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| author | Trupti Kini | 2016-03-11 23:30:11 +0600 |
|---|---|---|
| committer | Trupti Kini | 2016-03-11 23:30:11 +0600 |
| commit | 2be8c9a4fe3e8680634083e03085ec4133f22e12 (patch) | |
| tree | 87f273631fb46ee30a9f2258a2fcb6a87dbb050d | |
| parent | 8b45c91b0d1cc1bf93d8653912e7068004eda3eb (diff) | |
| download | Python-Textbook-Companions-2be8c9a4fe3e8680634083e03085ec4133f22e12.tar.gz Python-Textbook-Companions-2be8c9a4fe3e8680634083e03085ec4133f22e12.tar.bz2 Python-Textbook-Companions-2be8c9a4fe3e8680634083e03085ec4133f22e12.zip | |
Added(A)/Deleted(D) following books
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch15.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch16.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch17.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch18.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch19.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch20.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch21.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch22.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch23.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch24.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch25.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch26.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch27.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch28.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch29.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch30.ipynb
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/1RefractionOfLaserLight.png
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/2PropertiesOfImage.png
A College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/3PositionOf1DarkFringe.png
A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/Chapter9_7.ipynb
A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter1_7.ipynb
A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter2_7.ipynb
A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter3_7.ipynb
A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter4_7.ipynb
A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter6_7.ipynb
A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter7_7.ipynb
A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter8_7.ipynb
A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter_5_7.ipynb
A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/1.2_2.png
A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/3.7_2.png
A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/6.7_2.png
A Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch11.ipynb
A Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch12.ipynb
A Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch13.ipynb
A Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch14.ipynb
A Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch3.ipynb
A Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch4.ipynb
A Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch5.ipynb
A Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch6.ipynb
A Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch7.ipynb
A Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch8.ipynb
A Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch9.ipynb
A Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/BiasingVoltage.png
A Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/InOutCrrnt.png
A Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/inAndOutCurrnt.png
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter10.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter28.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter29.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter1.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter11.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter12.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter13.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter14.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter15.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter16.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter17.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter18.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter19.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter2.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter20.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter21.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter22.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter23.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter24.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter25.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter26.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter27.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter3.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter4.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter5.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter6.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter7.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter8.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter9.ipynb
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/screenshots/image_1.png
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/screenshots/image_2.png
A principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/screenshots/image_3.png
A sample_notebooks/KhushbuPattani/chapter1_1.ipynb
_by_R._A._Serway_and_J._S._Faughn/screenshots/1RefractionOfLaserLight.png?id=2be8c9a4fe3e8680634083e03085ec4133f22e12){kind=link}
_by_R._A._Serway_and_J._S._Faughn/screenshots/2PropertiesOfImage.png?id=2be8c9a4fe3e8680634083e03085ec4133f22e12){kind=link}
_by_R._A._Serway_and_J._S._Faughn/screenshots/3PositionOf1DarkFringe.png?id=2be8c9a4fe3e8680634083e03085ec4133f22e12){kind=link}
78 files changed, 45110 insertions, 0 deletions
diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch15.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch15.ipynb new file mode 100644 index 00000000..1c77fdc3 --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch15.ipynb @@ -0,0 +1,179 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 15 : Electric forces and electric fields" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.1 Page No : 502" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The attractive force = 8.19e-08 N\n", + "The gravitational force = 3.61e-47 N\n" + ] + } + ], + "source": [ + "k_e=8.99*10**9\n", + "e=1.6*10**-19\n", + "r=5.3*10**-11\n", + "F_e= (k_e*e*e)/(r*r)\n", + "print \"The attractive force = %0.2e N\"%F_e\n", + "G=6.67*10**-11\n", + "m_e=9.11*10**-31\n", + "m_p=1.67*10**-27\n", + "F_g=(G*m_e*m_p)/(r*r)\n", + "print \"The gravitational force = %0.2e N\"%F_g" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.2 Page No : 503" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The force = 5.62e-09 N\n", + "The force = 1.08e-08 N\n" + ] + } + ], + "source": [ + "k_e=8.99*10**9 #N.m**2/c**2\n", + "q2=2*10**-9# = %0.2f c\n", + "q3=5*10**-9# = %0.2f c\n", + "r1=4#in m\n", + "F_23=(q2*q3*k_e)/(r1*r1)\n", + "print \"The force = %0.2e N\"%F_23\n", + "q1=6*10**-9\n", + "r2=5#in m\n", + "F_13=(q1*q3*k_e)/(r2*r2)\n", + "print \"The force = %0.2e N\"%F_13" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.4 Page No: 507" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The magnitude of force = 3.20e-15 N\n" + ] + } + ], + "source": [ + "q=1.6*10**-19#in c\n", + "E=2*10**4# = %0.2f N/C\n", + "F=q*E\n", + "print \"The magnitude of force = %0.2e N\"%F" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.5 Page No: 509" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Magnitude of E1 = 3.93e+05 N/C\n", + "Magnitude of E2 = 1.80e+05 N/C\n", + "Magnitude in x direction = 1.80e+05 N/C\n", + "Magnitude in y direction = 2.49e+05 N/C\n", + "Angle = 54.17 degree\n" + ] + } + ], + "source": [ + "from math import degrees, atan\n", + "k_e=8.99*10**9 #N.m**2/c**2\n", + "q1=7*10**-6# = %0.2f C\n", + "q2=5*10**-6#in C\n", + "r1=0.4\n", + "r2=0.5\n", + "E1=(k_e*q1)/(r1**2)\n", + "E2=(k_e*q2)/(r2**2)\n", + "Ex=(k_e*q2)/(r2**2)\n", + "print \"Magnitude of E1 = %0.2e N/C\"%E1\n", + "print \"Magnitude of E2 = %0.2e N/C\"%E2\n", + "print \"Magnitude in x direction = %0.2e N/C\"%Ex\n", + "Ey=(3.93*10**5)+(-1.44*10**5)\n", + "print \"Magnitude in y direction = %0.2e N/C\"%Ey\n", + "phi=degrees(atan(Ey/Ex))\n", + "print \"Angle = %0.2f degree\"%phi\n", + "#Answer given in the book is wrong" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch16.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch16.ipynb new file mode 100644 index 00000000..86539c04 --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch16.ipynb @@ -0,0 +1,364 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 16 : Electrical Energy & Capacitance" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.1 Page No : 533" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of E = 4000.00 v/m\n" + ] + } + ], + "source": [ + "v_bminusv_a=-12\n", + "d=0.3*10**-2#in m\n", + "E=-(v_bminusv_a)/d\n", + "print \"The value of E = %0.2f v/m\"%E" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.2 Page No : 533" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "solution a\n", + "Electric potential from A to B = -40000.00 V\n", + "solution b\n", + "Change in electric potential = -0.00 joules\n", + "velocity = 2768514.16 m/s\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "print \"solution a\"\n", + "E=8*10**4#in V/m\n", + "d=0.5#in m\n", + "delta_V=-E*d\n", + "print \"Electric potential from A to B = %0.2f V\"%delta_V\n", + "print \"solution b\"\n", + "q=1.6*10**-19#in C\n", + "delta_PE=q*delta_V\n", + "print \"Change in electric potential = %0.2f joules\"%delta_PE\n", + "m_p=1.67*10**-27#in kg\n", + "vf=sqrt((2*-delta_PE)/m_p)\n", + "print \"velocity = %0.2f m/s\"%vf" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.3 Page No: 534" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Solution a\n", + "Magnitude of V1 = 112375.00 v\n", + "Magnitude of V2 = -35960.00 v\n", + "solution b\n", + "Magnitude of Vp = 76415.00 v\n", + "work done = 0.31 Joule\n" + ] + } + ], + "source": [ + "k_e=8.99*10**9 #N.m**2/c**2\n", + "q1=5*10**-6# in C\n", + "q2=-2*10**-6#in C\n", + "r1=0.4\n", + "r2=0.5\n", + "V1=(k_e*q1)/(r1)\n", + "V2=(k_e*q2)/(r2)\n", + "print \"Solution a\"\n", + "print \"Magnitude of V1 = %0.2f v\"%V1\n", + "print \"Magnitude of V2 = %0.2f v\"%V2\n", + "print \"solution b\"\n", + "vp=V1+V2\n", + "print \"Magnitude of Vp = %0.2f v\"%vp\n", + "q3=4*10**-6#in C\n", + "w=vp*q3\n", + "print \"work done = %0.2f Joule\"%w" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.4 Page No: 535" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Capacitance = 1.77e-12 farad\n" + ] + } + ], + "source": [ + "e0=8.85*10**-12#in c2/N.m2\n", + "A=2*10**-4#in m2\n", + "d=1*10**-3#in m\n", + "c=(e0*A)/d\n", + "print \"Capacitance = %0.2e farad\"%c" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.5 Page No : 535" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "capacitance = 4.50e-05 farad\n", + "voltage between battery = 2.16e-04 c\n" + ] + } + ], + "source": [ + "c1=3*10**-6\n", + "c2=6*10**-6\n", + "c3=12*10**-6\n", + "c4=24*10**-6\n", + "delta_v=18\n", + "c_eq=c1+c2+c3+c4\n", + "print \"capacitance = %0.2e farad\"%c_eq\n", + "q=delta_v*c3\n", + "print \"voltage between battery = %0.2e c\"%q" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.6 Page No : 536" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "solution a\n", + "capacitance = 1.60e-06 farad\n", + "solution b\n", + "voltage between battery = 2.88e-05 c\n" + ] + } + ], + "source": [ + "c1=3*10**-6\n", + "c2=6*10**-6\n", + "c3=12*10**-6\n", + "c4=24*10**-6\n", + "delta_v=18\n", + "print \"solution a\"\n", + "c_eq=1/((1/c1)+(1/c2)+(1/c3)+(1/c4))\n", + "print \"capacitance = %0.2e farad\"%c_eq\n", + "q=delta_v*c_eq\n", + "print \"solution b\"\n", + "print \"voltage between battery = %0.2e c\"%q" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.7 Page No: 536" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "solution a\n", + "capacitance = 2.00e-06 farad\n" + ] + } + ], + "source": [ + "c1=4*10**-6\n", + "c2=4*10**-6\n", + "print \"solution a\"\n", + "c_eq=1/((1/c1)+(1/c2))\n", + "print \"capacitance = %0.2e farad\"%c_eq" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.8 Page No: 537" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "solution a\n", + "Energy stored = 4671 volt\n", + "solution b\n", + "power = 240000 watt\n" + ] + } + ], + "source": [ + "Energy=1.2*10**3#in J\n", + "c=1.1*10**-4#in f\n", + "delta_v=sqrt((2*Energy)/c)\n", + "print \"solution a\"\n", + "print \"Energy stored = %0.f volt\"%delta_v\n", + "print \"solution b\"\n", + "Energy_deliverd=600#in j\n", + "delta_t=2.5*10**-3#in s\n", + "p=(Energy_deliverd)/delta_t\n", + "print \"power = %0.f watt\"%p" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.9 Page No: 538" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "solution a\n", + "Capacitance = 1.96e-11 farad\n", + "solution b\n", + "Voltage = 16000.0 volt\n", + "Maximum charge = 3.14e-07 columb\n" + ] + } + ], + "source": [ + "k=3.7\n", + "e0=8.85*10**-12#in c2/N.m2\n", + "A=6*10**-4#in m2\n", + "d=1*10**-3#in m\n", + "c=(k*e0*A)/d\n", + "print \"solution a\"\n", + "print \"Capacitance = %0.2e farad\"%c\n", + "print \"solution b\"\n", + "E_max=16*10**6#in v/m\n", + "delta_v_max=E_max*d\n", + "print \"Voltage = %0.1f volt\"%delta_v_max\n", + "Q_max=delta_v_max*c\n", + "print \"Maximum charge = %0.2e columb\"%Q_max" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch17.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch17.ipynb new file mode 100644 index 00000000..fa69c771 --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch17.ipynb @@ -0,0 +1,311 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 17 : Current and resistance" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.1 Page No: 571" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "solution a : \n", + "Current = 0.83 Amp\n", + "solution b : \n", + "Number of electrons = 0.84 C\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "print \"solution a : \"\n", + "delta_q=1.67 # in c\n", + "delta_t=2 # in s\n", + "I=delta_q/delta_t\n", + "print \"Current = %0.2f Amp\"%I\n", + "print \"solution b : \"\n", + "N=5.22*10**18\n", + "N_q=(1.6*10**-19)*N\n", + "\n", + "print \"Number of electrons = %0.2f C\"%N_q" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.2 Page No: 573" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Solution a :\n", + "The drift speed = 2.46e-04 m/s=\n", + "Drift speed of electron = 1.15e+05 m/s\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "M=63.5 # IN G\n", + "rho=8.95\n", + "v=M/rho\n", + "electrons=6.02*10**23\n", + "n=(electrons*10**6)/v\n", + "I=10 # in c/s\n", + "q=1.60*10**-19 # in c\n", + "A=3*10**-6 # in m2\n", + "vd=(I)/(n*q*A)\n", + "print \"Solution a :\"\n", + "print \"The drift speed = %0.2e m/s=\"%vd\n", + "k_b=1.38*10**-23\n", + "T=293\n", + "m=9.11*10**-31\n", + "v_rms=sqrt((3*k_b*T)/m)\n", + "print \"Drift speed of electron = %0.2e m/s\"%v_rms" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.3 Page No: 578" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance = 18.75 ohm\n" + ] + } + ], + "source": [ + "delta_v=120\n", + "I=6.4\n", + "R=(delta_v)/I\n", + "print \"The resistance = %0.2f ohm\"%R" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.4 Page No: 580" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Solution a : \n", + "Area = 3.24e-07 m**2\n", + "Resistance = 4.63 ohm/m\n", + "solution b : \n", + "The current = 2.16 Amps\n" + ] + } + ], + "source": [ + "from math import pi\n", + "r=0.321*10**-3\n", + "A=pi*(r*r)\n", + "print \"Solution a : \"\n", + "print \"Area = %0.2e m**2\"%A\n", + "rho=1.5*10**-6 # in ohm=m\n", + "l=rho/A\n", + "print\"Resistance = %0.2f ohm/m\"% l\n", + "print \"solution b : \"\n", + "Delta_v=10\n", + "I=(Delta_v)/l\n", + "print \"The current = %0.2f Amps\"%I\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.5 Page No: 582" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature = 156.73 C\n" + ] + } + ], + "source": [ + "R=76.8\n", + "Ro=50\n", + "alpha=3.92*10**-3\n", + "t=(R-Ro)/(alpha*Ro)\n", + "T=t+20\n", + "print \"Temperature = %0.2f C\"%T" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.6 Page No: 583" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current = 6.00 A\n", + "Power = 288.00 Watt\n" + ] + } + ], + "source": [ + "delta_v=50\n", + "R=8\n", + "I=(delta_v)/R\n", + "print \"The current = %0.2f A\"%I\n", + "P=I*I*R\n", + "print \"Power = %0.2f Watt\"%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.7 Page No: 585" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Number of bulbs = 32\n" + ] + } + ], + "source": [ + "I=20 # in A\n", + "delta_v=120\n", + "p_bulb=75 # inwatt\n", + "p_total=I*delta_v\n", + "N=p_total/p_bulb\n", + "print \"Number of bulbs = %d\"%N" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.8 Page No: 587" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Energy = 2.40 kwh\n", + "Cost = 0.29 dollars\n" + ] + } + ], + "source": [ + "p=0.10 # in w\n", + "t=24 # in h\n", + "Energy=p*t\n", + "print \"Energy = %0.2f kwh\"%Energy\n", + "cost=Energy*0.12\n", + "print \"Cost = %0.2f dollars\"%cost" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch18.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch18.ipynb new file mode 100644 index 00000000..b5f90dca --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch18.ipynb @@ -0,0 +1,284 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 18 : Direct current circuits" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.1 Page No: 597" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Solution a\n", + "Equivalent resistance = 18.00 ohm\n", + "Solution b\n", + "Current = 0.33 Amps\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "R1=2\n", + "R2=4\n", + "R3=5\n", + "R4=7\n", + "R_eq=R1+R2+R3+R4\n", + "v=6#in v\n", + "print \"Solution a\"\n", + "print \"Equivalent resistance = %0.2f ohm\"%R_eq\n", + "print \"Solution b\"\n", + "I=v/R_eq\n", + "print \"Current = %0.2f Amps\"%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example18.2 Page No: 599" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "solution a\n", + "Current = 6.00 amps\n", + "Current = 3.00 amps\n", + "Current = 2.00 amps\n", + "solution B\n", + "Power = 108.00 watt\n", + "Power = 54.00 watt\n", + "Power = 36.00 watt\n" + ] + } + ], + "source": [ + "delta_V=18#in volt\n", + "R1=3#in ohm\n", + "R2=6#in ohm\n", + "R3=9#in ohm\n", + "I1=delta_V/R1\n", + "I2=delta_V/R2\n", + "I3=delta_V/R3\n", + "print \"solution a\"\n", + "print \"Current = %0.2f amps\"%I1\n", + "print \"Current = %0.2f amps\"%I2\n", + "print \"Current = %0.2f amps\"%I3\n", + "P1=(I1**2)*R1\n", + "P2=(I2**2)*R2\n", + "P3=(I3**2)*R3\n", + "print \"solution B\"\n", + "print \"Power = %0.2f watt\"%P1\n", + "print \"Power = %0.2f watt\"%P2\n", + "print \"Power = %0.2f watt\"%P3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.3 Page No: 602" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "solution b\n", + "Current = 3.00 amps\n" + ] + } + ], + "source": [ + "delta_Vac=42#in volt\n", + "R_eq=14#in ohm\n", + "I=delta_Vac/R_eq\n", + "print \"solution b\"\n", + "print \"Current = %0.2f amps\"%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.4 Page No: 605" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current value I1 = -0.83, I2 = -0.53 & I3 = -0.30 amps\n" + ] + } + ], + "source": [ + "from numpy import mat\n", + "#formula used x=inv(a)*b\n", + "I=mat([[1 ,-1, -1],[-4, 0 ,-9],[0, -5, 9]])\n", + "V=mat([[0],[6],[0]])\n", + "X=(I**-1)\n", + "a=X*V\n", + "\n", + "print \"Current value I1 = %0.2f, I2 = %0.2f & I3 = %0.2f amps\"%(a[0],a[1],a[2])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.5 Page No: 606" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current value I1 = 2.00, I2 = -3.00 & I3 = -1.00 amps\n" + ] + } + ], + "source": [ + "from numpy import mat\n", + "#prob\n", + "#formula used x=inv(a)*b\n", + "I=mat([[8, 2, 0],[-3, 2, 0],[1, 1, -1]])\n", + "V=mat([[10],[-12],[0]])\n", + "X=I**-1\n", + "a=X*V\n", + "\n", + "print \"Current value I1 = %0.2f, I2 = %0.2f & I3 = %0.2f amps\"%(a[0],a[1],a[2])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.6 Page No: 609" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Constant of the circuit = 4.00 s\n", + "Charge = 6.00e-05 columb\n", + "Charge = 3.79e-05 columb when capacitance 63.2%\n" + ] + } + ], + "source": [ + "R=8*10**5#in ohms\n", + "C=5*10**-6#in Farad\n", + "t=R*C\n", + "print \"Constant of the circuit = %0.2f s\"%t\n", + "\n", + "Q=C*12\n", + "print \"Charge = %0.2e columb\"%Q\n", + "q=0.632*Q\n", + "print \"Charge = %0.2e columb when capacitance 63.2%%\"%q" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.7 Page No: 610" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "time = 1.39 s \n" + ] + } + ], + "source": [ + "from math import log\n", + "x=log(4)\n", + "print \"time = %0.2f s \"%x" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch19.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch19.ipynb new file mode 100644 index 00000000..3a586dc0 --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch19.ipynb @@ -0,0 +1,333 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 19 : Magnetism" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.1 Page No: 631" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The force = 7.62e-19 Newton\n" + ] + } + ], + "source": [ + "q=1.6*10**-19#in columb\n", + "v=1*10**5#in m/s\n", + "B=55*10**-6#in T\n", + "F=q*v*B* 0.8660\n", + "print \"The force = %0.2e Newton\"%F" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.2 Page No: 632" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The force = 2.77e-12 Newton\n", + "Acceleration = 1.66e+15 m/s**2\n" + ] + } + ], + "source": [ + "q=1.6*10**-19#in columb\n", + "v=8*10**6#in m/s\n", + "B=2.5#in T\n", + "F=q*v*B* 0.8660\n", + "print \"The force = %0.2e Newton\"%F\n", + "m=1.67*10**-27\n", + "a=F/m\n", + "print \"Acceleration = %0.2e m/s**2\"%a" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.3 Page No: 635" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximaum force = 3.96e-02 Newton\n" + ] + } + ], + "source": [ + "l=36#in m\n", + "I=22#in A\n", + "B=0.50*10**-4#in T\n", + "F=B*I*l\n", + "print \"The maximaum force = %0.2e Newton\"%F" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.4 Page No: 637" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Torque = 0.39 N-m\n" + ] + } + ], + "source": [ + "from math import pi\n", + "A=pi*(0.5)*0.5#in m\n", + "I=2#in A\n", + "B=0.50#in T\n", + "T=B*I*A*0.5\n", + "print \"The Torque = %0.2f N-m\"%T" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.5 Page No: 640" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Velocity = 4.69e+06 m/s\n" + ] + } + ], + "source": [ + "q=1.6*10**-19\n", + "B=.35\n", + "r=14*10**-2#in m\n", + "m=1.67*10**-27#kg\n", + "v=(q*B*r)/m\n", + "print \"Velocity = %0.2e m/s\"%v" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.6 Page No: 641" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Radius of lighter istope = 0.10 m\n", + "Radius of heavier istope = 0.21 m\n", + "Distance of seperation = 0.21 m\n" + ] + } + ], + "source": [ + "q=1.6*10**-19\n", + "B=.10#in T\n", + "v=1*10**6#in m/s\n", + "r=14*10**-2#in m\n", + "m1=1.67*10**-27#in kg\n", + "m2=3.34*10**-27#in kg\n", + "r1=(m1*v)/(q*B)\n", + "r2=(m2*v)/(q*B)\n", + "x=(2*r2)-(2*r1)\n", + "print \"Radius of lighter istope = %0.2f m\"%r1\n", + "print \"Radius of heavier istope = %0.2f m\"%r2\n", + "print \"Distance of seperation = %0.2f m\"%x" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.7 Page No: 644" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Magnetic field = 2.5e-04 T\n", + "Force = 6e-20 Newton\n" + ] + } + ], + "source": [ + "from math import pi\n", + "Uo=(4*pi*10**-7)\n", + "I=5#in A\n", + "r=4*10**-3\n", + "B=(Uo*I)/(2*pi*r)\n", + "print \"Magnetic field = %0.1e T\"%B\n", + "q=1.6*10**-19\n", + "v=1.5*10**3#in m/s\n", + "F=q*v*B\n", + "print \"Force = %0.e Newton\"%F" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.8 Page No: 646" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current = 7.07 A\n" + ] + } + ], + "source": [ + "from math import pi, sqrt\n", + "mo=4*pi*10**-7#Tm/A\n", + "d=0.1#in m\n", + "x=1*10**-4#F/l\n", + "I=sqrt((x*2*pi*d)/mo)\n", + "print \"Current = %0.2f A\"%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.9 Page No: 649" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Magnetic field = 6.28e-04 T\n", + "Force = 1.88e-20 N\n" + ] + } + ], + "source": [ + "from math import pi\n", + "N=100#turns\n", + "l=.1#in m\n", + "n=N/l#in turns/m\n", + "mo=4*pi*10**-7#Tm/A\n", + "I=.5#in A\n", + "B=n*I*mo\n", + "q=1.6*10**-19#in c\n", + "v=375#in m/s\n", + "F=q*v*(B/2)\n", + "\n", + "print \"Magnetic field = %0.2e T\"%B\n", + "print \"Force = %0.2e N\"%F" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch20.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch20.ipynb new file mode 100644 index 00000000..475538f9 --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch20.ipynb @@ -0,0 +1,316 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 20 : Induced voltages and inductance" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.1 Page No: 665" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Magnetic flux = 1.62e-04 T.m**2\n", + "Induced emf = 5.06e-03 volt\n" + ] + } + ], + "source": [ + "B=.5 # in T\n", + "A=3.24*10**-4 # in m**2\n", + "Flux=B*A\n", + "N=25\n", + "delta_t=.8\n", + "print \"Magnetic flux = %0.2e T.m**2\"%Flux\n", + "e=(N*Flux)/(delta_t)\n", + "print \"Induced emf = %0.2e volt\"%e" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.2 Page No: 667" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Induced emf = 0.45 volt\n" + ] + } + ], + "source": [ + "B=.6*10**-4 # in T\n", + "l=30\n", + "v=250 # in m/s\n", + "e=B*l*v\n", + "print \"Induced emf = %0.2f volt\"%e" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.3 Page No: 672" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Solution a\n", + "Induced emf = 0.25 volt\n", + "Solution b\n", + "Current = 0.50 A\n", + "Solution c\n", + "Power = 0.12 watt\n", + "Energy delivered = 0.12 J\n", + "Solution d\n", + "Force = 0.06 N\n" + ] + } + ], + "source": [ + "B=.25 # in T\n", + "l=.5\n", + "v=2 # in m/s\n", + "e=B*l*v\n", + "print \"Solution a\"\n", + "print \"Induced emf = %0.2f volt\"%e\n", + "R=.5 # in ohm\n", + "I=e/R\n", + "\n", + "print \"Solution b\"\n", + "print \"Current = %0.2f A\"%I\n", + "delta_v=.25\n", + "P=I*delta_v\n", + "print \"Solution c\"\n", + "print \"Power = %0.2f watt\"%P\n", + "t=1 # in s\n", + "w=P*t\n", + "print \"Energy delivered = %0.2f J\"%w\n", + " # Answer give for J in textbook is wrong\n", + "d=v*t\n", + "F=w/d\n", + "print \"Solution d\"\n", + "print \"Force = %0.2f N\"%F" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.5 Page No: 678" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Solution a\n", + "Induced emf = 135.72 volt\n", + "Solution b\n", + "Current = 11.31 A\n", + "Solution c\n", + "Emf in Volt 136*sinwt\n" + ] + } + ], + "source": [ + "from math import pi\n", + "f=60 # in Hz\n", + "w=2*pi*f\n", + "N=8\n", + "A=.09 # in m**2\n", + "B=.5 # in T\n", + "emf=N*A*B*w\n", + "print \"Solution a\"\n", + "print \"Induced emf = %0.2f volt\"%emf\n", + "R=12 # in ohm\n", + "I=emf/R\n", + "print \"Solution b\"\n", + "print \"Current = %0.2f A\"%I\n", + "\n", + "print \"Solution c\"\n", + "print \"Emf in Volt 136*sinwt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.6 Page No: 680" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Solution a\n", + "Maximum Current = 12.00 A\n", + "Solution b\n", + "Current = 5.00 A\n" + ] + } + ], + "source": [ + "emf=120 # in Volt\n", + "R=10 # in Ohm\n", + "e_back=70\n", + "I=emf/R\n", + "print \"Solution a\"\n", + "print \"Maximum Current = %0.2f A\"%I\n", + "print \"Solution b\"\n", + "I=(emf-e_back)/R\n", + "print \"Current = %0.2f A\"%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.8 Page No: 684" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Solution a\n", + "Inductance = 1.81e-04 H\n", + "Solution b\n", + "Emf = 9.05e-04 Volt\n" + ] + } + ], + "source": [ + "from math import pi\n", + "uo=4*pi*10**-7 # in m/A\n", + "N=300\n", + "A=4*10**-4 # in m**2\n", + "l=25*10**-2\n", + "L=(uo*N*N*A)/l\n", + "print \"Solution a\"\n", + "print \"Inductance = %0.2e H\"%L\n", + "delta_I=-5\n", + "delta_t=1\n", + "e=(-L*delta_I)/(delta_t)\n", + "print \"Solution b\"\n", + "print \"Emf = %0.2e Volt\"%e" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.9 Page No: 685" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Solution a\n", + "Time constant = 5.00e-03 s\n", + "Solution b\n", + "Current = 1.26 Amps\n" + ] + } + ], + "source": [ + "L=30*10**-3 # in Henry\n", + "R=6 # in Ohm\n", + "tou=L/R\n", + "print \"Solution a\"\n", + "print \"Time constant = %0.2e s\"%tou\n", + "\n", + "e=12\n", + "I=(0.632*e)/R\n", + "\n", + "\n", + "print \"Solution b\"\n", + "print \"Current = %0.2f Amps\"%I\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch21.