),\"N-mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max shear stress is 81.36 N/mm**2\n",
+ "Strain Energy stored in the shaft is 263181.37 N-mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.22,Page No.254"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=12 #mm #Diameter of helical spring\n",
+ "D=150 #mm #Mean Diameter\n",
+ "R=D*2**-1 #mm #Radius of helical spring\n",
+ "n=10 #no.of turns\n",
+ "G=80*10**3 #N/mm**2 \n",
+ "W=450 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max shear stress \n",
+ "q_s=16*W*R*(pi*d**3)**-1 #N/mm**2\n",
+ "\n",
+ "#Strain Energy stored\n",
+ "U=32*W**2*R**3*n*(G*d**4)**-1 #N-mm\n",
+ "\n",
+ "#Deflection Produced\n",
+ "dell=64*W*R**3*n*(G*d**4)**-1 #mm\n",
+ "\n",
+ "#Stiffness Spring\n",
+ "k=W*dell**-1 #N/mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Max shear stress is\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Strain Energy stored is\",round(U,2),\"N-mm\"\n",
+ "print\"Deflection Produced is\",round(dell,2),\"mm\"\n",
+ "print\"Stiffness spring is\",round(k,2),\"N/mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max shear stress is 99.47 N/mm**2\n",
+ "Strain Energy stored is 16479.49 N-mm\n",
+ "Deflection Produced is 73.24 mm\n",
+ "Stiffness spring is 6.14 N/mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.23,Page No.255"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "K=5 #N/mm #Stiffness\n",
+ "L=100 #mm #Solid Length\n",
+ "q_s=60 #N/mm**2 #Max shear stress\n",
+ "W=200 #N #Max Load\n",
+ "G=80*10**3 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#K=W*dell**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#d=0.004*R**3*n ........(1) #mm #Diameter of wire\n",
+ "#n=L*d**-1 ........(2)\n",
+ "\n",
+ "#From Shearing stress\n",
+ "#q_s=16*W*R*(pi*d**3)**-1 \n",
+ "#After substituting values and further simplifying we get\n",
+ "#d**4=0.004*R**3*n .................(4)\n",
+ "\n",
+ "#From Equation 1,2,3\n",
+ "#d**4=0.004*(0.0785*d**3)**3*100*d**-1\n",
+ "#after further simplifying we get\n",
+ "d=5168.101**0.25\n",
+ "n=100*d**-1\n",
+ "R=(d**4*(0.004*n)**-1)**0.3333\n",
+ "\n",
+ "#Result\n",
+ "print\"Diameter of Wire is\",round(d,2),\"mm\"\n",
+ "print\"No.of turns is\",round(n,2)\n",
+ "print\"Mean Radius of spring is\",round(R,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of Wire is 8.48 mm\n",
+ "No.fo turns is 11.79\n",
+ "Mean Radius of spring is 47.83 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 54
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.24,Page No.255"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "m=5*10**5 #Wagon Weighing\n",
+ "v=18*1000*36000**-1 \n",
+ "d=300 #mm #Diameter of Beffer springs\n",
+ "n=18 #no.of turns\n",
+ "G=80*10**3 #N/mm**2\n",
+ "dell=225\n",
+ "R=100 #mm #Mean Radius\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Energy of Wagon\n",
+ "E=m*v**2*(9.81*2)**-1 #N-mm\n",
+ "\n",
+ "#Load applied\n",
+ "W=dell*G*d**4*(64*R**3*n)**-1 #N \n",
+ "\n",
+ "#Energy each spring can absorb is\n",
+ "E2=W*dell*2**-1 #N-mm\n",
+ "\n",
+ "#No.of springs required to absorb energy of Wagon\n",
+ "n2=E*E2**-1 *10**7\n",
+ "\n",
+ "#Result\n",
+ "print\"No.of springs Required for Buffer is\",round(n2,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "No.of springs Required for Buffer is 4.47\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.25,Page No.259"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b=180 #mm #width of flange\n",
+ "d=10 #mm #Depth of flange\n",
+ "t=10 #mm #Thickness of flange\n",
+ "D=400 #mm #Overall Depth \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "I_xx=1*12**-1*(b*D**3-(b-t)*(D-2*d)**3)\n",
+ "I_yy=1*12**-1*((D-2*d)*t**3+2*t*b**3)\n",
+ "\n",
+ "#If warping is neglected\n",
+ "J=I_xx+I_yy #mm**4\n",
+ "\n",
+ "#Since b/d>1.6,we get\n",
+ "J2=1*3**-1*d**3*b*(1-0.63*d*b**-1)*2+1*3**-1*t**3*(D-2*d)*(1-0.63*t*b**-1)\n",
+ "\n",
+ "#Over Estimation of torsional Rigidity would have been \n",
+ "T=J*J2**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Error in assessing torsional Rigidity if the warping is neglected is\",round(T,2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Error in assessing torsional Rigidity if the warping is neglected is 808.28\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.26,Page No.261"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=100 #mm #Outer Diameter\n",
+ "d2=95 #mm #Inner Diameter\n",
+ "T=2*10**6 #N-mm #Torque\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar Modulus\n",
+ "\n",
+ "#Shear stress\n",
+ "q_max=T*J**-1*d1*2**-1 #N/mm**2 \n",
+ "\n",
+ "#Now theta*L**-1=T*(G*J)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#Let theta*L**-1=X\n",
+ "X=T*J**-1\n",
+ "\n",
+ "#Now Treating it as very thin walled tube\n",
+ "d=(d1+d2)*2**-1 #mm\n",
+ "\n",
+ "r=d*2**-1 \n",
+ "t=(d1-d2)*2**-1\n",
+ "q_max2=T*(2*pi*r**2*t)**-1 #N/mm**2\n",
+ "\n",
+ "X2=T*(2*pi*r**3*t)**-1 \n",
+ "\n",
+ "#Result\n",
+ "print\"When it is treated as hollow shaft:Max shear stress\",round(q_max,2),\"N/mm**2\"\n",
+ "print\" :Angle of Twist per unit Length\",round(X,3)\n",
+ "print\"When it is very thin Walled Tube :Max shear stress\",round(q_max2,2),\"N/mm**2\"\n",
+ "print\" :Angle of twist per Unit Length\",round(X2,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "When it is treated as hollow shaft:Max shear stress 54.91 N/mm**2\n",
+ " :Angle of Twist per unit Length 1.098\n",
+ "When it is very thin Walled Tube :Max shear stress 53.57 N/mm**2\n",
+ " :Angle of twist per Unit Length 1.099\n"
+ ]
+ }
+ ],
+ "prompt_number": 72
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.7_3.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.7_3.ipynb new file mode 100644 index 00000000..b75d886a --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.7_3.ipynb @@ -0,0 +1,1328 @@ +{
+ "metadata": {
+ "name": "chapter no.7.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter No.7:Compound Stresses And Strains"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.1,Page No.269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "sigma1=30 #N/mm**2 #Stress in tension\n",
+ "d=20 #mm #Diameter \n",
+ "sigma2=90 #N/mm**2 #Max compressive stress\n",
+ "sigma3=25 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#In TEnsion\n",
+ "\n",
+ "#Corresponding stress in shear\n",
+ "P=sigma1*2**-1 #N/mm**2\n",
+ "\n",
+ "#Tensile force\n",
+ "F=pi*4**-1*d**2*sigma1\n",
+ "\n",
+ "#In Compression\n",
+ "\n",
+ "#Correspong shear stress\n",
+ "P2=sigma2*2**-1 #N/mm**2\n",
+ "\n",
+ "#Correspong compressive(axial) stress\n",
+ "p=2*sigma3 #N/mm**2 \n",
+ "\n",
+ "#Corresponding Compressive force\n",
+ "P3=p*pi*4**-1*d**2 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Failure Loads are:\",round(F,2),\"N\"\n",
+ "print\" :\",round(P3,2),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Failure Loads are: 9424.78 N\n",
+ " : 15707.96 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.2,Page No.270"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=25 #mm #Diameter of circular bar\n",
+ "F=20*10**3 #N #Axial Force\n",
+ "theta=30 #Degree #angle \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Axial stresses\n",
+ "p=F*(pi*4**-1*d**2)**-1 #N/mm**2\n",
+ "\n",
+ "#Normal Stress\n",
+ "p_n=p*(cos(30*pi*180**-1))**2\n",
+ "\n",
+ "#Tangential Stress\n",
+ "p_t=p*2**-1*sin(2*theta*pi*180**-1)\n",
+ "\n",
+ "#Max shear stress occurs on plane where theta2=45 \n",
+ "theta2=45\n",
+ "sigma_max=p*2**-1*sin(2*theta2*pi*180**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses developed on a plane making 30 degree is:\",round(p_n,2),\"N/mm**2\"\n",
+ "print\" :\",round(p_t,2),\"N/mm**2\"\n",
+ "print\"stress on max shear stress is\",round(sigma_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses developed on a plane making 30 degree is: 30.56 N/mm**2\n",
+ " : 17.64 N/mm**2\n",
+ "stress on max shear stress is 20.37 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.3,Page No.272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "theta=30 #degree\n",
+ "\n",
+ "#Stresses acting on material\n",
+ "p1=120 #N/mm**2\n",
+ "p2=80 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Normal Stress\n",
+ "P_n=(p1+p2)*2**-1+(p1-p2)*2**-1*cos(2*theta*pi*180**-1) #N/mm**2\n",
+ "\n",
+ "#Tangential stress\n",
+ "P_t=(p1-p2)*2**-1*sin(2*theta*pi*180**-1)\n",
+ "\n",
+ "#Resultant stress\n",
+ "P=(P_n**2+P_t**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Inclination to the plane\n",
+ "phi=arctan(P_n*P_t**-1)*(180*pi**-1)\n",
+ "\n",
+ "#Angle made by resultant with 120 #N/mm**2 stress\n",
+ "phi2=phi+theta #Degree\n",
+ "\n",
+ "#Result\n",
+ "print\"Normal Stress is\",round(P_n,2),\"N/mm**2\"\n",
+ "print\"Tangential Stress is\",round(P_t,2),\"N/mm**2\"\n",
+ "print\"Angle made by resultant\",round(phi2,2),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Normal Stress is 110.0 N/mm**2\n",
+ "Tangential Stress is 17.32 N/mm**2\n",
+ "Angle made by resultant 111.05 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.4,Page No.272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Direct Stresses\n",
+ "P1=60 #N/mm**2 \n",
+ "P2=100 #N/mm**2\n",
+ "\n",
+ "Theta=25 #Degree #Angle\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Normal Stress\n",
+ "P_n=(P1-P2)*2**-1+(P1+P2)*2**-1*cos(2*Theta*pi*180**-1) #N/mm**2\n",
+ "\n",
+ "#Tangential Stress\n",
+ "P_t=(P1+P2)*2**-1*sin(Theta*2*pi*180**-1) #N/mm**2\n",
+ "\n",
+ "#Resultant stress\n",
+ "P=(P_n**2+P_t**2)**0.5 #N/mm**2\n",
+ "\n",
+ "theta2=arctan(P_n*P_t**-1)*(180*pi**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses on the plane AC is:\",round(P_n,2),\"N/mm**2\"\n",
+ "print\" \",round(P_t,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses on the plane AC is: 31.42 N/mm**2\n",
+ " 61.28 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.6,Page No.278"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Stresses acting on material\n",
+ "p_x=180 #N/mm**2 \n",
+ "p_y=120 #N/mm**2\n",
+ "\n",
+ "q=80 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "theta=arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1) #degrees\n",
+ "theta2=theta*2**-1 #Degrees\n",
+ "theta3=theta+180 #Degrees\n",
+ "theta4=theta3*2**-1 #Degrees\n",
+ "\n",
+ "#Stresses\n",
+ "p_1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "p_2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of Principal stress is:\",round(p_1,2),\"N/mm**2\"\n",
+ "print\" \",round(p_2,2),\"N/mm**2\"\n",
+ "print\"Magnitude of max shear stress is\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of Principal stress is: 235.44 N/mm**2\n",
+ " 64.56 N/mm**2\n",
+ "Magnitude of max shear stress is 85.44 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.7,Page No.279"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#stresses\n",
+ "p_x=60 #N/mm**2\n",
+ "p_y=-40 #N/mm**2\n",
+ "\n",
+ "q=10 #N/mm**2 #shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Principal Stresses\n",
+ "p1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "p2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Inclination of principal stress to plane\n",
+ "theta=arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1)#Degrees\n",
+ "theta2=(theta)*2**-1 #degrees\n",
+ "\n",
+ "theta3=(theta+180)*2**-1 #degrees\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses are:\",round(p1,2),\"N/mm**2\"\n",
+ "print\" :\",round(p2,2),\"N/mm**2\"\n",
+ "print\"Max shear stresses\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses are: 60.99 N/mm**2\n",
+ " : -40.99 N/mm**2\n",
+ "Max shear stresses 50.99 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.8,Page No.280"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#stresses\n",
+ "p_x=-120 #N/mm**2\n",
+ "p_y=-80 #N/mm**2\n",
+ "\n",
+ "q=-60 #N/mm**2 #shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Principal Stresses\n",
+ "p1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "p2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Inclination of principal stress to plane\n",
+ "theta=arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1)#Degrees\n",
+ "theta2=(theta)*2**-1 #degrees\n",
+ "\n",
+ "theta3=(theta+180)*2**-1 #degrees\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses are:\",round(p1,2),\"N/mm**2\"\n",
+ "print\" :\",round(p2,2),\"N/mm**2\"\n",
+ "print\"Max shear stresses\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses are: -36.75 N/mm**2\n",
+ " : -163.25 N/mm**2\n",
+ "Max shear stresses 63.25 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.9,Page No.282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#stresses\n",
+ "p_x=-40 #N/mm**2\n",
+ "p_y=80 #N/mm**2\n",
+ "\n",
+ "q=48 #N/mm**2 #shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=((((p_x-p_y)*2**-1)**2)+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Inclination of principal stress to plane\n",
+ "theta=arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1)#Degrees\n",
+ "theta2=(theta)*2**-1 #degrees\n",
+ "\n",
+ "theta3=(theta+180)*2**-1 #degrees\n",
+ "\n",
+ "#Normal Corresponding stress\n",
+ "p_n=(p_x+p_y)*2**-1+(p_x-p_y)*2**-1*cos(2*(theta2+45)*pi*180**-1)+q*sin(2*(theta2+45)*pi*180**-1) #Degrees\n",
+ "\n",
+ "#Resultant stress\n",
+ "p=((p_n**2+q_max**2)**0.5) #N/mm**2\n",
+ "\n",
+ "phi=arctan(p_n*q_max**-1)*(180*pi**-1) #Degrees\n",
+ "\n",
+ "#Inclination to the plane\n",
+ "alpha=round((theta2+45),2)+round(phi ,2)#Degree\n",
+ "\n",
+ "#Answer in book is incorrect of alpha ie41.25\n",
+ "\n",
+ "#Result\n",
+ "print\"Planes of max shear stress:\",round(p_n,2),\"N/mm**2\"\n",
+ "print\" \",round(q_max,2),\"N/mm*2\"\n",
+ "print\"Resultant Stress is\",round(p,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Planes of max shear stress: 20.0 N/mm**2\n",
+ " 76.84 N/mm*2\n",
+ "Resultant Stress is 79.4 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.10,Page No.283"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Stresses\n",
+ "p_x=50*cos(35*pi*180**-1)\n",
+ "q=50*sin(35*pi*180**-1)\n",
+ "p_y=0\n",
+ "\n",
+ "theta=40 #Degrees #Plane AB inclined to vertical\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Normal Stress on AB\n",
+ "p_n=(p_x+p_y)*2**-1+(p_x-p_y)*2**-1*cos(2*theta*pi*180**-1)+q*sin(2*theta*pi*180**-1)\n",
+ "\n",
+ "#Tangential Stress on AB\n",
+ "p_t=(p_x-p_y)*2**-1*sin(2*theta*pi*180**-1)-q*cos(2*theta*pi*180**-1) #N/mm**2\n",
+ "\n",
+ "#Resultant stress\n",
+ "p=(p_n**2+p_t**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Angle of resultant\n",
+ "phi=arctan(p_n*p_t**-1)*(180*pi**-1) #degrees\n",
+ "phi2=phi+theta #Degrees\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of resultant stress is\",round(p,2),\"N/mm**2\"\n",
+ "print\"Direction of Resultant stress is\",round(phi2,2),\"Degrees\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of resultant stress is 54.44 N/mm**2\n",
+ "Direction of Resultant stress is 113.8 Degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.12,Page No.285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Direct stresses\n",
+ "p_x=120 #N/mm**2 #Tensile stress\n",
+ "p_y=-100 #N/mm**2 #Compressive stress\n",
+ "p1=160 #N/mm**2 #Major principal stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let q be the shearing stress\n",
+ "\n",
+ "#p1=(p_x+p_y)*2**-1+((((p_x+p_y)*2**-1)**2)+q**2)**0.5\n",
+ "#After further simplifying we get\n",
+ "q=(p1-((p_x+p_y)*2**-1))**2-((p_x-p_y)*2**-1)**2 #N/mm**2\n",
+ "q2=(q)**0.5 #N/mm**2\n",
+ "\n",
+ "#Minimum Principal stress\n",
+ "p2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shearing stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Shearing stress of material\",round(q,2),\"N/mm**2\"\n",
+ "print\"Min Principal stress\",round(p2,2),\"N/mm**2\"\n",
+ "print\"Max shearing stress\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shearing stress of material 10400.0 N/mm**2\n",
+ "Min Principal stress -140.0 N/mm**2\n",
+ "Max shearing stress 150.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.14,Page No.291"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=40*10**3 #N #Shear Force\n",
+ "M=20*10**6 #Bending Moment\n",
+ "\n",
+ "#Rectangular section\n",
+ "b=100 #mm #Width\n",
+ "d=200 #mm #Depth\n",
+ "\n",
+ "x=20 #mm #Distance from Top surface upto point\n",
+ "y=80 #mm #Distance from point to Bottom\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "I=1*12**-1*b*d**3 #mm**4 #M.I\n",
+ "\n",
+ "#At 20 mm Below top Fibre\n",
+ "f_x=M*I**-1*y #N/mm**2 #Stress\n",
+ "\n",
+ "#Assuming sagging moment ,f_x is compressive p_x=f_x=-24 #N/mm**2\n",
+ "p_x=f_x=-24 #N/mm**2\n",
+ "\n",
+ "#Shearing stress\n",
+ "q=F*(b*I)**-1*(b*x*(b-x*2**-1)) #N/mm**2\n",
+ "\n",
+ "#Direct stresses\n",
+ "\n",
+ "p_y=0 #N/mm**2\n",
+ "\n",
+ "p1=(p_x+p_y)*2**-1+(((p_x+p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "p2=(p_x+p_y)*2**-1-(((p_x+p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Directions of principal stresses at a point below 20mm is:\",round(p1,2),\"N/mm**2\"\n",
+ "print\" \",round(p2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Directions of principal stresses at a point below 20mm is: 0.05 N/mm**2\n",
+ " -24.05 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.15,Page No.292"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=4000 #mm #Span\n",
+ "W1=W2=W3=2*10**3 #N #Load\n",
+ "\n",
+ "#SEction of beam\n",
+ "b=100 #mm #Width\n",
+ "d=240 #mm #Dept\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A and R_B be the reactions\n",
+ "R_A=R_B=(W1+W2+W3)*2**-1 #KN\n",
+ "\n",
+ "#Now at the section 1.5m from left support A\n",
+ "#Shear Force\n",
+ "F=R_A-W1 #KN\n",
+ "\n",
+ "#B.M\n",
+ "M=R_A*1.5-W1*0.5 #KN-m\n",
+ "\n",
+ "#M.I\n",
+ "I=1*12**-1*b*d**3 #mm**4\n",
+ "\n",
+ "#Bending stress\n",
+ "#f=M*I**-1*y\n",
+ "#After Sub values and further simplifying we get\n",
+ "#f=3.04*10**-2*y\n",
+ "\n",
+ "#As it varies Linearly\n",
+ "\n",
+ "#at distance 0 From NA \n",
+ "f1=0\n",
+ "#at distance 60 mm from NA\n",
+ "f2=1.823 #N/mm**2\n",
+ "#at distance 120 mm from NA\n",
+ "f3=3.646 #N/mm**2\n",
+ "\n",
+ "#Shearing stress\n",
+ "q=F*b*d*2**-1*d*4**-1*(b*I)**-1\n",
+ "\n",
+ "#At 60 mm above NA\n",
+ "q2=F*b*d*4**-1*(d*2**-1-d*8**-1)*(b*I)**-1\n",
+ "\n",
+ "#At 120 mm above NA\n",
+ "q3=0 \n",
+ "\n",
+ "#At NA element is under pure shear\n",
+ "p1=q #N/mm**2\n",
+ "p2=-q #N/mm**2 \n",
+ "\n",
+ "#Inclination of principal plane to vertical\n",
+ "#theta=2*q*0**-1\n",
+ "#Further simplifying we get\n",
+ "#theta=infinity\n",
+ "\n",
+ "#therefore\n",
+ "theta=90*2**-1 #degrees\n",
+ "theta2=270*2**-1 #degrees\n",
+ "\n",
+ "#At 60 mm From NA\n",
+ "p_x=-1.823 #N/mm**2 \n",
+ "p_y=0\n",
+ "q=0.0469 #N/mm**2\n",
+ "\n",
+ "#principal planes\n",
+ "P1=(p_x+p_y)*2**-1+(((p_x+p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2=(p_x+p_y)*2**-1-(((p_x+p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Principal planes inclination to hte plane of p_x is given by\n",
+ "theta3=(arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1))\n",
+ "theta4=theta3*2**-1#degrees\n",
+ "\n",
+ "theta5=theta3+180 #Degrees\n",
+ "\n",
+ "#At 120 mm From N-A\n",
+ "p_x2=3.646 #N/mm**2\n",
+ "p_y2=0 #N/mm**2\n",
+ "q2=0 #N/mm**2\n",
+ "\n",
+ "P3=p_x2 #N/mm**2\n",
+ "P4=0 #N/mm**2\n",
+ "\n",
+ "#Answer for P2 at 60 mm from NA is incorrect\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Planes at 60 mm from NA:\",round(p_x,2),\"N/mm**2\"\n",
+ "print\" \",round(p_y,2),\"N/mm**2\"\n",
+ "print\"Principal Stresses at 60 mm From NA\",round(P1,4),\"N/mm**2\"\n",
+ "print\" \",round(P2,4),\"N/mm**2\"\n",
+ "print\"Principal Planes at 60 mm from NA:\",round(p_x2,4),\"N/mm**2\"\n",
+ "print\" \",round(p_y2,4),\"N/mm**2\"\n",
+ "print\"Principal Stresses at 60 mm From NA\",round(P3,4),\"N/mm**2\"\n",
+ "print\" \",round(P4,4),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Planes at 60 mm from NA: -1.82 N/mm**2\n",
+ " 0.0 N/mm**2\n",
+ "Principal Stresses at 60 mm From NA 0.0012 N/mm**2\n",
+ " -1.8242 N/mm**2\n",
+ "Principal Planes at 60 mm from NA: 3.646 N/mm**2\n",
+ " 0.0 N/mm**2\n",
+ "Principal Stresses at 60 mm From NA 3.646 N/mm**2\n",
+ " 0.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.16,Page No.295"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=8000 #mm #Span of beam\n",
+ "w=40*10**6 #N/mm #udl\n",
+ "\n",
+ "#I-section\n",
+ "\n",
+ "#Flanges\n",
+ "b=100 #mm #Width\n",
+ "t=10 #mm #Thickness\n",
+ "\n",
+ "D=400 #mm #Overall Depth\n",
+ "t2=10 #mm #thickness of web\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A and R_B be the Reactions at A & B respectively\n",
+ "R_A=w*2**-1*L*10**-9 #KN\n",
+ "\n",
+ "#Shear force at 2m for left support\n",
+ "F=R_A-2*w*10**-6 #KN\n",
+ "\n",
+ "#Bending Moment\n",
+ "M=R_A*2-2*w*10**-6 #KN-m\n",
+ "\n",
+ "#M.I\n",
+ "I=1*12**-1*b*D**3-1*12**-1*(b-t)*(D-2*t2)**3 #mm**4\n",
+ "\n",
+ "#Bending stress at 100 mm above N_A\n",
+ "f=M*10**6*I**-1*b\n",
+ "\n",
+ "#Shear stress \n",
+ "q=F*10**3*(t*I)**-1*(b*t*(D-t)*2**-1 +t2*(b-t2)*145) #N/mm**2\n",
+ "\n",
+ "p_x=-197.06 #N/mm**2 \n",
+ "p_y=0 #N/mm**2\n",
+ "q=21.38 #N/mm**2\n",
+ "\n",
+ "#Principal Stresses\n",
+ "\n",
+ "P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses are:\",round(P1,2),\"N/mm**2\"\n",
+ "print\" \",round(P2,2),\"N/mm**2\"\n",
+ "print\"Max shear stress\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses are: 2.29 N/mm**2\n",
+ " -199.35 N/mm**2\n",
+ "Max shear stress 100.82 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.18,Page No.298"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=100 #mm #Diameter of shaft\n",
+ "M=3*10**6 #N-mm #B.M\n",
+ "T=6*10**6 #N-mm #Twisting Moment\n",
+ "mu=0.3\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max principal Stress\n",
+ "\n",
+ "P1=16*(pi*d**3)**-1*(M+(M**2+T**2)**0.5) #N/mm**2 \n",
+ "P2=16*(pi*d**3)**-1*(M-(M**2+T**2)**0.5) #N/mm**2 \n",
+ "\n",
+ "#Direct stress\n",
+ "P=round(P1,2)-mu*round(P2,2) #N/mm**2 \n",
+ "\n",
+ "#Result\n",
+ "print\"Principal stresses are:\",round(P1,2),\"N/mm**2\"\n",
+ "print\" :\",round(P2,2),\"N/mm**2\"\n",
+ "print\"Stress Producing the same strain is\",round(P,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal stresses are: 49.44 N/mm**2\n",
+ " : -18.89 N/mm**2\n",
+ "Stress Producing the same strain is 55.11 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.19,Page No.299"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=75 #mm #diameter \n",
+ "P=30*10**6 #W #Power transmitted\n",
+ "W=6 #N-mm/sec #Load\n",
+ "L=1000 #mm \n",
+ "N=300 #r.p.m\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#B.M\n",
+ "M=W*L*4**-1 #N-mm\n",
+ "T=P*60*(2*pi*N)**-1 #Torque transmitted\n",
+ "\n",
+ "#M.I\n",
+ "I=pi*64**-1*d**4 #mm**4\n",
+ "\n",
+ "#Bending stress\n",
+ "f_A=M*I**-1*(d*2**-1) #N/mm**2\n",
+ "\n",
+ "#At A\n",
+ "p_x=f_A\n",
+ "p_y=0\n",
+ "\n",
+ "#Polar Modulus\n",
+ "J=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Shearing stress\n",
+ "q=T*J**-1*(d*2**-1) #N/mm**2\n",
+ "\n",
+ "#Principal Stresses\n",
+ "P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Bending stress\n",
+ "p_x2=0\n",
+ "p_y2=0\n",
+ "\n",
+ "#Shearing stress\n",
+ "q2=T*J**-1*d*2**-1 #N/mm**2\n",
+ "\n",
+ "#Principal stresses\n",
+ "P3=(p_x2+p_y2)*2**-1+(((p_x2-p_y2)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "P4=(p_x2+p_y2)*2**-1-(((p_x2-p_y2)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max2=(((p_x2-p_y2)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Answer for Principal Stresses P1,P2 and Max stress i.e q_max is incorrect in Book\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses at vertical Diameter:P1\",round(P1,2),\"N/mm**2\"\n",
+ "print\" :P2\",round(P2,2),\"N/mm**2\"\n",
+ "print\"Max stress at vertical Diameter : \",round(q_max,2),\"N/mm**2\"\n",
+ "print\"Principal Stresses at Horizontal Diameter:P3\",round(P3,2),\"N/mm**2\"\n",
+ "print\" :P4\",round(P4,2),\"N/mm**2\"\n",
+ "print\"Max stress at Horizontal Diameter : \",round(q_max2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses at vertical Diameter:P1 11.55 N/mm**2\n",
+ " :P2 -11.51 N/mm**2\n",
+ "Max stress at vertical Diameter : 11.53 N/mm**2\n",
+ "Principal Stresses at Horizontal Diameter:P3 11.53 N/mm**2\n",
+ " :P4 -11.53 N/mm**2\n",
+ "Max stress at Horizontal Diameter : 11.53 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.20,Page No.302"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=100 #mm #External Diameter\n",
+ "d2=50 #mm #Internal Diameter\n",
+ "N=500 #mm #r.p.m\n",
+ "P=60*10**6 #N-mm/sec #Power\n",
+ "p=100 #N/mm**2 #principal stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#M.I\n",
+ "I=pi*(d1**4-d2**4)*64**-1 #mm**4\n",
+ "\n",
+ "#Bending Stress\n",
+ "#f=M*I*d1*2**-1 #N/mm**2\n",
+ "\n",
+ "#Principal Planes\n",
+ "#p_x=32*M*(pi*(d1**4-d2**4))*d1\n",
+ "#p_y=0\n",
+ "\n",
+ "#Shear stress\n",
+ "#q=T*J**-1*(d1*2**-1)\n",
+ "#After sub values and further simplifying we get\n",
+ "#q=16*T*d1*(pi*(d1**4-d2**4))*d1\n",
+ "\n",
+ "#Principal stresses\n",
+ "#P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "#After sub values and further simplifying we get\n",
+ "#P1=16*(pi*(d1**4-d2**4))*d1*(M+(M**2+t**2)**0.5) ...............(1)\n",
+ "\n",
+ "#P=2*pi*N*T*60**-1\n",
+ "#After sub values and further simplifying we get\n",
+ "T=P*60*(2*pi*N)**-1*10**-6 #N-mm\n",
+ "\n",
+ "#Again Sub values and further simplifying Equation 1 we get\n",
+ "M=(337.533)*(36.84)**-1 #KN-m\n",
+ "\n",
+ "#Min Principal stress\n",
+ "#P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "#Sub values and further simplifying we get\n",
+ "P2=16*(pi*(d1**4-d2**4))*d1*(M-(M**2+T**2)**0.5)*10**-11\n",
+ "\n",
+ "#Result\n",
+ "print\"Bending Moment safely applied to shaft is\",round(M,2),\"KN-m\"\n",
+ "print\"Min Principal Stress is\",round(P2,3),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Bending Moment safely applied to shaft is 9.16 KN-m\n",
+ "Min Principal Stress is -0.336 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.21,Page No.303"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=150 #mm #Diameter\n",
+ "T=20*10**6 #N #Torque\n",
+ "M=12*10**6 #N-mm #B.M\n",
+ "F=200*10**3 #N #Axial Thrust\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#M.I\n",
+ "I=(pi*64**-1*d**4)\n",
+ "\n",
+ "#Bending stress \n",
+ "f_A=M*I**-1*(d*2**-1) #N/mm**2\n",
+ "f_B=-f_A #N/mm**2\n",
+ "\n",
+ "#Axial thrust due to thrust\n",
+ "sigma=F*(pi*4**-1*d**2)**-1\n",
+ "\n",
+ "#At A\n",
+ "p_x=f_A-sigma #N/mm**2\n",
+ "\n",
+ "#At B\n",
+ "p_x2=f_B-sigma #N/mm**2\n",
+ "\n",
+ "p_y=0 #At A and B\n",
+ "\n",
+ "#Polar Modulus\n",
+ "J=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Shearing stress at A and B\n",
+ "q=T*J**-1*(d*2**-1) #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Principal Stresses\n",
+ "#At A\n",
+ "P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max1=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#At B\n",
+ "P1_2=(p_x2+p_y)*2**-1+(((p_x2-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2_2=(p_x2+p_y)*2**-1-(((p_x2-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max2=(((p_x2-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"MAx Principal Stresses:P1\",round(P1,2),\"N/mm**2\"\n",
+ "print\" :P2\",round(P2,2),\"N/mm**2\"\n",
+ "print\"Min Principal Stresses:P1_2\",round(P1_2,2),\"N/mm**2\"\n",
+ "print\" :P2_2\",round(P2_2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "MAx Principal Stresses:P1 45.1 N/mm**2\n",
+ " :P2 -20.2 N/mm**2\n",
+ "Min Principal Stresses:P1_2 14.65 N/mm**2\n",
+ " :P2_2 -62.18 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.22,Page No.311"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#strains\n",
+ "e_A=500 #microns\n",
+ "e_B=250 #microns\n",
+ "e_C=-150 #microns\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "theta=45 #Degrees\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "e_x=e_A=500\n",
+ "e_45=e_B=250\n",
+ "e_y=e_C=-150 \n",
+ "\n",
+ "#e_45=(e_x+e_y)*2**-1+(e_x-e_y)*2**-1*cos(2*theta)+rho_x_y*2**-1*sin(2*theta)\n",
+ "#After sub values and further simplifying we get\n",
+ "rho_x_y=(e_45-(e_x+e_y)*2**-1-(e_x-e_y)*2**-1*cos(2*theta*pi*180**-1))*(sin(2*theta*pi*180**-1))**-1*2\n",
+ "\n",
+ "#Principal strains are given by\n",
+ "e1=(e_x+e_y)*2**-1+(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 #microns\n",
+ "e2=(e_x+e_y)*2**-1-(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 #microns\n",
+ "\n",
+ "#Principal Stresses\n",
+ "sigma1=E*(e1+mu*e2)*(1-mu**2)**-1*10**-6 #N/mm**2\n",
+ "sigma2=E*(e2+mu*e1)*(1-mu**2)**-1*10**-6 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Strains are:e1\",round(e1,2),\"N/mm**2\"\n",
+ "print\" :e2\",round(e2,2),\"N/mm**2\"\n",
+ "print\"Principal Stresses are:sigma1\",round(sigma1,2),\"N/mm**2\"\n",
+ "print\" :sigma2\",round(sigma2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Strains are:e1 508.54 N/mm**2\n",
+ " :e2 -158.54 N/mm**2\n",
+ "Principal Stresses are:sigma1 101.31 N/mm**2\n",
+ " :sigma2 -1.31 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.23,Page No.313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Strains\n",
+ "e_A=600 #microns\n",
+ "e_B=-450 #microns\n",
+ "e_C=100 #micron\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "theta=240\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "e_x=e_A=600\n",
+ "\n",
+ "#e_A=(e_x+e_y)*2**-1+(e_x-e_y)*2**-1*cos(theta)+rho_x_y*2**-1*sin(theta)\n",
+ "#After sub values and further simplifying we get\n",
+ "#-450=(e_x+e_y)*2**-1-(e_x-e_y)*2**-1*(0.5)-0.866*2**-1*rho_x_y .....................(1)\n",
+ "\n",
+ "#e_C=(e_x+e_y)*2**-1+(e_x-e_y)*2**-1*cos(2*theta)+rho_x_y*2**-1*sin(2*theta)\n",
+ "#After sub values and further simplifying we get\n",
+ "#100=(e_x+e_y)*2**-1-0.5*(e_x-e_y)*2**-1*(0.5)-0.866*2**-1*rho_x_y .....................(2)\n",
+ "\n",
+ "#Adding Equation 1 and 2 we get equations as\n",
+ "#-350=e_x+e_y-(e_x-e_y)*2**-1 ...............(3)\n",
+ "#Further simplifying we get\n",
+ "\n",
+ "e_y=(-700-e_x)*3**-1 #micron \n",
+ "\n",
+ "rho_x_y=(e_C-(e_x+e_y)*2**-1-(e_x-e_y)*2**-1*cos(2*theta*pi*180**-1))*(sin(2*theta*pi*180**-1))**-1*2 #micron\n",
+ "\n",
+ "#Principal strains\n",
+ "e1=(e_x+e_y)*2**-1-(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 #microns\n",
+ "e2=(e_x+e_y)*2**-1+(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 #microns\n",
+ "\n",
+ "#Principal Stresses\n",
+ "sigma1=E*(e1+mu*e2)*(1-mu**2)**-1*10**-6 #N/mm**2\n",
+ "sigma2=E*(e2+mu*e1)*(1-mu**2)**-1*10**-6 #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses are:sigma1\",round(sigma1,2),\"N/mm**2\"\n",
+ "print\" :sigma2\",round(sigma2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses are:sigma1 -69.49 N/mm**2\n",
+ " :sigma2 117.11 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.8_3.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.8_3.ipynb new file mode 100644 index 00000000..08f57912 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.8_3.ipynb @@ -0,0 +1,1527 @@ +{
+ "metadata": {
+ "name": "chapter no.8.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter No.8:Thin And Thick Cyclinders And Spheres"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.1,Page No.322"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=3000 #mm #Length\n",
+ "d1=1000 #mm #Internal diameter\n",
+ "t=15 #mm #Thickness\n",
+ "P=1.5 #N/mm**2 #Fluid Pressure\n",
+ "E=2*10**5 #n/mm**2 #Modulus of elasticity\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Hoop stress\n",
+ "f1=P*d1*(2*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Longitudinal Stress\n",
+ "f2=P*d1*(4*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(f1-f2)*2**-1 #N/mm**2\n",
+ "\n",
+ "#Diametrical Strain\n",
+ "#Let e1=dell_d*d**-1 .....................(1)\n",
+ "e1=(f1-mu*f2)*E**-1 \n",
+ "\n",
+ "#Sub values in equation 1 and further simplifying we get\n",
+ "dell_d=e1*d1 #mm\n",
+ "\n",
+ "#Longitudinal strain\n",
+ "#e2=dell_L*L**-1 ......................(2)\n",
+ "e2=(f2-mu*f1)*E**-1 \n",
+ "\n",
+ "#Sub values in equation 2 and further simplifying we get\n",
+ "dell_L=e2*L #mm\n",
+ "\n",
+ "#Change in Volume \n",
+ "#Let Z=dell_V*V**-1 ................(3)\n",
+ "Z=2*e1+e2\n",
+ "\n",
+ "#Sub values in equation 3 and further simplifying we get\n",
+ "dell_V=Z*pi*4**-1*d1**2*L\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Intensity of shear stress\",round(q_max,2),\"N/mm**2\"\n",
+ "print\"Change in the Dimensions of the shell is:dell_d\",round(dell_d,2),\"mm\"\n",
+ "print\" :dell_L\",round(dell_L,2),\"mm\"\n",
+ "print\" :dell_V\",round(dell_V,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Intensity of shear stress 12.5 N/mm**2\n",
+ "Change in the Dimensions of the shell is:dell_d 0.21 mm\n",
+ " :dell_L 0.15 mm\n",
+ " :dell_V 1119192.38 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.2,Page No.323"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=2000 #mm #Length\n",
+ "d=200 #mm # diameter\n",
+ "t=10 #mm #Thickness\n",
+ "dell_V=25000 #mm**3 #Additional volume\n",
+ "E=2*10**5 #n/mm**2 #Modulus of elasticity\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let p be the pressure developed\n",
+ "\n",
+ "#Circumferential Stress\n",
+ "\n",
+ "#f1=p*d*(2*t)**-1 #N/mm**2\n",
+ "#After sub values and further simplifying\n",
+ "#f1=10*p\n",
+ "\n",
+ "#f1=p*d*(4*t)**-1 #N/mm**2\n",
+ "#After sub values and further simplifying\n",
+ "#f1=5*p\n",
+ "\n",
+ "#Diameterical strain = Circumferential stress\n",
+ "#Let X=dell_d*d**-1 ................................(1)\n",
+ "#X=e1=(f1-mu*f2)*E**-1 \n",
+ "#After sub values and further simplifying\n",
+ "#e1=8.5*p*E**-1\n",
+ "\n",
+ "#Longitudinal strain\n",
+ "#Let Y=dell_L*L**-1 ......................................(2)\n",
+ "#Y=e2=(f2-mu*f1)*E**-1 \n",
+ "#After sub values and further simplifying\n",
+ "#e2=2*p*E**-1\n",
+ "\n",
+ "#Volumetric strain\n",
+ "#Let X=dell_V*V**-1 \n",
+ "#X=2*e1+e2\n",
+ "#After sub values and further simplifying\n",
+ "#X=19*p*E**-1\n",
+ "#After further simplifying we get\n",
+ "p=dell_V*(pi*4**-1*d**2*L)**-1*E*19**-1 #N/mm**2\n",
+ "\n",
+ "#Hoop Stress\n",
+ "f1=p*d*(2*t)**-1\n",
+ "\n",
+ "X=e1=8.5*p*E**-1\n",
+ "#Sub value of X in equation 1 we get\n",
+ "dell_d=8.5*p*E**-1*d\n",
+ "\n",
+ "Y=e2=2*p*E**-1\n",
+ "#Sub value of Y in equation 2 we get\n",
+ "dell_L=2*p*E**-1*L\n",
+ "\n",
+ "#Result\n",
+ "print\"Pressure Developed is\",round(p,2),\"N/mm**2\"\n",
+ "print\"Hoop stress Developed is\",round(f1,2),\"N/mm**2\"\n",
+ "print\"Change in diameter is\",round(dell_d,2),\"mm\"\n",
+ "print\"Change in Length is\",round(dell_L,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Pressure Developed is 4.19 N/mm**2\n",
+ "Hoop stress Developed is 41.88 N/mm**2\n",
+ "Change in diameter is 0.04 mm\n",
+ "Change in Length is 0.08 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.3,Page No.324"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=750 #mm #Diameter of water supply pipes\n",
+ "h=50*10**3 #mm #Water head\n",
+ "sigma=20 #N/mm**2 #Permissible stress\n",
+ "rho=9810*10**-9 #N/mm**3\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Pressure of water\n",
+ "P=rho*h #N/mm**2\n",
+ "\n",
+ "#Stress\n",
+ "#sigma=p*d*(2*t)**-1\n",
+ "#After further simplifying\n",
+ "t=P*d*(2*sigma)**-1 #mm \n",
+ "\n",
+ "#Result\n",
+ "print\"Thickness of seamless pipe is\",round(t,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness of seamless pipe is 9.197 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.4,Page No.326"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=2500 #mm #Diameter of riveted boiler\n",
+ "P=1 #N/mm**2 #Pressure\n",
+ "rho1=0.7 #Percent efficiency\n",
+ "rho2=0.4 #Circumferential joints\n",
+ "sigma=150 #N/mm**2 #Permissible stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Equating Bursting force to longitudinal joint strength ,we get\n",
+ "#p*d*L=rho1*2*t*L*sigma\n",
+ "#After rearranging and further simplifying we get\n",
+ "t=P*d*(2*sigma*rho1)**-1 #mm\n",
+ "\n",
+ "#Considering Longitudinal force\n",
+ "#pi*d**2*4**-1*P=rho2*pi*d*t*sigma\n",
+ "#After rearranging and further simplifying we get\n",
+ "t2=P*d*(4*sigma*rho2)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Thickness of plate required is\",round(t,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness of plate required is 11.9 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.5,Page No.326"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Boiler Dimensions\n",
+ "t=16 #mm #Thickness\n",
+ "p=2 #N/mm**2 #internal pressure\n",
+ "f=150 #N/mm**2 #Permissible stress\n",
+ "rho1=0.75 #Longitudinal joints\n",
+ "rho2=0.45 #circumferential joints\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Equating Bursting force to longitudinal joint strength ,we get\n",
+ "d1=rho1*2*t*f*p**-1 #mm\n",
+ "\n",
+ "#Considering circumferential strength \n",
+ "d2=4*rho2*t*f*p**-1 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Largest diameter of Boiler is\",round(d1,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Largest diameter of Boiler is 1800.0 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 52
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.6,Page No.329"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=250 #mm #Diameter iron pipe\n",
+ "t=10 #mm #Thickness\n",
+ "d2=6 #mm #Diameter of steel\n",
+ "p=80 #N/mm**2 #stress\n",
+ "P=3 #N/mm**2 #Pressure\n",
+ "E_c=1*10**5 #N/mm**2\n",
+ "mu=0.3 #poissoin's ratio\n",
+ "E_s=2*10**5 #N/mm**2\n",
+ "n=1 #No.of wires\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "L=6 #mm #Length of cyclinder\n",
+ "\n",
+ "#Force Exerted by steel wire at diameterical section\n",
+ "F=p*2*pi*d2**2*1*4**-1 #N\n",
+ "\n",
+ "#Initial stress in cyclinder\n",
+ "f_c=F*(2*t*d2)**-1 #N/mm**2\n",
+ "\n",
+ "#LEt due to fluid pressure alone stresses developed in steel wire be F_w and in cyclinder f1 and f2\n",
+ "f2=P*d*(4*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Considering the equilibrium of half the cyclinder, 6mm long we get\n",
+ "#F_w*2*pi*4**-1*d2**2*n+f1*2*t*d2=P*d*d2\n",
+ "#After further simplifying we get\n",
+ "#F_w+2.122*f1=79.58 . ......................................(1)\n",
+ "\n",
+ "#Equating strain in wire to circumferential strain in cyclinder \n",
+ "#F_w=(f1-mu*f2)*E_s*E_c**-1 #N/mm**2\n",
+ "#After further simplifying we get\n",
+ "#F_w=2*f1-11.25 ....................................(2)\n",
+ "\n",
+ "#Sub in equation in1 we get\n",
+ "f1=(79.58+11.25)*(4.122)**-1 #N/mm**2\n",
+ "F_w=2*f1-11.25 #N/mm**2\n",
+ "\n",
+ "#Final stresses\n",
+ "#1) In steel Wire\n",
+ "sigma=F_w+p #N/mm**2\n",
+ "\n",
+ "#2) In Cyclinder\n",
+ "sigma2=f1-f_c\n",
+ "\n",
+ "#Result\n",
+ "print\"Final Stresses developed in:cyclinder is\",round(sigma,2),\"N/mm**2\"\n",
+ "print\" :Steel is\",round(sigma2,2),\"N/mm**2\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Final Stresses developed in:cyclinder is 112.82 N/mm**2\n",
+ " :Steel is -15.66 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.7,Page No.332"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=750 #mm #Diameter of shell\n",
+ "t=8 #mm #THickness\n",
+ "p=2.5 #N/mm**2\n",
+ "E=2*10**5 #N/mm**2\n",
+ "mu=0.25 #Poissoin's ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Hoop stress\n",
+ "f1=f2=p*d*(4*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Change in Diameter\n",
+ "dell_d=d*p*d*(1-mu)*(4*t*E)**-1 #mm\n",
+ "\n",
+ "#Change in Volume\n",
+ "dell_V=3*p*d*(1-mu)*(4*t*E)**-1*pi*6**-1*d**3\n",
+ "\n",
+ "#Answer for Change in diameter is incorrect in book\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress introduced is\",round(f1,2),\"N/mm**2\"\n",
+ "print\"Change in Diameter is\",round(dell_d,2),\"N/mm**2\"\n",
+ "print\"Change in Volume is\",round(dell_V,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress introduced is 58.59 N/mm**2\n",
+ "Change in Diameter is 0.16 N/mm**2\n",
+ "Change in Volume is 145608.33 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.8,Page No.333"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=600 #mm #Diameter of sherical shell\n",
+ "t=10 #mm #Thickness\n",
+ "f=80 #N/mm**2 #Permissible stress\n",
+ "rho=0.75 #Efficiency joint\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max Pressure\n",
+ "p=f*4*t*rho*d**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Pressure is\",round(p,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Pressure is 4.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 60
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.9,Page No.333"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1000 #mm #Length of shell\n",
+ "d=200 #mm #Diameter\n",
+ "t=6 #mm #Thickness\n",
+ "p=1.5 #N/mm**2 #Internal Pressure\n",
+ "E=2*10**5 #N/mm**2\n",
+ "mu=0.25 #Poissoin's Ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Change in Volume of sphere\n",
+ "dell_V_s=3*p*d*(1-mu)*(4*t*E)**-1*pi*6**-1*d**3\n",
+ "\n",
+ "#Hoop stress\n",
+ "f1=p*d*(2*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Longitudinal stress\n",
+ "f2=p*d*(4*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Principal strain\n",
+ "e1=(f1-mu*f2)*E**-1\n",
+ "e2=(f2-mu*f1)*E**-1\n",
+ "\n",
+ "V_c=1000 #mm**3\n",
+ "\n",
+ "#Change in Volume of cyclinder\n",
+ "dell_V_c=(2*e1+e2)*pi*4**-1*d**2*L\n",
+ "\n",
+ "#Total Change in Diameter\n",
+ "dell_V=dell_V_s+dell_V_c #mm**3\n",
+ "\n",
+ "#Result\n",
+ "print\"Change in Volume is\",round(dell_V,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Change in Volume is 8443.03 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.10,Page No.337"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=400 #mm #Internal Diameter\n",
+ "t=100 #mm #Thickness\n",
+ "p=80 #N/mm**2 #Fluid pressure\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Internal Radius\n",
+ "r1=d1*2**-1 #mm\n",
+ "\n",
+ "#Outer Radius\n",
+ "r_o=r1+t #mm\n",
+ "\n",
+ "p1=80 #N/mm**2\n",
+ "p2=0\n",
+ "\n",
+ "#Now From Lame's Euation\n",
+ "#p_x=b*(x**2)**-1-a\n",
+ "#at x=200 #mm \n",
+ "p_x=80 #N/mm**2\n",
+ "#80=b*(200**2)**-1-a ..........................(1)\n",
+ "\n",
+ "#at x=300 #mm\n",
+ "#p_x2=0\n",
+ "#0=b*(300**2)**-1-a ...........................(2)\n",
+ "\n",
+ "#Sub equation 2 from 1\n",
+ "#80=b*(200**2)**-1-b*(300**2)**-1\n",
+ "#After Further simplifying we get\n",
+ "b=(50000)**-1*(200**2*300**2*80)\n",
+ "\n",
+ "#From equation 2 we get\n",
+ "a=b*(300**2)**-1\n",
+ "\n",
+ "#Variation of radial pressure p_x;\n",
+ "#p_x=b*(x**2)**-1-a\n",
+ "#After sub values and further simplifying we get\n",
+ "\n",
+ "#Radial pressure Variation\n",
+ "#At \n",
+ "x=200 #mm\n",
+ "p_x=b*(x**2)**-1-a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x2=250 #mm\n",
+ "p_x2=b*(x2**2)**-1-a #N/mm**2\n",
+ "\n",
+ "#At \n",
+ "x3=300 #mm\n",
+ "p_x3=b*(x3**2)**-1-a #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Hoop stress Distribution\n",
+ "#Variation of F_x\n",
+ "\n",
+ "#At \n",
+ "x=200 #mm\n",
+ "F_x=b*(x**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x2=250 #mm\n",
+ "F_x2=b*(x2**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x3=300 #mm\n",
+ "F_x3=b*(x3**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Hoop stress is\",round(F_x,2),\"N/mm**2\"\n",
+ "print\"Min Hoop stress is\",round(F_x3,2),\"N/mm**2\"\n",
+ "print\"Plot of Hoop stress\"\n",
+ "\n",
+ "#Plotting Variation of hoop stress\n",
+ "\n",
+ "X1=[x,x2,x3]\n",
+ "Y1=[p_x,p_x2,p_x3]\n",
+ "Y2=[-F_x,-F_x2,-F_x3]\n",
+ "Z1=[0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Y2,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Radial Stress Distribution & Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Hoop stress is 208.0 N/mm**2\n",
+ "Min Hoop stress is 128.0 N/mm**2\n",
+ "Plot of Hoop stress\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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skdWFW7ZsQffu3XH8+HGMGzcOY8aMAaA9wnfKlCkIDg7GmDFjEBcXJ7WK4uLi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\nbm5uYtOmTUII3b27hBBCoVCIHTt2CCG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+ "text": [
+ "<matplotlib.figure.Figure at 0x50e8310>"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.11,Page No.338"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=300 #mm #Outside diameter \n",
+ "d2=200 #mm #Internal Diameter\n",
+ "p=14 #N/mm**2 #internal Fluid pressure\n",
+ "t=50 #mm #Thickness\n",
+ "r_o=150 #mm #Outside Diameter\n",
+ "r2=100 #mm #Internal Diameter\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Lame's Equation\n",
+ "#p_x=b*(x**2)**-1-a #N/mm**2 ...................(1)\n",
+ "#F_x=b*(x**2)**-1+a #N/mm**2 ...................(2)\n",
+ "\n",
+ "#At \n",
+ "x=r2=100 #mm\n",
+ "p_x=14 #N/mm**2\n",
+ "\n",
+ "#Sub value of p_x in equation 1 we get\n",
+ "#14=(100)**-1*b-a ............................(3)\n",
+ "\n",
+ "#At\n",
+ "x2=r_o=150 #mm\n",
+ "p_x2=0 #N/mm**2\n",
+ "\n",
+ "#Sub value in equation 1 we get\n",
+ "#0=b*(150**2)**-1-a ......................(4)\n",
+ "\n",
+ "#From Equations 3 and 4 we get\n",
+ "#14=b*(100**2)**-1-b*(100**2)**-1\n",
+ "#After sub values and further simplifying we get\n",
+ "b=14*100**2*150**2*(150**2-100**2)**-1\n",
+ "\n",
+ "#From equation 4 we get\n",
+ "a=b*(150**2)**-1\n",
+ "\n",
+ "#Hoop Stress\n",
+ "#F_x=b*(x**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At \n",
+ "x=100 #mm\n",
+ "F_x=b*(x**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x2=125 #mm\n",
+ "F_x2=b*(x2**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x3=150 #mm\n",
+ "F_x3=b*(x3**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#If thin Cyclindrical shell theory is used,hoop stress is uniform and is given by\n",
+ "F=p*d2*(2*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Percentage error in estimating max hoop tension\n",
+ "E=(F_x-F)*F_x**-1*100 #%\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Hoop Stress Developed in the cross-section is\",round(F,2),\"N/mm**2\"\n",
+ "print\"Plot of Variation of hoop stress\"\n",
+ "\n",
+ "#Plotting Variation of hoop stress\n",
+ "\n",
+ "X1=[x,x2,x3]\n",
+ "Y1=[F_x,F_x2,F_x3]\n",
+ "Z1=[0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Radial Stress Distribution & Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Hoop Stress Developed in the cross-section is 28.0 N/mm**2\n",
+ "Plot of Variation of hoop stress\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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29vaQJAknTpx4bAP379/HiBEjMGzYMMycOdNwT41GA1dXV2RkZGDgwIEcMiIi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\nSUVORK5CYII=\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x58c2e30>"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.12,Page No.339"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=300 #mm #Outside diameter \n",
+ "d2=200 #mm #Internal Diameter\n",
+ "p=12 #N/mm**2 #internal Fluid pressure\n",
+ "F_max=16 #N/mm**2 #Tensile stress\n",
+ "r_o=150 #mm #Outside Diameter\n",
+ "r2=100 #mm #Internal Diameter\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let p_o be the External Pressure applied.\n",
+ "#From LLame's theorem\n",
+ "#p_x=b*(x**2)**-1-a ..............(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#Now At\n",
+ "x=100 #mm\n",
+ "p_x=12 #N/mm**2\n",
+ "#sub in equation 1 we get\n",
+ "#12=b*(100**2)**-1-a . ..................(3)\n",
+ "\n",
+ "#The Max Hoop stress occurs at least value of x where\n",
+ "x=r1=100 #mm\n",
+ "#16=b*(100**2)**-1+a .......................(4)\n",
+ "\n",
+ "#From Equations 1 and 2 we get\n",
+ "#28=b*(100**2)**-1+b*(100**2)**-1\n",
+ "#After furhter Simplifying we get\n",
+ "b=28*100**2*2**-1\n",
+ "\n",
+ "#sub in equation 1 we get\n",
+ "a=-(12-(b*(100**2)**-1))\n",
+ "\n",
+ "#Thus At\n",
+ "x2=150 #mm\n",
+ "p_o=b*(x2**2)**-1-a\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum External applied is\",round(p_o,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum External applied is 4.22 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.13,Page No.340"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=160 #mm #Internal Diameter \n",
+ "r1=80 #mm #External Diameter\n",
+ "p1=40 #N/mm**2 #Internal Diameter\n",
+ "P_max=120 #N/mm**2 #Allowable stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Lame's Equation we have\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#At \n",
+ "x=r1=80 #N/mm**2 \n",
+ "#Sub in equation 1 we get\n",
+ "#120=b*(80**2)**-1+a ........................(3)\n",
+ "\n",
+ "#The hoop tension at inner edge is max stress\n",
+ "#Hence\n",
+ "#120=b*(80**2)**-1+a .............................(4)\n",
+ "\n",
+ "#From Equation 3 and 4 we get\n",
+ "b=160*80**2*2**-1 \n",
+ "\n",
+ "#Sub in equation 3 we get\n",
+ "a=-(40-(b*(80**2)**-1))\n",
+ "\n",
+ "#Let External radius be r_o.Since at External Surface is Zero,we get\n",
+ "#0=b*(r_o)**-1-a\n",
+ "#After Further simplifying we get\n",
+ "r_o=(b*a**-1)**0.5\n",
+ "\n",
+ "#Thickness of Cyclinder \n",
+ "t=r_o-r1 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Thickness Required is\",round(t,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness Required is 33.14 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.14,Page No.341"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=300 #mm #Outside diameter \n",
+ "d1=180 #mm #Internal Diameter\n",
+ "p=12 #N/mm**2 #internal Fluid pressure\n",
+ "p_o=6 #N/mm**2 #External Pressure\n",
+ "r_o=150 #mm #Outside Diameter\n",
+ "r=90 #mm #Internal Diameter\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Lame's Equation we have\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#At \n",
+ "x=r1=90 #N/mm**2 \n",
+ "p=42 #N/mm**2\n",
+ "#Sub in equation 1 we get\n",
+ "#42=b*(90**2)**-1-a ..............................(3)\n",
+ "\n",
+ "#At \n",
+ "x=r_o=150 #mm\n",
+ "p2=6 #N/mm**2\n",
+ "#sub in equation 1 we get\n",
+ "#6=b*(150**2)**-1-a ..............................(4)\n",
+ "\n",
+ "#From equations 3 and 4 weget\n",
+ "#36=b*(90**2)**-1-b2(150**2)**-1\n",
+ "#After further simplifying we get\n",
+ "b=36*90**2*150**2*(150**2-90**2)**-1\n",
+ "\n",
+ "#Sub value of b in equation 4 we get\n",
+ "a=b*(150**2)**-1-p_o\n",
+ "\n",
+ "#At \n",
+ "x=r1=90 #mm\n",
+ "F_x=b*(x**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At \n",
+ "x2=r_o=150 #mm \n",
+ "F_x2=b*(x2**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#Now if External pressure is doubled i.e p_o2=12 #N/mm**2 We have\n",
+ "p_o2=12 #N/mm**2\n",
+ "#sub in equation 4 we get\n",
+ "#12=b2*(150**2)**-1-a2 ..........................(5)\n",
+ "\n",
+ "#Max Hoop stress is to be 70.5 #N/mm**2,which occurs at x=r1=90 #mm\n",
+ "#Sub in equation 4 we get\n",
+ "#70.5=b*(90**2)**-1+a2 ................................(6)\n",
+ "\n",