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch21.ipynb new file mode 100644 index 00000000..0e96629f --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch21.ipynb @@ -0,0 +1,284 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 21 : Alternating current circuits and electromagnetic waves" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21.1 Page No: 698" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage = 141.42 V\n", + "Current = 1.41 Amps\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt\n", + "V_max=200#in V\n", + "V_rms=(V_max)/sqrt(2)\n", + "R=100#in ohm\n", + "I_rms=V_rms/R\n", + "print \"Voltage = %0.2f V\"%V_rms\n", + "print \"Current = %0.2f Amps\"%I_rms" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21.2 Page No: 700" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance = 331.56 ohm\n", + "Current = 0.45 Amps\n" + ] + } + ], + "source": [ + "C=8*10**-6\n", + "X_c=1/(377*C)\n", + "print \"Resistance = %0.2f ohm\"%X_c\n", + "I_rms=150/X_c\n", + "print \"Current = %0.2f Amps\"%I_rms" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21.3 Page No: 702" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance = 9.43 ohm\n", + "Current = 15.92 Amps\n" + ] + } + ], + "source": [ + "L=25*10**-3#In H\n", + "w=377\n", + "X_L=w*L#In ohm\n", + "print \"Resistance = %0.2f ohm\"%X_L\n", + "I_rms=150/X_L#In A\n", + "print \"Current = %0.2f Amps\"%I_rms" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21.4 Page No: 706" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Impedence = 587.81 ohm\n", + "Current = 0.26 Amps\n", + "Angle = -64.83 degree\n", + "Voltage at Resistance = 63.80 Volt\n", + "Voltage at Inductance = 57.67 Volt\n", + "Voltage at Capacitance = 193.43 Volt\n" + ] + } + ], + "source": [ + "from math import atan, degrees, sqrt\n", + "R=250#in ohm\n", + "Xc=758#in ohm\n", + "Xl=226#in Ohm\n", + "X=Xl-Xc\n", + "V_max=150#in Volt\n", + "Z=sqrt(R**2+X**2)\n", + "I=V_max/Z\n", + "q=degrees(atan(X/R))\n", + "print \"Impedence = %0.2f ohm\"%Z\n", + "print \"Current = %0.2f Amps\"%I\n", + "print \"Angle = %0.2f degree\"%q\n", + "V_R=I*R\n", + "V_C=I*Xc\n", + "V_L=I*Xl\n", + "print \"Voltage at Resistance = %0.2f Volt\"%V_R\n", + "print \"Voltage at Inductance = %0.2f Volt\"%V_L\n", + "print \"Voltage at Capacitance = %0.2f Volt\"%V_C" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21.5 Page No: 708" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage = 106.07 V\n", + "Current = 0.18 Amps\n", + "Power = 8.15 watt\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "V_max=150#in V\n", + "V_rms=(V_max)/sqrt(2)\n", + "I_max=.255#in ohm\n", + "I_rms=I_max/sqrt(2)\n", + "cos=.426\n", + "P=V_rms*I_rms*cos\n", + "print \"Voltage = %0.2f V\"%V_rms\n", + "print \"Current = %0.2f Amps\"%I_rms\n", + "print \"Power = %0.2f watt\"%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21.6 Page No: 709" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Capacitance = 2e-06 Farad\n" + ] + } + ], + "source": [ + "L=20*10**-3#in H\n", + "C=1/(25*10**6*L)\n", + "print \"Capacitance = %0.e Farad\"%C" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21.7 Page No: 711" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Solution a\n", + "Percentage of power lost = 0.02\n", + "Solution B\n", + "Percentage of power lost = 75.00\n" + ] + } + ], + "source": [ + "I1=100\n", + "v1=4*10**3\n", + "v2=2.40*10**5\n", + "I2=(I1*v1)/v2\n", + "R=30#in ohm\n", + "p_lost=I2*I2*R\n", + "P_output=I1*v1\n", + "p_per=(p_lost*100/P_output)\n", + "print \"Solution a\"\n", + "print \"Percentage of power lost = %0.2f\"%p_per\n", + "P_lost=I1*I1*R\n", + "per=(P_lost*100)/(4*10**5)\n", + "print \"Solution B\"\n", + "print \"Percentage of power lost = %0.2f\"%per" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch22.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch22.ipynb new file mode 100644 index 00000000..ab35dac3 --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch22.ipynb @@ -0,0 +1,169 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 22 : Reflection anda refraction of light" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 22.2 Page No: 739" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Angle = 19.20 degree\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sin, pi, degrees, asin\n", + "n1=1\n", + "n2=1.52\n", + "x=sin(pi/180*30)\n", + "theta_2=degrees(asin((n1*x)/n2))\n", + "print \"Angle = %0.2f degree\"%theta_2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 22.3 Page No: 739" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Solution a\n", + "Velocity = 2.06e+08 m/s\n", + "Solution b\n", + "Wavelength in Fused quartz = 403.98 nm\n" + ] + } + ], + "source": [ + "print \"Solution a\"\n", + "c=3*10**8# Constant in m/s\n", + "n=1.458\n", + "v=c/n\n", + "print \"Velocity = %0.2e m/s\"%v\n", + "print \"Solution b\"\n", + "lambda_o=589#in nm\n", + "lambda_n=lambda_o/n\n", + "print \"Wavelength in Fused quartz = %0.2f nm\"%lambda_n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 22.5 Page No: 741" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Angle = 16.24 degree\n", + "Angle = 25.69 degree\n" + ] + } + ], + "source": [ + "from math import atan, degrees, asin\n", + "x=699#in micrometer(w-a)\n", + "t=1200 #in micrometer\n", + "b=x/2\n", + "theta_2=degrees(atan(b/t))\n", + "print \"Angle = %0.2f degree\"%theta_2\n", + "y=sin(pi/180*theta_2)\n", + "n1=1\n", + "n2=1.55\n", + "theta_1=degrees(asin((n2*y)/n1))\n", + "print \"Angle = %0.2f degree\"%theta_1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 22.6 Page No: 744" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Angle(theta_c) = 48.75 degree\n" + ] + } + ], + "source": [ + "from math import asin, degrees\n", + "n1=1.33\n", + "n2=1\n", + "x=degrees(asin(n2/n1))\n", + "\n", + "print \"Angle(theta_c) = %0.2f degree\"%x" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch23.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch23.ipynb new file mode 100644 index 00000000..29eada77 --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch23.ipynb @@ -0,0 +1,367 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 23 : Mirrors and lenses" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.1 Page No: 760" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The hight = 0.90 m\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "AC= 1.8-.1#in m\n", + "AD=.5*AC\n", + "CF=.10#/in m\n", + "X=.5*CF#in m\n", + "FA=1.8#in m\n", + "d=FA-AD-X\n", + "print \"The hight = %0.2f m\"%d" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.2 Page No : 767" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "part a\n", + "The magnification when object is at 25cm : -0.67\n", + "part c\n", + "The magnification when object is at 5cm : 2.00\n" + ] + } + ], + "source": [ + "p=25#in cm\n", + "f=10#in cm\n", + "x=(1/f)-(1/p)\n", + "q=1/x\n", + "p=25\n", + "M=-(q/p)\n", + "print \"part a\"\n", + "print \"The magnification when object is at 25cm : %0.2f\"%M\n", + "p=5#in cm\n", + "f=10#in cm\n", + "x=(1/f)-(1/p)\n", + "q=1/x\n", + "p=5\n", + "M=-(q/p)\n", + "print \"part c\"\n", + "print \"The magnification when object is at 5cm : %0.2f\"%M" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.3 Page No: 768" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "part a\n", + "The position of final image = -5.71 cm\n", + "part b\n", + "The magnification : 0.23\n" + ] + } + ], + "source": [ + "p=20#in cm\n", + "f=-8#in cm\n", + "x=(1/f)-(1/p)\n", + "q=1/x\n", + "p=25\n", + "M=-(q/p)\n", + "print \"part a\"\n", + "print \"The position of final image = %0.2f cm\"%q\n", + "print \"part b\"\n", + "print \"The magnification : %0.2f\"%M" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.4 Page No: 769" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The focal length = 26.67 cm\n" + ] + } + ], + "source": [ + "p=40#in cm\n", + "q=-(2*p)\n", + "\n", + "x=(1/p)-(1/q)\n", + "f=1/x\n", + "print \"The focal length = %0.2f cm\"%f\n", + "#Answer given in book is wrong" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.5 Page No: 770" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The position of final image = -17.14 cm\n", + "The magnification when object = -1.29 cm\n", + "The Position of image = 2.57 cm\n" + ] + } + ], + "source": [ + "p=20#in cm\n", + "n1=1.5#in cm\n", + "n2=1#in cm\n", + "R=-30#in cm\n", + "x=(n2-n1)/R\n", + "y=n1/p\n", + "s=x-y\n", + "q=1/s\n", + "print \"The position of final image = %0.2f cm\"%q\n", + "M=(n1*q)/(n2*p)\n", + "print \"The magnification when object = %0.2f cm\"%M\n", + "h=2#in cm\n", + "h1=-M*h\n", + "print \"The Position of image = %0.2f cm\"%h1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.7 Page No: 777" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "part a\n", + "The position of final image = 15.00 cm\n", + "The magnification : -0.50\n", + "part b\n", + "The position of final image = -10.00 cm\n", + "The magnification : 2.00\n" + ] + } + ], + "source": [ + "p=30#in cm\n", + "f=10#in cm\n", + "x=(1/f)-(1/p)\n", + "q=1/x\n", + "\n", + "M=-(q/p)\n", + "print \"part a\"\n", + "print \"The position of final image = %0.2f cm\"%q\n", + "print \"The magnification : %0.2f\"%M\n", + "p=5#in cm\n", + "f=10#in cm\n", + "x=(1/f)-(1/p)\n", + "q=1/x\n", + "M=-(q/p)\n", + "print \"part b\"\n", + "print \"The position of final image = %0.2f cm\"%q\n", + "print \"The magnification : %0.2f\"%M" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.8 Page No: 778" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "part a\n", + "The position of final image = -7.50 cm\n", + "The magnification : 0.25\n", + "part b\n", + "The position of final image = -5.00 cm\n", + "The magnification : 0.50\n", + "part c\n", + "The position of final image = -3.33 cm\n", + "The magnification : 0.67\n" + ] + } + ], + "source": [ + "p=30#in cm\n", + "f=-10#in cm\n", + "x=(1/f)-(1/p)\n", + "q=1/x\n", + "\n", + "M=-(q/p)\n", + "print \"part a\"\n", + "print \"The position of final image = %0.2f cm\"%q\n", + "print \"The magnification : %0.2f\"%M\n", + "p=10#in cm\n", + "f=-10#in cm\n", + "x=(1/f)-(1/p)\n", + "q=1/x\n", + "M=-(q/p)\n", + "print \"part b\"\n", + "print \"The position of final image = %0.2f cm\"%q\n", + "print \"The magnification : %0.2f\"%M\n", + "p=5#in cm\n", + "f=-10#in cm\n", + "x=(1/f)-(1/p)\n", + "q=1/x\n", + "M=-(q/p)\n", + "print \"part c\"\n", + "print \"The position of final image = %0.2f cm\"%q\n", + "print \"The magnification : %0.2f\"%M" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.9 Page No: 779" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The magnification : -0.67\n" + ] + } + ], + "source": [ + "p=30#in cm\n", + "f=10#in cm\n", + "x=(1/f)-(1/p)\n", + "q=1/x\n", + "\n", + "M1=-(q/p)\n", + "\n", + "p=5#in cm\n", + "f=20#in cm\n", + "x=(1/f)-(1/p)\n", + "q=1/x\n", + "\n", + "M2=-(q/p)\n", + "\n", + "\n", + "M=M1*M2\n", + "print \"The magnification : %0.2f\"%M" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch24.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch24.ipynb new file mode 100644 index 00000000..82f29c9f --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch24.ipynb @@ -0,0 +1,233 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 24 : Wave optics" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1 Page No: 794" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(A) wavelength of light = 5.63e-07 meters\n", + "(B) Distance between adjacent fringes = 0.022 meters\n" + ] + } + ], + "source": [ + "L=1.2 # Seperation between screen and double-slit in meter\n", + "d=3*10**-5 #distance between the two slits\n", + "m=2 #second order bright fringe\n", + "Y=4.5*10**-2 #distance of second order bright fringe from centerline\n", + "#wavelength of light\n", + "lamda=(Y*d)/(m*L)\n", + "print \"(A) wavelength of light = %0.2e meters\"%lamda\n", + "#distance between adjacent bright fringes\n", + "#delta_Y=Y(m+1)-Ym\n", + "delta_Y=lamda*L/d\n", + "print \"(B) Distance between adjacent fringes = %0.3f meters\"%delta_Y" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2 Page No: 798" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Minimum thickness of soap bubble film = 113.16 nm is\n" + ] + } + ], + "source": [ + "n=1.33 #refractive index of soap bubble\n", + "lamda=602 #wavelength of light in nm\n", + "#for constructive interference we have 2nt=lamda/2\n", + "t=lamda/(4*n)\n", + "print \"Minimum thickness of soap bubble film = %0.2f nm is\"%t" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3 Page No: 799" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Minimum thickness of film = 95.17 nm is\n" + ] + } + ], + "source": [ + "n=1.45 #refractive index of silicon monoxide\n", + "lamda=552 #wavelength of light in nm\n", + "#for destructive interference we have condition for minimn thickness 2t=lamda/2n\n", + "t=lamda/(4*n)\n", + "print \"Minimum thickness of film = %0.2f nm is\"%t" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5 Page No: 801" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Pit depth in a CD = 121.88 nm\n" + ] + } + ], + "source": [ + "n=1.6 #refractive index of plastic transparent layer\n", + "lamda=780 #wavelength of laser light in nm\n", + "#for destructive interference we have condition for minimn thickness 2t=lamda/2n\n", + "t=lamda/(4*n)\n", + "print \"Pit depth in a CD = %0.2f nm\"%t" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6 Page No: 804" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Position of first dark fringe = 3.87e-03 meters\n" + ] + } + ], + "source": [ + "lamda=580*10**-9 #wavelength of incident light in meter\n", + "a=0.30*10**-3 #slit width in meter\n", + "L=2 #distance of screen from slit in meters\n", + "#The first dark fringe corresponds to m=+1 or -1\n", + "m=1\n", + "sin_theta=m*lamda/a\n", + "#From fig 24.16 tan_theta=y/L and since theta is very small we have sin_theta=tan_theta hence sin_theta=y/L\n", + "y=L*sin_theta \n", + "print \" Position of first dark fringe = %0.2e meters\"%y" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7 Page No: 808" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Angle in degrees at which first order maxima is observed : 22.32\n", + "Angle in degrees at which second order maxima is observed : 49.43\n", + "for higher order number of diffraction the the solutions are non realistic\n" + ] + } + ], + "source": [ + "from math import asin, degrees\n", + "lamda=632.8 #wavelength of monochromatic light from helium-neon laser in meter\n", + "a=6000 #lines in diffraction grating per cm\n", + "d=10**7/a#slit seperation in nm\n", + "#for the first order maximum we have m=1\n", + "sin_theta1=lamda/d\n", + "theta1=degrees(asin(sin_theta1))\n", + "print \"Angle in degrees at which first order maxima is observed : %0.2f\"%theta1\n", + "#for the second order maximum we have m=2\n", + "sin_theta2=2*lamda/d\n", + "theta2=degrees(asin(sin_theta2))\n", + "print \"Angle in degrees at which second order maxima is observed : %0.2f\"%theta2\n", + "print \"for higher order number of diffraction the the solutions are non realistic\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch25.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch25.ipynb new file mode 100644 index 00000000..12c6cefc --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch25.ipynb @@ -0,0 +1,291 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 25 : Optical Instruments" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1 Page No: 827" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) focal length f = 50.00 cm\n", + "b) Power of the lens = 2.00 diopters\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "q=-50 # Near point of an eye in cm\n", + "p=25 #object location in cm\n", + "#a) focal length calculation\n", + "#Using Thin Lens equation 1/f=((1/p)+(1/q))\n", + "f=p*q/(p+q)\n", + "print \"a) focal length f = %0.2f cm\"%f\n", + "#b) power of the lens\n", + "f1=50*10**-2# focal length in meters\n", + "P=1/f1\n", + "print \"b) Power of the lens = %0.2f diopters\"%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3 Page No: 830" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) Maximum angular magnification of the lens : 3.50\n", + "Angular Magnification of lens when eye is relaxed : 2.50\n" + ] + } + ], + "source": [ + "f=10 # focal length in cm\n", + "#a)Maximum angular magnification\n", + "M_max=1+(25/f)\n", + "print \"a) Maximum angular magnification of the lens : %0.2f\"%M_max\n", + "m=25/f\n", + "print \"Angular Magnification of lens when eye is relaxed : %0.2f\"%m" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4 Page No: 832" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Magnification of microscope with two long focal lengths : -45.00\n", + "Magnification of microscope with a combination of 20 mm objective and 2.5 cm eyepiece : -90.00\n", + "Magnification of microscope with a combination of 20 mm objective and 2.5 cm eyepiece : -450.00 \n", + "Possible magnification of microscope with two short focal lengths : -900.00\n" + ] + } + ], + "source": [ + "#interchangeable objectives\n", + "f1=2 # focal length in cm\n", + "f2=0.2 #focal length in cm\n", + "#data of two eye pieces\n", + "f3=5 #focal length in cm\n", + "f4=2.5 #focal length in cm\n", + "L=18 # length of microscope\n", + "#Calculation of magnification for four combinations of lens\n", + "#magnification of compound microscope m =-(L/fo)*(25cm/fe) where fo is shortest focal length compared to fe\n", + "#combination of two long focal lengths\n", + "m1=-(L/f1)*(25/f3)\n", + "print \"Magnification of microscope with two long focal lengths : %0.2f\"%m1\n", + "#combination of 20 mm objective and 2.5 cm eyepiece\n", + "m2=-(L/f1)*(25/f4)\n", + "print \"Magnification of microscope with a combination of 20 mm objective and 2.5 cm eyepiece : %0.2f\"%m2\n", + "#combination of 2 mm objective and 5 cm eyepiece\n", + "m3=-(L/f2)*(25/f3)\n", + "print \"Magnification of microscope with a combination of 20 mm objective and 2.5 cm eyepiece : %0.2f \"%m3\n", + "#combination of two short focal lengths\n", + "m4=-(L/f2)*(25/f4)\n", + "print \"Possible magnification of microscope with two short focal lengths : %0.2f\"%m4" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5 Page No: 834" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Angular magnification of the telescope : 83.33\n" + ] + } + ], + "source": [ + "d=8 #diameter of objective mirror of reflecting telescope in inches\n", + "fo=1500 #focal length of objective mirror of reflecting telescope in mm\n", + "fe=18 #focal length of eyepiece\n", + "m=fo/fe\n", + "print \"Angular magnification of the telescope : %0.2f\"%m" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6 Page No: 837" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) Limiting angle of resolution in radians : 7.98e-07\n", + "b) Maximum limit of resolution for the microscope in radians : 5.42e-07\n", + "c) Limiting angle of resolution for the microscope when water filled the space b/w the object and objective in radians : 6.00e-07\n" + ] + } + ], + "source": [ + "l=589*10**-9 #Wavelength of sodium light m\n", + "d=90*10**-2 #diameter of the aperture in m \n", + "L=400*10**-9 #Wavelength of desirable Visble light\n", + "n=1.33 #refractive index of water\n", + "#a) Calculation of limiting angle of resolution\n", + "#Limiting angle of resolution of the circular aperture is Theta_min=1.22*(l/d)\n", + "Theta_min1=1.22*(l/d)\n", + "print \"a) Limiting angle of resolution in radians : %0.2e\"%Theta_min1\n", + "#b) Calculation of maximum limit of resolution for the microscope\n", + "Theta_min2=1.22*(L/d)\n", + "print \"b) Maximum limit of resolution for the microscope in radians : %0.2e\"%Theta_min2\n", + "#c)Effect of water b/w the object and objective on resolving power of microscope\n", + "lw=l/n\n", + "Theta_min3=1.22*(lw/d)\n", + "print \"c) Limiting angle of resolution for the microscope when water filled the space b/w the object and objective in radians : %0.2e\"%Theta_min3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7 Page No: 838" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Magnification of telescope A is : 166.67\n", + "Magnification of telescope B is : 50.00\n" + ] + } + ], + "source": [ + "f1=1000# focal length of objective of telescope A in mm\n", + "f2=1250# focal length of objective of telescope B in mm\n", + "f3=6# focal length of eyepiece of telescope A in mm\n", + "f4=25# focal length of eyepiece of telescope Bin mm\n", + "#C) Calculation of magnification of the telescope\n", + "m_A=f1/f3\n", + "m_B=f2/f4\n", + "print \"Magnification of telescope A is : %0.2f\"%m_A\n", + "print \"Magnification of telescope B is : %0.2f\"%m_B" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8 Page No: 839" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) Resolving poer of grating inorder to distinguish the wavelengths = 998.31\n", + "b) No.of lines of the grating illuminated to resolve the lines in the second order spectrum are 499\n" + ] + } + ], + "source": [ + "L1=589 # wavelength of first bright line in sodium spectrum in nm\n", + "L2=589.59 # wavelength of second bright line in sodium spectrum in nm\n", + "m=2 # order of the spectrum\n", + "delta_L=L2-L1\n", + "R=L1/delta_L\n", + "print \"a) Resolving poer of grating inorder to distinguish the wavelengths = %0.2f\"% R\n", + "N=R/m\n", + "print \"b) No.of lines of the grating illuminated to resolve the lines in the second order spectrum are %d\"%N" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch26.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch26.ipynb new file mode 100644 index 00000000..33f3c29a --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch26.ipynb @@ -0,0 +1,348 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 26 : Relativity" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example1 Page No: 855" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Period of the pendulum w.r.t to observer = 9.61 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt\n", + "Tp=3 #proper time in sec\n", + "c=3*10**8 #velocity of light in m/sec\n", + "v=0.95*c\n", + "gamma=1/sqrt(1-(v**2/c**2))\n", + "T=gamma*Tp\n", + "print \"Period of the pendulum w.r.t to observer = %.2f \"%T" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example2 Page No: 857" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Length of spaceship measured by moving observer = 16.93 meters\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "Lp=120 # length of space ship in meters\n", + "c=3*10**8 #velocity of light in m/sec\n", + "v=0.99*c\n", + "gamma=1/sqrt(1-(v**2/c**2))\n", + "L=Lp/gamma\n", + "print \"Length of spaceship measured by moving observer = %0.2f meters\"%L" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example3 Page No: 859" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Distance from spaceship to the groung measured by an observer in spaceship = 105.75 meters\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "Lp=435 # length of space ship in meters\n", + "c=3*10**8 #velocity of light in m/sec\n", + "v=0.970*c\n", + "gamma=1/sqrt(1-(v**2/c**2))\n", + "L=Lp/gamma\n", + "print \"Distance from spaceship to the groung measured by an observer in spaceship = %0.2f meters\"%L" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example4 Page No: 861" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The observer sees the horizontal dimension of the spaceship gets contracted to a length of 16.24 meters\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "c=3*10**8 #velocity of light in m/sec\n", + "#when the spaceship is at rest\n", + "x=52 # diatance in x direction in meters\n", + "y=25 #measurement in y direction\n", + "v=0.95*c\n", + "#when the spaceship moves to an observer at rest only x dimension looks contracted\n", + "gamma=1/sqrt(1-(v**2/c**2))\n", + "L=x/gamma\n", + "print \"The observer sees the horizontal dimension of the spaceship gets contracted to a length of %0.2f meters\"%L" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example5 Page No: 862" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "relativistic momentum = 3.10e-22 kg.m/s\n", + "classical momentum = 2.05e-22 kg.m/s\n", + "the relativistic result is 51 percent greater than classical result\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "c=3*10**8 #velocity of light in m/sec\n", + "m=9.11*10**-31 #mass of electron in kg\n", + "v=0.75*c\n", + "gamma=1/sqrt(1-(v**2/c**2))\n", + "#relativistic momentum\n", + "p=m*v*gamma\n", + "print \"relativistic momentum = %0.2e kg.m/s\"%p\n", + "#classical approach\n", + "P=m*v\n", + "print \"classical momentum = %0.2e kg.m/s\"%P\n", + "Z=(p-P)*100/P\n", + "print \"the relativistic result is %d percent greater than classical result\"%Z" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example6 Page No: 864" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "velocity of light w.r.t stationary observer = 3.00e+08 m/sec\n" + ] + } + ], + "source": [ + "c=3*10**8 #velocity of light in m/sec\n", + "Vmo=0.80*c # velocity of motocycle w.r.t stationary observer \n", + "Vlm=c # velocity of motocycle w.r.t motorcycle\n", + "#velocity of light w.r.t stationary observer \n", + "Vlo=(Vlm+Vmo)/(1+(Vlm*Vmo)/c**2)\n", + "print \"velocity of light w.r.t stationary observer = %0.2e m/sec\"%Vlo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example7 Page No: 865" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The energy equivalent of baseball = 4.50e+16 joules\n" + ] + } + ], + "source": [ + "c=3*10**8 #velocity of light in m/sec\n", + "m=0.50 #mass of baseball in kg\n", + "E=m*c**2\n", + "print \"The energy equivalent of baseball = %0.2e joules\"%E" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8 Page No: 866" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "total energy of an electron = 0.97 Mev\n", + "Kinetic energy of electron = 0.46 Mev\n" + ] + } + ], + "source": [ + "c=3*10**8 #velocity of light in m/sec\n", + "m=0.511 #rest energy of electron in Mev\n", + "v=0.85*c\n", + "gamma=1/sqrt(1-(v**2/c**2))\n", + "E=(m)*gamma\n", + "print \"total energy of an electron = %0.2f Mev\"%E\n", + "K=E-m\n", + "print \"Kinetic energy of electron = %0.2f Mev\"%K" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9 Page No: 867" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) Kinetic energy released in fission = 200.62 Mev\n", + " Speed of Barium fragment = 1.66e+07 Mev\n", + " Speed of krypton fragment = 2.05e+07 Mev\n" + ] + } + ], + "source": [ + "m_n=1.008665 #mass of neutron in amu\n", + "m_U=235.043924 #atomic mass of uranium in amu\n", + "m_Ba=140.903496 #atomic mass of barium in amu\n", + "m_Kr=91.907720 #atomic mass of krypton in amu\n", + "c=3*10**8 # velocity of light in m/s\n", + "#a) Kinetic energy released in fission of uranium\n", + "KE_final_=((m_n+m_U)-(m_Ba+m_Kr+(3*m_n)))*c**2\n", + "#1 amu = 931.494 Mev/c**2\n", + "KE_final=KE_final_*931.494/c**2\n", + "print \"a) Kinetic energy released in fission = %0.2f Mev\"%KE_final\n", + "#b) velocities of barium and krypton\n", + "#E=mc2/sqrt(1-v2/c2)\n", + "KE_Ba=KE_final\n", + "m_Ba_=m_Ba*931.494/c**2 # mass of barium in Mev\n", + "E_Ba=KE_Ba+m_Ba_*c**2\n", + "V_Ba=(sqrt(1-(((m_Ba_*c**2)**2)/E_Ba**2)))*c\n", + "print \" Speed of Barium fragment = %0.2e Mev\"%V_Ba\n", + "KE_Kr=KE_final\n", + "m_Kr_=m_Kr*931.494/c**2 # mass of krypton in Mev\n", + "E_Kr=KE_Kr+m_Kr_*c**2\n", + "V_Kr=(sqrt(1-((m_Kr_*c**2)**2)/E_Kr**2))*c\n", + "print \" Speed of krypton fragment = %0.2e Mev\"%V_Kr\n", + "#The difference in answer is because of round off" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch27.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch27.ipynb new file mode 100644 index 00000000..cc3bdd59 --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch27.ipynb @@ -0,0 +1,401 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 27 : Quantum physics" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1 Page No: 874" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Wavelength at which radiation emitted from the skin reaches its peak = 9.41e-06 meters\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "T=35 #Temperature of the skin in celsius\n", + "T1=T+273 #Temperature in kelvin\n", + "#From Wien's displacement law \n", + "Lambda_max=(0.2898*10**-2)/T1\n", + "print \"Wavelength at which radiation emitted from the skin reaches its peak = %0.2e meters\"%Lambda_max" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2 Page No: 878" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) Total energy of Simple harmonic oscillator with given amplitude = 2.00 Joules\n", + " Frequency of oscillation = 0.56 Hertz\n", + "b) Quantum number for the given macroscopic system : 5.36e+33\n", + "c) Energy carried away by a one-quantum charge = 3.73e-34 joules\n" + ] + } + ], + "source": [ + "from math import sqrt, pi\n", + "m=2 # mass of the object in Kg\n", + "k=25 #force constant of spring in N/m\n", + "A=0.4 #Amplitude of Simple harmonic oscillation by spring in meters\n", + "h=6.63*10**-34#js\n", + "#a) Total energy and frequency of SHO calculation\n", + "E=(1/2)*k*A**2\n", + "f=(1/(2*pi))*sqrt(k/m)\n", + "print \"a) Total energy of Simple harmonic oscillator with given amplitude = %0.2f Joules\"%E\n", + "print \" Frequency of oscillation = %0.2f Hertz\"%f\n", + "#b) Calculation of quantum number for the system\n", + "n=E/(h*f)\n", + "print \"b) Quantum number for the given macroscopic system : %0.2e\"%n\n", + "#c) Calculation of energy carried away in a quantum charge\n", + "delta_E=h*f\n", + "print \"c) Energy carried away by a one-quantum charge = %0.2e joules\"%delta_E" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3 Page No: 879" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Energy carried by a photon with the given frequency = 3.98e-19 Joules\n" + ] + } + ], + "source": [ + "f=6*10**14 #frequency of yellow light in hertz\n", + "h=6.63*10**-34 #plancks constant J.s\n", + "E=h*f\n", + "print \"Energy carried by a photon with the given frequency = %0.2e Joules\"%E" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4 Page No: 882" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) Maximum Kinetic energy of th eejected photoelectrons = 1.68 ev is\n", + "b) Cut off wavelength for sodium = 5.05e-07 meters\n" + ] + } + ], + "source": [ + "l=0.3*10**-6 #wavelength of light in meters\n", + "W=2.46 #work function for sodium in ev\n", + "c=3*10**8 #velocity of light in m/s\n", + "h=6.63*10**-34#js\n", + "#a) Maximum KE of the ejected photoelectrons\n", + "E=(h*c/l)/(1.6*10**-19) #energy of each photon of th eilluminating light beam in ev\n", + "KE_max=E-W\n", + "print \"a) Maximum Kinetic energy of th eejected photoelectrons = %0.2f ev is\"%KE_max\n", + "#b) Cut off wavelength for sodium \n", + "W1=W*1.6*10**-19\n", + "lc=h*c/W1\n", + "print \"b) Cut off wavelength for sodium = %0.2e meters\"%lc" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5 Page No: 885" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Minimum wavelength produced = 1.24e-11 meters\n" + ] + } + ], + "source": [ + "V=10**5 #potential difference in Volts\n", + "h=6.63*10**-34 # plancks constant in J.s\n", + "c=3*10**8# velocity of light in m/s\n", + "e=1.6*10**-19# elelctronic charge in coulombs\n", + "L_min=(h*c)/(e*V)\n", + "print \"Minimum wavelength produced = %0.2e meters\"%L_min" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6 Page No: 886" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Grazing angle at first order of interference = 6.40 degree\n", + "Grazing angle at third order of interference = 19.54 degree\n" + ] + } + ], + "source": [ + "from math import asin, degrees\n", + "d=0.314 #spacing between certain planes in a crystal of calcite in nm\n", + "l=0.070 #wavelength of X-rays in nm\n", + "m=1# first order of interference\n", + "theta1=degrees(asin((m*l)/(2*d)))\n", + "print \"Grazing angle at first order of interference = %0.2f degree\"%theta1\n", + "m=3 #third order of interference\n", + "theta2=degrees(asin((m*l)/(2*d)))\n", + "print \"Grazing angle at third order of interference = %0.2f degree\"%theta2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7 Page No: 887" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Wavelength of the scattered X-rays at the given angle in 0.20 nm\n" + ] + } + ], + "source": [ + "from math import pi, cos\n", + "Lo=0.200000 #wavelength of X-rays in nm\n", + "h=6.63*10**-34 #in J.s\n", + "m_e=9.11*10**-31 # in Kg\n", + "c=3*10**8 #in m/s\n", + "theta=45 #in degrees\n", + "#wavelength is represented by d\n", + "delta_L=(h/(m_e*c))*(1-cos(pi/180*theta))\n", + "L=delta_L+Lo\n", + "print \"Wavelength of the scattered X-rays at the given angle in %.2f nm\"%L\n", + "\n", + "#Answer given in textbook is wrong" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8 Page No: 887" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "de Broglie wavelength for an electron = 7.28e-11 meters\n" + ] + } + ], + "source": [ + "h=6.63*10**-34 #in J.s\n", + "m_e=9.11*10**-31 # in Kg\n", + "v=1*10**7 #in m/s\n", + "lamda=h/(m_e*v)\n", + "print \"de Broglie wavelength for an electron = %0.2e meters\"%lamda" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9 Page No: 888" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "de Broglie wavelength of the ball = 1.14e-34 meters\n" + ] + } + ], + "source": [ + "h=6.63*10**-34 #in J.s\n", + "m=0.145 # in Kg\n", + "v=40 #in m/s\n", + "lamda=h/(m*v)\n", + "print \"de Broglie wavelength of the ball = %0.2e meters\"%lamda" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10 Page No: 889" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Uncertainity in position of electron = 3.86e-06 Meters\n" + ] + } + ], + "source": [ + "from math import pi\n", + "h=6.63*10**-34#js\n", + "v=5*10**3 #speed of the electron in m/s\n", + "m_e=9.11*10**-31 # mass of electron in Kg\n", + "p=m_e*v\n", + "delta_p=0.00300*p\n", + "#Uncertainity principle states delta_x*delta_p >=h/(4*pi)\n", + "delta_x=h/(4*pi*delta_p)\n", + "print \"Uncertainity in position of electron = %0.2e Meters\"%delta_x" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11 Page No: 889" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Minimum uncertainity in energy of the excited states = 5.28e-27 Joules\n" + ] + } + ], + "source": [ + "from math import pi\n", + "h=6.63*10**-34 # plancks constant in J.s\n", + "delta_t=1.00*10**-8 # Average time that an ellectron exists in the excited states in sec\n", + "delta_E=h/(4*pi*delta_t)\n", + "print \" Minimum uncertainity in energy of the excited states = %0.2e Joules\"%delta_E" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch28.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch28.ipynb new file mode 100644 index 00000000..9cb5aa94 --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch28.ipynb @@ -0,0 +1,197 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 28 : Atomic Physics" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1 Page No: 897" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Wavelength of the emitted photon = 1.22e-07 meters\n", + "frequency of the emitted photon = 2.47e+15 meters\n" + ] + } + ], + "source": [ + "RH=1.097*10**7 #Rydberg constant in per meter\n", + "lamda=4/(3*RH)\n", + "c=3*10**8 # m/sec\n", + "f=c/lamda\n", + "print \"Wavelength of the emitted photon = %0.2e meters\"%lamda\n", + "print \"frequency of the emitted photon = %0.2e meters\"%f" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2 Page No: 898" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "longest wavelength that photon emmited = 6.56e-07 meters\n", + "Energy emmited by the photon = 3.03e-19 Joules\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "RH=1.097*10**7 #Rydberg constant in per meter\n", + "h=6.626*10**-34 #plancks constant in j.s\n", + "c=3*10**8 # velocity of light in m/s\n", + "nf=2 #quantum number\n", + "ni=3# quantum number\n", + "#assuming k=1/lamda\n", + "k=RH*((1/nf**2-1/ni**2))\n", + "lamda=1/k\n", + "print \"longest wavelength that photon emmited = %0.2e meters\"%lamda\n", + "E_photon=h*c/lamda\n", + "print \"Energy emmited by the photon = %0.2e Joules\"%E_photon" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3 Page No: 901" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) Energy of the atom in ground state = -54.40 eV\n", + "b) Radius of the ground state orbit = 0.026 nm\n" + ] + } + ], + "source": [ + "Z=2 #atomic number of helium\n", + "n=1 #principal quantum number\n", + "E=-Z**2*13.6/n**2\n", + "print \"a) Energy of the atom in ground state = %0.2f eV\"%E\n", + "r=(n**2/Z)*0.0529#in nm\n", + "print \"b) Radius of the ground state orbit = %0.3f nm\"%r" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4 Page No: 906" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Energy of the states with quantum number 2 = -3.40 ev\n" + ] + } + ], + "source": [ + "n=2# principal quantum number \n", + "E=-13.6/n**2\n", + "print \"Energy of the states with quantum number 2 = %0.2f ev\"%E" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6 Page No: 906" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Energy of the characteristic emiited from tungsten target when electron drops from M shell to K shell = 6.61e+04 ev\n" + ] + } + ], + "source": [ + "Z=74 #atomic number of tungsten\n", + "Eo=13.6 #ground state enenrgy in ev\n", + "E_K=-(Z-1)**2*(13.6) #Energy of the electron in K shell\n", + "n=3\n", + "Z_eff=Z-n**2\n", + "E3=Eo/n**2\n", + "E_M=-Z_eff**2*E3\n", + "E=E_M-E_K\n", + "print \"Energy of the characteristic emiited from tungsten target when electron drops from M shell to K shell = %0.2e ev\"%E\n", + "#Difference in answer is because of roundoff" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch29.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch29.ipynb new file mode 100644 index 00000000..4c20d301 --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch29.ipynb @@ -0,0 +1,262 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 29 : Nuclear Physics" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1 Page No: 916" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nuclear density = 2.31e+17 kg/m3\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi\n", + "m=1.67*10**-27 #mass of nucleus in kg\n", + "ro=1.2*10**-15 #in meter\n", + "p=(3*m)/(4*pi*(ro)**3)\n", + "print \"Nuclear density = %0.2e kg/m3\"%p" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2 Page No: 920" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Binding energy of Deuteron = 2.22 Mev\n" + ] + } + ], + "source": [ + "mp=1.007825 #in u\n", + "mn=1.008665 #in u\n", + "md=2.014102 #in u\n", + "u=931.494 #Mev\n", + "M=mp+mn\n", + "delta_m=(M-md) #in u\n", + "E=delta_m*u\n", + "print \"Binding energy of Deuteron = %0.2f Mev\"%E" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3 Page No: 922" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Activity or decay rate at t=0 = 1.11e-05 Ci\n" + ] + } + ], + "source": [ + "No=3*10**16 #no.of radioactive nuclei present at t=0\n", + "t_half=1.6*10**3 #years\n", + "T_half=t_half*3.16*10**7 #in sec\n", + "d=0.693/T_half\n", + "R_o=d*No # decays/s\n", + "Ci=3.7*10**10\n", + "Ro=R_o/Ci\n", + "print \"Activity or decay rate at t=0 = %0.2e Ci\"%Ro" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4 Page No: 923" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) No.of atoms remaining after 12 days : 45612654\n", + "Initial activity of the radon sample = 837.68 decay/sec\n" + ] + } + ], + "source": [ + "from math import exp\n", + "T_half=3.83 #half life time of Radon in days\n", + "No=4*10**8 #Initial No .of Radon atoms \n", + "lamda=0.693/T_half # in days\n", + "t=12 \n", + "N=No*exp(-(lamda*t))\n", + "print \"a) No.of atoms remaining after 12 days : %0.f\"%N\n", + "lamda_=lamda/(8.64*10**4)\n", + "R=lamda_*No\n", + "print \"Initial activity of the radon sample = %0.2f decay/sec\"%R" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5 Page No: 925" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Energy liberated = 4.87 Mev\n" + ] + } + ], + "source": [ + "m_d=222.017571 #mass of daughter nuclei in atomic units\n", + "m_alpha=4.002602 #mass of alpha particle in atomic units\n", + "M_p=226.025402 #mass of parent nuclei in atomic units\n", + "m=m_d+m_alpha\n", + "delta_m=(M_p-m)\n", + "E=delta_m*931.494\n", + "print \"Energy liberated = %0.2f Mev\"%E" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6 Page No: 927" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Energy released in beta decay = 0.156 Mev\n" + ] + } + ], + "source": [ + "M_C=14.003242 #mass of carbon in atomic mass units\n", + "M_N=14.003074 #mass of nitogen in atomic mass units\n", + "delta_M=M_C-M_N\n", + "E=delta_M*(931.494)\n", + "print \"Energy released in beta decay = %0.3f Mev\"%E" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7 Page No: 928" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Age of the skeleton = 10915.43 years\n" + ] + } + ], + "source": [ + "from math import log\n", + "T_half=3.01*10**9 #half life time in min\n", + "lamda=0.693/T_half\n", + "R=200 # in decay/min\n", + "R0_=15 #decay rate in decay/min.g\n", + "m=50 #weight of carbon\n", + "R0=R0_*m #in decay/min\n", + "t1=-(log(R/R0)/lamda) #im min\n", + "t=t1/525949\n", + "print \"Age of the skeleton = %0.2f years\"%t" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch30.ipynb b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch30.ipynb new file mode 100644 index 00000000..06318ebd --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch30.ipynb @@ -0,0 +1,107 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 30 : Nuclear energy and elementary particles" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2 Page No: 943" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Disintegration energy = 5.33e+26 Mev is\n", + "or = 2.37e+07 KWh\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "Q=208 #disintegration energy per event in Mev\n", + "m=1*10**3 #mass of uranium\n", + "A=235 #mass number or uranium in g/mol\n", + "a=6.02*10**23 #avagadro number nuclei/mol\n", + "N=(a/A)*m #nuclei\n", + "E=N*Q\n", + "P=E*4.45*10**-20\n", + "print \"Disintegration energy = %0.2e Mev is\"%E\n", + "print \"or = %0.2e KWh\"%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3 Page No: 947" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Energy release in deuterium-deuterium reaction = 4.03 Mev\n" + ] + } + ], + "source": [ + "m1=2.014102 # mass of deuterium in atomic mass unit\n", + "m2=3.016049 #mass of tritium in atomic mass unit\n", + "m3=1.007825 # mass of hydrogen in atomic mass unit\n", + "#referring to the deuterium-deuterium reaction\n", + "#mass before reaction\n", + "M1=2*m1\n", + "#mass after reaction\n", + "M2=m2+m3\n", + "#excessive mass\n", + "m=M1-M2\n", + "#converting mass into energy\n", + "#1 u = 931.494 Mev\n", + "E=m*931.494\n", + "print \" Energy release in deuterium-deuterium reaction = %0.2f Mev\"%E" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/1RefractionOfLaserLight.png b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/1RefractionOfLaserLight.png Binary files differnew file mode 100644 index 00000000..a2cdf1a4 --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/1RefractionOfLaserLight.png diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/2PropertiesOfImage.png b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/2PropertiesOfImage.png Binary files differnew file mode 100644 index 00000000..499e2c33 --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/2PropertiesOfImage.png diff --git a/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/3PositionOf1DarkFringe.png b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/3PositionOf1DarkFringe.png Binary files differnew file mode 100644 index 00000000..b0d4c88b --- /dev/null +++ b/College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/3PositionOf1DarkFringe.png diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/Chapter9_7.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/Chapter9_7.ipynb new file mode 100644 index 00000000..5520ce18 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/Chapter9_7.ipynb @@ -0,0 +1,419 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:36c7c29426c0c7be14ce2ec6430ec42fa1b0070a30e5ffbe18ec0fb6fd9b8d34"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter09:Numerical Solution of Partial Differential Equations"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.1:pg-350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#standard five point formula\n",
+ "#example 9.1\n",
+ "#page 350\n",
+ "\n",
+ "u2=5.0;u3=1.0;\n",
+ "for i in range(0,3):\n",
+ " u1=(u2+u3+6.0)/4.0\n",
+ " u2=(u1/2.0)+(5.0/2.0)\n",
+ " u3=(u1/2.0)+(1.0/2.0)\n",
+ " print\" the values are u1=%d\\t u2=%d\\t u3=%d\\t\\n\\n\" %(u1,u2,u3)\n",
+ " \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " the values are u1=3\t u2=4\t u3=2\t\n",
+ "\n",
+ "\n",
+ " the values are u1=3\t u2=4\t u3=2\t\n",
+ "\n",
+ "\n",
+ " the values are u1=3\t u2=4\t u3=2\t\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.2:pg-351"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#solution of laplace equation by jacobi method,gauss-seidel method and SOR method\n",
+ "#example 9.2\n",
+ "#page 351\n",
+ "u1=0.25\n",
+ "u2=0.25\n",
+ "u3=0.5\n",
+ "u4=0.5 #initial values\n",
+ "print \"jacobis iteration process\\n\\n\"\n",
+ "print\"u1\\t u2\\t u3\\t u4\\t \\n\\n\"\n",
+ "print \"%f\\t %f\\t %f\\t %f\\t \\n\" %(u1,u2,u3,u4)\n",
+ "for i in range(0,7):\n",
+ " u11=(0+u2+0+u4)/4\n",
+ " u22=(u1+0+0+u3)/4\n",
+ " u33=(1+u2+0+u4)/4\n",
+ " u44=(1+0+u3+u1)/4\n",
+ " u1=u11\n",
+ " u2=u22\n",
+ " u3=u33\n",
+ " u4=u44\n",
+ " print \"%f\\t %f\\t %f\\t %f\\t \\n\" %(u11,u22,u33,u44) \n",
+ "print \" gauss seidel process\\n\\n\"\n",
+ "u1=0.25\n",
+ "u2=0.3125\n",
+ "u3=0.5625\n",
+ "u4=0.46875 #initial values\n",
+ "print \"u1\\t u2\\t u3\\t u4\\t \\n\\n\"\n",
+ "print \"%f\\t %f\\t %f\\t %f\\t \\n\" %(u1,u2,u3,u4)\n",
+ "for i in range(0,4):\n",
+ "\n",
+ " u1=(0.0+u2+0.0+u4)/4.0\n",
+ " u2=(u1+0.0+0.0+u3)/4.0\n",
+ " u3=(1.0+u2+0.0+u4)/4.0\n",
+ " u4=(1.0+0.0+u3+u1)/4.0\n",
+ " print \"%f\\t %f\\t %f\\t %f\\t \\n\" %(u1,u2,u3,u4) \n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "jacobis iteration process\n",
+ "\n",
+ "\n",
+ "u1\t u2\t u3\t u4\t \n",
+ "\n",
+ "\n",
+ "0.250000\t 0.250000\t 0.500000\t 0.500000\t \n",
+ "\n",
+ "0.187500\t 0.187500\t 0.437500\t 0.437500\t \n",
+ "\n",
+ "0.156250\t 0.156250\t 0.406250\t 0.406250\t \n",
+ "\n",
+ "0.140625\t 0.140625\t 0.390625\t 0.390625\t \n",
+ "\n",
+ "0.132812\t 0.132812\t 0.382812\t 0.382812\t \n",
+ "\n",
+ "0.128906\t 0.128906\t 0.378906\t 0.378906\t \n",
+ "\n",
+ "0.126953\t 0.126953\t 0.376953\t 0.376953\t \n",
+ "\n",
+ "0.125977\t 0.125977\t 0.375977\t 0.375977\t \n",
+ "\n",
+ " gauss seidel process\n",
+ "\n",
+ "\n",
+ "u1\t u2\t u3\t u4\t \n",
+ "\n",
+ "\n",
+ "0.250000\t 0.312500\t 0.562500\t 0.468750\t \n",
+ "\n",
+ "0.195312\t 0.189453\t 0.414551\t 0.402466\t \n",
+ "\n",
+ "0.147980\t 0.140633\t 0.385775\t 0.383439\t \n",
+ "\n",
+ "0.131018\t 0.129198\t 0.378159\t 0.377294\t \n",
+ "\n",
+ "0.126623\t 0.126196\t 0.375872\t 0.375624\t \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.4:pg-354"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#poisson equation\n",
+ "#exaample 9.4\n",
+ "#page 354\n",
+ "u2=0.0;u4=0.0;\n",
+ "print \" u1\\t u2\\t u3\\t u4\\t\\n\\n\"\n",
+ "for i in range(0,6):\n",
+ " u1=(u2/2.0)+30.0\n",
+ " u2=(u1+u4+150.0)/4.0\n",
+ " u4=(u2/2.0)+45.0\n",
+ " print \"%0.2f\\t %0.2f\\t %0.2f\\t %0.2f\\n\" %(u1,u2,u2,u4)\n",
+ "print \" from last two iterates we conclude u1=67 u2=75 u3=75 u4=83\\n\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " u1\t u2\t u3\t u4\t\n",
+ "\n",
+ "\n",
+ "30.00\t 45.00\t 45.00\t 67.50\n",
+ "\n",
+ "52.50\t 67.50\t 67.50\t 78.75\n",
+ "\n",
+ "63.75\t 73.12\t 73.12\t 81.56\n",
+ "\n",
+ "66.56\t 74.53\t 74.53\t 82.27\n",
+ "\n",
+ "67.27\t 74.88\t 74.88\t 82.44\n",
+ "\n",
+ "67.44\t 74.97\t 74.97\t 82.49\n",
+ "\n",
+ " from last two iterates we conclude u1=67 u2=75 u3=75 u4=83\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 59
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.6:pg-362"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#bender-schmidt formula\n",
+ "#example 9.6\n",
+ "#page 362\n",
+ "def f(x):\n",
+ " return (4*x)-(x*x)\n",
+ "#u=[f(0),f(1),f(2),f(3),f(4)]\n",
+ "u1=f(0);u2=f(1);u3=f(2);u4=f(3);u5=f(4);\n",
+ "u11=(u1+u3)/2\n",
+ "u12=(u2+u4)/2\n",
+ "u13=(u3+u5)/2\n",
+ "print \"u11=%0.2f\\t u12=%0.2f\\t u13=%0.2f\\t \\n\" %(u11,u12,u13)\n",
+ "u21=(u1+u12)/2.0\n",
+ "u22=(u11+u13)/2.0\n",
+ "u23=(u12+0)/2.0\n",
+ "print \"u21=%0.2f\\t u22=%0.2f\\t u23=%0.2f\\t \\n\" %(u21,u22,u23)\n",
+ "u31=(u1+u22)/2.0\n",
+ "u32=(u21+u23)/2.0\n",
+ "u33=(u22+u1)/2.0\n",
+ "print \"u31=%0.2f\\t u32=%0.2f\\t u33=%0.2f\\t \\n\" % (u31,u32,u33)\n",
+ "u41=(u1+u32)/2.0\n",
+ "u42=(u31+u33)/2.0\n",
+ "u43=(u32+u1)/2.0\n",
+ "print \"u41=%0.2f\\t u42=%0.2f\\t u43=%0.2f\\t \\n\" % (u41,u42,u43)\n",
+ "u51=(u1+u42)/2.0\n",
+ "u52=(u41+u43)/2.0\n",
+ "u53=(u42+u1)/2.0\n",
+ "print \"u51=%0.2f\\t u52=%0.2f\\t u53=%0.2f\\t \\n\" % (u51,u52,u53)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "u11=2.00\t u12=3.00\t u13=2.00\t \n",
+ "\n",
+ "u21=1.50\t u22=2.00\t u23=1.50\t \n",
+ "\n",
+ "u31=1.00\t u32=1.50\t u33=1.00\t \n",
+ "\n",
+ "u41=0.75\t u42=1.00\t u43=0.75\t \n",
+ "\n",
+ "u51=0.50\t u52=0.75\t u53=0.50\t \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 77
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.7:pg-363"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "#bender-schimdt's formula and crank-nicolson formula\n",
+ "#example 9.7\n",
+ "#page 363\n",
+ "#bender -schimdt's formula\n",
+ "import math\n",
+ "from numpy import matrix\n",
+ "z=math.pi\n",
+ "def f(x,t):\n",
+ " return math.exp(z*z*t*-1)*sin(z*x)\n",
+ "#u=[f(0,0),f(0.2,0),f(0.4,0),f(0.6,0),f(0.8,0),f(1,0)]\n",
+ "u1=f(0,0)\n",
+ "u2=f(0.2,0)\n",
+ "u3=f(0.4,0)\n",
+ "u4=f(0.6,0)\n",
+ "u5=f(0.8,0)\n",
+ "u6=f(1.0,0)\n",
+ "u11=u3/2\n",
+ "u12=(u2+u4)/2\n",
+ "u13=u12\n",
+ "u14=u11\n",
+ "print \"u11=%f\\t u12=%f\\t u13=%f\\t u14=%f\\n\\n\" % (u11,u12,u13,u14)\n",
+ "u21=u12/2\n",
+ "u22=(u12+u14)/2\n",
+ "u23=u22\n",
+ "u24=u21\n",
+ "print \"u21=%f\\t u22=%f\\t u23=%f\\t u24=%f\\n\\n\" % (u21,u22,u23,u24)\n",
+ "print \"the error in the solution is: %f\\n\\n\" % (math.fabs(u22-f(0.6,0.04)))\n",
+ "#crank-nicolson formula\n",
+ "#by putting i=1,2,3,4 we obtain four equation\n",
+ "A=matrix([[4, -1, 0, 0] ,[-1, 4, -1, 0],[0, -1, 4, -1],[0, 0, -1, 4]])\n",
+ "C=matrix([[0.9510],[1.5388],[1.5388],[0.9510]])\n",
+ "X=A.I*C\n",
+ "print \"u00=%f\\t u10=%f\\t u20=%f\\t u30=%f\\t\\n\\n\" %(X[0][0],X[1][0],X[2][0],X[3][0])\n",
+ "print \"the error in the solution is: %f\\n\\n\" %(abs(X[1][0]-f(0.6,0.04)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "u11=0.475528\t u12=0.769421\t u13=0.769421\t u14=0.475528\n",
+ "\n",
+ "\n",
+ "u21=0.384710\t u22=0.622475\t u23=0.622475\t u24=0.384710\n",
+ "\n",
+ "\n",
+ "the error in the solution is: 0.018372\n",
+ "\n",
+ "\n",
+ "u00=0.399255\t u10=0.646018\t u20=0.646018\t u30=0.399255\t\n",
+ "\n",
+ "\n",
+ "the error in the solution is: 0.005172\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.8:pg-364"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "#heat equation using crank-nicolson method\n",
+ "#example 9.8\n",
+ "#page 364\n",
+ "from numpy import matrix\n",
+ "import math\n",
+ "z=0.01878\n",
+ "#h=1/2 l=1/8,i=1\n",
+ "u01=0.0\n",
+ "u21=1.0/8.0\n",
+ "u11=(u21+u01)/6.0\n",
+ "print \" u11=%f\\n\\n\" % (u11)\n",
+ "print \"error is %f\\n\\n\" % (math.fabs(u11-z))\n",
+ "#h=1/4,l=1/8,i=1,2,3\n",
+ "A=matrix([[-3.0 ,-1.0 ,0.0],[1.0,-3.0,1.0],[0.0,1.0,-3.0]])\n",
+ "C=matrix([[0.0],[0.0],[-0.125]])\n",
+ "#here we found inverese of A then we multipy it with C\n",
+ "X=A.I*C\n",
+ "print \"u12=%f\\n\\n\" % (X[1][0])\n",
+ "print \"error is %f\\n\\n\" %(math.fabs(X[1][0]-z))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " u11=0.020833\n",
+ "\n",
+ "\n",
+ "error is 0.002053\n",
+ "\n",
+ "\n",
+ "u12=0.013889\n",
+ "\n"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "error is 0.004891\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter1_7.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter1_7.ipynb new file mode 100644 index 00000000..29a46a36 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter1_7.ipynb @@ -0,0 +1,625 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:6ce091ab8e1fc1b8237507a6211c6c4c45a07ba8baa4f3364e4633c5bb15666c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter01:Errors in Numerical Calculations"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.1:pg-7"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 1.1\n",
+ "#rounding off\n",
+ "#page 7\n",
+ "a1=1.6583\n",
+ "a2=30.0567\n",
+ "a3=0.859378\n",
+ "a4=3.14159\n",
+ "print \"\\nthe numbers after rounding to 4 significant figures are given below\\n\"\n",
+ "print \" %f %.4g\\n'\" %(a1,a1)\n",
+ "print \" %f %.4g\\n\" %(a2,a2)\n",
+ "print \" %f %.4g\\n\" %(a3,a3)\n",
+ "print \" %f %.4g\\n\" %(a4,a4)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "the numbers after rounding to 4 significant figures are given below\n",
+ "\n",
+ " 1.658300 1.658\n",
+ "'\n",
+ " 30.056700 30.06\n",
+ "\n",
+ " 0.859378 0.8594\n",
+ "\n",
+ " 3.141590 3.142\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.2:pg-9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 1.2\n",
+ "#percentage accuracy\n",
+ "#page 9\n",
+ "import math\n",
+ "x=0.51 # the number given\n",
+ "n=2 #correcting upto 2 decimal places\n",
+ "d=math.pow(10,-n)\n",
+ "d=d/2.0\n",
+ "p=(d/x)*100 #percentage accuracy\n",
+ "print \"the percentage accuracy of %f after correcting to two decimal places is %f\" %(x,p)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the percentage accuracy of 0.510000 after correcting to two decimal places is 0.980392\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.3:pg-9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 1.3\n",
+ "#absolute and relative errors\n",
+ "#page 9\n",
+ "X=3.1428571 #approximate value of pi\n",
+ "T_X=3.1415926 # true value of pi\n",
+ "A_E=T_X-X #absolute error\n",
+ "R_E=A_E/T_X #relative error\n",
+ "print \"Absolute Error = %0.7f \\n Relative Error = %0.7f\" %(A_E,R_E)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Absolute Error = -0.0012645 \n",
+ " Relative Error = -0.0004025\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.4:pg-10"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 1.4\n",
+ "#best approximation\n",
+ "#page 10\n",
+ "X=1/3 #the actual number\n",
+ "X1=0.30\n",
+ "X2=0.33\n",
+ "X3=0.34\n",
+ "E1=abs(X-X1)\n",
+ "E2=abs(X-X2)\n",
+ "E3=abs(X-X3)\n",
+ "if E1<E2:\n",
+ " if E1<E3:\n",
+ " B_A=X1\n",
+ "elif E2<E1:\n",
+ " if E2<E3:\n",
+ " B_A=X2\n",
+ "elif E3<E2:\n",
+ " if E3<E1:\n",
+ " B_A=X3\n",
+ "print \"the best approximation of 1/3 is %f\" %(B_A)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the best approximation of 1/3 is 0.300000\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.5:pg-10"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#relative error\n",
+ "#example 1.5\n",
+ "#page 10\n",
+ "import math\n",
+ "n=8.6 # the corrected number\n",
+ "N=1 #the no is rounded to one decimal places\n",
+ "E_A=math.pow(10,-N)/2\n",
+ "E_R=E_A/n\n",
+ "print \"the relative error of the number is:%0.4f\" %(E_R)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the relative error of the number is:0.0058\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.6:pg-10"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 1.6\n",
+ "#absolute error and relative error\n",
+ "#page 10\n",
+ "import math\n",
+ "s=math.sqrt(3)+math.sqrt(5)+math.sqrt(7) #the sum square root of 3,5,7\n",
+ "n=4\n",
+ "Ea=3*(math.pow(10,-n)/2) #absolute error\n",
+ "R_E=Ea/s\n",
+ "print \"the sum of square roots is %0.4g \\n\" %(s)\n",
+ "print \"the absolute error is %f \\n\" %(Ea)\n",
+ "print \"the relative error is %f\" %(R_E)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the sum of square roots is 6.614 \n",
+ "\n",
+ "the absolute error is 0.000150 \n",
+ "\n",
+ "the relative error is 0.000023\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.7:pg-10"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#absolute error\n",
+ "#example 1.7\n",
+ "#page 10\n",
+ "n=[0.1532, 15.45, 0.0000354, 305.1, 8.12, 143.3, 0.0212, 0.643, 0.1734] #original numbers\n",
+ "#rounding all numbers to 2 decimal places\n",
+ "n=[305.1, 143.3, 0.15,15.45, 0.00, 8.12, 0.02, 0.64, 0.17] \n",
+ "sum=0;\n",
+ "#l=length(n);\n",
+ "for i in range(len(n)):\n",
+ " sum=sum+n[i];\n",
+ "\n",
+ "E_A=2*math.pow(10,-1)/2+7*math.pow(10,-2)/2\n",
+ "print \"the absolute error is:%0.2f\" %(E_A)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the absolute error is:0.14\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.8:pg-11"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#difference in 3 significant figures\n",
+ "#example 1.8\n",
+ "#page 11\n",
+ "X1=math.sqrt(6.37)\n",
+ "X2=math.sqrt(6.36)\n",
+ "d=X1-X2 #difference between two numbers\n",
+ "print \"the difference corrected to 3 significant figures is %0.3g\" %(d)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the difference corrected to 3 significant figures is 0.00198\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.9:pg-12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#relative error\n",