+ "#Adding equation 5 and 6\n",
+ "#82.5=b2*(150**2)**-1+b*(90**2)**-1\n",
+ "#After furhter simplifying we get\n",
+ "b2=82.5*150**2*90**2*(150**2+90**2)**-1\n",
+ "\n",
+ "#Sub in equation 5 we get\n",
+ "a2=b2*(150**2)**-1-12 \n",
+ "\n",
+ "#If p_i is the internal pressure required then from Lame's theorem\n",
+ "p_i=b2*(r1**2)**-1-a2\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses int the material are:F_x\",round(F_x,2),\"N/mm**2\"\n",
+ "print\" :F_x2\",round(F_x2,2),\"N/mm**2\"\n",
+ "print\"Internal Pressure that can be maintained is\",round(p_i,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses int the material are:F_x 70.5 N/mm**2\n",
+ " :F_x2 34.5 N/mm**2\n",
+ "Internal Pressure that can be maintained is 50.82 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.15,Page No.344"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "r1=200 #mm #Inner Radius\n",
+ "r2=250 #mm #Radius at common surface\n",
+ "r3=300 #mm #Outer radius\n",
+ "p=6 #N/mm**2 #Inital pressure\n",
+ "p2=80 #N/mm**2 #Pressure\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Inner Cyclinder:\n",
+ "\n",
+ "#From Lame's Equation we have\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#At \n",
+ "x=r1=200 #mm\n",
+ "p_x=0\n",
+ "#0=b1*(250**2)**-1-a1 .................(3)\n",
+ "\n",
+ "#At x=r2=250 #mm\n",
+ "p_x2=6 #N/mm**2\n",
+ "#6=b1*(250**2)-a1 ...................(4)\n",
+ "\n",
+ "#From Equation 3 and 4 we get\n",
+ "b1=6*200**2*250**2*(200**2-250**2)**-1\n",
+ "\n",
+ "#From equation 3 we get\n",
+ "a1=b1*(200**2)**-1\n",
+ "\n",
+ "F_200=b1*(200**2)**-1+a1\n",
+ "F_250=b1*(250**2)**-1+a1\n",
+ "\n",
+ "#For outer cyclinder \n",
+ "#From Lame's Equation we have\n",
+ "#p_x2=b2*(x**2)**-1-a2 ..........................(5)\n",
+ "#F_x2=b2*(x**2)**-1+a2 ...........................(6)\n",
+ "\n",
+ "\n",
+ "#At \n",
+ "x2=r2=250 #mm\n",
+ "p_x2=6 #N/mm**2\n",
+ "#6=b2*(250**2)**-1-a2 ...........................(7) \n",
+ "\n",
+ "#At\n",
+ "x3=300 #mm\n",
+ "#p_x2=0\n",
+ "#0=b2**2*(300**2)**-1-a2 .................................(8)\n",
+ "\n",
+ "#from equation 7 and 8 we get\n",
+ "b2=6*250**2*300**2*(300**2-250**2)**-1\n",
+ "\n",
+ "#sub in equation 8 we get\n",
+ "a2=b2*(300**2)**-1\n",
+ "\n",
+ "F_250_2=b2*(250**2)**-1+a2\n",
+ "F_300_2=b2*(300**2)**-1+a2\n",
+ "\n",
+ "#When Fluid is admitted\n",
+ "#Let Lame's equation be\n",
+ "#p_x3=b3*(x**2)**-1-a3 ..........................(5)\n",
+ "#F_x3=b3*(x**2)**-1+a3 ...........................(6)\n",
+ "\n",
+ "\n",
+ "#At x=200\n",
+ "p_x3=80 #N/mm**2\n",
+ "#80=b3*(200**2)**-1-a3 ................................(7)\n",
+ "\n",
+ "#At x=300 #mm\n",
+ "#p_x=0\n",
+ "#0=b3*(300**2)**-1-a3 ..............................(8)\n",
+ "\n",
+ "#from Equation 7 and 8 we get\n",
+ "b3=80*200**2*300**2*(300**2-200**2)**-1\n",
+ "\n",
+ "#From Equation 8 we get\n",
+ "a3=b3*(300**2)**-1\n",
+ "\n",
+ "#Hoop stresses \n",
+ "F_200_3=b3*(200**2)**-1+a3 #N/mm**2\n",
+ "F_250_3=b3*(250**2)**-1+a3 #N/mm**2\n",
+ "F_300_3=b3*(300**2)**-1+a3 #N/mm**2\n",
+ "\n",
+ "#Pressure at common surface\n",
+ "p_250=b3*(250**2)**-1-a3 #N/mm**2\n",
+ "\n",
+ "#final stress\n",
+ "f_200=F_200+F_200_3 #N/mm**2\n",
+ "f_250=F_250+F_250_3 #N/mm**2\n",
+ "f_300=F_250_2+F_250_3 #N/mm**2\n",
+ "f_300_2=F_300_2+F_300_3 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"final Hoop stress are:f_200\",round(f_200,2),\"N/mm**2\"\n",
+ "print\" :f_250\",round(f_250,2),\"N/mm**2\"\n",
+ "print\" :f_300\",round(f_300,2),\"N/mm**2\"\n",
+ "print\" :f_300_2\",round(f_300_2,2),\"N/mm**2\"\n",
+ "print\"Variation of Hoop stress and Radial stress\"\n",
+ "\n",
+ "#Final stresses\n",
+ "#Variation of hoop stress \n",
+ " \n",
+ "X1=[x,x2,x3,x3]\n",
+ "Y1=[f_200,f_250,f_300,f_300_2]\n",
+ "Z1=[0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Due to Fluid\n",
+ "#Variation of hoop stress \n",
+ " \n",
+ "X1=[x,x2,x3]\n",
+ "Y1=[F_200_3,F_250_3,F_300_3]\n",
+ "Z1=[0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "final Hoop stress are:f_200 174.67 N/mm**2\n",
+ " :f_250 128.83 N/mm**2\n",
+ " :f_300 189.43 N/mm**2\n",
+ " :f_300_2 155.27 N/mm**2\n",
+ "Variation of Hoop stress and Radial stress\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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MHz4cr7zyCmbPno1///vf1fIBnZeXp/l+8+bNmhVR0dHR+P7771FYWIjMzExcvHgRnTp1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\nVw+AnLO4d++eQY0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+ "text": [
+ "<matplotlib.figure.Figure at 0x5604510>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x56e5cb0>"
+ ]
+ }
+ ],
+ "prompt_number": 89
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.16,Page No.348"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "do=200 #mm #Inner Diameter\n",
+ "r_o=100 #mm #Inner radius\n",
+ "d1=300 #mm #outer diameter\n",
+ "r1=150 #mm #Outer radius\n",
+ "d2=250 #mm #Junction Diameter\n",
+ "r2=125 #mm #Junction radius\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "p=30 #N/mm**2 #radial pressure\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#from Lame's Equation we get\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#Then from Boundary condition \n",
+ "#p_x=0 at x=100 #mm\n",
+ "#0=b1*(100**2)**-1-a1 .....................(3)\n",
+ "\n",
+ "#p_x2=30 #N/mm**2 at x2=125 #mm\n",
+ "#30=b1*(125**2)**-1-a1 ................................(4)\n",
+ "\n",
+ "#From equation 3 and 4 we get\n",
+ "b1=30*125**2*100**2*(100**2-125**2)**-1\n",
+ "\n",
+ "#From Equation 3 we get\n",
+ "a1=b1*(100**2)**-1\n",
+ "\n",
+ "#therefore Hoop stress in inner cyclinder at junction\n",
+ "F_2_1=b1*(125**2)**-1+a1 #N/mm**2\n",
+ "\n",
+ "#Outer Cyclinder\n",
+ "#p_x=b*(x**2)**-1-a ..........................(5)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(6)\n",
+ "\n",
+ "#Now at x=125 #mm\n",
+ "#p_x3=30 #N/mm**2\n",
+ "#30=b2*(125**2)**-1-a2 ..................................(7)\n",
+ "\n",
+ "#At x=150 #mm\n",
+ "#p_x4=0\n",
+ "#0=b2*(150**2)**-1-a2 ...................................(8)\n",
+ "\n",
+ "#From equations 7 and 8\n",
+ "b2=30*150**2*125**2*(150**2-125**2)**-1\n",
+ "\n",
+ "#From eqauation 8 we get\n",
+ "a2=b2*(150**2)**-1\n",
+ "\n",
+ "#Hoop stress at junction \n",
+ "F_2_0=b2*(125**2)**-1+a2 #N/mm**2\n",
+ "\n",
+ "rho_r=(F_2_0-F_2_1)*E**-1*r2\n",
+ "\n",
+ "#Result\n",
+ "print\"Shrinkage Allowance is\",round(rho_r,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shrinkage Allowance is 0.189 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.17,Page No.350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=500 #mm #Outer Diameter\n",
+ "r_o=250 #mm #Outer Radius\n",
+ "d1=300 #mm #Inner Diameter\n",
+ "r1=150 #mm #Inner Radius\n",
+ "d2=400 #mm #Junction Diameter\n",
+ "E=2*10**5 #N/mm**2 #Modulus ofElasticity\n",
+ "alpha=12*10**-6 #Per degree celsius\n",
+ "dell_d=0.2 #mm\n",
+ "dell_r=0.1 #mm\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let p be the radial pressure developed at junction\n",
+ "#Let Lame's Equation for internal cyclinder be\n",
+ "#p_x=b*(x**2)**-1-a ................................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...............................(2)\n",
+ "\n",
+ "#At \n",
+ "x=150 #mm \n",
+ "p_x=0\n",
+ "#Sub in equation 1 we get\n",
+ "#0=b*(150**2)**-1-a .........................(3)\n",
+ "\n",
+ "#At \n",
+ "x2=200 #mm\n",
+ "#p_x2=p\n",
+ "#p=b*(200**2)**-1-a ......................(4)\n",
+ " \n",
+ "#From Equation 3 and 4\n",
+ "#p=b*(200**2)**-1-b(150**2)**-1\n",
+ "#after further simplifying we get\n",
+ "#b=-51428.571*p\n",
+ "\n",
+ "#sub in equation 3 we get\n",
+ "#a1=-2.2857*p\n",
+ "\n",
+ "#therefore hoop stress at junction is\n",
+ "#F_2_1=-21428.571*p*(200**2)**-1-2.2857*p\n",
+ "#after Further simplifying we geet\n",
+ "#F_2_1=3.5714*p\n",
+ "\n",
+ "#Let Lame's Equation for cyclinder be \n",
+ "#p_x=b*(x**2)**-1-a .........................5\n",
+ "#F_x=b*(x**2)**-1+a .............................6\n",
+ "\n",
+ "#At \n",
+ "x=200 #mm\n",
+ "#p_x=p2\n",
+ "#p2=b2*(20**2)**-1-a2 ...................7\n",
+ "\n",
+ "#At\n",
+ "x2=200 #mm\n",
+ "p_x2=0\n",
+ "#0=b2*(250**2)**-1-a2 ....................8\n",
+ "\n",
+ "#from equation 7 and 8 we get\n",
+ "#p2=b2*(200**2)**-1-b2*(250**2)**-1\n",
+ "#After further simplifying we get\n",
+ "#p2=b2*(250**2-200**2)*(200**2*250**2)**-1\n",
+ "#b2=111111.11*p\n",
+ "\n",
+ "#from equation 7\n",
+ "#a2=b2*(250**2)**-1\n",
+ "#further simplifying we get\n",
+ "#a2=1.778*p\n",
+ "\n",
+ "#At the junctionhoop stress in outer cyclinder \n",
+ "#F_2_0=b2*(200**2)**-1+a2\n",
+ "#After further simplifying we get\n",
+ "#F_2_0=4.5556*p\n",
+ "\n",
+ "#Considering circumferential strain,the compatibility condition\n",
+ "#rho_r*r2**-1=1*E**-1*(F_2_1+F_2_0)\n",
+ "#where F_2_1 is compressive and F_2_0 is tensile\n",
+ "#furter simplifying we get\n",
+ "p=0.1*200**-1*2*10**5*(3.5714+4.5556)**-1\n",
+ "\n",
+ "#Let T be the rise in temperature required\n",
+ "#dell_d=d*alpha*T\n",
+ "#After sub values and further simplifying we get\n",
+ "d=250 #mm\n",
+ "T=dell_d*(d*alpha)**-1 #Per degree celsius\n",
+ "\n",
+ "#Result\n",
+ "print\"Radial Pressure Developed at junction\",round(p,2),\"N/mm**2\"\n",
+ "print\"Min Temperatureto outer cyclinder\",round(T,2),\"Per degree Celsius\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radial Pressure Developed at junction 12.3 N/mm**2\n",
+ "Min Temperatureto outer cyclinder 66.67 Per degree Celsius\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.18,Page No.355"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=400 #mm #Outer Diameter\n",
+ "r_o=200 #mm #Outer radius\n",
+ "t=50 #mm #Thickness\n",
+ "r1=150 #mm #Internal Radius\n",
+ "p=50 #N/mm**2 #Internal Pressure\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#The Radial Pressure and hoop stress at any radial distance x are given by\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#Now at\n",
+ "x=150 #N/mm**2\n",
+ "p_x1=50 #N/mm**2\n",
+ "#Sub in equation 1 we get\n",
+ "#50=2*b*(150**3)**-1-a ...........................(3)\n",
+ "\n",
+ "#At x=200 #mm\n",
+ "p_x2=0\n",
+ "#0=2*b*(200**2)**-1-a ....................(4)\n",
+ "\n",
+ "#From equation 3 and 4 we get\n",
+ "#50=2*b*(150**3)**-1-2*b*(200**3)**-1\n",
+ "#After further simplifying we get\n",
+ "b=50*150**3*200**3*(200**3-150**3)**-1*2**-1\n",
+ "\n",
+ "#Sub in equation 3 we get\n",
+ "a=b*(200**3)**-1\n",
+ "\n",
+ "#Now At\n",
+ "x=150 #mm\n",
+ "F_x=b*(x**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x2=160 #mm\n",
+ "F_x2=b*(x2**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x3=170 #mm\n",
+ "F_x3=b*(x3**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x4=180 #mm\n",
+ "F_x4=b*(x4**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x5=190 #mm\n",
+ "F_x5=b*(x5**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x6=200 #mm\n",
+ "F_x6=b*(x6**3)**-1+a\n",
+ "\n",
+ "#Result\n",
+ "print\"Plot of Variation of hoop stress\"\n",
+ "\n",
+ "#Plotting Variation of hoop stress\n",
+ "\n",
+ "X1=[x,x2,x3,x4,x5,x6]\n",
+ "Y1=[F_x,F_x2,F_x3,F_x4,F_x5,F_x6]\n",
+ "Z1=[0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Plot of Variation of hoop stress\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x56b4bf0>"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.9_3.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.9_3.ipynb new file mode 100644 index 00000000..cecacb12 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.9_3.ipynb @@ -0,0 +1,779 @@ +{
+ "metadata": {
+ "name": "chapter no.9.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 9:Columns And Struts"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.1,Page No.377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "L=5000 #mm #Length of strut\n",
+ "dell=10 #mm #Deflection\n",
+ "W=10 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Central Deflection of a simply supported beam with central concentrated load is\n",
+ "#dell=W*L**3*(48*E*I)**-1 \n",
+ "\n",
+ "#Let E*I=X\n",
+ "X=W*L**3*(48*dell)**-1 #mm\n",
+ "\n",
+ "#Euler's Load\n",
+ "#Let Euler's Load be P\n",
+ "P=pi**2*X*(L**2)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Critical Load of Bar is\",round(P,2),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Critical Load of Bar is 1028.08 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.2,Page No.377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=2000 #mm #Length of square column\n",
+ "E=12*10**3 #N/mm**2 #Modulus of Elasticity\n",
+ "sigma=12 #N/mm*2 #stress\n",
+ "W1=95*10**3 #N #Load1\n",
+ "W2=200*10**3 #N #Load2\n",
+ "FOS=3\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Euler's Formula\n",
+ "#P=pi**2*E*I*(L**2)**-1 .........(1)\n",
+ "\n",
+ "#Working Load\n",
+ "#W=P*(FOS)**-1\n",
+ "\n",
+ "#Part-1\n",
+ "\n",
+ "#At W1=95*10**3 #N\n",
+ "#W1=P*(3*L**2)**-1\n",
+ "\n",
+ "#Let 'a' be the side of the square\n",
+ "#I=1*12**-1*a**4\n",
+ "\n",
+ "#sub value of I in Equation 1 and further rearranging we get\n",
+ "a=(W1*3*12*L**2*(pi**2*E)**-1)**0.25 #mm\n",
+ "\n",
+ "#From Consideration of direct crushing\n",
+ "#sigma*a**2=W1\n",
+ "#After Reaaranging the above equation we get\n",
+ "a2=(W1*(sigma)**-1)**0.5 #mm\n",
+ "\n",
+ "#required size is 103.67*103.67 i.e a*a\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#At W2=200*10**3 #N\n",
+ "#W2=P*(3*L**2)**-1\n",
+ "#After substituting values and further Rearranging the above equation we get\n",
+ "a3=(W2*3*12*L**2*(pi**2*E)**-1)**0.25 #mm\n",
+ "\n",
+ "#From consideration of direct compression,size required is\n",
+ "a4=(W2*sigma**-1)**0.5\n",
+ "\n",
+ "#required size is 129.10*129.10 i.e a4*a4\n",
+ "\n",
+ "#Result\n",
+ "print\"For W1 Load Required size is\",round(a*a,2),\"mm**2\"\n",
+ "print\"For W2 Load Required size is\",round(a4*a4,2),\"mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "For W1 Load Required size is 10747.38 mm**2\n",
+ "For W1 Load Required size is 16666.67 mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.3,Page No.378"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flange \n",
+ "b=100 #mm #Width\n",
+ "\n",
+ "D=80 #mm #Overall Depth\n",
+ "t=10 #mm #Thickness of web and flanges\n",
+ "L=3000 #mm #Length of strut\n",
+ "E=200*10**3 #N/mm**2 #Modulus of Elasticity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let centroid be at depth y_bar from top fibre\n",
+ "y_bar=(b*t*t*2**-1+(D-t)*t*((D-t)*2**-1+t))*(b*t+(D-t)*t)**-1 #mm \n",
+ "\n",
+ "#M.I at x-x axis\n",
+ "I_x=1*12**-1*b*t**3+b*t*(y_bar-t*2**-1)**2+1*12**-1*t*((D-t))**3+t*((D-t))*((((D-t)*2**-1)+t)-y_bar)**2\n",
+ "\n",
+ "#M.I at y-y axis\n",
+ "I_y=1*12**-1*t*b**3+1*12**-1*(D-t)*t**3 #mm**3\n",
+ "\n",
+ "#Least M.I\n",
+ "I=I_y\n",
+ "\n",
+ "#Since both ends are hinged\n",
+ "#Feective Length=Actual Length\n",
+ "L=l=3000 #mm\n",
+ "\n",
+ "#Buckling Load \n",
+ "P=pi**2*E*I*(l**2)**-1*10**-3 #KN\n",
+ "\n",
+ "#Result\n",
+ "print\"The Buckling Load for strut of tee section\",round(P,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Buckling Load for strut of tee section 184.05 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.4,Page No.379"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=400 #mm #Overall Depth\n",
+ "\n",
+ "#Flanges\n",
+ "b=300 #mm #Width\n",
+ "t=50 #mm #Thickness\n",
+ "\n",
+ "t2=30 #mm #Web Thickness\n",
+ "\n",
+ "dell=10 #mm #Deflection\n",
+ "w=40 #N/mm #Load\n",
+ "FOS=1.75 #Factor of safety\n",
+ "E=2*10**5 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#M.I at x-x axis\n",
+ "I_x=1*12**-1*(b*D**3-(b-t2)*b**3) #mm**4\n",
+ "\n",
+ "#Central Deflection\n",
+ "#dell=5*w*L**4*(384*E*I)**-1\n",
+ "#After sub values in above equation and further simplifying we get\n",
+ "L=(dell*384*E*I_x*(5*w)**-1)**0.25\n",
+ "\n",
+ "#M.I aty-y axis\n",
+ "I=I_y=1*12**-1*t*b**3+1*12**-1*b*t2**3+1*12**-1*t*b**3 #mm**4\n",
+ "\n",
+ "#Both the Ends of column are hinged\n",
+ "\n",
+ "#Crippling Load\n",
+ "P=pi**2*E*I*(L**2)**-1 #N\n",
+ "\n",
+ "#Safe Load\n",
+ "S=P*(FOS)**-1*10**-3 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe Load if I-section is used as column with both Ends hhinged\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe Load if I-section is used as column with both Ends hhinged 4123.29 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.5,Page No.381"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=200 #mm #External Diameter\n",
+ "t=20 #mm #hickness\n",
+ "d=200-2*t #mm #Internal Diameter\n",
+ "E=1*10**5 #N/mm**2\n",
+ "a=1*(1600)**-1 #Rankine's Constant\n",
+ "L=4.5 #m #Length\n",
+ "sigma=550 #N/mm**2 #Stress\n",
+ "FOS=2.5\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=pi*D**4*64**-1-pi*d**4*64**-1\n",
+ "\n",
+ "#Both Ends are fixed\n",
+ "\n",
+ "#Effective Length\n",
+ "l=1*2**-1*L*10**3 #mm\n",
+ "\n",
+ "#Euler's Critical Load\n",
+ "P_E=pi**2*E*I*(l**2)**-1\n",
+ "\n",
+ "A=pi*4**-1*(D**2-d**2) #mm*2\n",
+ "\n",
+ "k=(I*A**-1)**0.5\n",
+ "\n",
+ "#Rankine's Critical Load\n",
+ "P_R=sigma*A*(1+a*(l*k**-1)**2)**-1\n",
+ "\n",
+ "X=P_E*P_R**-1 \n",
+ "\n",
+ "#Safe Load using Rankine's Formula\n",
+ "S=P_R*(FOS)**-1*10**-3 #KN\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe Load by Rankine's Formula is\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe Load by Rankine's Formula is 1404.36 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.6,Page No.382"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=3000 #mm #Length of column\n",
+ "W=800*10**3 #N #Load\n",
+ "a=1*1600**-1 #Rankine's constant\n",
+ "FOS=4 #Factor of safety\n",
+ "sigma=550 #N/mm**2 #stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Effective Length\n",
+ "l=L*2**-1 #mm \n",
+ "\n",
+ "#Let d1=outer diameter & d2=inner diameter\n",
+ "#d1=5*8**-1*d2\n",
+ "\n",
+ "#M.I\n",
+ "#I=pi*64**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "#Area of section\n",
+ "#A=pi4**-1*(d1**2-d2**2) #mm**2\n",
+ "\n",
+ "#k=(I*A**-1) \n",
+ "#substituting values in above equation \n",
+ "#k=1*16**-1*(d1**2-d2**2)\n",
+ "#after simplifying further we get\n",
+ "#k=0.2948119.d1\n",
+ "\n",
+ "#X=l*k**-1\n",
+ "#substituting values in above equation and after simplifying further we get\n",
+ "#X=5087.9898*d1**-1\n",
+ "\n",
+ "#Crtitcal Load\n",
+ "P=W*FOS #N\n",
+ "\n",
+ "#From Rankine's Load\n",
+ "#P2=sigma*A*(1+a*(X)**2)**-1\n",
+ "#substituting values in above equation and after simplifying further we get\n",
+ "#d1**4-12156618*d1**4-1.96691*10**8=0\n",
+ "#Solving Quadratic Equation we get\n",
+ "#d1**2-12156618*d1-196691000=0\n",
+ "a=1\n",
+ "b=-12156.618\n",
+ "c=-196691000\n",
+ "\n",
+ "Y=b**2-4*a*c\n",
+ "\n",
+ "d1_1=((-b+Y**0.5)*(2*a)**-1)**0.5 #mm\n",
+ "d1_2=((-b-Y**0.5)*(2*a)**-1) #mm\n",
+ "\n",
+ "d2=5*8**-1*d1_1\n",
+ "\n",
+ "#Result\n",
+ "print\"Section of cast iron hollow cylindrical column is:d1_1\",round(d1_1,2),\"mm\"\n",
+ "print\" :d2 \",round(d2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Section of cast iron hollow cylindrical column is:d1_1 146.16 mm\n",
+ " :d2 91.35 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.7,Page No.383"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Let X=(P*A**-1) #Average Stress at Failure \n",
+ "Lamda_1=70 #Slenderness Ratio\n",
+ "Lamda_2=170 #Slenderness Ratio\n",
+ "X1=200 #N/mm**2 \n",
+ "X2=69 #N/mm**2 \n",
+ "\n",
+ "#Rectangular section\n",
+ "b=60 #mm #width\n",
+ "t=20 #mm #Thickness\n",
+ "\n",
+ "L=1250 #mm #Length of strut\n",
+ "FOS=4 #Factor of safety\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Slenderness ratio\n",
+ "#Lamda=L*k**-1\n",
+ "\n",
+ "#The Rankine's Formula for strut\n",
+ "#P=sigma*A*(1+a*(L*k**-1)**-1\n",
+ "\n",
+ "#From test result 1,\n",
+ "#After sub values in above equation we get and further simplifying we get\n",
+ "#sigma_1=200+980000*a ...................(1)\n",
+ "\n",
+ "#From test result 2,\n",
+ "#After sub values in above equation we get and further simplifying we get\n",
+ "#sigma_2=69+1994100*a ...................(2)\n",
+ "\n",
+ "#Substituting it in equation (1) we get\n",
+ "a=131*1014100**-1 \n",
+ "\n",
+ "#Substituting a in equation 1\n",
+ "sigma_1=200+980000*a #N/mm**2\n",
+ "\n",
+ "#Effective Length \n",
+ "l=1*2**-1*L #mm\n",
+ "\n",
+ "#Least of M.I\n",
+ "I=1*12**-1*b*t**3 #mm**4\n",
+ "\n",
+ "#Area \n",
+ "A=b*t #mm**2 \n",
+ "\n",
+ "k=(I*A**-1)**0.5\n",
+ "\n",
+ "#Slenderness ratio\n",
+ "Lamda=l*k**-1\n",
+ "\n",
+ "#From Rankine's Ratio\n",
+ "P=sigma_1*A*(1+a*(Lamda)**2)**-1\n",
+ "\n",
+ "#Safe Load\n",
+ "S=P*(FOS)**-1*10**-3 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Constant in the Formula is:a \",round(a,6)\n",
+ "print\" :sigma_1\",round(sigma_1,2)\n",
+ "print\"Safe Load is\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Constant in the Formula is:a 0.000129\n",
+ " :sigma_1 326.6\n",
+ "Safe Load is 38.98 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.8,Page No.385"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=200 #mm #Depth\n",
+ "b=140 #mm #width\n",
+ "\n",
+ "#Plate\n",
+ "b2=160 #mm #Width\n",
+ "t2=10 #mm #Thickness\n",
+ "\n",
+ "L=l=4000 #mm #Length\n",
+ "FOS=4 #Factor of safety\n",
+ "sigma=315 #N/mm**2 #stress\n",
+ "a2=1*7500**-1 \n",
+ "I_xx=26.245*10**6 #mm**4 #M.I at x-x\n",
+ "I_yy=3.288*10**6 #mm**4 #M.I at y-y\n",
+ "a=3671 #mm**2 #Area\n",
+ "k_x=84.6#mm\n",
+ "k_y=29.9 #mm\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Total Area\n",
+ "A=a+2*t2*b2 #mm**2\n",
+ "\n",
+ "#M.I\n",
+ "I=I_yy+2*12**-1*t2*b2**3 #mm**4\n",
+ "\n",
+ "k=(I*A**-1)**0.5 #mm\n",
+ "\n",
+ "#Let X=L*k**-1\n",
+ "X=L*k**-1\n",
+ "\n",
+ "#Appliying Rankine's Formula\n",
+ "P=sigma*A*(1+a2*(X)**2)**-1 #N\n",
+ "\n",
+ "#Safe Load\n",
+ "S=P*(FOS)**-1*10**-3 #KN\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe axial Load is\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe axial Load is 220.93 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.9,Page No.389"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=200*10**3 #N/mm**2 #Modulus of elasticity\n",
+ "sigma=330 #N/mm**2 #Stress\n",
+ "a=1*7500**-1 #Rankine's constant\n",
+ "A=5205 #mm**2 #area of column\n",
+ "I_xx=59.431*10**6 #mm**4 #M.I at x-x axis\n",
+ "I_yy=8.575*10**6 #mm**24#M.I at y-y axis\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Total M.I\n",
+ "I=I_xx+I_yy #mm**4\n",
+ "\n",
+ "#Area of compound Section \n",
+ "A2=2*A #mm**2\n",
+ "\n",
+ "k=(I*A2**-1)**0.5 #mm\n",
+ "\n",
+ "#Equating Euler's Load to Rankine's Load we get\n",
+ "#pi**2*E*I*(L**2)**-1=sigma*A*(1+a*(L*k)**2)**-1\n",
+ "#After Substitt=uting values and further simplifying we get\n",
+ "L=(39076198*(1-0.7975432)**-1)**0.5*10**-3 #m\n",
+ "\n",
+ "#Result\n",
+ "print\"Length of column for which Rankine's formula and Euler's Formula give the same result is\",round(L,2),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Length of column for which Rankine's formula and Euler's Formula give the same result is 13.89 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.10,Page No.387"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "sigma=326 #N/mm**2 #stress\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "FOS=2 #Factor of safety\n",
+ "a=1*7500**-1 #Rankine's constant\n",
+ "D=350 #mm #Overall Depth \n",
+ "\n",
+ "#Cover plates\n",
+ "b1=500 #mm #width\n",
+ "t1=10 #mm #Thickness\n",
+ "\n",
+ "d=220 #mm #Distance between two channels\n",
+ "\n",
+ "L=6000 #mm #Length of column\n",
+ "\n",
+ "A=5366 #mm**2 #Area of Column section \n",
+ "I_xx=100.08*10**6 #mm**4 #M.I of x-x axis\n",
+ "I_yy=4.306*10**6 #mm**4 #M.I of y-y axis\n",
+ "C_yy=23.6 #mm #Centroid at y-y axis\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Symmetric axes are the centroidal axes is\n",
+ "\n",
+ "#M.I of Channel at x-x axis\n",
+ "I_xx_1=2*I_xx+2*(1*12**-1*b1*t1**3+b1*t1*(D*2**-1+t1*2**-1)**2)\n",
+ "\n",
+ "#M.I of Channel at y-y axis\n",
+ "I_yy_1=2*(I_yy+A*(d*2**-1+C_yy)**2)+2*12**-1*t1*b1**3\n",
+ "\n",
+ "#As I_yy<I_xx\n",
+ "#So\n",
+ "I=I_yy_1 #mm**4 \n",
+ "\n",
+ "A2=2*A+2*t1*b1 #Area of channel\n",
+ "\n",
+ "k=(I*A2**-1)**0.5 #mm\n",
+ "\n",
+ "#Critical Load\n",
+ "P=sigma*A2*(1+a*(L*k**-1)**2)**-1 \n",
+ "\n",
+ "#Safe Load\n",
+ "S=P*2**-1*10**-3 #KN\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe Load carrying Capacity is\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe Load carrying Capacity is 2717.35 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 67
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.11,Page No.390"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "I=4.085*10**8 #mm**4 #M.I\n",
+ "A=20732.0 #mm**2 #area of column\n",
+ "f_y=250 #N/mm**2 \n",
+ "L=6000 #mm #Length of column\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "k=(I*A**-1)**0.5 #mm\n",
+ "lamda=L*k**-1 #Slenderness ratro\n",
+ "\n",
+ "#From Indian standard table\n",
+ "lamda_1=40 \n",
+ "sigma_a_c_1=139 #N/mm**2\n",
+ "lamda_2=50 \n",
+ "sigma_a_c_2=132 #N/mm**2 \n",
+ "\n",
+ "#Linearly interpolating between these values for lambda=42.744\n",
+ "\n",
+ "sigma_a_c_3=sigma_a_c_1-2.744*10**-1*(sigma_a_c_1-sigma_a_c_2)\n",
+ "\n",
+ "#Safe Load carrying capacity of column\n",
+ "P=sigma_a_c_3*A*10**-3\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe Load carrying capacity is\",round(P,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe Load carrying capacity is 2841.93 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/B.M.D_1_2.JPG b/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/B.M.D_1_2.JPG Binary files differnew file mode 100644 index 00000000..9bbe723e --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/B.M.D_1_2.JPG diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_1_2.jpg b/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_1_2.jpg Binary files differnew file mode 100644 index 00000000..0ccd99c0 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_1_2.jpg diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_2_2.jpg b/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_2_2.jpg Binary files differnew file mode 100644 index 00000000..3be9f649 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_2_2.jpg diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter10_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter10_2.ipynb new file mode 100644 index 00000000..c3f165d1 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter10_2.ipynb @@ -0,0 +1,938 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:08446d35c254f7719d0bb2f900fc25303d15f487208d5c6fccc1dc48e1acf8aa"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 10 Magnets and Earth's Magnetism"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.1 Page no 558"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "F=0.8*10**-3*9.8 #N\n",
+ "d=0.1 #m\n",
+ "u=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "m=math.sqrt(F*d**2/(u*5))\n",
+ "m1=5*m\n",
+ "\n",
+ "#Result\n",
+ "print\"Strength of pole M1 is\", round(m,2),\"Am\"\n",
+ "print\"Strength of pole M2 is\",round(m1,1),\"Am\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Strength of pole M1 is 12.52 Am\n",
+ "Strength of pole M2 is 62.6 Am\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.2 Page no 559"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "F=14.4*10**-4 #N\n",
+ "d=0.05 #m\n",
+ "F1=1.6*10**-4\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "u=4*math.pi*10**-7\n",
+ "m=math.sqrt((F*4*math.pi*d**2)/u)\n",
+ "d1=1/(math.sqrt((F1*4*math.pi)/(u*m**2)))\n",
+ "\n",
+ "#Result\n",
+ "print \"Distance is\",d1,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance is 0.15 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.5 Page no 560"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "M=8\n",
+ "d=0.2\n",
+ "\n",
+ "#Calculation\n",
+ "B=u*2*M/(4*math.pi*d**3)\n",
+ "Beqa=B/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Magnetic induction at axial point\", B*10**4,\"*10**-4 T\"\n",
+ "print\"(ii) Magnetic induction at equatorial point is\",Beqa*10**4,\"*10**-4 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnetic induction at axial point 2.0 *10**-4 T\n",
+ "(ii) Magnetic induction at equatorial point is 1.0 *10**-4 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.6 Page no 560"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=6.4*10**6 #m\n",
+ "B=0.4*10**-4 #T\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "M=(B*4*math.pi*d**3)/u\n",
+ "\n",
+ "#Result\n",
+ "print\"Earth's dipole moment is\", round(M*10**-23,2)*10**23,\"Am**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Earth's dipole moment is 1.05e+23 Am**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.7 Page no 560"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=0.40\n",
+ "d=0.5\n",
+ "\n",
+ "#Calculation\n",
+ "Beqa=u*M/(4*math.pi*d**3)\n",
+ "Baxial=2*Beqa\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of axial field is\", Baxial,\"T\"\n",
+ "print\"Magnitude of equatorial field is\",Beqa,\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of axial field is 6.4e-07 T\n",
+ "Magnitude of equatorial field is 3.2e-07 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.8 Page no 560"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19\n",
+ "f=6.8*10**15\n",
+ "n=1\n",
+ "r=0.53*10**-10\n",
+ "\n",
+ "#Calculation \n",
+ "import math\n",
+ "I=e*f\n",
+ "M=n*I*math.pi*r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Equivalent magnetic moment is\", round(M*10**24,1)*10**-24,\"Am**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equivalent magnetic moment is 9.6e-24 Am**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.9 Page no 561"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=50\n",
+ "r=0.2 #m\n",
+ "I=12 #A\n",
+ "\n",
+ "#Calculation\n",
+ "B=(u*n*I)/(2.0*r)\n",
+ "M=n*I*math.pi*r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Magnetic field at the centre of the coil is\", round(B*10**3,3),\"*10**-3 T\"\n",
+ "print\"(ii) Magnetic moment is\",round(M,1),\"Am**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnetic field at the centre of the coil is 1.885 *10**-3 T\n",
+ "(ii) Magnetic moment is 75.4 Am**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 58
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.10 Page no 561"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=7.5*10**-4 #m**2\n",
+ "I=12 #A\n",
+ "\n",
+ "#Calculation\n",
+ "M=A*I\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the magnetic moment is\", M*10**3,\"*10**-3 Am**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the magnetic moment is 9.0 *10**-3 Am**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 62
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.11 Page no 561"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=100\n",
+ "I=0.1 #A\n",
+ "r=0.05\n",
+ "B=1.5 #T\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "M=n*I*math.pi*r**2\n",
+ "W=2*M*B\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the coil is\", round(M,4),\"Am**2\"\n",
+ "print\"Workdone is\",round(W,4),\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the coil is 0.0785 Am**2\n",
+ "Workdone is 0.2356 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 70
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.12 Page no 561"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=10\n",
+ "I=3\n",
+ "A=7.85*10**-3\n",
+ "B=10**-2 #T\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "M=n*I*A\n",
+ "U1=-M*B*math.cos(0)\n",
+ "Uf=-M*B*math.cos(90)\n",
+ "w=-U1\n",
+ "t=M*B*math.sin(90*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Work done is\", round(t*10**3,1),\"*10**-3 Nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Work done is 2.4 *10**-3 Nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 81
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.13 Page no 562"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=4.8*10**-2 #J/T\n",
+ "a=30 #degree\n",
+ "B=3*10**-2 #t\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "t=M*B*math.sin(a*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Torque acting on the needle is\", round(t*10**4,1),\"*10**-4 Nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Torque acting on the needle is 7.2 *10**-4 Nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 86
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.14 Page no 562"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=0.2 #T\n",
+ "a=30 #degree\n",
+ "t=0.06 #Nm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "M=t/(B*math.sin(a*3.14/180.0))\n",
+ "U=M*B*math.cos(1*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Magnetic moment of the magnet is\", round(M,1),\"Am**2\"\n",
+ "print\"(ii) Orientation of the magnet is\", round(U,0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnetic moment of the magnet is 0.6 Am**2\n",
+ "(ii) Orientation of the magnet is 0.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 97
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.15 Page no 562"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "a=30 #degree\n",
+ "B=800*10**-4 #T\n",
+ "t=0.016 #Nm\n",
+ "A=2*10**-4 #m**2\n",
+ "n=1000 #turns\n",
+ "\n",
+ "#Calculation\n",
+ "M=t/(B*math.sin(a*3.14/180.0))\n",
+ "W=2*M*B\n",
+ "I=M/(n*A)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Magnetic moment of the magnet is\", round(M,2),\"Am**2\"\n",
+ "print\"(b) Work done is\", round(W,3),\"J\"\n",
+ "print\"(c) Current flowing through the solenoid is\", round(I,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Magnetic moment of the magnet is 0.4 Am**2\n",
+ "(b) Work done is 0.064 J\n",
+ "(c) Current flowing through the solenoid is 2.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 108
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.16 Page no 563"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "t=6.70\n",
+ "n=10.0\n",
+ "I=7.5*10**-6 #Kgm**2\n",
+ "M=6.7*10**-2 #Am**2\n",
+ "\n",
+ "#Calculation\n",
+ "T=t/n\n",
+ "B=(4*math.pi**2*I)/(M*T**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the magnetic field is\", round(B,2),\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the magnetic field is 0.01 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 113
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.18 Page no 569"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "t=1.2*10**-3 #nm\n",
+ "M=60\n",
+ "H=40*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=t/(M*H)\n",
+ "a=math.asin(A)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of the declination is\", round(a,0),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of the declination is 30.0 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 126
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.19 Page no 569"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=math.sqrt(3)\n",
+ "\n",
+ "#calculation\n",
+ "import math\n",
+ "a=math.atan(V)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of dip is\", round(a,0),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of dip is 60.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 131
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.20 Page no 569"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "H=0.28 #G\n",
+ "V=0.40 #G\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=V/H\n",
+ "a=math.atan(A)*180/3.14\n",
+ "R=math.sqrt(H**2+V**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Angle of dip is\", round(a,0),\"Degree\"\n",
+ "print\"(ii) Earth's total magnetic field is\", round(R,2),\"G\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Angle of dip is 55.0 Degree\n",
+ "(ii) Earth's total magnetic field is 0.49 G\n"
+ ]
+ }
+ ],
+ "prompt_number": 140
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.22 Page no 570"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "H=0.40\n",
+ "a=18 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "R=H/(math.cos(a*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of earth's magnetic field is\", round(R,2),\"G\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of earth's magnetic field is 0.42 G\n"
+ ]
+ }
+ ],
+ "prompt_number": 145
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.24 Page no 571"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=45 #Degree\n",
+ "b=60 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=math.tan(a*3.14/180.0)/(math.cos(b*3.14/180.0))\n",
+ "a=math.atan(A)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Apparant dip is\", round(a,1),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Apparant dip is 63.4 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 152
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.25 Page no 574"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=1.6 #Am**2\n",
+ "d=0.20 #m\n",
+ "u=10**-7 #N/A**2\n",
+ "\n",
+ "#Calculation\n",
+ "H=u*2*M/(d**3)\n",
+ "\n",
+ "#Result\n",
+ "print\"Horizontal component of the earth's magnetic field is\", H,\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Horizontal component of the earth's magnetic field is 4e-05 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 155
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.26 Page no 574"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=0.05 #m\n",
+ "d=0.12 #m\n",
+ "H=0.34*10**-4 #T\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "u=4*math.pi*10**-7\n",
+ "M=(4*math.pi*H*(d**2+l**2)**1.5)/u\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic moment of the magnet is\", round(M,3),\"J/T\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic moment of the magnet is 0.747 J/T\n"
+ ]
+ }
+ ],
+ "prompt_number": 162
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.27 Page no 577"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=7*10**-2 #m\n",
+ "H=2*10**-5 #T\n",
+ "n=50\n",
+ "\n",
+ "#calculation\n",
+ "import math\n",
+ "l=(2*r*H*math.tan(45*180/3.14))/u*n\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of current is\", round(l*10**-3,3),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of current is 0.043 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 172
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.28 Page no 577"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "K=0.095 #A\n",
+ "n=50\n",
+ "r=10*10**-2 #m\n",
+ "\n",
+ "#Calculation\n",
+ "H=K*u*n/(2.0*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"Horizontal component of earth's magnetic field is\", round(H*10**4,3),\"*10**-4 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Horizontal component of earth's magnetic field is 0.298 *10**-4 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 178
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.30 Page no 577"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=30 #degree\n",
+ "b=45 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "m=(2*math.tan(a*3.14/180.0))/(math.tan(b*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"Ratio of number of turns of the tangent galvanometers\", round(m,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Ratio of number of turns of the tangent galvanometers 1.155\n"
+ ]
+ }
+ ],
+ "prompt_number": 188
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter28_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter28_2.ipynb new file mode 100644 index 00000000..afb57de5 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter28_2.ipynb @@ -0,0 +1,59 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:511d2d405e1ede92783e0ff6e1c085ebc325e49ab2eff49fe438f3081d70cb4d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter28 Digital Electronics"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Example 28.3 page no 1497"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=1\n",
+ "\n",
+ "#Calculation\n",
+ "A=a*2**5+a*2**4+a*2**0\n",
+ "\n",
+ "#Result\n",
+ "print\"equivilant decimal is\",A"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "equivilant decimal is 49\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter29_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter29_2.ipynb new file mode 100644 index 00000000..9f1a9134 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter29_2.ipynb @@ -0,0 +1,526 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:8b87017ad47964520c2ead85c01092623a6e1bffcb1688b462701d086beba4f8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter29 Communication System"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.1 page no 1550"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "c=3*10**8\n",
+ "f=30.0*10**6\n",
+ "f1=300*10**6\n",
+ "f2=3000*10**6\n",
+ "\n",
+ "#Calculation\n",
+ "l=c/f\n",
+ "l1=l/2.0\n",
+ "l2=c/f1\n",
+ "l3=l2/2.0\n",
+ "l4=c/f2\n",
+ "l5=l4/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) length of half wave dipole antenna at 30 MHz is\",l1,\"m\"\n",
+ "print\"(ii) length of half wave dipole antenna at 300 MHz is\",l3,\"m\"\n",
+ "print\"(iii) length of half wave dipole antenna at 3000 MHz is\",15,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) length of half wave dipole antenna at 30 MHz is 5.0 m\n",
+ "(ii) length of half wave dipole antenna at 300 MHz is 0.5 m\n",
+ "(iii) length of half wave dipole antenna at 3000 MHz is 15 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.2 page no 1550"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "c=3*10**8 #m/s\n",
+ "f=3*10**4\n",
+ "f1=6*10**6\n",
+ "f2=5*10**7\n",
+ "\n",
+ "#Calculation\n",
+ "l=c/(4.0*f)\n",
+ "l1=c/(4.0*f1)\n",
+ "l2=c/(4.0*f2)\n",
+ "\n",
+ "#Result \n",
+ "print\"(i) length of quarter wave dipole antenna at 3*10**4 is\",l,\"m\"\n",
+ "print\"(ii) length of quarter wave dipole antenna at 6*10**6 is\",l1,\"m\"\n",
+ "print\"(iii) length of quarter wave dipole antena at 5*10**7 is\",l2,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) length of quarter wave dipole antenna at 3*10**4 is 2500.0 m\n",
+ "(ii) length of quarter wave dipole antenna at 6*10**6 is 12.5 m\n",
+ "(iii) length of quarter wave dipole antena at 5*10**7 is 1.5 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.3 page no 1551"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "AM=16 #mV\n",
+ "AM1=4 #mV\n",
+ "\n",
+ "#Calculation\n",
+ "Vmax=AM/2.0\n",
+ "Vmin=AM1/2.0\n",
+ "Ma=(Vmax-Vmin)/(Vmax+Vmin)\n",
+ "\n",
+ "#Result\n",
+ "print\" The modulation factor is\",Ma"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The modulation factor is 0.6\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.4 page no 1551"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Es=50 #V\n",
+ "Ec=100.0 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Ma=Es/Ec\n",
+ "\n",
+ "#Result\n",
+ "print\"The modulation factor\",Ma"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The modulation factor 0.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.6 page no 1552"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Pc=500 #watts\n",
+ "\n",
+ "#Calculation\n",
+ "Ps=(1/2.0)*(Pc)\n",
+ "Pt=Pc+Ps\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) sideband power is\",Ps,\"W\"\n",
+ "print\"(ii) power of AM wave is\",Pt,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) sideband power is 250.0 W\n",
+ "(ii) power of AM wave is 750.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.7 page no 1552"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Pc=50\n",
+ "Ma=0.8\n",
+ "Ma1=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "Ps=(1/2.0)*Ma**2*Pc\n",
+ "Ps1=(1/2.0)*Ma1**2*Pc\n",
+ "\n",
+ "#Result\n",
+ "print\"total sideband at 80% is\",Ps,\"KW\"\n",
+ "print\"total sideband at 10% is\",Ps1,\"KW\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "total sideband at 80% is 16.0 KW\n",
+ "total sideband at 10% is 0.25 KW\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.8 page no 1552"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Fc=500 #KHz\n",
+ "Fs=1 #KHz\n",
+ "\n",
+ "#Calculation\n",
+ "A1=Fc+Fs\n",
+ "A2=Fc-Fs\n",
+ "B=A1-A2\n",
+ "\n",
+ "#Result\n",
+ "print\"sideband frequancies are\",A1,\"KHz and\",A2,\"KHz\"\n",
+ "print\"bandwidth required is\",B,\"KHz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "sideband frequancies are 501 KHz and 499 KHz\n",
+ "bandwidth required is 2 KHz\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.9 page no 1552"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "F=10**10 #Hz\n",
+ "D=8*10**3 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "B=2/100.0*10**10\n",
+ "C=B/D\n",
+ "\n",
+ "#Result\n",
+ "print\"No. of telephones channels are\",C*10**-4,\"10**4\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "No. of telephones channels are 2.5 10**4\n"
+ ]
+ }
+ ],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.10 page no 1553"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=800.0*10**-7\n",
+ "C=3.0*10**8\n",
+ "f1=4.5*10**6 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "f=C/L\n",
+ "d=(1/100.0)*f\n",
+ "E=d/L\n",
+ "G=d/f1\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) number of channels for audio signal is\",round(E*10**-14,1),\"*10**8\"\n",
+ "print\"(ii) number of channels for video tv signal is\",round(G*10**-3,1),\"*10**5\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) number of channels for audio signal is 4.7 *10**8\n",
+ "(ii) number of channels for video tv signal is 8.3 *10**5\n"
+ ]
+ }
+ ],
+ "prompt_number": 80
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.11 Page no 1565"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=6400*10**3 #m\n",
+ "h=160\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "d=math.sqrt(2*R*h)\n",
+ "h2=4*h\n",
+ "\n",
+ "#Result\n",
+ "print\"Height is\", h2,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Height is 640 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 85
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.12 Page no 1565"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=6.4*10**6 #m\n",
+ "h=110\n",
+ "\n",
+ "#Calculation \n",
+ "import math\n",
+ "d=(math.sqrt(2*R*h))*10**-3\n",
+ "P=math.pi*d**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Population covered is\", round(P*10**-3,1),\"*10**6\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Population covered is 4.4 *10**6\n"
+ ]
+ }
+ ],
+ "prompt_number": 93
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.13 Page no 1565"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=6.4*10**6 #m\n",
+ "hr=50 #m\n",
+ "ht=32 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "d=math.sqrt(2*R*ht)+math.sqrt(2*R*hr)\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum distance is\", round(d*10**-3,1),\"Km\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum distance is 45.5 Km\n"
+ ]
+ }
+ ],
+ "prompt_number": 99
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.14 Page no 1566"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=300\n",
+ "R=6.4*10**6 #m\n",
+ "N=10**12\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "d=math.sqrt(2*R*h)\n",
+ "fc=9*N**0.5\n",
+ "\n",
+ "#Result\n",
+ "print\"fc=\", fc*10**-6,\"MHz\"\n",
+ "print\"5 MHz comes via ionospheric propogation.and 100 MHz signal comes via satellite transmission.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "fc= 9.0 MHz\n",
+ "5 MHz comes via ionospheric propogation.and 100 MHz signal comes via satellite transmission.\n"
+ ]
+ }
+ ],
+ "prompt_number": 107
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter11_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter11_2.ipynb new file mode 100644 index 00000000..11322719 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter11_2.ipynb @@ -0,0 +1,467 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:f9cf5cb53006209575af5d70cc318cffdaba99568e9075844fb3e2f1810bf06f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 11 Classification of magnetic materials"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.1 Page no 614"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19\n",
+ "f=6.8*10**15\n",
+ "n=1\n",
+ "r=0.53*10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=e*f\n",
+ "M=n*I*math.pi*r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Equivalent magnetic moment is\", round(M*10**24,1)*10**-24,\"Am**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equivalent magnetic moment is 9.6e-24 Am**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.2 Page no 615"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=240\n",
+ "R=474.0\n",
+ "r=12.5*10**-2\n",
+ "N=500\n",
+ "ur=5000\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=E/R\n",
+ "I1=2*math.pi*r\n",
+ "H=(N*I)/I1\n",
+ "u=4*math.pi*10**-7\n",
+ "B=u*ur*H\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The magnetising force is\", round(H,0),\"AT/m\"\n",
+ "print\"(ii) The magnetic flux density is\",round(B,2),\"Wb/m**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The magnetising force is 322.0 AT/m\n",
+ "(ii) The magnetic flux density is 2.03 Wb/m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.3 Page no 615"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r1=11\n",
+ "r2=12\n",
+ "B=2.5 #T\n",
+ "a=3000\n",
+ "I=0.70 #A\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=((r1+r2)/2.0)*10**-2\n",
+ "n=a/(2*math.pi*r)\n",
+ "ur=B*2*math.pi*r/(4*math.pi*10**-7*a*I)\n",
+ "\n",
+ "#Result\n",
+ "print\"Relative permeability is\", round(ur,1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Relative permeability is 684.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.4 Page no 616"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=0.5 #m\n",
+ "N=500\n",
+ "I1=0.15 #A\n",
+ "a=5000\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "H=(N*I1)/I\n",
+ "B=4*math.pi*10**-7*H\n",
+ "B1=B*a\n",
+ "I3=(B1-(H*4*math.pi*10**-7))/(4.0*math.pi*10**-7)\n",
+ "\n",
+ "#Result\n",
+ "print round(I3*10**-5,1),\"*10**5 A/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "7.5 *10**5 A/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.5 Page no 616"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=0.6\n",
+ "H=360.0\n",
+ "\n",
+ "#Calculation\n",
+ "u=B/H\n",
+ "x=(u-1*4*math.pi*10**-7)/(4.0*math.pi*10**-7)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Permeability is\",round(u*10**3,2),\"*10**-3 T/A m\"\n",
+ "print\"(ii) Susceptibility of the material is\",round(x,0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Permeability is 1.67 *10**-3 T/A m\n",
+ "(ii) Susceptibility of the material is 1325.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.6 Page no 616"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=8.0*10**22 #Am**2\n",
+ "R=64*10**5 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=(3*M)/(4.0*math.pi*R**3)\n",
+ "\n",
+ "#Result\n",
+ "print\"Earth's magnetisation is\", round(I,1),\"A/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Earth's magnetisation is 72.9 A/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 52
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.7 Page no 616"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "N=1800\n",
+ "l=0.6\n",
+ "I=0.9 #A\n",
+ "ur=500\n",
+ "n1=6.02*10**26\n",
+ "a=55.85\n",
+ "y=7850\n",
+ "\n",
+ "#Calculation\n",
+ "n=N/l\n",
+ "H=n*I\n",
+ "I1=(ur-1)*H\n",
+ "B=4*math.pi*10**-7*ur*H\n",
+ "x=(y*n1)/a\n",
+ "X=I1/x\n",
+ "\n",
+ "#Result\n",
+ "print\"Average magnetic moment per iron atom is\", round(X*10**23,2)*10**-23,\" A m**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average magnetic moment per iron atom is 1.59e-23 A m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 63
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.8 Page no 617"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=8.4 #g\n",
+ "d=7200.0\n",
+ "f=50 #Hz\n",
+ "E=3.2*10**4\n",
+ "t=30*60.0\n",
+ "\n",
+ "#Calculation\n",
+ "V=M/d\n",
+ "P=E/t\n",
+ "E1=P/(V*f)\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy dissipated per unit volume is\", round(E1,0),\"J/m**3/cycle\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy dissipated per unit volume is 305.0 J/m**3/cycle\n"
+ ]
+ }
+ ],
+ "prompt_number": 70
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.9 Page no 617"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "H=4*10**3 #A/m\n",
+ "a=60\n",
+ "b=0.12\n",
+ "\n",
+ "#Calculation\n",
+ "n=a/b\n",
+ "I=H/n\n",
+ "\n",
+ "#Result\n",
+ "print\"Current should be sent through the solenoid is\", I,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current should be sent through the solenoid is 8.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 73
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.10 Page no 617"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "x=1.68*10**-4\n",
+ "T1=293\n",
+ "T2=77.4\n",
+ "\n",
+ "#Calculation\n",
+ "x1=(x*T1)/T2\n",
+ "\n",
+ "#Result\n",
+ "print\"Susceptibility is\", round(x1*10**4,2),\"*10**-4\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Susceptibility is 6.36 *10**-4\n"
+ ]
+ }
+ ],
+ "prompt_number": 79
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.11 Page no 617"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=10**-6 #m\n",
+ "d=7.9 #g\n",
+ "a=6.023*10**23\n",
+ "n=55.0\n",
+ "M1=9.27*10**-24\n",
+ "\n",
+ "#Calculation\n",
+ "V=l**2\n",
+ "M=V*d\n",
+ "N=(a*M)/n\n",
+ "Mmax=N*M1\n",
+ "Imax=Mmax/V*10**-4\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of iron atom is\",round(N*10**-10,2),\"*10**10 atoms\"\n",
+ "print\"Magnetisation of the dipole is\",round(Imax*10**5,0),\"*10**5 A/m\"\n",
+ "print\"Maximum possible dipole moment is\",round(Mmax*10**13,0)*10**-13,\"A m**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of iron atom is 8.65 *10**10 atoms\n",
+ "Magnetisation of the dipole is 8.0 *10**5 A/m\n",
+ "Maximum possible dipole moment is 8e-13 A m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 113
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter12_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter12_2.ipynb new file mode 100644 index 00000000..c7cfe98a --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter12_2.ipynb @@ -0,0 +1,1049 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:58b6e0817b83a6d5c266f15dbe5c66e891c1c2ed8d1ef60dafaf43f98ff2d620"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 12 Electromagnetic induction"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.1 Page no 665"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=20 #mWb\n",
+ "a1=-20 #mWb\n",
+ "t=2*10**-3 #s\n",
+ "N=100\n",
+ "\n",
+ "#Calculation\n",
+ "a2=(a1-a)*10**-3\n",
+ "e=(-N*a2)/t\n",
+ "\n",
+ "#Result\n",
+ "print\"Average e.m.f induced in the coil is\", e,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average e.m.f induced in the coil is 2000.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.2 Page no 665"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=5*10**-2 #m\n",
+ "N=1\n",
+ "B=0.35\n",
+ "t=0.12 #S\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=math.pi*r**2\n",
+ "a1=B*A\n",
+ "a2=-B*a1\n",
+ "e=(N*a1)/t\n",
+ "\n",
+ "#Result\n",
+ "print round(e,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "0.02 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.3 Page no 665"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=10**-2 #m**2\n",
+ "a=45 #degree\n",
+ "B1=0.1 #T\n",
+ "R=0.5 #ohm\n",
+ "t=0.7 #S\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a1=B1*A*math.cos(a*3.14/180.0)\n",
+ "a2=0\n",
+ "a3=a1-a2\n",
+ "e=a3/t\n",
+ "I=e/R\n",
+ "\n",
+ "#Result\n",
+ "print\"Current during this time interval is\", round(I*10**3,1),\"*10**-3 A\"\n",
+ "print\"Magnitude of induced emf is\",round(e*10**3,0),\"*10**-3 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current during this time interval is 2.0 *10**-3 A\n",
+ "Magnitude of induced emf is 1.0 *10**-3 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.4 Page no 666"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=1.2*10**-3 #A\n",
+ "N=1.0\n",
+ "R=10 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "e=I*R\n",
+ "a=e/N\n",
+ "\n",
+ "#Result\n",
+ "print\"Necessary rate is\", a*10**2,\"*10**-2 Wb/second\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Necessary rate is 1.2 *10**-2 Wb/second\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.5 Page no 666"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=10**-1 #m\n",
+ "B=3.0*10**-5 #T\n",
+ "t=0.25 #S\n",
+ "N=500\n",
+ "R=2 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a1=B*math.pi*r**2*math.cos(0*3.14/180.0)\n",
+ "a2=B*math.pi*r**2*math.cos(180*3.14/180.0)\n",
+ "a3=a1-a2\n",
+ "e=(N*a3)/t\n",
+ "I=e/R\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the emf is\", round(e*10**3,1),\"*10**-3 V\"\n",
+ "print\"Current induced in the coil is\",round(I*10**3,1),\"*1)**-3 A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the emf is 3.8 *10**-3 V\n",
+ "Current induced in the coil is 1.9 *1)**-3 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.6 Page no 666"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=10**-2 #V\n",
+ "B=5*10**-5 #T\n",
+ "r=0.5 #m\n",
+ "N=1\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=math.pi*r**2\n",
+ "n=(e*N)/(math.pi*r**2*B)\n",
+ "\n",
+ "#Result\n",
+ "print\"Rate of rotation of the blade is\", round(n,1),\"revolutions/second\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate of rotation of the blade is 254.6 revolutions/second\n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.7 Page no 667"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=12\n",
+ "b=7\n",
+ "t=2\n",
+ "\n",
+ "#Calculation\n",
+ "e=((a*t)+b)*10**-3\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Magnitude of induced emf is\", e*10**3,\"mV\"\n",
+ "print\"(ii) The current induced in the coil will be anticlockwise\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnitude of induced emf is 31.0 mV\n",
+ "(ii) The current induced in the coil will be anticlockwise\n"
+ ]
+ }
+ ],
+ "prompt_number": 59
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.8 Page no 673"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=1 #T\n",
+ "l=0.5 #m\n",
+ "v=40 #m/s\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "e=B*l*v*math.sin(60*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"emf induced in the conductor is\", round(e,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "emf induced in the conductor is 17.32\n"
+ ]
+ }
+ ],
+ "prompt_number": 63
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.9 Page no 673"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "g=9.8\n",
+ "h=10\n",
+ "B=1.7*10**-5\n",
+ "l=1 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "v=math.sqrt(2*g*h)\n",
+ "e=B*l*v\n",
+ "\n",
+ "#Result\n",
+ "print\"Potential difference between its end is\", e*10**4,\"*10**4 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Potential difference between its end is 2.38 *10**4 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.10 Page no 673"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=72 *(5/18.0) #Km/h\n",
+ "B=40*10**-6 #T\n",
+ "A=40\n",
+ "l=2 #m\n",
+ "t=1.0\n",
+ "N=1\n",
+ "\n",
+ "#Calculation\n",
+ "A=l*v\n",
+ "a=B*A\n",
+ "e=N*a/t\n",
+ "\n",
+ "#Result\n",
+ "print\"e.m.f generated in the axle of the car\", e*10**3,\"mV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "e.m.f generated in the axle of the car 1.6 mV\n"
+ ]
+ }
+ ],
+ "prompt_number": 77
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.11 Page no 673"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "w=1000/60.0\n",
+ "r=0.3\n",
+ "B=0.5 #T\n",
+ "\n",
+ "#Calculation\n",
+ "v=w*r\n",
+ "vav=v/2.0\n",
+ "e=B*r*vav\n",
+ "\n",
+ "#Result\n",
+ "print\"e.m.f induced is\",e,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "e.m.f induced is 0.375 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 84
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.12 Page no "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.5 #m\n",
+ "n=2 #r.p.s\n",
+ "B=0.4*10**-4 #T\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "w=2*math.pi*n\n",
+ "e=0.5*B*r**2*w\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of induced e.m.f between the axle and rim is\", round(e*10**5,2)*10**-5,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of induced e.m.f between the axle and rim is 6.28e-05 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 91
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.13 Page no 674"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=1 #m\n",
+ "B=1\n",
+ "f=50\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "e=math.pi*R**2*B*f\n",
+ "\n",
+ "#Result\n",
+ "print\"e.m.f between the centre and the matallic ring is\", round(e,1),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "e.m.f between the centre and the matallic ring is 157.1 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 96
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.14 Page no 679"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=500\n",
+ "a=1.4*10**-4 #Wb\n",
+ "l=2.5 #A\n",
+ "\n",
+ "#Calculation\n",
+ "L=(N*a)/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Inductance of the coil is\", L*10**3,\"mH\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Inductance of the coil is 28.0 mH\n"
+ ]
+ }
+ ],
+ "prompt_number": 100
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.15 Page no 679"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=130*10**-3 #H\n",
+ "I1=20 #mA\n",
+ "I2=28 #mA\n",
+ "t=140.0*10**-3 #S\n",
+ "\n",
+ "#Calculation\n",
+ "l=I2-I1\n",
+ "e=(-L*l)/t\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of induced e.m.f is\", round(e,2),\"*10**-3 V\"\n",
+ "print\"Direction oppose the increase in current\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of induced e.m.f is -7.43 *10**-3 V\n",
+ "Direction oppose the increase in current\n"
+ ]
+ }
+ ],
+ "prompt_number": 106
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.16 Page no 679"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=4000\n",
+ "l=0.6 #m\n",
+ "r=16*10**-4 #m\n",
+ "\n",
+ "#Calculation\n",
+ "u=4*math.pi*10**-7\n",
+ "L=(u*N**2*((math.pi*r)/4.0))/l\n",
+ "Liron=N*L\n",
+ "\n",
+ "#Result\n",
+ "print\"Inductance of the solenoid is\", round(Liron,0),\"H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Inductance of the solenoid is 168.0 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 115
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.17 Page no 679"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=10.0 #H\n",
+ "e=300 #V\n",
+ "t=10**-2 #S\n",
+ "\n",
+ "#Calculation\n",
+ "dl=(e*t)/L\n",
+ "a=e*t\n",
+ "\n",
+ "#Result\n",
+ "print\"Charge in magnetic flux is\", a,\"Wb\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge in magnetic flux is 3.0 Wb\n"
+ ]
+ }
+ ],
+ "prompt_number": 119
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.18 Page no 680"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=10*10**-3\n",
+ "I=4*10**-3\n",
+ "N=200.0\n",
+ "\n",
+ "#Calculation\n",
+ "N1=L*I\n",
+ "a=N1/N\n",
+ "\n",
+ "#Result\n",
+ "print\"Total flux linked with the coil is\", N1,\"Wb\"\n",
+ "print\"Magnetic flux through the cross section of the coil is\",a,\"Wb\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total flux linked with the coil is 4e-05 Wb\n",
+ "Magnetic flux through the cross section of the coil is 2e-07 Wb\n"
+ ]
+ }
+ ],
+ "prompt_number": 125
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.19 Page no 680"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=500*10**-3\n",
+ "I1=20*10**-3 #A\n",
+ "I2=10*10**-3 #A\n",
+ "\n",
+ "#Calculation\n",
+ "U1=0.5*L*I1**2\n",
+ "U2=0.5*L*I2**2\n",
+ "\n",
+ "#Result\n",
+ "print \"Magnetic energy stored in the coil is\",U1*10**6,\"*10**-4 J\"\n",
+ "print\"New value of energy is\",U2,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic energy stored in the coil is 100.0 *10**-4 J\n",
+ "New value of energy is 2.5e-05 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 134
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.20 Page no 680"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=12\n",
+ "R=30.0 #ohm\n",
+ "L=0.22 \n",
+ "\n",
+ "#Calculation\n",
+ "I0=E/R\n",
+ "I=I0/2.0\n",
+ "P=E*I\n",
+ "dl=(E-(I*R))/L\n",
+ "du=L*I*dl\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Energy being delivered by the battery is\", P,\"W\"\n",
+ "print\"(ii) ENergy being stored in the magnetic field of inductor is\",du,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Energy being delivered by the battery is 2.4 W\n",
+ "(ii) ENergy being stored in the magnetic field of inductor is 1.2 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 144
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.21 Page no 680"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=2.0 #H\n",
+ "i=2 #A\n",
+ "\n",
+ "#Calculation\n",
+ "U=0.5*L*i**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Amount of energy spent during the period is\", U,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Amount of energy spent during the period is 4.0 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 147
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.22 Page no 686"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1500 #V\n",
+ "dl=3 #A\n",
+ "dt=0.001 #s\n",
+ "\n",
+ "#Calculation\n",
+ "M=(e*dt)/dl\n",
+ "\n",
+ "#Result\n",
+ "print\"Mumtual induction between the two coils is\", M,\"H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mumtual induction between the two coils is 0.5 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 150
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.23 Page no 686"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N2=1000\n",
+ "I1=5.0 #A\n",
+ "a2=0.4*10**-4 #Wb\n",
+ "dl=-24 #A\n",
+ "dt=0.02 #S\n",
+ "\n",
+ "#Calculation\n",
+ "M=(N2*a2)/I1\n",
+ "eb=(-M*dl)/dt\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Mutual induction between A and B is\", M,\"H\"\n",
+ "print\"(ii) e.m.f induced by the coil is\", eb"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Mutual induction between A and B is 0.008 H\n",
+ "(ii) e.m.f induced by the coil is 9.6\n"
+ ]
+ }
+ ],
+ "prompt_number": 155
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.24 Page no 687"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=1200\n",
+ "A=12*10**-4 #m**2\n",
+ "r=0.15 #m\n",
+ "N2=300\n",
+ "a=0.05\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "u=4*math.pi*10**-7\n",
+ "L=(u*N**2*A)/(2*math.pi*r)\n",
+ "M=(u*N*N2*A)/(2*math.pi*r)\n",
+ "dl=2/a\n",
+ "e=M*dl\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Self inductance of the toroid is\", round(L*10**3,1),\"*10**-3 H\"\n",
+ "print\"(ii) Induced e.m.f. in the second coil is\",round(e,3),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Self inductance of the toroid is 2.3 *10**-3 H\n",
+ "(ii) Induced e.m.f. in the second coil is 0.023 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 168
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.25 Page no 688"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=2.0\n",
+ "a1=20*10**-2\n",
+ "x=0.15\n",
+ "A2=0.3*10**-2\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "u=4*math.pi*10**-7\n",
+ "B1=(u*I*a1**2)/(2.0*(a1**2+x**2)**1.5)\n",
+ "a=B1*math.pi*A2**2\n",
+ "M=a/I\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Flux linking the bigger loop is\", round(a*10**11,1)\n",
+ "print\"(ii) Mutual induction between the two loops is\",round(M*10**11,2),\"!0**-11 H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Flux linking the bigger loop is 9.1\n",
+ "(ii) Mutual induction between the two loops is 4.55 !0**-11 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 181
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.26 Page no 688"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=0.5 #m\n",
+ "n=20 #turns\n",
+ "r=50 #cm\n",
+ "A1=40*10**-4 #m**2\n",
+ "n1=25\n",
+ "A2=25*10**-4 #m**2\n",
+ "\n",
+ "#Calculation\n",
+ "u=4*math.pi*10**-7\n",
+ "N=n*r\n",
+ "N2=n1*r\n",
+ "M=(u*N*N2*A2)/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Mutual induction of the system is\",round(M*10**3,2),\"*10**-3 H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mutual induction of the system is 7.85 *10**-3 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 188
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter13_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter13_2.ipynb new file mode 100644 index 00000000..8800aa9c --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter13_2.ipynb @@ -0,0 +1,1642 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:2cfb52e11f318cb94f2cd1051460c5732d92997d5b28187a631460f42051172b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 13 Alternating currents"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.1 Page no 721"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I0=141.4 #A\n",
+ "w=314\n",
+ "t=3*10**-3 #s\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "f=w/(2*math.pi)\n",
+ "T=1/f\n",
+ "I=-I0*t*math.sin(314*180/3.14)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The maximum value is\",I0,\"A\"\n",
+ "print\"(ii) Frequency is\",round(f,0),\"Hz\"\n",
+ "print\"(iii) Time period is\",round(T,2),\"S\"\n",
+ "print\"(iv) The instantaneous value is\", round(I*10**3,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The maximum value is 141.4 A\n",
+ "(ii) Frequency is 50.0 Hz\n",
+ "(iii) Time period is 0.02 S\n",
+ "(iv) The instantaneous value is 411.54 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.3 Page no 721"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=220 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "E0=math.sqrt(2)*E\n",
+ "Emean=2*E0/math.pi\n",
+ "\n",
+ "#Result\n",
+ "print\"Average e.m.f during a positive half cycle is\", round(Emean,0),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average e.m.f during a positive half cycle is 198.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.4 Page no 721"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=2\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=math.sqrt(A**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"r.m.s Value of current is\", I,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "r.m.s Value of current is 2.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.5 Page no 722"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I0=120 #A\n",
+ "a=360.0\n",
+ "b=96\n",
+ "c=120.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "t=1/a\n",
+ "I=I0*math.sin(math.pi/3.0)\n",
+ "a1=b/c\n",
+ "a2=math.asin(a1)\n",
+ "t=a2/(c*math.pi)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Instantaneous value after 1/360 second is\",round(I,2),\"A\"\n",
+ "print\"(ii) Time taken to reach 96 A for the first time is\", round(t,5),\"S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Instantaneous value after 1/360 second is 103.92 A\n",
+ "(ii) Time taken to reach 96 A for the first time is 0.00246 S\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.7 Page no 726"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E0=60\n",
+ "R=20.0 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Ev=E0/(math.sqrt(2))\n",
+ "Iv=Ev/R\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) A.C ammeter will\",round(Iv,2),\"A\"\n",
+ "print\"(ii) Average value of a.c over one cycle is zero\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) A.C ammeter will 2.12 A\n",
+ "(ii) Average value of a.c over one cycle is zero\n"
+ ]
+ }
+ ],
+ "prompt_number": 73
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.8 Page no 726"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E0=250 #V\n",
+ "I0=10 #A\n",
+ "\n",
+ "#Calculation\n",
+ "P=E0*I0\n",
+ "P1=P/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Peak power is\", P,\"W\"\n",
+ "print\"(ii) Average power is\",P1,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Peak power is 2500 W\n",
+ "(ii) Average power is 1250.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.9 Page no 726"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ev=120.0\n",
+ "P=1000 #W\n",
+ "Ev1=240\n",
+ "\n",
+ "#Calculation\n",
+ "Iv=P/Ev\n",
+ "I0=math.sqrt(2)*Iv\n",
+ "R=Ev/Iv\n",
+ "P=Ev1**2/R\n",
+ "\n",
+ "#Result\n",
+ "print\"Resistance is\", R,\"ohm \\nPeak current is\",P,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance is 14.4 ohm \n",
+ "Peak current is 4000.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 88
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.11 Page no 729"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Xl=220 #ohm\n",
+ "L=0.7 #H\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "f=Xl/(2*math.pi*L)\n",
+ "\n",
+ "#Result\n",
+ "print\"Frequency is\", round(f,0),\"HZ\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Frequency is 50.0 HZ\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.12 Page no 729"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=50 #Hz\n",
+ "I=1.4\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "E=2*math.pi*f*I*2*math.cos(2*math.pi*f)\n",
+ "Ev=E/math.sqrt(2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Potential difference across the coil is\", round(E,0),\"cos 100*math.pi*t\"\n",
+ "print\"(ii) r.m.s value of p.d across the coil is\", round(Ev,1),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Potential difference across the coil is 880.0 cos 100*math.pi*t\n",
+ "(ii) r.m.s value of p.d across the coil is 622.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.13 Page no 729"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=50 #Hz\n",
+ "L=2 \n",
+ "Ev=12 #V\n",
+ "L1=6\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xl=2*math.pi*f*L\n",
+ "Iv=Ev/Xl\n",
+ "Xl1=2*math.pi*f*L1\n",
+ "Iv1=Ev/Xl1\n",
+ "\n",
+ "#Result\n",
+ "print\"Current flows when the inductance is changed to 6 H\", round(Iv1,4),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current flows when the inductance is changed to 6 H 0.0064 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.14 Page no 729"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ev=200 #V\n",
+ "I0=0.9 #A\n",
+ "f=50 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "E0=math.sqrt(2)*Ev\n",
+ "Xl=E0/I0\n",
+ "L=Xl/(2*math.pi*f)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of inductance is\", round(L,0),\"H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of inductance is 1.0 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.15 Page no 732"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=318*10**-6 #F\n",
+ "Ev=230 #V\n",
+ "f=50 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xc=1/(2*math.pi*f*C)\n",
+ "Iv=Ev/Xc\n",
+ "E0=math.sqrt(2)*Ev\n",
+ "I0=math.sqrt(2)*Iv\n",
+ "w=2*math.pi*f\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The capacitive reactance is\",round(Xc,0),\"ohm\"\n",
+ "print\"(ii) r.m.s value of circuit current is\",round(Iv,0),\"A\"\n",
+ "print\"(iii) Equation for voltage is\",round(E0,2),\"sin 314 t\"\n",
+ "print\"Equation for current is\",round(I0,2),\"sin (314 t+pi/2)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The capacitive reactance is 10.0 ohm\n",
+ "(ii) r.m.s value of circuit current is 23.0 A\n",
+ "(iii) Equation for voltage is 325.27 sin 314 t\n",
+ "Equation for current is 32.5 sin (314 t+pi/2)\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.16 Page no 732"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=1 #H\n",
+ "Xl=3142.0 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "f=Xl/(2*math.pi*L)\n",
+ "C=1/(2.0*math.pi*f*Xl)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Value of frequency is\", round(f,0),\"ohm\"\n",
+ "print\"(ii) Capacity of a condenser is\",round(C*10**6,2),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Value of frequency is 500.0 ohm\n",
+ "(ii) Capacity of a condenser is 0.1 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 59
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.17 Page no 732"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=50*10**-6 #F\n",
+ "V=230 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "q=C*V*math.sqrt(2)\n",
+ "E=0.5*C*(V*math.sqrt(2))**2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Maximum charge on the capacitor is\", round(q*10**3,2),\"*10**-3 C\"\n",
+ "print\"(ii) The maximum energy stored in the capacitor is\", round(E,2),\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Maximum charge on the capacitor is 16.26 *10**-3 C\n",
+ "(ii) The maximum energy stored in the capacitor is 2.65 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 69
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.18 Page no 732"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I0=10 #A\n",
+ "w=314\n",
+ "L=5\n",
+ "\n",
+ "#Calculation\n",
+ "E=0.5*L*I0**2\n",
+ "E0=w*L*I0\n",
+ "C=(E*2)/(E0**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Capacitance of the capacitor is\",round(C*10**6,2),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance of the capacitor is 2.03 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 76
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.19 Page no 737"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=50\n",
+ "L=31.8*10**-3 #H\n",
+ "R=7.0 #ohm\n",
+ "Ev=230 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xl=2*math.pi*f*L\n",
+ "Z=math.sqrt(R**2+Xl**2)\n",
+ "Iv=Ev/Z\n",
+ "T=Xl/R\n",
+ "a=math.atan(T)*180/3.14\n",
+ "a1=math.cos(a)*3.14/180.0\n",
+ "P=Iv**2*R\n",
+ "t=55*math.pi/(180.0*3.14)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Circuit current is\",round(Iv,2),\"A\"\n",
+ "print\"(ii) Phase angle is\", round(a,0),\"lag\"\n",
+ "print\"(iii) Power factor is\", round(a1*10**3,3),\"lag\"\n",
+ "print\"(iv) Power consumed is\",round(P,0),\"W\"\n",
+ "print\"Time lag between voltage maximum and current maximum is\",round(t*10**1,2),\"*10**-3 S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Circuit current is 18.85 A\n",
+ "(ii) Phase angle is 55.0 lag\n",
+ "(iii) Power factor is 0.554 lag\n",
+ "(iv) Power consumed is 2488.0 W\n",
+ "Time lag between voltage maximum and current maximum is 3.06 *10**-3 S\n"
+ ]
+ }
+ ],
+ "prompt_number": 121
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.20 Page no 737"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=400 #W\n",
+ "Ev=250 #V\n",
+ "Iv=2.5 #A\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=P/(Ev*Iv)\n",
+ "Z=Ev/Iv\n",
+ "R=Z*a\n",
+ "Xl=math.sqrt(Z**2-R**2)\n",
+ "L=Xl/(2*math.pi*f)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The power factor is\",a,\"lag\"\n",
+ "print\"(ii) Resistance of the coil is\", R,\"ohm\"\n",
+ "print\"(iii) Inductance of the coil is\", round(L,3),\"H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The power factor is 0.64 lag\n",
+ "(ii) Resistance of the coil is 64.0 ohm\n",
+ "(iii) Inductance of the coil is 0.245 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 134
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.21 Page no 738"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vr=150 #V\n",
+ "R=75.0 #ohm\n",
+ "f=50 #Hz\n",
+ "L=318*10**-3 #H\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Iv=Vr/R\n",
+ "Xl=2*math.pi*f*L\n",
+ "Vl=Iv*Xl\n",
+ "Z=math.sqrt(R**2+Xl**2)\n",
+ "Ev=Iv*Z\n",
+ "a=Xl/R\n",
+ "a1=math.atan(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print \"(i) The supply voltage is\",round(Ev,0),\"V\"\n",
+ "print\"(ii) The phase angle is\",round(a1,2),\"degree lag\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The supply voltage is 250.0 V\n",
+ "(ii) The phase angle is 53.13 degree lag\n"
+ ]
+ }
+ ],
+ "prompt_number": 149
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.22 Page no 738"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=60 #W\n",
+ "Ev=100.0 #V\n",
+ "Ev1=220 #v\n",
+ "f=50 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "Iv=P/Ev\n",
+ "Vr=Ev1-Ev\n",
+ "R=Vr/Iv\n",
+ "Xl=Vl/Iv\n",
+ "Vl=math.sqrt(Ev1**2-Ev**2)\n",
+ "L=Xl/(2*math.pi*f)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The value of non inductive resistance is\", R,\"ohm\"\n",
+ "print\"(ii) Pure inductance is\",round(L,2),\"H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The value of non inductive resistance is 200.0 ohm\n",
+ "(ii) Pure inductance is 1.04 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 163
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.23 Page no 739"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f1=50.0\n",
+ "L=1\n",
+ "E=100 #V\n",
+ "I=1.0 #A\n",
+ "Iv=0.5 #A\n",
+ "f=0\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xl=2*math.pi*f*L\n",
+ "R=E/I\n",
+ "Z=Ev/Iv\n",
+ "Xl1=math.sqrt(Z**2-R**2)\n",
+ "L=Xl1/(2.0*math.pi*f1)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of resistance is\",R ,\"ohm\"\n",
+ "print\"The value of impedence is\",Z,\"ohm\"\n",
+ "print\"Inductance of the coil is\",round(L,2),\"H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of resistance is 100.0 ohm\n",
+ "The value of impedence is 200.0 ohm\n",
+ "Inductance of the coil is 0.55 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 183
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.24 Page no 742"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=10**6 #A\n",
+ "f=50 #Hz\n",
+ "C=79.5\n",
+ "R=30 #ohm\n",
+ "Ev=100 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xc=I/(2*math.pi*f*C)\n",
+ "Z=math.sqrt(R**2+Xc**2)\n",
+ "Iv=Ev/Z\n",
+ "a=Xc/R\n",
+ "a1=math.atan(a)*180/3.14\n",
+ "w=2*math.pi*f\n",
+ "I0=math.sqrt(2)*Iv\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Circuit impedence is\", round(Z,0),\"ohm\"\n",
+ "print\"(ii) Circuit current is\",round(Iv,0),\"A\"\n",
+ "print\"(iii) Phase angle is\",round(a1,0),\"degree lead\"\n",
+ "print\"(iv) Equation for the instantaneous value of current is\",round(I0,3),\"sin(\",round(w,0),\"t+\",round(a1,0),\"degree)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Circuit impedence is 50.0 ohm\n",
+ "(ii) Circuit current is 2.0 A\n",
+ "(iii) Phase angle is 53.0 degree lead\n",
+ "(iv) Equation for the instantaneous value of current is 2.827 sin( 314.0 t+ 53.0 degree)\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.25 Page no 742"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=80 #W\n",
+ "V=100.0 #v\n",
+ "V1=200 #V\n",
+ "f=50 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Iv=P/V\n",
+ "Vc=math.sqrt(V1**2-V**2)\n",
+ "Xc=Vc/Iv\n",
+ "C=1/(2.0*math.pi*f*Xc)\n",
+ "\n",
+ "#Result\n",
+ "print\"Capcitance of a capacitor is\", round(C*10**6,1),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capcitance of a capacitor is 14.7 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.26 Page no 742"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ev=200 #V\n",
+ "Iv=10.0\n",
+ "f=50 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "z=Ev/Iv\n",
+ "R=z*math.cos(30*3.14/180.0)\n",
+ "Xc=z*math.sin(30*3.14/180.0)\n",
+ "C=1/(2.0*math.pi*f*Xc)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Value of resistance is\", round(R,2),\"ohm\"\n",
+ "print\"(ii) Capacitive reactance is\", round(Xc,0),\"ohm\"\n",
+ "print\"(iii) Capacitance of the circuit is\", round(C*10**6,0),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Value of resistance is 17.32 ohm\n",
+ "(ii) Capacitive reactance is 10.0 ohm\n",
+ "(iii) Capacitance of the circuit is 318.0 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.27 Page no 743"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Iv=5 #A\n",
+ "R=10 #ohm\n",
+ "Ev=60 #V\n",
+ "C=400 #micro F\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Vr=Iv*R\n",
+ "Vc=math.sqrt(Ev**2-Vr**2)\n",
+ "Xc=Vc/Iv\n",
+ "f=1/(2.0*math.pi*C*Xc)\n",
+ "a=Vc/Vr\n",
+ "a1=math.atan(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of supplied frequency is\", round(f*10**6,0),\"Hz\"\n",
+ "print\"Phase angle between circuit current and supply voltage is\",round(a1,1),\"degree lead\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of supplied frequency is 60.0 Hz\n",
+ "Phase angle between circuit current and supply voltage is 33.6 degree lead\n"
+ ]
+ }
+ ],
+ "prompt_number": 65
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.28 Page no 743"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=200 #ohm\n",
+ "C=15*10**-6 #F\n",
+ "Ev=220 #V\n",
+ "f=50 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xc=1/(2*math.pi*f*C)\n",
+ "Z=math.sqrt(R**2+Xc**2)\n",
+ "Iv=Ev/Z\n",
+ "Vr=Iv*R\n",
+ "Vc=Iv*Xc\n",
+ "V=Vr+Vc\n",
+ "Vrc=math.sqrt(Vr**2+Vc**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The current in the circuit is\", round(Iv,3),\"A\"\n",
+ "print\"(b) Voltage across the resistor and capacitor is\",Vrc,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " (a) The current in the circuit is 0.754 A\n",
+ "(b) Voltage across the resistor and capacitor is 220.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 83
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.29 Page no 746"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=10.0 #ohm\n",
+ "R2=5.0 #ohm\n",
+ "R3=15 #ohm\n",
+ "Ev=200\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "R=R1+R2+R3\n",
+ "X=R3-(R1+R3)\n",
+ "Z=math.sqrt(R**2+R1**2)\n",
+ "Iv=Ev/Z\n",
+ "T=X/R\n",
+ "a=-math.atan(T)*180/3.14\n",
+ "b=math.cos(a*3.14/180.0)\n",
+ "P=Iv**2*R\n",
+ "print\"(i) Circuit current is\",round(Iv,2) ,\"A\"\n",
+ "print\"(ii) Circuit phase angle is\",round(a,2),\"degree lead\"\n",
+ "print\"(iii)Phase angle between applied voltage and circuit current \",round(b,3),\"lead\"\n",
+ "print\"(iv)Power consumed is\",P,\"W\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Circuit current is 6.32 A\n",
+ "(ii) Circuit phase angle is 18.44 degree lead\n",
+ "(iii)Phase angle between applied voltage and circuit current 0.949 lead\n",
+ "(iv)Power consumed is 1200.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 116
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.30 Page no. 746"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "F=50 #HZ\n",
+ "L=0.06 \n",
+ "C=6.8\n",
+ "l=10**6\n",
+ "R=2.5\n",
+ "Ev=230 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math \n",
+ "Xl=2*math.pi*F*L\n",
+ "Xc=l/(2*math.pi*F*C)\n",
+ "Z=math.sqrt(R**2+(Xl-Xc)**2)\n",
+ "Iv=Ev/Z\n",
+ "a=(Xl-Xc)/R\n",
+ "a2=-math.atan(a)*180.0/3.14\n",
+ "P=R/Z\n",
+ "P1=Ev*Iv*P\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Circuit impedence is\",round(Z,1),\"ohm\"\n",
+ "print\"(ii) Circuit current is\",round(Iv,3),\"A\"\n",
+ "print\"(iii) Phase angle is \",round(a2,1),\"degree lead\" \n",
+ "print\"(iv) Power factor is\",round(P,4),\"lead\"\n",
+ "print\"(v) Power consumed is\",round(P1,2),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Circuit impedence is 449.3 ohm\n",
+ "(ii) Circuit current is 0.512 A\n",
+ "(iii) Phase angle is 89.7 degree lead\n",
+ "(iv) Power factor is 0.0056 lead\n",
+ "(v) Power consumed is 0.66 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 146
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.31 Page no. 746"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=65 #degree\n",
+ "b=20 #degree\n",
+ "w=3000\n",
+ "L=0.01\n",
+ "E0=400 #V\n",
+ "I=10\n",
+ "f=50\n",
+ "\n",
+ "#calculation\n",
+ "import math\n",
+ "a=a-b\n",
+ "Xl=w*L\n",
+ "Z=E0/(I*math.sqrt(2))\n",
+ "R=Z/math.sqrt(2)\n",
+ "Xc=Xl-R\n",
+ "C=1/(w*Xc*10**-6)\n",
+ "\n",
+ "#Result\n",
+ "print\" The value of C is\" ,round(C,1),\"microF\"\n",
+ "print\" The value of R is\",R,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The value of C is 33.3 microF\n",
+ " The value of R is 20.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 161
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.32 Page no 747"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=50 #Hz\n",
+ "L=0.03\n",
+ "R=8 #ohm\n",
+ "Ev=240 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xl=2*math.pi*f*L\n",
+ "Z=math.sqrt(R**2+Xl**2)\n",
+ "Iv=Ev/Z\n",
+ "P=Iv**2*R\n",
+ "a=R/Z\n",
+ "Xc=2*Xl\n",
+ "C=1/(2*math.pi*f*Xc)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The value of current is\",round(Iv,2) ,\"A\"\n",
+ "print\" The value of power is\",round(P,0),\"W\"\n",
+ "print \" Power factor is\",round(a,2),\"lag\"\n",
+ "print\"(ii) The vaue of capacitance is\", round(C*10**6,0),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The value of current is 19.41 A\n",
+ " The value of power is 3015.0 W\n",
+ " Power factor is 0.65 lag\n",
+ "(ii) The vaue of capacitance is 169.0 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 186
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.33 Page no 747"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vr=65 #V\n",
+ "R=100.0 #ohm\n",
+ "Vl=204\n",
+ "Vc=415\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Iv=Vr/R\n",
+ "Xl=Vl/Iv\n",
+ "L=Xl/(2*math.pi*f)\n",
+ "Xc=Vc/Iv\n",
+ "C=1/(2*math.pi*f*Xc)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Circuit current is\", Iv,\"A\"\n",
+ "print\"(ii) Inductance is\",round(L,0),\"H\"\n",
+ "print\"(iii) The value of capacitance is\",round(C*10**6,0),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Circuit current is 0.65 A\n",
+ "(ii) Inductance is 1.0 H\n",
+ "(iii) The value of capacitance is 5.0 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 201
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.34 Page no 752"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=100*10**-12 #F\n",
+ "L=100*10**-6 #H\n",
+ "Ev=10\n",
+ "R=100.0 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "fr=1/(2*math.pi*math.sqrt(L*C))\n",
+ "Iv=Ev/R\n",
+ "Vl=Iv*2*math.pi*fr*L\n",
+ "Vc=Iv/(2.0*math.pi*fr*C)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Resonant frequency is\", round(fr*10**-6,2),\"*10**6 HZ\"\n",
+ "print\"(ii) Current at resonance is\", Iv,\"A\"\n",
+ "print\"(iii) Voltage across L and C is\", Vc,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Resonant frequency is 1.59 *10**6 HZ\n",
+ "(ii) Current at resonance is 0.1 A\n",
+ "(iii) Voltage across L and C is 100.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 218
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.35 Page no 752"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=50 #Hz\n",
+ "L=0.5\n",
+ "Ev=100 #v\n",
+ "R=4 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "C=1/(4*math.pi**2*f**2*L)\n",
+ "Ir=Ev/R\n",
+ "Vr=Ir*2*math.pi*f*L\n",
+ "Vc=Ir/(2*math.pi*f*C)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The capacitance is\", round(C*10**6,2),\"micro F\"\n",
+ "print\"(ii) The voltage across inductance and capacitance is\", round(Vc,0),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The capacitance is 20.26 micro F\n",
+ "(ii) The voltage across inductance and capacitance is 3927.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 229
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.36 Page no 752"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=50 #Hz\n",
+ "L=0.318 #H\n",
+ "Iv=2.3\n",
+ "R=100 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "C=1/((2*math.pi*f)**2*L)\n",
+ "Vl=Iv*2*math.pi*f*C*10**4\n",
+ "P=Iv**2*R\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The value of capacitor is\", round(C*10**6,1),\"micro F\"\n",
+ "print\"(ii) Voltage across the inductor is\", round(Vl,0),\"V\"\n",
+ "print\"(iii)Total power consumed is\",P,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The value of capacitor is 31.9 micro F\n",
+ "(ii) Voltage across the inductor is 230.0 V\n",
+ "(iii)Total power consumed is 529.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 245
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.37 Page no 753"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E0=283 #V\n",
+ "f=50 #Hz\n",
+ "R=3.0 #ohm\n",
+ "L=25.48*10**-3 #h\n",
+ "C=796*10**-6 #F\n",
+ "Xl=8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xc=1/(2*math.pi*f*C)\n",
+ "Z=math.sqrt(R**2+(Xl-Xc)**2)\n",
+ "a=math.atan(Xc/R)*180/3.14\n",
+ "Iv=(E0/math.sqrt(2))/Z\n",
+ "P=Iv**2*R\n",
+ "a1=math.cos(a*180/3.14)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The inpedence of the circuit is\", round(Z,0),\"ohm\"\n",
+ "print\"(b) The phase difference is\", round(a,1),\"degree\"\n",
+ "print\"(c) The power dissipated is\", round(P,0),\"W\"\n",
+ "print\"(d) Power factor is\", round(a1,1),\"lag\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The inpedence of the circuit is 5.0 ohm\n",
+ "(b) The phase difference is 53.1 degree\n",
+ "(c) The power dissipated is 4804.0 W\n",
+ "(d) Power factor is 0.8 lag\n"
+ ]
+ }
+ ],
+ "prompt_number": 270
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.38 Page no 753"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=25.48*10**-3 #H\n",
+ "C=796*10**-6\n",
+ "R=3.0 #ohm\n",
+ "E0=283\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "fr=1/(2.0*math.pi*math.sqrt(L*C))\n",
+ "Iv=(E0/math.sqrt(2))/R\n",
+ "P=Iv**2*R\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Frequency of the source is\", round(fr,1),\"Hz\"\n",
+ "print\"(b) The value of impedence is\",R,\"ohm\"\n",
+ "print\"The value of current is\",round(Iv,1),\"A\"\n",
+ "print\"The power dissipated is\",round(P,0),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Frequency of the source is 35.3 Hz\n",
+ "(b) The value of impedence is 3.0 ohm\n",
+ "The value of current is 66.7 A\n",
+ "The power dissipated is 13348.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 287
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.39 Page no 757"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=1200*10**-12 #F\n",
+ "E=500\n",
+ "L=0.075 #H\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "q0=C*E\n",
+ "I0=q0/(math.sqrt(L*C))\n",
+ "f=1/(2*math.pi*math.sqrt(L*C))\n",
+ "T=1/f\n",
+ "U=q0**2/(2.0*C)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The initial charge onthe capcitor is\",q0,\"c\"\n",
+ "print\"(ii) The maximum current is\",round(I0*10**3,0),\"mA\"\n",
+ "print\"(iii) The value of frequency is\", round(f*10**-3,0),\"*10**3 Hz\"\n",
+ "print\"Time period is\", round(T*10**5,0),\"*10**-5 S\"\n",
+ "print\"(iv) Total energy is\",U*10**4,\"*10**-4 J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The initial charge onthe capcitor is 6e-07 c\n",
+ "(ii) The maximum current is 63.0 mA\n",
+ "(iii) The value of frequency is 17.0 *10**3 Hz\n",
+ "Time period is 6.0 *10**-5 S\n",
+ "(iv) Total energy is 1.5 *10**-4 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 315
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.40 Page no 758"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=8*10**-6 #H\n",
+ "C=0.02*10**-6 #F\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/(2*math.pi*math.sqrt(L*C))\n",
+ "w=c/f\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength is\", round(w*10**-2,2),\"*10**2 m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength is 7.54 *10**2 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 321
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter14_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter14_2.ipynb new file mode 100644 index 00000000..7bb51839 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter14_2.ipynb @@ -0,0 +1,629 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:1bd22700738f4a2a80468f70e18e63c26ad56b1a125cfb69784af1d3eb280a8a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 14 Electrical devices"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.1 Page no 787"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=100\n",
+ "A=10**-2 #m**2\n",
+ "B=0.5 #T\n",
+ "f=500/60.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "w=2*math.pi*f\n",
+ "E0=N*A*B*w\n",
+ "E=E0*math.sin(60*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum emf produced in the coil is\", round(E0,2),\"V\"\n",
+ "print\"Instantaneous value of e.m.f. is\",round(E,1),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum emf produced in the coil is 26.18 V\n",
+ "Instantaneous value of e.m.f. is 22.7 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.2 Page no 787"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=50\n",
+ "A=2.5\n",
+ "B=0.3 #T\n",
+ "w=60\n",
+ "R=500 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "E0=N*B*A*w\n",
+ "I0=E0/R\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Maximum current drawn from the gnerator is\",I0,\"A\"\n",
+ "print\"(ii) The maximum flux through the coil is zero and when induced current is maximum, flux through the coil is zero\"\n",
+ "print\"(iii) Yes the generator will work because the basic condition for induced an .m.f. is that there should be relative motion between coil and magnetic field\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Maximum current drawn from the gnerator is 4.5 A\n",
+ "(ii) The maximum flux through the coil is zero and when induced current is maximum, flux through the coil is zero\n",
+ "(iii) Yes the generator will work because the basic condition for induced an .m.f. is that there should be relative motion between coil and magnetic field\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.3 Page no 787"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=150\n",
+ "A=2*10**-2 #m**2\n",
+ "B=0.15 #T\n",
+ "f=60\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "w=2*math.pi*f\n",
+ "E0=N*A*B*w\n",
+ "\n",
+ "#Result\n",
+ "print\"Peak value of e.m.f is\", round(E0,0),\"V\"\n",
+ "print\"Average value of induced e.m.f is zero\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Peak value of e.m.f is 170.0 V\n",
+ "Average value of induced e.m.f is zero\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.4 Page no 787"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=100\n",
+ "A=3\n",
+ "B=0.04 #T\n",
+ "w=60\n",
+ "R=500 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "E0=N*A*B*w\n",
+ "I0=E0/R\n",
+ "P=E0*I0\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum power dissipated in the coil is\", P,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum power dissipated in the coil is 1036.8 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.5 Page no 788"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=100\n",
+ "A=0.10 #m**2\n",
+ "f=0.5 #Hz\n",
+ "B=0.01 #T\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "w=2*math.pi*f\n",
+ "E0=N*A*B*w\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum voltage generated in the coil is\", round(E0,3),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum voltage generated in the coil is 0.314 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.6 Page no 792"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=240 #V\n",
+ "I=5 #A\n",
+ "R=4 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Eb=V-(I*R)\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of back e.m.f is\", Eb,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of back e.m.f is 220 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.7 Page no 792"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=20 #A\n",
+ "R=2 #ohm\n",
+ "n=0.5 \n",
+ "P=2000 #W\n",
+ "\n",
+ "#Calculation\n",
+ "P1=P/n\n",
+ "V=P1/I\n",
+ "Eb=V-(I*R)\n",
+ "\n",
+ "#Result\n",
+ "print\"The back e.m.f is\", Eb,\"V \\nSupply voltage is\",V,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The back e.m.f is 160.0 V \n",
+ "Supply voltage is 200.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.8 Page no 793"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=100 #V\n",
+ "I=6 #A\n",
+ "V1=0.7\n",
+ "\n",
+ "#Calculation\n",
+ "Pin=V*I\n",
+ "R=(V1*Pin)/I**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Armature resistance is\", round(R,2),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Armature resistance is 11.67 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.10 Page no 793"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=200 #V\n",
+ "I=5 #A\n",
+ "R=8.5 #ohm \n",
+ "\n",
+ "#Calculation\n",
+ "Eb=V-(I*R)\n",
+ "Pi=V*I\n",
+ "P0=Eb*I\n",
+ "n=(P0*100)/Pi\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Back e.m.f of motor is\", Eb,\"V\"\n",
+ "print\"(ii) Power input is\",Pi,\"W\"\n",
+ "print\"(iii) Output power is\",P0,\"W\"\n",
+ "print\"(iv) Efficiency of motor is\",n,\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Back e.m.f of motor is 157.5 V\n",
+ "(ii) Power input is 1000 W\n",
+ "(iii) Output power is 787.5 W\n",
+ "(iv) Efficiency of motor is 78.75 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 61
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.11 Page no 796"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vp=200 #V\n",
+ "n=200.0\n",
+ "Ip=2 #A\n",
+ "\n",
+ "#Calculation\n",
+ "Vs=Vp*n\n",
+ "Is=(Ip*V)/Vs\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Voltage developed in the secondary is\", Vs,\"V\"\n",
+ "print\"(ii) The current in the secondary is\",Is ,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Voltage developed in the secondary is 40000.0 V\n",
+ "(ii) The current in the secondary is 0.01 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 69
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.12 Page no 796"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vp=220.0 #V\n",
+ "Is=5 #A\n",
+ "n=20\n",
+ "\n",
+ "#Calculation\n",
+ "Vs=Vp*n\n",
+ "Ip=(Vs*Is)/Vp\n",
+ "P=Vs*Is\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Voltage across secondary is\",Vs,\"V\"\n",
+ "print\"(ii) The current in primary is\",Ip,\"A\"\n",
+ "print\"(iii) The power output is\",P*10**-3,\"K W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Voltage across secondary is 4400.0 V\n",
+ "(ii) The current in primary is 100.0 A\n",
+ "(iii) The power output is 22.0 K W\n"
+ ]
+ }
+ ],
+ "prompt_number": 79