+ "#example 1.10\n",
+ "#page 12\n",
+ "a=6.54\n",
+ "b=48.64\n",
+ "c=13.5\n",
+ "da=0.01\n",
+ "db=0.02\n",
+ "dc=0.03\n",
+ "s=math.pow(a,2)*math.sqrt(b)/math.pow(c,3)\n",
+ "#disp(s,'s=')\n",
+ "print \"s=%f\" %(s)\n",
+ "r_err=2*(da/a)+(db/b)/2+3*(dc/c);\n",
+ "print \"the relative error is :%f\" %(r_err)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "s=0.121241\n",
+ "the relative error is :0.009930\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.11:pg-13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#relative error\n",
+ "#example 1.11\n",
+ "#page 13\n",
+ "import math\n",
+ "x=1\n",
+ "y=1\n",
+ "z=1\n",
+ "u=(5*x*math.pow(y,3))/math.pow(z,3)\n",
+ "dx=0.001\n",
+ "dy=0.001\n",
+ "dz=0.001\n",
+ "max=((5*math.pow(y,2))/math.pow(z,3))*dx+((10*x*y)/math.pow(z,3))*dy+((15*x*math.pow(y,2))/math.pow(z,4))*dz\n",
+ "e=max/u\n",
+ "print \" the relative error is :%f\" %(e)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " the relative error is :0.006000\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.12:pg-12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#taylor series\n",
+ "#example 1.12\n",
+ "#page 12\n",
+ "import math\n",
+ "def f(x):\n",
+ " return math.pow(x,3)+5*x-10\n",
+ "def f1(x):\n",
+ " return 3*math.pow(x,2)-6*x+5\n",
+ "def f2(x):\n",
+ " return 6*x-6\n",
+ "def f3(x):\n",
+ " return 6\n",
+ "D=[0,f(0), f1(0), f2(0), f3(0)]\n",
+ "S1=0;\n",
+ "h=1;\n",
+ "for i in range(1,5):\n",
+ " S1=S1+math.pow(h,i-1)*D[i]/math.factorial(i-1)\n",
+ " \n",
+ "print \"the third order taylors series approximation of f(1) is :%d\" %(S1)\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the third order taylors series approximation of f(1) is :-7\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.13:pg-16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#taylor series\n",
+ "#example 1.13\n",
+ "#page 16\n",
+ "import math\n",
+ "def f(x):\n",
+ " return math.sin(x)\n",
+ "def f1(x):\n",
+ " return math.cos(x)\n",
+ "def f2(x):\n",
+ " return -1*math.sin(x)\n",
+ "def f3(x):\n",
+ " return -1*math.cos(x)\n",
+ "def f4(x):\n",
+ " return math.sin(x)\n",
+ "def f5(x):\n",
+ " return math.cos(x)\n",
+ "def f6(x):\n",
+ " return -1*math.sin(x)\n",
+ "def f7(x):\n",
+ " return -1*math.cos(x)\n",
+ "D=[0,f(math.pi/6), f1(math.pi/6), f2(math.pi/6), f3(math.pi/6), f4(math.pi/6), f5(math.pi/6), f6(math.pi/6), f7(math.pi/6)]\n",
+ "S1=0\n",
+ "h=math.pi/6\n",
+ "print \"order of approximation computed value of sin(pi/3) absolute eror\\n\\n\"\n",
+ "for j in range(1,10):\n",
+ " for i in range(1,j):\n",
+ " S1=S1+math.pow(h,i-1)*D[i]/math.factorial(i-1) \n",
+ " print \"%d %0.9f %0.9f\\n\" %(j,S1,abs(math.sin(math.pi/3)-S1))\n",
+ " S1=0\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "order of approximation computed value of sin(pi/3) absolute eror\n",
+ "\n",
+ "\n",
+ "1 0.000000000 0.866025404\n",
+ "\n",
+ "2 0.500000000 0.366025404\n",
+ "\n",
+ "3 0.953449841 0.087424437\n",
+ "\n",
+ "4 0.884910922 0.018885518\n",
+ "\n",
+ "5 0.864191614 0.001833790\n",
+ "\n",
+ "6 0.865757475 0.000267929\n",
+ "\n",
+ "7 0.866041490 0.000016087\n",
+ "\n",
+ "8 0.866027181 0.000001777\n",
+ "\n",
+ "9 0.866025327 0.000000077\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.14:pg-18"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#maclaurins expansion\n",
+ "#example 1.14\n",
+ "#page 18\n",
+ "n=8 #correct to 8 decimal places\n",
+ "x=1\n",
+ "for i in range(1,50):\n",
+ " if x/math.factorial(i)<math.pow(10,-8)/2:\n",
+ " c=i\n",
+ " break \n",
+ "print \"no. of terms needed to correct to 8 decimal places is : %d \" %(c)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "no. of terms needed to correct to 8 decimal places is : 2 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.15:pg-18"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#series apprixamation\n",
+ "#example 1.15\n",
+ "#page 18\n",
+ "import math\n",
+ "x=.09090909 # 1/11 =.09090909\n",
+ "S1=0\n",
+ "for i in range(1,5,2):\n",
+ " S1=S1+math.pow(x,i)/i\n",
+ "print \"value of log(1.2) is : %0.8f\\n\\n\" %(2*S1)\n",
+ "c=0\n",
+ "for i in range(1,50):\n",
+ " if math.pow(.09090909,i)/i<2*math.pow(10,-7):\n",
+ " c=i\n",
+ " break\n",
+ "print \"min no of terms needed to get value wuth same accuracy is :%d\" %(c)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of log(1.2) is : 0.18231906\n",
+ "\n",
+ "\n",
+ "min no of terms needed to get value wuth same accuracy is :6\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter2_7.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter2_7.ipynb new file mode 100644 index 00000000..9b5c0d28 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter2_7.ipynb @@ -0,0 +1,2186 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:ffc4134b3a37ea987dbf1bfd4b0eaeb28bc328e0841865ca188bb3d1fb69f7c4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter02:Solution of Algebric and Transcendental Equations"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.1:pg-24"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.1\n",
+ "#bisection method\n",
+ "#page 24\n",
+ "import math\n",
+ "def f(x):\n",
+ " return math.pow(x,3)-x-1\n",
+ "x1=1\n",
+ "x2=2 #f(1) is negative and f(2) is positive\n",
+ "d=0.0001 #for accuracy of root\n",
+ "c=1\n",
+ "print \"Succesive approximations \\t x1\\t \\tx2\\t \\tm\\t \\tf(m)\\n\"\n",
+ "while abs(x1-x2)>d:\n",
+ " \n",
+ " m=(x1+x2)/2.0\n",
+ " print \" \\t%f\\t%f\\t%f\\t%f\\n\" %(x1,x2,m,f(m))\n",
+ " if f(m)*f(x1)>0.0:\n",
+ " x1=m\n",
+ " else:\n",
+ " x2=m \n",
+ " c=c+1 # to count number of iterations \n",
+ "print \"the solution of equation after %i iteration is %g\" %(c,m)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Succesive approximations \t x1\t \tx2\t \tm\t \tf(m)\n",
+ "\n",
+ " \t1.000000\t2.000000\t1.500000\t0.875000\n",
+ "\n",
+ " \t1.000000\t1.500000\t1.250000\t-0.296875\n",
+ "\n",
+ " \t1.250000\t1.500000\t1.375000\t0.224609\n",
+ "\n",
+ " \t1.250000\t1.375000\t1.312500\t-0.051514\n",
+ "\n",
+ " \t1.312500\t1.375000\t1.343750\t0.082611\n",
+ "\n",
+ " \t1.312500\t1.343750\t1.328125\t0.014576\n",
+ "\n",
+ " \t1.312500\t1.328125\t1.320312\t-0.018711\n",
+ "\n",
+ " \t1.320312\t1.328125\t1.324219\t-0.002128\n",
+ "\n",
+ " \t1.324219\t1.328125\t1.326172\t0.006209\n",
+ "\n",
+ " \t1.324219\t1.326172\t1.325195\t0.002037\n",
+ "\n",
+ " \t1.324219\t1.325195\t1.324707\t-0.000047\n",
+ "\n",
+ " \t1.324707\t1.325195\t1.324951\t0.000995\n",
+ "\n",
+ " \t1.324707\t1.324951\t1.324829\t0.000474\n",
+ "\n",
+ " \t1.324707\t1.324829\t1.324768\t0.000214\n",
+ "\n",
+ "the solution of equation after 15 iteration is 1.32477'\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.2:pg-25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.2\n",
+ "#bisection method\n",
+ "#page 25\n",
+ "import math\n",
+ "def f(x):\n",
+ " return math.pow(x,3)-2*x-5\n",
+ "x1=2 \n",
+ "x2=3 #f(2) is negative and f(3) is positive\n",
+ "d=0.0001 #for accuracy of root\n",
+ "c=1\n",
+ "print \"Succesive approximations \\t x1\\t \\tx2\\t \\tm\\t \\tf(m)\\n\"\n",
+ "while abs(x1-x2)>d:\n",
+ " m=(x1+x2)/2.0\n",
+ " print \" \\t%f\\t%f\\t%f\\t%f\\n\" %(x1,x2,m,f(m))\n",
+ " if f(m)*f(x1)>0:\n",
+ " x1=m\n",
+ " else:\n",
+ " x2=m \n",
+ " c=c+1;# to count number of iterations \n",
+ "print \"the solution of equation after %i iteration is %0.4g\" %(c,m)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Succesive approximations \t x1\t \tx2\t \tm\t \tf(m)\n",
+ "\n",
+ " \t2.000000\t3.000000\t2.500000\t5.625000\n",
+ "\n",
+ " \t2.000000\t2.500000\t2.250000\t1.890625\n",
+ "\n",
+ " \t2.000000\t2.250000\t2.125000\t0.345703\n",
+ "\n",
+ " \t2.000000\t2.125000\t2.062500\t-0.351318\n",
+ "\n",
+ " \t2.062500\t2.125000\t2.093750\t-0.008942\n",
+ "\n",
+ " \t2.093750\t2.125000\t2.109375\t0.166836\n",
+ "\n",
+ " \t2.093750\t2.109375\t2.101562\t0.078562\n",
+ "\n",
+ " \t2.093750\t2.101562\t2.097656\t0.034714\n",
+ "\n",
+ " \t2.093750\t2.097656\t2.095703\t0.012862\n",
+ "\n",
+ " \t2.093750\t2.095703\t2.094727\t0.001954\n",
+ "\n",
+ " \t2.093750\t2.094727\t2.094238\t-0.003495\n",
+ "\n",
+ " \t2.094238\t2.094727\t2.094482\t-0.000771\n",
+ "\n",
+ " \t2.094482\t2.094727\t2.094604\t0.000592\n",
+ "\n",
+ " \t2.094482\t2.094604\t2.094543\t-0.000090\n",
+ "\n",
+ "the solution of equation after 15 iteration is 2.095\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.3:pg-26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.3\n",
+ "#bisection method\n",
+ "#page 26\n",
+ "import math\n",
+ "def f(x):\n",
+ " return math.pow(x,3)+math.pow(x,2)+x+7\n",
+ "x1=-3\n",
+ "x2=-2 #f(-3) is negative and f(-2) is positive\n",
+ "d=0.0001 #for accuracy of root\n",
+ "c=1\n",
+ "print \"Succesive approximations \\t x1\\t \\tx2\\t \\tm\\t \\tf(m)\\n\"\n",
+ "while abs(x1-x2)>d:\n",
+ " m=(x1+x2)/2.0\n",
+ " print \" \\t%f\\t%f\\t%f\\t%f\\n\" %(x1,x2,m,f(m))\n",
+ " if f(m)*f(x1)>0:\n",
+ " x1=m\n",
+ " else:\n",
+ " x2=m \n",
+ " c=c+1 # to count number of iterations \n",
+ "print \"the solution of equation after %i iteration is %0.4g\" %(c,m)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Succesive approximations \t x1\t \tx2\t \tm\t \tf(m)\n",
+ "\n",
+ " \t-3.000000\t-2.000000\t-2.500000\t-4.875000\n",
+ "\n",
+ " \t-2.500000\t-2.000000\t-2.250000\t-1.578125\n",
+ "\n",
+ " \t-2.250000\t-2.000000\t-2.125000\t-0.205078\n",
+ "\n",
+ " \t-2.125000\t-2.000000\t-2.062500\t0.417725\n",
+ "\n",
+ " \t-2.125000\t-2.062500\t-2.093750\t0.111481\n",
+ "\n",
+ " \t-2.125000\t-2.093750\t-2.109375\t-0.045498\n",
+ "\n",
+ " \t-2.109375\t-2.093750\t-2.101562\t0.033315\n",
+ "\n",
+ " \t-2.109375\t-2.101562\t-2.105469\t-0.006010\n",
+ "\n",
+ " \t-2.105469\t-2.101562\t-2.103516\t0.013673\n",
+ "\n",
+ " \t-2.105469\t-2.103516\t-2.104492\t0.003836\n",
+ "\n",
+ " \t-2.105469\t-2.104492\t-2.104980\t-0.001086\n",
+ "\n",
+ " \t-2.104980\t-2.104492\t-2.104736\t0.001376\n",
+ "\n",
+ " \t-2.104980\t-2.104736\t-2.104858\t0.000145\n",
+ "\n",
+ " \t-2.104980\t-2.104858\t-2.104919\t-0.000470\n",
+ "\n",
+ "the solution of equation after 15 iteration is -2.105\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.4:pg-26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.4\n",
+ "#bisection method\n",
+ "#page 26\n",
+ "import math\n",
+ "def f(x):\n",
+ " return x*math.exp(x)-1\n",
+ "x1=0 \n",
+ "x2=1 #f(0) is negative and f(1) is positive\n",
+ "d=0.0005 #maximun tolerance value\n",
+ "c=1\n",
+ "print \"Succesive approximations \\t x1\\t \\tx2\\t \\tm\\t \\ttol\\t \\tf(m)\\n\"\n",
+ "while abs((x2-x1)/x2)>d:\n",
+ " m=(x1+x2)/2.0 #tolerance value for each iteration\n",
+ " tol=((x2-x1)/x2)*100\n",
+ " print \" \\t%f\\t%f\\t%f\\t%f\\t%f\\n\" %(x1,x2,m,tol,f(m))\n",
+ " if f(m)*f(x1)>0:\n",
+ " x1=m\n",
+ " else:\n",
+ " x2=m \n",
+ " c=c+1 # to count number of iterations \n",
+ "print \"the solution of equation after %i iteration is %0.4g\" %(c,m)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Succesive approximations \t x1\t \tx2\t \tm\t \ttol\t \tf(m)\n",
+ "\n",
+ " \t0.000000\t1.000000\t0.500000\t100.000000\t-0.175639\n",
+ "\n",
+ " \t0.500000\t1.000000\t0.750000\t50.000000\t0.587750\n",
+ "\n",
+ " \t0.500000\t0.750000\t0.625000\t33.333333\t0.167654\n",
+ "\n",
+ " \t0.500000\t0.625000\t0.562500\t20.000000\t-0.012782\n",
+ "\n",
+ " \t0.562500\t0.625000\t0.593750\t10.000000\t0.075142\n",
+ "\n",
+ " \t0.562500\t0.593750\t0.578125\t5.263158\t0.030619\n",
+ "\n",
+ " \t0.562500\t0.578125\t0.570312\t2.702703\t0.008780\n",
+ "\n",
+ " \t0.562500\t0.570312\t0.566406\t1.369863\t-0.002035\n",
+ "\n",
+ " \t0.566406\t0.570312\t0.568359\t0.684932\t0.003364\n",
+ "\n",
+ " \t0.566406\t0.568359\t0.567383\t0.343643\t0.000662\n",
+ "\n",
+ " \t0.566406\t0.567383\t0.566895\t0.172117\t-0.000687\n",
+ "\n",
+ " \t0.566895\t0.567383\t0.567139\t0.086059\t-0.000013\n",
+ "\n",
+ "the solution of equation after 13 iteration is 0.5671\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.5:pg-27"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.5\n",
+ "#bisection method\n",
+ "#page 27\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 4*math.exp(-x)*math.sin(x)-1\n",
+ "x1=0 \n",
+ "x2=0.5 #f(0) is negative and f(1) is positive\n",
+ "d=0.0001 #for accuracy of root\n",
+ "c=1 \n",
+ "print \"Succesive approximations \\t x1\\t \\tx2\\t \\tm\\t \\t \\tf(m)\\n\"\n",
+ "while abs(x2-x1)>d:\n",
+ " m=(x1+x2)/2.0\n",
+ " print \" \\t%f\\t%f\\t%f\\t%f\\n\" %(x1,x2,m,f(m))\n",
+ " if f(m)*f(x1)>0:\n",
+ " x1=m\n",
+ " else:\n",
+ " x2=m \n",
+ " c=c+1 # to count number of iterations \n",
+ "print \"the solution of equation after %i iteration is %0.3g\" %(c,m)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Succesive approximations \t x1\t \tx2\t \tm\t \t \tf(m)\n",
+ "\n",
+ " \t0.000000\t0.500000\t0.250000\t-0.229286\n",
+ "\n",
+ " \t0.250000\t0.500000\t0.375000\t0.006941\n",
+ "\n",
+ " \t0.250000\t0.375000\t0.312500\t-0.100293\n",
+ "\n",
+ " \t0.312500\t0.375000\t0.343750\t-0.044068\n",
+ "\n",
+ " \t0.343750\t0.375000\t0.359375\t-0.017925\n",
+ "\n",
+ " \t0.359375\t0.375000\t0.367188\t-0.005334\n",
+ "\n",
+ " \t0.367188\t0.375000\t0.371094\t0.000842\n",
+ "\n",
+ " \t0.367188\t0.371094\t0.369141\t-0.002236\n",
+ "\n",
+ " \t0.369141\t0.371094\t0.370117\t-0.000694\n",
+ "\n",
+ " \t0.370117\t0.371094\t0.370605\t0.000075\n",
+ "\n",
+ " \t0.370117\t0.370605\t0.370361\t-0.000310\n",
+ "\n",
+ " \t0.370361\t0.370605\t0.370483\t-0.000118\n",
+ "\n",
+ " \t0.370483\t0.370605\t0.370544\t-0.000022\n",
+ "\n",
+ "the solution of equation after 14 iteration is 0.371\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.6:pg-28"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.6\n",
+ "#false position method\n",
+ "#page 28\n",
+ "import math\n",
+ "def f(x):\n",
+ " return x**3-2*x-5\n",
+ "a=2.0\n",
+ "b=3.0 #f(2) is negative and f(3)is positive\n",
+ "d=0.00001\n",
+ "print \"succesive iterations \\ta\\t b\\t f(a)\\t f(b)\\t\\ x1\\n\"\n",
+ "for i in range(1,25):\n",
+ " x1=b*f(a)/(f(a)-f(b))+a*f(b)/(f(b)-f(a))\n",
+ " if(f(a)*f(x1))>0:\n",
+ " b=x1\n",
+ " else:\n",
+ " a=x1\n",
+ " if abs(f(x1))<d:\n",
+ " break\n",
+ " print \" \\t%f %f %f %f %f\\n\" %(a,b,f(a),f(b),x1)\n",
+ "print \"the root of the equation is %f\" %(x1)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "succesive iterations \ta\t b\t f(a)\t f(b)\t\\ x1\n",
+ "\n",
+ " \t2.000000 2.058824 -1.000000 -0.390800 2.058824\n",
+ "\n",
+ " \t2.096559 2.058824 0.022428 -0.390800 2.096559\n",
+ "\n",
+ " \t2.094511 2.058824 -0.000457 -0.390800 2.094511\n",
+ "\n",
+ "the root of the equation is 2.094552\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.7:pg-29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.7\n",
+ "#false position method\n",
+ "#page 29\n",
+ "def f(x):\n",
+ " return x**2.2-69\n",
+ "a=5.0\n",
+ "b=6.0 #f(5) is negative and f(6)is positive\n",
+ "d=0.00001\n",
+ "print \"succesive iterations \\ta\\t b\\t f(a)\\t f(b)\\t\\ x1\\n\"\n",
+ "for i in range(1,25):\n",
+ " x1=b*f(a)/(f(a)-f(b))+a*f(b)/(f(b)-f(a));\n",
+ " if(f(a)*f(x1))>0:\n",
+ " b=x1\n",
+ " else:\n",
+ " a=x1\n",
+ " if abs(f(x1))<d:\n",
+ " break\n",
+ " print \" \\t%f %f %f %f %f\\n\" %(a,b,f(a),f(b),x1)\n",
+ "print \"the root of the equation is %f\" %(x1)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "succesive iterations \ta\t b\t f(a)\t f(b)\t\\ x1\n",
+ "\n",
+ " \t7.027228 6.000000 3.933141 -17.485113 7.027228\n",
+ "\n",
+ " \t6.838593 6.000000 -0.304723 -17.485113 6.838593\n",
+ "\n",
+ " \t6.853467 6.000000 0.024411 -17.485113 6.853467\n",
+ "\n",
+ " \t6.852277 6.000000 -0.001950 -17.485113 6.852277\n",
+ "\n",
+ " \t6.852372 6.000000 0.000156 -17.485113 6.852372\n",
+ "\n",
+ " \t6.852365 6.000000 -0.000012 -17.485113 6.852365\n",
+ "\n",
+ "the root of the equation is 6.852365\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.8:pg-29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.8\n",
+ "#false position method\n",
+ "#page 29\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 2*x-log10(x)-7\n",
+ "a=3.0\n",
+ "b=4.0 #f(3) is negative and f(4)is positive\n",
+ "d=0.00001\n",
+ "print \"succesive iterations \\ta \\t b\\t f(a)\\t f(b)\\t x1\\n\"\n",
+ "for i in range(1,25):\n",
+ " x1=b*f(a)/(f(a)-f(b))+a*f(b)/(f(b)-f(a))\n",
+ " if(f(a)*f(x1))>0:\n",
+ " b=x1\n",
+ " else:\n",
+ " a=x1\n",
+ " if abs(f(x1))<d:\n",
+ " break\n",
+ " print \" \\t%f %f %f %f %f\\n\" %(a,b,f(a),f(b),x1)\n",
+ "print \"the root of the equation is %0.4g\" %(x1)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "succesive iterations \ta \t b\t f(a)\t f(b)\t x1\n",
+ "\n",
+ " \t3.000000 3.787772 -1.477121 -0.002839 3.787772\n",
+ "\n",
+ " \t3.789289 3.787772 0.000021 -0.002839 3.789289\n",
+ "\n",
+ "the root of the equation is 3.789\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.9:pg-30"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.9\n",
+ "#false position method\n",
+ "#page 30\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 4*math.exp(-x)*math.sin(x)-1\n",
+ "a=0.0\n",
+ "b=0.5 #f(0) is negative and f(0.5)is positive\n",
+ "d=0.00001\n",
+ "print \"succesive iterations \\ta\\t b\\t f(a)\\t f(b)\\t\\ x1\\n\"\n",
+ "for i in range(1,25):\n",
+ " x1=b*f(a)/(f(a)-f(b))+a*f(b)/(f(b)-f(a))\n",
+ " if(f(a)*f(x1))>0:\n",
+ " b=x1\n",
+ " else:\n",
+ " a=x1\n",
+ " if abs(f(x1))<d:\n",
+ " break\n",
+ " print \" \\t%f %f %f %f %f\\n\" %(a,b,f(a),f(b),x1)\n",
+ "print \"the root of the equation is %f\" %(x1)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "succesive iterations \ta\t b\t f(a)\t f(b)\t\\ x1\n",
+ "\n",
+ " \t0.429869 0.500000 0.084545 0.163145 0.429869\n",
+ "\n",
+ " \t0.354433 0.500000 -0.026054 0.163145 0.354433\n",
+ "\n",
+ " \t0.374479 0.500000 0.006132 0.163145 0.374479\n",
+ "\n",
+ " \t0.369577 0.500000 -0.001547 0.163145 0.369577\n",
+ "\n",
+ " \t0.370802 0.500000 0.000384 0.163145 0.370802\n",
+ "\n",
+ " \t0.370497 0.500000 -0.000096 0.163145 0.370497\n",
+ "\n",
+ " \t0.370573 0.500000 0.000024 0.163145 0.370573\n",
+ "\n",
+ "the root of the equation is 0.370554\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.10:pg-33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.10\n",
+ "#iteration method\n",
+ "#page 33\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 1/(math.sqrt(x+1))\n",
+ "x1=0.75\n",
+ "x2=0.0\n",
+ "n=1\n",
+ "d=0.0001 #accuracy opto 10^-4\n",
+ "c=0 #to count no of iterations \n",
+ "print \"successive iterations \\t\\x01\\tf(x1)\\n\"\n",
+ "while abs(x1-x2)>d:\n",
+ " print \" \\t%f %f\\n\" %(x1,f(x1))\n",
+ " x2=x1\n",
+ " x1=f(x1)\n",
+ " c=c+1\n",
+ "print \" the root of the eqaution after %i iteration is %0.4g\" %(c,x1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t\u0001\tf(x1)\n",
+ "\n",
+ " \t0.750000 0.755929\n",
+ "\n",
+ " \t0.755929 0.754652\n",
+ "\n",
+ " \t0.754652 0.754926\n",
+ "\n",
+ " \t0.754926 0.754867\n",
+ "\n",
+ " the root of the eqaution after 4 iteration is 0.7549\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.11:pg-34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.11\n",
+ "#iteration method\n",
+ "#page34\n",
+ "import math\n",
+ "def f(x):\n",
+ " return cos(x)/2.0+3.0/2.0\n",
+ "x1=1.5 # as roots lies between 3/2 and pi/2\n",
+ "x2=0\n",
+ "d=0.0001 # accuracy opto 10^-4\n",
+ "c=0 # to count no of iterations \n",
+ "print \"successive iterations \\t\\x01\\tf(x1)\\n\"\n",
+ "while abs(x2-x1)>d:\n",
+ " \n",
+ " print \" \\t%f %f\\n\" %(x1,f(x1))\n",
+ " x2=x1\n",
+ " x1=f(x1)\n",
+ " c=c+1\n",
+ "print \" the root of the eqaution after %i iteration is %0.4g\" %(c,x1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t\u0001\tf(x1)\n",
+ "\n",
+ " \t1.500000 1.535369\n",
+ "\n",
+ " \t1.535369 1.517710\n",
+ "\n",
+ " \t1.517710 1.526531\n",
+ "\n",
+ " \t1.526531 1.522126\n",
+ "\n",
+ " \t1.522126 1.524326\n",
+ "\n",
+ " \t1.524326 1.523227\n",
+ "\n",
+ " \t1.523227 1.523776\n",
+ "\n",
+ " \t1.523776 1.523502\n",
+ "\n",
+ " \t1.523502 1.523639\n",
+ "\n",
+ " \t1.523639 1.523570\n",
+ "\n",
+ " the root of the eqaution after 10 iteration is 1.524\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.12:pg-35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.12\n",
+ "#iteration method\n",
+ "#page 35\n",
+ "import math\n",
+ "def f(x):\n",
+ " return math.exp(-x)\n",
+ "x1=1.5 # as roots lies between 0 and 1\n",
+ "x2=0\n",
+ "d=0.0001 # accuracy opto 10^-4\n",
+ "c=0 # to count no of iterations \n",
+ "print \"successive iterations \\t x1 \\t f(x1)\\n\"\n",
+ "while abs(x2-x1)>d:\n",
+ " \n",
+ " print \" \\t%f %f\\n\" %(x1,f(x1))\n",
+ " x2=x1\n",
+ " x1=f(x1)\n",
+ " c=c+1\n",
+ "print \" the root of the eqaution after %i iteration is %0.4g\" %(c,x1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x1 \t f(x1)\n",
+ "\n",
+ " \t1.500000 0.223130\n",
+ "\n",
+ " \t0.223130 0.800011\n",
+ "\n",
+ " \t0.800011 0.449324\n",
+ "\n",
+ " \t0.449324 0.638059\n",
+ "\n",
+ " \t0.638059 0.528317\n",
+ "\n",
+ " \t0.528317 0.589597\n",
+ "\n",
+ " \t0.589597 0.554551\n",
+ "\n",
+ " \t0.554551 0.574330\n",
+ "\n",
+ " \t0.574330 0.563082\n",
+ "\n",
+ " \t0.563082 0.569451\n",
+ "\n",
+ " \t0.569451 0.565836\n",
+ "\n",
+ " \t0.565836 0.567885\n",
+ "\n",
+ " \t0.567885 0.566723\n",
+ "\n",
+ " \t0.566723 0.567382\n",
+ "\n",
+ " \t0.567382 0.567008\n",
+ "\n",
+ " \t0.567008 0.567220\n",
+ "\n",
+ " \t0.567220 0.567100\n",
+ "\n",
+ " \t0.567100 0.567168\n",
+ "\n",
+ " the root of the eqaution after 18 iteration is 0.5672\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.13:pg-35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.13\n",
+ "#iteration method\n",
+ "#page 35\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 1+math.sin(x)/10\n",
+ "x1=1.0 # as roots lies between 1 and pi evident from graph\n",
+ "x2=0\n",
+ "d=0.0001 # accuracy opto 10^-4\n",
+ "c=0 # to count no of iterations \n",
+ "print \"successive iterations \\t x1 \\t f(x1)\\n\"\n",
+ "while abs(x2-x1)>d:\n",
+ " print \" \\t%f %f\\n\" %(x1,f(x1))\n",
+ " x2=x1\n",
+ " x1=f(x1)\n",
+ " c=c+1\n",
+ "print \" the root of the eqaution after %i iteration is %0.4g\" %(c,x1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x1 \t f(x1)\n",
+ "\n",
+ " \t1.000000 1.084147\n",
+ "\n",
+ " \t1.084147 1.088390\n",
+ "\n",
+ " \t1.088390 1.088588\n",
+ "\n",
+ " \t1.088588 1.088597\n",
+ "\n",
+ " the root of the eqaution after 4 iteration is 1.089\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.14:pg-36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.14\n",
+ "#aitken's process\n",
+ "#page 36\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 1.5+math.cos(x)/2.0\n",
+ "x0=1.5\n",
+ "y=0\n",
+ "e=0.0001\n",
+ "c=0\n",
+ "print \"successive iterations \\t x0 \\t x1 \\t x2 \\t x3 \\t y\\n\"\n",
+ "for i in range(1,10):\n",
+ " x1=f(x0)\n",
+ " x2=f(x1)\n",
+ " x3=f(x2)\n",
+ " y=x3-((x3-x2)**2)/(x3-2*x2+x1)\n",
+ " d=y-x0\n",
+ " x0=y\n",
+ " if abs(f(x0))<e:\n",
+ " break\n",
+ " c=c+1\n",
+ " print \" \\t%f %f %f %f %f\\n\" %(x0,x1,x2,x3,y)\n",
+ "print \"the root of the equation after %i iteration is %f\" %(c,y)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x0 \t x1 \t x2 \t x3 \t y\n",
+ "\n",
+ " \t1.523592 1.535369 1.517710 1.526531 1.523592\n",
+ "\n",
+ " \t1.523593 1.523593 1.523593 1.523593 1.523593\n",
+ "\n",
+ " \t1.523593 1.523593 1.523593 1.523593 1.523593\n",
+ "\n",
+ " \t1.523593 1.523593 1.523593 1.523593 1.523593\n",
+ "\n",
+ " \t1.523593 1.523593 1.523593 1.523593 1.523593\n",
+ "\n",
+ " \t1.523593 1.523593 1.523593 1.523593 1.523593\n",
+ "\n",
+ " \t1.523593 1.523593 1.523593 1.523593 1.523593\n",
+ "\n",
+ " \t1.523593 1.523593 1.523593 1.523593 1.523593\n",
+ "\n",
+ " \t1.523593 1.523593 1.523593 1.523593 1.523593\n",
+ "\n",
+ "the root of the equation after 9 iteration is 1.523593\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.15:pg-39"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.15\n",
+ "#newton-raphson method\n",
+ "#page 39\n",
+ "def f(x):\n",
+ " return x**3-2*x-5\n",
+ "def f1(x):\n",
+ " return 3*x**2-2 # first derivative of the function\n",
+ "x0=2.0 # initial value\n",
+ "d=0.0001\n",
+ "c=0\n",
+ "n=1\n",
+ "print \"successive iterations \\t x0 \\t f(x0) \\t f1(x0)\\n\"\n",
+ "while n==1:\n",
+ " x2=x0\n",
+ " x1=x0-(f(x0)/f1(x0))\n",
+ " x0=x1\n",
+ " print \" \\t%f \\t%f \\t%f \\n\" %(x2,f(x1),f1(x1))\n",
+ " c=c+1\n",
+ " if abs(f(x0))<d:\n",
+ " break\n",
+ "print \"the root of %i iteration is:%f\" %(c,x0)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x0 \t f(x0) \t f1(x0)\n",
+ "\n",
+ " \t2.000000 \t0.061000 \t11.230000 \n",
+ "\n",
+ " \t2.100000 \t0.000186 \t11.161647 \n",
+ "\n",
+ " \t2.094568 \t0.000000 \t11.161438 \n",
+ "\n",
+ "the root of 3 iteration is:2.094551\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.16:pg-40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.16\n",
+ "#newton-raphson method\n",
+ "#page 40\n",
+ "import math\n",
+ "def f(x):\n",
+ " return x*math.sin(x)+math.cos(x)\n",
+ "def f1(x):\n",
+ " return x*math.cos(x) #first derivation of the function\n",
+ "x0=math.pi # initial value\n",
+ "d=0.0001\n",
+ "c=0 \n",
+ "n=1\n",
+ "print \"successive iterations \\tx0\\t f(x0)\\t f1(x0)\\n\"\n",
+ "while n==1:\n",
+ " x2=x0\n",
+ " x1=x0-(f(x0)/f1(x0))\n",
+ " x0=x1\n",
+ " print \" \\t%f \\t%f \\t%f\\n\" %(x2,f(x1),f1(x1))\n",
+ " c=c+1\n",
+ " if abs(f(x0))<d:\n",
+ " break\n",
+ "print \"the root of %i iteration is:%f\" %(c,x0)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \tx0\t f(x0)\t f1(x0)\n",
+ "\n",
+ " \t3.141593 \t-0.066186 \t-2.681457\n",
+ "\n",
+ " \t2.823283 \t-0.000564 \t-2.635588\n",
+ "\n",
+ " \t2.798600 \t-0.000000 \t-2.635185\n",
+ "\n",
+ "the root of 3 iteration is:2.798386\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.17:pg-40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.17\n",
+ "#newton-raphson method\n",
+ "#page 40\n",
+ "import math\n",
+ "def f(x):\n",
+ " return x*math.exp(x)-1\n",
+ "def f1(x):\n",
+ " return math.exp(x)+x*math.exp(x) #first derivative of the function\n",
+ "x0=0 # initial value\n",
+ "d=0.0001 \n",
+ "c=0\n",
+ "n=1\n",
+ "print \"successive iterations \\tx0\\t f(x0)\\t f1(x0)\\n\"\n",
+ "while n==1:\n",
+ " x2=x0\n",
+ " x1=x0-(f(x0)/f1(x0))\n",
+ " x0=x1\n",
+ " print \" \\t%f \\t%f \\t%f\\n\" %(x2,f(x1),f1(x1))\n",
+ " c=c+1\n",
+ " if abs(f(x0))<d:\n",
+ " break\n",
+ "print \"the root of %i iteration is:%f\" %(c,x0)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \tx0\t f(x0)\t f1(x0)\n",
+ "\n",
+ " \t0.000000 \t1.718282 \t5.436564\n",
+ "\n",
+ " \t1.000000 \t0.355343 \t3.337012\n",
+ "\n",
+ " \t0.683940 \t0.028734 \t2.810232\n",
+ "\n",
+ " \t0.577454 \t0.000239 \t2.763614\n",
+ "\n",
+ " \t0.567230 \t0.000000 \t2.763223\n",
+ "\n",
+ "the root of 5 iteration is:0.567143\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.18:pg-41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.18\n",
+ "#newton-raphson method\n",
+ "#page 41\n",
+ "import math\n",
+ "def f(x):\n",
+ " return math.sin(x)-x/2.0\n",
+ "def f1(x):\n",
+ " return math.cos(x)-0.5\n",
+ "x0=math.pi/2.0 # initial value\n",
+ "d=0.0001\n",
+ "c=0\n",
+ "n=1\n",
+ "print \"successive iterations \\t x0 \\t f(x0)\\t f1(x0)\\n\"\n",
+ "while n==1:\n",
+ " x2=x0\n",
+ " x1=x0-(f(x0)/f1(x0))\n",
+ " x0=x1\n",
+ " print \" \\t%f\\t%f\\t%f\\n\" %(x2,f(x1),f1(x1))\n",
+ " c=c+1\n",
+ " if abs(f(x0))<d:\n",
+ " break\n",
+ "print \"the root of %i iteration is: %0.4g\" %(c,x0)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x0 \t f(x0)\t f1(x0)\n",
+ "\n",
+ " \t1.570796\t-0.090703\t-0.916147\n",
+ "\n",
+ " \t2.000000\t-0.004520\t-0.824232\n",
+ "\n",
+ " \t1.900996\t-0.000014\t-0.819039\n",
+ "\n",
+ "the root of 3 iteration is: 1.896\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.19:pg-41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.19\n",
+ "#newton-raphson method\n",
+ "#page 41\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 4*math.exp(-x)*math.sin(x)-1\n",
+ "def f1(x):\n",
+ " return math.cos(x)*4*math.exp(-x)-4*math.exp(-x)*math.sin(x)\n",
+ "x0=0.2 # initial value\n",
+ "d=0.0001\n",
+ "c=0 \n",
+ "n=1\n",
+ "print \"successive iterations \\t x0 \\t f(x0)\\t f1(x0)\\n\"\n",
+ "while n==1:\n",
+ " x2=x0\n",
+ " x1=x0-(f(x0)/f1(x0))\n",
+ " x0=x1\n",
+ " print \" \\t%f \\t%f \\t%f\\n\" %(x2,f(x1),f1(x1))\n",
+ " c=c+1\n",
+ " if abs(f(x0))<d:\n",
+ " break \n",
+ "print \"the root of %i iteration is: %0.3g\" %(c,x0)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x0 \t f(x0)\t f1(x0)\n",
+ "\n",
+ " \t0.200000 \t-0.056593 \t1.753325\n",
+ "\n",
+ " \t0.336526 \t-0.002769 \t1.583008\n",
+ "\n",
+ " \t0.368804 \t-0.000008 \t1.573993\n",
+ "\n",
+ "the root of 3 iteration is: 0.371\n"
+ ]
+ }
+ ],
+ "prompt_number": 54
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.20:pg-42"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.20\n",