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.13 Page no 797"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=120*10**3 #W\n",
+ "R=0.4 #ohm\n",
+ "Ev=240.0 #V\n",
+ "Ev1=24000.0 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Iv=P/Ev\n",
+ "P=Iv**2*R\n",
+ "Iv1=P/Ev1\n",
+ "P1=Iv1**2*R\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Power loss at 240 V is\", P*10**-3,\"K W\"\n",
+ "print\"(ii) Power loss at 24000 V is\", round(P1,0),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Power loss at 240 V is 100.0 K W\n",
+ "(ii) Power loss at 24000 V is 7.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 94
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14 Page no 797"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Np=5000\n",
+ "Vp=2200 #V\n",
+ "Vs=220 #V\n",
+ "Pout=8 #K W\n",
+ "n=0.9\n",
+ "\n",
+ "#Calculation\n",
+ "Ns=(Vs*Np)/Vp\n",
+ "Pin=Pout/n\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The number of turns in the secondary is\", Ns\n",
+ "print\"(ii) Input power is\",round(Pin,1),\"K W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The number of turns in the secondary is 500\n",
+ "(ii) Input power is 8.9 K W\n"
+ ]
+ }
+ ],
+ "prompt_number": 99
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.15 Page no 797"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vp=220.0 #V\n",
+ "Vs=22 #V\n",
+ "Z=220 #ohm\n",
+ "Is=0.1\n",
+ "\n",
+ "#Calclation\n",
+ "Ip=(Vs*Is)/Vp\n",
+ "\n",
+ "#Result\n",
+ "print\"Current drawn is\", Ip,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current drawn is 0.01 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 104
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.16 Page no 798"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vs=24 #v\n",
+ "R=9.6 #ohm\n",
+ "Vp=120.0\n",
+ "\n",
+ "#Calculation\n",
+ "Is=Vs/R\n",
+ "Ip=(Vs*Is)/Vp\n",
+ "P1=Vs*Is\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Current in the secondary coil is\", Is,\"A\"\n",
+ "print\"(ii) Current in primary coil is\",Ip ,\"A\"\n",
+ "print\"Power used is\",P1,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Current in the secondary coil is 2.5 A\n",
+ "(ii) Current in primary coil is 0.5 A\n",
+ "Power used is 60.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 116
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter15_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter15_2.ipynb new file mode 100644 index 00000000..111cd390 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter15_2.ipynb @@ -0,0 +1,381 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:977816390da876a89acf9dab4a43ac1149d2a8f7e4f57b74a89090971103e376"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 15 Electromagnetic waves"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.1 Page no 836"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=8.854*10**-12 #C**2/N/m**2\n",
+ "A=10**-4 #m**2\n",
+ "E=3*10**6 #V/ms\n",
+ "\n",
+ "#Calculation\n",
+ "Id=e*A*E\n",
+ "\n",
+ "#Result\n",
+ "print\"Displacement current is\", round(Id*10**9,1)*10**-9,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Displacement current is 2.7e-09 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.2 Page no 836"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Id=1 #A\n",
+ "C=10.0**-6 #F\n",
+ "\n",
+ "#Calculation\n",
+ "V=Id/C\n",
+ "\n",
+ "#Result\n",
+ "print\"Instantaneous current is\", V,\"V/S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Instantaneous current is 1000000.0 V/S\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.3 Page no 836"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=0.15 #A\n",
+ "R=0.12 #m\n",
+ "r=0.065 #m\n",
+ "r1=0.15 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=math.pi*R**2\n",
+ "u=4*math.pi*10**-7\n",
+ "B=(u*I*r)/(2*math.pi*R**2)\n",
+ "B1=(u*I)/(2*math.pi*r1)\n",
+ "Bmax=(u*I)/(2*math.pi*R)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) (a) Magnetic field on the axis is zero\"\n",
+ "print\"(b) Magnetic field at r=6.5 cm is\",round(B*10**7,2)*10**-7 ,\"T\"\n",
+ "print\"(c) Magnetic field at r=15 cm is\", B1,\"T\"\n",
+ "print\"(ii) Distance is\", Bmax,\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) (a) Magnetic field on the axis is zero\n",
+ "(b) Magnetic field at r=6.5 cm is 1.35e-07 T\n",
+ "(c) Magnetic field at r=15 cm is 2e-07 T\n",
+ "(ii) Distance is 2.5e-07 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.5 Page no 837"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.05 #m\n",
+ "E=10**12 #V/m/s\n",
+ "e=8.854*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Id=e*math.pi*r**2*E\n",
+ "\n",
+ "#Result\n",
+ "print\"Displacement current is\", round(Id,4),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Displacement current is 0.0695 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.7 Page no 846 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=100 #v\n",
+ "c=3.0*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=E/c\n",
+ "u=4.0*math.pi*10**-7\n",
+ "H=B/u\n",
+ "U=e*E**2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Value of B is\", round(B*10**7,2)*10**-7,\" T\"\n",
+ "print\"(ii) Value of H is\",round(H,3),\"A/m\"\n",
+ "print\"(iii) Energy density is\",U,\"J/m**3"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Value of B is 3.33e-07 T\n",
+ "(ii) Value of H is 0.265 A/m\n",
+ "(iii) Energy density is 8.854e-08\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.8 Page no 847"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E0=8*10**-4 #v\n",
+ "c=3.0*10**8\n",
+ "w=6*10**6\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B0=E0/c\n",
+ "f=w/(2.0*math.pi)\n",
+ "l=c/f\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of the wave is\", round(l*10**-4,4),\"m\"\n",
+ "print\"Frequency is\",round(f*10**-6,3),\"*10**10 Hz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of the wave is 0.0314 m\n",
+ "Frequency is 0.955 *10**10 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.9 Page no 847"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=6.3 #V/m\n",
+ "c=3.0*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "B=E/c\n",
+ "\n",
+ "#Result\n",
+ "print\"B=\", B,\"K^ Tesla\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "B= 2.1e-08 K^ Tesla\n"
+ ]
+ }
+ ],
+ "prompt_number": 59
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.10 Page no 847"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=18 #W/cm**2\n",
+ "A=20 #cm**2\n",
+ "t=30*60\n",
+ "\n",
+ "#Calculation\n",
+ "U=f*A*t\n",
+ "P=U/c\n",
+ "F=P/t\n",
+ "P1=2*P\n",
+ "F1=P1/t\n",
+ "\n",
+ "#Result\n",
+ "print\"Average force exerted on the surface is\", F1,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average force exerted on the surface is 2.4e-06 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 65
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.11 Page no 847"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=3.0 #m\n",
+ "n=0.025\n",
+ "P=100 #w\n",
+ "C=3*10**8\n",
+ "\n",
+ "#Calculation \n",
+ "import math\n",
+ "A=4*math.pi*r**2\n",
+ "I=(n*P)/A\n",
+ "E0=math.sqrt((2*I)/(e*C))\n",
+ "B0=E0/C\n",
+ "\n",
+ "#Result\n",
+ "print\"Peak value of electric field is\", round(E0,2),\"V/m\"\n",
+ "print\"Peak value of magnetic field is\",round(B0*10**8,2)*10**-8,\"T\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Peak value of electric field is 4.08 V/m\n",
+ "Peak value of magnetic field is 1.36e-08 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 76
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter16_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter16_2.ipynb new file mode 100644 index 00000000..f74b4c1b --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter16_2.ipynb @@ -0,0 +1,369 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:badd30ad9356c0b0c69f3c0e035f97e954d74c789af08b85cb880fbe7e95ff6b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 16 Reflection of light"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Exaple 16.1 Page no 890"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "HF=2\n",
+ "EF=1.9\n",
+ "\n",
+ "#Calculation\n",
+ "L=0.5*HF\n",
+ "DB=0.5*EF\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum height of mirror is\", L,\"m\"\n",
+ "print\"Bottom edge of the mirror is\",DB,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum height of mirror is 1.0 m\n",
+ "Bottom edge of the mirror is 0.95 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.2 Page no 890"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-15.0 #cm\n",
+ "f=-10 #cm\n",
+ "o=2.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/((1.0/f)-(1.0/u))\n",
+ "m=v/u\n",
+ "I=-m*o\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image is\", v,\"cm\"\n",
+ "print\"Size of the image is\",I,\"cm\"\n",
+ "print\"Nature of the image isreal and inverted \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the image is -30.0 cm\n",
+ "Size of the image is -4.0 cm\n",
+ "Nature of the image isreal and inverted \n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.3 Page no 891"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-10.0 #cm\n",
+ "f=-15.0\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/((1/f)-(1/u))\n",
+ "m=-v/u\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Image position is\", v,\"cm\"\n",
+ "print\"(ii) Magnification is\", m"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Image position is 30.0 cm\n",
+ "(ii) Magnification is 3.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.4 Page no 891"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=12.0\n",
+ "v=4.0\n",
+ "\n",
+ "#Calculation\n",
+ "u=1/((1/f)-(1/v))\n",
+ "m=-v/u\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Object position is\", u,\"cm\"\n",
+ "print\"(ii) Magnification is\", round(m,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Object position is -6.0 cm\n",
+ "(ii) Magnification is 0.67\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.5 Page no 891"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=36 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "f=-R/2.0\n",
+ "u=(2*f)/3.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the object is\", u,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the object is -12.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.6 Page no 892"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=20 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=R/2.0\n",
+ "u=-f\n",
+ "v=-u/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the object is\",u,\"cm\"\n",
+ "print\"Position of the image is\",v,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the object is -10.0 cm\n",
+ "Position of the image is 5.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.7 Page no 892"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=-15.0\n",
+ "\n",
+ "#Calculation\n",
+ "u=(1/(1/f)/3.0)*4\n",
+ "v=u/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of object is\" ,u,\"cm\"\n",
+ "print\"When the image is virtual\",v,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of object is -20.0 cm\n",
+ "When the image is virtual -10.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.8 Page no 892"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=30 #ohm\n",
+ "u=-10.0 \n",
+ "h1=5\n",
+ "\n",
+ "#Calculation\n",
+ "f=R/2.0\n",
+ "v=1/((1/f)-(1/u))\n",
+ "h2=(-v*h1)/u\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image is\", v,\"cm\"\n",
+ "print\"Size of the image is\",h2,\"cm\"\n",
+ "print\"The image is virtual\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the image is 6.0 cm\n",
+ "Size of the image is 3.0 cm\n",
+ "The image is virtual\n"
+ ]
+ }
+ ],
+ "prompt_number": 58
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.9 Page no 893"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=-10.0 #cm\n",
+ "u=-25.0 #cm\n",
+ "h1=3\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/((1/f)-(1/u))\n",
+ "h2=(-v*h1)/u\n",
+ "A=h2**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Area enclosed by the image of the wire is\", A,\"cm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Area enclosed by the image of the wire is 4.0 cm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter17_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter17_2.ipynb new file mode 100644 index 00000000..6ddbfa96 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter17_2.ipynb @@ -0,0 +1,1299 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:7117da667e9242c9a19d8eb5f355fd755219357cacc372784f1e3ef0c539e46b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 17 Refraction of the light"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.1 Page no 918"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u1=1.50\n",
+ "u2=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "sinr=u1*math.sin(50*3.14/180.0)/u2\n",
+ "a=math.asin(sinr)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of refraction is\", round(a,1),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of refraction is 59.8 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.2 Page no 918"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u1=1.0\n",
+ "u2=1.526\n",
+ "i=45 #degree\n",
+ "#Calculation\n",
+ "sinr=(u1*math.sin(i*3.14/180.0))/u2\n",
+ "r=math.asin(sinr)*180/3.14\n",
+ "d=i-r\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of deviation is\", round(d,2),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of deviation is 17.39 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.3 Page no 918"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "c=3.0*10**8\n",
+ "u=1.5\n",
+ "f=6*10**14 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "v=c/u\n",
+ "l=c/f\n",
+ "lm=v/f\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Wavelength of light in air is\", l,\"m\"\n",
+ "print\"(ii) Wavelength of light in glass is\",round(lm*10**7,1)*10**-7,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Wavelength of light in air is 5e-07 m\n",
+ "(ii) Wavelength of light in glass is 3.3e-07 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.4 Page no 919"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ug=1.5\n",
+ "uw=1.3\n",
+ "vw=2.25*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "vg=(uw*vw)/ug\n",
+ "\n",
+ "#Result\n",
+ "print\"Speed of the light in glass is\", vg*10**-8,\"*10**8 m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Speed of the light in glass is 1.95 *10**8 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.5 Page no 919"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.6\n",
+ "t=8\n",
+ "t1=4.5\n",
+ "u1=1.5\n",
+ "t2=6\n",
+ "u2=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "d=t*(1-(1/u))\n",
+ "d1=t1*(1-(1/u1))\n",
+ "d2=t2*(1-(1/u2))\n",
+ "D=d+d1+d2\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of mark from the bottom is\", round(D,0),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of mark from the bottom is 6.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.6 Page no 919"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "uw=1.33\n",
+ "uo=1.20\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "uow=uw/uo\n",
+ "sinr=(math.sin(30*3.14/180.0))/uow\n",
+ "r=math.asin(sinr)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of refraction in water is\", round(r,1),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of refraction in water is 26.8 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.7 Page no 920"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=2.0*10**8 #m/s\n",
+ "c=3*10**8 #m/s\n",
+ "d=6.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "ug=c/v\n",
+ "a=d/ug\n",
+ "D=d-a\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance through which ink dot appears to be raised is\", D,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance through which ink dot appears to be raised is 2.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 54
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.8 Page no 924"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ug=1.5\n",
+ "uw=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "u1=ug/uw\n",
+ "sinC=1/u1\n",
+ "C=math.asin(sinC)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Critical angle is\", round(C,2),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Critical angle is 62.49 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 60
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.9 Page no 924"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=1.5*10**8\n",
+ "c=3.0*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=v/c\n",
+ "C=math.asin(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of critical angle is\", round(C,0),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of critical angle is 30.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 65
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.10 Page no 924"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "uw=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "a=1/uw\n",
+ "b=math.sin(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of refraction is\", round(b,0),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of refraction is 39.0 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.11 Page no 924"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=4\n",
+ "b=6.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=a/b\n",
+ "B=math.atan(A)*180/3.14\n",
+ "ur=1/(math.sin(B*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"Refrective index of the liquid is\", round(ur,1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Refrective index of the liquid is 1.8\n"
+ ]
+ }
+ ],
+ "prompt_number": 79
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.12 Page no 925"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=52 #Degree\n",
+ "b=33 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "u2=(math.sin(a*3.14/180.0))/(math.sin(b*3.14/180.0))\n",
+ "C=1/u2\n",
+ "A=math.asin(C)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of refrection is\", round(A,1),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of refrection is 43.7 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 88
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.13 Page no 932"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-240.0\n",
+ "R=15.0 #cm\n",
+ "u1=1.33\n",
+ "u2=1.5\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/((((u2-u1)/R)+(u1/u))/u2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image is\", round(v,0),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the image is 259.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 94
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.14 Page no 932"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-9.0 #cm\n",
+ "y=1\n",
+ "y1=1.5\n",
+ "R=-15.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/(((y-y1)/R)-(y1/-u))\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of distance is\",v,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of distance is -7.5 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.15 Page no 933 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-15 #cm\n",
+ "y1=1\n",
+ "y2=1.5\n",
+ "R=-7.5 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/(((y1-y2)/R)-(y2/-u))\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image is\",v,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the image is -30.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.16 Page no 933"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-60.0 #cm\n",
+ "R=25.0 #cm\n",
+ "y1=1\n",
+ "y2=1.5\n",
+ "\n",
+ "#Calcution\n",
+ "v=1/((((y2-y1)/R)+(y1/u))/y2)\n",
+ "P=(y2-y1)/(R*10**-2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image is\", v,\"cm\"\n",
+ "print\"Power of the refracting surface is\", P,\"dioptre\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the image is 450.0 cm\n",
+ "Power of the refracting surface is 2.0 dioptre\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.17 Page no 934"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u1=1\n",
+ "u2=1.5\n",
+ "R=1\n",
+ "\n",
+ "#Calculation\n",
+ "x=(u1+u2)/(u2-u1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance of the object is\", x,\"R\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance of the object is 5.0 R\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.18 Page no 934"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=7.5 #cm\n",
+ "u1=1\n",
+ "u2=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/(((u1-u2)/R))\n",
+ "\n",
+ "#Result\n",
+ "print\"It gets focused at\", round(v,1),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "It gets focused at -22.7 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.19 Page no 935"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u1=1\n",
+ "u2=1.5\n",
+ "u=-10\n",
+ "v=-40 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "R=-v*(u2-u1)/(u1+u2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Curvature given to the bounding surface is\", R,\"cm (Convex)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Curvature given to the bounding surface is 8.0 cm (Convex)\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.20 Page no 935"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u1=1\n",
+ "u2=1.5\n",
+ "v=100 #cm\n",
+ "R=20.0 #cm\n",
+ "a=3\n",
+ "b=200.0\n",
+ "\n",
+ "#Calculation\n",
+ "u1=(u2-u1)/R\n",
+ "u2=-1/(u1-(a/b))\n",
+ "d=-u2+R\n",
+ "\n",
+ "#Result\n",
+ "print\"The object distance from the centre of curvature is\", d,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The object distance from the centre of curvature is 120.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.21 Page no 952"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ug=1.5\n",
+ "R1=50.0 #cm\n",
+ "R2=-50.0 #cm\n",
+ "uw=9/8.0\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/((ug-1)*((1/R1)+(1/R1)))\n",
+ "f1=1/((uw-1)*((1/R1)+(1/R1)))\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Focal length in air is\", f,\"cm\"\n",
+ "print\"(ii) Focal lenth in water is\", f1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Focal length in air is 50.0 cm\n",
+ "(ii) Focal lenth in water is 200.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 50
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.22 Page no 953"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "fa=20 #cm\n",
+ "ug=9/8.0\n",
+ "uw=3/2.0\n",
+ "\n",
+ "#Calculation\n",
+ "a=(uw-1)/(ug-1)\n",
+ "fw=a*fa\n",
+ "f=fw-fa\n",
+ "\n",
+ "#Result\n",
+ "print\"Change in focal length is\", f,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Change in focal length is 60.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 54
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.23 Page no 953"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.56\n",
+ "R1=20.0 #cm\n",
+ "u1=-10.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/((u-1)*(2/R1))\n",
+ "v=1/((1/u1)+(1/f))\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image formed is\", round(v,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the image formed is -22.73\n"
+ ]
+ }
+ ],
+ "prompt_number": 63
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.24 Page no 953"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.47\n",
+ "\n",
+ "#Calculation\n",
+ "u1=u\n",
+ "\n",
+ "#Result\n",
+ "print\"The liquid is not water because refractive index of water is 1.33\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The liquid is not water because refractive index of water is 1.33\n"
+ ]
+ }
+ ],
+ "prompt_number": 65
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.25 Page no 954"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=18 #cm\n",
+ "u=1.5\n",
+ "\n",
+ "#Calculation\n",
+ "R=(u-1)*f\n",
+ "\n",
+ "#Result\n",
+ "print\"Radius of the curvature is\", R,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of the curvature is 9.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.26 Page no 954"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-25.0 #cm\n",
+ "f=10.0 #cm\n",
+ "h1=5\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/((1/f)+(1/u))\n",
+ "h2=(v*h1)/u\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image is\", round(v,2),\"cm\"\n",
+ "print\"Size of the image is\",round(h2,2),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the image is 16.67 cm\n",
+ "Size of the image is -3.33 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.27 Page no 954"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=-15.0 #cm\n",
+ "v=-10.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "u=1/((1/v)-1/f)\n",
+ "\n",
+ "#Result\n",
+ "print\"The object is placed at a distance of\", u,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The object is placed at a distance of -30.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 81
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.28 Page no 954"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=-20.0 #cm\n",
+ "u=-60.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/((1/v)-(1/u))\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length of the lens is\", f,\"cm\"\n",
+ "print\"The lens is diverging\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length of the lens is -30.0 cm\n",
+ "The lens is diverging\n"
+ ]
+ }
+ ],
+ "prompt_number": 85
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.29 Page no 955"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-10.0 #cm\n",
+ "m=-3.0\n",
+ "\n",
+ "#Calculation\n",
+ "v=m*u\n",
+ "f=1/((1/v)-(1/u))\n",
+ "\n",
+ "#Result\n",
+ "print\"Image formed at\",v,\"cm\"\n",
+ "print\"Focal length is\",f,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Image formed at 30.0 cm\n",
+ "Focal length is 7.5 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 94
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.30 Page no 955"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P1=6\n",
+ "P2=-2.0\n",
+ "\n",
+ "#Calculation\n",
+ "P=P1+P2\n",
+ "f=1/P\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length of the combination is\", f*10**2,\"cm\"\n",
+ "print\"Power of the combinationis\",P,\"D\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length of the combination is 25.0 cm\n",
+ "Power of the combinationis 4.0 D\n"
+ ]
+ }
+ ],
+ "prompt_number": 102
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.31 Page no 955"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f1=20.0 #cm\n",
+ "f2=-40.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/((1/f1)+(1/f2))\n",
+ "P=1/f\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length is\", f,\"cm\"\n",
+ "print\"Power is\",P*10**2,\"D\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length is 40.0 cm\n",
+ "Power is 2.5 D\n"
+ ]
+ }
+ ],
+ "prompt_number": 107
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.32 Page no 955"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=2.0\n",
+ "b=1\n",
+ "\n",
+ "#Calculation\n",
+ "u=(b/a)+b\n",
+ "\n",
+ "#Result\n",
+ "print\"Refractive index of the material is\", u"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Refractive index of the material is 1.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 109
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.33 Page no 955"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=-0.2 #m\n",
+ "v=0.3 #m\n",
+ "\n",
+ "#Calculation\n",
+ "u=1/((1/v)-(1/f))\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the point is\", u,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the point is 0.12 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 113
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.35 Page no 957"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u1=-30.0 #cm\n",
+ "f1=10.0\n",
+ "u2=10\n",
+ "f2=-10.0\n",
+ "\n",
+ "#calculation\n",
+ "v1=1/((1/u1)+(1/f1))\n",
+ "v2=1/((1/u2)+(1/f2))\n",
+ "v3=-u1\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image for first lens is\", v1,\"cm\"\n",
+ "print\"Position of the image for second lens is\", round(v2*10**-2,0),\"cm\"\n",
+ "print\"Position of the image for third lens is\", v3,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the image for first lens is 15.0 cm\n",
+ "Position of the image for second lens is -0.0 cm\n",
+ "Position of the image for third lens is 30.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 127
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter18_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter18_2.ipynb new file mode 100644 index 00000000..46d58979 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter18_2.ipynb @@ -0,0 +1,549 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:b23935cae4f05cd3030e505584dd90917a65a9efe1b245c771989ab0310cdeb5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 18 Dispersion of light"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.1 Page no 986"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=60 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=math.sqrt(2)*math.sin(30*3.14/180.0)\n",
+ "b=math.asin(a)*180/3.14\n",
+ "c=(b*2)-A\n",
+ "i=(A+c)/2.0\n",
+ "r=A/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Angle of minimum deviation is\", round(c,0),\"Degree\"\n",
+ "print\"(ii) Angle of incidence is\", round(i,0),\"Degree\"\n",
+ "print\"(iii) The angle of refraction is\", r,\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Angle of minimum deviation is 30.0 Degree\n",
+ "(ii) Angle of incidence is 45.0 Degree\n",
+ "(iii) The angle of refraction is 30.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.2 Page no 986"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=51 #Degree\n",
+ "A=60 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "b=(A+a)/2.0\n",
+ "c=A/2.0\n",
+ "u=(math.sin(b*3.14/180.0))/(math.sin(c*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The refracting angle of the prism is\", A,\"Degree\"\n",
+ "print\"(ii) The refractive index of the material is\",round(u,4)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The refracting angle of the prism is 60 Degree\n",
+ "(ii) The refractive index of the material is 1.6485\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.3 Page no 987"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "i1=48 #Degree\n",
+ "A=60 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=A/2.0\n",
+ "u=math.sin(i1*3.14/180.0)/math.sin(r*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Refractive index of the material is\", round(u,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Refractive index of the material is 1.49\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.4 Page no 987"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=2.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=math.sqrt(a)/a\n",
+ "i=math.asin(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of incidence is\", round(i,0),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of incidence is 45.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.5 Page no 987"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.5\n",
+ "a=6 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "A=a/(u-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of the prism is\", A,\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of the prism is 12.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.6 Page no 987"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ug=1.5\n",
+ "r1=30 #Degree\n",
+ "ua=1.0\n",
+ "A=60 #Degree\n",
+ "A1=90 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "sin=(ug*math.sin(r1*3.14/180.0))/ua\n",
+ "i1=math.asin(sin)*180/3.14\n",
+ "a=(2*i1)-A\n",
+ "sin1=1/ug\n",
+ "r1=math.asin(sin1)*180/3.14\n",
+ "r2=A-r1\n",
+ "sin2=(ug*math.sin(r2*3.14/180.0))\n",
+ "i2=math.asin(sin2)*180/3.14\n",
+ "A3=A1+i2-A\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The angle of incidence for minimum deviation is\", round(i1,0),\"Degree\"\n",
+ "print\"(ii) The angle of minimum deviation is\", round(a,0)\n",
+ "print\"(iii) The angle of emergence of light at maximum deviation is\", round(i2,0),\"Degree\"\n",
+ "print\"(iv) Angle of maximum deviation is\", round(A3,0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The angle of incidence for minimum deviation is 49.0 Degree\n",
+ "(ii) The angle of minimum deviation is 37.0\n",
+ "(iii) The angle of emergence of light at maximum deviation is 28.0 Degree\n",
+ "(iv) Angle of maximum deviation is 58.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.7 Page no 991"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "uv=1.68\n",
+ "ur=1.56\n",
+ "A=18 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "A1=A*(uv-ur)\n",
+ "\n",
+ "#Result\n",
+ "print\"Angular dispersion is\", A1,\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angular dispersion is 2.16 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 71
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.8 Page no 991"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "av=3.32 #Degree\n",
+ "ar=3.22 #Degree\n",
+ "a=3.27 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "w=(av-ar)/a\n",
+ "\n",
+ "#Result\n",
+ "print\"Dispersive power of the flint glass is\", round(w,4)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Dispersive power of the flint glass is 0.0306\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.9 Page no 991"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ur=1.5155\n",
+ "uv=1.5245\n",
+ "\n",
+ "#Calculation\n",
+ "u=(ur+uv)/2.0\n",
+ "w=(uv-ur)/(u-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Dispersive power of the crown glass is\", round(w,4)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Dispersive power of the crown glass is 0.0173\n"
+ ]
+ }
+ ],
+ "prompt_number": 79
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.10 Page no 991"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "w=0.031\n",
+ "ur=1.645\n",
+ "ub=1.665\n",
+ "\n",
+ "#Calculation\n",
+ "u=1+((ub-ur))/w\n",
+ "\n",
+ "#Result\n",
+ "print\"Refrective index for yellow colour is\", round(u,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Refrective index for yellow colour is 1.645\n"
+ ]
+ }
+ ],
+ "prompt_number": 83
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.11 Page no 992"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=5 #Degree\n",
+ "uv=1.523\n",
+ "ur=1.515\n",
+ "uv1=1.688\n",
+ "ur1=1.650\n",
+ "\n",
+ "#Calculation\n",
+ "u=(uv+ur)/2.0\n",
+ "u1=(uv1+ur1)/2.0\n",
+ "A1=-((u-1)*A)/(u1-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of flint line is\",round(A1,2),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of flint line is -3.88 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 91
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.12 Page no 992"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "w=0.021\n",
+ "u=1.53\n",
+ "w1=0.045\n",
+ "u1=1.65\n",
+ "A1=4.20 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "A=-(w1*A1*(u1-1))/(w*(u-1))\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of the prism is\", round(A,2),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of the prism is -11.04 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.13 Page no 992"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=72 #Degree\n",
+ "ab=56.4 #Degree\n",
+ "ar=53 #Degree\n",
+ "ay=54.6 #Degree\n",
+ "az=54\n",
+ "A11=60 #Degree\n",
+ "ab1=52.8 \n",
+ "A12=50.6\n",
+ "A13=51.9\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A1=(A+ay)/2.0\n",
+ "A2=A/2.0\n",
+ "ub=(math.sin(A1*3.14/180.0))/(math.sin(A2*3.14/180.0))\n",
+ "A3=(A+ar)/2.0\n",
+ "ur=(math.sin(A3*3.14/180.0))/(math.sin(A2*3.14/180.0))\n",
+ "A4=(A+az)/2.0\n",
+ "uy=(math.sin(A4*3.14/180.0))/(math.sin(A2*3.14/180.0))\n",
+ "w=(ub-ur)/(uy-1)\n",
+ "\n",
+ "#For flint glass prism\n",
+ "A5=(A11+ab1)/2.0\n",
+ "A51=A11/2.0\n",
+ "ub1=(math.sin(A5*3.14/180.0))/(math.sin(A51*3.14/180.0))\n",
+ "A6=(A11+A12)/2.0\n",
+ "ur1=(math.sin(A6*3.14/180.0))/(math.sin(A51*3.14/180.0))\n",
+ "A7=(A11+A13)/2.0\n",
+ "uy1=(math.sin(A7*3.14/180.0))/(math.sin(A51*3.14/180.0))\n",
+ "w1=(ub1-ur1)/(uy1-1)\n",
+ "w2=w/w1\n",
+ "\n",
+ "#Result\n",
+ "print\"The ratio of dispersive power of crown glass and flint glass prism is\", round(w2,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of dispersive power of crown glass and flint glass prism is 0.64\n"
+ ]
+ }
+ ],
+ "prompt_number": 122
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter19_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter19_2.ipynb new file mode 100644 index 00000000..3a7eac21 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter19_2.ipynb @@ -0,0 +1,841 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:91c393a3f1616e8ab4ec5b337712a2b4f8e8dca0635755aebf9e9a0db573e23b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 19 Optical instruments"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.1 Page no 1013"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=-75.0\n",
+ "u=0\n",
+ "\n",
+ "#Calculation\n",
+ "f=v\n",
+ "P=100/f\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length is\", f,\"cm\"\n",
+ "print\"Power of the lens is\",round(P,2),\"D\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length is -75.0 cm\n",
+ "Power of the lens is -1.33 D\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.2 Page no 1014"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-25.0 #cm\n",
+ "v=-150.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/((1/v)-1/u)\n",
+ "P=100/f\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length of the lens is\", f,\"cm\"\n",
+ "print\"Power of the lens is\",round(P,2),\"D\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length of the lens is 30.0 cm\n",
+ "Power of the lens is 3.33 D\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.3 Page no 1014"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-25.0 #cm\n",
+ "v=-50.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/((1/v)-1/u)\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length is\", f,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length is 50.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.4 Page no 1014"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=-80.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=v\n",
+ "P=100/f\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Power of the lens is\", P,\"D\"\n",
+ "print\"(b) No the corrective lens is concave and it reduces the size of the image. Because it bring the object at the far point of the eye\"\n",
+ "print\"(c) The myopic person may have a normal near point. He must keep the book at a distance greater than 25 cm.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Power of the lens is -1.25 D\n",
+ "(b) No the corrective lens is concave and it reduces the size of the image. Because it bring the object at the far point of the eye\n",
+ "(c) The myopic person may have a normal near point. He must keep the book at a distance greater than 25 cm.\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.5 Page no 1015"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=-75.0 #cm\n",
+ "u=-25.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/((1/v)-1/u)\n",
+ "P=100/f\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Power of the lens is\", round(P,2),\"D\"\n",
+ "print\"(b) The corrective lens produce a virtual imageof an object at 25 cm. The angular size of this image is the same as the object\"\n",
+ "print\"(c) A hypermetropic eye may have normal far point.Hence the person prefers not to use the spectacles for distant object\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Power of the lens is 2.67 D\n",
+ "(b) The corrective lens produce a virtual imageof an object at 25 cm. The angular size of this image is the same as the object\n",
+ "(c) A hypermetropic eye may have normal far point.Hence the person prefers not to use the spectacles for distant object\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.6 Pageno 1015"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=-0.8 #d\n",
+ "v1=-15.0 #cm \n",
+ "v2=-100.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=100/P\n",
+ "u1=1/((1/v1)-1/f)\n",
+ "u2=1/((1/v2)-(1/f))\n",
+ "\n",
+ "#Result\n",
+ "print\"The person can see objects lying between\",round(-u1,0),\"cm and\",-u2,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The person can see objects lying between 17.0 cm and 500.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.7 Page no 1016"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-25 #cm\n",
+ "p=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "f=100/p\n",
+ "v=1/((1/f)+1/u)\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance is\",round(v,0),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance is -1.0 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.8 Page no 1016"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-25.0 #cm\n",
+ "v=-90.0 #cm\n",
+ "\n",
+ "#calculation\n",
+ "f=1/((1/v)-1/u)\n",
+ "f1=(1/2.0)*10**2\n",
+ "u=1/((1/v)-1/f1)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) focal length is\",round(f,1),\"cm\"\n",
+ "print\"(ii) Distance is\",round(u,1),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) focal length is 34.6 cm\n",
+ "(ii) Distance is -32.1 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.9 Page no 1022"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "D=25\n",
+ "f=5.0 #cm\n",
+ "\n",
+ "#calculation\n",
+ "M=1+(D/f)\n",
+ "M1=D/f\n",
+ "\n",
+ "#Result\n",
+ "print\"The magnifying power if the final image is formed at the least distance is\",M\n",
+ "print\"The magnifying power if image is formed at infinity is\",M1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The magnifying power if the final image is formed at the least distance is 6.0\n",
+ "The magnifying power if image is formed at infinity is 5.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 61
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.10 Page no 1023"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=4.80 #cm\n",
+ "a=1.20\n",
+ "v=-24.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "D=f/(a-1)\n",
+ "u=1/((1/v)-1/f)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The least distance of distinct vision is\",D,\"cm\"\n",
+ "print\"(ii) Distance from the lens is\",-u,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The least distance of distinct vision is 24.0 cm\n",
+ "(ii) Distance from the lens is 4.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 69
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.11 Page no 1023"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v0=15.0 #cm\n",
+ "f0=3.0 #cm\n",
+ "D=25\n",
+ "fe=9\n",
+ "\n",
+ "#Calculation\n",
+ "u0=1/((1/v0)-1/f0)\n",
+ "M=-(v0*D)/(u0*fe)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnifying power is\", round(M,1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnifying power is 11.1\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.12 Page no 1024"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P1=1.5 #D\n",
+ "P2=20.0 #D\n",
+ "u=-25.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f2=100/P2\n",
+ "M=1+(D/f2)\n",
+ "f1=100/P1\n",
+ "v=1/((1/f1)+1/u)\n",
+ "M1=1-(v/f2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The maximum magnifying power together with his glasses\", M\n",
+ "print\"(ii) The maximum magnifying power without glasses\",M1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The maximum magnifying power together with his glasses 6.0\n",
+ "(ii) The maximum magnifying power without glasses 9.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 83
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.13 Page no 1024"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=16\n",
+ "d=-2.5 #cm\n",
+ "f0=0.4 #cm\n",
+ "D=25\n",
+ "\n",
+ "#Calculation\n",
+ "v0=l+d\n",
+ "u0=1/((1/v0)-1/f0)\n",
+ "M=-v0*D/(u0*d)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnifying power of the microscope is\", M"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnifying power of the microscope is -327.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 88
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.14 Page no 1025"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f0=1.0\n",
+ "u0=-1.1 #cm\n",
+ "D=25\n",
+ "fe=5.0\n",
+ "ve=25.0\n",
+ "\n",
+ "#Calculation\n",
+ "v0=1/((1/f0)+1/u0)\n",
+ "d=v0+fe\n",
+ "M=-(v0*D)/(u0*fe)\n",
+ "ue=-1/((1/ve)+1/fe)\n",
+ "D1=v0-ue\n",
+ "M1=-(v0/u0)*(1+(D/fe))\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Distance between the lenses when image is at infinity\", d,\"cm\"\n",
+ "print\"Magnifying power is\",M\n",
+ "print\"(ii) Distance between the lenses when image is at distinct vision\",round(D1,2),\"cm\"\n",
+ "print\"Magnifying Power is\",M1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Distance between the lenses when image is at infinity 16.0 cm\n",
+ "Magnifying power is 50.0\n",
+ "(ii) Distance between the lenses when image is at distinct vision 15.17 cm\n",
+ "Magnifying Power is 60.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 106
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.15 Page no 1032"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f0=200 #cm\n",
+ "fe=5.0 #cm\n",
+ "D=25.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "M=(f0/fe)*(1+(fe/D))\n",
+ "M1=f0/fe\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Magnifying power when image is formed at near point is\", M\n",
+ "print\"(ii) Magnifying power when image is formed at infinity\",M1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnifying power when image is formed at near point is 48.0\n",
+ "(ii) Magnifying power when image is formed at infinity 40.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 110
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.16 Page no 1033"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "fe=3\n",
+ "M=4\n",
+ "\n",
+ "#Calculation\n",
+ "f0=fe*M\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length of the lenses is\" ,f0,\"cm and\",fe,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length of the lenses is 12 cm and 3 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 112
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.17 Page no 1033"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u0=-200.0 #cm\n",
+ "f0=30.0 #cm\n",
+ "fe=3\n",
+ "\n",
+ "#Calculation\n",
+ "v0=1/((1/f0)+1/u0)\n",
+ "a=v0+fe\n",
+ "\n",
+ "#Result\n",
+ "print\"Separation between the objective and eyepiece is\", round(a,1),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Separation between the objective and eyepiece is 38.3 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 118
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.18 Page no 1033"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ve=24.0\n",
+ "fe=8.0\n",
+ "f0=250.0\n",
+ "a=10\n",
+ "\n",
+ "#Calculation\n",
+ "ue=1/((1/ve)-(1/fe))\n",
+ "D=f0-ue\n",
+ "d=a/2.0\n",
+ "A=d/f0\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Distance between objective and eyepiece is\", D,\"cm\"\n",
+ "print\"(ii) Angle subtended by the sun at the objective is\",A,\"rad\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Distance between objective and eyepiece is 262.0 cm\n",
+ "(ii) Angle subtended by the sun at the objective is 0.02 rad\n"
+ ]
+ }
+ ],
+ "prompt_number": 126
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.19 Page no 1034"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=-20\n",
+ "R=-120\n",
+ "\n",
+ "#Calculation\n",
+ "f0=R/2.0\n",
+ "fe=f0/M\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length of eyepiece is\", fe,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length of eyepiece is 3.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 131
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.20 Page no 1034"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "fa=180\n",
+ "f=3.5\n",
+ "fe=5.0\n",
+ "\n",
+ "#Calculation\n",
+ "d=fa+(2*f)+(2*f)+fe\n",
+ "M=-fa/fe\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnifying power of thetelescope is\", M"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnifying power of thetelescope is -36.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 135
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.21 Page no 1034"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u0=-200.0 #cm\n",
+ "fa=50.0 #cm\n",
+ "ve=-25.0 #cm\n",
+ "fe=5.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "v0=1/((1/fa)+1/u0)\n",
+ "M0=v0/u0\n",
+ "ue=1/((1/ve)-1/fe)\n",
+ "Me=ve/ue\n",
+ "D=v0-ue\n",
+ "M=M0*Me\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Saparation between the objective and eyepiece is\", round(D,2),\"cm\"\n",
+ "print\"(ii) Magnification is\",M"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Saparation between the objective and eyepiece is 70.83 cm\n",
+ "(ii) Magnification is -2.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 147
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter1_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter1_2.ipynb new file mode 100644 index 00000000..f6180b5a --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter1_2.ipynb @@ -0,0 +1,543 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:9880f2d8505e271317a099910ead6c2116ce86fa0e83f56feb35ac33a1b96b23"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 1 Electric charge"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.1 Page no 9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=4.5*10**-19 #C\n",
+ "e=1.6*10**-19 #C\n",
+ "\n",
+ "#Calculation\n",
+ "n=q/e\n",
+ "\n",
+ "#Result\n",
+ "print\"n= \",round(n,1),\"This value of charge is not possible\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "n= 2.8 This value of charge is not possible\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.2 Page no 9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=3.2*10**-7 #C\n",
+ "e=1.6*10**-19 #C\n",
+ "\n",
+ "#Calculation\n",
+ "n=q/e\n",
+ "\n",
+ "#Result\n",
+ "print\"The required number of electrons is \",n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The required number of electrons is 2e+12\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.3 Page no 9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=19.2*10**-19\n",
+ "e=1.6*10**-19\n",
+ "me=9*10**-31 #Kg\n",
+ "\n",
+ "#Calculation\n",
+ "n=q/e\n",
+ "M=n*me\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The value of n=\",n,\"\\n(ii) Charge on silk=\",-q*10**19,\"*10**-19\"\n",
+ "print\"(iii) Mass=\",M,\"Therefore mass transferred is negligibly small\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The value of n= 12.0 \n",
+ "(ii) Charge on silk= -19.2 *10**-19\n",
+ "(iii) Mass= 1.08e-29 Therefore mass transferred is negligibly small\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.4 Page no 9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=16\n",
+ "n=6.023*10**23 #C\n",
+ "\n",
+ "#Calculation\n",
+ "W=2+a\n",
+ "A=((n*100)/W)*10\n",
+ "\n",
+ "#Result\n",
+ "print\"Total number of electrons in 100 g of water \", round(A,-23)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total number of electrons in 100 g of water 3.35e+25\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.5 Page no 9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=10**9\n",
+ "e=1.6*10**-19 #C\n",
+ "Q=1\n",
+ "\n",
+ "#Calculation\n",
+ "q=n*e\n",
+ "t=Q/q\n",
+ "\n",
+ "#Result\n",
+ "print (t*10**-9),\"10**9 S\"\n",
+ "print\"Time required is about 198 years\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "6.25 10**9 S\n",
+ "Time required is about 198 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.6 Page no 13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q1=20 #micro C\n",
+ "q2=-5 #micro C\n",
+ "a=9*10**9\n",
+ "r=0.1 \n",
+ "\n",
+ "#Calculation\n",
+ "q=q1+q2\n",
+ "q3=q/2.0\n",
+ "F=(a*q3*q3)/r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Force is \",round(F*10**-13,3),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force is 5.062 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 1.10 Page no 14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q=5*10**-6\n",
+ "r=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "F=(m*q*q)/r**2\n",
+ "C=2*F*math.cos(30)*(180/3.14)\n",
+ "\n",
+ "#Result\n",
+ "print\"Force on each charge is \", round(C,1)*10**-1,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force on each charge is 39.79 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.11 Page no 14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q=1\n",
+ "r=0.24\n",
+ "A=20\n",
+ "B=12.0\n",
+ "m1=10**-4\n",
+ "g=9.8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "F=(m*q**2)/r**2\n",
+ "AD=math.sqrt(A**2-B**2)\n",
+ "C=AD/B\n",
+ "F1=(1/C)*m1*g\n",
+ "Q=math.sqrt(F1/F)\n",
+ "\n",
+ "#Result\n",
+ "print\"Charge on each sphere\", round(Q*10**8,1),\"10**-8\",\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge on each sphere 6.9 10**-8 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 79
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.12 Page no 15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "F=3.7*10**-9 #C\n",
+ "q=1.6*10**-19 #c\n",
+ "m=9*10**9\n",
+ "r=5*10**-10\n",
+ "\n",
+ "#Calculation \n",
+ "import math\n",
+ "n=math.sqrt(F*r**2/(m*q**2))\n",
+ "\n",
+ "#Result\n",
+ "print round(n,0),\"electrons are missing from each icon\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "2.0 electrons are missing from each icon\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.14 Page no 16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19\n",
+ "m=9*10**9\n",
+ "G=6.67*10**-11\n",
+ "me=9.11*10**-31\n",
+ "mp=1.67*10**-27\n",
+ "r=10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "F0=(m*e**2)/(G*me*mp)\n",
+ "F1=(m*e**2)/(G*mp*mp)\n",
+ "F2=m*e**2/r**2\n",
+ "A1=F2/me\n",
+ "A2=F2/mp\n",
+ "\n",
+ "#Result\n",
+ "print\"(a)(i)strength of an electrons and protons\", round(F0*10**-39,1)*10**39\n",
+ "print\" (ii)Strength of two protons \",round(F1*10**-36,1)*10**36\n",
+ "print\"(b) Acceleration of electron is \",round(A1*10**-22,1)*10**22,\"m/s**2\"\n",
+ "print\" Acceleration of proton is \",round(A2*10**-19,1)*10**19,\"m/s*2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)(i)strength of an electrons and protons 2.3e+39\n",
+ " (ii)Strength of two protons 1.2e+36\n",
+ "(b) Acceleration of electron is 2.5e+22 m/s**2\n",
+ " Acceleration of proton is 1.4e+19 m/s*2\n"
+ ]
+ }
+ ],
+ "prompt_number": 112
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.16 Page no 19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9 #C\n",
+ "q1=10*10**-6\n",
+ "q2=5*10**-6\n",
+ "r=0.05\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "F1=m*q1*q2/r**2\n",
+ "F2=m*q1*q2/r**2\n",
+ "F3=math.sqrt(F1**2+F2**2+(2*F1*F2*math.cos(120)*180/3.14))\n",
+ "\n",
+ "#Result\n",
+ "print\"Resultant charge is \", round(F3*10**-1,0),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resultant charge is 176.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 132
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.17 Page no 20 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q1=1.2*10**-8\n",
+ "q2=1\n",
+ "r=0.03\n",
+ "r1=0.04\n",
+ "q3=1.6*10**-8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "F1=m*q1*q2/r**2\n",
+ "F2=m*q3*q2/r1**2\n",
+ "F3=math.sqrt(F1**2+F2**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Total force is \", F3*10**-5,\"10**5\",\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total force is 1.5 10**5 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 149
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.18 Page no 20"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q1=1\n",
+ "q2=100\n",
+ "r=10\n",
+ "q3=75 #C\n",
+ "r1=5\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "F=m*q1*q2/r**2 #along BA\n",
+ "F1=m*q1*q2/r**2 #along AC\n",
+ "F2=m*q3/(math.sqrt(r**2-r1**2)**2)\n",
+ "F3=math.sqrt(F1**2+F2**2)\n",
+ "X=F1/F2\n",
+ "\n",
+ "#Result\n",
+ "print\"Force experienced by 1 C Charge is \",round(F3*10**-9,2),\"N\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force experienced by 1 C Charge is 12.73 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 168
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter20_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter20_2.ipynb new file mode 100644 index 00000000..e96a7bf3 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter20_2.ipynb @@ -0,0 +1,330 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:49e10d509d6c3c83253662b249f2d9cebaf084cb6d339d2868de883d5e7038f4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 20 Photometry"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.1 Page no 1055"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=2.5*10**5 #lm/m**2\n",
+ "r=1.5*10**11 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l=E*r**2\n",
+ "a=4*math.pi*l\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Luminous intensity is\", l,\"cd\"\n",
+ "print\"(ii) Luminous flux of the sun is\",round(a*10**-28,3)*10**28,\"lm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Luminous intensity is 5.625e+27 cd\n",
+ "(ii) Luminous flux of the sun is 7.069e+28 lm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.2 Page no 1055"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I2=150\n",
+ "I1=75.0\n",
+ "E1=20\n",
+ "\n",
+ "#Calculation\n",
+ "E2=(I2*E1)/I1\n",
+ "\n",
+ "#Result\n",
+ "print\"Illumination is\", E2,\"lux\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Illumination is 40.0 lux\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.3 Page no 1056"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=35\n",
+ "e=5.0 #lumen/watt\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=4*math.pi*I\n",
+ "P=a/e\n",
+ "\n",
+ "#Result\n",
+ "print\"Power of the lamp is\", round(P,0),\"Watt\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power of the lamp is 88.0 Watt\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.4 Page no 1056"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=1260\n",
+ "r=8 #m\n",
+ "a1=6\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=a/(4.0*math.pi)\n",
+ "Ea=I/r**2\n",
+ "LB=math.sqrt(r**2+a1**2)\n",
+ "cos=r/LB\n",
+ "Eb=(I*cos)/LB**2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The illumination at a point immediately below the lamp is\", round(Ea,2),\"lux\"\n",
+ "print\"(ii) The illumination on the working plane is\",round(Eb,1),\"lux\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The illumination at a point immediately below the lamp is 1.57 lux\n",
+ "(ii) The illumination on the working plane is 0.8 lux\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.5 Page no 1056"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=6.0 #m\n",
+ "I=250 #cd\n",
+ "PQ=8\n",
+ "\n",
+ "#Calculation\n",
+ "Ep=I/r**2\n",
+ "LQ=math.sqrt(r**2+PQ**2)\n",
+ "cos=r/LQ\n",
+ "EQ=(I*cos)/LQ**2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Illumination at a point P is\", round(Ep,2),\"lux\"\n",
+ "print\"(ii) illumination at a point Q is\",EQ,\"lux\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Illumination at a point P is 6.94 lux\n",
+ "(ii) illumination at a point Q is 1.5 lux\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.6 Page no 1057"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "t1=2.5 #second\n",
+ "r1=0.5\n",
+ "r2=1\n",
+ "\n",
+ "#Calculation\n",
+ "t2=(t1*r2**2)/r1**2\n",
+ "\n",
+ "#Result\n",
+ "print\"exposure time is\",t2,\"s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "exposure time is 10.0 s\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.7 Page no 1058"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "i2=60\n",
+ "r2=105.0\n",
+ "r1=70\n",
+ "\n",
+ "#Calculation\n",
+ "i1=(i2*r1**2)/r2**2\n",
+ "\n",
+ "#Result\n",
+ "print\"The luminous intensity of the first lamp is\",round(i1,2),\"cd\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The luminous intensity of the first lamp is 26.67 cd\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.8 Page no 1059"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ra=60\n",
+ "rb=45.0\n",
+ "a=40.0\n",
+ "\n",
+ "#Calculation\n",
+ "ia1=(ra**2)/(rb**2)\n",
+ "ia=(ra**2)/(a**2)\n",
+ "i=ia-ia1\n",
+ "A=(i*100)/ia\n",
+ "\n",
+ "#Result\n",
+ "print\"percentage of light is absorbed by the glass is\",round(A,0),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "percentage of light is absorbed by the glass is 21.0 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 52
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter21_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter21_2.ipynb new file mode 100644 index 00000000..5e2ded02 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter21_2.ipynb @@ -0,0 +1,383 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:5eb9e6d48b48ecf2d0c9ee8abbe7a462b6b60df5a09da8ebed0ac004de2a0383"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 21 Huygen Principle and interference "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.1 Page no 1090"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Goven\n",
+ "d=5*10**-3 #m\n",
+ "D=1.0 #m\n",
+ "b=0.1092*10**-3\n",
+ "\n",
+ "#Calculation\n",
+ "l=(d*b)/D\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of light used is\", l*10**10,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of light used is 5460.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.2 Page no 1090"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=6200*10**-10 #m\n",
+ "D=0.8\n",
+ "b=2.8*10**-3/4.0\n",
+ "\n",
+ "#Calculation\n",
+ "d=(l*D)/b\n",
+ "\n",
+ "#Result\n",
+ "print\"Separation of the two slit is\", round(d*10**3,1),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Separation of the two slit is 0.7 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.3 Page no 1090"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=62\n",
+ "l=5893\n",
+ "l1=4358.0\n",
+ "\n",
+ "#Calculation\n",
+ "n=(a*l)/l1\n",
+ "\n",
+ "#Result\n",
+ "print\"Fringes required is\", round(n,0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Fringes required is 84.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.4 Page no 1091"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=6000*10**-10 #m\n",
+ "D=0.800 #m\n",
+ "d=0.200*10**-3 #m\n",
+ "\n",
+ "#Calculation\n",
+ "x2=(3*l*D)/(2.0*d)\n",
+ "x21=(2*D*l)/d\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Distance of the second dark fringe is\", x2*10**2,\"cm\"\n",
+ "print\"(ii) Distance of the second dark fringe is\", x21*10**2,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Distance of the second dark fringe is 0.36 cm\n",
+ "(ii) Distance of the second dark fringe is 0.48 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.6 Page no 1091"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Imax=16\n",
+ "Imin=4\n",
+ "\n",
+ "#Calculation\n",
+ "r=Imax/Imin\n",
+ "\n",
+ "#Result\n",
+ "print\"Deduce the ratio of intensity is\", r,\":1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deduce the ratio of intensity is 4 :1\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.7 Page no 1092"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "b=2\n",
+ "u=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "b1=b/u\n",
+ "\n",
+ "#Result\n",
+ "print\"Fringe width is\", round(b1,1),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Fringe width is 1.5 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.8 Page no 1092"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "b2=0.4\n",
+ "b1=0.6\n",
+ "l1=5000\n",
+ "\n",
+ "#Calculation\n",
+ "l2=(b2*2*l1)/b1\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of the light is\", round(l2,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of the light is 6667.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.9 Page no 1092"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=0.125*10**-3 #m\n",
+ "l=4500*10**-10 #m\n",
+ "D=1 #m\n",
+ "\n",
+ "#Calculation\n",
+ "x2=(2*D*l)/d\n",
+ "d1=2*x2\n",
+ "\n",
+ "#Result\n",
+ "print\"Separation between the fringes is\", d1*10**3,\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Separation between the fringes is 14.4 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.10 Page no 1092"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Imax=121\n",
+ "Imin=81.0\n",
+ "\n",
+ "#Calculation\n",
+ "a=Imax/Imin\n",
+ "\n",
+ "#Result\n",
+ "print\"The ratio of intensity at the maxima and minima is\",round(a,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of intensity at the maxima and minima is 1.49\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.13 Page no 1093"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=5.0 #m\n",
+ "d=1 #mm\n",
+ "\n",
+ "#Calculation\n",
+ "a=d/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Width of each slit is\", a,\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Width of each slit is 0.2 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter22_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter22_2.ipynb new file mode 100644 index 00000000..e91e3a27 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter22_2.ipynb @@ -0,0 +1,900 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:c280b16f38bc8dbf3b2a0607bdd0cfd4670bde51a104a8d547b07a434c49c7f5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 22 Diffraction and polarisation"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 22.1 Page no 1124"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "D=1 #m\n",
+ "l=5*10**-7 #m\n",
+ "d=0.1*10**-3 #m\n",
+ "\n",
+ "#Calculation\n",
+ "W=(2*D*l)/d\n",
+ "\n",
+ "#Result\n",
+ "print\"Width of the central maximum is\", W*10**2,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Width of the central maximum is 1.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.2 Page no 1124"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "D=1.60 #m\n",
+ "l=6328*10**-10 #m\n",
+ "w=4.0*10**-3\n",
+ "\n",
+ "#Calculation\n",
+ "d=(2*D*l)/w\n",
+ "\n",
+ "#Result\n",
+ "print\"Width of the slit is\", round(d*10**3,2),\"mm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Width of the slit is 0.51 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.3 Page no 1124"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=7500*10**-10\n",
+ "d=1.0*10**-6\n",
+ "c=20\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=l/d\n",
+ "b=math.asin(a)*180/3.14\n",
+ "A=2*b\n",
+ "x=c*math.tan(a*3.14/180.0)\n",
+ "w=2*x\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Width of central maximum is\", round(A,0),\"Degree\"\n",
+ "print\"(ii) Width of central maximum is\",round(w*10**2,0),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Width of central maximum is 97.0 Degree\n",
+ "(ii) Width of central maximum is 52.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.4 Page no 1125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=6.3*10**-7 #m\n",
+ "a=3.6 #Degree\n",
+ "n=10\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "d=(n*l)/math.sin(a*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Slit width is\", round(d*10**3,1),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Slit width is 0.1 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.5 Page no 1125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=5500*10**-10\n",
+ "d=0.01\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=l/d\n",
+ "b=math.asin(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angular deflection is\", round(b,4),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angular deflection is 0.0032 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.6 Page no 1125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "lr=660\n",
+ "d=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "l1=(2*lr)/d\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of lambda is\",l1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of lambda is 440.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.7 Page no 1126"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l1=5890*10**-10 #m\n",
+ "l2=5896*10**-10\n",
+ "d=2.0*10**-6 #m\n",
+ "D=2 #m\n",
+ "\n",
+ "#Calculation\n",
+ "x=(3*D*(l2-l1))/(2*d)\n",
+ "\n",
+ "#Result\n",
+ "print\"Spacing between the first maxima of two sodium lines is\",x*10**4,\"*10**-4 m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Spacing between the first maxima of two sodium lines is 9.0 *10**-4 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.8 Page no 1126"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=3*10**-3 #m\n",
+ "l=500*10**-9 #m\n",
+ "\n",
+ "#Calculation\n",
+ "Z=d**2/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance is\",Z,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance is 18.0 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.9 Page no 1126"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=2*10**-3 #m\n",
+ "l=600*10**-9 #m\n",
+ "\n",
+ "#Calculation\n",
+ "Z=d**2/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance is\",round(Z,2),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance is 6.67 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.10 Page no 1126"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=2*10**-3 #m\n",
+ "l=5000*10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "Z=d**2/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Fresnel Distance is\",Z,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Fresnel Distance is 8.0 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.11 Page no 1129"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=5.50*10**-7 #m\n",
+ "D=5.1\n",
+ "\n",
+ "#Calculation\n",
+ "a=(1.22*l)/D\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum angular separation is\", round(a*10**7,1)*10**-7,\"rad\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum angular separation is 1.3e-07 rad\n"
+ ]
+ }
+ ],
+ "prompt_number": 60
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.12 Page no 1130"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=6*10**-7 #m\n",
+ "D=0.6\n",
+ "l1=10**10 #m\n",
+ "r=10.0**4*9.46*10**15 #m\n",
+ "\n",
+ "#Calculation\n",
+ "a=(1.22*l)/D\n",
+ "a1=l1/r\n",
+ "\n",
+ "#Result\n",
+ "print round(a1*10**10,2)*10**-10,\"rad\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "1.06e-10 rad\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.13 Page no 1130"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=6000*10**-8\n",
+ "D=254.0\n",
+ "\n",
+ "#Calculation\n",
+ "a=(1.22*l)/D\n",
+ "\n",
+ "#Result\n",
+ "print\"Limt of resolution of a telescope is\",round(a*10**7,1)*10**-7,\"Radian\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Limt of resolution of a telescope is 2.9e-07 Radian\n"
+ ]
+ }
+ ],
+ "prompt_number": 70
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.14 Page no 1130"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "D=600.0 #cm\n",
+ "l=5.5*10**-5 #cm\n",
+ "d=3.8*10**10 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "a=(1.22*l)/D\n",
+ "x=d*a\n",
+ "\n",
+ "#Result\n",
+ "print\"Separation of two points is\", round(x,0),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Separation of two points is 4250.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.15 Page no 1130"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=10**-4 #cm\n",
+ "l=5.8*10**-5 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "Na=l/(2.0*d)\n",
+ "\n",
+ "#Result\n",
+ "print\"Numerical aperature of a microscope is\", Na"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Numerical aperature of a microscope is 0.29\n"
+ ]
+ }
+ ],
+ "prompt_number": 80
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.16 Page no 1131"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1\n",
+ "l=600*10**-9 #,\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "rp=(2*u*math.sin(30*3.14/180.0))/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Resolving power of a microscope is\", round(rp*10**-6,2),\"*10**6\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resolving power of a microscope is 1.67 *10**6\n"
+ ]
+ }
+ ],
+ "prompt_number": 86
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.17 Page no 1133"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l1=15*10**-10 #m\n",
+ "l=6563*10**-10\n",
+ "c=3*10**8 #m/s\n",
+ "\n",
+ "#Calculation\n",
+ "v=(c*l1)/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Speed of star is\", round(v*10**-5,2),\"*10**5 m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Speed of star is 6.86 *10**5 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 91
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.18 Page no 1133"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l1=0.032\n",
+ "l=100.0\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "v=-(l1*c)/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Velocity of star is\",v*10**-4,\"*10**4 m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Velocity of star is -9.6 *10**4 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.21 Page no 1142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=60 #Degree\n",
+ "a1=90\n",
+ "\n",
+ "import math\n",
+ "A=math.tan(a*3.14/180.0)\n",
+ "ap=a1-a\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Refractive index of the medium is\", round(A,3)\n",
+ "print\"(ii) The refracting angle is\",ap,\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Refractive index of the medium is 1.73\n",
+ "(ii) The refracting angle is 30 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 118
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.22 Page no 1142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "ap=math.atan(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of incidence is\", round(ap,0),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of incidence is 53.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 122
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.23 Page no 1142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.33\n",
+ "a=90\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "ap=math.atan(u)*180/3.14\n",
+ "A=a-ap\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle between the sun and the horizon is\", round(A,0),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle between the sun and the horizon is 37.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 127
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.24 Page no 1142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.5\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "ap=math.atan(u)*180/3.14\n",
+ "r=90-ap\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of refraction is\", round(r,1),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of refraction is 33.7 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 132
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.25 Page no 1143"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=30 #Degree\n",
+ "I=3 #A\n",
+ "I0=4.0\n",
+ "I1=1\n",
+ "\n",
+ "#Calculation\n",
+ "a=I/I0\n",
+ "a1=I1/I0\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Fraction of maximum light transferred for 30 degree is\", a\n",
+ "print\"(ii) Fraction of maximum light transferred for 60 degree is\", a1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Fraction of maximum light transferred for 30 degree is 0.75\n",
+ "(ii) Fraction of maximum light transferred for 60 degree is 0.25\n"
+ ]
+ }
+ ],
+ "prompt_number": 135
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.26 Page no 1143"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ap=60 #Degree\n",
+ "u=3\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=1/math.sqrt(u)\n",
+ "C=math.asin(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Critical angle for this medium is\", round(C,2),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Critical angle for this medium is 35.28 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 140
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter23_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter23_2.ipynb new file mode 100644 index 00000000..7d0fa841 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter23_2.ipynb @@ -0,0 +1,893 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:38e55bd383948f67d919af3879ad291116d41c75f201f86aa7c1c2e80cc59941"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Cahpter 23 Dual nature of radiation and matter"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.1 Page no 1200"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.62*10**-34 #J\n",
+ "c=3*10**8 #m/s\n",
+ "l=4.0*10**-7 #m\n",
+ "\n",
+ "#Calculation\n",
+ "E=((h*c)/l)/1.6*10**-19\n",
+ "p=h/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of energy is\", round(E*10**38,1),\"ev\"\n",
+ "print\"Momentum of photon is\",p,\"kg m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of energy is 3.1 ev\n",
+ "Momentum of photon is 1.655e-27 kg m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.2 Page no 1200"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=75*1.6*10**-19 #J\n",
+ "h=6.62*10**-34 #J s\n",
+ "\n",
+ "#Calculation\n",
+ "f=E/h\n",
+ "l=(12400/E)*1.6*10**-19\n",
+ "f=c/(l*10**10)\n",
+ "\n",
+ "#Result\n",
+ "print\"Frequency of the photon is\", round(f*10**5,0)*10**15,\"Hz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Frequency of the photon is 1.8e+16 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.3 Page no 1200"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.62*10**-34 #Js\n",
+ "f=880*10**3 #Hz\n",
+ "E1=10*10**3\n",
+ "\n",
+ "#Calculation\n",
+ "E=h*f\n",
+ "n=E1/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of photons emitted per second is\", round(n*10**-31,3)*10**31"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of photons emitted per second is 1.717e+31\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.4 Page no 1200"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "w=1.8\n",
+ "h=6.63*10**-34\n",
+ "l=5000*10**-10\n",
+ "m=9.0*10**-31\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "W=12400/w\n",
+ "h1=(((h*c)/l)-(w*1.6*10**-19))\n",
+ "h2=h1/1.6*10**-19\n",
+ "vmax=math.sqrt((2*h1)/m)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Threshold wavelength is\",round(W,0),\"A\"\n",
+ "print\"(ii) Maximum K.E of emitted photoelectrons is\", round(h2*10**38,3),\"ev\"\n",
+ "print\"(iii) Maximum velocity is\",round(vmax*10**-5,0),\"*10**5 m/s\"\n",
+ "print\"(iv) If the intensity of light is doubled, K.E of emitted electrons will remain unchanged\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Threshold wavelength is 6889.0 A\n",
+ "(ii) Maximum K.E of emitted photoelectrons is 0.686 ev\n",
+ "(iii) Maximum velocity is 5.0 *10**5 m/s\n",
+ "(iv) If the intensity of light is doubled, K.E of emitted electrons will remain unchanged\n"
+ ]
+ }
+ ],
+ "prompt_number": 65
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.5 Page no 1201"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=2*10**-4\n",
+ "I=30*10**-2\n",
+ "t=1\n",
+ "E=6.62*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "n=(I*A)/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Rate at which photons strike the surface is\",round(n*10**-13,2)*10**13,\"photons/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate at which photons strike the surface is 9.06e+13 photons/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 71
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.6 Page no 1201"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.62*10**-34 #Js\n",
+ "c=3*10**8\n",
+ "l=4500*10**-10 #m\n",
+ "w=2.3\n",
+ "\n",
+ "#Calculation\n",
+ "E=(h*c)/l\n",
+ "E1=(E/1.6*10**-19)*10**38\n",
+ "K=E1-w\n",
+ "f0=(w*1.6*10**-19)/h\n",
+ "p=h/l\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The energy of photon is\", round(E1,1),\"ev\"\n",
+ "print\"(ii) The maximum kinetic energy of emitted electrons is\",round(K,1),\"ev\"\n",
+ "print\"(iii) Threshold frequency for sodium is\",round(f0*10**-14,1)*10**14,\"Hz\"\n",
+ "print\"(iv) Momentum of a photon is\",round(p*10**27,1)*10**-27,\"Kg m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The energy of photon is 2.8 ev\n",
+ "(ii) The maximum kinetic energy of emitted electrons is 0.5 ev\n",
+ "(iii) Threshold frequency for sodium is 5.6e+14 Hz\n",
+ "(iv) Momentum of a photon is 1.5e-27 Kg m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 100
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.7 Page no 1202"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=36.0*10**-8 #m\n",
+ "w0=2*1.6*10**-19 #J\n",
+ "h=6.62*10**-34 #Js\n",
+ "c=3*10**8\n",
+ "e=1.6*10**-19\n",
+ "m=9.0*10**-31\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l0=(h*c)/w0\n",
+ "E=(h*c)/l\n",
+ "E1=(E/1.6*10**-19)*10**38\n",
+ "K=E1-2\n",
+ "v0=K\n",
+ "vmax=math.sqrt(e*v0*2/m)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Threshold wavelength is\",round(l0*10**10,0),\"A\"\n",
+ "print\"(ii) Maximum kinetic energy of emitted photoelectrons is\", round(K,3),\"ev\"\n",
+ "print\"(iii) Stopping potential is\",round(v0,3),\"Volts\"\n",
+ "print\"(iv) Velocity is \",round(vmax*10**-5,2),\"*10**5 m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Threshold wavelength is 6206.0 A\n",
+ "(ii) Maximum kinetic energy of emitted photoelectrons is 1.448 ev\n",
+ "(iii) Stopping potential is 1.448 Volts\n",
+ "(iv) Velocity is 7.18 *10**5 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.8 Page no 1202"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.62*10**-34\n",
+ "c=3*10**8\n",
+ "l0=24.8*10**-8\n",
+ "a=1.2\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "w0=(h*c)/l0\n",
+ "w01=(w0/1.6*10**-19)*10**38\n",
+ "h1=w01+a\n",
+ "C=h1*e\n",
+ "l=(h*c)/C\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of incident light is\", round(l*10**10,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of incident light is 2000.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.9 Page no 1203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v1=16.5\n",
+ "V0=6.6 #V\n",
+ "f0=4.6*10**15 #Hz\n",
+ "f=2.2*10**15 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "h=(e*(v1-V0))/((f0-f))\n",
+ "\n",
+ "#Result\n",
+ "print\"Planck's constant is\", h"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Planck's constant is 6.6e-34\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.10 Page no 1203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.62*10**-34\n",
+ "f0=44*10**13 #Hz\n",
+ "a=11.5*10**14\n",
+ "b=4.4*10**14\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "w0=((h*f0)/1.6*10**-19)*10**38\n",
+ "h=3/(a-b)\n",
+ "h1=h*e\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Work function of the material is\", round(w0,2),\"ev\"\n",
+ "print\"(ii) Plank's constant is\", round(h1*10**34,2)*10**-34"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Work function of the material is 1.82 ev\n",
+ "(ii) Plank's constant is 6.76e-34\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.11 Page no 1204"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.6*10**-34\n",
+ "c=3*10**8\n",
+ "l=2000*10**-10\n",
+ "w0=4.2*1.6*10**-19\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "K=((h*c)/l)-w0\n",
+ "v0=K/e\n",
+ "l1=(h*c)/w0\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Potential difference is\", v0,\"V\"\n",
+ "print\"(ii) Wavelength of incident light is\", round(l1*10**10,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Potential difference is 1.9875 V\n",
+ "(ii) Wavelength of incident light is 2946.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.12 Page no 1204"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.6*10**-34\n",
+ "c=3*10**8\n",
+ "w0=2.39*1.6*10**-19\n",
+ "f1=4000.0 #A\n",
+ "f2=6000 #A\n",
+ "m=9.1*10**-31\n",
+ "e=1.9*10**-19\n",
+ "d=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l=(h*c)/w0\n",
+ "K=(12400/f1)-2.39\n",
+ "vmax=math.sqrt((2*K*1.6*10**-19)/m)\n",
+ "B=(m*vmax)/(e*d)\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum value of B is\", round(B*10**5,2)*10**-5,\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum value of B is 2.39e-05 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.13 Page no 1204"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "w0=4.4\n",
+ "\n",
+ "#Calculation\n",
+ "l=12400/w0\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of visible light is\", round(l,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of visible light is 2818.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.14 Page no 1205"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.625*10**-34\n",
+ "c=3*10**8\n",
+ "l=5600*10**-10\n",
+ "a=5\n",
+ "\n",
+ "#Calculation\n",
+ "E=(h*c)/l\n",
+ "n=a/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of visible photons emitted per second is\", round(n*10**-19,2)*10**19"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of visible photons emitted per second is 1.41e+19\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.15 Page no 1211"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=100\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l=12.27/math.sqrt(v)\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of an electron is\", l,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of an electron is 1.227 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.16 Page no 1212"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.62*10**-34\n",
+ "m=9*10**-31\n",
+ "v=10**5\n",
+ "mp=1.67*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "l=h/(m*v)\n",
+ "lp=h/(mp*v)\n",
+ "\n",
+ "#Result\n",
+ "print\"De-Broglie wavelength of electrons is\", round(l*10**10,1)*10**-10,\"m\"\n",
+ "print\"De-Broglie wavelength of protons is\",round(lp*10**10,4)*10**-10 ,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "De-Broglie wavelength of electrons is 7.36e-09 m\n",
+ "De-Broglie wavelength of protons is 3.96e-12 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 62
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.17 Page no 1212"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=500*1.6*10**-19\n",
+ "mp=1.67*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l=h/(math.sqrt(2*mp*E))\n",
+ "\n",
+ "#Result\n",
+ "print\"De-Broglie wavelength is\", round(l*10**12,2)*10**-12,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "De-Broglie wavelength is 1.28e-12 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.18 Page no 1212"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=150.0\n",
+ "mn=1.675*10**-27 #Kg\n",
+ "En=150*1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "le=12.27/math.sqrt(v)\n",
+ "ln=h/math.sqrt(2*mn*En)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) De-Broglie wavelength of electron is\",round(le,0),\"A\"\n",
+ "print\"(ii) De-Broglie wavelength of neutron is\", round(ln*10**10,4),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) De-Broglie wavelength of electron is 1.0 A\n",
+ "(ii) De-Broglie wavelength of neutron is 0.0233 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.19 Page no 1213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=2.0*10**-10 #m\n",
+ "h=6.62*10**-34\n",
+ "\n",
+ "#Calculation\n",
+ "p=h/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Momentum of electrons is\", p,\"Kg m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Momentum of electrons is 3.31e-24 Kg m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 81
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.20 Page no 1213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=1.4*10**-10 #m\n",
+ "h=6.63*10**-34\n",
+ "l1=2.0*10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "E=h*c*(1/l-1/l1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy of the scattered electron is\", round(E*10**16,2)*10**-16,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy of the scattered electron is 4.26e-16 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 87
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.22 Page no 1213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "me=9.11*10**-31 #Kg\n",
+ "lp=1.813*10**-4\n",
+ "vp=3\n",
+ "\n",
+ "#Calculation\n",
+ "mp=me/(lp*vp)\n",
+ "\n",
+ "#Result\n",
+ "print\"The particle's mass is\", round(mp*10**27,3)*10**-27,\"Kg. The particle is proton\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The particle's mass is 1.675e-27 Kg. The particle is proton\n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.23 Page no 1214"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=0.82*10**-10 #m\n",
+ "h=6.6*10**-34\n",
+ "m=9.1*10**-31\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "le=math.sqrt((h*l)/(2*c*m))\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength associated with the photoelectrons is\", round(le*10**10,4),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength associated with the photoelectrons is 0.0996 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 103
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter24_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter24_2.ipynb new file mode 100644 index 00000000..b7102dac --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter24_2.ipynb @@ -0,0 +1,874 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:6d1662d2dadbe072b20c80081401408d705c47c14e10e838032934acc7c20ff4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 24 Atoms"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.1 Page no 1264"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "k=7.68*10**6*1.6*10**-19 #J\n",
+ "e=1.6*10**-19\n",
+ "Z=29\n",
+ "m=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "r=(m*2*Z*e**2)/k\n",
+ "\n",
+ "#Result\n",
+ "print\"The distance of the closest approach is\",round(r*10**14,1)*10**-14,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The distance of the closest approach is 1.1e-14 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.2 Page no 1265"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=10 #degree\n",
+ "e=1.6*10**-19\n",
+ "Z=79\n",
+ "m=9*10**9\n",
+ "a=5.0*1.6*10**-13\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "b=(Z*e**2*(1/(math.tan(5*3.14/180.0)))*m)/a\n",
+ "\n",
+ "#Result\n",
+ "print\"Impact parameter is\", round(b*10**13,1)*10**-13,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Impact parameter is 2.6e-13 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.3 Page no 1265"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Z=79\n",
+ "m=9*10**9\n",
+ "e=1.6*10**-19\n",
+ "r=4.0*10**-14\n",
+ "\n",
+ "#Calculation\n",
+ "K=(m*2*Z*e**2)/(r*1.6*10**-13)\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy is\", round(K,2),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy is 5.69 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.4 Page no 1265"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=2.1*10**7 #m/s\n",
+ "a=4.8*10**7 #C/Kg\n",
+ "Z=79\n",
+ "e=1.6*10**-19\n",
+ "m=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "r0=(2*m*Z*e*a)/v**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance of the closest approach is\", round(r0*10**14,1)*10**-14,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance of the closest approach is 2.5e-14 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.6 Page no 1266"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Z=79\n",
+ "e=1.6*10**-19 #C\n",
+ "v=1.6*10**-12\n",
+ "m=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "b=(m*Z*e**2*(1/(math.tan(45*3.14/180.0))))/v\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Scattering angle is 180 degree\"\n",
+ "print\"(b) The value of scattering angle decreases\"\n",
+ "print\"(c) Impact parameter is\", round(b*10**14,1)*10**-14,\"m\"\n",
+ "print\"(d) The scattering of particle takes place due to charge on the nucleus\",\n",
+ "print\"(e) Scattering angle is increase with decrease in impact parameter\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Scattering angle is 180 degree\n",
+ "(b) The value of scattering angle decreases\n",
+ "(c) Impact parameter is 1.1e-14 m\n",
+ "(d) The scattering of particle takes place due to charge on the nucleus (e) Scattering angle is increase with decrease in impact parameter\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.7 Page no 1280"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=8.854*10**-12\n",
+ "h=6.62*10**-34\n",
+ "m=9*10**-31\n",
+ "e1=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r1=((e*h**2)/(math.pi*m*e1**2))*10**10\n",
+ "v1=e1**2/(2*e*h)\n",
+ "n=2*r1\n",
+ "\n",
+ "#Result\n",
+ "print\"Radius of first orbit is\", round(r1,2),\"A\"\n",
+ "print\"Velocity of electron is\",round(v1*10**-6,1),\"*10**6 m/s\"\n",
+ "print\"Size of hydrogen atom is\",round(n,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of first orbit is 0.54 A\n",
+ "Velocity of electron is 2.2 *10**6 m/s\n",
+ "Size of hydrogen atom is 1.07 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 62
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.8 Page no 1281"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=1.0\n",
+ "n1=2.0\n",
+ "n2=3.0\n",
+ "a=0.53*10**-10\n",
+ "Z=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "r1=(a*n)/Z\n",
+ "r2=(a*n1**2)/Z\n",
+ "r3=(a*n2**2)/Z\n",
+ "E1=(-13.6*Z**2)/n**2\n",
+ "E2=(-13.6*Z**2)/n1**2\n",
+ "E3=(-13.6*Z**2)/n2**2\n",
+ "E=E3-E1\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Radii of three lowest allowed orbits is\", round(r1*10**10,2),\"A,\",round(r2*10**10,2),\"A and\",r3*10**10,\"A\"\n",
+ "print\"(ii) Energy of three lowest allowed orbits is\",E1,\"ev,\",E2,\"ev and\",E3,\"ev\"\n",
+ "print\"Energy of the photon is\",E,\"ev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Radii of three lowest allowed orbits is 0.18 A, 0.71 A and 1.59 A\n",
+ "(ii) Energy of three lowest allowed orbits is -122.4 ev, -30.6 ev and -13.6 ev\n",
+ "Energy of the photon is 108.8 ev\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.9 Page no 1281"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=2.0\n",
+ "n1=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "E2=-13.6/n**2\n",
+ "E3=-13.6/n1**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Energies of two energy level is\",E2,\"ev and\",round(E3,2),\"ev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energies of two energy level is -3.4 ev and -1.51 ev\n"
+ ]
+ }
+ ],
+ "prompt_number": 86
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.10 Page no 1282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rh=1.097*10**7\n",
+ "\n",
+ "#Calculation\n",
+ "l=9/(8.0*Rh)\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of second line is\",round(l*10**10,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of second line is 1026.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 90
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.11 Page no 1282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rh=1.097*10**7\n",
+ "\n",
+ "#Calculation\n",
+ "l=4/Rh\n",
+ "\n",
+ "#Result\n",
+ "print\"Shortest wavelength is\",round(l*10**10,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shortest wavelength is 3646.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.12 Page no 1282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rh=1.097*10**7\n",
+ "\n",
+ "#Calculation\n",
+ "l=4/(3.0*Rh)\n",
+ "\n",
+ "#Result\n",
+ "print\"Longest wavelength is\",round(l*10**10,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Longest wavelength is 1215.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 99
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.13 Page no 1282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=1.0\n",
+ "h=6.62*10**-34\n",
+ "c=3*10**8\n",
+ "f=1.6*10**-19\n",
+ "Z=2\n",
+ "\n",
+ "#Calculation\n",
+ "E1=(-13.6*Z**2)/n**2\n",
+ "l=-(h*c)/(E1*f)\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum wavelength is\", round(l*10**10,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum wavelength is 228.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 104
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.14 Page no 1283"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=1\n",
+ "Z=1.0\n",
+ "a=0.53*10**-10\n",
+ "Z1=4.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "rh=(a*n)/Z**2\n",
+ "n1=math.sqrt((a*Z1/rh))\n",
+ "Eh=(-13.6*Z**2)/n**2\n",
+ "Ebe=(-13.6*Z1**2)/n1**2\n",
+ "E=Ebe/Eh\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy of two states is\",E"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy of two states is 4.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 112
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.15 Page no 1283"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Z=2\n",
+ "e=1.6*10**-19\n",
+ "e1=8.854*10**-12\n",
+ "n=3\n",
+ "h=6.62*10**-34\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "v=(Z*e**2)/(2*e1*n*h)\n",
+ "a=v/c\n",
+ "\n",
+ "#Result\n",
+ "print\"Speed of the electron is\",round(a,3 )"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Speed of the electron is 0.005\n"
+ ]
+ }
+ ],
+ "prompt_number": 116
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.16 Page no 1284"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=10**-10\n",
+ "R=10**-15\n",
+ "Rs=7*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "R1=r/R\n",
+ "Re=R1*Rs\n",
+ "\n",
+ "#Result\n",
+ "print\"Radius of the earth's orbit is\",Re,\"m. Thus the earth would be much farther away from the sun\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of the earth's orbit is 7e+13 m. Thus the earth would be much farther away from the sun\n"
+ ]
+ }
+ ],
+ "prompt_number": 118
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.17 Page no 1284"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=-13.6*1.9*10**-19 #J\n",
+ "m=9*10**9\n",
+ "e=1.6*10**-19\n",
+ "n=1\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "r=(-e**2*m)/(2.0*E)\n",
+ "v=c/(137*n)\n",
+ "\n",
+ "#Result\n",
+ "print\"Orbital radius is\", round(r*10**11,1)*10**-11,\"m\"\n",
+ "print\"Velocity of the electron is\",round(v*10**-6,1),\"*10**6 m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Orbital radius is 4.5e-11 m\n",
+ "Velocity of the electron is 2.2 *10**6 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 131
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.18 Page no 1284"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=2.2*10**6\n",
+ "r=5.3*10**-11\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "f=v/(2*math.pi*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"Initial frequency of light is\",round(f*10**-15,1)*10**15,\"Hz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Initial frequency of light is 6.6e+15 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 135
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.19 Page no 1285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=10 #Kg\n",
+ "T=2*60*60 #S\n",
+ "rn=8*10**6 #m\n",
+ "h=6.62*10**-34\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "vn=(2*math.pi*rn)/T\n",
+ "n=(2*math.pi*rn*vn)/h\n",
+ "\n",
+ "#Result\n",
+ "print\"Quantum number is\",round(n*10**-44,1)*10**45"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Quantum number is 5.3e+45\n"
+ ]
+ }
+ ],
+ "prompt_number": 139
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.20 Page no 1285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E2=18.70\n",
+ "E1=16.70\n",
+ "h=6.62*10**-34\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "E=E2-E1\n",
+ "l=(h*c)/(E*1.6*10**-19)\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength is\", round(l*10**9,0),\"nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength is 621.0 nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 148
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.21 Page no 1285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n1=2\n",
+ "n2=3\n",
+ "lb=6563\n",
+ "a=20\n",
+ "b=108.0\n",
+ "\n",
+ "#Calculation\n",
+ "l1=(lb*a)/b\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of first member is\",round(l1,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of first member is 1215.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 151
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.22 Page no 1285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rh=1.097*10**7 #/m\n",
+ "h=6.63*10**-34\n",
+ "c=3*10**8\n",
+ "n=2.0\n",
+ "n1=4.0\n",
+ "\n",
+ "#Calculation\n",
+ "E=(h*c*Rh*(1/n**2-1/n1**2))/1.6*10**-19\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum energy is\", round(E*10**38,2),\"ev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum energy is 2.56 ev\n"
+ ]
+ }
+ ],
+ "prompt_number": 158
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.23 Page no 1286"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rh=1.097*10**7\n",
+ "n2=4.0\n",
+ "n1=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "lm=1/(Rh*(1/n1**2-1/n2**2))\n",
+ "lm1=9/Rh\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength is\", round(lm1*10**9,1),\"nm. This wavelength is in infrared part\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength is 820.4 nm. This wavelength is in infrared part\n"
+ ]
+ }
+ ],
+ "prompt_number": 167
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter25_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter25_2.ipynb new file mode 100644 index 00000000..95708b68 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter25_2.ipynb @@ -0,0 +1,1188 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:efdc68d7aa35d22a94f64e5e8f01516d21c5f3fcea362a5520b7b1d532197f7c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 25 Nuclei"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.1 Page no 1312"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R0=1.2*10**-15 #m\n",