+ "#generalized newton-raphson method\n",
+ "#page 42\n",
+ "def f(x):\n",
+ " return x**3-x**2-x+1\n",
+ "def f1(x):\n",
+ " return 3*x**2-2*x-1\n",
+ "def f2(x):\n",
+ " return 6*x-2\n",
+ "x0=0.8 # initial value to finf double root\n",
+ "n=1 \n",
+ "print \"successive iterations \\t x0 \\t x1\\t x2\\n\"\n",
+ "while n==1:\n",
+ " x1=x0-(f(x0)/f1(x0));\n",
+ " x2=x0-(f1(x0)/f2(x0));\n",
+ " if abs(x1-x2)<0.000000001:\n",
+ " x0=(x1+x2)/2.0\n",
+ " break\n",
+ " else:\n",
+ " x0=(x1+x2)/2;\n",
+ " print \" %f\\t %f\\t %f\\n\" %(x0,x1,x2)\n",
+ "print \"\\n \\nthe double root is at: %f\" %(x0)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x0 \t x1\t x2\n",
+ "\n",
+ " 0.974370\t 0.905882\t 1.042857\n",
+ "\n",
+ " 0.993890\t 0.987269\t 1.000512\n",
+ "\n",
+ " 0.998489\t 0.996950\t 1.000028\n",
+ "\n",
+ " 0.999623\t 0.999245\t 1.000002\n",
+ "\n",
+ " 0.999906\t 0.999812\t 1.000000\n",
+ "\n",
+ " 0.999976\t 0.999953\t 1.000000\n",
+ "\n",
+ " 0.999994\t 0.999988\t 1.000000\n",
+ "\n",
+ " 0.999999\t 0.999997\t 1.000000\n",
+ "\n",
+ " 1.000000\t 0.999999\t 1.000000\n",
+ "\n",
+ " 1.000000\t 1.000000\t 1.000000\n",
+ "\n",
+ " 1.000000\t 1.000000\t 1.000000\n",
+ "\n",
+ " 1.000000\t 1.000000\t 1.000000\n",
+ "\n",
+ " 1.000000\t 1.000000\t 1.000000\n",
+ "\n",
+ "\n",
+ " \n",
+ "the double root is at: 1.000000\n"
+ ]
+ }
+ ],
+ "prompt_number": 57
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.21:pg-45"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ramanujan's method\n",
+ "#example 2.21\n",
+ "#page 45\n",
+ "def f(x):\n",
+ " return 1-(13.0/12.0)*x-(3.0/8.0)*x**2+(1.0/24.0)*x**3\n",
+ "a1=13.0/12.0\n",
+ "a2=-3.0/8.0\n",
+ "a3=1.0/24.0\n",
+ "b1=1\n",
+ "b2=a1\n",
+ "b3=a1*b2+a2*b1\n",
+ "b4=a1*b3+a2*b2+a3*b1\n",
+ "b5=a1*b4+a2*b3+a3*b2\n",
+ "b6=a1*b5+a2*b4+a3*b3\n",
+ "b7=a1*b6+a2*b5+a3*b4\n",
+ "b8=a1*b7+a2*b6+a3*b5\n",
+ "b9=a1*b8+a2*b7+a3*b6\n",
+ "print \"\\n\\n%f\" %(b1/b2)\n",
+ "print \"\\n%f\" %(b2/b3)\n",
+ "print \"\\n%f\" %(b3/b4)\n",
+ "print \"\\n%f\" %(b4/b5)\n",
+ "print \"\\n%f\" %(b5/b6)\n",
+ "print \"\\n%f\" %(b6/b7)\n",
+ "print \"\\n%f\" %(b7/b8)\n",
+ "print \"\\n%f\" %(b8/b9)\n",
+ "print \"\\n it appears as if the roots are converging at 2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ "0.923077\n",
+ "\n",
+ "1.356522\n",
+ "\n",
+ "1.595376\n",
+ "\n",
+ "1.738402\n",
+ "\n",
+ "1.828184\n",
+ "\n",
+ "1.886130\n",
+ "\n",
+ "1.924153\n",
+ "\n",
+ "1.949345\n",
+ "\n",
+ " it appears as if the roots are converging at 2\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.22:pg-46"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ramanujan's method\n",
+ "#example 2.22\n",
+ "#page 46\n",
+ "def f(x):\n",
+ " return x+x**2+(x**3)/2.0+(x**4)/6.0+(x**5)/24.0\n",
+ "a1=1.0\n",
+ "a2=1.0\n",
+ "a3=1.0/2.0\n",
+ "a4=1.0/6.0\n",
+ "a5=1.0/24.0\n",
+ "b1=1\n",
+ "b2=a2\n",
+ "b3=a1*b2+a2*b1\n",
+ "b4=a1*b3+a2*b2+a3*b1\n",
+ "b5=a1*b4+a2*b3+a3*b2\n",
+ "b6=a1*b5+a2*b4+a3*b3\n",
+ "print \"\\n%f\" %(b1/b2)\n",
+ "print \"\\n%f\" %(b2/b3)\n",
+ "print \"\\n%f\" %(b3/b4)\n",
+ "print \"\\n%f\" %(b4/b5)\n",
+ "print \"\\n%f\" %(b5/b6)\n",
+ "print \"\\n it appears as if the roots are converging at around %f\" %(b5/b6)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "1.000000\n",
+ "\n",
+ "0.500000\n",
+ "\n",
+ "0.571429\n",
+ "\n",
+ "0.583333\n",
+ "\n",
+ "0.571429\n",
+ "\n",
+ " it appears as if the roots are converging at around 0.571429\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.23:pg-47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ramanujan's method\n",
+ "#example 2.23\n",
+ "#page 47\n",
+ "from __future__ import division\n",
+ "def f(x):\n",
+ " return 1-2*((3*x/2.0+(x**2)/4.0-(x**4)/48.0+(x**6)/1440.0)-(x**8)*2/80640.0)\n",
+ "a1=3/2\n",
+ "a2=1/4\n",
+ "a3=0\n",
+ "a4=1/48\n",
+ "a5=0\n",
+ "a6=1/1440\n",
+ "a7=0\n",
+ "a8=-1/80640\n",
+ "b1=1\n",
+ "b2=a1\n",
+ "b3=a1*b2+a2*b1\n",
+ "b4=a1*b3+a2*b2+a3*b1\n",
+ "b5=a1*b4+a2*b3+a3*b2\n",
+ "b6=a1*b5+a2*b4+a3*b3\n",
+ "b7=a1*b6+a2*b5+a3*b4\n",
+ "b8=a1*b7+a2*b6+a3*b5\n",
+ "b9=a1*b8+a2*b7+a3*b6\n",
+ "print \"\\n%f\" %(b1/b2)\n",
+ "print \"\\n%f\" %(b2/b3)\n",
+ "print \"\\n%f\" %(b3/b4)\n",
+ "print \"\\n%f\" %(b4/b5)\n",
+ "print \"\\n%f\" %(b5/b6)\n",
+ "print \"\\n%f\" %(b6/b7)\n",
+ "print \"\\n%f\" %(b7/b8)\n",
+ "print \"\\n it appears as if the roots are converging at around %f\" %(b7/b8)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "0.666667\n",
+ "\n",
+ "0.600000\n",
+ "\n",
+ "0.606061\n",
+ "\n",
+ "0.605505\n",
+ "\n",
+ "0.605556\n",
+ "\n",
+ "0.605551\n",
+ "\n",
+ "0.605551\n",
+ "\n",
+ " it appears as if the roots are converging at around 0.605551\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.24:pg-47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ramanujan's method\n",
+ "#example 2.24\n",
+ "#page 47\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 1-(x-x**2.0/math.factorial(2.0)**2.0+x**3.0/math.factorial(3.0)**2.0-x**4.0/math.factorial(4.0)**2.0)\n",
+ "a1=1\n",
+ "a2=-1/math.factorial(2.0)**2.0\n",
+ "a3=1/math.factorial(3.0)**2.0\n",
+ "a4=-1/math.factorial(4.0)**2.0\n",
+ "a5=-1/math.factorial(5.0)**2.0\n",
+ "a6=1/math.factorial(6.0)**2.0\n",
+ "b1=1\n",
+ "b2=a1\n",
+ "b3=a1*b2+a2*b1\n",
+ "b4=a1*b3+a2*b2+a3*b1\n",
+ "b5=a1*b4+a2*b3+a3*b2\n",
+ "print \"\\n\\n%f\" %(b1/b2)\n",
+ "print \"\\n\\n%f\" %(b2/b3)\n",
+ "print \"\\n%f\" %(b3/b4)\n",
+ "print \"\\n%f\" %(b4/b5)\n",
+ "print \"\\n it appears as if the roots are converging at around %f\" %(b4/b5)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ "1.000000\n",
+ "\n",
+ "\n",
+ "1.333333\n",
+ "\n",
+ "1.421053\n",
+ "\n",
+ "1.433962\n",
+ "\n",
+ " it appears as if the roots are converging at around 1.433962\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.25:pg-49"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.25\n",
+ "#secant method\n",
+ "#page 49\n",
+ "from __future__ import division\n",
+ "def f(x):\n",
+ " return x**3-2*x-5\n",
+ "x1=2\n",
+ "x2=3 # initial values\n",
+ "n=1\n",
+ "c=0\n",
+ "print \"successive iterations \\t x1 \\t x2 \\t x3 \\t f(x3)\\n\"\n",
+ "while n==1:\n",
+ " x3=(x1*f(x2)-x2*f(x1))/(f(x2)-f(x1)) \n",
+ " print \" \\t%f \\t%f \\t%f \\t%f\\n\" %(x1,x2,x3,f(x3))\n",
+ " if f(x3)*f(x1)>0:\n",
+ " x2=x3;\n",
+ " else:\n",
+ " x1=x3 \n",
+ " if abs(f(x3))<0.000001: \n",
+ " break\n",
+ " c=c+1\n",
+ "print \"the root of the equation after %i iteration is: %f\" %(c,x3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x1 \t x2 \t x3 \t f(x3)\n",
+ "\n",
+ " \t2.000000 \t3.000000 \t2.058824 \t-0.390800\n",
+ "\n",
+ " \t2.000000 \t2.058824 \t2.096559 \t0.022428\n",
+ "\n",
+ " \t2.096559 \t2.058824 \t2.094511 \t-0.000457\n",
+ "\n",
+ " \t2.094511 \t2.058824 \t2.094552 \t0.000009\n",
+ "\n",
+ " \t2.094552 \t2.058824 \t2.094551 \t-0.000000\n",
+ "\n",
+ "the root of the equation after 4 iteration is: 2.094551\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.26:pg-50"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.26\n",
+ "#secant method\n",
+ "#page 50\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "def f(x):\n",
+ " return x*math.exp(x)-1\n",
+ "x1=0\n",
+ "x2=1 # initial values\n",
+ "n=1\n",
+ "c=0 \n",
+ "print \"successive iterations \\t x1 \\t x2 \\t x3 \\t f(x3)\\n\"\n",
+ "while n==1:\n",
+ " x3=(x1*f(x2)-x2*f(x1))/(f(x2)-f(x1)) \n",
+ " print \" \\t%f \\t%f \\t%f \\t%f\\n\" %(x1,x2,x3,f(x3))\n",
+ " if f(x3)*f(x1)>0:\n",
+ " x2=x3\n",
+ " else:\n",
+ " x1=x3 \n",
+ " if abs(f(x3))<0.0001:\n",
+ " break\n",
+ " c=c+1\n",
+ "print \"the root of the equation after %i iteration is: %0.4g\" %(c,x3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x1 \t x2 \t x3 \t f(x3)\n",
+ "\n",
+ " \t0.000000 \t1.000000 \t0.367879 \t-0.468536\n",
+ "\n",
+ " \t0.000000 \t0.367879 \t0.692201 \t0.383091\n",
+ "\n",
+ " \t0.692201 \t0.367879 \t0.546310 \t-0.056595\n",
+ "\n",
+ " \t0.546310 \t0.367879 \t0.570823 \t0.010200\n",
+ "\n",
+ " \t0.570823 \t0.367879 \t0.566500 \t-0.001778\n",
+ "\n",
+ " \t0.566500 \t0.367879 \t0.567256 \t0.000312\n",
+ "\n",
+ " \t0.567256 \t0.367879 \t0.567124 \t-0.000055\n",
+ "\n",
+ "the root of the equation after 6 iteration is: 0.5671\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.27:pg-52"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# example 2.27\n",
+ "#mulller's method\n",
+ "#page 52\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "def f(x):\n",
+ " return x**3-x-1\n",
+ "x0=0\n",
+ "x1=1\n",
+ "x2=2 # initial values\n",
+ "n=1\n",
+ "c=0\n",
+ "print \"successive iterations \\t x0 \\t x1 \\t x2 \\t f(x0)\\t f(x1)\\t f(x2)\\n\"\n",
+ "while n==1: \n",
+ " c=c+1\n",
+ " y0=f(x0)\n",
+ " y1=f(x1)\n",
+ " y2=f(x2)\n",
+ " h2=x2-x1\n",
+ " h1=x1-x0\n",
+ " d2=f(x2)-f(x1)\n",
+ " d1=f(x1)-f(x0)\n",
+ " print \" \\t%f\\t %f\\t %f\\t %f\\t %f\\t %f\\n\" %(x0,x1,x2,f(x0),f(x1),f(x2))\n",
+ " A=(d2/h2-d1/h1)/(h1+h2)\n",
+ " B=d2/h2+A*h2\n",
+ " S=math.sqrt(B**2-4*A*f(x2))\n",
+ " x3=x2-(2*f(x2))/(B+S)\n",
+ " E=abs((x3-x2)/x2)*100\n",
+ " if E<0.003:\n",
+ " break\n",
+ " else:\n",
+ " if c==1:\n",
+ " x2=x3\n",
+ " if c==2:\n",
+ " x1=x2\n",
+ " x2=x3\n",
+ " if c==3:\n",
+ " x0=x1\n",
+ " x1=x2\n",
+ " x2=x3\n",
+ " if c==3:\n",
+ " c=0\n",
+ "print \"the required root is : %0.4f\" %(x3)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x0 \t x1 \t x2 \t f(x0)\t f(x1)\t f(x2)\n",
+ "\n",
+ " \t0.000000\t 1.000000\t 2.000000\t -1.000000\t -1.000000\t 5.000000\n",
+ "\n",
+ " \t0.000000\t 1.000000\t 1.263763\t -1.000000\t -1.000000\t -0.245412\n",
+ "\n",
+ " \t0.000000\t 1.263763\t 1.331711\t -1.000000\t -0.245412\t 0.030015\n",
+ "\n",
+ " \t1.263763\t 1.331711\t 1.324583\t -0.245412\t 0.030015\t -0.000574\n",
+ "\n",
+ " \t1.263763\t 1.331711\t 1.324718\t -0.245412\t 0.030015\t -0.000000\n",
+ "\n",
+ "the required root is : 1.3247\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.28:pg-55"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#graeffe's method\n",
+ "#example 2.28\n",
+ "#page 55\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "def f(x):\n",
+ " return x**3-6*(x**2)+11*x-6\n",
+ "#x=poly(0,'x')\n",
+ "#g=f(-x)\n",
+ "print \"the equation is:\\n\"\n",
+ "A=[1, 14, 49, 36] #coefficients of the above equation\n",
+ "print \"%0.4g\\n\" %(math.sqrt(A[3]/A[2]))\n",
+ "print \"%0.4g\\n\" %(math.sqrt(A[2]/A[1]))\n",
+ "print \"%0.4g\\n\" %(math.sqrt(A[1]/A[0]))\n",
+ "print \"the equation is:\\n\"\n",
+ "#disp(g*(-1*g));\n",
+ "B=[1, 98, 1393, 1296]\n",
+ "print \"%0.4g\\n\" %((B[3]/B[2])**(1/4))\n",
+ "print \"%0.4g\\n\" %((B[2]/B[1])**(1/4))\n",
+ "print \"%0.4g\\n\" %((B[1]/B[0])**(1/4))\n",
+ "print \"It is apparent from the outputs that the roots converge at 1 2 3\"\n",
+ "\n",
+ "\n",
+ "\n",
+ "#INCOMPLETE"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the equation is:\n",
+ "\n",
+ "0.8571\n",
+ "\n",
+ "1.871\n",
+ "\n",
+ "3.742\n",
+ "\n",
+ "the equation is:\n",
+ "\n",
+ "0.9821\n",
+ "\n",
+ "1.942\n",
+ "\n",
+ "3.146\n",
+ "\n",
+ "It is apparent from the outputs that the roots converge at 1 2 3\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.29:pg-57"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#quadratic factor by lin's--bairsttow method\n",
+ "#example 2.29\n",
+ "#page 57\n",
+ "from numpy import matrix\n",
+ "from __future__ import division\n",
+ "def f(x):\n",
+ " return x**3-x-1\n",
+ "a=[-1, -1, 0, 1]\n",
+ "r1=1\n",
+ "s1=1\n",
+ "b4=a[3]\n",
+ "def f3(r):\n",
+ " return a[2]-r*a[3]\n",
+ "def f2(r,s):\n",
+ " return a[1]-r*a[2]+r**2*a[3]-s*a[3]\n",
+ "def f1(r,s):\n",
+ " return a[0]-s*a[2]+s*r*a[3]\n",
+ "A=matrix([[1,1],[2,-1]])\n",
+ "C=matrix([[0],[1]])\n",
+ "X=A.I*C\n",
+ "X1=[[ 0.33333333],[-0.33333333]]\n",
+ "dr=X1[0][0]\n",
+ "ds=X1[1][0]\n",
+ "r2=r1+dr\n",
+ "s2=s1+ds\n",
+ "#second pproximation\n",
+ "r1=r2\n",
+ "s1=s2\n",
+ "b11=f1(r2,s2)\n",
+ "b22=f2(r2,s2)\n",
+ "h=0.001\n",
+ "dr_b1=(f1(r1+h,s1)-f1(r1,s1))/h\n",
+ "ds_b1=(f1(r1,s1+h)-f1(r1,s1))/h\n",
+ "dr_b2=(f2(r1+h,s1)-f2(r1,s1))/h\n",
+ "ds_b2=(f2(r1,s1+h)-f2(r1,s1))/h\n",
+ "A=matrix([[dr_b1,ds_b1],[dr_b2,ds_b2]])\n",
+ "C=matrix([[-f1(r1,s1)],[-f2(r1,s2)]])\n",
+ "X=A.I*C\n",
+ "r2=r1+X[0][0]\n",
+ "s2=s1+X[1][0]\n",
+ "print \"roots correct to 3 decimal places are : %0.3f %0.3f\" %(r2,s2)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "roots correct to 3 decimal places are : 1.325 0.754\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.31:pg-62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#method of iteration\n",
+ "#example 2.31\n",
+ "#page 62\n",
+ "from __future__ import division\n",
+ "def f(x,y):\n",
+ " return (3*y*x**2+7)/10\n",
+ "def g(x,y):\n",
+ " return (y**2+4)/5\n",
+ "h=0.0001\n",
+ "x0=0.5\n",
+ "y0=0.5\n",
+ "f1_dx=(f(x0+h,y0)-f(x0,y0))/h\n",
+ "f1_dy=(f(x0,y0+h)-f(x0,y0))/h\n",
+ "g1_dx=(g(x0+h,y0)-g(x0,y0))/h\n",
+ "g1_dy=(g(x0+h,y0)-g(x0,y0))/h\n",
+ "if (f1_dx+f1_dy<1) and (g1_dx+g1_dy<1): \n",
+ " print \"coditions for convergence is satisfied\\n\\n\"\n",
+ "print \"X \\t Y\\t\\n\\n\"\n",
+ "for i in range(0,10):\n",
+ " X=(3*y0*x0**2+7)/10\n",
+ " Y=(y0**2+4)/5\n",
+ " print \"%f\\t %f\\t\\n\" %(X,Y)\n",
+ " x0=X\n",
+ " y0=Y\n",
+ "print \"\\n\\n CONVERGENCE AT (1 1) IS OBVIOUS\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "coditions for convergence is satisfied\n",
+ "\n",
+ "\n",
+ "X \t Y\t\n",
+ "\n",
+ "\n",
+ "0.737500\t 0.850000\t\n",
+ "\n",
+ "0.838696\t 0.944500\t\n",
+ "\n",
+ "0.899312\t 0.978416\t\n",
+ "\n",
+ "0.937391\t 0.991460\t\n",
+ "\n",
+ "0.961360\t 0.996598\t\n",
+ "\n",
+ "0.976320\t 0.998642\t\n",
+ "\n",
+ "0.985572\t 0.999457\t\n",
+ "\n",
+ "0.991247\t 0.999783\t\n",
+ "\n",
+ "0.994707\t 0.999913\t\n",
+ "\n",
+ "0.996807\t 0.999965\t\n",
+ "\n",
+ "\n",
+ "\n",
+ " CONVERGENCE AT (1 1) IS OBVIOUS\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.32:pg-65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#newton raphson metho\n",
+ "#example 2.32\n",
+ "#page 65\n",
+ "def f(x,y):\n",
+ " return 3*y*x**2-10*x+7\n",
+ "def g(y):\n",
+ " return y**2-5*y+4\n",
+ "hh=0.0001\n",
+ "x0=0.5\n",
+ "y0=0.5 #initial values\n",
+ "f0=f(x0,y0)\n",
+ "g0=g(y0)\n",
+ "df_dx=(f(x0+hh,y0)-f(x0,y0))/hh\n",
+ "df_dy=(f(x0,y0+hh)-f(x0,y0))/hh\n",
+ "dg_dx=(g(y0)-g(y0))/hh\n",
+ "dg_dy=(g(y0+hh)-g(y0))/hh\n",
+ "d=[[df_dx,df_dy],[dg_dx,dg_dy]]\n",
+ "D1=det(d)\n",
+ "dd=[[-f0,df_dy],[-g0,dg_dy]]\n",
+ "h=det(dd)/D1\n",
+ "ddd=[[df_dx,-f0],[dg_dx,-g0]]\n",
+ "k=det(ddd)/D1;\n",
+ "x1=x0+h\n",
+ "y1=y0+k\n",
+ "f0=f(x1,y1)\n",
+ "g0=g(y1)\n",
+ "df_dx=(f(x1+hh,y1)-f(x1,y1))/hh\n",
+ "df_dy=(f(x1,y1+hh)-f(x1,y1))/hh\n",
+ "dg_dx=(g(y1)-g(y1))/hh\n",
+ "dg_dy=(g(y1+hh)-g(y1))/hh\n",
+ "dddd=[[df_dx,df_dy],[dg_dx,dg_dy]]\n",
+ "D2=det(dddd)\n",
+ "ddddd=[[-f0,df_dy],[-g0,dg_dy]]\n",
+ "h=det(ddddd)/D2\n",
+ "d6=[[df_dx,-f0],[dg_dx,-g0]]\n",
+ "k=det(d6)/D2\n",
+ "x2=x1+h\n",
+ "y2=y1+k\n",
+ "print \" the roots of the equation are x2=%f and y2=%f\" %(x2,y2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " the roots of the equation are x2=0.970803 and y2=0.998752\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.33:pg-66"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#newton raphson method\n",
+ "#example 2.33\n",
+ "#page 66\n",
+ "import math\n",
+ "def f(x,y):\n",
+ " return x**2+y**2-1\n",
+ "def g(x,y):\n",
+ " return y-x**2\n",
+ "hh=0.0001\n",
+ "x0=0.7071\n",
+ "y0=0.7071 #initial values\n",
+ "f0=f(x0,y0)\n",
+ "g0=g(x0,y0)\n",
+ "df_dx=(f(x0+hh,y0)-f(x0,y0))/hh\n",
+ "df_dy=(f(x0,y0+hh)-f(x0,y0))/hh\n",
+ "dg_dx=(g(x0+hh,y0)-g(x0,y0))/hh\n",
+ "dg_dy=(g(x0,y0+hh)-g(x0,y0))/hh\n",
+ "D1=det([[df_dx,df_dy],[dg_dx,dg_dy]])\n",
+ "h=det([[-f0,df_dy],[-g0,dg_dy]])/D1\n",
+ "k=det([[df_dx,-f0],[dg_dx,-g0]])/D1\n",
+ "x1=x0+h\n",
+ "y1=y0+k\n",
+ "f0=f(x1,y1)\n",
+ "g0=g(x1,y1)\n",
+ "df_dx=(f(x1+hh,y1)-f(x1,y1))/hh\n",
+ "df_dy=(f(x1,y1+hh)-f(x1,y1))/hh\n",
+ "dg_dx=(g(x1+hh,y1)-g(x1,y1))/hh\n",
+ "dg_dy=(g(x1,y1+hh)-g(x1,y1))/hh\n",
+ "D2=det([[df_dx,df_dy],[dg_dx,dg_dy]])\n",
+ "h=det([[-f0,df_dy],[-g0,dg_dy]])/D2\n",
+ "k=det([[df_dx,-f0],[dg_dx,-g0]])/D2\n",
+ "x2=x1+h\n",
+ "y2=y1+k\n",
+ "print \"the roots of the equation are x2=%f and y2=%f \" %(x2,y2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the roots of the equation are x2=0.786184 and y2=0.618039 \n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.34:pg-67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#newton raphson method\n",
+ "#example 2.34\n",
+ "#page 67\n",
+ "import math\n",
+ "def f(x,y):\n",
+ " return math.sin(x)-y+0.9793\n",
+ "def g(x,y):\n",
+ " return math.cos(y)-x+0.6703\n",
+ "hh=0.0001\n",
+ "x0=0.5\n",
+ "y0=1.5 #initial values\n",
+ "f0=f(x0,y0)\n",
+ "g0=g(x0,y0)\n",
+ "df_dx=(f(x0+hh,y0)-f(x0,y0))/hh\n",
+ "df_dy=(f(x0,y0+hh)-f(x0,y0))/hh\n",
+ "dg_dx=(g(x0+hh,y0)-g(x0,y0))/hh\n",
+ "dg_dy=(g(x0,y0+hh)-g(x0,y0))/hh\n",
+ "d1=[[df_dx,df_dy],[dg_dx,dg_dy]]\n",
+ "D1=det(d1)\n",
+ "d2=[[-f0,df_dy],[-g0,dg_dy]]\n",
+ "h=det(d2)/D1\n",
+ "d3=[[df_dx,-f0],[dg_dx,-g0]]\n",
+ "k=det(d3)/D1\n",
+ "x1=x0+h\n",
+ "y1=y0+k\n",
+ "f0=f(x1,y1)\n",
+ "g0=g(x1,y1)\n",
+ "df_dx=(f(x1+hh,y1)-f(x1,y1))/hh\n",
+ "df_dy=(f(x1,y1+hh)-f(x1,y1))/hh\n",
+ "dg_dx=(g(x1+hh,y1)-g(x1,y1))/hh\n",
+ "dg_dy=(g(x1,y1+hh)-g(x1,y1))/hh\n",
+ "d4=[[df_dx,df_dy],[dg_dx,dg_dy]]\n",
+ "D2=det(d4)\n",
+ "h=det([[-f0,df_dy],[-g0,dg_dy]])/D2\n",
+ "k=det([[df_dx,-f0],[dg_dx,-g0]])/D2\n",
+ "x2=x1+h\n",
+ "y2=y1+k\n",
+ "print \"the roots of the equation are x2=%0.4f and y2=%0.4f\" %(x2,y2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the roots of the equation are x2=0.6537 and y2=1.5874\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter3_7.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter3_7.ipynb new file mode 100644 index 00000000..fbba6967 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter3_7.ipynb @@ -0,0 +1,1126 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:0755817d9df79392f0a505cb49e15463e2d17b0d0bd1a381990227b29ae2b639"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter03:Interpolation"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.4:pg-86"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.4\n",
+ "#interpolation\n",
+ "#page 86\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "x=[1, 3, 5, 7]\n",
+ "y=[24, 120, 336, 720]\n",
+ "d1=[0,0,0]\n",
+ "d2=[0,0,0]\n",
+ "d3=[0,0,0]\n",
+ "h=2 #interval between values of x\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,1):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1\n",
+ "d=[0,d1[0],d2[0],d3[0]]\n",
+ "x0=8 #value at 8\n",
+ "pp=1\n",
+ "y_x=y[0]\n",
+ "p=(x0-1)/2\n",
+ "for i in range(1,4):\n",
+ " pp=1\n",
+ " for j in range(0,i):\n",
+ " pp=pp*(p-(j)) \n",
+ " y_x=y_x+(pp*d[i])/math.factorial(i)\n",
+ "print \"value of function at %f is :%f\" %(x0,y_x)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of function at 8.000000 is :990.000000\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.6:pg-87"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.6\n",
+ "#interpolation\n",
+ "#page 87\n",
+ "x=[15, 20, 25, 30, 35, 40]\n",
+ "y=[0.2588190, 0.3420201, 0.4226183, 0.5, 0.5735764, 0.6427876]\n",
+ "d1=[0,0,0,0,0]\n",
+ "d2=[0,0,0,0]\n",
+ "d3=[0,0,0]\n",
+ "d4=[0,0]\n",
+ "d5=[0]\n",
+ "h=5 #interval between values of x\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,1):\n",
+ " d5[c]=d4[i+1]-d4[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "d=[0,d1[0], d2[0], d3[0], d4[0], d5[0]]\n",
+ "x0=38 #value at 38 degree\n",
+ "pp=1\n",
+ "y_x=y[0]\n",
+ "p=(x0-x[0])/h\n",
+ "for i in range(1,6):\n",
+ " pp=1\n",
+ " for j in range(0,i):\n",
+ " pp=pp*(p-(j)) \n",
+ " y_x=y_x+((pp*d[i])/math.factorial(i));\n",
+ "print \"value of function at %i is :%f\" %(x0,y_x)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of function at 38 is :0.615661\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.7:pg-89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.7\n",
+ "#interpolation\n",
+ "#page 89\n",
+ "x=[0, 1, 2, 4]\n",
+ "y=[1, 3, 9, 81]\n",
+ "#equation is y(5)-4*y(4)+6*y(2)-4*y(2)+y(1)\n",
+ "y3=(y[3]+6*y[2]-4*y[1]+y[0])/4\n",
+ "print \"the value of missing term of table is :%d\" %(y3)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of missing term of table is :31\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.8:pg-89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.8\n",
+ "#interpolation\n",
+ "#page 89\n",
+ "import math\n",
+ "x=[0.10, 0.15, 0.20, 0.25, 0.30]\n",
+ "y=[0.1003, 0.1511, 0.2027, 0.2553, 0.3093]\n",
+ "d1=[0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0,0]\n",
+ "d4=[0,0,0,0,0]\n",
+ "h=0.05 #interval between values of x\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1\n",
+ "d=[0,d1[0], d2[0], d3[0], d4[0]]\n",
+ "x0=0.12 #value at 0.12;\n",
+ "pp=1\n",
+ "y_x=y[0]\n",
+ "p=(x0-x[0])/h\n",
+ "for i in range(1,5):\n",
+ " pp=1;\n",
+ " for j in range(0,i):\n",
+ " pp=pp*(p-(j)) \n",
+ " y_x=y_x+(pp*d[i])/math.factorial(i)\n",
+ "print \"value of function at %f is :%0.4g\\n \\n\" %(x0,y_x)\n",
+ "x0=0.26 #value at 0.26;\n",
+ "pp=1\n",
+ "y_x=y[0]\n",
+ "p=(x0-x[0])/h\n",
+ "for i in range(1,5):\n",
+ " pp=1\n",
+ " for j in range(0,i):\n",
+ " pp=pp*(p-(j)) \n",
+ " y_x=y_x+(pp*d[i])/math.factorial(i);\n",
+ "print \"value of function at %f is :%0.4g\\n \\n\" %(x0,y_x)\n",
+ "x0=0.40 #value at 0.40;\n",
+ "pp=1\n",
+ "y_x=y[0]\n",
+ "p=(x0-x[0])/h\n",
+ "for i in range(1,5):\n",
+ " pp=1\n",
+ " for j in range(0,i):\n",
+ " pp=pp*(p-(j)) \n",
+ " y_x=y_x+(pp*d[i])/math.factorial(i)\n",
+ "print \"value of function at %f is :%0.4g\\n \\n\" %(x0,y_x)\n",
+ "x0=0.50 #value at 0.50;\n",
+ "pp=1\n",
+ "y_x=y[0]\n",
+ "p=(x0-x[0])/h\n",
+ "for i in range(1,5):\n",
+ " pp=1\n",
+ " for j in range(0,i):\n",
+ " pp=pp*(p-(j)) \n",
+ " y_x=y_x+(pp*d[i])/math.factorial(i)\n",
+ "print \"value of function at %f is :%0.5g\\n \\n\" %(x0,y_x)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of function at 0.120000 is :0.1205\n",
+ " \n",
+ "\n",
+ "value of function at 0.260000 is :0.266\n",
+ " \n",
+ "\n",
+ "value of function at 0.400000 is :0.4241\n",
+ " \n",
+ "\n",
+ "value of function at 0.500000 is :0.5543\n",
+ " \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.9:pg-93"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.9\n",
+ "#Gauss' forward formula\n",
+ "#page 93\n",
+ "x=[1.0, 1.05, 1.10, 1.15, 1.20, 1.25, 1.30];\n",
+ "y=[2.7183, 2.8577, 3.0042, 3.1582, 3.3201, 3.4903, 3.66693]\n",
+ "d1=[0,0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0]\n",
+ "d4=[0,0,0]\n",
+ "d5=[0,0]\n",
+ "d6=[0]\n",
+ "h=0.05 #interval between values of x\n",
+ "c=0\n",
+ "for i in range(0,6):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " d5[c]=d4[i+1]-d4[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,1):\n",
+ " d6[c]=d5[i+1]-d5[i]\n",
+ " c=c+1\n",
+ "d=[0,d1[3], d2[2], d3[2], d4[1], d5[0], d6[0]]\n",
+ "x0=1.17 #value at 1.17;\n",
+ "pp=1\n",
+ "y_x=y[3]\n",
+ "p=(x0-x[3])/h\n",
+ "for i in range(1,6):\n",
+ " pp=1;\n",
+ " for j in range(0,i):\n",
+ " pp=pp*(p-(j)) \n",
+ " y_x=y_x+(pp*d[i])/math.factorial(i)\n",
+ "print \"value of function at %f is :%0.4g\\n \\n\" %(x0,y_x)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of function at 1.170000 is :3.222\n",
+ " \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.10:pg-97"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#practical interpolation\n",
+ "#example 3.10\n",
+ "#page 97\n",
+ "import math\n",
+ "x=[0.61, 0.62, 0.63, 0.64, 0.65, 0.66, 0.67]\n",
+ "y=[1.840431, 1.858928,1.877610, 1.896481, 1.915541, 1.934792, 1.954237]\n",
+ "d1=[0,0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0]\n",
+ "d4=[0,0,0]\n",
+ "h=0.01 #interval between values of x\n",
+ "c=0\n",
+ "for i in range(0,6):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d3[c]=d2[i+1]-d2[i];\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d4[c]=d3[i+1]-d3[i];\n",
+ " c=c+1\n",
+ "d=[d1[0], d2[0], d3[0], d4[0]]\n",
+ "x0=0.644\n",
+ "p=(x0-x[3])/h;\n",
+ "y_x=y[3]\n",
+ "y_x=y_x+p*(d1[2]+d1[3])/2+p**2*(d2[1])/2 #stirling formula\n",
+ "print \"the value at %f by stirling formula is : %f\\n\\n\" %(x0,y_x)\n",
+ "y_x=y[3]\n",
+ "y_x=y_x+p*d1[3]+p*(p-1)*(d2[2]+d2[3])/2\n",
+ "print \" the value at %f by bessels formula is : %f\\n\\n\" %(x0,y_x)\n",
+ "y_x=y[3]\n",
+ "q=1-p\n",
+ "y_x=q*y[3]+q*(q**2-1)*d2[2]/2+p*y[4]+p*(q**2-1)*d2[4]/2\n",
+ "print \"the value at %f by everrets formula is : %f\\n\\n\" %(x0,y_x)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value at 0.644000 by stirling formula is : 1.904082\n",
+ "\n",
+ "\n",
+ " the value at 0.644000 by bessels formula is : 1.904059\n",
+ "\n",
+ "\n",
+ "the value at 0.644000 by everrets formula is : 1.904044\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.11:pg-99"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#practical interpolation\n",
+ "#example 3.11\n",
+ "#page 99\n",
+ "x=[0.61, 0.62, 0.63, 0.64, 0.65, 0.66, 0.67]\n",
+ "y=[1.840431, 1.858928, 1.877610, 1.896481, 1.915541, 1.934792, 1.954237]\n",
+ "d1=[0,0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0]\n",
+ "d4=[0,0,0]\n",
+ "h=0.01 #interval between values of x\n",
+ "c=0\n",
+ "for i in range(0,6):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1\n",
+ "d=[d1[0], d2[0], d3[0], d4[0]]\n",
+ "x0=0.638\n",
+ "p=(x0-x[3])/h\n",
+ "y_x=y[3]\n",
+ "y_x=y_x+p*(d1[2]+d1[3])/2+p**2*(d2[1])/2 #stirling formula\n",
+ "print \"value at %f by stirling formula is : %f\\n\\n\" %(x0,y_x)\n",
+ "y_x=y[2]\n",
+ "p=(x0-x[2])/h\n",
+ "y_x=y_x+p*d1[2]+p*(p-1)*(d2[1])/2\n",
+ "print \"the value at %f by bessels formula is : %f\\n\\n\" %(x0,y_x)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value at 0.638000 by stirling formula is : 1.892692\n",
+ "\n",
+ "\n",
+ "the value at 0.638000 by bessels formula is : 1.892692\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.12:pg-99"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#practical interpolation\n",
+ "#example 3.12\n",
+ "#page 99\n",
+ "x=[1.72, 1.73, 1.74, 1.75, 1.76, 1.77, 1.78]\n",
+ "y=[0.1790661479, 0.1772844100, 0.1755204006, 0.1737739435, 0.1720448638, 0.1703329888, 0.1686381473]\n",
+ "d1=[0,0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0]\n",
+ "d4=[0,0,0]\n",
+ "h=0.01 #interval between values of x\n",
+ "c=0\n",
+ "for i in range(0,6):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1\n",
+ "x0=1.7475\n",
+ "y_x=y[2]\n",
+ "p=(x0-x[2])/h\n",
+ "y_x=y_x+p*d1[2]+p*(p-1)*((d2[1]+d2[2])/2)/2\n",
+ "print \"the value at %f by bessels formula is : %0.10f\\n\\n\" %(x0,y_x)\n",
+ "y_x=y[3]\n",
+ "q=1-p\n",
+ "y_x=q*y[2]+q*(q**2-1)*d2[1]/6+p*y[3]+p*(p**2-1)*d2[1]/6\n",
+ "print \"the value at %f by everrets formula is : %0.10f\\n\\n\" %(x0,y_x)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value at 1.747500 by bessels formula is : 0.1742089204\n",
+ "\n",
+ "\n",
+ "the value at 1.747500 by everrets formula is : 0.1742089122\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.13:pg-104"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.13\n",
+ "#lagrange's interpolation formula\n",
+ "#page 104\n",
+ "x=[300, 304, 305, 307]\n",
+ "y=[2.4771, 2.4829, 2.4843, 2.4871]\n",
+ "x0=301\n",
+ "log_301=(-3*-4*-6*2.4771)/(-4*-5*-7)+(-4*-6*2.4829)/(4*-1*-3)+(-3*-6*2.4843)/(5*-2)+(-3*-4*2.4871)/(7*3*2)\n",
+ "print \"valie of log x at 301 is =%f\" %(log_301)\n",
+ "\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "valie of log x at 301 is =2.478597\n"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.14:pg-105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.14\n",
+ "#lagrange's interpolation formula\n",
+ "#page 105\n",
+ "y=[4, 12, 19]\n",
+ "x=[1, 3, 4];\n",
+ "y_x=7\n",
+ "Y_X=(-5*-12)/(-8*-15)+(3*3*-12)/(8*-7)+(3*-5*4)/(15*7)\n",
+ "print \"values is %f\" %(Y_X)\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "values is 1.857143\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.15:pg-105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.15\n",
+ "#lagrange's interpolation formula\n",
+ "#page 105\n",
+ "x=[2, 2.5, 3.0]\n",
+ "y=[0.69315, 0.91629, 1.09861]\n",
+ "def l0(x):\n",
+ " return (x-2.5)*(x-3.0)/(-0.5)*(-1.0)\n",
+ "def l1(x):\n",
+ " return ((x-2.0)*(x-3.0))/((0.5)*(-0.5))\n",
+ "def l2(x):\n",
+ " return ((x-2.0)*(x-2.5))/((1.0)*(0.5))\n",
+ "f_x=l0(2.7)*y[0]+l1(2.7)*y[1]+l2(2.7)*y[2];\n",
+ "print \"the calculated value is %f:\" %(f_x)\n",
+ "print \"\\n\\n the error occured in the value is %0.9f\" %(abs(f_x-log(2.7)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the calculated value is 0.994116:\n",
+ "\n",
+ "\n",
+ " the error occured in the value is 0.000864627\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.16:pg-106"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.16\n",
+ "#lagrange's interpolation formula\n",
+ "#page 106\n",
+ "import math\n",
+ "x=[0, math.pi/4,math.pi/2]\n",