+ "A=208\n",
+ "A1=16\n",
+ "\n",
+ "#calculation\n",
+ "R=R0*A**0.33\n",
+ "R1=R0*A1**0.33\n",
+ "\n",
+ "#Result\n",
+ "print\"Nuclear radius of lead is\", round(R*10**15,1),\"fm\"\n",
+ "print\"Nuclear radius of oxygen is\", round(R1*10**15,0),\"fm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Nuclear radius of lead is 7.0 fm\n",
+ "Nuclear radius of oxygen is 3.0 fm\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.2 page no1312"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "me=9.1*10**-31\n",
+ "c=3*10**8\n",
+ "e=1.6*10**-19\n",
+ "mp=1.673*10**-27\n",
+ "mn=1.675*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "E=(me*c**2)/e\n",
+ "E1=(mp*c**2)/e\n",
+ "E2=(mn*c**2)/e\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Equivalent energy of electron is\",round(E*10**-6,2),\"Mev\"\n",
+ "print\"(ii) Equivalent energy of proton is\",round(E1*10**-6,1),\"Mev\"\n",
+ "print\"(iii) Equivalent energy of neutron is\",round(E2*10**-6,1),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Equivalent energy of electron is 0.51 Mev\n",
+ "(ii) Equivalent energy of proton is 941.1 Mev\n",
+ "(iii) Equivalent energy of neutron is 942.2 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.3 Page no 1312"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=10**-3 #m\n",
+ "c=3*10**8 #m/s\n",
+ "a=3.6*10**6 #J\n",
+ "\n",
+ "#Calculation\n",
+ "E=(m*c**2)/a\n",
+ "\n",
+ "#Result\n",
+ "print E*10**-7,\"*10**7 KWh\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "2.5 *10**7 KWh\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.4 Page no 1313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Z=17\n",
+ "A=35\n",
+ "Z1=92\n",
+ "A1=235\n",
+ "Z2=4\n",
+ "A2=9\n",
+ "\n",
+ "#Calculation\n",
+ "n=A-Z\n",
+ "n1=A1-Z1\n",
+ "n2=A2-Z2\n",
+ "\n",
+ "#Calculation\n",
+ "print\"Number of neutron in 17Cl35 is\",n\n",
+ "print\"Number of neutron in 92U235 is\",n1\n",
+ "print\"Number of neutron in 4Be9 is\",n2"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of neutron in 17Cl35 is 18\n",
+ "Number of neutron in 92U235 is 143\n",
+ "Number of neutron in 4Be9 is 5\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.5 Page no 1313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A2=235\n",
+ "A1=16.0\n",
+ "R1=3*10**-15 #m\n",
+ "\n",
+ "#Calculation\n",
+ "R=(A2/A1)**0.33\n",
+ "R2=R*R1\n",
+ "\n",
+ "#Result\n",
+ "print\"Nuclear radius is\", round(R2*10**15,3),\"fermi\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Nuclear radius is 7.281 fermi\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.6 Page no 1313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "me=55.85\n",
+ "u=1.66*10**-27 #Kg\n",
+ "R=1.2*10**-15 \n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "m=me*u\n",
+ "a=(3*u)/(4.0*math.pi*R**3)\n",
+ "\n",
+ "#Result\n",
+ "print\"Nuclear density is\", round(a*10**-17,2)*10**17,\"Kg/m**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Nuclear density is 2.29e+17 Kg/m**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.7 Page no 1317"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=4.001509 #a.m.u\n",
+ "N=1.008666\n",
+ "N1=1.007277\n",
+ "a=1.66*10**-27\n",
+ "c=3*10**8\n",
+ "e=1.6*10**-19\n",
+ "n=4.0\n",
+ "\n",
+ "#Calculation\n",
+ "A=2*N1+2*N\n",
+ "M1=A-M\n",
+ "Eb=M1*a*c**2/e\n",
+ "B=Eb/n\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Mass defect is\",M1,\"a.m.u\"\n",
+ "print\"(ii) Binding energy is\",round(Eb*10**-6,1),\"Mev\"\n",
+ "print\" Binding energy per nucleon is\",round(B*10**-6,2),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Mass defect is 0.030377 a.m.u\n",
+ "(ii) Binding energy is 28.4 Mev\n",
+ " Binding energy per nucleon is 7.09 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 61
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.8 Page no 1317"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ma=1.00893\n",
+ "m1=1.00813\n",
+ "m2=2.01473\n",
+ "a=931.5\n",
+ "a1=4.00389\n",
+ "\n",
+ "#Calculation\n",
+ "m=ma+m1-m2\n",
+ "Eb=m*a\n",
+ "m3=2*ma+2*m1-a1\n",
+ "Eb1=m3*a\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Binding energy when one neutron and one proton combined together is\", round(Eb,2),\"Mev\"\n",
+ "print\"(ii) Binding energy when two neutrons and two protons are combined is\",round(Eb1,1) ,\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Binding energy when one neutron and one proton combined together is 2.17 Mev\n",
+ "(ii) Binding eergy when two neutrons and two protons are combined is 28.2 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.10 Page no 1318"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=1.66*10**-27 #Kg\n",
+ "c=3*10**8\n",
+ "mp=1.00727\n",
+ "mn=1.00866\n",
+ "mo=15.99053\n",
+ "\n",
+ "#Calculation\n",
+ "E=(a*c**2)/1.6*10**-19\n",
+ "m1=8*mp+8*mn-mo\n",
+ "a1=m1*E\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy equivalent of one atomic mass unit is\", round(a1*10**32,1),\"Mev/c**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy equivalent of one atomic mass unit is 127.8 Mev/c**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.11 Page no 1318"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "mp=1.007825\n",
+ "mn=1.008665\n",
+ "m=39.962589\n",
+ "a2=931.5\n",
+ "Z=40.0\n",
+ "\n",
+ "#Calculation\n",
+ "E=20*mp+20*mn\n",
+ "m1=E-m\n",
+ "Eb=m1*a2\n",
+ "B=Eb/Z\n",
+ "\n",
+ "#Result\n",
+ "print\"Binding energy per nucleon is\", round(B,3),\"Mev/nucleon\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Binding energy per nucleon is 8.551 Mev/nucleon\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.12 Page no 1330"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "t=5000 #Days\n",
+ "t1=2000.0\n",
+ "a=0.693 \n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "dt=(a*t)/t1\n",
+ "N=math.log10(dt)\n",
+ "l=a*N/(t1)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The fraction remaining after 5000 days is\", round(N,3)\n",
+ "print\"(ii) The activity of sample after 5000 days is\",round(l*10**5,1),\"*10**8 Bq\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The fraction remaining after 5000 days is 0.239\n",
+ "(ii) The activity of sample after 5000 days is 8.3 *10**8 Bq\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.13 Page no 1330"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=3.67*10**10 #dis/second\n",
+ "r=226.0\n",
+ "A=6.023*10**23\n",
+ "\n",
+ "#Calculation\n",
+ "n=A/r\n",
+ "l=N/n\n",
+ "D=0.693/l\n",
+ "a=D/(3600.0*24.0*365.0)\n",
+ "\n",
+ "#Result\n",
+ "print\" Half life of radium is\",round(a,0),\"years\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Half life of radium is 1596.0 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.14 page no 1330"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N0=475\n",
+ "N=270.0\n",
+ "t=5.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=N0/N\n",
+ "l=math.log(a)/t\n",
+ "T=1/l\n",
+ "T1=0.693/l\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The decay constant is\",round(l,3),\"/minute\"\n",
+ "print\"(ii) Mean life is\",round(T,2),\"minute\"\n",
+ "print\"(iii) Half life is\",round(T1,2),\"minute\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The decay constant is 0.113 /minute\n",
+ "(ii) Mean life is 8.85 minute\n",
+ "(iii) Half life is 6.13 minute\n"
+ ]
+ }
+ ],
+ "prompt_number": 55
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.15 page no 1331"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "t=1500\n",
+ "N=0.01\n",
+ "N0=0.999\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "T=t*math.log(N)/math.log(0.5)\n",
+ "T1=t*math.log(N0)/math.log(0.5)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Years will reduce to 1 centigram is\",round(T,1),\"years\"\n",
+ "print\"(ii) Years will lose 1 mg is\",round(T1,2),\"years\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Years will reduce to 1 centigram is 9965.8 years\n",
+ "(ii) Years will lose 1 mg is 2.17 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 61
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.16 page no 1331"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=2*10**12\n",
+ "b=9.0*10**12\n",
+ "T=80\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "c=math.log(a/b)\n",
+ "t=-(c*T)/0.693\n",
+ "\n",
+ "#Result\n",
+ "print\"Time required is\",round(t,0),\"second\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time required is 174.0 second\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.17 page no 1332"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "T=6.0\n",
+ "A=6.023*10**23\n",
+ "W=99.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l=0.693/T\n",
+ "N0=A*10**-12/W\n",
+ "A0=l*N0\n",
+ "N=N0*(1/math.log10(l))\n",
+ "A1=-(l*N)\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\" Activity in the beginning and after one hour\",round(A1*10**-8,3),\"/h\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Activity in the beginning and after one hour 7.496 /h\n"
+ ]
+ }
+ ],
+ "prompt_number": 80
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.18 page no 1332"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "T=30.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l=0.693/T\n",
+ "T1=1/l\n",
+ "t=math.log(4)/l\n",
+ "t1=math.log(8)/l\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) average life is\",round(l,4),\"/day\"\n",
+ "print\"(ii) The time taken for 3/4 of the original no. to disintegrate is\",round(T1,2),\"days\"\n",
+ "print\"(iii) Time taken is\",round(t,0),\"days\"\n",
+ "print\"(iv) Time taken is\",round(t1,0),\"days\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) average life is 0.0231 /day\n",
+ "(ii) The time taken for 3/4 of the original no. to disintegrate is 43.29 days\n",
+ "(iii) Time taken is 60.0 days\n",
+ "(iv) Time taken is 90.0 days\n"
+ ]
+ }
+ ],
+ "prompt_number": 93
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.19 page no 1332"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=1620.0\n",
+ "l1=405.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "T=(1/l)+(1/l1)\n",
+ "t=math.log(4)/T\n",
+ "\n",
+ "#Result\n",
+ "print\"The time during which three-fourths of a sample will decay is\",round(t,0),\"years\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The time during which three-fourths of a sample will decay is 449.0 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 99
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.20 page no 1333"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=3.7*10**10 #disintegrations/s\n",
+ "A=6.02*10**23\n",
+ "B=234\n",
+ "\n",
+ "#Calculation\n",
+ "D=(C*B)/A\n",
+ "\n",
+ "#Result \n",
+ "print\"Mass ofuranium atoms disintegrated per second is\",round(D*10**11,3)*10**-11,\"g\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mass ofuranium atoms disintegrated per second is 1.438e-11 g\n"
+ ]
+ }
+ ],
+ "prompt_number": 105
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.21 page no 1333"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=0.075 #kg /mol\n",
+ "m=1.2*10**-6 #kg\n",
+ "A=6.0*10**23 #/mol\n",
+ "t=9.6*10**18\n",
+ "N=170\n",
+ "\n",
+ "#Calculation\n",
+ "n=(A*m)/M\n",
+ "l=N/t\n",
+ "T=0.693/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of K-40 atoms in the sample is\", n\n",
+ "print\"Half life of K-40 is\", round(T/(24.0*3600.0*365)*10**-9,3),\"*10**9 years\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of K-40 atoms in the sample is 9.6e+18\n",
+ "Half life of K-40 is 1.241 *10**9 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 122
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.22 Page no 1337"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "mp=232.03714\n",
+ "mn=228.02873\n",
+ "m0=4.002603\n",
+ "a=931.5\n",
+ "A=232.0\n",
+ "e=1.6*10**-19\n",
+ "m=1.66*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "M=mp-mn-m0\n",
+ "Q=M*a\n",
+ "K=(A-4)*Q/A\n",
+ "S=math.sqrt((2*K*e)/(4.0*m))\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Kinetic energy is\", round(K,1),\"Mev\"\n",
+ "print\"(ii) Speed of particle is\", round(S*10**-4,1),\"*10**7 m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Kinetic energy is 5.3 Mev\n",
+ "(ii) Speed of particle is 1.6 *10**7 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 139
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.23 Page no 1337"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "b=238\n",
+ "c=206\n",
+ "d=92\n",
+ "e=82\n",
+ "\n",
+ "#Calculation\n",
+ "a=(b-c)/4.0\n",
+ "A=-d+(2*a)+e\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The emission of alpha particle will reduce the mass number by 4a and charge number by 2a\"\n",
+ "print\"(ii) Number of alpha particle is\", a\n",
+ "print\"Number of beta particle is\",A"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The emission of alpha particle will reduce the mass number by 4a and charge number by 2a\n",
+ "(ii) Number of alpha particle is 8.0\n",
+ "Number of beta particle is 6.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 145
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.24 Page no 1338"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=218\n",
+ "b=84\n",
+ "\n",
+ "#Calculation\n",
+ "A=a-4\n",
+ "Z=b-2\n",
+ "\n",
+ "#Result\n",
+ "print\"Atomic number of new element formed is\", A\n",
+ "print\"Mass number of new element formed is\",Z"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Atomic number of new element formed is 214\n",
+ "Mass number of new element formed is 82\n"
+ ]
+ }
+ ],
+ "prompt_number": 148
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.27 Page no 1340"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "mp=10.016125\n",
+ "mn=4.003874\n",
+ "mp1=13.007490\n",
+ "mn1=1.008146\n",
+ "a=931.5\n",
+ "\n",
+ "#Calculation\n",
+ "Mr=mp+mn\n",
+ "Mp=mp1+mn1\n",
+ "Md=Mr-Mp\n",
+ "A=a*Md\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy released in the reaction is\",round(A,3),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy released in the reaction is 4.064 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 154
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.28 Page no 1345"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=10**6 #J/s\n",
+ "E=200*10**6*1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "N=a/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of fission per second is\", round(N*10**-16,2)*10**16"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of fission per second is 3.13e+16\n"
+ ]
+ }
+ ],
+ "prompt_number": 159
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.29 Page no 1345"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=3*10**8 #W\n",
+ "E=200*10**6*1.6*10**-19\n",
+ "a=235\n",
+ "m=6.023*10**23\n",
+ "\n",
+ "#Calculation\n",
+ "E1=P*3600\n",
+ "N=E1/E\n",
+ "M1=(a*N)/m\n",
+ "\n",
+ "#Result\n",
+ "print\"Mass of uranium fissioned per hour is\", round(M1,2),\"g\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mass of uranium fissioned per hour is 13.17 g\n"
+ ]
+ }
+ ],
+ "prompt_number": 166
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.30 Page no 1345"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=6.023*10**26\n",
+ "a=235.0\n",
+ "t=30 #Days\n",
+ "E=200*10**6*1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "N=(2/a)*m\n",
+ "A=N/(t*24*60.0*60.0)\n",
+ "P=E*A\n",
+ "\n",
+ "#Result\n",
+ "print\"Power output is\", round(P*10**-6,1),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power output is 63.3 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 173
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.31 Page no 1348"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=1.0076\n",
+ "mp=4.0039\n",
+ "a=931.5*10**6 #ev\n",
+ "\n",
+ "#Calculation\n",
+ "Mr=4*m\n",
+ "Md=Mr-mp\n",
+ "E=Md*a*1.6*10**-19\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy released is\", round(E*10**13,2)*10**-13,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy released is 3.95e-12 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 181
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.32 Page no 1349"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=6*10**-3 #Kg\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "E=a*c**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy liberated is\", E,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy liberated is 5.4e+14 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 184
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter26_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter26_2.ipynb new file mode 100644 index 00000000..98e47e10 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter26_2.ipynb @@ -0,0 +1,425 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:e41de143240c9df8b907c856d0ba61f830495897881ab9f990dfa099753c5c2e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 26 Semiconductors"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.1 Page no 1414"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=0.47\n",
+ "ue=0.39 #m**2/volt sec\n",
+ "uh=0.19 #m**2/volt sec\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "a1=1/a\n",
+ "ni=a1/(e*(ue+uh))\n",
+ "\n",
+ "#Result\n",
+ "print\"Intrinsic carrier conceentration is\", round(ni*10**-19,1)*10**19,\"/m**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Intrinsic carrier conceentration is 2.3e+19 /m**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.2 Page no 1414"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=0.01\n",
+ "e=1.6*10**-19\n",
+ "ue=0.39\n",
+ "\n",
+ "#Calculation\n",
+ "a1=1/a\n",
+ "Nd=a1/(e*ue)\n",
+ "\n",
+ "#Result\n",
+ "print\"Donor concentration is\", round(Nd*10**-21,1)*10**21,\"/m**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Donor concentration is 1.6e+21 /m**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.3 Page no 1414"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ni=2.5*10**19 #/m**3\n",
+ "e=1.6*10**19\n",
+ "ue=0.36 #m**2/volt sec\n",
+ "uh=0.17 \n",
+ "\n",
+ "#Calculation\n",
+ "a=ni*e*(ue+uh)\n",
+ "\n",
+ "#Result\n",
+ "print\"Conductivity is\", a*10**-38,\"S/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Conductivity is 2.12 S/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.4 Page no 1414"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ne=8*10**13 #/cm**3\n",
+ "nh=5*10**12 #/cm**3\n",
+ "ue=23000 #cm**2/vs\n",
+ "e=1.6*10**-19\n",
+ "uh=100 #cm**2/vs\n",
+ "\n",
+ "#Calculation\n",
+ "a=e*((ne*ue)+(nh*uh))\n",
+ "A1=1/a\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Since electron density is greater than the hole density, the semiconductor is n-type\"\n",
+ "print\"(ii) Resistivity of the sample is\", round(A1,3),\"ohm cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Since electron density is greater than the hole density, the semiconductor is n-type\n",
+ "(ii) Resistivity of the sample is 3.396 ohm cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.5 Page no 1415"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ni=1.5*10**16 #/m**3\n",
+ "nh=4.5*10**22 #/m**3\n",
+ "\n",
+ "#Calculation\n",
+ "ne=ni**2/nh\n",
+ "\n",
+ "#Result\n",
+ "print\"ne in the doped semiconductor is\",ne*10**-9,\"*10**9 /m**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "ne in the doped semiconductor is 5.0 *10**9 /m**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.6 Page no 1415 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=5890.0 #A\n",
+ "\n",
+ "#Calculation\n",
+ "E=12400/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum energy is\",round(E,1),\"ev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum energy is 2.1 ev\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.7 Page no 1415"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=6*10**19\n",
+ "b=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "A=a*b\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of holes is\",A"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of holes is 6e+12\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.8 Page no 1415"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=0.65\n",
+ "a=10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "l=(12400*a)/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum wavelength of electromagnetic radiation is\",round(l*10**6,1)*10**-6,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum wavelength of electromagnetic radiation is 1.9e-06 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.9 Page no 1416"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=5 #/ohm/cm\n",
+ "ue=3900 #cm**2/vs\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "Nd=a/(ue*e)\n",
+ "\n",
+ "#Result\n",
+ "print\"Number density of donor atom is\",round(Nd*10**-15,2)*10**15,\"/cm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number density of donor atom is 8.01e+15 /cm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.10 Page no 1416"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ni=1.5*10**16 #/m**3\n",
+ "a=5*10**28\n",
+ "b=10.0**6\n",
+ "\n",
+ "#Calculation\n",
+ "Ne=a/b\n",
+ "nh=ni**2/Ne\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of Electrons is\",Ne,\"/m**3\"\n",
+ "print\"Number of holes is\",nh*10**-9,\"*10**9 /m**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of Electrons is 5e+22 /m**3\n",
+ "Number of holes is 4.5 *10**9 /m**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 50
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.11 Page no 1416"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=4.0*10*-8 #m\n",
+ "\n",
+ "#Calculation\n",
+ "a=2/1.6*10**-19\n",
+ "E=-a/d\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric field is\", round(E*10**22,0),\"*10**7 V/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric field is 4.0 *10**7 V/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 61
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter27_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter27_2.ipynb new file mode 100644 index 00000000..01e54494 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter27_2.ipynb @@ -0,0 +1,921 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:eeaae422be8c264750ed4950e51451be86906b82350f05dcb46e0f610199fb25"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 27 Semiconductor devices"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.1 Page no 1446"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=1.5 #V\n",
+ "Vd=0.5 #V\n",
+ "P=0.1 #W\n",
+ "\n",
+ "#Calculation\n",
+ "Imax=P/Vd\n",
+ "V=E-Vd\n",
+ "R1=V/Imax\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of resistance is\",R1,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of resistance is 5.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.2 Page no 1446"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=2 #V\n",
+ "R=10.0 #ohm\n",
+ "R1=20.0\n",
+ "\n",
+ "#Calculation\n",
+ "I=V/R\n",
+ "I1=V/R1\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Current drawn from battery is\", I,\"A\"\n",
+ "print\"(ii) Current drawn from point B is\",I1,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Current drawn from battery is 0.2 A\n",
+ "(ii) Current drawn from point B is 0.1 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.3 Page no 1446"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "Vl=15 #V\n",
+ "Rl=2.0*10**3\n",
+ "Iz=10 #mA\n",
+ "\n",
+ "#Calculation\n",
+ "Il=(Vl/Rl)*10**3\n",
+ "Ir=Iz+Il\n",
+ "Vr=Ir*10**-2*R1\n",
+ "V=Vr+Vl\n",
+ "\n",
+ "#Result\n",
+ "print\"Voltage is\", V,\"V\"\n",
+ "print\"Zener rating required is\",Ir,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage is 18.5 V\n",
+ "Zener rating required is 17.5 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.4 Page no 1447"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=10.0\n",
+ "V=230 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Vrpm=math.sqrt(2)*V\n",
+ "Vsm=Vrpm/N\n",
+ "Vdc=Vsm/math.pi\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The output dc voltage is\", round(Vdc,2),\"V\"\n",
+ "print\"(ii) Peak inverse voltage is\",round(Vsm,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The output dc voltage is 10.35 V\n",
+ "(ii) Peak inverse voltage is 32.53 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.5 Page no 1447"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vm=50 #V\n",
+ "rf=20.0\n",
+ "Rl=800 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Im=(Vm/(rf+Rl))*10**3\n",
+ "Idc=Im/math.pi\n",
+ "Irms=Im/2.0\n",
+ "P=(Irms/1000.0)**2*(rf+Rl)\n",
+ "P1=(Idc/1000.0)**2*Rl\n",
+ "V=Idc*Rl*10**-3\n",
+ "A=P1*100/P\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Im=\",round(Im,0),\"mA \\nIdc=\",round(Idc,1),\"mA \\nIrms=\",round(Irms,1),\"mA\"\n",
+ "print\"(ii) a.c power input is\",round(P,3),\"watt \\nd.c. power is\",round(P1,3),\"watt\"\n",
+ "print \"(iii) d.c. output voltage is\",round(V,2),\"Volts\"\n",
+ "print\"(iv) Efficiency of rectification is\", round(A,1),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Im= 61.0 mA \n",
+ "Idc= 19.4 mA \n",
+ "Irms= 30.5 mA\n",
+ "(ii) a.c power input is 0.762 watt \n",
+ "d.c. power is 0.301 watt\n",
+ "(iii) d.c. output voltage is 15.53 Volts\n",
+ "(iv) Efficiency of rectification is 39.5 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 67
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.6 Page no 1448"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "rf=20 #ohm\n",
+ "Rl=980\n",
+ "V=50 #v\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Vm=V*math.sqrt(2)\n",
+ "Im=(Vm/(rf+Rl))*10**3\n",
+ "Idc=(2*Im)/(math.pi)\n",
+ "Irms=Im/math.sqrt(2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) load current is\",round(Im,1),\"mA\"\n",
+ "print\"(ii) Mean load currant is\",round(Idc,0),\"mA\"\n",
+ "print\"(iii) R.M.S value of load current is\",Irms,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) load current is 70.7 mA\n",
+ "(ii) Mean load currant is 45.0 mA\n",
+ "(iii) R.M.S value of load current is 50.0 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 85
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.7 page no 1448"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=5.0\n",
+ "A=230 #V\n",
+ "B=2\n",
+ "Rl=100\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "V1=A/N\n",
+ "V2=V1*math.sqrt(2)\n",
+ "Vm=V2/B\n",
+ "Idc=2*Vm/(math.pi*Rl)\n",
+ "Vdc=Idc*Rl\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) d.c voltage output is\",round(Vdc,1),\"V\"\n",
+ "print\"(ii) peak inverse voltage is\",round(V2,0),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) d.c voltage output is 20.7 V\n",
+ "(ii) peak inverse voltage is 65.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 122
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.8 page no 1448"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Il=4.0 #mA\n",
+ "Vz=6 #V\n",
+ "E=10.0 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Lz=5*Il\n",
+ "L=Il+Lz\n",
+ "Rs=E-Vz\n",
+ "Rs1=Rs/(L*10**-3)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of series resister Rs\",round(Rs1,0),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of series resister Rs 167.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 123
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.9 page no 1449"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vf=0.3 #V\n",
+ "If=4.3*10**-3 #A\n",
+ "Vc=0.35\n",
+ "Va=0.25\n",
+ "Ic=6*10**-3\n",
+ "Ia=3*10**-3\n",
+ "\n",
+ "#Calculation\n",
+ "Rdc=Vf/If\n",
+ "Vf1=Vc-Va\n",
+ "If1=Ic-Ia\n",
+ "Rac=Vf1/If1\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) D.C. resistance is\",round(Rdc,2),\"ohm\"\n",
+ "print\"(ii) A.C. resistance is\",round(Rac,2),\"ohm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) D.C. resistance is 69.77 ohm\n",
+ "(ii) A.C. resistance is 33.33 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 139
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.10 page no 1462"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=0.9\n",
+ "Ie=1 #mA\n",
+ "\n",
+ "#Calculation\n",
+ "Ic=A*Ie\n",
+ "Ib=Ie-Ic\n",
+ "\n",
+ "#Result\n",
+ "print\"Base current is\",Ib,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Base current is 0.1 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 144
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.11 page no 1462"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=50\n",
+ "Ib=0.02 #mA\n",
+ "\n",
+ "#Calculation\n",
+ "Ic=B*Ib\n",
+ "Ie=Ib+Ic\n",
+ "\n",
+ "#Result\n",
+ "print\"Ie =\",Ie,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Ie = 1.02 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 147
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.12 page no 1462"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=49\n",
+ "Ie=12 #mA\n",
+ "Ib=240 #microA\n",
+ "\n",
+ "#Calculation\n",
+ "A=(B/1+B)*10**-2\n",
+ "Ic=A*Ie\n",
+ "Ic1=B*Ib\n",
+ "\n",
+ "#Result\n",
+ "print\" The value of Ic using A is\",Ic,\"mA\"\n",
+ "print\" The value of Ic using B is\",Ic1*10**-3,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The value of Ic using A is 11.76 mA\n",
+ " The value of Ic using B is 11.76 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 165
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.13 page no 1463"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=45.0\n",
+ "Ic=1 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Ib=Ic/B\n",
+ "\n",
+ "#Result\n",
+ "print\" The base current for common emitter connection is\",round(Ib,3),\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The base current for common emitter connection is 0.022 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 169
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.14 page no 1463"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vcc=8 #V\n",
+ "V=0.5 #V\n",
+ "Rc=800.0 #ohm\n",
+ "a=0.96\n",
+ "\n",
+ "#Calculation\n",
+ "Vce=Vcc-V\n",
+ "Ic=V/Rc*10**3\n",
+ "B=a/(1-a)\n",
+ "Ib=Ic/B\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Collector-emitter voltage is\",Vce,\"V\"\n",
+ "print\"(ii) Base current is\",round(Ib,3),\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Collector-emitter voltage is 7.5 V\n",
+ "(ii) Base current is 0.026 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 184
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.15 page no 1463"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=10\n",
+ "b=2\n",
+ "c=3\n",
+ "\n",
+ "#Calculation\n",
+ "Vce=a-b\n",
+ "Ic=c-b\n",
+ "Ro=Vce/Ic\n",
+ "\n",
+ "#Result\n",
+ "print\"The output resistance is\",Ro,\"k ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The output resistance is 8 k ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 188
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.16 page no 1464"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ic=4.0 #mA\n",
+ "Ib=30 #micro A\n",
+ "Ib1=20 #micro A\n",
+ "Vce=10 #V\n",
+ "c=4.5 #mA\n",
+ "d=3.0 #mA\n",
+ "\n",
+ "#Calculation\n",
+ "Ib2=Ib-Ib1\n",
+ "Ic1=c-d\n",
+ "Bac=Ic1/Ib2*10**3\n",
+ "Bdc=c/Ib*10**3\n",
+ "\n",
+ "#Result \n",
+ "print\"The value of Bac of the transister is\",Bdc\n",
+ "print\"The value of Bdc of the transister is\",Bdc"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Bac of the transister is 150.0\n",
+ "The value of Bdc of the transister is 150.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 249
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.17 page no 1464"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ri=665.0 #ohm\n",
+ "Ib=15.0 #micro A\n",
+ "Ic=2 #mA\n",
+ "Ro=5*10**3 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Bac=Ic/Ib*10**3\n",
+ "Av=Bac*(Ro/Ri)\n",
+ "\n",
+ "#Result\n",
+ "print\" The voltage gain is\", round(Av,0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The voltage gain is 1003.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 240
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.18 page no 1464"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given \n",
+ "Vbb=2.0 #v\n",
+ "Rc=2000 #ohm\n",
+ "B=100\n",
+ "Vbe=0.6 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Ic=Vbb/Rc*10**3\n",
+ "Ib=Ic/B\n",
+ "Ib1=10*Ib\n",
+ "Rb=(Vbb-Vbe)/Ib\n",
+ "Ic=B*Ib1\n",
+ "\n",
+ "#Result \n",
+ "print\"d.c. collector current is\",Ic,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "d.c. collector current is 10.0 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 236
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.19 page no 1465"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=10**10\n",
+ "e=1.6*10**-19\n",
+ "t=10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "Ie=(N*e)/t*10**3\n",
+ "Ib=(2/100.0)*Ie\n",
+ "Ic=Ie-Ib\n",
+ "c=Ic/Ie\n",
+ "B=Ic/Ib\n",
+ "#Result\n",
+ "print\"The current transfer ratio\",c\n",
+ "print\"current amplification factor is\",B"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current transfer ratio 0.98\n",
+ "current amplification factor is 49.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 257
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.20 page no 1465"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=200\n",
+ "b=50\n",
+ "c=17\n",
+ "d=5\n",
+ "e=4000\n",
+ "\n",
+ "#Calculation\n",
+ "Ib=(a-b)*10**-3\n",
+ "Ic=c-d\n",
+ "B=Ic/Ib\n",
+ "D=e/B\n",
+ "Ap=B**2*D\n",
+ "\n",
+ "#Result\n",
+ "print\" The value of current gain is\",B\n",
+ "print\" The value of resistance gain is\",D \n",
+ "print\" The value of power gain is\",Ap*10**-5,\"*10**5\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The value of current gain is 80.0\n",
+ " The value of resistance gain is 50.0\n",
+ " The value of power gain is 3.2 *10**5\n"
+ ]
+ }
+ ],
+ "prompt_number": 279
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.21 page no 1469"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L1=58.6*10**-6 #H\n",
+ "C1=300.0*10**-12 #F\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "f=1/((2.0*math.pi)*math.sqrt(L1*C1))\n",
+ "\n",
+ "#Result\n",
+ "print\"Frequency of oscillation is\", round(f*10**-3,0),\"KHz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Frequency of oscillation is 1200.0 KHz\n"
+ ]
+ }
+ ],
+ "prompt_number": 294
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.22 Page no 1469"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vbe=0.8 #V\n",
+ "Vcc=5 #V\n",
+ "Rc=1 #K ohm\n",
+ "b=250.0\n",
+ "Rb=100 #K ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Ic=Vcc/Rc\n",
+ "Ib=(Ic/b)*10**3\n",
+ "Vi=(Ib*Rb)+Vbe\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The minimum base current is\",Ib,\"micro A\"\n",
+ "print\"(ii) The input voltage is\",round(Vi*10**-3,0),\"V\"\n",
+ "print\"(iii) Between 0 V and 0.6 V,the transistor will switched off. Between 2.8 V and 5.0 V the transistor will switched on\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The minimum base current is 20.0 micro A\n",
+ "(ii) The input voltage is 2.0 V\n",
+ "(iii) Between 0 V and 0.6 V,the transistor will switched off. Between 2.8 V and 5.0 V the transistor will switched on\n"
+ ]
+ }
+ ],
+ "prompt_number": 309
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter2_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter2_2.ipynb new file mode 100644 index 00000000..e9ce1a12 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter2_2.ipynb @@ -0,0 +1,682 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:ef2be2de13061356088f1dea63be1d10f7d8030c93430c7778acfc0d827cf022"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 2 Electric field"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.1 Page no 47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=1000\n",
+ "d=10.0*10**-3\n",
+ "m=4.8*10**-15\n",
+ "g=10\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "E=V/d\n",
+ "q=m*g/E\n",
+ "n=q/e\n",
+ "\n",
+ "#Result\n",
+ "print\"The number of electrons on the drop \", n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The number of electrons on the drop 3.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2 Page no 47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19\n",
+ "E=3*10**4\n",
+ "m=9.0*10**-31\n",
+ "y1=4*10**-2\n",
+ "m2=1.67*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "a=e*E/m\n",
+ "t=math.sqrt((2*y1)/a)\n",
+ "a2=e*E/m2\n",
+ "t2=math.sqrt((2*y1)/a2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Time t1=\", round(t*10**9,1)*10**-9,\"S\",\"\\nTime t2=\",round(t2*10**7,2)*10**-7,\"S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time t1= 3.9e-09 S \n",
+ "Time t2= 1.67e-07 S\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.3 Page no 48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=1 #m\n",
+ "m=9*10**9\n",
+ "q=500*10**-6\n",
+ "r1=0.3 #m\n",
+ "\n",
+ "#Calculation\n",
+ "E=m*q/r**2\n",
+ "E2=m*q/r1**2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Electric field intensity from the centre of the sphere \",E*10**-6,\"10**6\",\"N/C\"\n",
+ "print\"(ii) Electric field intensity at the surface of the sphere is \",E2*10**-7,\"10**7 N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Electric field intensity from the centre of the sphere 4.5 10**6 N/C\n",
+ "(ii) Electric field intensity at the surface of the sphere is 5.0 10**7 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.4 Page no 48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=2*10**-8\n",
+ "E=2*10**4\n",
+ "m=80*10**-6\n",
+ "g=9.8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=q*E/(m*g)\n",
+ "b=math.atan(a)*180/3.14\n",
+ "T=(q*E/(math.sin(b*3.14/180.0)))*10**-4\n",
+ "\n",
+ "#Result\n",
+ "print\"The angle is \", round(b,0),\"degree\"\n",
+ "print\"Tension in the thread of the pendulum is \", round(T*10**8,2),\"*10**-4 N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The angle is 27.0 degree\n",
+ "Tension in the thread of the pendulum is 8.8 *10**-4 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.5 Page no 48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "r=0.707\n",
+ "q=5*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "E=m*q/r**2 #along AO\n",
+ "E2=m*q/r**2 #along BO\n",
+ "E3=m*q/r**2 #along OD\n",
+ "E11=E+E2\n",
+ "E12=E2+E3\n",
+ "I=(2*E11*r)*10**-4\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric field at the centre of the sphere is \",round(I,2),\"*10**4 N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric field at the centre of the sphere is 25.46 *10**4 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.6 Page no 49"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=5*10**-9\n",
+ "x=0.15 #m\n",
+ "r=0.1 #m\n",
+ "a=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "E=(a*q*x)/((r**2+x**2))**1.5\n",
+ "\n",
+ "#Result\n",
+ "print\"Intensity of the electric field is \", round(E,0),\"N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Intensity of the electric field is 1152.0 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.7 Page no 49"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=10**-3\n",
+ "F=1\n",
+ "v0=20\n",
+ "v=0\n",
+ "\n",
+ "#Calculation\n",
+ "a=-F/m\n",
+ "s=v**2-v0**2/(2.0*a)\n",
+ "\n",
+ "#Result\n",
+ "print\"The distance is \", s,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The distance is 0.2 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.9 Page no 50"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q1=1/3.0*10**-7\n",
+ "r=5*10**-2\n",
+ "F=58.8*10**-3\n",
+ "\n",
+ "#Calculation\n",
+ "q2=F*r**2/(q1*m)\n",
+ "\n",
+ "#Result\n",
+ "print\"Charge is \", q2,\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge is 4.9e-07 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.10 Page no 61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=4*10**5 #V/m\n",
+ "q=3.2*10**-19\n",
+ "a=2.4*10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "p=q*a\n",
+ "W=p*E*(1-(math.cos(180*180/3.14)))\n",
+ "\n",
+ "#Result\n",
+ "print\"Work done is \", W"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Work done is 2.79670959474e-23\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.11 Page no 61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=16*10**-19\n",
+ "a=3.9*10**-12\n",
+ "E=10**5\n",
+ "\n",
+ "#Calculation\n",
+ "p=q*a\n",
+ "U=-p*E\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The electric dipole moment \", p,\"Cm\"\n",
+ "print\"(ii) Potential energy of dipole in the stable equilibrium position \",U,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The electric dipole moment 6.24e-30 Cm\n",
+ "(ii) Potential energy of dipole in the stable equilibrium position -6.24e-25 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.12 Page no 61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=20*10**-6\n",
+ "a=10**-2\n",
+ "m=9*10**9\n",
+ "r=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "p=q*a\n",
+ "E=m*2*p/r**3\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric field intensity is \", E*10**-5,\"*10**5 N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric field intensity is 36.0 *10**5 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.13 Page no 62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=4*10**5\n",
+ "q=1*10**-6\n",
+ "a=3*10**-2\n",
+ "\n",
+ "#Calculation\n",
+ "t=q*a*E\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum torque on the dipole is \", t*10**2,\"*10**-2 Nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum torque on the dipole is 1.2 *10**-2 Nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 55
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.14 Page no 62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=1*10**-6\n",
+ "a=2*10**-2\n",
+ "E=10**5\n",
+ "\n",
+ "#Calculation\n",
+ "p=q*a\n",
+ "W=2*p*E\n",
+ "\n",
+ "#Result\n",
+ "print\"Work done in the rotation is \", W*10**3,\"*10**-3 J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Work done in the rotation is 4.0 *10**-3 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 60
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.15 Page no 62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=2*10**-6\n",
+ "a=0.1\n",
+ "m=9*10**9\n",
+ "r=0.5\n",
+ "\n",
+ "#Calculation\n",
+ "p=q*a\n",
+ "E=m*p/r**3\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric field intensity is \",E*10**-4,\"*10**4 N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric field intensity is 1.44 *10**4 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 65
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.16 Page no 62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "qa=2.5*10**-7\n",
+ "qb=-2.5*10**-7\n",
+ "a=15\n",
+ "b=15\n",
+ "\n",
+ "#Calculation\n",
+ "q=qa+qb\n",
+ "C=(a+b)*10**-2\n",
+ "E=qa*C\n",
+ "\n",
+ "#Result\n",
+ "print\"Total charge is \", q,\"\\nElectric dipole moment of the system is \",E,\"Cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total charge is 0.0 \n",
+ "Electric dipole moment of the system is 7.5e-08 Cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 71
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.17 Page no 63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "p=2*10**-8\n",
+ "m=9*10**9\n",
+ "r=1\n",
+ "\n",
+ "#Calculation\n",
+ "E=(m*p*math.sqrt(3*(math.cos**2(60)*180/3.14))+1)/r**3\n",
+ "print E"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "ename": "TypeError",
+ "evalue": "'int' object is not callable",
+ "output_type": "pyerr",
+ "traceback": [
+ "\u001b[1;31m---------------------------------------------------------------------------\u001b[0m\n\u001b[1;31mTypeError\u001b[0m Traceback (most recent call last)",
+ "\u001b[1;32m<ipython-input-77-6c8b884fd561>\u001b[0m in \u001b[0;36m<module>\u001b[1;34m()\u001b[0m\n\u001b[0;32m 5\u001b[0m \u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m 6\u001b[0m \u001b[1;31m#Calculation\u001b[0m\u001b[1;33m\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[1;32m----> 7\u001b[1;33m \u001b[0mE\u001b[0m\u001b[1;33m=\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0mm\u001b[0m\u001b[1;33m*\u001b[0m\u001b[0mp\u001b[0m\u001b[1;33m*\u001b[0m\u001b[0mmath\u001b[0m\u001b[1;33m.\u001b[0m\u001b[0msqrt\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;36m3\u001b[0m\u001b[1;33m*\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0mmath\u001b[0m\u001b[1;33m.\u001b[0m\u001b[0mcos\u001b[0m\u001b[1;33m**\u001b[0m\u001b[1;36m2\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;36m60\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m*\u001b[0m\u001b[1;36m180\u001b[0m\u001b[1;33m/\u001b[0m\u001b[1;36m3.14\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m+\u001b[0m\u001b[1;36m1\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m/\u001b[0m\u001b[0mr\u001b[0m\u001b[1;33m**\u001b[0m\u001b[1;36m3\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0m\u001b[0;32m 8\u001b[0m \u001b[1;32mprint\u001b[0m \u001b[0mE\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n",
+ "\u001b[1;31mTypeError\u001b[0m: 'int' object is not callable"
+ ]
+ }
+ ],
+ "prompt_number": 77
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.18 Page no 63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "p=5*10**-8\n",
+ "m=9*10**9\n",
+ "r=0.15\n",
+ "\n",
+ "#Calculation\n",
+ "E=m*2*p/r**3\n",
+ "E1=m*p/r**3\n",
+ "\n",
+ "print\"(i) Electric field along AB is \", round(E*10**-5,2),\"*10**5 N/C\"\n",
+ "print\"(ii) Electric field along BA is \", round(E1*10**-5,2),\"*10**5 N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Electric field along AB is 2.67 *10**5 N/C\n",
+ "(ii) Electric field along BA is 1.33 *10**5 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 89
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter3_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter3_2.ipynb new file mode 100644 index 00000000..8aff7044 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter3_2.ipynb @@ -0,0 +1,1016 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:b7451f8a5d88007ceae262e7f31a9c2ef46acd4d68ca53e0c7b4ce898e623cd8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3 Electrostatic potential and flux"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.1 Page no 89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=300*10**-6 #c\n",
+ "V=6\n",
+ "\n",
+ "#Calculation\n",
+ "W=q*V\n",
+ "\n",
+ "#Result\n",
+ "print\"Work done is \", W*10**3,\"*10**-3 J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Work done is 1.8 *10**-3 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.2 Page no 89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given \n",
+ "Va=-10 #V\n",
+ "W=300 #J\n",
+ "q=3.0 #C\n",
+ "\n",
+ "#Calculation\n",
+ "V=(W/q)+Va\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of V is \", V,\"Volts\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of V is 90.0 Volts\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3 Page no 89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q=16*10**-10 #C\n",
+ "r=0.1\n",
+ "r1=0.06\n",
+ "q1=12*10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "Vb=m*q/r\n",
+ "Vb1=m*q/r1\n",
+ "V=Vb1-Vb\n",
+ "W=q1*V\n",
+ "\n",
+ "#Result\n",
+ "print\"Workdone is \", W*10**8,\"*10**-8 J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Workdone is 11.52 *10**-8 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.4 Page no 89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=3.4*10**-14 #m\n",
+ "n=47\n",
+ "q=1.6*10**-19 #C\n",
+ "m=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "V=m*n*q/r\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric potential at the surface of silver nucleus is \", round(V*10**-6,2),\"*10**6 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric potential at the surface of silver nucleus is 1.99 *10**6 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.5 Page no 90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q=4*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "V=2*q*m\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric potential is \", V*10**-3,\"*10**3 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric potential is 72.0 *10**3 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.9 Page no 91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q=250*10**-6\n",
+ "r=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "V=m*q/r\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric potential at the centre is \", V*10**-7,\"*10**7 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric potential at the centre is 2.25 *10**7 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.10 Page no 96"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=3*10**-16\n",
+ "g=9.8\n",
+ "d=5*10**-3\n",
+ "q=16.0*10**-18\n",
+ "\n",
+ "#Calculation\n",
+ "V=(m*g*d/q)*10\n",
+ "\n",
+ "#Result\n",
+ "print\"Voltage needed to balance an oil drop is \",round(V,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage needed to balance an oil drop is 9.19 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.12 Page no 96"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=1.6*10**-19 #C\n",
+ "V=3000 #V\n",
+ "r=5*10**-2 #m\n",
+ "g=9.8\n",
+ "\n",
+ "#Calculation\n",
+ "E=V/r\n",
+ "m=q*E/g\n",
+ "\n",
+ "#Result\n",
+ "print\"The mass of the particle is \", round(m*10**16,1),\"*10**-16 Kg\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mass of the particle is 9.8 *10**-16 Kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.13 Page no 100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**-9\n",
+ "q1=3*10**-9\n",
+ "q2=3*10**-9\n",
+ "q3=10**9\n",
+ "r=0.2\n",
+ "\n",
+ "#Calculation\n",
+ "W=m*((q1*q3/r)+(q2*q3/r))\n",
+ "\n",
+ "#Result\n",
+ "print\"Workdone is \", W,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Workdone is 2.7e-07 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.14 Page no 100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q=1.6*10**-19\n",
+ "r=10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "U=m*q**2/r\n",
+ "K=U/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Kinetic energy is \",K,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Kinetic energy is 1.152e-18 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.15 Page no 100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**-31\n",
+ "V=10**6\n",
+ "q=1.6*10**-19\n",
+ "a=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "K=m*V**2\n",
+ "r=a*q**2/K\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance of the closest approach is \", r,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance of the closest approach is 2.56e-10 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 57
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.17 Page no 101"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.53*10**-10 #m\n",