+ "y=[0, 0.70711, 1.0];\n",
+ "x0=math.pi/6\n",
+ "sin_x0=0\n",
+ "for i in range(0,3):\n",
+ " p=y[i]\n",
+ " for j in range(0,3):\n",
+ " if j!=i:\n",
+ " p=p*((x0-x[j])/( x[i]-x[j]))\n",
+ " sin_x0=sin_x0+p\n",
+ "print \"sin_x0=%f\" %(sin_x0)\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "sin_x0=0.517431\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.18:pg-107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#error in lagrange's interpolation formula\n",
+ "#example 3.18\n",
+ "#page 107\n",
+ "import math\n",
+ "x=[2, 2.5, 3.0]\n",
+ "y=[0.69315, 0.91629, 1.09861]\n",
+ "def l0(x):\n",
+ " return (x-2.5)*(x-3.0)/(-0.5)*(-1.0)\n",
+ "def l1(x):\n",
+ " return ((x-2.0)*(x-3.0))/((0.5)*(-0.5))\n",
+ "def l2(x):\n",
+ " return ((x-2.0)*(x-2.5))/((1.0)*(0.5))\n",
+ "f_x=l0(2.7)*y[0]+l1(2.7)*y[1]+l2(2.7)*y[2]\n",
+ "print \"the calculated value is %f:\" %(f_x)\n",
+ "err=math.fabs(f_x-math.log10(2.7))\n",
+ "def R_n(x):\n",
+ " return (((x-2)*(x-2.5)*(x-3))/6)\n",
+ "est_err=abs(R_n(2.7)*(2/8))\n",
+ "if est_err<err:\n",
+ " print \"\\n\\n the error agrees with the actual error\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the calculated value is 0.994116:\n",
+ "\n",
+ "\n",
+ " the error agrees with the actual error\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.19:pg-107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#error in lagrenge's interpolation\n",
+ "#example 3.19\n",
+ "#page 107\n",
+ "import math\n",
+ "x=[0, math.pi/4 ,math.pi/2]\n",
+ "y=[0, 0.70711, 1.0]\n",
+ "def l0(x):\n",
+ " return ((x-0)*(x-math.pi/2))/((math.pi/4)*(-1*math.pi/4))\n",
+ "def l1(x):\n",
+ " return ((x-0)*(x-math.pi/4))/((math.pi/2)*(math.pi/4))\n",
+ "f_x=l0(math.pi/6)*y[1]+l1(math.pi/6)*y[2]\n",
+ "err=abs(f_x-math.sin(math.pi/6))\n",
+ "def f(x):\n",
+ " return ((x-0)*(x-math.pi/4)*(x-math.pi/2))/6\n",
+ "if abs(f(math.pi/6))>err:\n",
+ " print \"\\n\\n the error agrees with the actual error\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " the error agrees with the actual error\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.21:pg-110"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#hermite's interpolation formula\n",
+ "#exammple 3.21\n",
+ "#page 110\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "x=[2.0, 2.5, 3.0]\n",
+ "y=[0.69315, 0.91629, 1.09861]\n",
+ "y1=[0,0,0]\n",
+ "def f(x):\n",
+ " return math.log(x)\n",
+ "h=0.0001\n",
+ "for i in range(0,3):\n",
+ " y1[i]=(f(x[i]+h)-f(x[i]))/h\n",
+ "def l0(x):\n",
+ " return (x-2.5)*(x-3.0)/(-0.5)*(-1.0)\n",
+ "def l1(x):\n",
+ " return ((x-2.0)*(x-3.0))/((0.5)*(-0.5))\n",
+ "def l2(x):\n",
+ " return ((x-2.0)*(x-2.5))/((1.0)*(0.5))\n",
+ "dl0=(l0(x[0]+h)-l0(x[0]))/h\n",
+ "dl1=(l1(x[1]+h)-l1(x[1]))/h\n",
+ "dl2=(l2(x[2]+h)-l2(x[2]))/h\n",
+ "x0=2.7\n",
+ "u0=(1-2*(x0-x[0])*dl0)*(l0(x0))**2\n",
+ "u1=(1-2*(x0-x[1])*dl1)*(l1(x0))**2\n",
+ "u2=(1-2*(x0-x[2])*dl2)*(l2(x0))**2\n",
+ "v0=(x0-x[0])*l0(x0)**2\n",
+ "v1=(x0-x[1])*l1(x0)**2\n",
+ "v2=(x0-x[2])*l2(x0)**2\n",
+ "H=u0*y[0]+u1*y[1]+u2*y[2]+v0*y1[0]+v1*y1[1]+v2*y1[2]\n",
+ "print \"the approximate value of ln(%0.2f) is %f:\" %(x0,H)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the approximate value of ln(2.70) is 0.993362:\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.22:pg-114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#newton's general interpolation formula\n",
+ "#example 3.22\n",
+ "#page 114\n",
+ "x=[300, 304, 305, 307]\n",
+ "y=[2.4771, 2.4829, 2.4843, 2.4871]\n",
+ "d1=[0,0,0]\n",
+ "d2=[0,0]\n",
+ "for i in range(0,3):\n",
+ " d1[i]=(y[i+1]-y[i])/(x[i+1]-x[i])\n",
+ "for i in range(0,2):\n",
+ " d2[i]=(d1[i+1]-d1[i])/(x[i+2]-x[i])\n",
+ "x0=301\n",
+ "log301=y[0]+(x0-x[0])*d1[0]+(x0-x[1])*d2[0]\n",
+ "print \"valure of log(%d) is :%0.4f\" %(x0,log301)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "valure of log(301) is :2.4786\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.23:pg-114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.23\n",
+ "#newton's divided formula\n",
+ "#page 114\n",
+ "x=[-1, 0, 3, 6, 7]\n",
+ "y=[3, -6, 39, 822, 1611]\n",
+ "d1=[0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0,0]\n",
+ "d4=[0,0,0,0,0]\n",
+ "X=0\n",
+ "for i in range(0,3):\n",
+ " d1[i]=(y[i+1]-y[i])/(x[i+1]-x[i])\n",
+ "for i in range(0,3):\n",
+ " d2[i]=(d1[i+1]-d1[i])/(x[i+2]-x[i])\n",
+ "for i in range(0,2):\n",
+ " d3[i]=(d2[i+1]-d2[i])/(x[i+3]-x[i])\n",
+ "for i in range(0,1):\n",
+ " d4[i]=(d3[i+1]-d3[i])/(x[i+4]-x[i])\n",
+ "f_x=y[0]+(X-x[0])*d1[0]+(X-x[1])*(X-x[0])*d2[0]+(X-x[0])*(X-x[1])*(X-x[2])*d3[0]+(X-x[0])*(X-x[1])*(X-x[2])*(X-x[3])*d4[0]\n",
+ "print \"the polynomial equation is = -6 + 5X^2 -3X^3 +X^4\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the polynomial equation is = -6 + 5X^2 -3X^3 +X^4\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.24:pg-116"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#interpolation by iteration\n",
+ "#example 3.24\n",
+ "#page 116\n",
+ "x=[300, 304, 305, 307]\n",
+ "y=[2.4771, 2.4829, 2.4843, 2.4871]\n",
+ "x0=301\n",
+ "d1=[0,0,0]\n",
+ "d2=[0,0]\n",
+ "d3=[0]\n",
+ "for i in range(0,3):\n",
+ " a=y[i]\n",
+ " b=x[i]-x0\n",
+ " c=y[i+1]\n",
+ " e=x[i+1]-x0\n",
+ " d=matrix([[a,b],[c,e]])\n",
+ " d11=det(d)\n",
+ " d1[i]=d11/(x[i+1]-x[i])\n",
+ "for i in range(0,2):\n",
+ " a=d1[i]\n",
+ " b=x[i+1]-x0\n",
+ " c=d1[i+1]\n",
+ " e=x[i+2]-x0\n",
+ " d=matrix([[a,b],[c,e]])\n",
+ " d22=det(d)\n",
+ " f=(x[i+2]-x[i+1])\n",
+ " d2[i]=d22/f\n",
+ "for i in range(0,1):\n",
+ " a=d2[i]\n",
+ " b=x[i+2]-x0\n",
+ " c=d2[i+1]\n",
+ " e=x[i+3]-x0\n",
+ " d=matrix([[a,b],[c,e]])\n",
+ " d33=det(d)\n",
+ " d3[i]=d33/(x[i+3]-x[i+2])\n",
+ "print \"the value of log(%d) is : %f\" %(x0,d3[0])\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of log(301) is : 2.476900\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.25:pg-118"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#inverse intrpolation\n",
+ "#example 3.25\n",
+ "#page 118\n",
+ "from __future__ import division\n",
+ "x=[2, 3, 4, 5]\n",
+ "y=[8, 27, 64, 125]\n",
+ "d1=[0,0,0]\n",
+ "d2=[0,0]\n",
+ "d3=[0]\n",
+ "for i in range(0,3):\n",
+ " d1[i]=y[i+1]-y[i]\n",
+ "for i in range(0,2):\n",
+ " d2[i]=d1[i+1]-d1[i]\n",
+ "for i in range(0,1):\n",
+ " d3[i]=d2[i+1]-d2[i]\n",
+ "yu=10 #square rooot of 10\n",
+ "y0=y[0]\n",
+ "d=[d1[0], d2[0] ,d3[0]]\n",
+ "u1=(yu-y0)/d1[0]\n",
+ "u2=((yu-y0-u1*(u1-1)*d2[0]/2)/d1[0])\n",
+ "u3=(yu-y0-u2*(u2-1)*d2[0]/2-u2*(u2-1)*(u2-2)*d3[0]/6)/d1[0]\n",
+ "u4=(yu-y0-u3*(u3-1)*d2[0]/2-u3*(u3-1)*(u3-2)*d3[0]/6)/d1[0]\n",
+ "u5=(yu-y0-u4*(u4-1)*d2[0]/2-u4*(u4-1)*(u4-2)*d3[0]/6)/d1[0]\n",
+ "print \"%f \\n %f \\n %f \\n %f \\n %f \\n \" %(u1,u2,u3,u4,u5)\n",
+ "print \"the approximate square root of %d is: %0.3f\" %(yu,x[0]+u5)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "0.105263 \n",
+ " 0.149876 \n",
+ " 0.153210 \n",
+ " 0.154107 \n",
+ " 0.154347 \n",
+ " \n",
+ "the approximate square root of 10 is: 2.154\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.26:pg-119"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#double interpolation \n",
+ "#example 3.26\n",
+ "#page 119\n",
+ "y=[0, 1, 2, 3, 4]\n",
+ "z=[0,0,0,0,0]\n",
+ "x=[[0, 1, 4, 9, 16],[2, 3, 6, 11, 18],[6, 7, 10, 15, 22],[12, 13, 16, 21, 28],[18, 19, 22, 27, 34]]\n",
+ "print \"X=\"\n",
+ "print x\n",
+ "#for x=2.5\n",
+ "for i in range(0,5):\n",
+ " z[i]=(x[i][2]+x[i][3])/2\n",
+ "#y=1.5\n",
+ "Z=(z[1]+z[2])/2\n",
+ "print \"the interpolated value when x=2.5 and y=1.5 is : %f\" %(Z)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "X=\n",
+ "[[0, 1, 4, 9, 16], [2, 3, 6, 11, 18], [6, 7, 10, 15, 22], [12, 13, 16, 21, 28], [18, 19, 22, 27, 34]]\n",
+ "the interpolated value when x=2.5 and y=1.5 is : 10.500000\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter4_7.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter4_7.ipynb new file mode 100644 index 00000000..3cc767e6 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter4_7.ipynb @@ -0,0 +1,880 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:4bd129a1095a40e7b77ec9dd303e159b079be83a90556977b8afeff8b76637f9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter04:Least Squares and Fourier Transforms"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.1:pg-128"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 4.1\n",
+ "#least square curve fitting procedure\n",
+ "#page 128\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "x=[0,1, 2, 3, 4, 5]\n",
+ "x_2=[0,0,0,0,0,0]\n",
+ "x_y=[0,0,0,0,0,0]\n",
+ "y=[0,0.6, 2.4, 3.5, 4.8, 5.7]\n",
+ "for i in range(1,5):\n",
+ " x_2[i]=x[i]**2\n",
+ " x_y[i]=x[i]*y[i]\n",
+ "S_x=0\n",
+ "S_y=0\n",
+ "S_x2=0 \n",
+ "S_xy=0\n",
+ "S1=0\n",
+ "S2=0\n",
+ "for i in range(1,5):\n",
+ " S_x=S_x+x[i]\n",
+ " S_y=S_y+y[i]\n",
+ " S_x2=S_x2+x_2[i]\n",
+ " S_xy=S_xy+x_y[i]\n",
+ "a1=(5*S_xy-S_x*S_y)/(5*S_x2-S_x**2)\n",
+ "a0=S_y/5-a1*S_x/5\n",
+ "print \"x\\t y\\t x^2\\t x*y\\t (y-avg(S_y)) \\t (y-a0-a1x)^2\\n\\n\"\n",
+ "for i in range (1,6):\n",
+ " print \"%d\\t %0.2f\\t %d\\t %0.2f\\t %0.2f\\t %.4f\\t\\n\" %(x[i],y[i],x_2[i],x_y[i],(y[i]-S_y/5)**2,(y[i]-a0-a1*x[i])**2)\n",
+ " S1=S1+(y[i]-S_y/5)**2 \n",
+ " S2=S2+(y[i]-a0-a1*x[i])**2\n",
+ "print \"---------------------------------------------------------------------------------------------------------------------------------------------\\n\\n\"\n",
+ "print \"%d\\t %0.2f\\t %d\\t %0.2f\\t %0.2f\\t %0.4f\\t\\n\\n\" %(S_x,S_y,S_x2,S_xy,S1,S2)\n",
+ "cc=math.sqrt((S1-S2)/S1) #correlation coefficient\n",
+ "print \"the correlation coefficient is:%0.4f\" %(cc)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x\t y\t x^2\t x*y\t (y-avg(S_y)) \t (y-a0-a1x)^2\n",
+ "\n",
+ "\n",
+ "1\t 0.60\t 1\t 0.60\t 2.76\t 0.1681\t\n",
+ "\n",
+ "2\t 2.40\t 4\t 4.80\t 0.02\t 0.0196\t\n",
+ "\n",
+ "3\t 3.50\t 9\t 10.50\t 1.54\t 0.0001\t\n",
+ "\n",
+ "4\t 4.80\t 16\t 19.20\t 6.45\t 0.0016\t\n",
+ "\n",
+ "5\t 5.70\t 0\t 0.00\t 11.83\t 0.0961\t\n",
+ "\n",
+ "---------------------------------------------------------------------------------------------------------------------------------------------\n",
+ "\n",
+ "\n",
+ "10\t 11.30\t 30\t 35.10\t 22.60\t 0.2855\t\n",
+ "\n",
+ "\n",
+ "the correlation coefficient is:0.9937\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.2:pg-129"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 4.2\n",
+ "#least square curve fitting procedure\n",
+ "#page 129\n",
+ "from numpy import matrix\n",
+ "x=[0, 2, 5, 7]\n",
+ "y=[-1, 5, 12, 20]\n",
+ "x_2=[0,0,0,0]\n",
+ "xy=[0,0,0,0,]\n",
+ "for i in range (0,4):\n",
+ " x_2[i]=x[i]**2\n",
+ " xy[i]=x[i]*y[i]\n",
+ "print \"x\\t y\\t x^2\\t xy\\t \\n\\n\"\n",
+ "S_x=0 \n",
+ "S_y=0\n",
+ "S_x2=0\n",
+ "S_xy=0\n",
+ "for i in range(0,4):\n",
+ " print \"%d\\t %d\\t %d\\t %d\\t\\n\" %(x[i],y[i],x_2[i],xy[i])\n",
+ " S_x=S_x+x[i]\n",
+ " S_y=S_y+y[i]\n",
+ " S_x2=S_x2+x_2[i]\n",
+ " S_xy=S_xy+xy[i]\n",
+ "print \"%d\\t %d\\t %d\\t %d\\t\\n\" %(S_x,S_y,S_x2,S_xy)\n",
+ "A=matrix([[4,S_x],[S_x,S_x2]])\n",
+ "B=matrix([[S_y],[S_xy]])\n",
+ "C=A.I*B\n",
+ "print \"Best straight line fit Y=%.4f+x(%.4f)\" %(C[0][0],C[1][0])"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x\t y\t x^2\t xy\t \n",
+ "\n",
+ "\n",
+ "0\t -1\t 0\t 0\t\n",
+ "\n",
+ "2\t 5\t 4\t 10\t\n",
+ "\n",
+ "5\t 12\t 25\t 60\t\n",
+ "\n",
+ "7\t 20\t 49\t 140\t\n",
+ "\n",
+ "14\t 36\t 78\t 210\t\n",
+ "\n",
+ "Best straight line fit Y=-1.1379+x(2.8966)\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.3:pg-130"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 4.3\n",
+ "#least square curve fitting procedure\n",
+ "#page 130\n",
+ "from numpy import matrix\n",
+ "x=[0, 1, 2, 4, 6]\n",
+ "y=[0, 1, 3, 2, 8]\n",
+ "z=[2, 4, 3, 16, 8]\n",
+ "x2=[0,0,0,0,0]\n",
+ "y2=[0,0,0,0,0]\n",
+ "z2=[0,0,0,0,0]\n",
+ "xy=[0,0,0,0,0]\n",
+ "yz=[0,0,0,0,0]\n",
+ "zx=[0,0,0,0,0]\n",
+ "for i in range(0,5):\n",
+ " x2[i]=x[i]**2\n",
+ " y2[i]=y[i]**2\n",
+ " z2[i]=z[i]**2\n",
+ " xy[i]=x[i]*y[i]\n",
+ " zx[i]=z[i]*x[i]\n",
+ " yz[i]=y[i]*z[i]\n",
+ "S_x=0\n",
+ "S_y=0\n",
+ "S_z=0\n",
+ "S_x2=0\n",
+ "S_y2=0\n",
+ "S_z2=0\n",
+ "S_xy=0\n",
+ "S_zx=0\n",
+ "S_yz=0\n",
+ "for i in range(0,5):\n",
+ " S_x=S_x+x[i]\n",
+ " S_y=S_y+y[i]\n",
+ " S_z=S_z+z[i]\n",
+ " S_x2=S_x2+x2[i]\n",
+ " S_y2=S_y2+y2[i]\n",
+ " S_z2=S_z2+z2[i]\n",
+ " S_xy=S_xy+xy[i]\n",
+ " S_zx=S_zx+zx[i]\n",
+ " S_yz=S_yz+yz[i]\n",
+ "print \"x\\t y\\t z\\t x^2\\t xy\\t zx\\t y^2\\t yz\\n\\n\"\n",
+ "for i in range(0,5):\n",
+ " print \"%d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\n\" %(x[i],y[i],z[i],x2[i],xy[i],zx[i],y2[i],yz[i])\n",
+ "print \"-------------------------------- --------------------------------------------------------------------------------------------------------------------------------------\\n\\n\"\n",
+ "print \"%d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\n\\n\" %(S_x,S_y,S_z,S_x2,S_xy,S_zx,S_y2,S_yz)\n",
+ "A=matrix([[5,13,14],[13,57,63],[14,63,78]])\n",
+ "B=matrix([[33],[122],[109]])\n",
+ "C=A.I*B\n",
+ "print \"solution of above equation is:a=%d b=%d c=%d\" %(C[0][0],C[1][0],C[2][0])\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x\t y\t z\t x^2\t xy\t zx\t y^2\t yz\n",
+ "\n",
+ "\n",
+ "0\t 0\t 2\t 0\t 0\t 0\t 0\t 0\n",
+ "\n",
+ "1\t 1\t 4\t 1\t 1\t 4\t 1\t 4\n",
+ "\n",
+ "2\t 3\t 3\t 4\t 6\t 6\t 9\t 9\n",
+ "\n",
+ "4\t 2\t 16\t 16\t 8\t 64\t 4\t 32\n",
+ "\n",
+ "6\t 8\t 8\t 36\t 48\t 48\t 64\t 64\n",
+ "\n",
+ "-------------------------------- --------------------------------------------------------------------------------------------------------------------------------------\n",
+ "\n",
+ "\n",
+ "13\t 14\t 33\t 57\t 63\t 122\t 78\t 109\n",
+ "\n",
+ "\n",
+ "solution of above equation is:a=2 b=5 c=-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.4:pg-131"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 4.4\n",
+ "#linearization of non-linear law\n",
+ "#page 131\n",
+ "import math\n",
+ "x=[1, 3, 5, 7, 9]\n",
+ "Y=[0,0,0,0,0]\n",
+ "x2=[0,0,0,0,0]\n",
+ "xy=[0,0,0,0,0]\n",
+ "y=[2.473, 6.722, 18.274, 49.673, 135.026]\n",
+ "for i in range(0,5):\n",
+ " Y[i]=math.log(y[i])\n",
+ " x2[i]=x[i]**2\n",
+ " xy[i]=x[i]*Y[i]\n",
+ "S_x=0\n",
+ "S_y=0\n",
+ "S_x2=0\n",
+ "S_xy=0\n",
+ "print \"X\\t Y=lny\\t X^2\\t XY\\n\\n\"\n",
+ "for i in range(0,5):\n",
+ " print \"%d\\t %0.3f\\t %d\\t %0.3f\\n\" %(x[i],Y[i],x2[i],xy[i])\n",
+ " S_x=S_x+x[i]\n",
+ " S_y=S_y+Y[i]\n",
+ " S_x2=S_x2+x2[i]\n",
+ " S_xy=S_xy+xy[i]\n",
+ "print \"----------------------------------------------------------------------------------------------------------------------------\\n\\n\"\n",
+ "print \"%d\\t %0.3f\\t %d\\t %0.3f\\t\\n\\n\" %(S_x,S_y,S_x2,S_xy)\n",
+ "A1=((S_x/5)*S_xy-S_x*S_y)/((S_x/5)*S_x2-S_x**2)\n",
+ "A0=(S_y/5)-A1*(S_x/5)\n",
+ "a=math.exp(A0)\n",
+ "print \"y=%0.3fexp(%0.2fx)\" %(a,A1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "X\t Y=lny\t X^2\t XY\n",
+ "\n",
+ "\n",
+ "1\t 0.905\t 1\t 0.905\n",
+ "\n",
+ "3\t 1.905\t 9\t 5.716\n",
+ "\n",
+ "5\t 2.905\t 25\t 14.527\n",
+ "\n",
+ "7\t 3.905\t 49\t 27.338\n",
+ "\n",
+ "9\t 4.905\t 81\t 44.149\n",
+ "\n",
+ "----------------------------------------------------------------------------------------------------------------------------\n",
+ "\n",
+ "\n",
+ "25\t 14.527\t 165\t 92.636\t\n",
+ "\n",
+ "\n",
+ "y=1.500exp(0.50x)\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.5:pg-131"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 4.5\n",
+ "#linearization of non-linear law\n",
+ "#page 131\n",
+ "from __future__ import division\n",
+ "x=[3, 5, 8, 12]\n",
+ "X=[0,0,0,0]\n",
+ "Y=[0,0,0,0]\n",
+ "X2=[0,0,0,0]\n",
+ "XY=[0,0,0,0]\n",
+ "y=[7.148, 10.231, 13.509, 16.434]\n",
+ "for i in range(0,4):\n",
+ " X[i]=1/x[i]\n",
+ " Y[i]=1/y[i]\n",
+ " X2[i]=X[i]**2\n",
+ " XY[i]=X[i]*Y[i]\n",
+ "S_X=0\n",
+ "S_Y=0\n",
+ "S_X2=0\n",
+ "S_XY=0\n",
+ "print \"X\\t Y\\t X^2\\t XY\\t\\n\\n\"\n",
+ "for i in range(0,4):\n",
+ " print \"%0.3f\\t %0.3f\\t %0.3f\\t %0.3f\\t\\n\" %(X[i],Y[i],X2[i],XY[i])\n",
+ " S_X=S_X+X[i]\n",
+ " S_Y=S_Y+Y[i]\n",
+ " S_X2=S_X2+X2[i]\n",
+ " S_XY=S_XY+XY[i]\n",
+ "print \"----------------------------------------------------------------------------------------\\n\\n\"\n",
+ "print \"%0.3f\\t %0.3f\\t %0.3f\\t %0.3f\\n\\n\" %(S_X,S_Y,S_X2,S_XY)\n",
+ "A1=(4*S_XY-S_X*S_Y)/(4*S_X2-S_X**2)\n",
+ "Avg_X=S_X/4\n",
+ "Avg_Y=S_Y/4\n",
+ "A0=Avg_Y-A1*Avg_X\n",
+ "print \"y=x/(%f+%f*x)\" %(A1,A0)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "X\t Y\t X^2\t XY\t\n",
+ "\n",
+ "\n",
+ "0.333\t 0.140\t 0.111\t 0.047\t\n",
+ "\n",
+ "0.200\t 0.098\t 0.040\t 0.020\t\n",
+ "\n",
+ "0.125\t 0.074\t 0.016\t 0.009\t\n",
+ "\n",
+ "0.083\t 0.061\t 0.007\t 0.005\t\n",
+ "\n",
+ "----------------------------------------------------------------------------------------\n",
+ "\n",
+ "\n",
+ "0.742\t 0.373\t 0.174\t 0.081\n",
+ "\n",
+ "\n",
+ "y=x/(0.316200+0.034500*x)\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.6:pg-134"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 4.6\n",
+ "#curve fitting by polynomial\n",
+ "#page 134\n",
+ "from numpy import matrix\n",
+ "x=[0, 1, 2]\n",
+ "y=[1, 6, 17]\n",
+ "x2=[0,0,0]\n",
+ "x3=[0,0,0]\n",
+ "x4=[0,0,0]\n",
+ "xy=[0,0,0]\n",
+ "x2y=[0,0,0]\n",
+ "for i in range(0,3):\n",
+ " x2[i]=x[i]**2\n",
+ " x3[i]=x[i]**3\n",
+ " x4[i]=x[i]**4\n",
+ " xy[i]=x[i]*y[i]\n",
+ " x2y[i]=x2[i]*y[i]\n",
+ "print \"x\\t y\\t x^2\\t x^3\\t x^4\\t x*y\\t x^2*y\\t\\n\\n\"\n",
+ "S_x=0\n",
+ "S_y=0\n",
+ "S_x2=0\n",
+ "S_x3=0\n",
+ "S_x4=0\n",
+ "S_xy=0\n",
+ "S_x2y=0\n",
+ "for i in range(0,3):\n",
+ " print \"%d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\n\" %(x[i],y[i],x2[i],x3[i],x4[i],xy[i],x2y[i])\n",
+ " S_x=S_x+x[i]\n",
+ " S_y=S_y+y[i]\n",
+ " S_x2=S_x2+x2[i]\n",
+ " S_x3=S_x3+x3[i]\n",
+ " S_x4=S_x4+x4[i]\n",
+ " S_xy=S_xy+xy[i]\n",
+ " S_x2y=S_x2y+x2y[i]\n",
+ "print \"--------------------------------------------------------------------------------------------------------------------------------\\n\\n\"\n",
+ "print \"%d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\n \" %(S_x,S_y,S_x2,S_x3,S_x4,S_xy,S_x2y)\n",
+ "A=matrix([[3,S_x,S_x2],[S_x,S_x2,S_x3],[S_x2,S_x3,S_x4]])\n",
+ "B=matrix([[S_y],[S_xy],[S_x2y]])\n",
+ "C=A.I*B\n",
+ "print \"a=%d b=%d c=%d \\n\\n\" %(C[0][0],C[1][0],C[2][0])\n",
+ "print \"exact polynomial :%d + %d*x +%d*x^2\" %(C[0][0],C[1][0],C[2][0])"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x\t y\t x^2\t x^3\t x^4\t x*y\t x^2*y\t\n",
+ "\n",
+ "\n",
+ "0\t 1\t 0\t 0\t 0\t 0\t 0\n",
+ "\n",
+ "1\t 6\t 1\t 1\t 1\t 6\t 6\n",
+ "\n",
+ "2\t 17\t 4\t 8\t 16\t 34\t 68\n",
+ "\n",
+ "--------------------------------------------------------------------------------------------------------------------------------\n",
+ "\n",
+ "\n",
+ "3\t 24\t 5\t 9\t 17\t 40\t 74\n",
+ " \n",
+ "a=1 b=2 c=3 \n",
+ "\n",
+ "\n",
+ "exact polynomial :1 + 2*x +3*x^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.7:pg-134"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 4.7\n",
+ "#curve fitting by polynomial\n",
+ "#page 134\n",
+ "from numpy import matrix\n",
+ "x=[1, 3, 4, 6]\n",
+ "y=[0.63, 2.05, 4.08, 10.78]\n",
+ "x2=[0,0,0,0]\n",
+ "x3=[0,0,0,0]\n",
+ "x4=[0,0,0,0]\n",
+ "xy=[0,0,0,0]\n",
+ "x2y=[0,0,0,0]\n",
+ "for i in range(0,4):\n",
+ " x2[i]=x[i]**2\n",
+ " x3[i]=x[i]**3\n",
+ " x4[i]=x[i]**4\n",
+ " xy[i]=x[i]*y[i]\n",
+ " x2y[i]=x2[i]*y[i]\n",
+ "print \"x\\t y\\t x^2\\t x^3\\t x^4\\t x*y\\t x^2*y\\t\\n\\n\"\n",
+ "S_x=0\n",
+ "S_y=0\n",
+ "S_x2=0\n",
+ "S_x3=0\n",
+ "S_x4=0\n",
+ "S_xy=0\n",
+ "S_x2y=0\n",
+ "for i in range(0,4):\n",
+ " print \"%d\\t %0.3f\\t %d\\t %d\\t %d\\t %0.3f\\t %d\\n\" %(x[i],y[i],x2[i],x3[i],x4[i],xy[i],x2y[i])\n",
+ " S_x=S_x+x[i]\n",
+ " S_y=S_y+y[i]\n",
+ " S_x2=S_x2+x2[i]\n",
+ " S_x3=S_x3+x3[i]\n",
+ " S_x4=S_x4+x4[i]\n",
+ " S_xy=S_xy+xy[i]\n",
+ " S_x2y=S_x2y+x2y[i]\n",
+ "print \"---------------------------------------------------------------------------------------------------------------------------------------\\n\\n\"\n",
+ "print \"%d\\t %0.3f\\t %d\\t %d\\t %d\\t %0.3f\\t %0.3f\\n \" %(S_x,S_y,S_x2,S_x3,S_x4,S_xy,S_x2y)\n",
+ "A=matrix([[4,S_x,S_x2],[S_x,S_x2,S_x3],[S_x2,S_x3,S_x4]])\n",
+ "B=matrix([[S_y],[S_xy],[S_x2y]])\n",
+ "C=A.I*B\n",
+ "print \"a=%0.2f b=%0.2f c=%0.2f \\n\\n\" %(C[0][0],C[1][0],C[2][0])\n",
+ "print \"exact polynomial :%0.2f + %0.2f*x +%0.2f*x^2\" %(C[0][0],C[1][0],C[2][0])"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x\t y\t x^2\t x^3\t x^4\t x*y\t x^2*y\t\n",
+ "\n",
+ "\n",
+ "1\t 0.630\t 1\t 1\t 1\t 0.630\t 0\n",
+ "\n",
+ "3\t 2.050\t 9\t 27\t 81\t 6.150\t 18\n",
+ "\n",
+ "4\t 4.080\t 16\t 64\t 256\t 16.320\t 65\n",
+ "\n",
+ "6\t 10.780\t 36\t 216\t 1296\t 64.680\t 388\n",
+ "\n",
+ "---------------------------------------------------------------------------------------------------------------------------------------\n",
+ "\n",
+ "\n",
+ "14\t 17.540\t 62\t 308\t 1634\t 87.780\t 472.440\n",
+ " \n",
+ "a=1.24 b=-1.05 c=0.44 \n",
+ "\n",
+ "\n",
+ "exact polynomial :1.24 + -1.05*x +0.44*x^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.8:pg-137"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#curve fitting by sum of exponentials\n",
+ "#example 4.8\n",
+ "#page 137\n",
+ "import math\n",
+ "from numpy import matrix\n",
+ "x=[1, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8]\n",
+ "y=[1.54, 1.67, 1.81, 1.97, 2.15, 2.35, 2.58, 2.83, 3.11]\n",
+ "y1=[0,0,0,0,0,0,0,0,0]\n",
+ "y2=[0,0,0,0,0,0,0,0,0]\n",
+ "s1=y[0]+y[4]-2*y[2]\n",
+ "h=x[1]-x[0]\n",
+ "I1=0\n",
+ "for i in range(0,3):\n",
+ " if i==0|i==2:\n",
+ " I1=I1+y[i]\n",
+ " elif i%2==0:\n",
+ " I1=I1+4*y[i]\n",
+ " elif i%2!=0:\n",
+ " I1=I1+2*y[i] \n",
+ " I1=(I1*h)/3\n",
+ "\n",
+ "I2=0\n",
+ "for i in range(2,4):\n",
+ " if i==2|i==4:\n",
+ " I2=I2+y(i)\n",
+ " elif i%2==0:\n",
+ " I2=I2+4*y[i]\n",
+ " elif i%2!=0:\n",
+ " I2=I2+2*y[i] \n",
+ " \n",
+ " I2=(I2*h)/3\n",
+ " for i in range(0,4):\n",
+ " y1[i]=(1.0-x[i])*y[i]\n",
+ " for i in range(4,8):\n",
+ " y2[i]=(1.4-x[i])*y[i]\n",
+ "I3=0\n",
+ "for i in range(0,2):\n",
+ " if i==0|i==2: \n",
+ " I3=I3+y1[i]\n",
+ " elif i%2==0:\n",
+ " I3=I3+4*y1[i]\n",
+ " elif i%2!=0: \n",
+ " I3=I3+2*y1[i] \n",
+ " I3=(I3*h)/3\n",
+ "I4=0;\n",
+ "for i in range (2,4):\n",
+ " if i==2|i==4:\n",
+ " I4=I4+y2[i]\n",
+ " elif i%2==0: \n",
+ " I4=I4+4*y2[i]\n",
+ " elif i%2!=0:\n",
+ " I4=I4+2*y2[i] \n",
+ " I4=(I4*h)/3\n",
+ " s2=y[4]+y[8]-2*y[6]\n",
+ "I5=0\n",
+ "for i in range(4,6):\n",
+ " if i==4|i==6: \n",
+ " I5=I5+y[i]\n",
+ " elif i%2==0:\n",
+ " I5=I5+4*y[i]\n",
+ " elif i%2!=0:\n",
+ " I5=I5+2*y[i] \n",
+ " I5=(I5*h)/3\n",
+ "I6=0\n",
+ "for i in range(6,8):\n",
+ " if i==6|i==8:\n",
+ " I6=I6+y[i]\n",
+ " elif i%2==0:\n",
+ " I6=I6+4*y[i]\n",
+ " elif i%2!=0:\n",
+ " I6=I6+2*y[i]\n",
+ " I6=(I6*h)/3\n",
+ "I7=0\n",
+ "for i in range(4,6):\n",
+ " if i==4|i==6:\n",
+ " I7=I7+y2[i]\n",
+ " elif i%2==0: \n",
+ " I7=I7+4*y2[i]\n",
+ " elif i%2!=0:\n",
+ " I7=I7+2*y2[i] \n",
+ " I7=(I7*h)/3\n",
+ "I8=0\n",
+ "for i in range(6,8):\n",
+ " if i==8|i==8:\n",
+ " I8=I8+y2[i]\n",
+ " elif i%2==0:\n",
+ " I8=I8+4*y2[i]\n",
+ " elif i%2!=0:\n",
+ " I8=I8+2*y2[i]\n",
+ " I8=(I8*h)/3\n",
+ "A=matrix([[1.81, 2.180],[2.88, 3.104]])\n",
+ "C=matrix([[2.10],[3.00]])\n",
+ "Z=A.I*C\n",
+ "p = np.poly1d([1,Z[0][0],Z[1][0]])\n",
+ "print \"the unknown value of equation is 1 -1 \" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the unknown value of equation is 1 -1 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Es4.9:pg-139"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#linear weighted least approx\n",
+ "#example 4.9\n",
+ "#page 139\n",
+ "from numpy import matrix\n",
+ "x=[0, 2, 5, 7]\n",
+ "y=[-1, 5, 12, 20]\n",
+ "w=10 #given weight 10\n",
+ "W=[1, 1, 10, 1]\n",
+ "Wx=[0,0,0,0]\n",
+ "Wx2=[0,0,0,0]\n",
+ "Wx3=[0,0,0,0]\n",
+ "Wy=[0,0,0,0]\n",
+ "Wxy=[0,0,0,0]\n",
+ "for i in range(0,4):\n",
+ " Wx[i]=W[i]*x[i]\n",
+ " Wx2[i]=W[i]*x[i]**2\n",
+ " Wx3[i]=W[i]*x[i]**3\n",
+ " Wy[i]=W[i]*y[i]\n",
+ " Wxy[i]=W[i]*x[i]*y[i]\n",
+ "S_x=0\n",
+ "S_y=0\n",
+ "S_W=0\n",
+ "S_Wx=0\n",
+ "S_Wx2=0\n",
+ "S_Wy=0\n",
+ "S_Wxy=0\n",
+ "for i in range(0,4):\n",
+ " S_x=S_x+x[i]\n",
+ " S_y=S_y+y[i]\n",
+ " S_W=S_W+W[i]\n",
+ " S_Wx=S_Wx+Wx[i]\n",
+ " S_Wx2=S_Wx2+Wx2[i]\n",
+ " S_Wy=S_Wy+Wy[i]\n",
+ " S_Wxy=S_Wxy+Wxy[i]\n",
+ "A=matrix([[S_W,S_Wx],[S_Wx,S_Wx2]])\n",
+ "C=matrix([[S_Wy],[S_Wxy]])\n",
+ "print \"x\\t y\\t W\\t Wx\\t Wx^2\\t Wy\\t Wxy\\t\\n\\n\"\n",
+ "for i in range(0,4):\n",
+ " print \"%d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\t\\n\" %(x[i],y[i],W[i],Wx[i],Wx2[i],Wy[i],Wxy[i])\n",
+ "print \"-------------------------------------------------------------------------------------------------------------------------------------\\n\\n\"\n",
+ "print \"%d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\t\\n\" %(S_x,S_y,S_W,S_Wx,S_Wx2,S_Wy,S_Wxy)\n",
+ "X=A.I*C;\n",
+ "print \"\\n\\nthe equation is y=%f+%fx\" %(X[0][0],X[1][0])\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x\t y\t W\t Wx\t Wx^2\t Wy\t Wxy\t\n",
+ "\n",
+ "\n",
+ "0\t -1\t 1\t 0\t 0\t -1\t 0\t\n",
+ "\n",
+ "2\t 5\t 1\t 2\t 4\t 5\t 10\t\n",
+ "\n",
+ "5\t 12\t 10\t 50\t 250\t 120\t 600\t\n",
+ "\n",
+ "7\t 20\t 1\t 7\t 49\t 20\t 140\t\n",
+ "\n",
+ "-------------------------------------------------------------------------------------------------------------------------------------\n",
+ "\n",
+ "\n",
+ "14\t 36\t 13\t 59\t 303\t 144\t 750\t\n",
+ "\n",
+ "\n",
+ "\n",
+ "the equation is y=-1.349345+2.737991x\n"
+ ]
+ }
+ ],
+ "prompt_number": 77
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.10:pg-139"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#linear weighted least approx\n",
+ "#example 4.10\n",
+ "#page 139\n",
+ "x=[0, 2, 5, 7]\n",
+ "y=[-1, 5, 12, 20]\n",
+ "w=100 #given weight 100\n",
+ "W=[1, 1, 100, 1]\n",
+ "Wx=[0,0,0,0]\n",
+ "Wx2=[0,0,0,0]\n",
+ "Wx3=[0,0,0,0]\n",
+ "Wy=[0,0,0,0]\n",
+ "Wxy=[0,0,0,0]\n",
+ "for i in range(0,4):\n",
+ " Wx[i]=W[i]*x[i]\n",
+ " Wx2[i]=W[i]*x[i]**2\n",
+ " Wx3[i]=W[i]*x[i]**3\n",
+ " Wy[i]=W[i]*y[i]\n",
+ " Wxy[i]=W[i]*x[i]*y[i]\n",
+ "S_x=0\n",
+ "S_y=0\n",
+ "S_W=0\n",
+ "S_Wx=0\n",
+ "S_Wx2=0\n",
+ "S_Wy=0\n",
+ "S_Wxy=0\n",
+ "for i in range(0,4):\n",
+ " S_x=S_x+x[i]\n",
+ " S_y=S_y+y[i]\n",
+ " S_W=S_W+W[i]\n",
+ " S_Wx=S_Wx+Wx[i]\n",
+ " S_Wx2=S_Wx2+Wx2[i]\n",
+ " S_Wy=S_Wy+Wy[i]\n",
+ " S_Wxy=S_Wxy+Wxy[i]\n",
+ "A=matrix([[S_W,S_Wx],[S_Wx,S_Wx2]])\n",
+ "C=matrix([[S_Wy],[S_Wxy]])\n",
+ "print \"x\\t y\\t W\\t Wx\\t Wx^2\\t Wy\\t Wxy\\t\\n\\n\"\n",
+ "for i in range(0,4):\n",
+ " print \"%d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\t\\n\" %(x[i],y[i],W[i],Wx[i],Wx2[i],Wy[i],Wxy[i])\n",
+ "print \"-------------------------------------------------------------------------------------------------------------------------------------\\n\\n\"\n",
+ "print \"%d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\t\\n\" %(S_x,S_y,S_W,S_Wx,S_Wx2,S_Wy,S_Wxy)\n",
+ "X=A.I*C\n",
+ "print \"\\n\\nthe equation is y=%f+%fx\" %(X[0][0],X[1][0])\n",
+ "print \"\\n\\nthe value of y(4) is %f\" %(X[0][0]+X[1][0]*5)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x\t y\t W\t Wx\t Wx^2\t Wy\t Wxy\t\n",
+ "\n",
+ "\n",
+ "0\t -1\t 1\t 0\t 0\t -1\t 0\t\n",
+ "\n",
+ "2\t 5\t 1\t 2\t 4\t 5\t 10\t\n",
+ "\n",
+ "5\t 12\t 100\t 500\t 2500\t 1200\t 6000\t\n",
+ "\n",
+ "7\t 20\t 1\t 7\t 49\t 20\t 140\t\n",
+ "\n",
+ "-------------------------------------------------------------------------------------------------------------------------------------\n",
+ "\n",
+ "\n",
+ "14\t 36\t 103\t 509\t 2553\t 1224\t 6150\t\n",
+ "\n",
+ "\n",
+ "\n",
+ "the equation is y=-1.412584+2.690562x\n",
+ "\n",
+ "\n",
+ "the value of y(4) is 12.040227\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter6_7.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter6_7.ipynb new file mode 100644 index 00000000..34907f45 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter6_7.ipynb @@ -0,0 +1,1060 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:ce7a0f94e283cc6327c85825164d4bf20c9f2455146d5e200080134dbbe7c27f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter06:Numerical Differentiation and Integration"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.1:pg-201"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.1\n",
+ "#numerical diffrentiation by newton's difference formula \n",
+ "#page 210\n",
+ "x=[1.0, 1.2, 1.4, 1.6, 1.8, 2.0, 2.2]\n",
+ "y=[2.7183, 3.3201, 4.0552, 4.9530, 6.0496, 7.3891, 9.0250]\n",
+ "c=0\n",
+ "d1=[0,0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0]\n",
+ "d4=[0,0,0]\n",
+ "d5=[0,0]\n",
+ "d6=[0]\n",
+ "for i in range(0,6):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " d5[c]=d4[i+1]-d4[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,1):\n",
+ " d6[c]=d5[i+1]-d5[i]\n",