+ "q1=1.6*10**-19 #C\n",
+ "q2=-1.6*10**-19 #C\n",
+ "a=9*10**9\n",
+ "r1=1.06*10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "U=a*q1*q2/r\n",
+ "Ue=U/q1\n",
+ "K=-Ue/2.0\n",
+ "E=Ue+K\n",
+ "U1=(a*q1*q2/r1)/q1\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Potential energy of the system is \", round(Ue,1),\"eV\"\n",
+ "print\"(ii) Minimum amount of work required to free the elctrons ia \",round(E,1),\"ev\"\n",
+ "print\"(iii) Potential energyof the system is \",round(E,1) ,\"ev and work requiredto free the electrons is \",round(-E,1),\"eV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Potential energy of the system is -27.2 eV\n",
+ "(ii) Minimum amount of work required to free the elctrons ia -13.6 ev\n",
+ "(iii) Potential energyof the system is -13.6 ev and work requiredto free the electrons is 13.6 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 75
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.18 Page no 102"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=9*10**9\n",
+ "q1=7*10**-6 #C\n",
+ "q2=-2*10**-6\n",
+ "r=0.18\n",
+ "r1=0.09\n",
+ "A=9*10**5\n",
+ "\n",
+ "#Calculation\n",
+ "U=a*q1*q2/r\n",
+ "W=0-U\n",
+ "U1=(q1*A/r1)+(q2*A/r1)+U\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Electrostatic potential energy is \", round(U,1),\"J\"\n",
+ "print\"(b) Work required to seperate two charges is \",round(W,1),\"J\"\n",
+ "print\"(c) Electrostatic energy is \", U1,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Electrostatic potential energy is -0.7 J\n",
+ "(b) Work required to seperate two charges is 0.7 J\n",
+ "(c) Electrostatic energy is 49.3 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 93
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.20 Page no 103"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "p=6*10**-6\n",
+ "E=10**6\n",
+ "a=1\n",
+ "\n",
+ "#Calculation,\n",
+ "U1=-p*E*a\n",
+ "U2=(p*E*(math.cos(60)*180/3.14))*10**-2\n",
+ "U3=U2-U1\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat released by substance is \", round(U3,0),\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat released by substance is 3.0 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 103
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.21 Page no 109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=10**-7\n",
+ "e=8.854*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "a=q/e\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric flux through the surface of the cube is \", round(a*10**-4,2),\"Nm**2C-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric flux through the surface of the cube is 1.13 Nm**2C-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 108
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.22 Page no 109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=8.85*10**-6 \n",
+ "e=8.85*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "a=q/e\n",
+ "b=a/6.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric flux through each face is \", round(b*10**-5,2),\"Nm**2C-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric flux through each face is 1.67 Nm**2C-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 115
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.23 Page no 109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E0=2*10**3 #N/C\n",
+ "S=0.2\n",
+ "\n",
+ "#Calculation\n",
+ "a=(3/5.0)*E0*S\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric flux of the field is \", a,\"Nm**2C-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric flux of the field is 240.0 Nm**2C-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 118
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.24 Page no 109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.2\n",
+ "m=9*10**9\n",
+ "b=50\n",
+ "\n",
+ "import math\n",
+ "E=250*r\n",
+ "a=E*4*math.pi*r**2\n",
+ "q=b*r**2/m\n",
+ "\n",
+ "#Result\n",
+ "print\"Charge contained in a sphere is \", round(q*10**10,2)*10**-10,\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge contained in a sphere is 2.22e-10 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 128
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.25 Page no 110"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=0.1 #m\n",
+ "A=800\n",
+ "e=8.854*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "b=A*a**2.5*(math.sqrt(2)-1)\n",
+ "q=e*b\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The flux through the cube is \", round(b,2),\"Nm**2C-1\"\n",
+ "print\"The charge within the cube is \",round(q*10**12,2)*10**-12,\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The flux through the cube is 1.05 Nm**2C-1\n",
+ "The charge within the cube is 9.28e-12 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 139
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.26 Page no 111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=200\n",
+ "a=0.05\n",
+ "e=8.854*10**-12\n",
+ "d=3.14\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "b=E*math.pi*a**2\n",
+ "c=2*b\n",
+ "q=e*d\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Net outward flux through each flat face is \", round(b,2),\"Nm**2C-1\"\n",
+ "print\"(b) Flux through the side of cylinder is zero \"\n",
+ "print\"(c) Net outward flux through the cylinder is \", round(c,2),\"Nm**2C-1\"\n",
+ "print\"(d) The net charge in the cylinder is \",round(q*10**11,2)*10**-11,\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Net outward flux through each flat face is 1.57 Nm**2C-1\n",
+ "(b) Flux through the side of cylinder is zero \n",
+ "(c) Net outward flux through the cylinder is 3.14 Nm**2C-1\n",
+ "(d) The net charge in the cylinder is 2.78e-11 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 156
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.28 Page no 114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=5.8*10**-6 #C\n",
+ "r=8*10**-2 #m\n",
+ "e=8.854*10**-12\n",
+ "l=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "E=q/(2*math.pi*e*r*l)\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric field is \", round(E*10**-5,1),\"*10**5 N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric field is 4.3 *10**5 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.29 Page no 114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=9*10**4 #N/C\n",
+ "r=2*10**-2 #m\n",
+ "m=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "a=r*E/(2.0*m)\n",
+ "print\"Linear charge density is \", a,\"Cm-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Linear charge density is 1e-07 Cm-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.30 Page no 115"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=10*10**-6 #C\n",
+ "r=0.1 #m\n",
+ "a=8.85*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "E=q/(4.0*math.pi*a*r**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Electric field intensity at a point 10 cm from the centre\", round(E*10**-6,0),\"*10**6 N/C\"\n",
+ "print\"(ii) Since the point is lying inside the shell, electric intensity at this point is zero\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Electric field intensity at a point 10 cm from the centre 9.0 *10**6 N/C\n",
+ "(ii) Since the point is lying inside the shell, electric intensity at this point is zero\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.31 Page no 118"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Z=79\n",
+ "e=1.6*10**-19\n",
+ "e0=8.854*10**-12\n",
+ "R=6.2*10**-15\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "q=Z*e\n",
+ "E=q/(4.0*math.pi*e0*R**2)\n",
+ "b=E/4.0\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The magnitude of the electric field at the surface of nucleus is \", round(E*10**-21,0)*10**21,\"N/C\"\n",
+ "print\"(ii) The magnitude of the electric field at a distance 2R from the centre of the nucleus is \",round(b*10**-21,2),\"*10**21 N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The magnitude of the electric field at the surface of nucleus is 3e+21 N/C\n",
+ "(ii) The magnitude of the electric field at a distance 2R from the centre of the nucleus is 0.74 *10**21 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.32 Page no 119"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=8.854*10**-12\n",
+ "A=0.5\n",
+ "F=1.8*10**-12 #N\n",
+ "E=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "q=(2*e*A**2*F)/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Total charge on the sheet is \", round(q*10**6,0),\"micro C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total charge on the sheet is 50.0 micro C\n"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.33 Page no 119"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=5*10**-6\n",
+ "e=8.854*10**-12\n",
+ "r=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "b=-(a*math.pi*r**2*(math.cos(60)*180/3.14))/(2*e)\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric flux through a circular area is \", round(b*10**-5,2),\"*10**3 Nm**2C-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric flux through a circular area is 4.84 *10**3 Nm**2C-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 54
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter4_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter4_2.ipynb new file mode 100644 index 00000000..5987ff7f --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter4_2.ipynb @@ -0,0 +1,1326 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:2ebe494197bc592ac147978ecceacaa802bf7a7b9283aeec109e01967ce4cfa8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 4 Capacitance "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.1 Page no 159"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "r=6.4*10**6 #m\n",
+ "\n",
+ "#Calculation\n",
+ "C=r/m\n",
+ "\n",
+ "#Result\n",
+ "print\"The capacitance of the earth is \", round(C*10**6,0),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The capacitance of the earth is 711.0 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.2 Page no 160"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "c=50*10**-12\n",
+ "V=10**4\n",
+ "\n",
+ "#Calculation\n",
+ "r=(m*c)*10**2\n",
+ "q=(c*V)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Radius of a isolated sphere is \",r,\"cm\"\n",
+ "print\"(ii) Charge of a isolated sphere is \", q*10**6,\"micro C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Radius of a isolated sphere is 45.0 cm\n",
+ "(ii) Charge of a isolated sphere is 0.5 micro C\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.3 Page no 160"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=3*10**-3 #m\n",
+ "m=9*10**9\n",
+ "q1=27*10**-12 #C\n",
+ "\n",
+ "#Calculation\n",
+ "R=3*r\n",
+ "C=R/m\n",
+ "V=q1/C\n",
+ "\n",
+ "#Result\n",
+ "print\"Capacitance of the bigger drop is \", C*10**12,\"pico F \\npotential of the bigger drop is \",V,\"Volts\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance of the bigger drop is 1.0 pico F \n",
+ "potential of the bigger drop is 27.0 Volts\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4 Page no 162"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "ra=0.09\n",
+ "rb=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "C=ra*rb/(m*(rb-ra))\n",
+ "\n",
+ "#Result\n",
+ "print\"Capacitance of the capacitor is \", C*10**12,\"pico F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance of the capacitor is 100.0 pico F\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.5 Page no 162"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=2 #cm\n",
+ "d=1.2\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "R=(d/r)*10**2\n",
+ "rab=(R*2)\n",
+ "x=r**2+4*rab\n",
+ "y=math.sqrt(x)\n",
+ "\n",
+ "#Result\n",
+ "print\"ra+rb=\", y,\"cm \\nra-rb=\",r ,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "ra+rb= 22.0 cm \n",
+ "ra-rb= 2 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 64
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.6 Page no 164"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=10**-3 #m\n",
+ "c=1 #F\n",
+ "e=8.854*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "A=c*d/e\n",
+ "\n",
+ "#Result\n",
+ "print\"Area is \", round(A*10**-8,1),\"*10**8 m**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Area is 1.1 *10**8 m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 69
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.7 Page no 164"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=0.02 #m**2\n",
+ "r=0.5 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "d=A/(4.0*math.pi*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance is \", round(d*10**3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance is 3.18 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.8 Page no 164"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=8.854*10**-12\n",
+ "K=6\n",
+ "A=30\n",
+ "d=2.0*10**-3\n",
+ "E=500\n",
+ "\n",
+ "#Calculation\n",
+ "C=e*K*A/d\n",
+ "V=E*d*10**3\n",
+ "q=C*V\n",
+ "\n",
+ "#Result\n",
+ "print\"Capacitance of a parallel plate \", round(q*10**3,3),\"micro C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance of a parallel plate 0.797 micro C\n"
+ ]
+ }
+ ],
+ "prompt_number": 87
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.9 Page no 165"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=300*10**-12\n",
+ "V=10*10**3\n",
+ "A=0.01\n",
+ "d=1*10**-3\n",
+ "\n",
+ "#Calculation\n",
+ "q=C*V\n",
+ "a=q/A\n",
+ "E=V/d\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Charge on each plate is \", q,\"C\"\n",
+ "print\"(ii) Electric flux density is \", a*10**4,\"10**-4 C/m**2\"\n",
+ "print\"(iii) Potential gradient is \", E,\"V/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Charge on each plate is 3e-06 C\n",
+ "(ii) Electric flux density is 3.0 10**-4 C/m**2\n",
+ "(iii) Potential gradient is 10000000.0 V/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 101
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.10 Page no 165"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A2=500 #cm**2\n",
+ "A1=100 #cm**2\n",
+ "d1=0.05 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "d2=A2*d1/A1\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance between the plates of second capacitor is \", d2,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance between the plates of second capacitor is 0.25 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 104
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.11 page no 167"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "c1=0.5 #micro F\n",
+ "c2=0.3 #micro F\n",
+ "c3=0.2 #micro F\n",
+ "\n",
+ "#Calculation\n",
+ "Cp=c1+c2+c3 \n",
+ "Cs=(1/c1)+(1/c2)+(1/c3)\n",
+ "\n",
+ "#Result\n",
+ "print\" The ratio ofmaximum capacitance to minimum capacitance is \",round (Cs,1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The ratio ofmaximum capacitance to minimum capacitance is 10.3\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.12 Page no 168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "c1=15.0 #micro F\n",
+ "c2=20.0 #micro F\n",
+ "V=10**-6\n",
+ "v1=600 #V \n",
+ "\n",
+ "#Calculation\n",
+ "Cs=c1*c2/(c1+c2)\n",
+ "Q=Cs*V*v1\n",
+ "Pd=(Q/c1)*10**6\n",
+ "Pd1=(Q/c2)*10**6\n",
+ "\n",
+ "#Result\n",
+ "print\"(i)charge on each capacitor is\",round(Q *10**3,2),\"10**-3 C\"\n",
+ "print\"(ii)P.D across15 micro Fcapacitor is\",round (Pd,1),\"V\"\n",
+ "print\" P.D across 20 micro F is\",round (Pd1,0),\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i)charge on each capacitor is 5.14 10**-3 C\n",
+ "(ii)p.D across15 micro Fcapacitor is 342.9 V\n",
+ " P.D across 20 micro F is 257.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.13 page no.168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ca=18 #micro F\n",
+ "Cb=4 #micro F\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "C=Ca*Cb\n",
+ "C12=math.sqrt(Ca**2-4*C)\n",
+ "C2=2*C12\n",
+ "\n",
+ "#Result\n",
+ "print\"The capacitance of capacitor C1 is\", C12,\"micro F\"\n",
+ "print\"The capacitance of capacitor C2 is\",C2,\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The capacitance of capacitor C1 is 6.0 micro F\n",
+ "The capacitance of capacitor C2 is 12.0 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.14 Page no 168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=750*10**-6\n",
+ "C1=15*10**-6\n",
+ "V2=20.0 #V\n",
+ "C3=8*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "V1=q/C1\n",
+ "V=V1+V2\n",
+ "q3=C3*V2\n",
+ "q2=q-q3\n",
+ "C2=q2/V2\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of V1 is \", V1,\"V\"\n",
+ "print\"The value of V is \",V,\"V\"\n",
+ "print\"The value of capacitance is\",C2*10**6,\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of V1 is 50.0 V\n",
+ "The value of V is 70.0 V\n",
+ "The value of capacitance is 29.5 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 52
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.15 Page no 169"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C2=9.0 #micro F\n",
+ "C3=9.0\n",
+ "C4=9.0\n",
+ "C1=3\n",
+ "V=10 #V\n",
+ "\n",
+ "#Calculation\n",
+ "C=1/((1/C2)+(1/C3)+(1/C4))\n",
+ "Cab=C1+C\n",
+ "q=Cab*V\n",
+ "\n",
+ "#Result\n",
+ "print\"Equivalent capacitance between point A and B is \", Cab,\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equivalent capacitance between point A and B is 6.0 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 60
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.17 Page no 169"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Cab=10 #micro F\n",
+ "C1=8.0 #micro F\n",
+ "C2=8.0\n",
+ "C3=8\n",
+ "C4=8\n",
+ "C5=12\n",
+ "V=400\n",
+ "\n",
+ "#Calculation\n",
+ "Cbc=((C1*C2)/(C1+C2))+C3+C4\n",
+ "Cac=Cab*Cbc/(Cab+Cbc)\n",
+ "Ccd=C1+C5\n",
+ "Cad=Cac*Ccd/(Cac+Ccd)\n",
+ "q=Cad*V\n",
+ "Vcd=q/Ccd\n",
+ "q1=C5*Vcd\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The equivalent capacitance between A and D is \", Cad,\"micro f\"\n",
+ "print\"(ii) The charge on 12 micro F capacitor is \",q1*10**-3,\"mC\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The equivalent capacitance between A and D is 5.0 micro f\n",
+ "(ii) The charge on 12 micro F capacitor is 1.2 mC\n"
+ ]
+ }
+ ],
+ "prompt_number": 84
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.20 Page no 171"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C1=5 #micro F\n",
+ "C2=6 #micro F\n",
+ "V=10 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Cp=C1+C2\n",
+ "q=Cp*V\n",
+ "\n",
+ "#Result\n",
+ "print\"Charge supplied by battery is \", q,\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge supplied by battery is 110 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 88
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.21 Page no 171"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C1=2 #micro F\n",
+ "C2=2 #micro F\n",
+ "C3=2\n",
+ "C4=2\n",
+ "\n",
+ "#Calculation\n",
+ "Cs=C1*C2/(C1+C2)\n",
+ "Cab=C3*C4/(C3+C4)\n",
+ "\n",
+ "#Result\n",
+ "print\"The capacitance of the Capacitors\", Cab,\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The capacitance of the Capacitors 1 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 93
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.22 Page no 171"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C1=10.0 #micro F\n",
+ "C2=10.0\n",
+ "C3=10.0\n",
+ "C4=10*10**-3\n",
+ "V=500 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Cs=1/((1/C1)+(1/C2)+(1/C3))\n",
+ "Cab=Cs+(C4*10**3)\n",
+ "Q=(C1*(500/3.0))*10**-3\n",
+ "Q1=C4*V\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The equivalent capacitance of the network is\",round(Cab,1),\"micro F\"\n",
+ "print \"(b) The charge on 12 micro F Capacitor is\",Q1,\"*10**-3 C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The equivalent capacitance of the network is 13.3 micro F\n",
+ "(b) The charge on 12 micro F Capacitor is 5.0 *10**-3 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 115
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.23 Page no 172"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C4=6 #micro F\n",
+ "C5=12 \n",
+ "C1=8.0\n",
+ "C7=1\n",
+ "\n",
+ "#Calculation\n",
+ "Cs=C4*C5/(C4+C5)\n",
+ "C11=(C1*Cs)/(C1+Cs)\n",
+ "Cs1=C1*C7/(C1+C7)\n",
+ "Cp=C11+Cs1\n",
+ "C=1/(1-(1/Cp))\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of capacitance C is \", round(C,2),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of capacitance C is 1.39 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 129
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.24 Page no 175"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "K=5\n",
+ "l=0.2\n",
+ "c=10**-9 #F\n",
+ "b=15.4\n",
+ "a=15\n",
+ "pd=5000 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "C=(K*l*c)/(41.1*math.log10(b/a))\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The capacitance of cylindrical capacitor is \", round(C*10**9,1)*10**-9,\"F\"\n",
+ "print\"(ii) The potential of the inner cylinder is equal to p.d. between two cylinders i.e potentila of inner cylinder is\",pd,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The capacitance of cylindrical capacitor is 2.1e-09 F\n",
+ "(ii) The potential of the inner cylinder is equal to p.d. between two cylinders i.e potentila of inner cylinder is 5000 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 138
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.25 Page no 179"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=5*10**-6\n",
+ "V=100\n",
+ "C1=3*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "q=C*V\n",
+ "Cp=C+C1\n",
+ "pd=q/Cp\n",
+ "\n",
+ "#Result\n",
+ "print\"P.D across the capacitor is \", pd,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "P.D across the capacitor is 62.5 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 143
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.26 Page no 179 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=250 #V\n",
+ "C1=6 #micro F\n",
+ "C2=4\n",
+ "Cp=10*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "pd=V*C1/(C1+C2)\n",
+ "q=pd*C2*10**-6\n",
+ "q1=2*q\n",
+ "pd1=q1/Cp\n",
+ "q2=C2*pd1\n",
+ "q3=C1*pd1\n",
+ "\n",
+ "#Result\n",
+ "print\"New potentila difference is \", pd1,\"V\"\n",
+ "print\"Charge on 4 micro F capacitor is \",q2,\"micro C\"\n",
+ "print\"Charge on 6 micro F capacitor is \",q3,\"micro C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "New potentila difference is 120.0 V\n",
+ "Charge on 4 micro F capacitor is 480.0 micro C\n",
+ "Charge on 6 micro F capacitor is 720.0 micro C\n"
+ ]
+ }
+ ],
+ "prompt_number": 156
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.28 Page no 180"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C1=16*10**-6 # F\n",
+ "C2=4 #micro F\n",
+ "V1=100 #V\n",
+ "Cp=20*10**-6 #f\n",
+ "\n",
+ "#Calculation\n",
+ "q=C1*V1\n",
+ "U1=0.5*C1*V1**2\n",
+ "V=q/Cp\n",
+ "U2=0.5*Cp*V**2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Potential difference across the capacitor is \", V,\"Volts\"\n",
+ "print\"(ii) The electrostatic energies before and after the capacitors are connected \",U2,\"J\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Potential difference across the capacitor is 80.0 Volts\n",
+ "(ii) The electrostatic energies before and after the capacitors are connected 0.064 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.29 Page no 180"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "V=3.0*10**6\n",
+ "r=2\n",
+ "\n",
+ "#Calculation\n",
+ "q=(V*r)/m\n",
+ "E=0.5*q*V\n",
+ "\n",
+ "#Result\n",
+ "print\"The heat generated is \", E,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The heat generated is 1000.0 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.30 Page no 180"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=12 #V\n",
+ "C=1.35*10**-10 #C\n",
+ "\n",
+ "#Calculation\n",
+ "q=C\n",
+ "\n",
+ "#Result\n",
+ "print\"Extra Charge supplied by battery is \", q,\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Extra Charge supplied by battery is 1.35e-10 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.31 Page no 181"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=100*10**-6 #F\n",
+ "V=500 #V\n",
+ "\n",
+ "#Calculation\n",
+ "q=V/2.0\n",
+ "E=0.5*(0.5*C*V**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Charge in the new stored energy is \", E,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge in the new stored energy is 6.25 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.32 Page no 187"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=2*10**-3 #m**2\n",
+ "d=0.01 #m\n",
+ "t=6*10**-3 #m\n",
+ "K=3\n",
+ "a=8.854*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "C=a*A/(d-t*(1-(1/3.0)))\n",
+ "\n",
+ "#Result\n",
+ "print\"The capacitance of the capacitor is \", round(C*10**12,2)*10**-12,\"F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The capacitance of the capacitor is 2.95e-12 F\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.33 Page no 188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=8.854*10**-12\n",
+ "A=2\n",
+ "t1=0.5*10**-3\n",
+ "t2=1.5*10**-3\n",
+ "t3=0.3*10**-3\n",
+ "K1=2.0\n",
+ "K2=4.0\n",
+ "K3=6.0\n",
+ "\n",
+ "#Calculation\n",
+ "C=(e*A)/((t1/K1)+(t2/K2)+(t3/K3))\n",
+ "\n",
+ "#Result\n",
+ "print\"The capacitance of the capacitor is \", round(C*10**6,3)*10**-6,\"F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The capacitance of the capacitor is 2.6e-08 F\n"
+ ]
+ }
+ ],
+ "prompt_number": 52
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.34 Page no 188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=3 #mm\n",
+ "b=4.0 #mm\n",
+ "K1=5\n",
+ "\n",
+ "#Calaculation\n",
+ "K2=1/((a**2/b)-a/b)*K1\n",
+ "\n",
+ "#Result\n",
+ "print\"The relative permittivity of the additional dielectric is \", round(K2,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The relative permittivity of the additional dielectric is 3.33\n"
+ ]
+ }
+ ],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.35 Page no 188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=5\n",
+ "t=2\n",
+ "K=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "D=d+(t-t/K)\n",
+ "\n",
+ "#Result\n",
+ "print\"New seperaion between the plates are \", round(D,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "New seperaion between the plates are 6.33 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 62
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.36 Page no 189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=4\n",
+ "t=2\n",
+ "K=4.0\n",
+ "C1=50*10**-12 #f\n",
+ "V0=200 #V\n",
+ "\n",
+ "#Calculation\n",
+ "C=(d-t+(t/K))/d\n",
+ "q=C1*V0\n",
+ "V=V0*C\n",
+ "U=0.5*q*V\n",
+ "E=0.5*q*(V0-V)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Final charge on ach plate is \", q,\"C\"\n",
+ "print\"(ii) P.D batween the plates is \", V,\"volts\"\n",
+ "print\"(iii)Final energy in the capacitor is \", U,\"J\"\n",
+ "print\"(iv) Energy loss is \", E,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Final charge on ach plate is 1e-08 C\n",
+ "(ii) P.D batween the plates is 125.0 volts\n",
+ "(iii)Final energy in the capacitor is 6.25e-07 J\n",
+ "(iv) Energy loss is 3.75e-07 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.39 Page no 193"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=25*10**5\n",
+ "E=5.0*10**7\n",
+ "\n",
+ "#Calculation\n",
+ "r=V/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum radius of the spherical shell is \", r*100,\"cm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum radius of the spherical shell is 5.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter5_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter5_2.ipynb new file mode 100644 index 00000000..639e5486 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter5_2.ipynb @@ -0,0 +1,1741 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:76fc8177d7b8f6f9c96106003a3d3f70d960561edd9596ed73b03964026f4fb8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 5 Electric current and resistance "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.1 Page no 257"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=10**17\n",
+ "e=1.6*10**-19 #C\n",
+ "t=1.0 #S\n",
+ "\n",
+ "#Calculation\n",
+ "I=n*e/t\n",
+ "\n",
+ "#Result\n",
+ "print\"The magnitude of current in the wire is \",I*10**2,\"10**-2 A and direction is from left to right\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The magnitude of current in the wire is 1.6 10**-2 A and direction is from left to right\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.2 Page no 257"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "I=0.5\n",
+ "T=1\n",
+ "e=1.6*10**-19\n",
+ "t=60 #minute\n",
+ "\n",
+ "#Calculation\n",
+ "n=I*T/e\n",
+ "q=I*t**2\n",
+ "n1=q/e\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The number of electrons passing a cross section of the bulb is \",round(n*10**-18,1)*10**18,\"electrons/S\"\n",
+ "print\"(ii) The number of electrons is \",round(n1*10**-22,1)*10**22,\"electrons/hour\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The number of electrons passing a cross section of the bulb is 3.1e+18 electrons/S\n",
+ "(ii) The number of electrons is 1.1e+22 electrons/hour\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.3 Page no 257"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19 #C\n",
+ "f=6.8*10**15 #rev/sec\n",
+ "r=0.51*10**-10 #m\n",
+ "\n",
+ "#Calculation\n",
+ "I=e*f\n",
+ "\n",
+ "#Result\n",
+ "print\"The equivalent current is \", I*10**3,\"*10**-3 A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The equivalent current is 1.088 *10**-3 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5 Page no 267"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=10 #A\n",
+ "A=1 #m*m**2\n",
+ "e=1.6*10**-19 #C\n",
+ "n=10**28 #m**-3\n",
+ "\n",
+ "#Calculation\n",
+ "Vd=I/(n*A*e)\n",
+ "\n",
+ "#Result\n",
+ "print\"Drift velocity of the conduction electrons are \", Vd,\"m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Drift velocity of the conduction electrons are 6.25e-09 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.6 Page no 267"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=10 #A\n",
+ "A=4*10**-6 #m**2\n",
+ "e=1.6*10**-19 #C\n",
+ "n=8*10**28 #m**-3\n",
+ "l=4\n",
+ "\n",
+ "#Calculation\n",
+ "Vd=I/(n*A*e)\n",
+ "t=l/Vd\n",
+ "\n",
+ "#Result\n",
+ "print\"Time required by an electron is \", t*10**-4,\"*10**4 S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time required by an electron is 2.048 *10**4 S\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.7 page no 267"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=6.023*10**23\n",
+ "m=63.5*10**-3\n",
+ "d=9*10**3\n",
+ "A=10**-7 #m**2\n",
+ "e=1.6*10**-19 #C\n",
+ "I=1.5 #a\n",
+ "K=1.38*10**-23 #J/K\n",
+ "T=300 #K\n",
+ "Vd=1.1*10**-3\n",
+ "C=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "n=a*d/m\n",
+ "Vd=I/(n*A*e)\n",
+ "V=math.sqrt((3*K*T*a)/m)\n",
+ "V1=Vd/V\n",
+ "E=Vd/C\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Thermal speeds of copper atoms at ordinary temperatures are \", round(V1*10**6,2),\"*10**-6 m/s\"\n",
+ "print\"(ii) Speed of propagation of electric fild is \", round(E*10**12,1)*10**-12"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Thermal speeds of copper atoms at ordinary temperatures are 3.2 *10**-6 m/s\n",
+ "(ii) Speed of propagation of electric fild is 3.7e-12\n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.8 Page no 268"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=5\n",
+ "l=0.1\n",
+ "Vd=2.5*10**-4\n",
+ "\n",
+ "#Calculation\n",
+ "E=V/l\n",
+ "u=Vd/E\n",
+ "\n",
+ "#Result\n",
+ "print\"The electron mobility is \", u,\"m**2/V/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electron mobility is 5e-06 m**2/V/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 57
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.9 Page no 268"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=2.4\n",
+ "A=0.30*10**-6\n",
+ "m=9.1*10**-31\n",
+ "n=8.4*10**28\n",
+ "e=1.6*10**-19\n",
+ "E=7.5\n",
+ "\n",
+ "#Calculation\n",
+ "J=I/A\n",
+ "t=m*J/(n*e**2*E)\n",
+ "\n",
+ "#Result\n",
+ "print\"Average relaxation time is \", round(t*10**16,2)*10**-16,\"S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average relaxation time is 4.51e-16 S\n"
+ ]
+ }
+ ],
+ "prompt_number": 64
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.10 Page no 268"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.12*10**-2 #m\n",
+ "I=10\n",
+ "r1=0.08*10**-2 #m\n",
+ "I=10 #A\n",
+ "e=1.6*10**-19 #C\n",
+ "n=8.4*10**28\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=math.pi*(r**2)\n",
+ "J=I/A\n",
+ "A1=math.pi*r1**2\n",
+ "Vd=I/(e*n*A1)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Current density in the alluminium wire is \",round(J*10**-6,1),\"*10**6 A/m**2\"\n",
+ "print\"(ii) Drift velocity of electrons in the copper wire is \",round(Vd*10**4,1),\"*10**-4 m/S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Current density in the alluminium wire is 2.2 *10**6 A/m**2\n",
+ "(ii) Drift velocity of electrons in the copper wire is 3.7 *10**-4 m/S\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.11 Page no 269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "D=0.13*10**-2\n",
+ "R=3.4 #ohms\n",
+ "l=10.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=(math.pi/4.0)*D**2\n",
+ "a=R*A/l\n",
+ "b=1/a\n",
+ "\n",
+ "#Result\n",
+ "print\"Conductivity of a material is \",round(b*10**-6,1),\"*10**6 S/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Conductivity of a material is 2.2 *10**6 S/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 93
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.12 Page no 269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A1=25.0 #mm**2\n",
+ "l2=1 #m\n",
+ "R2=1/58.0\n",
+ "A2=1\n",
+ "l1=1000\n",
+ "\n",
+ "#Calculation\n",
+ "R=(l1/l2)*(A2/A1)\n",
+ "R1=R*R2\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of resistance is \", round(R1,2),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of resistance is 0.69 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 98
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.13 Page no 269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=4.5\n",
+ "A1=1\n",
+ "A2=2.0\n",
+ "l2=3\n",
+ "l1=1.0\n",
+ "\n",
+ "#Calculation\n",
+ "R=(l2/l1)*(A1/A2)\n",
+ "R2=R*R1\n",
+ "\n",
+ "#Result\n",
+ "print\"The resistance of another wire is \", R2,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resistance of another wire is 6.75 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 103
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.14 Page no 269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=1\n",
+ "r1=0.5\n",
+ "R1=0.15 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A1=(math.pi/4.0)*r**2\n",
+ "A2=(math.pi/4.0)*r1**2\n",
+ "l=A1/A2\n",
+ "R=l*l\n",
+ "R2=R*R1\n",
+ "\n",
+ "#Result\n",
+ "print\"New resistance of the wire is \", R2,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "New resistance of the wire is 2.4 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 115
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.15 Page no 270"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=1.5 #ohm\n",
+ "e=1.6*10**-19 #C\n",
+ "t=1 #second\n",
+ "V=3 #V\n",
+ "\n",
+ "#Calculation\n",
+ "I=V/R\n",
+ "n=I*t/e\n",
+ "\n",
+ "#Result\n",
+ "print \"Number of electrons flowing through it in 1 S is \",n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of electrons flowing through it in 1 S is 1.25e+19\n"
+ ]
+ }
+ ],
+ "prompt_number": 118
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.16 Page no 270"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ne=2.8*10**18\n",
+ "np=1.2*10**18\n",
+ "e=1.6*10**-19\n",
+ "t=1 #S\n",
+ "V=220\n",
+ "\n",
+ "#Calculation\n",
+ "q=(ne+np)*e\n",
+ "I=q/t\n",
+ "R=V/I\n",
+ "\n",
+ "#Result\n",
+ "print\"Effective resistance of the tube is \", round(R,0),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Effective resistance of the tube is 344.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 124
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.17 Page no 270"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=84 #g\n",
+ "d=10.5 #g/cm**3\n",
+ "a=1.6*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "V=m/d\n",
+ "s=V**(1/3.0)\n",
+ "R=a/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Resistance between the opposite faces is \", R,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance between the opposite faces is 8e-07 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 130
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.18 Page no 270"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=1.001\n",
+ "A=1.001\n",
+ "\n",
+ "#Calculation\n",
+ "R=l*A\n",
+ "R1=R-1\n",
+ "A=R1*100\n",
+ "\n",
+ "#Result\n",
+ "print\"Percentage change in its resistance is \", round(A,1),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percentage change in its resistance is 0.2 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 137
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.19 Page no 271"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=0.45 #Kg\n",
+ "R=0.0014 #ohm\n",
+ "a=1.78*10**-8 #ohm\n",
+ "d=8.93*10**3 #Kg/m**3\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l=math.sqrt(R*m/(a*d))\n",
+ "r=math.sqrt(m/(math.pi*d*1.99))\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of length is\",round(l,2),\"m\"\n",
+ "print\"The value of radius is \",round(r*10**3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of length is 1.99 m\n",
+ "The value of radius is 2.84 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 148
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.20 Page no 271"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R15=80 #ohm\n",
+ "a=0.004\n",
+ "\n",
+ "#Calculation\n",
+ "R0=R15/(1+15*a)\n",
+ "R50=R0*(1+a*50)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of resistance at 50 degree C is \", round(R50,2),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of resistance at 50 degree C is 90.57 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 154
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.21 Page no 271"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R20=20 #ohm\n",
+ "P=60 #W\n",
+ "V=120.0 #Volts\n",
+ "a=5*10**-3\n",
+ "\n",
+ "#Calculation\n",
+ "I=P/V\n",
+ "Rt=V/I\n",
+ "t=(((Rt/R20)-1)/a)+R20\n",
+ "\n",
+ "#Result\n",
+ "print\"Normal working temperature of the lamp is \", t,\"degree C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Normal working temperature of the lamp is 2220.0 degree C\n"
+ ]
+ }
+ ],
+ "prompt_number": 160
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.22 Page no 271"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R0=5 #ohm\n",
+ "R100=5.23 #ohm\n",
+ "Rt=5.795 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "t=((Rt-R0)/(R100-R0))*100\n",
+ "\n",
+ "#Result\n",
+ "print\"The temperature of the bath is \", round(t,2),\"degree C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The temperature of the bath is 345.65 degree C\n"
+ ]
+ }
+ ],
+ "prompt_number": 165
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.23 Page no 272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=15*10**-4 #m**2\n",
+ "a=7.6*10**-8 # ohm m\n",
+ "l=2000 #m\n",
+ "b=0.005 #degree/C\n",
+ "\n",
+ "#Calculation\n",
+ "R0=a*l/A\n",
+ "R50=R0*(1+(b*50))\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of resistance is \", round(R50,3),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of resistance is 0.127 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 172
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.24 Page no 272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=0.004\n",
+ "ac=0.0007\n",
+ "R0=100\n",
+ "\n",
+ "#Calculation\n",
+ "R=ac*R0/a\n",
+ "\n",
+ "#Result\n",
+ "print\"The resistance of a copper filament is \", R,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resistance of a copper filament is 17.5 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 175
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.28 Page no 281"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=4.0 #ohm\n",
+ "R2=4.0 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Rab=1/((1/R1)+(1/R2))\n",
+ "\n",
+ "#Result\n",
+ "print\"The equivalent resisatance is \", Rab,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The equivalent resisatance is 2.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 178
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.29 Page no 281"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=15 #ohm\n",
+ "R2=30 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "R=R1*R2/(R1+R2)\n",
+ "\n",
+ "#Result\n",
+ "print\"The equivqlent resistance between A and B is \", R,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The equivqlent resistance between A and B is 10 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 181
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.31 Page no 282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=5 #ohm\n",
+ "R2=9 #ohm\n",
+ "R3=14 #ohm\n",
+ "R4=11\n",
+ "R5=7\n",
+ "R6=18\n",
+ "R7=13\n",
+ "R8=22\n",
+ "V=22\n",
+ "\n",
+ "#Calculation\n",
+ "Rec=(R1+R2)*R3/(R1+R2+R3)\n",
+ "Rbe=(R4+R5)*R6/(R4+R5+R6)\n",
+ "Rae=(R7+R2)*R8/(R7+R2+R8)\n",
+ "I=V/Rae\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of current in the branch AF is \", I,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of current in the branch AF is 2 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 187
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.32 Page no 282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=12 #ohm\n",
+ "R2=6 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Rdg=R1*R2/(R1+R2)\n",
+ "Rch=R1*R2/(R1+R2)\n",
+ "Rab=Rdg+Rch\n",
+ "\n",
+ "#Result\n",
+ "print\"The equivalent resistance is \", Rab,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The equivalent resistance is 8 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 191
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.33 Page no 283"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rab=500.0 #ohm\n",
+ "Rl=500 #ohm\n",
+ "Rbc=1500 #ohm\n",
+ "E=50 #Volts\n",
+ "Rac=2000.0 #ohm\n",
+ "V=40\n",
+ "\n",
+ "#Calculation\n",
+ "R=Rbc*Rl/(Rbc+Rl)\n",
+ "I=E/(Rab+R)\n",
+ "Pd=I*Rab\n",
+ "Rl1=E-Pd\n",
+ "I1=E/Rac\n",
+ "R12=V/I1\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Potential difference across the road is \", round(Rl1,2),\"V\"\n",
+ "print\"(ii) Resistance at BC is \", R12,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Potential difference across the road is 21.43 V\n",
+ "(ii) Resistance at BC is 1600.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 206
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.35 Page no 284"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=5 #ohm\n",
+ "R2=5.0 #ohm\n",
+ "R=6\n",
+ "\n",
+ "#Calculation\n",
+ "n=(1/(R-R1)*R2)\n",
+ "\n",
+ "#Result\n",
+ "print\"There are\", n,\"resistance are in parallel\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "There are 5.0 resistance are in parallel\n"
+ ]
+ }
+ ],
+ "prompt_number": 210
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.36 Page no 284"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=20.0 #ohm\n",
+ "R2=10.0 #ohm\n",
+ "R4=10\n",
+ "\n",
+ "#Calculation\n",
+ "Rbd=(R1*R2)/(R1+R2)\n",
+ "Rae=R2+Rbd+R4\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of resistance is \", round(Rae,2),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of resistance is 26.67 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 218
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.37 Page no 284"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=2.0 #ohm\n",
+ "R2=3 #ohm\n",
+ "R3=2.8\n",
+ "E=6 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Rab=R1*R2/(R1+R2)\n",
+ "Rt=Rab+R3\n",
+ "I=E/Rt\n",
+ "Vab=I*Rab\n",
+ "I1=Vab/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"The steady state current is \", I1,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The steady state current is 0.9 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 226
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.38 Page no 285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=3 #ohm\n",
+ "R2=3\n",
+ "R3=6\n",
+ "\n",
+ "#Calculation\n",
+ "Rad=(R1+R2)*R3/(R1+R2+R3)\n",
+ "Rae=(Rad+R1)*R3/(Rad+R1+R3)\n",
+ "Raf=(Rae+R1)*R3/(Rae+R1+R3)\n",
+ "Rab=(Raf+R1)*R2/(Rae+R1+R2)\n",
+ "\n",
+ "#Result\n",
+ "print\"the effective resistance between the point A and B is\", Rab,\"Ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the effective resistance between the point A and B is 2 Ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 234
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 39 Page no 285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R2=50.0 #ohm\n",
+ "R3=50.0 #ohm\n",
+ "R4=75.0 #ohm\n",
+ "E=4.75\n",
+ "R1=100\n",
+ "\n",
+ "#Calculation\n",
+ "Rbc=1/((1/R2)+(1/R3)+(1/R4))\n",
+ "R=R1+Rbc\n",
+ "I=E/R\n",
+ "R11=I*R1\n",
+ "Vbc=E-(I*R1)\n",
+ "I2=Vbc/R2\n",
+ "I3=Vbc/R3\n",
+ "I4=Vbc/R4\n",
+ "\n",
+ "#Result\n",
+ "print\"Equivalent resistance of the circuit is \", R,\"ohm\"\n",
+ "print\"Current in R2 is\",I2,\"A\"\n",
+ "print\"Current in R3 is\",I3,\"A\"\n",
+ "print\"Current in R4 is\",I4,\"A\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equivalent resistance of the circuit is 118.75 ohm\n",
+ "Current in R2 is 0.015 A\n",
+ "Current in R3 is 0.015 A\n",
+ "Current in R4 is 0.01 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 251
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.40 Page no 292"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=19\n",
+ "I1=0.5\n",
+ "I2=2 #A\n",
+ "r=2 \n",
+ "\n",
+ "#Calculation\n",
+ "E=V+I1*r\n",
+ "\n",
+ "#Result\n",
+ "print\"E.M.F is \", E,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "E.M.F is 20.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 254
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.41 Page no 292"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=1.5\n",
+ "a=1.5\n",
+ "r1=0.5 #ohm\n",
+ "r2=0.25\n",
+ "R=2.25 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "E=V+a\n",
+ "r=r1+r2\n",
+ "Rt=r+R\n",
+ "I=E/Rt\n",
+ "pd=V-(I*r1)\n",
+ "pd1=V-(I*r2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The circuit current is \",I,\"A\"\n",
+ "print\"(ii) P.D across the terminals of each cell is \",pd,\"V and \",pd1,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The circuit current is 1.0 A\n",
+ "(ii) P.D across the terminals of each cell is 1.0 V and 1.25 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 268
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.42 Page no 292"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=10\n",
+ "E=1.5\n",
+ "R=4 #ohm\n",
+ "r=0.1\n",
+ "a=8\n",
+ "\n",
+ "#Calculation\n",
+ "Emf=n*E\n",
+ "Rt=R+(n*r)\n",
+ "I=Emf/Rt\n",
+ "Emf1=(a*E)-(2*E)\n",
+ "I1=Emf1/Rt\n",
+ "I11=I-I1\n",
+ "\n",
+ "#Result\n",
+ "print\"Reduction in current is \", I11,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Reduction in current is 1.2 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 277
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.43 Page no 292"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Emf=2\n",
+ "Emf1=1.9\n",
+ "Emf2=1.8\n",
+ "R1=0.05\n",
+ "R2=0.06\n",
+ "R3=0.07\n",
+ "R0=5 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Emft=Emf+Emf1+Emf2\n",
+ "R=R1+R2+R3\n",
+ "Rt=R+R0\n",
+ "I=Emft/Rt\n",
+ "\n",
+ "#Result\n",
+ "print\"The reading of the ammeter is \", round(I,1),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The reading of the ammeter is 1.1 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 283
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.44 Page no 293"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=6.0 #ohm\n",
+ "R2=3\n",
+ "I=0.8 #A\n",
+ "a=24\n",
+ "\n",
+ "#Calculation\n",
+ "I1=I*(R1+R2)/R1\n",
+ "I11=I1-I\n",
+ "Rp=R1*R2/(R1+R2)\n",
+ "Rt=Rp+8\n",
+ "r=(a/I1)-10\n",
+ "V=I1*Rt\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Current in 6 ohm resistance is \", I11,\"A\"\n",
+ "print\"(ii) Internal resistance of the battery is \", r,\"ohm\"\n",
+ "print\"(iii) The terminal potential difference of the battery is \", V,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Current in 6 ohm resistance is 0.4 A\n",
+ "(ii) Internal resistance of the battery is 10.0 ohm\n",
+ "(iii) The terminal potential difference of the battery is 12.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 295
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.45 Page no 294"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=2 #ohm\n",
+ "R2=4\n",
+ "R3=6\n",
+ "E=8\n",
+ "r=1\n",
+ "\n",
+ "#Calculation\n",
+ "Rac=(R1+R2)*R3/(R1+R2+R3)\n",
+ "I=E/(Rac+r)\n",
+ "I1=I/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Internal resistance is \", I1,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Internal resistance is 1.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 299
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.46 Page no 294"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=1\n",
+ "R=2\n",
+ "\n",
+ "#Calculation\n",
+ "r=(E*R)-E\n",
+ "print\"The internal resisatnce of aech cell is \",r,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The internal resisatnce of aech cell is 1 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.47 Page no 294"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=15.0 # ohm\n",
+ "R2=15.0\n",
+ "E=2\n",
+ "V=1.6\n",
+ "\n",
+ "#Calculation\n",
+ "R=R1*R2/(R1+R2)\n",
+ "r=((E/V)-1)*R*4\n",
+ "\n",
+ "#Result\n",
+ "print\"Internal resisatnce of each cell is \", r,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Internal resisatnce of each cell is 7.5 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.48 Page no 295"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I1=1 #A\n",
+ "E=1.5\n",
+ "I2=0.6\n",
+ "R2=2.33 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "R=2*E/I1\n",
+ "R1=2*E/I2\n",
+ "r=R1-2*R2\n",
+ "\n",
+ "#Result\n",
+ "print\"Internal resisatnce of each battery is \", r,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Internal resisatnce of each battery is 0.34 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.49 Page no 295"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=4 #ohm\n",
+ "R2=4 #ohm\n",
+ "R3=12\n",
+ "R4=6.0\n",
+ "E=16\n",
+ "r=1 #ohm\n",
+ "\n",
+ "#calculation\n",
+ "Rab=R1*R2/(R1+R2)\n",
+ "Rcd=R3*R4/(R3+R4)\n",
+ "R=Rab+Rcd+1\n",
+ "I=E/(R+r)\n",
+ "I1=I/2.0\n",
+ "I3=I*R4/(R3+R4)\n",
+ "I4=I*R3/(R3+R4)\n",
+ "Vab=4*I1\n",
+ "Vbc=I*1\n",
+ "Vcd=12*I3\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) equivalent resistance of the network is \", R,\"ohm\"\n",
+ "print\"(ii) Circuit current is\", I,\"A , Current in R1 is\",I1,\"A , Current in R3 is\",round(I3,2),\"A , Current in R4 is \",round(I4,2)\n",
+ "print \"Voltage drop Vab is\",Vab,\"V \\nVbc is\",Vbc,\"V \\nVcd is\",Vcd,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) equivalent resistance of the network is 7.0 ohm\n",
+ "(ii) Circuit current is 2.0 A , Current in R1 is 1.0 A , Current in R3 is 0.67 A , Current in R4 is 1.33\n",
+ "Voltage drop Vab is 4.0 V \n",
+ "Vbc is 2.0 V \n",
+ "Vcd is 8.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter6_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter6_2.ipynb new file mode 100644 index 00000000..f1c6f4b7 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter6_2.ipynb @@ -0,0 +1,624 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:1bababaf98233cb133c1893aef0743afe5d48d8c1bcfe7a7487181ba9a8fde89"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 6 Electrical measurements"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.1 Page no 332"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=4\n",
+ "b=2.0\n",
+ "c=8\n",
+ "d=5\n",
+ "e=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "I1=((a*c)+(b*e))/((b*c)+(d*e))\n",
+ "I2=(a-(2*I1))/e\n",
+ "V=(I1-I2)*5\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Current through each battery is\", round(I1,2),\"A and\",round(I2,2),\"A\"\n",
+ "print\"(ii) Terminal voltage is\",round(V,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Current through each battery is 1.23 A and 0.52 A\n",
+ "(ii) Terminal voltage is 3.55 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.4 Page no 334"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=10\n",
+ "b=5.0\n",
+ "c=9.0\n",
+ "d=19.0\n",
+ "\n",
+ "#Calculation\n",
+ "I2=(a-c)/((b*a)-(d*c))\n",
+ "I1=(1-(5*I2))/c\n",
+ "I=I1+I2\n",
+ "pd=I*10\n",
+ "\n",
+ "#Result\n",
+ "print\"Current through each cell is\", round(I,2),\"A\"\n",
+ "print\"Potential difference across 10 ohm wire is\",round(pd,3),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current through each cell is 0.11 A\n",
+ "Potential difference across 10 ohm wire is 1.074 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6 Page no 335"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=-3\n",
+ "b=4.0\n",
+ "c=3\n",
+ "\n",
+ "#Calculation\n",
+ "I1=a/(b+(c**2))\n",
+ "I2=-1-c*I1\n",
+ "I3=-(I1+I2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Current through each cell is\", round(I1,2),\"A ,\",round(I2,2),\"A and\",round(I3,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current through each cell is -0.23 A , -0.31 A and 0.54 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.7 Page no 336"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=15\n",
+ "b=4\n",
+ "c=12.0\n",
+ "d=10\n",
+ "\n",
+ "#Calculation\n",
+ "R=(a*b)/c\n",
+ "X=(d*R)/(d-R)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of resistance is\", X,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of resistance is 10.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.8 Page no 336"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=4 #ohm\n",
+ "R2=3 #ohm\n",
+ "R3=2.0\n",
+ "R11=2.4 #ohm\n",
+ "E=6\n",
+ "\n",
+ "#Calculation\n",
+ "X=(R1*R2)/R3\n",
+ "R4=R2+X\n",
+ "R5=R1+R3\n",
+ "Rt=((R4*R5)/(R4+R5))+R11\n",
+ "I=E/Rt\n",
+ "\n",
+ "#Result\n",
+ "print\"the value of unknown resistance is\", X,\"ohm\"\n",
+ "print\"The current drawn by the circuit is\",I,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of unknown resistance is 6.0 ohm\n",
+ "The current drawn by the circuit is 1.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 59
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.9 Page no 337"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=10\n",
+ "b=7.0\n",
+ "c=5\n",
+ "d=4\n",
+ "e=8.0\n",
+ "\n",
+ "#Calculation\n",
+ "I1=(a+a)/(b+1)\n",
+ "I3=(c+(4*I1))/e\n",
+ "I2=(-a+(6*I3)+I1)/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Current I1=\",I1,\"A \\nI2=\",I2,\"A \\nI3=\",I3,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current I1= 2.5 A \n",
+ "I2= 1.875 A \n",
+ "I3= 1.875 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 70
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.11 Page no 338"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=28\n",
+ "b=5.0\n",
+ "c=2\n",
+ "\n",
+ "#Calculation\n",
+ "Rak=a/(b*c)\n",
+ "\n",
+ "#Result\n",
+ "print\"Total resistance from one end of vacant edge to other end is\", Rak,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total resistance from one end of vacant edge to other end is 2.8 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.12 Page no 345"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=10\n",
+ "l2=68.5\n",
+ "l1=58.3\n",
+ "\n",
+ "#Calculation\n",
+ "X=R*(l2/l1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of X is\", round(X,1),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of X is 11.7 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.13 Page no 346"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=2 #ohm\n",
+ "R1=2.4 #ohm\n",
+ "V=4 #V\n",
+ "E=1.5\n",
+ "\n",
+ "#Calculation\n",
+ "R11=R+R1\n",
+ "I=V/R11\n",
+ "Vab=I*R\n",
+ "K=Vab\n",
+ "l=E/K\n",
+ "\n",
+ "#Result\n",
+ "print\"Length for zero galvanometer deflection is\", l,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Length for zero galvanometer deflection is 0.825 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 85
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.15 Page no 347"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l1=33.7\n",
+ "l2=51.9\n",
+ "\n",
+ "#Calculation\n",
+ "S1=l1/(100-l1)\n",
+ "s11=l2/(100-l2)\n",
+ "s=((s11*12)/S1)-12\n",
+ "R=s*S1\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of R is\", round(R,2),\"ohm \\nValue of S is\",round(s,1),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of R is 6.85 ohm \n",
+ "Value of S is 13.5 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.16 Page no 347"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=0.4\n",
+ "b=0.6\n",
+ "lab=10\n",
+ "\n",
+ "#Calculation\n",
+ "K=a/b\n",
+ "Vab=K*lab\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Potentila gradient along AB is\",round(K,2),\"V/m\"\n",
+ "print \"(ii) P.D between point A and B is\",round(Vab,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Potentila gradient along AB is 0.67 V/m\n",
+ "(ii) P.D between point A and B is 6.67 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 103
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.17 Page no 348"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=990 #ohm\n",
+ "R=10.0 #ohm\n",
+ "E=2\n",
+ "l=1000 #mm\n",
+ "l1=400 #mm\n",
+ "\n",
+ "#Calculation\n",
+ "Rt=R1+R\n",
+ "I=E/Rt\n",
+ "pd=I*R\n",
+ "K=pd/l\n",
+ "pd1=K*l1\n",
+ "\n",
+ "#Result\n",
+ "print\"e.m.f. generated by the thermocouple is\", pd1,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "e.m.f. generated by the thermocouple is 0.008 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 111
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.18 Page no 348"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "AB=600 #cm\n",
+ "AC=500.0 #cm\n",
+ "l=40*10**-3 #A\n",
+ "E=2\n",
+ "r=10\n",
+ "\n",
+ "#Calculation\n",
+ "R=2*AB/(AC*l)\n",
+ "K=2/AC\n",
+ "AC1=AC-r\n",
+ "pd=K*AC1\n",
+ "Iv=(E-pd)/r\n",
+ "R1=pd/Iv\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The resistance of the whole wire is\", R,\"ohm\"\n",
+ "print\"(ii) Reading of voltmeter is\", pd,\"V\"\n",
+ "print\"(iii) Resistance of the voltmeter is\",R1,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The resistance of the whole wire is 60.0 ohm\n",
+ "(ii) Reading of voltmeter is 1.96 V\n",
+ "(iii) Resistance of the voltmeter is 490.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 124
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.20 Page no 350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=6\n",
+ "b=2\n",
+ "\n",
+ "#Calculation\n",
+ "R1=a/((b*b)-1)\n",
+ "R2=b*R1\n",
+ "\n",
+ "#Result\n",
+ "print\"Resistance R1 is\", R1,\"ohm \\nR2 is\",R2,\"Ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance R1 is 2 ohm \n",
+ "R2 is 4 Ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 130
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.21 Page no 350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=20 #ohm\n",
+ "L=10 #m\n",
+ "pd=10**-3 #V/m\n",
+ "V=10**-2 #Volts\n",
+ "\n",
+ "#Calculation\n",
+ "I=V/R\n",
+ "R11=(2/I)-R\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of resistance is\", R11,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of resistance is 3980.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 134
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter7_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter7_2.ipynb new file mode 100644 index 00000000..74f3f441 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter7_2.ipynb @@ -0,0 +1,735 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:47b2c0fcc74d4ba925e8938987dfe5c551c445c65c0145d8b28ae4df323cfc30"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 7 Heating effect of electric curent"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.1 Page no 373"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=240 #V\n",
+ "P=60\n",
+ "P1=100\n",
+ "\n",
+ "#Calculation\n",
+ "R=V**2/P\n",
+ "R1=V**2/P1\n",
+ "\n",
+ "#Result\n",
+ "print\"Resistance of a bulb for 60 W is\", R,\"ohm and for 100 W is\",R1,\"ohm\"\n",
+ "print\"Hence The resistance of 60 W,240 V bulb is more than that of 100 W,240 V bulb\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance of a bulb for 60 W is 960 ohm and for 100 W is 576 ohm\n",
+ "Hence The resistance of 60 W,240 V bulb is more than that of 100 W,240 V bulb\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.2 Page no 373"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=230 #v\n",
+ "P=100\n",
+ "t=20*60\n",
+ "V1=115 #V\n",
+ "\n",
+ "#Calculation\n",
+ "R=V**2/P\n",
+ "E=(V1**2*t)/R\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat and light energy is\", E,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat and light energy is 30000 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.3 Page no 374"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=500 #W\n",
+ "V=200.0 #V\n",
+ "V1=240\n",
+ "\n",
+ "#Calculation\n",
+ "I=P/V\n",
+ "R=V1-V\n",
+ "R1=R/I\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of R=\",R1,\"ohm\"\n",
+ "print\"Current in a circuit is\",I,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R= 16.0 ohm\n",
+ "Current in a circuit is 2.5 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.4 Page no 374"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P1=100.0 #W\n",
+ "P=1100.0 #W\n",
+ "V=250\n",
+ "\n",
+ "#Calculation\n",
+ "P2=P-P1\n",
+ "R=V**2/P2\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of unknown resistance is\", R,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of unknown resistance is 62.5 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.6 Page no 374"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=220\n",
+ "P=200.0\n",
+ "P1=100\n",
+ "\n",
+ "#Calculation\n",
+ "R1=V**2/P\n",
+ "R2=V**2/P1\n",
+ "H=R1/R2\n",
+ "\n",
+ "#Result\n",
+ "print\"The ratio of heats genetated in them is\", H"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of heats genetated in them is 0.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.7 Page no 375"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=1\n",
+ "c=1\n",
+ "a=100 #W\n",
+ "b=15\n",
+ "t=7.5 #second\n",
+ "P=1 #KW\n",
+ "C=860 #Kcal\n",
+ "\n",
+ "#Calculation\n",
+ "A=m*c*(a-b)\n",
+ "B=P*t/60.0\n",
+ "D=B*C\n",
+ "n=A*a/D\n",
+ "\n",
+ "#Result\n",
+ "print\"Efficiency of the kettle is\", round(n,1),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Efficiency of the kettle is 79.1 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.8 Page no 375"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P1=9 #W\n",
+ "R1=8\n",
+ "R2=12.0\n",
+ "\n",
+ "#Calculation\n",
+ "P2=(P1*R1)/R2\n",
+ "\n",
+ "#Result\n",
+ "print\"Power dissipated in 12 ohm resistor is\", P2,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power dissipated in 12 ohm resistor is 6.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.9 Page no 375"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "H1=10\n",
+ "a=5.0\n",
+ "b=4.2\n",
+ "\n",
+ "#Calculation\n",
+ "I1=(H1*b)/(a*4)\n",
+ "A=I1*4/b\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat generated in 4 ohm resistor is\", A,\"cal/sec\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat generated in 4 ohm resistor is 2.0 cal/sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.10 Page no 375"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=12 #V\n",
+ "I=1 #A\n",
+ "r=0.5 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "P1=E*I\n",
+ "P2=I**2*r\n",
+ "P=P1-P2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Rate of consumption of chemical energy is\", P1,\"W\"\n",
+ "print\"(ii) Rate Of energy dissipated inside the battery is\",P2,\"W\"\n",
+ "print\"(iv) Rate of energy dissipated in the resistor is\", P,\"W\"\n",
+ "print\"(v) Power output of the source is\",P,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Rate of consumption of chemical energy is 12 W\n",
+ "(ii) Rate Of energy dissipated inside the battery is 0.5 W\n",
+ "(iv) Rate of energy dissipated in the resistor is 11.5 W\n",
+ "(v) Power output of the source is 11.5 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 55
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.11 Page no 376"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=110 #W\n",
+ "P1=100 #W\n",
+ "n=5\n",
+ "V=220 #V\n",
+ "t=2 #hours\n",
+ "n1=4\n",
+ "P2=1120 #W\n",
+ "m=1.5 #per KWh\n",
+ "\n",
+ "#Calculation\n",
+ "W=n*P1\n",
+ "W1=V*t\n",
+ "W2=n1*P\n",
+ "W3=W+W1+W2+P2\n",
+ "E=(W3*t)*10**-3\n",
+ "E2=E*30\n",
+ "B=m*E2\n",
+ "\n",
+ "#Result\n",
+ "print\"Electricity bill for the month of september is\", B,\"Rs\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electricity bill for the month of september is 225.0 Rs\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.12 Page no 376"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=220 #V\n",
+ "P=60.0 #W\n",
+ "P1=85 #w\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "R=V**2/P\n",
+ "V1=math.sqrt(P1*R)\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum voltage is\", round(V1,1),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum voltage is 261.9 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 75
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.13 Page no 376"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=200 #V\n",
+ "P=500.0 #W\n",
+ "V1=160 #v\n",
+ "\n",
+ "#Calculation\n",
+ "R=V**2/P\n",
+ "H=V1**2/R\n",
+ "P1=P-H\n",
+ "H1=P1*100/P\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat percentage is\", H1,\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat percentage is 36.0 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.14 Page no 376"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P1=500 #W\n",
+ "P2=100\n",
+ "\n",
+ "#Calculation\n",
+ "R=P1/P2\n",
+ "\n",
+ "#Result\n",
+ "print\"Since P1'=5P2', 100W bulb will glow brighter\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Since P1'=5P2', 100W bulb will glow brighter\n"
+ ]
+ }
+ ],
+ "prompt_number": 86
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.15 Page no 377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=900\n",
+ "w=100.0\n",
+ "c=1\n",
+ "a=80\n",
+ "b=4.2\n",
+ "V=210 #V\n",
+ "x=12\n",
+ "y=60\n",
+ "\n",
+ "#Calculation\n",
+ "Hout=(m+w)*c*a\n",
+ "Hin=(V*x*y)/b\n",
+ "Hin1=90/w*Hin\n",
+ "I=Hout/Hin1\n",
+ "\n",
+ "#Result\n",
+ "print\"Strength of the current is\", round(I,3),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Strength of the current is 2.469 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.16 Page no 377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=0.8\n",
+ "\n",
+ "#Calculation\n",
+ "H=a**2\n",
+ "H1=(1-H)*100\n",
+ "\n",
+ "#Result\n",
+ "print\"Decreased percentage is\", H1,\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Decreased percentage is 36.0 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 99
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.17 Page no 377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=14\n",
+ "b=60\n",
+ "c=24\n",
+ "d=7.0\n",
+ "\n",
+ "#Calculation\n",
+ "t=a*b/60.0\n",
+ "t1=(c/d)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Time in series is\", t,\"minute\"\n",
+ "print\"(ii) Time in parallel is\",round(t1,2),\"minute\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Time in series is 14.0 minute\n",
+ "(ii) Time in parallel is 3.43 minute\n"
+ ]
+ }
+ ],
+ "prompt_number": 108
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.19 Page no 378"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=0.5\n",
+ "R=100\n",
+ "t=30\n",
+ "a=4.2\n",
+ "m=200 #g\n",
+ "w=10 #g\n",
+ "\n",
+ "#Calculation\n",
+ "H=I**2*R*t*60/a\n",
+ "A=H/(m+w)\n",
+ "\n",
+ "#Result\n",
+ "print\"The rise of temperature is\", round(A,2),\"degree C\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The rise of temperature is 51.02 degree C\n"
+ ]
+ }
+ ],
+ "prompt_number": 113
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.20 Page no 379"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "c=4.2 #KJ/Kg/C\n",
+ "m=0.2 #Kg\n",
+ "a=90\n",
+ "b=20\n",
+ "t=30\n",
+ "V=230\n",
+ "\n",
+ "#calculation\n",
+ "d=a-b\n",
+ "H=c*m*d\n",
+ "P=H/t\n",
+ "I=P/V\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of current is\", round(I*10**3,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of current is 8.52 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 120
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter8_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter8_2.ipynb new file mode 100644 index 00000000..baadbaea --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter8_2.ipynb @@ -0,0 +1,643 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:db58e543ef86cd601814ac49a8404db7a1403e7140977a41ff4c6b1fc2ae61b9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 8 Magnetic field due to electric current"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.1 Page no 417"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=1.6*10**-19 #c\n",
+ "B=0.1 #T\n",
+ "v=5.0*10**6 #m/s\n",
+ "a=90 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Fm=q*v*B*math.sin(a)\n",
+ "\n",
+ "#Result\n",
+ "print\"Force on the proton is\", round(Fm*10**14,1)*10**-14,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force on the proton is 7.2e-14 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.2 Page no 417"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=1.0*10**29 #m**-3\n",
+ "e=1.6*10**-19 #C\n",
+ "A=2*10**-6 #m**2\n",
+ "I=5 #A\n",
+ "B=0.15 #T\n",
+ "a=90 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Vd=I/(n*e*A)\n",
+ "Fm=e*Vd*B*math.sin(a)\n",
+ "\n",
+ "#Result\n",
+ "print\"Force acting on each electron is\", round(Fm*10**24,2)*10**-24,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force acting on each electron is 3.35e-24 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.3 Page no 417"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=2*1.6*10**-19 #C\n",
+ "v=6*10**5 #m/s\n",
+ "B=0.2 #T\n",
+ "a=90 #degree\n",
+ "m=6.65*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Fm=q*v*B*math.sin(a)\n",
+ "a=Fm/m\n",
+ "\n",
+ "#Result\n",
+ "print\"Force on alpha particle is\", round(Fm*10**14,2)*10**-14,\"N\"\n",
+ "print\"Acceleration of alpha particle is\",round(a*10**-12,2)*10**12,\"m/s**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force on alpha particle is 3.43e-14 N\n",
+ "Acceleration of alpha particle is 5.16e+12 m/s**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.4 Page no 426"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=60 #degree\n",
+ "u=4*3.14*10**-7 #T/A m\n",
+ "Bc=2\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=(Bc/2.0)/(math.tan(60)*180/3.14)\n",
+ "B1=(10**-7*math.tan(60)*(math.sin(60*180/3.14)+math.sin(60*180/3.14)))*10\n",
+ "B=3*B1\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic fieldat the centroid of the triangle is\", round(B*10**7,0),\"*10**-7 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic fieldat the centroid of the triangle is 10.0 *10**-7 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 64
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.5 Page no 426"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=20\n",
+ "I=1 #A\n",
+ "r=0.08 #m\n",
+ "u=4*3.14*10**-7 #T/A m\n",
+ "\n",
+ "#Calculation\n",
+ "B=u*n*I/(2*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the magnetic field is\", B*10**4,\"*10*4 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the magnetic field is 1.57 *10*4 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.6 Page no 426"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=10**-7\n",
+ "I=10*10**-2 #A\n",
+ "r=0.5\n",
+ "\n",
+ "#Calculation\n",
+ "B=u*I/r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic field on Y axis is\", B,\"K^ T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic field on Y axis is 4e-08 K^ T\n"
+ ]
+ }
+ ],
+ "prompt_number": 71
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.7 Page no 426"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=5 #A\n",
+ "l=0.01 #m\n",
+ "a=45 #degree\n",
+ "r=2 #m\n",
+ "u=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=(u*I*l*math.sin(a)*180/3.14)/r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic field is\", round(B*10**8,1)*10**-10,\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic field is 6.1e-10 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.8 Page no 427"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=4*3.14*10**-7 #T/A m\n",
+ "n=20\n",
+ "I=12 #A\n",
+ "r=0.1 #m\n",
+ "\n",
+ "#Calculation\n",
+ "B=u*n*I/(2*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic field at the centre of coil is\", round(B*10**3,1),\"*10**-3 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic field at the centre of coil is 1.5 *10**-3 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 87
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.12 Page no 429"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.02 #m\n",
+ "\n",
+ "#Calculation\n",
+ "B=u*I/(4*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"The magnitude of magnetic field is\", round(B*10**4,2),\"*10**-4 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The magnitude of magnetic field is 1.88 *10**-4 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 92
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.13 Page no 429"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=4*10**6\n",
+ "r=0.5*10**-10\n",
+ "e=1.6*10**-19\n",
+ "t=1\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "f=v/(2.0*math.pi*r)\n",
+ "I=f*e/t\n",
+ "B=u*I/(2*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic field produced by the electrons is\", round(B,1),\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic field produced by the electrons is 25.6 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 98
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.15 Page no 430"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=100\n",
+ "I=5 #A\n",
+ "r=0.1 #m\n",
+ "x=0.05\n",
+ "\n",
+ "#Calculation\n",
+ "B=u*n*I/(2*r)\n",
+ "B1=(u*n*I*r**2)/(2.0*(r**2+x**2)**1.5)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Magnetic field at the centre of the coil is\",B*10**3,\"*10**-3 T\"\n",
+ "print\"(ii) The magnetic field at the point on the axis of the coil is\",round(B1*10**3,2),\"*10**-3 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnetic field at the centre of the coil is 3.14 *10**-3 T\n",
+ "(ii) The magnetic field at the point on the axis of the coil is 2.25 *10**-3 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 109
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.18 Page no 431"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=5*10**-2\n",
+ "I=50\n",
+ "e=1.6*10**-19\n",
+ "B1=10**7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=u*I/(2*math.pi*a)\n",
+ "F=e*B1*B\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Force on electron when velocity is towards the wire\", round(F*10**16,1)*10**-16,\"N\"\n",
+ "print\"(ii) Force on electron when velocity is parallel to the wire\", round(F*10**16,1)*10**-16,\"N\"\n",
+ "print\"(iii) Force on electron when velocity is perpendicular to the wire is zero\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Force on electron when velocity is towards the wire 3.2e-16 N\n",
+ "(ii) Force on electron when velocity is parallel to the wire 3.2e-16 N\n",
+ "(iii) Force on electron when velocity is perpendicular to the wire is zero\n"
+ ]
+ }
+ ],
+ "prompt_number": 118
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.20 Page no 432"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19\n",
+ "f=6.8*10**15\n",
+ "r=0.51*10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=e*f\n",
+ "B=(u*I)/(2*r)\n",
+ "M=1*I*math.pi*r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"The effective dipole moment is\",round(M*10**24,0)*10**-24,\"Am**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The effective dipole moment is 9e-24 Am**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 127
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.22 Page no 439"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=5*850/1.23\n",
+ "I=5.57 #A\n",
+ "\n",
+ "#calculation\n",
+ "u=4*math.pi*10**-7\n",
+ "B=u*n*I\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of magnetic field is\", round(B*10**3,1),\"*10**-3 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of magnetic field is 24.2 *10**-3 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.23 Page no 439"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r1=20\n",
+ "r2=25\n",
+ "I=2 #a\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=(r1+r2)/2.0\n",
+ "l=(2*math.pi*r)*10**-2\n",
+ "n=1500/l\n",
+ "B=u*n*I\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Magnetic field inside the toroid is\", round(B,3),\"T\"\n",
+ "print\"(ii) magnetic field outside the toroid is zero\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnetic field inside the toroid is 0.003 T\n",
+ "(ii) magnetic field outside the toroid is zero\n"
+ ]
+ }
+ ],
+ "prompt_number": 138
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.25 Page no 440"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=2 #A\n",
+ "R=5*10**-2 #m\n",
+ "r=3*10**-2 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=u*I*r/(2*math.pi*R**2)\n",
+ "\n",
+ "#Result\n",
+ "print round(B*10**6,1),\"*10**-6 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "4.8 *10**-6 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 142
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter9_2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter9_2.ipynb new file mode 100644 index 00000000..7669d0e6 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter9_2.ipynb @@ -0,0 +1,1740 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:02dc05916beb2a686e89acb729415599e9656d95fc169884e1d4a92b0e8ee888"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 9 Motion of charged particles in electric and magnetic motion"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.1 Page no 472"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=90 #V\n",
+ "d=2.0*10**-2\n",
+ "e=1.8*10**11\n",
+ "x=5*10**-2\n",
+ "v=10**7\n",
+ "\n",
+ "#Calculation\n",
+ "E=V/d\n",
+ "a=e*E\n",
+ "t=x/v\n",
+ "y=0.5*a*t**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Transverse deflection produced by electric field is\", round(y*10**2,1),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Transverse deflection produced by electric field is 1.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.2 Page no 473"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=500\n",
+ "d=2*10**-2 #m\n",
+ "v=3*10**7\n",
+ "x=6*10**-2\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "E=V/d\n",
+ "a=E*e\n",
+ "t=x/v\n",
+ "v1=a*t\n",
+ "T=v1/v\n",
+ "A=math.atan(T)*180.0/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle is\", round(A,1),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle is 16.7 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.3 Page no 474"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "x=10*10**-2\n",
+ "v=3*10**7\n",
+ "S=1.76*10**-3\n",
+ "a=1800\n",
+ "\n",
+ "#Calculation\n",
+ "t=x/v\n",
+ "e=S*2/(a*t**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Specific charge of the electron is\", e,\"C/Kg\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Specific charge of the electron is 1.76e+11\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.4 Page no 478"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**-31\n",
+ "v=3*10**7\n",
+ "q=1.6*10**-19 #C\n",
+ "B=6*10**-4\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=m*v/(q*B)\n",
+ "f=q*B/(2.0*math.pi*m)\n",
+ "E=(0.5*m*v**2)/1.6*10**-16\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy is\", round(E*10**32,2),\"Kev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy is 2.53 Kev\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.5 Page no 479"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**-31\n",
+ "e=1.6*10**-19\n",
+ "V=100\n",
+ "B=0.004\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=math.sqrt(2*m*e*V)/(e*B)\n",
+ "\n",
+ "#Result\n",
+ "print\"Radius of the path is\", round(r*10**3,1),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of the path is 8.4 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 50
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.6 Page no 479"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=1.67*10**-27\n",
+ "v=4*10**5\n",
+ "a=60\n",
+ "q=1.6*10**-19\n",
+ "B=0.3\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=(m*v*math.sin(a*3.14/180.0))/q*B\n",
+ "P=v*math.cos(a*3.14/180.0)*((2*math.pi*m)/(q*B))\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Radius of the helical path is\",round(r*10**3,1),\"cm\"\n",
+ "print\"(ii) Pitch of helix is\", round(P*10**2,2),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Radius of the helical path is 1.1 cm\n",
+ "(ii) Pitch of helix is 4.38 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 70
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.7 Page no 479"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=5*10**6 #ev\n",
+ "e=1.6*10**-19\n",
+ "m=1.6*10**-27\n",
+ "B=1.5\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "v=math.sqrt((2*M*e)/m)\n",
+ "F=q*v*B*math.sin(90*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the force is\", round(F*10**12,2)*10**-12,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the force is 7.59e-12 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.8 Page no 480"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=1.67*10**-27 #Kg\n",
+ "v=4*10**5\n",
+ "B=0.3 #T\n",
+ "q=1.6*10**-19 #C\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=m*v*math.sin(60*3.14/180.0)/(q*B)\n",
+ "P=2*math.pi*r*1/(math.tan(60*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"Pitch of the helix is\", round(P*10**2,2),\"cm\"\n",
+ "print\"Radius of helical path is\",round(r*10**2,3),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Pitch of the helix is 4.38 cm\n",
+ "Radius of helical path is 1.205 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9 Page no 480"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=3.2*10**-19\n",
+ "B=1.2\n",
+ "r=0.45\n",
+ "m=6.8*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "v=(q*B*r)/m\n",
+ "f=v/(2.0*math.pi*r)\n",
+ "K=(0.5*m*v**2)/(1.6*10**-19)\n",
+ "V=K/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Required potentila difference is\", round(V*10**-6,0),\"*10**6 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Required potentila difference is 7.0 *10**6 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.10 Page no 480"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=4\n",
+ "u=10**-7\n",
+ "a=0.2 #m\n",
+ "v=4*10**6\n",
+ "q=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "B=(u*2*I)/a\n",
+ "F=q*v*B\n",
+ "\n",
+ "#Result\n",
+ "print\"Force is\", F,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force is 2.56e-18 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.11 Page no 481"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19\n",
+ "a=10**6\n",
+ "\n",
+ "#Calculation\n",
+ "q=2*e\n",
+ "F=q*a\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude force acting on the particle is\", F"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude force acting on the particle is 3.2e-13\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.13 Page no 482"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=3.4*10**4 #V/m\n",
+ "B=2*10**-3 #Wb/m**2\n",
+ "m=9.1*10**-31\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "v=E/B\n",
+ "r=(m*v)/(e*B)\n",
+ "\n",
+ "#Result\n",
+ "print\"Radius of the circular path is\", round(r*10**2,1),\"*10**-2 m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of the circular path is 4.8 *10**-2 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.14 Page no 482"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=600 #V\n",
+ "d=3*10**-3 #m\n",
+ "v=2*10**6 #m/s\n",
+ "\n",
+ "#Calculation\n",
+ "B=V/(d*v)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of magnetic field is\", B,\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of magnetic field is 0.1 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.15 Page no 487"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=1.6*10**-19 #c\n",
+ "B=2 #T\n",
+ "m=1.66*10**-27 #Kg\n",
+ "K=5*10**6\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "f=(q*B)/(2.0*math.pi*m)\n",
+ "v=math.sqrt((2*K*q)/m)\n",
+ "r=(m*v)/(q*B)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The frequency needed for applied alternating voltage is\", round(f*10**-7,0),\"*10**7 HZ\"\n",
+ "print\"(ii) Radius of the cyclotron is\",round(r,2),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The frequency needed for applied alternating voltage is 3.0 *10**7 HZ\n",
+ "(ii) Radius of the cyclotron is 0.16 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.16 Page no 487"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=1.7 #T\n",
+ "q=1.6*10**-19 #c\n",
+ "r=0.5\n",
+ "m=1.66*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "K=((B**2*q**2*r**2)/(2.0*m))/q\n",
+ "\n",
+ "#Result\n",
+ "print\"Kinetic energy of proton is\", round(K*10**-6,0),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Kinetic energy of proton is 35.0 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.17 Page no 487"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=0.8\n",
+ "q=3.2*10**-19 #C\n",
+ "d=1.2\n",
+ "m=4*1.66*10**-27 #Kg\n",
+ "a=1.60*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=d/2.0\n",
+ "K=(B**2*q**2*r**2)/(2.0*m*a)\n",
+ "v=(q*B*r)/m\n",
+ "f=(q*B)/(2.0*math.pi*m)\n",
+ "\n",
+ "#Result\n",
+ "print\"Frequency of alternating voltage is\", round(f*10**-7,2),\"*10**7 HZ\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Frequency of alternating voltage is 0.61 *10**7 HZ\n"
+ ]
+ }
+ ],
+ "prompt_number": 61
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.18 Page no 488"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=1.6*10**-19 #C\n",
+ "r=0.6 #m\n",
+ "m=1.67*10**-27 #Kg\n",
+ "f=10**7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=(2*math.pi*m*f)/q\n",
+ "K=((B**2*q**2*r**2)/(2.0*m))/1.6*10**-13\n",
+ "\n",
+ "#Result\n",
+ "print\"Kinetic energy of the protons is\", round(K*10**26,1),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Kinetic energy of the protons is 7.4 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 67
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.19 Page no 493"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=5 #A\n",
+ "l=0.06 #m\n",
+ "B=0.02 #T\n",
+ "a=90\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "F=I*B*l*math.sin(a*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Force is\", round(F,3),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force is 0.006 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 71
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.20 Page no 494"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=0.2 #Kg\n",
+ "I=2 #A\n",
+ "l=1.5 #m\n",
+ "g=9.8\n",
+ "\n",
+ "#Calculation\n",
+ "B=(m*g)/(I*l)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the magnetic field is\", round(B,2),\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the magnetic field is 0.65 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 76
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.21 Page no 494"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r=0.002 #m\n",
+ "m=0.05\n",
+ "g=9.8\n",
+ "\n",
+ "#Calculation\n",
+ "u=4*math.pi*10**-7\n",
+ "f=u/(2*math.pi*r)\n",
+ "f1=m*g\n",
+ "I=math.sqrt(f1*f**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Current in each wire is\", I,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current in each wire is 70.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.22 Page no 494"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.04 #m\n",
+ "I1=20\n",
+ "I2=16\n",
+ "l=0.15\n",
+ "r1=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "u=4*math.pi*10**-7\n",
+ "F1=(u*I1*I2*l)/(2.0*math.pi*r)\n",
+ "F2=(u*I1*I2*l)/(2.0*math.pi*r1)\n",
+ "F=F1-F2\n",
+ "\n",
+ "#Result\n",
+ "print\"Net force on the loop is\", F*10**4,\"*10**-4 N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Net force on the loop is 1.44 *10**-4 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 89
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.23 Page no 495"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=0.3 #Kg\n",
+ "a=30 #degree\n",
+ "B=0.15 #T\n",
+ "g=9.8 #m/s**2\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=(m*g*math.tan(a*3.14/180.0))/B\n",
+ "\n",
+ "#Result\n",
+ "print\"value of current is\", round(I,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of current is 11.31 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 94
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.24 Page no 495"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=3*10**-5 #T\n",
+ "I=1 #A\n",
+ "\n",
+ "#Calculation\n",
+ "F=I*B*math.sin(90)\n",
+ "\n",
+ "#Result\n",
+ "print\"The direction of the force is downward i.e\", round(F*10**5,0),\"*10**-5 N/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The direction of the force is downward i.e 3.0 *10**-5 N/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 99
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Example 9.25 Page no 495"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=1.2*10**-3\n",
+ "B=0.6 #T\n",
+ "g=9.8 #m/s**2\n",
+ "r=0.05\n",
+ "b=3.8\n",
+ "\n",
+ "#Calculation\n",
+ "I=(m*g)/B\n",
+ "R=r*b\n",
+ "V=I*R\n",
+ "\n",
+ "#Result\n",
+ "print\"Potentila difference is\", round(V*10**3,1),\"*10**-3 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Potentila difference is 3.7 *10**-3 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 105
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.26 Page no 496"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I2=10 #A\n",
+ "r=0.1 #m\n",
+ "l=2 #m\n",
+ "I1=2\n",
+ "I2=10\n",
+ "r=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "u=4*math.pi*10**-7\n",
+ "F=u*I1*I2*I1/(2.0*math.pi*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"Force on small conductor\", F,\"N\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force on small conductor 8e-05 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 109
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.27 Page no 500"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=10**-3 #m**\n",
+ "n=10\n",
+ "I=2 #A\n",
+ "B=0.1 #T\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "t=n*I*A*B*math.cos(0)\n",
+ "t1=n*I*A*B*math.cos(60*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Torque when magnetic field is parallel to the field\", round(t*10**3,0),\"*10**-3 Nm\"\n",
+ "print\"(ii) Torque when magnetic field is perpendicular to the field is zero\"\n",
+ "print\"(iii) Torque when magnetic field is 60 degree to the field is\",round(t1*10**3,1),\"*10**-3 Nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Torque when magnetic field is parallel to the field 2.0 *10**-3 Nm\n",
+ "(ii) Torque when magnetic field is perpendicular to the field is zero\n",
+ "(iii) Torque when magnetic field is 60 degree to the field is 1.0 *10**-3 Nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 121
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.28 Page no 500"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=7\n",
+ "I=10\n",
+ "B=100*10**-4\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=math.pi*r**2\n",
+ "t=I*A*B\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of maximum torque is\", round(t*10**-1,2),\"*10**-3 Nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of maximum torque is 1.54 *10**-3 Nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 127
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.29 Page no 501"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=10\n",
+ "I=0.06\n",
+ "r=0.05\n",
+ "n=1000\n",
+ "I2=25\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=math.pi*r**2\n",
+ "M=N*I*A\n",
+ "u=4*math.pi*10**-7\n",
+ "B=u*n*I2\n",
+ "t=M*B*math.sin(45*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Torgue is\", round(t*10**4,2),\"*10**-4 Nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Torgue is 1.05 *10**-4 Nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 134
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.30 Page no 501"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=100\n",
+ "l=3.2 \n",
+ "r=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "u=4*math.pi*10**-7\n",
+ "B=(u*n*l)/(2.0*r)\n",
+ "M=n*l*math.pi*r**2\n",
+ "t=M*B*math.sin(0)\n",
+ "t1=(M*B*math.sin(90*3.14/180.0))*10**3\n",
+ "w=math.sqrt((2*M*B*10**3)/r)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Field at the centre of the coil is\", round(B*10**3,0),\"*10**-3 T\"\n",
+ "print\"(b) Magnetic moment of the coil is\",round(M,0),\"Am**2\"\n",
+ "print\"(c) Magnitude of the torque on the coil in the initial position is\",t\n",
+ "print\" Magnitude of the torque on the coil in the final position is\",round(t1,0),\"Nm\"\n",
+ "print \"(d) Angular speed acquired by the coil is\",round(w,0),\"rad/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Field at the centre of the coil is 2.0 *10**-3 T\n",
+ "(b) Magnetic moment of the coil is 10.0 Am**2\n",
+ "(c) Magnitude of the torque on the coil in the initial position is 0.0\n",
+ " Magnitude of the torque on the coil in the final position is 20.0 Nm\n",
+ "(d) Angular speed acquired by the coil is 20.0 rad/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.31 Page no 505"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=125\n",
+ "I=20*10**-3 #A\n",
+ "B=0.5 #T\n",
+ "A=400*10**-6 #m**2\n",
+ "K=40*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "T=n*I*B*A\n",
+ "a=T/K\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Torque exerted is\", T*10**4,\"*10**-4 Nm\"\n",
+ "print\"(ii) Angular deflection of the coil is\", a,\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Torque exerted is 5.0 *10**-4 Nm\n",
+ "(ii) Angular deflection of the coil is 12.5 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.32 Page no 505"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "K=3*10**-9 #Nm/deg\n",
+ "a=36\n",
+ "n=60\n",
+ "B=9*10**-3 #T\n",
+ "A=5*10**-5 #m**2\n",
+ "\n",
+ "#Calculation\n",
+ "I=(K*a)/(n*B*A)\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum current is\", I*10**3,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum current is 4.0 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.33 Page no 506"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=30\n",
+ "B=0.25 #T\n",
+ "A=1.5*10**-3\n",
+ "K=10**-3\n",
+ "\n",
+ "#Calculation\n",
+ "S=(n*B*A)/K\n",
+ "\n",
+ "#Result\n",
+ "print\"Current sensitivity of the galvanometer is\", S,\"degree/A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current sensitivity of the galvanometer is 11.25 degree/A\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.35 Page no 509"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ig=0.015 #A\n",
+ "G=5\n",
+ "I=1\n",
+ "V=15\n",
+ "\n",
+ "#Calculation\n",
+ "S=(Ig*G)/(I-Ig)\n",
+ "R=G*S/(G+S)\n",
+ "R1=(V/Ig)-G\n",
+ "R2=R1+G\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Resistance of ammeter of range 0-1 A is\", R,\"ohm\"\n",
+ "print\"(ii) Resistance of ammeter of range 0-15 A is\", R2,\"ohm\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Resistance of ammeter of range 0-1 A is 0.075 ohm\n",
+ "(ii) Resistance of ammeter of range 0-15 A is 1000.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.36 Page no 510"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=75 #mV\n",
+ "Ig=0.025 #A\n",
+ "I=25 #mA\n",
+ "I1=100\n",
+ "V1=750\n",
+ "\n",
+ "#Calculation\n",
+ "G=V/I\n",
+ "S=(Ig*G)/(I1-Ig)\n",
+ "R=(V1/Ig)-G\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Resistance for an ammeter of range 0-100 A is\", round(S,5),\"ohm\"\n",
+ "print\"(ii) Resistance for an ammeter of range 0-750 A is\", round(R,5),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Resistance for an ammeter of range 0-100 A is 0.00075 ohm\n",
+ "(ii) Resistance for an ammeter of range 0-750 A is 29997.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 58
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.37 Page no 510"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rg=60\n",
+ "R1=3.0\n",
+ "rs=0.02\n",
+ "\n",
+ "#Calculation\n",
+ "Rt=Rg+R1\n",
+ "I=R1/Rt\n",
+ "Rm=(Rg*rs)/(Rg+rs)\n",
+ "R2=Rm+R1\n",
+ "I1=R1/R2\n",
+ "I2=R1/R1\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The value of current is\", round(I,3),\"A\"\n",
+ "print\"(ii) The value of current is\", round(I1,2),\"A\"\n",
+ "print\"(iii) The value of current is\",I2,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The value of current is 0.048 A\n",
+ "(ii) The value of current is 0.99 A\n",
+ "(iii) The value of current is 1.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.38 Page no 511"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=100\n",
+ "v=1\n",
+ "a=1980\n",
+ "\n",
+ "#Calculation\n",
+ "Rm=a/(V-v)\n",
+ "\n",
+ "#Result\n",
+ "print\"Resistance of the voltmeter is\", Rm,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance of the voltmeter is 20 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 77
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.39 Page no 511"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=1200.0 #ohm\n",
+ "R2=600 #ohm\n",
+ "Vab=5 #V\n",
+ "V=35\n",
+ "\n",
+ "#Calculation\n",
+ "Rp=(R1*R2)/(R1+R2)\n",
+ "I=Vab/Rp\n",
+ "pd=V-Vab\n",
+ "R=pd/I\n",
+ "\n",
+ "#Result\n",
+ "print\"value of unknown resistance is\", R,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of unknown resistance is 2400.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 84
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.40 Page no 511"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=400 #ohm\n",
+ "R2=800.0\n",
+ "R3=10\n",
+ "V=6\n",
+ "R11=10000.0\n",
+ "R22=400\n",
+ "\n",
+ "#Calculation\n",
+ "Rt=R1+R2+R3\n",
+ "I=V/Rt\n",
+ "Rp=(R11*R22)/(R11+R22)\n",
+ "R=Rp+800\n",
+ "I1=V/R\n",
+ "Vab=I1*Rp\n",
+ "\n",
+ "#Result\n",
+ "print\"Hence the voltmeter will read\", round(Vab,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Hence the voltmeter will read 1.95 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.41 Page no 512"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=2 #V\n",
+ "R=2000.0 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "I=V/R\n",
+ "pd=I*R\n",
+ "\n",
+ "#Result\n",
+ "print\"Reading of ammeter is\", I*10**3,\"mA \\nReading of voltmeter is\",pd,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Reading of ammeter is 1.0 mA \n",
+ "Reading of voltmeter is 2.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 101
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.42 Page no 512"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=3\n",
+ "G=100\n",
+ "R=200.0\n",
+ "n=30\n",
+ "\n",
+ "#Calculation\n",
+ "Ig=E/(G+R)\n",
+ "K=(Ig/n)*10**6\n",
+ "\n",
+ "#Result\n",
+ "print\"Figure of merit of the galvanometer is\", round(K,1),\"micro A/division\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Figure of merit of the galvanometer is 333.3 micro A/division\n"
+ ]
+ }
+ ],
+ "prompt_number": 108
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.43 Page no 513"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V1=60 #ohm\n",
+ "V2=30\n",
+ "R=300.0\n",
+ "R1=1200\n",
+ "R2=400 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "V=V1-V2\n",
+ "I=V/R\n",
+ "R11=(R1*R)/(R1+R)\n",
+ "I=V1/(R11+R2)\n",
+ "V11=I*R11\n",
+ "\n",
+ "#Result\n",
+ "print\"Voltmeter will read\", V11,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltmeter will read 22.5 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 115
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.44 Page no 513"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=20.0 #K ohm\n",
+ "R2=1 #K ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Vr=(R*R2)/(R+R2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Voltmeter resistance is\", R,\"K ohm\"\n",
+ "print\"(ii) Voltmeter resistance is\",R2,\"K ohm\"\n",
+ "print\"(iii) Voltmeter resistance is\",round(Vr,2),\"K ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Voltmeter resistance is 20.0 K ohm\n",
+ "(ii) Voltmeter resistance is 1 K ohm\n",
+ "(iii) Voltmeter resistance is 0.95 K ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 123
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.45 Page no 514"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "s=20*10**-6\n",
+ "n=30\n",
+ "I=1 #A\n",
+ "G=25 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Ig=s*n\n",
+ "S=Ig*G/(1-Ig)\n",
+ "Ra=G*S/(G+S)\n",
+ "\n",
+ "#Result\n",
+ "print\"Resistance of ammeter is\",Ra,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance of ammeter is 0.015 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 128
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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