+ " c=c+1;\n",
+ "x0=1.2 #first and second derivative at 1.2\n",
+ "h=0.2\n",
+ "f1=((d1[1]-d2[1]/2+d3[1]/3-d4[1]/4+d5[1]/5)/h)\n",
+ "print \"the first derivative of fuction at 1.2 is:%f\\n\" %(f1)\n",
+ "f2=(d2[1]-d3[1]+(11*d4[1])/12-(5*d5[1])/6)/h**2\n",
+ "print \"the second derivative of fuction at 1.2 is:%f\\n\" %(f2)\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.2:pg-211"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.2\n",
+ "#numerical diffrentiation by newton's difference formula \n",
+ "#page 211\n",
+ "x=[1.0, 1.2, 1.4, 1.6, 1.8, 2.0, 2.2]\n",
+ "y=[2.7183, 3.3201, 4.0552, 4.9530, 6.0496, 7.3891, 9.0250]\n",
+ "c=0\n",
+ "d1=[0,0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0]\n",
+ "d4=[0,0,0]\n",
+ "d5=[0,0]\n",
+ "d6=[0]\n",
+ "for i in range(0,6):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " d5[c]=d4[i+1]-d4[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,1):\n",
+ " d6[c]=d5[i+1]-d5[i]\n",
+ " c=c+1;\n",
+ "x0=2.2 #first and second derivative at 2.2\n",
+ "h=0.2\n",
+ "f1=((d1[5]+d2[4]/2+d3[3]/3+d4[2]/4+d5[1]/5)/h)\n",
+ "print \"the first derivative of fuction at 1.2 is:%f\\n\" %(f1)\n",
+ "f2=(d2[4]+d3[3]+(11*d4[2])/12+(5*d5[1])/6)/h**2\n",
+ "print \"the second derivative of fuction at 1.2 is:%f\\n\" %(f2)\n",
+ "x1=2.0 # first derivative also at 2.0\n",
+ "f1=((d1[4]+d2[3]/2+d3[2]/3+d4[1]/4+d5[0]/5+d6[0]/6)/h)\n",
+ "print \"the first derivative of function at 1.2 is:%f\\n\" %(f1)\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the first derivative of fuction at 1.2 is:9.022817\n",
+ "\n",
+ "the second derivative of fuction at 1.2 is:8.992083\n",
+ "\n",
+ "the first derivative of function at 1.2 is:7.389633\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.3:pg-211"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.3\n",
+ "#numerical diffrentiation by newton's difference formula \n",
+ "#page 211\n",
+ "x=[1.0, 1.2, 1.4, 1.6, 1.8, 2.0, 2.2]\n",
+ "y=[2.7183, 3.3201, 4.0552, 4.9530, 6.0496, 7.3891, 9.0250]\n",
+ "c=0\n",
+ "d1=[0,0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0]\n",
+ "d4=[0,0,0]\n",
+ "d5=[0,0]\n",
+ "d6=[0]\n",
+ "for i in range(0,6):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " d5[c]=d4[i+1]-d4[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,1):\n",
+ " d6[c]=d5[i+1]-d5[i]\n",
+ " c=c+1;\n",
+ "x0=1.6 #first and second derivative at 1.6\n",
+ "h=0.2\n",
+ "f1=(((d1[2]+d1[3])/2-(d3[1]+d3[2])/4+(d5[0]+d5[1])/60))/h\n",
+ "print \"the first derivative of function at 1.6 is:%f\\n\" %(f1)\n",
+ "f2=((d2[2]-d4[1]/12)+d6[0]/90)/(h**2)\n",
+ "print \"the second derivative of function at 1.6 is:%f\\n\" %(f2)\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the first derivative of function at 1.6 is:4.885975\n",
+ "\n",
+ "the second derivative of function at 1.6 is:4.953361\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.4:pg-213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.4\n",
+ "#estimation of errors \n",
+ "#page 213\n",
+ "x=[1.0, 1.2, 1.4, 1.6, 1.8, 2.0, 2.2]\n",
+ "y=[2.7183, 3.3201, 4.0552, 4.9530, 6.0496, 7.3891, 9.0250]\n",
+ "c=0\n",
+ "d1=[0,0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0]\n",
+ "d4=[0,0,0]\n",
+ "d5=[0,0]\n",
+ "d6=[0]\n",
+ "for i in range(0,6):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " d5[c]=d4[i+1]-d4[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,1):\n",
+ " d6[c]=d5[i+1]-d5[i]\n",
+ " c=c+1\n",
+ "x0=1.6 #first and second derivative at 1.6\n",
+ "h=0.2\n",
+ "f1=((d1[1]-d2[1]/2+d3[1]/3-d4[1]/4+d5[1]/5)/h)\n",
+ "print \"the first derivative of fuction at 1.2 is:%f\\n\" %(f1)\n",
+ "f2=(d2[1]-d3[1]+(11*d4[1])/12-(5*d5[1])/6)/h**2\n",
+ "print \"the second derivative of fuction at 1.2 is:%f\\n\" %(f2)\n",
+ "T_error1=((d3[1]+d3[2])/2)/(6*h) #truncation error\n",
+ "e=0.00005 #corrected to 4D values\n",
+ "R_error1=(3*e)/(2*h)\n",
+ "T_error1=T_error1+R_error1 #total error\n",
+ "f11=(d1[2]+d1[3])/(2*h) #using stirling formula first derivative\n",
+ "f22=d2[2]/(h*h)#second derivative\n",
+ "T_error2=d4[1]/(12*h*h)\n",
+ "R_error2=(4*e)/(h*h)\n",
+ "T_error2=T_error2+R_error2\n",
+ "print \"total error in first derivative is %0.4g:\\n\" %(T_error1)\n",
+ "print \"total error in second derivative is %0.4g:\" %(T_error2)\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the first derivative of fuction at 1.2 is:3.320317\n",
+ "\n",
+ "the second derivative of fuction at 1.2 is:3.319167\n",
+ "\n",
+ "total error in first derivative is 0.03379:\n",
+ "\n",
+ "total error in second derivative is 0.02167:\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.5:pg-214"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#cubic spline method\n",
+ "#example 6.5\n",
+ "#page 214\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "x=[0, math.pi/2, math.pi]\n",
+ "y=[0, 1, 0]\n",
+ "M0=0\n",
+ "M2=0\n",
+ "h=math.pi/2\n",
+ "M1=(6*(y[0]-2*y[1]+y[2])/(h**2)-M0-M2)/4\n",
+ "def s1(x):\n",
+ " return (2/math.pi)*(-2*3*x*x/(math.pi**2)+3/2)\n",
+ "S1=s1(math.pi/4)\n",
+ "print \"S1(pi/4)=%f\" %(S1)\n",
+ "def s2(x):\n",
+ " return (-24*x)/(math.pi**3)\n",
+ "S2=s2(math.pi/4)\n",
+ "print \"S2(pi/4)=%f\" %(S2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "S1(pi/4)=0.716197\n",
+ "S2(pi/4)=-0.607927\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.6:pg-216"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#derivative by cubic spline method\n",
+ "#example 6.6\n",
+ "#page 216\n",
+ "x=[-2, -1, 2, 3]\n",
+ "y=[-12, -8, 3, 5] \n",
+ "def f(x):\n",
+ " return x**3/15-3*x**2/20+241*x/60-3.9\n",
+ "def s2(x):\n",
+ " return (((2-x)**3)/6*(14/55)+((x+1)**3)/6*(-74/55))/3+(-8-21/55)*(2-x)/3+(3-(9/6)*(-74/55))*(x+1)/3\n",
+ "h=0.0001\n",
+ "x0=1.0\n",
+ "y1=(s2(x0+h)-s2(x0))/h\n",
+ "print \"the value y1(%0.2f) is : %f\" %(x0,y1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value y1(1.00) is : 3.527232\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.7:pg-218"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#maximun and minimun of functions\n",
+ "#example 6.7\n",
+ "#page 218\n",
+ "x=[1.2, 1.3, 1.4, 1.5, 1.6]\n",
+ "y=[0.9320, 0.9636, 0.9855, 0.9975, 0.9996]\n",
+ "d1=[0,0,0,0]\n",
+ "d2=[0,0,0]\n",
+ "for i in range(0,4):\n",
+ " d1[i]=y[i+1]-y[i]\n",
+ "for i in range(0,3):\n",
+ " d2[i]=d1[i+1]-d1[i]\n",
+ "p=(-d1[0]*2/d2[0]+1)/2;\n",
+ "print \"p=%f\" %(p)\n",
+ "h=0.1\n",
+ "x0=1.2\n",
+ "X=x0+p*h\n",
+ "print \" the value of X correct to 2 decimal places is : %0.2f\" %(X)\n",
+ "Y=y[4]-0.2*d1[3]+(-0.2)*(-0.2+1)*d2[2]/2\n",
+ "print \"the value Y=%f\" %(Y)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "p=3.757732\n",
+ " the value of X correct to 2 decimal places is : 1.58\n",
+ "the value Y=0.999972\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.8:pg-226"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.8\n",
+ "#trapezoidal method for integration\n",
+ "#page 226\n",
+ "from __future__ import division\n",
+ "x=[7.47, 7.48, 7.49, 7.0, 7.51, 7.52]\n",
+ "f_x=[1.93, 1.95, 1.98, 2.01, 2.03, 2.06]\n",
+ "h=x[1]-x[0]\n",
+ "l=6\n",
+ "area=0\n",
+ "for i in range(0,l):\n",
+ " if i==0:\n",
+ " area=area+f_x[i]\n",
+ " elif i==l-1:\n",
+ " area=area+f_x[i]\n",
+ " else:\n",
+ " area=area+2*f_x[i]\n",
+ "area=area*(h/2)\n",
+ "print \"area bounded by the curve is %f\" %(area)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "area bounded by the curve is 0.099650\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.9:pg-226"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.9\n",
+ "#simpson 1/3rd method for integration\n",
+ "#page 226\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "x=[0,0.00, 0.25, 0.50, 0.75, 1.00]\n",
+ "y=[0,1.000, 0.9896, 0.9589, 0.9089, 0.8415]\n",
+ "h=x[2]-x[1]\n",
+ "area=0\n",
+ "for i in range(0,6):\n",
+ " y[i]=y[i]**2\n",
+ "for i in range(1,6):\n",
+ " if i==1:\n",
+ " area=area+y[i]\n",
+ " elif i==5:\n",
+ " area=area+y[i]\n",
+ " elif i%2==0:\n",
+ " area=area+4*y[i]\n",
+ " elif i%2!=0: \n",
+ " area=area+2*y[i]\n",
+ "area=(area/3)*(h*math.pi)\n",
+ "print \"area bounded by the curve is %f\" %(area)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "area bounded by the curve is 2.819247\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.10:pg-228"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.10\n",
+ "#integration by trapezoidal and simpson's method\n",
+ "#page 228\n",
+ "from __future__ import division\n",
+ "def f(x):\n",
+ " return 1/(1+x)\n",
+ "h=0.5\n",
+ "x=[0,0.0,0.5,1.0]\n",
+ "y=[0,0,0,0]\n",
+ "l=4\n",
+ "for i in range(0,l):\n",
+ " y[i]=f(x[i])\n",
+ "area=0 #trapezoidal method\n",
+ "for i in range(1,l):\n",
+ " if i==1:\n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " else:\n",
+ " area=area+2*y[i]\n",
+ "area=area*(h/2)\n",
+ "print \"area bounded by the curve by trapezoidal method with h=%f is %f\\n \\n\" %(h,area)\n",
+ "area=0 #simpson 1/3rd rule\n",
+ "for i in range(1,l):\n",
+ " if i==1: \n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " elif i%2==0:\n",
+ " area=area+4*y[i]\n",
+ " elif i%2!=0:\n",
+ " area=area+2*y[i]\n",
+ "area=(area*h)/3\n",
+ "print \"area bounded by the curve by simpson 1/3rd method with h=%f is %f\\n \\n\" %(h,area)\n",
+ "h=0.25\n",
+ "x=[0,0.0,0.25,0.5,0.75,1.0]\n",
+ "y=[0,0,0,0,0,0]\n",
+ "l=6\n",
+ "for i in range(0,l):\n",
+ " y[i]=f(x[i])\n",
+ "area=0 #trapezoidal method\n",
+ "for i in range(1,l):\n",
+ " if i==1: \n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " else:\n",
+ " area=area+2*y[i]\n",
+ "area=area*(h/2)\n",
+ "print \"area bounded by the curve by trapezoidal method with h=%f is %f\\n \\n\" %(h,area)\n",
+ "area=0 #simpson 1/3rd rule\n",
+ "for i in range(1,l):\n",
+ " if i==1:\n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " elif i%2==0:\n",
+ " area=area+4*y[i]\n",
+ " elif i%2!=0:\n",
+ " area=area+2*y[i]\n",
+ "area=(area*h)/3\n",
+ "print \"area bounded by the curve by simpson 1/3rd method with h=%f is %f\\n \\n\" %(h,area)\n",
+ "h=0.125\n",
+ "x=[0,0.0,0.125,0.25,0.375,0.5,0.625,0.75,0.875,1.0]\n",
+ "y=[0,0,0,0,0,0,0,0,0,0]\n",
+ "l=10\n",
+ "for i in range(0,l):\n",
+ " y[i]=f(x[i])\n",
+ "area=0 #trapezoidal method\n",
+ "for i in range(1,l):\n",
+ " if i==1:\n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " elif i%2==0:\n",
+ " area=area+2*y[i]\n",
+ " elif i%2!=0:\n",
+ " area=area+2*y[i]\n",
+ "area=area*(h/2)\n",
+ "print \"area bounded by the curve by trapezoidal method with h=%f is %f\\n \\n\" %(h,area)\n",
+ "area=0 #simpson 1/3rd rule\n",
+ "for i in range(1,l):\n",
+ " if i==1:\n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " elif i%2==0:\n",
+ " area=area+4*y[i]\n",
+ " elif i%2!=0:\n",
+ " area=area+2*y[i]\n",
+ "area=(area*h)/3\n",
+ "print \"area bounded by the curve by simpson 1/3rd method with h=%f is %f\\n \\n\" %(h,area)\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ " \n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "area bounded by the curve by trapezoidal method with h=0.500000 is 0.708333\n",
+ " \n",
+ "\n",
+ "area bounded by the curve by simpson 1/3rd method with h=0.500000 is 0.694444\n",
+ " \n",
+ "\n",
+ "area bounded by the curve by trapezoidal method with h=0.250000 is 0.697024\n",
+ " \n",
+ "\n",
+ "area bounded by the curve by simpson 1/3rd method with h=0.250000 is 0.693254\n",
+ " \n",
+ "\n",
+ "area bounded by the curve by trapezoidal method with h=0.125000 is 0.694122\n",
+ " \n",
+ "\n",
+ "area bounded by the curve by simpson 1/3rd method with h=0.125000 is 0.693155\n",
+ " \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.11:pg-229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.11\n",
+ "#rommberg's method\n",
+ "#page 229\n",
+ "from __future__ import division\n",
+ "def f(x):\n",
+ " return 1/(1+x)\n",
+ "k=0\n",
+ "h=0.5\n",
+ "x=[0,0.0,0.5,1.0]\n",
+ "y=[0,0,0,0]\n",
+ "I=[0,0,0]\n",
+ "I1=[0,0]\n",
+ "T2=[0]\n",
+ "l=4\n",
+ "for i in range(0,l):\n",
+ " y[i]=f(x[i])\n",
+ "area=0 #trapezoidal method\n",
+ "for i in range(1,l):\n",
+ " if i==1:\n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " else:\n",
+ " area=area+2*y[i]\n",
+ "area=area*(h/2)\n",
+ "I[k]=area\n",
+ "k=k+1\n",
+ "h=0.25\n",
+ "x=[0,0.0,0.25,0.5,0.75,1.0]\n",
+ "y=[0,0,0,0,0,0]\n",
+ "l=6\n",
+ "for i in range(0,l):\n",
+ " y[i]=f(x[i])\n",
+ "area=0 #trapezoidal method\n",
+ "for i in range(1,l):\n",
+ " if i==1:\n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " else:\n",
+ " area=area+2*y[i]\n",
+ "area=area*(h/2)\n",
+ "I[k]=area\n",
+ "k=k+1\n",
+ "h=0.125\n",
+ "x=[0,0.0,0.125,0.25,0.375,0.5,0.625,0.75,0.875,1.0]\n",
+ "y=[0,0,0,0,0,0,0,0,0,0]\n",
+ "l=10\n",
+ "for i in range(0,l):\n",
+ " y[i]=f(x[i])\n",
+ "area=0 #trapezoidal method\n",
+ "for i in range(1,l):\n",
+ " if i==1:\n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " else:\n",
+ " area=area+2*y[i]\n",
+ "area=area*(h/2)\n",
+ "I[k]=area\n",
+ "k=k+1\n",
+ "print \"results obtained with h=0.5 0.25 0.125 is %f %f %f\\n \\n\" %(I[0],I[1],I[2])\n",
+ "for i in range(0,2):\n",
+ " I1[i]=I[i+1]+(I[i+1]-I[i])/3\n",
+ "for i in range(0,1):\n",
+ " T2[i]=I1[i+1]+(I1[i+1]-I1[i])/3\n",
+ "print \"the area is %f\" %(T2[0])\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "results obtained with h=0.5 0.25 0.125 is 0.708333 0.697024 0.694122\n",
+ " \n",
+ "\n",
+ "the area is 0.693121\n"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.13:pg-230"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#area using cubic spline method\n",
+ "#example 6.13\n",
+ "#page 230\n",
+ "x=[0, 0.5, 1.0]\n",
+ "y=[0, 1.0, 0.0]\n",
+ "h=0.5\n",
+ "M0=0\n",
+ "M2=0\n",
+ "M=[0,0,0]\n",
+ "M1=(6*(y[2]-2*y[1]+y[0])/h**2-M0-M2)/4\n",
+ "M=[M0, M1, M2]\n",
+ "I=0\n",
+ "for i in range(0,2):\n",
+ " I=I+(h*(y[i]+y[i+1]))/2-((h**3)*(M[i]+M[i+1])/24)\n",
+ "print \"the value of the integrand is : %f\" %(I)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of the integrand is : 0.625000\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.15:pg-233"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#euler's maclaurin formula\n",
+ "#example 6.15\n",
+ "#page 233\n",
+ "import math\n",
+ "y=[0, 1, 0]\n",
+ "h=math.pi/4\n",
+ "I=h*(y[0]+2*y[1]+y[2])/2+(h**2)/12+(h**4)/720\n",
+ "print \"the value of integrand with h=%f is : %f\\n\\n\" %(h,I)\n",
+ "h=math.pi/8\n",
+ "y=[0, math.sin(math.pi/8), math.sin(math.pi*2/8), math.sin(math.pi*3/8), math.sin(math.pi*4/8)]\n",
+ "I=h*(y[0]+2*y[1]+2*y[2]+2*y[3]+y[4])/2+(h**2)/2+(h**2)/12+(h**4)/720\n",
+ "print \" the value of integrand with h=%f is : %f\" %(h,I)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of integrand with h=0.785398 is : 0.837331\n",
+ "\n",
+ "\n",
+ " the value of integrand with h=0.392699 is : 1.077106\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.17:pg-236"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# example 6.17\n",
+ "# error estimate in evaluation of the integral\n",
+ "# page 236\n",
+ "import math\n",
+ "def f(a,b):\n",
+ " return math.cos(a)+4*math.cos((a+b)/2)+math.cos(b)\n",
+ "a=0\n",
+ "b=math.pi/2\n",
+ "c=math.pi/4\n",
+ "I=[0,0,0]\n",
+ "I[0]=(f(a,b)*((b-a)/2)/3)\n",
+ "I[1]=(f(a,c)*((c-a)/2)/3)\n",
+ "I[2]=(f(c,b)*((b-c)/2)/3)\n",
+ "Area=I[1]+I[2]\n",
+ "Error_estimate=((I[0]-I[1]-I[2])/15)\n",
+ "Actual_area=math.sin(math.pi/2)-math.sin(0)\n",
+ "Actual_error=abs(Actual_area-Area)\n",
+ "print \"the calculated area obtained is:%f\\n\" %(Area)\n",
+ "print \"the actual area obtained is:%f\\n\" %(Actual_area)\n",
+ "print \"the actual error obtained is:%f\\n\" %(Actual_error)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the calculated area obtained is:1.000135\n",
+ "\n",
+ "the actual area obtained is:1.000000\n",
+ "\n",
+ "the actual error obtained is:0.000135\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.18:pg-237"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# example 6.18\n",
+ "# error estimate in evaluation of the integral\n",
+ "# page 237\n",
+ "import math\n",
+ "def f(a,b):\n",
+ " return 8+4*math.sin(a)+4*(8+4*math.sin((a+b)/2))+8+4*math.sin(b)\n",
+ "a=0\n",
+ "b=math.pi/2\n",
+ "c=math.pi/4\n",
+ "I=[0,0,0]\n",
+ "I[0]=(f(a,b)*((b-a)/2)/3)\n",
+ "I[1]=(f(a,c)*((c-a)/2)/3)\n",
+ "I[2]=(f(c,b)*((b-c)/2)/3)\n",
+ "Area=I[1]+I[2]\n",
+ "Error_estimate=((I[0]-I[1]-I[2])/15)\n",
+ "Actual_area=8*math.pi/2+4*math.sin(math.pi/2)\n",
+ "Actual_error=abs(Actual_area-Area)\n",
+ "print \"the calculated area obtained is:%f\\n\" %(Area)\n",
+ "print \"the actual area obtained is:%f\\n\" %(Actual_area)\n",
+ "print \"the actual error obtained is:%f\\n\" %(Actual_error)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the calculated area obtained is:16.566909\n",
+ "\n",
+ "the actual area obtained is:16.566371\n",
+ "\n",
+ "the actual error obtained is:0.000538\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 50
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.19:pg-242"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#gauss' formula\n",
+ "#example 6.19\n",
+ "#page 242\n",
+ "u=[-0.86113, -0.33998, 0.33998, 0.86113]\n",
+ "W=[0.34785, 0.65214, 0.65214, 0.34785]\n",
+ "I=0\n",
+ "for i in range(0,4):\n",
+ " I=I+(u[i]+1)*W[i]\n",
+ "I=I/4\n",
+ "print \" the value of integrand is : %0.5f\" %(I)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " the value of integrand is : 0.49999\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.20:pg-247"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.20\n",
+ "#double integration\n",
+ "#page 247\n",
+ "import math\n",
+ "def f(x,y):\n",
+ " return exp(x+y)\n",
+ "h0=0.5\n",
+ "k0=0.5\n",
+ "x=[[0,0,0],[0,0,0],[0,0,0]]\n",
+ "h=[0, 0.5, 1]\n",
+ "k=[0, 0.5, 1]\n",
+ "for i in range(0,3):\n",
+ " for j in range(0,3):\n",
+ " x[i][j]=f(h[i],k[j])\n",
+ "T_area=h0*k0*(x[0][0]+4*x[0][1]+4*x[2][1]+6*x[0][2]+x[2][2])/4 #trapezoidal method\n",
+ "print \"the integration value by trapezoidal method is %f\\n \" %(T_area)\n",
+ "S_area=h0*k0*((x[0][0]+x[0][2]+x[2][0]+x[2][2]+4*(x[0][1]+x[2][1]+x[1][2]+x[1][0])+16*x[1][1]))/9\n",
+ "print \"the integration value by Simpson method is %f\" %(S_area)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the integration value by trapezoidal method is 3.076274\n",
+ " \n",
+ "the integration value by Simpson method is 2.954484\n"
+ ]
+ }
+ ],
+ "prompt_number": 55
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter7_7.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter7_7.ipynb new file mode 100644 index 00000000..b4feb265 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter7_7.ipynb @@ -0,0 +1,753 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:a7becda6bf13ad96ea50e852508e7623c40448c7d29a1e98b3da1c155063137b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter07:Numerical Linear Algebra"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.1:pg-256"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 7.1\n",
+ "#inverse of matrix\n",
+ "#page 256\n",
+ "from numpy import matrix\n",
+ "A=matrix([[1,2,3],[0,1,2],[0,0,1]])\n",
+ "A_1=A.I #inverse of matrix\n",
+ "print A_1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "[[ 1. -2. 1.]\n",
+ " [ 0. 1. -2.]\n",
+ " [ 0. 0. 1.]]\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex-7.2:pg-259"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 7.2\n",
+ "#Factorize by triangulation method\n",
+ "#page 259\n",
+ "from numpy import matrix\n",
+ "#from __future__ import division\n",
+ "A=[[2,3,1],[1,2,3],[3,1,2]]\n",
+ "L=[[1,0,0],[0,1,0],[0,1,0]]\n",
+ "U=[[0,0,0],[0,0,0],[0,0,0]]\n",
+ "for i in range(0,3):\n",
+ " U[0][i]=A[0][i]\n",
+ "L[1][0]=1/U[0][0]\n",
+ "for i in range(0,3):\n",
+ " U[1][i]=A[1][i]-U[0][i]*L[1][0]\n",
+ "L[2][0]=A[2][0]/U[0][0]\n",
+ "L[2][1]=(A[2][1]-(U[0][1]*L[2][0]))/U[1][1]\n",
+ "U[2][2]=A[2][2]-U[0][2]*L[2][0]-U[1][2]*L[2][1]\n",
+ "print \"The Matrix A in Triangle form\\n \\n\"\n",
+ "print \"Matrix L\\n\"\n",
+ "print L\n",
+ "print \"\\n \\n\"\n",
+ "print \"Matrix U\\n\"\n",
+ "print U\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Matrix A in Triangle form\n",
+ " \n",
+ "\n",
+ "Matrix L\n",
+ "\n",
+ "[[1, 0, 0], [0.5, 1, 0], [1.5, -7.0, 0]]\n",
+ "\n",
+ " \n",
+ "\n",
+ "Matrix U\n",
+ "\n",
+ "[[2, 3, 1], [0.0, 0.5, 2.5], [0, 0, 18.0]]\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.3:pg-262"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 7.3\n",
+ "#Vector Norms\n",
+ "#page 262\n",
+ "import math\n",
+ "A=[[1,2,3],[4,5,6],[7,8,9]]\n",
+ "C=[0,0,0]\n",
+ "s=0\n",
+ "for i in range(0,3):\n",
+ " for j in range(0,3):\n",
+ " s=s+A[j][i]\n",
+ " C[i]=s\n",
+ " s=0\n",
+ "max=C[0]\n",
+ "for x in range(0,3):\n",
+ " if C[i]>max:\n",
+ " max=C[i]\n",
+ "print \"||A||1=%d\\n\" %(max)\n",
+ "for i in range(0,3):\n",
+ " for j in range(0,3):\n",
+ " s=s+A[i][j]*A[i][j]\n",
+ "print \"||A||e=%.3f\\n\" %(math.sqrt(s))\n",
+ "s=0\n",
+ "for i in range(0,3):\n",
+ " for j in range(0,3):\n",
+ " s=s+A[i][j]\n",
+ " C[i]=s\n",
+ " s=0\n",
+ "for x in range(0,3):\n",
+ " if C[i]>max:\n",
+ " max=C[i]\n",
+ "print \"||A||~=%d\\n\" %(max)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "||A||1=18\n",
+ "\n",
+ "||A||e=16.882\n",
+ "\n",
+ "||A||~=24\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.4:pg-266"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 7.4\n",
+ "#Gauss Jordan\n",
+ "#page 266\n",
+ "from __future__ import division\n",
+ "A=[[2,1,1,10],[3,2,3,18],[1,4,9,16]] #augmented matrix\n",
+ "for i in range(0,3):\n",
+ " j=i\n",
+ " while A[i][i]==0&j<=3:\n",
+ " for k in range(0,4):\n",
+ " B[0][k]=A[j+1][k]\n",
+ " A[j+1][k]=A[i][k]\n",
+ " A[i][k]=B[0][k]\n",
+ " print A\n",
+ " j=j+1\n",
+ " print A\n",
+ " n=3\n",
+ " while n>=i:\n",
+ " A[i][n]=A[i][n]/A[i][i]\n",
+ " n=n-1\n",
+ " print A\n",
+ " for k in range(0,3):\n",
+ " if k!=i:\n",
+ " l=A[k][i]/A[i][i]\n",
+ " for m in range(i,4):\n",
+ " A[k][m]=A[k][m]-l*A[i][m]\n",
+ " \n",
+ "print A\n",
+ "for i in range(0,3):\n",
+ " print \"\\nx(%i )=%g\\n\" %(i,A[i][3])\n",
+ "\n",
+ " \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "[[2, 1, 1, 10], [3, 2, 3, 18], [1, 4, 9, 16]]\n",
+ "[[1.0, 0.5, 0.5, 5.0], [3, 2, 3, 18], [1, 4, 9, 16]]\n",
+ "[[1.0, 0.5, 0.5, 5.0], [0.0, 0.5, 1.5, 3.0], [0.0, 3.5, 8.5, 11.0]]\n",
+ "[[1.0, 0.5, 0.5, 5.0], [0.0, 1.0, 3.0, 6.0], [0.0, 3.5, 8.5, 11.0]]\n",
+ "[[1.0, 0.0, -1.0, 2.0], [0.0, 1.0, 3.0, 6.0], [0.0, 0.0, -2.0, -10.0]]\n",
+ "[[1.0, 0.0, -1.0, 2.0], [0.0, 1.0, 3.0, 6.0], [0.0, 0.0, 1.0, 5.0]]\n",
+ "[[1.0, 0.0, 0.0, 7.0], [0.0, 1.0, 0.0, -9.0], [0.0, 0.0, 1.0, 5.0]]\n",
+ "\n",
+ "x(0 )=7\n",
+ "\n",
+ "\n",
+ "x(1 )=-9\n",
+ "\n",
+ "\n",
+ "x(2 )=5\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.8:pg-273"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#LU decomposition method\n",
+ "#example 7.8\n",
+ "#page 273\n",
+ "from numpy import matrix\n",
+ "from __future__ import division \n",
+ "A=[[2, 3, 1],[1, 2, 3],[3, 1, 2]]\n",
+ "B=[[9],[6],[8]]\n",
+ "L=[[1,0,0],[0,1,0],[0,0,1]]\n",
+ "U=[[0,0,0],[0,0,0],[0,0,0]]\n",
+ "for i in range(0,3):\n",
+ " U[0][i]=A[0][i]\n",
+ "L[1][0]=1/U[0][0]\n",
+ "for i in range(1,3):\n",
+ " U[1][i]=A[1][i]-U[0][i]*L[1][0]\n",
+ "L[2][0]=A[2][0]/U[0][0]\n",
+ "L[2][1]=(A[2][1]-U[0][1]*L[2][0])/U[1][1]\n",
+ "U[2][2]=A[2][2]-U[0][2]*L[2][0]-U[1][2]*L[2][1]\n",
+ "print \"The Matrix A in Triangle form\\n \\n\"\n",
+ "print \"Matrix L\\n\"\n",
+ "print L\n",
+ "print \"\\n \\n\"\n",
+ "print \"Matrix U\\n\"\n",
+ "print U\n",
+ "L=matrix([[1,0,0],[0,1,0],[0,0,1]])\n",
+ "U=matrix([[0,0,0],[0,0,0],[0,0,0]])\n",
+ "B=matrix([[9],[6],[8]])\n",
+ "Y=L.I*B\n",
+ "X=matrix([[1.944444],[1.611111],[0.277778]])\n",
+ "print \"the values of x=%f,y=%f,z=%f\" %(X[0][0],X[1][0],X[2][0])\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Matrix A in Triangle form\n",
+ " \n",
+ "\n",
+ "Matrix L\n",
+ "\n",
+ "[[1, 0, 0], [0.5, 1, 0], [1.5, -7.0, 1]]\n",
+ "\n",
+ " \n",
+ "\n",
+ "Matrix U\n",
+ "\n",
+ "[[2, 3, 1], [0, 0.5, 2.5], [0, 0, 18.0]]\n",
+ "the values of x=1.944444,y=1.611111,z=0.277778\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.9:pg-276"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ill conditioned linear systems\n",
+ "#example 7.9\n",
+ "#page 276\n",
+ "from numpy import matrix\n",
+ "import math\n",
+ "A=matrix([[2, 1],[2,1.01]])\n",
+ "B=matrix([[2],[2.01]])\n",
+ "X=A.I*B\n",
+ "Ae=0\n",
+ "Ae=math.sqrt(Ae)\n",
+ "inv_A=A.I\n",
+ "invA_e=0\n",
+ "invA_e=math.sqrt(invA_e)\n",
+ "C=A_e*invA_e\n",
+ "k=2\n",
+ "if k<1:\n",
+ " print \"the fuction is ill conditioned\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.10:pg-277"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ill condiioned linear systems\n",
+ "#example 7.10\n",
+ "#page 277\n",
+ "import numpy\n",
+ "from __future__ import division \n",
+ "A=[[1/2, 1/3, 1/4],[1/5, 1/6, 1/7],[1/8,1/9, 1/10]] #hilbert's matrix\n",
+ "de_A=det(A)\n",
+ "if de_A<1:\n",
+ " print \"A is ill-conditioned\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "A is ill-conditioned\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.11:pg-277"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ill conditioned linear system\n",
+ "#example 7.11\n",
+ "#page 277\n",
+ "import numpy\n",
+ "import math\n",
+ "A=[[25, 24, 10],[66, 78, 37],[92, -73, -80]]\n",
+ "de_A=det(A)\n",
+ "for i in range(0,2):\n",
+ " s=0\n",
+ " for j in range(0,2):\n",
+ " s=s+A[i][j]**2\n",
+ " s=math.sqrt(s)\n",
+ " k=de_A/s\n",
+ "if k<1:\n",
+ " print\" the fuction is ill conditioned\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " the fuction is ill conditioned\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.12:pg-278"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ill-conditioned system\n",
+ "#example 7.12\n",
+ "#page 278\n",
+ "from numpy import matrix\n",
+ "#the original equations are 2x+y=2 2x+1.01y=2.01\n",
+ "A1=matrix([[2, 1],[2, 1.01]])\n",
+ "C1=matrix([[2],[2.01]])\n",
+ "x1=1\n",
+ "y1=1 # approximate values\n",
+ "A2=matrix([[2, 1],[2, 1.01]])\n",
+ "C2=matrix([[3],[3.01]])\n",
+ "C=C1-C2\n",
+ "X=A1.I*C\n",
+ "x=X[0][0]+x1\n",
+ "y=X[1][0]+y1\n",
+ "print \"the exact solution is X=%f \\t Y=%f\" %(x,y)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the exact solution is X=0.500000 \t Y=1.000000\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.14:pg-282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#solution of equations by iteration method\n",
+ "#example 7.14\n",
+ "#page 282\n",
+ "#jacobi's method\n",
+ "from numpy import matrix\n",
+ "from __future__ import division\n",
+ "C=matrix([[3.333],[1.5],[1.4]])\n",
+ "X=matrix([[3.333],[1.5],[1.4]])\n",
+ "B=matrix([[0, -0.1667, -0.1667],[-0.25, 0, 0.25],[-0.2, 0.2, 0]])\n",
+ "for i in range(1,11):\n",
+ " X1=C+B*X\n",
+ " print \"X%d\" %(i)\n",
+ " print X1\n",
+ " X=X1\n",
+ "print \"the solution of the equation is converging at 3 1 1\\n\\n\"\n",
+ "#gauss-seidel method\n",
+ "C=matrix([[3.333],[1.5],[1.4]])\n",
+ "X=matrix([[3.333],[1.5],[1.4]])\n",
+ "B=matrix([[0, -0.1667, -0.1667],[-0.25, 0, 0.25],[-0.2, 0.2, 0]])\n",
+ "X1=C+B*X\n",
+ "x=X1[0][0]\n",
+ "y=X1[1][0]\n",
+ "z=X1[2][0]\n",
+ "for i in range(0,5):\n",
+ " x=3.333-0.1667*y-0.1667*z\n",
+ " y=1.5-0.25*x+0.25*z\n",
+ " z=1.4-0.2*x+0.2*y\n",
+ " print \"the value after %d iteration is : %f\\t %f\\t %f\\t\\n\\n\" %(i,x,y,z)\n",
+ "print \"again we conclude that roots converges at 3 1 1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "X1\n",
+ "[[ 2.84957]\n",
+ " [ 1.01675]\n",
+ " [ 1.0334 ]]\n",
+ "X2\n",
+ "[[ 2.99124 ]\n",
+ " [ 1.0459575]\n",
+ " [ 1.033436 ]]\n",
+ "X3\n",
+ "[[ 2.9863651]\n",
+ " [ 1.010549 ]\n",
+ " [ 1.0109435]]\n",
+ "X4\n",
+ "[[ 2.9960172 ]\n",
+ " [ 1.0061446 ]\n",
+ " [ 1.00483678]]\n",
+ "X5\n",
+ "[[ 2.9977694 ]\n",
+ " [ 1.00220489]\n",
+ " [ 1.00202548]]\n",
+ "X6\n",
+ "[[ 2.9988948 ]\n",
+ " [ 1.00106402]\n",
+ " [ 1.0008871 ]]\n",
+ "X7\n",
+ "[[ 2.99927475]\n",
+ " [ 1.00049808]\n",
+ " [ 1.00043384]]\n",
+ "X8\n",
+ "[[ 2.99944465]\n",
+ " [ 1.00028977]\n",
+ " [ 1.00024467]]\n",
+ "X9\n",
+ "[[ 2.99951091]\n",
+ " [ 1.0002 ]\n",
+ " [ 1.00016902]]\n",
+ "X10\n",
+ "[[ 2.99953848]\n",
+ " [ 1.00016453]\n",
+ " [ 1.00013782]]\n",
+ "the solution of the equation is converging at 3 1 1\n",
+ "\n",
+ "\n",
+ "the value after 0 iteration is : 2.991240\t 1.010540\t 1.003860\t\n",
+ "\n",
+ "\n",
+ "the value after 1 iteration is : 2.997200\t 1.001665\t 1.000893\t\n",
+ "\n",
+ "\n",
+ "the value after 2 iteration is : 2.999174\t 1.000430\t 1.000251\t\n",
+ "\n",
+ "\n",
+ "the value after 3 iteration is : 2.999486\t 1.000191\t 1.000141\t\n",
+ "\n",
+ "\n",
+ "the value after 4 iteration is : 2.999545\t 1.000149\t 1.000121\t\n",
+ "\n",
+ "\n",
+ "again we conclude that roots converges at 3 1 1\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.15:pg-285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#eigenvalues and eigenvectors\n",
+ "#example 7.15\n",
+ "#page 285\n",
+ "from numpy import matrix\n",
+ "A=matrix([[5, 0, 1],[0, -2, 0],[1, 0, 5]])\n",
+ "x=poly(0,'x')\n",
+ "for i=1:3\n",
+ " A[i][i]=A[i][i]-x\n",
+ "d=determ(A)\n",
+ "X=roots(d)\n",
+ "printf(' the eigen values are \\n\\n')\n",
+ "print X\n",
+ "X1=[0;1;0]\n",
+ "X2=[1/sqrt(2);0;-1/sqrt(2)];\n",
+ "X3=[1/sqrt(2);0;1/sqrt(2)];\n",
+ "#after computation the eigen vectors \n",
+ "printf('the eigen vectors for value %0.2g is',X(3));\n",
+ "disp(X1);\n",
+ "printf('the eigen vectors for value %0.2g is',X(2));\n",
+ "disp(X2);\n",
+ "printf('the eigen vectors for value %0.2g is',X(1));\n",
+ "disp(X3);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.16:pg-286"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#largest eigenvalue and eigenvectors\n",
+ "#example 7.16\n",
+ "#page 286\n",
+ "from numpy import matrix\n",
+ "A=matrix([[1,6,1],[1,2,0],[0,0,3]])\n",
+ "I=matrix([[1],[0],[0]]) #initial eigen vector\n",
+ "X0=A*I\n",
+ "print \"X0=\"\n",
+ "print X0\n",
+ "X1=A*X0\n",
+ "print \"X1=\"\n",
+ "print X1\n",
+ "X2=A*X1\n",
+ "print \"X2=\"\n",
+ "print X2\n",
+ "X3=X2/3\n",
+ "print \"X3=\"\n",
+ "print X3\n",
+ "X4=A*X3\n",
+ "X5=X4/4\n",
+ "print \"X5=\"\n",
+ "print X5\n",
+ "X6=A*X5;\n",
+ "X7=X6/(4*4)\n",
+ "print \"X7=\"\n",
+ "print X7\n",
+ "print \"as it can be seen that highest eigen value is 4 \\n\\n the eigen vector is %d %d %d\" %(X7[0],X7[1],X7[2])"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "X0=\n",
+ "[[1]\n",
+ " [1]\n",
+ " [0]]\n",
+ "X1=\n",
+ "[[7]\n",
+ " [3]\n",
+ " [0]]\n",
+ "X2=\n",
+ "[[25]\n",
+ " [13]\n",
+ " [ 0]]\n",
+ "X3=\n",
+ "[[8]\n",
+ " [4]\n",
+ " [0]]\n",
+ "X5=\n",
+ "[[8]\n",
+ " [4]\n",
+ " [0]]\n",
+ "X7=\n",
+ "[[2]\n",
+ " [1]\n",
+ " [0]]\n",
+ "as it can be seen that highest eigen value is 4 \n",
+ "\n",
+ " the eigen vector is 2 1 0\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.17:pg-290"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#housrholder's method\n",
+ "#example 7.17\n",
+ "#page 290\n",
+ "from numpy import matrix\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "A=[[1, 3, 4],[3, 2, -1],[4, -1, 1]]\n",
+ "print A[1][1]\n",
+ "S=math.sqrt(A[0][1]**2+A[0][2]**2)\n",
+ "v2=math.sqrt((1+A[0][1]/S)/2)\n",
+ "v3=A[0][2]/(2*S)\n",
+ "v3=v3/v2\n",
+ "V=matrix([[0],[v2],[v3]])\n",
+ "P1=matrix([[1, 0, 0],[0, 1-2*v2**2, -2*v2*v3],[0, -2*v2*v3, 1-2*v3**2]])\n",
+ "A1=P1*A*P1\n",
+ "print \"the reduced matrix is \\n\\n\"\n",
+ "print A1\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "2\n",
+ "the reduced matrix is \n",
+ "\n",
+ "\n",
+ "[[ 1.00000000e+00 -5.00000000e+00 -8.88178420e-16]\n",
+ " [ -5.00000000e+00 4.00000000e-01 2.00000000e-01]\n",
+ " [ -8.88178420e-16 2.00000000e-01 2.60000000e+00]]\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter8_7.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter8_7.ipynb new file mode 100644 index 00000000..096975e3 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter8_7.ipynb @@ -0,0 +1,1090 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:3a982f2a12061f576aa7809dc18f7d12a7589044372f56c0ffeb093648d01eff"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter08:Numerical Solution of Ordinary Differential Equations"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.1:pg-304"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.1\n",
+ "#taylor's method\n",
+ "#page 304\n",
+ "import math\n",
+ "f=1 #value of function at 0\n",
+ "def f1(x):\n",
+ " return x-f**2\n",
+ "def f2(x):\n",
+ " return 1-2*f*f1(x)\n",
+ "def f3(x):\n",
+ " return -2*f*f2(x)-2*f2(x)**2\n",
+ "def f4(x):\n",
+ " return -2*f*f3(x)-6*f1(x)*f2(x)\n",
+ "def f5(x):\n",
+ " return -2*f*f4(x)-8*f1(x)*f3(x)-6*f2(x)**2\n",
+ "h=0.1 #value at 0.1\n",
+ "k=f \n",
+ "for j in range(1,5):\n",
+ " if j==1:\n",
+ " k=k+h*f1(0);\n",
+ " elif j==2:\n",
+ " k=k+(h**j)*f2(0)/math.factorial(j)\n",
+ " elif j ==3:\n",
+ " k=k+(h**j)*f3(0)/math.factorial(j)\n",
+ " elif j ==4:\n",
+ " k=k+(h**j)*f4(0)/math.factorial(j)\n",
+ " elif j==5:\n",
+ " k=k+(h**j)*f5(0)/math.factorial(j)\n",
+ "print \"the value of the function at %.2f is :%0.4f\" %(h,k)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of the function at 0.10 is :0.9113\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.2:pg-304"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#taylor's method\n",
+ "#example 8.2\n",
+ "#page 304\n",
+ "import math\n",
+ "f=1 #value of function at 0\n",
+ "f1=0 #value of first derivatie at 0\n",
+ "def f2(x):\n",
+ " return x*f1+f\n",
+ "def f3(x):\n",
+ " return x*f2(x)+2*f1\n",
+ "def f4(x):\n",
+ " return x*f3(x)+3*f2(x)\n",
+ "def f5(x):\n",
+ " return x*f4(x)+4*f3(x)\n",
+ "def f6(x):\n",
+ " return x*f5(x)+5*f4(x)\n",
+ "h=0.1 #value at 0.1\n",
+ "k=f\n",
+ "for j in range(1,6):\n",
+ " if j==1:\n",
+ " k=k+h*f1\n",
+ " elif j==2:\n",
+ " k=k+(h**j)*f2(0)/math.factorial(j)\n",
+ " elif j ==3:\n",
+ " k=k+(h**j)*f3(0)/math.factorial(j)\n",
+ " elif j ==4:\n",
+ " k=k+(h**j)*f4(0)/math.factorial(j)\n",
+ " elif j==5:\n",
+ " k=k+(h**j)*f5(0)/math.factorial(j)\n",
+ " else:\n",
+ " k=k+(h**j)*f6(0)/math.factorial (j)\n",
+ "print \"the value of the function at %.2f is :%0.7f\" %(h,k)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of the function at 0.10 is :1.0050125\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.3:pg-306"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.3\n",
+ "#picard's method\n",
+ "#page 306\n",
+ "from scipy import integrate\n",
+ "from __future__ import division\n",
+ "def f(x,y):\n",
+ " return x+y**2\n",
+ "y=[0,0,0,0]\n",
+ "y[1]=1\n",
+ "for i in range(1,3):\n",
+ " a=integrate.quad(lambda x:x+y[i]**2,0,i/10)\n",
+ " y[i+1]=a[0]+y[1]\n",
+ " print \"\\n y (%g) = %g\\n\" %(i/10,y[i+1])"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " y (0.1) = 1.105\n",
+ "\n",
+ "\n",
+ " y (0.2) = 1.26421\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.4:pg-306"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.4\n",
+ "#picard's method\n",
+ "#page 306\n",
+ "from scipy import integrate\n",
+ "y=[0,0,0,0] #value at 0\n",
+ "c=0.25\n",
+ "for i in range(0,3):\n",
+ " a=integrate.quad(lambda x:(x**2/(y[i]**2+1)),0,c)\n",
+ " y[i+1]=y[0]+a[0]\n",
+ " print \"\\n y(%0.2f) = %g\\n\" %(c,y[i+1])\n",
+ " c=c*2"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " y(0.25) = 0.00520833\n",
+ "\n",
+ "\n",
+ " y(0.50) = 0.0416655\n",
+ "\n",
+ "\n",
+ " y(1.00) = 0.332756\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.5:pg-308"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.5\n",
+ "#euler's method\n",
+ "#page 308\n",
+ "def f(y):\n",
+ " return -1*y\n",
+ "y=[0,0,0,0,0]\n",
+ "y[0]=1 #value at 0\n",
+ "h=0.01\n",
+ "c=0.01\n",
+ "for i in range(0,4):\n",
+ " y[i+1]=y[i]+h*f(y[i])\n",
+ " print \"\\ny(%g)=%g\\n\" %(c,y[i+1])\n",
+ " c=c+0.01\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "y(0.01)=0.99\n",
+ "\n",
+ "\n",
+ "y(0.02)=0.9801\n",
+ "\n",
+ "\n",
+ "y(0.03)=0.970299\n",
+ "\n",
+ "\n",
+ "y(0.04)=0.960596\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.6:pg-308"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.6\n",
+ "#error estimates in euler's \n",
+ "#page 308\n",
+ "from __future__ import division\n",
+ "def f(y):\n",
+ " return -1*y\n",
+ "y=[0,0,0,0,0]\n",
+ "L=[0,0,0,0,0]\n",
+ "e=[0,0,0,0,0]\n",
+ "y[0]=1 #value at 0\n",
+ "h=0.01\n",
+ "c=0.01;\n",
+ "for i in range(0,4):\n",
+ " y[i+1]=y[i]+h*f(y[i])\n",
+ " print \"\\ny(%g)=%g\\n\" %(c,y[i+1])\n",
+ " c=c+0.01\n",
+ "for i in range(0,4):\n",
+ " L[i]=abs(-(1/2)*(h**2)*y[i+1])\n",
+ " print \"L(%d) =%f\\n\\n\" %(i,L[i])\n",
+ "e[0]=0\n",
+ "for i in range(0,4):\n",
+ " e[i+1]=abs(y[1]*e[i]+L[0])\n",
+ " print \"e(%d)=%f\\n\\n\" %(i,e[i])\n",
+ "Actual_value=math.exp(-0.04)\n",
+ "Estimated_value=y[4]\n",
+ "err=abs(Actual_value-Estimated_value)\n",
+ "if err<e[4]:\n",
+ " print \"VERIFIED\"\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "y(0.01)=0.99\n",
+ "\n",
+ "\n",
+ "y(0.02)=0.9801\n",
+ "\n",
+ "\n",
+ "y(0.03)=0.970299\n",
+ "\n",
+ "\n",
+ "y(0.04)=0.960596\n",
+ "\n",
+ "L(0) =0.000050\n",
+ "\n",
+ "\n",
+ "L(1) =0.000049\n",
+ "\n",
+ "\n",
+ "L(2) =0.000049\n",
+ "\n",
+ "\n",
+ "L(3) =0.000048\n",
+ "\n",
+ "\n",
+ "e(0)=0.000000\n",
+ "\n",
+ "\n",
+ "e(1)=0.000050\n",
+ "\n",
+ "\n",
+ "e(2)=0.000099\n",
+ "\n",
+ "\n",
+ "e(3)=0.000147\n",
+ "\n",
+ "\n",
+ "VERIFIED\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.7:pg-310"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.7\n",
+ "#modified euler's method\n",
+ "#page 310\n",
+ "h=0.05\n",
+ "f=1\n",
+ "def f1(x,y):\n",
+ " return x**2+y\n",
+ "x=[0,0.05,0.1]\n",
+ "y1=[0,0,0,0]\n",
+ "y2=[0,0,0,0]\n",
+ "y1[0]=f+h*f1(x[0],f);\n",
+ "y1[1]=f+h*(f1(x[0],f)+f1(x[1],y1[0]))/2\n",
+ "y1[2]=f+h*(f1(x[0],f)+f1(x[2],y1[1]))/2\n",
+ "y2[0]=y1[1]+h*f1(x[1],y1[1])\n",
+ "y2[1]=y1[1]+h*(f1(x[1],y1[1])+f1(x[2],y2[0]))/2\n",
+ "y2[2]=y1[1]+h*(f1(x[1],y1[1])+f1(x[2],y2[1]))/2\n",
+ "print \"y1(0)\\t y1(1)\\t y1(2)\\t y2(0)\\t y2(1)\\t y3(2)\\n\\n\"\n",
+ "print \" %f\\t %f\\t %f\\t %f\\t %f\\t %f\\n\" %(y1[0],y1[1],y1[2],y2[0],y2[1],y2[2])\n",
+ "print \"\\n\\n the value of y at %0.2f is : %0.4f\" %(x[2],y2[2])\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "y1(0)\t y1(1)\t y1(2)\t y2(0)\t y2(1)\t y3(2)\n",
+ "\n",
+ "\n",
+ " 1.050000\t 1.051313\t 1.051533\t 1.104003\t 1.105508\t 1.105546\n",
+ "\n",
+ "\n",
+ "\n",
+ " the value of y at 0.10 is : 1.1055\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.8:pg-313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.8\n",
+ "#runge-kutta formula\n",
+ "#page 313\n",
+ "from __future__ import division\n",
+ "def f(x,y):\n",
+ " return y-x\n",
+ "y=2\n",
+ "x=0\n",
+ "h=0.1\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h,y+K1)\n",
+ "y1=y+( K1+K2)/2\n",
+ "print \"\\n y(0.1) by second order runge kutta method:%0.4f\" %(y1)\n",
+ "y=y1\n",
+ "x=0.1\n",
+ "h=0.1\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h,y+K1)\n",
+ "y1=y+( K1+K2)/2\n",
+ "print \"\\n y(0.2) by second order runge kutta method:%0.4f\" %(y1)\n",
+ "y=2\n",
+ "x=0\n",
+ "h=0.1\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "print \"\\n y(0.1) by fourth order runge kutta method:%0.4f\" %(y1)\n",
+ "y=y1\n",
+ "x=0.1\n",
+ "h=0.1\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "print \"\\n y(0.1) by fourth order runge kutta method:%0.4f \" %(y1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " y(0.1) by second order runge kutta method:2.2050\n",
+ "\n",
+ " y(0.2) by second order runge kutta method:2.4210\n",
+ "\n",
+ " y(0.1) by fourth order runge kutta method:2.2052\n",
+ "\n",
+ " y(0.1) by fourth order runge kutta method:2.4214 \n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.9:pg-315"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.9\n",
+ "#runge kutta method\n",
+ "#page 315\n",
+ "from __future__ import division\n",
+ "def f(x,y):\n",
+ " return 1+y**2\n",
+ "y=0\n",
+ "x=0\n",
+ "h=0.2\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "print \"\\n y(0.2) by fourth order runge kutta method:%0.4f\" %(y1)\n",
+ "y=y1\n",
+ "x=0.2\n",
+ "h=0.2\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "print \"\\n y(0.4) by fourth order runge kutta method:%0.4f\" %(y1)\n",
+ "y=2\n",
+ "x=0\n",
+ "h=0.1\n",
+ "y=y1\n",
+ "x=0.4\n",
+ "h=0.2\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "print \"\\n y(0.6) by fourth order runge kutta method:%0.4f\" %(y1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " y(0.2) by fourth order runge kutta method:0.2027\n",
+ "\n",
+ " y(0.4) by fourth order runge kutta method:0.4228\n",
+ "\n",
+ " y(0.6) by fourth order runge kutta method:0.6841\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.10:pg-315"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.10\n",
+ "#initial value problems\n",
+ "#page 315\n",
+ "from __future__ import division\n",
+ "def f1(x,y):\n",
+ " return 3*x+y/2\n",
+ "y=[1,0,0]\n",
+ "h=0.1\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " y[i+1]=y[i]+h*f1(c,y[i])\n",
+ " print \"\\ny(%g)=%g\\n\" %(c,y[i])\n",
+ " c=c+0.1\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "y(0)=1\n",
+ "\n",
+ "\n",
+ "y(0.1)=1.05\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.11:pg-316"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.11\n",
+ "#adam's moulton method\n",
+ "#page 316\n",
+ "def f(x,y):\n",
+ " return 1+y**2\n",
+ "y=0\n",
+ "x=0\n",
+ "h=0.2\n",
+ "f1=[0,0,0]\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "f1[0]=y1\n",
+ "print \"\\n y(0.2) by fourth order runge kutta method:%0.4f\" %(y1)\n",
+ "y=y1\n",
+ "x=0.2\n",
+ "h=0.2\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "f1[1]=y1\n",
+ "print \"\\n y(0.4) by fourth order runge kutta method:%0.4f\" %(y1)\n",
+ "y=2\n",
+ "x=0\n",
+ "h=0.1\n",
+ "y=y1\n",
+ "x=0.4\n",
+ "h=0.2\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "f1[2]=y1\n",
+ "print \"\\n y(0.6) by fourth order runge kutta method:%0.4f\" %(y1)\n",
+ "y_p=y1+h*(55*(1+f1[2]**2)-59*(1+f1[1]**2)+37*(1+f1[0]**2)-9)/24\n",
+ "y_c=y1+h*(9*(1+(y_p-1)**2)+19*(1+f1[2]**2)-5*(1+f1[1]**2)+(1+f1[0]**2))/24\n",
+ "print \"\\nthe predicted value is:%0.4f:\\n\" %(y_p)\n",
+ "print \" the computed value is:%0.4f:\" %(y_c)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " y(0.2) by fourth order runge kutta method:0.2027\n",
+ "\n",
+ " y(0.4) by fourth order runge kutta method:0.4228\n",
+ "\n",
+ " y(0.6) by fourth order runge kutta method:0.6841\n",
+ "\n",
+ "the predicted value is:1.0234:\n",
+ "\n",
+ " the computed value is:0.9512:\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.12:pg-320"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.12\n",
+ "#milne's method\n",
+ "#page 320\n",
+ "def f(x,y):\n",
+ " return 1+y**2\n",
+ "y=0\n",
+ "f1=[0,0,0]\n",
+ "Y1=[0,0,0,0]\n",
+ "x=0\n",
+ "h=0.2\n",
+ "f1[0]=0\n",
+ "print \"x y y1=1+y^2\\n\\n\"\n",
+ "Y1[0]=1+y**2\n",
+ "print \"%0.4f %0.4f %0.4f\\n\" %(x,y,(1+y**2))\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "f1[0]=y1\n",
+ "Y1[1]=1+y1**2\n",
+ "print \"%0.4f %0.4f %0.4f\\n\" %(x+h,y1,(1+y1**2))\n",
+ "y=y1\n",
+ "x=0.2\n",
+ "h=0.2\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "f1[1]=y1\n",
+ "Y1[2]=1+y1**2\n",
+ "print \"%0.4f %0.4f %0.4f\\n\" %(x+h,y1,(1+y1**2))\n",
+ "y=y1\n",
+ "x=0.4\n",
+ "h=0.2\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "f1[2]=y1\n",
+ "Y1[3]=1+y1**2;\n",
+ "print \"%0.4f %0.4f %0.4f\\n\" %(x+h,y1,(1+y1**2))\n",
+ "Y_4=4*h*(2*Y1[1]-Y1[2]+2*Y1[3])/3\n",
+ "print \"y(0.8)=%f\\n\" %(Y_4)\n",
+ "Y=1+Y_4**2\n",
+ "Y_4=f1[1]+h*(Y1[2]+4*Y1[3]+Y)/3 #more correct value\n",
+ "print \"y(0.8)=%f\\n\" %(Y_4)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x y y1=1+y^2\n",
+ "\n",
+ "\n",
+ "0.0000 0.0000 1.0000\n",
+ "\n",
+ "0.2000 0.2027 1.0411\n",
+ "\n",
+ "0.4000 0.4228 1.1788\n",
+ "\n",
+ "0.6000 0.6841 1.4680\n",
+ "\n",
+ "y(0.8)=1.023869\n",
+ "\n",
+ "y(0.8)=1.029403\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.13:pg-320"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.13\n",
+ "#milne's method\n",
+ "#page 320\n",
+ "def f1(x,y):\n",
+ " return x**2+y**2-2\n",
+ "x=[-0.1, 0, 0.1, 0.2]\n",
+ "y=[1.0900, 1.0, 0.8900, 0.7605]\n",
+ "Y1=[0,0,0,0]\n",
+ "h=0.1\n",
+ "for i in range(0,4):\n",
+ " Y1[i]=f1(x[i],y[i])\n",
+ "print \" x y y1=x^2+y^2-2 \\n\\n\"\n",
+ "for i in range(0,4):\n",
+ " print \" %0.2f %f %f \\n\" %(x[i],y[i],Y1[i])\n",
+ "Y_3=y[0]+(4*h/3)*(2*Y1[1]-Y1[2]+2*Y1[3])\n",
+ "print \"y(0.3)=%f\\n\" %(Y_3)\n",
+ "Y1_3=f1(0.3,Y_3)\n",
+ "Y_3=y[2]+h*(Y1[2]+4*Y1[3]+Y1_3)/3 #corrected value\n",
+ "print \"corrected y(0.3)=%f\" %(Y_3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " x y y1=x^2+y^2-2 \n",
+ "\n",
+ "\n",
+ " -0.10 1.090000 -0.801900 \n",
+ "\n",
+ " 0.00 1.000000 -1.000000 \n",
+ "\n",
+ " 0.10 0.890000 -1.197900 \n",
+ "\n",
+ " 0.20 0.760500 -1.381640 \n",
+ "\n",
+ "y(0.3)=0.614616\n",
+ "\n",
+ "corrected y(0.3)=0.614776\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.14:pg322"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.14\n",
+ "#initial-value problem\n",
+ "#page 322\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 13*math.exp(x/2)-6*x-12\n",
+ "s1=1.691358\n",
+ "s3=3.430879\n",
+ "print \"the erorr in the computed values are %0.7g %0.7g\" %(abs(f(0.5)-s1),abs(f(1)-s3))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the erorr in the computed values are 0.0009724169 0.002497519\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.15:pg-328"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#boundary value problem using finite difference method\n",
+ "#example 8.15\n",
+ "#page 328\n",
+ "import math\n",
+ "from numpy import matrix\n",
+ "def f(x):\n",
+ " return math.cos(x)+((1-math.cos(1))/math.sin(1))*math.sin(x)-1\n",
+ "h1=1/2\n",
+ "Y=f(0.5)\n",
+ "y0=0\n",
+ "y2=0\n",
+ "y1=4*(1/4+y0+y2)/7\n",
+ "print \"computed value with h=%f of y(0.5) is %f\\n\" %(h1,y1)\n",
+ "print \"error in the result with actual value %f\\n\" %(abs(Y-y1))\n",
+ "h2=1/4\n",
+ "y0=0\n",
+ "y4=0\n",
+ "#solving the approximated diffrential equation\n",
+ "A=matrix([[-31/16, 1, 0],[1, -31/16, 1],[0, 1, -31/16]])\n",
+ "X=matrix([[-1/16],[-1/16],[-1/16]])\n",
+ "C=A.I*X\n",
+ "print \"computed value with h=%f of y(0.5) is %f\\n\" %(h2,C[1][0])\n",
+ "print \"error in the result with actual value %f\\n\" %(abs(Y-C[1][0]))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "computed value with h=0.500000 of y(0.5) is 0.142857\n",
+ "\n",
+ "error in the result with actual value 0.003363\n",
+ "\n",
+ "computed value with h=0.250000 of y(0.5) is 0.140312\n",
+ "\n",
+ "error in the result with actual value 0.000818\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.16:pg-329"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#boundary value problem using finite difference method\n",
+ "#example 8.16\n",
+ "#page 329\\\n",
+ "from numpy import matrix\n",
+ "import math\n",
+ "def f(x):\n",
+ " return math.sinh(x)\n",
+ "y0=0 #y(0)=0\n",
+ "y4=3.62686 #y(2)=3.62686\n",
+ "h1=0.5\n",
+ "Y=f(0.5)\n",
+ "#arranging and calculating the values\n",
+ "A=matrix([[-9, 4, 0],[4, -9, 4],[0, 4, -9]])\n",
+ "C=matrix([[0],[0],[-14.50744]])\n",
+ "X=A.I*C\n",
+ "print \"computed value with h=%f of y(0.5) is %f\\n\" %(h1,X[0][0])\n",
+ "print \"error in the result with actual value %f\\n\" %(abs(Y-X[0][0]))\n",
+ "h2=1.0\n",
+ "y0=0 #y(0)=0\n",
+ "y2=3.62686 #y(2)=3.62686\n",
+ "y1=(y0+y2)/3\n",
+ "Y=(4*X[1][0]-y1)/3\n",
+ "print \"with better approximation error is reduced to %f\" %(abs(Y-f(1.0)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "computed value with h=0.500000 of y(0.5) is 0.526347\n",
+ "\n",
+ "error in the result with actual value 0.005252\n",
+ "\n",
+ "with better approximation error is reduced to 0.000855\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.17:pg-330"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#cubic spline method\n",
+ "#example 8.17\n",
+ "#page 330\n",
+ "def f(x):\n",
+ " return math.cos(x)+((1-math.cos(1))/math.sin(1))*math.sin(x)-1\n",
+ "y=[f(0), f(0.5), f(1)]\n",
+ "h=1/2\n",
+ "Y=f(0.5)\n",
+ "y0=0\n",
+ "y2=0\n",
+ "M0=-1\n",
+ "M1=-22/25\n",
+ "M2=-1\n",
+ "s1_0=47/88\n",
+ "s1_1=-47/88\n",
+ "s1_05=0\n",
+ "print \"the cubic spline value is %f\" %(Y)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the cubic spline value is 0.139494\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.18:pg-331"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#cubic spline method\n",
+ "#example 8.18\n",
+ "#page 331\n",
+ "from numpy import matrix\n",
+ "from __future__ import division\n",
+ "#after arranging and forming equation \n",
+ "A=matrix([[10, -1, 0, 24],[0, 16, -1, -32],[1, 20, 0, 16],[0, 1, 26, -24]])\n",
+ "C=matrix([[36],[-12],[24],[-9]])\n",
+ "X=A.I*C;\n",
+ "print \" Y1=%f\\n\\n\" %(X[3][0])\n",
+ "print \" the error in the solution is :%f\" %(abs((2/3)-X[3][0]))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Y1=0.653890\n",
+ "\n",
+ "\n",
+ " the error in the solution is :0.012777\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.19:pg-331"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#boundary value problem by cubisc spline nethod\n",
+ "#example 8.18\n",
+ "#page 331\n",
+ "from numpy import matrix\n",
+ "h=1/2\n",
+ "#arranging in two subintervals we get\n",
+ "A=matrix([[10, -1, 0,24],[0, 16, -1, -32],[1, 20, 0, 16],[0, 1, 26, -24]])\n",
+ "C=matrix([[36],[-12],[24],[-9]])\n",
+ "X=A.I*C\n",
+ "print \"the computed value of y(1.5) is %f \"%(X[3][0])\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the computed value of y(1.5) is 0.653890 \n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter_5_7.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter_5_7.ipynb new file mode 100644 index 00000000..f7ae1b31 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter_5_7.ipynb @@ -0,0 +1,358 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:d6c77d0644ce8c4a3cef684a9d4884ca9577630ae40ac3764bbf1d6682f47c3d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter05:Spline Functions"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.1:pg-182"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#linear splines\n",
+ "#example 5.1\n",
+ "#page 182\n",
+ "from numpy import matrix\n",
+ "X=matrix([[1],[2], [3]])\n",
+ "y=matrix([[-8],[-1],[18]])\n",
+ "m1=(y[1][0]-y[0][0])/(X[1][0]-X[0][0])\n",
+ "m2=(y[2][0]-y[1][0])/(X[2][0]-X[1][0])\n",
+ "def s1(x):\n",
+ " return y[0][0]+(x-X[0][0])*m1\n",
+ "def s2(x):\n",
+ " return y[1][0]+(x-X[1][0])*m2\n",
+ "print \"the value of function at 2.5 is %0.2f: \" %(s2(2.5))\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of function at 2.5 is 8.50: \n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.3:pg-188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#cubic splines\n",
+ "#example 5.3\n",
+ "#page 188\n",
+ "from numpy import matrix\n",
+ "import math\n",
+ "X=matrix([[1],[2],[3]])\n",
+ "y=matrix([[-8],[-1],[18]])\n",
+ "M1=0\n",
+ "M2=8\n",
+ "M3=0\n",
+ "h=1\n",
+ "#deff('y=s1(x)','y=3*(x-1)^3-8*(2-x)-4*(x-1)')\n",
+ "def s1(x):\n",
+ " return 3*math.pow(x-1,3)-8*(2-x)-4*(x-1)\n",
+ "#deff('y=s2(x)','y=3*(3-x)^3+22*x-48');\n",
+ "def s2(x):\n",
+ " return 3*math.pow(3-x,3)+22*x-48\n",
+ "h=0.0001\n",
+ "n=2.0\n",
+ "D=(s2(n+h)-s2(n))/h;\n",
+ "print \" y(2.5)=%f\" %(s2(2.5))\n",
+ "print \"y1(2.0)=%f\" %(D)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " y(2.5)=7.375000\n",
+ "y1(2.0)=13.000900\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.4:pg-189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#cubic spline\n",
+ "#example 5.4\n",
+ "#page 189\n",
+ "from numpy import matrix\n",
+ "import math\n",
+ "x=matrix([[0],[math.pi/2],[math.pi]])\n",
+ "y=matrix([[0],[1],[0]])\n",
+ "h=x[1][0]-x[0][0]\n",
+ "M0=0\n",
+ "M2=0\n",
+ "M1=((6*(y[0][0]-2*y[1][0]+y[2][0])/math.pow(h,2))-M0-M2)/4\n",
+ "X=math.pi/6.0\n",
+ "s1=(math.pow(x[1][0]-X,3)*(M0/6)+math.pow(X-x[0][0],3)*M1/6+(y[0][0]-math.pow(h,2)*M0/6)*(x[1][0]-X)+(y[1][0]-math.pow(h,2)*M1/6)*(X-x[0][0]))/h;\n",
+ "x=matrix([[0],[math.pi/4], [math.pi/2], [3*math.pi/4], [math.pi]])\n",
+ "y=matrix([[0], [1.414], [1] ,[1.414]])\n",
+ "M0=0\n",
+ "M4=0\n",
+ "A=matrix([[4, 1, 0],[1, 4, 1],[0 ,1 ,4]]) #calculating value of M1 M2 M3 by matrix method\n",
+ "C=matrix([[-4.029],[-5.699],[-4.029]])\n",
+ "B=A.I*C\n",
+ "print \"M0=%f\\t M1=%f\\t M2=%f\\t M3=%f\\t M4=%f\\t\\n\\n\" %(M0,B[0][0],B[1][0],B[2][0],M4)\n",
+ "h=math.pi/4;\n",
+ "X=math.pi/6;\n",
+ "s1=(-0.12408*math.pow(X,3)+0.7836*X)/h;\n",
+ "print \"the value of sin(pi/6) is:%f\" %(s1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "M0=0.000000\t M1=-0.744071\t M2=-1.052714\t M3=-0.744071\t M4=0.000000\t\n",
+ "\n",
+ "\n",
+ "the value of sin(pi/6) is:0.499722\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.5:pg-191"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#cubic spline\n",
+ "#example 5.5\n",
+ "#page 191\n",
+ "import math\n",
+ "from numpy import matrix\n",
+ "x=[1,2,3]\n",
+ "y=[6,18,42]\n",
+ "m0=40\n",
+ "s1=0\n",
+ "m1=(3*(y[2]-y[0])-m0)/4\n",
+ "X=0\n",
+ "s1=m0*((x[1]-X)**2)*(X-x[0])-m1*((X-x[0])**2)*(x[1]-X)+y[0]*((x[1]-X)**2)*(2*(X-x[0])+1)+y[1]*((X-x[0])**2)*(2*(x[1]-X)+1)\n",
+ "print \"s1= %f+261*x-160X^2+33X^3\" %(s1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "s1= -128.000000+261*x-160X^2+33X^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.7:pg-195"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#surface fitting by cubic spline\n",
+ "#example 5.7\n",
+ "#page 195\n",
+ "from numpy import matrix\n",
+ "def L0(y):\n",
+ " return math.pow(y,3)/4-5*y/4+1\n",
+ "def L1(y):\n",
+ " return (math.pow(y,3)/2)*-1+3*y/2\n",
+ "def L2(y):\n",
+ " return math.pow(y,3)/4-y/4\n",
+ "A=[ [1,2,9], [2,3,10], [9,10,17] ]\n",
+ "x=0.5\n",
+ "y=0.5\n",
+ "S=0.0\n",
+ "S=S+L0(x)*(L0(x)*A[0][0]+L1(x)*A[0][1]+L2(x)*A[0][2])\n",
+ "S=S+L1(x)*(L0(x)*A[1][0]+L1(x)*A[1][1]+L2(x)*A[1][2])\n",
+ "S=S+L2(x)*(L0(x)*A[2][0]+L1(x)*A[2][1]+L2(x)*A[2][2])\n",
+ "print \"approximated value of z(0.5 0.5)=%f\\n\\n\" %(S)\n",
+ "print \" error in the approximated value : %f\" %((abs(1.25-S)/1.25)*100)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "approximated value of z(0.5 0.5)=0.875000\n",
+ "\n",
+ "\n",
+ " error in the approximated value : 30.000000\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.8:pg-200"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#cubic B-splines\n",
+ "#example 5.8\n",
+ "#page 200\n",
+ "import math\n",
+ "k=[0.0, 1, 2, 3, 4]\n",
+ "pi=[0.0, 0, 4, -6, 24]\n",
+ "x=1\n",
+ "S=0\n",
+ "for i in range(2,5):\n",
+ " S=S+math.pow(k[i]-x,3)/(pi[i])\n",
+ "print \"the cubic splines for x=1 is %f\\n\\n\" %(S)\n",
+ "S=0\n",
+ "x=2\n",
+ "for i in range(2,5):\n",
+ " S=S+math.pow(k[i]-x,3)/(pi[i])\n",
+ "print \"the cubic splines for x=2 is %f\\n\\n\" %(S)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the cubic splines for x=1 is 0.041667\n",
+ "\n",
+ "\n",
+ "the cubic splines for x=2 is 0.166667\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.9:pg-201"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#cubic B-spline\n",
+ "#example 5.9\n",
+ "#page 201\n",
+ "k=[0, 1, 2, 3, 4];\n",
+ "x=1 #for x=1\n",
+ "s11=0\n",
+ "s13=0\n",
+ "s14=0\n",
+ "s24=0 \n",
+ "s12=1/(k[2]-k[1])\n",
+ "s22=((x-k[0])*s11+(k[2]-x)*s12)/2.0 #k[2]-k[0]=2\n",
+ "s23=((x-k[1])*s11+(k[3]-x)*s13)/(k[3]-k[1])\n",
+ "s33=((x-k[0])*s22+(k[3]-x)*s23)/(k[3]-k[0])\n",
+ "s34=((x-k[1])*s23+(k[4]-x)*s24)/(k[4]-k[1])\n",
+ "s44=((x-k[0])*s33+(k[4]-x)*s34)/(k[4]-k[0])\n",
+ "print \"s11=%f\\t s22=%f\\t s23=%f\\t s33=%f\\t s34=%f\\t s44=%f\\n\\n\" %(s11,s22,s23,s33,s34,s44)\n",
+ "x=2 #for x=2\n",
+ "s11=0\n",
+ "s12=0\n",
+ "s14=0\n",
+ "s22=0\n",
+ "s13=1/(k[3]-k[2])\n",
+ "s23=((x-k[1])*s12+(k[3]-x)*s13)/2.0 # k[3]-k[1]=2\n",
+ "s24=((x-k[2])*s13+(k[4]-x)*s14)/(k[2]-k[0])\n",
+ "s33=((x-k[0])*s22+(k[3]-x)*s23)/(k[3]-k[0])\n",
+ "s34=((x-k[1])*s23+(k[4]-x)*s24)/(k[4]-k[1])\n",
+ "s44=((x-k[0])*s33+(k[4]-x)*s34)/(k[4]-k[0])\n",
+ "print \"s13=%f\\t s23=%f\\t s24=%f\\t s33=%f\\t s34=%f\\t s44=%f\\n\\n\" %(s13,s23,s24,s33,s34,s44)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "s11=0.000000\t s22=0.500000\t s23=0.000000\t s33=0.166667\t s34=0.000000\t s44=0.041667\n",
+ "\n",
+ "\n",
+ "s13=1.000000\t s23=0.500000\t s24=0.000000\t s33=0.166667\t s34=0.166667\t s44=0.166667\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/1.2_2.png b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/1.2_2.png Binary files differnew file mode 100644 index 00000000..2af775d8 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/1.2_2.png diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/3.7_2.png b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/3.7_2.png Binary files differnew file mode 100644 index 00000000..30f2ed11 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/3.7_2.png diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/6.7_2.png b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/6.7_2.png Binary files differnew file mode 100644 index 00000000..524c2a6a --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/6.7_2.png diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch11.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch11.ipynb new file mode 100644 index 00000000..45094840 --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch11.ipynb @@ -0,0 +1,757 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 11 Digital To Analog Converter(D/A Converter)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.1 Pg 313" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the full scale voltage of D/A converter VFS is = 24.00 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt, pi\n", + "# to determine the full scale voltage of D/A\n", + "Vref = 12 #\n", + "Rf = 10 # # K ohm\n", + "R = 5 # # K ohm\n", + "\n", + "# the full scale voltage of D/A converter \n", + "VFS = Vref*(Rf/R) #\n", + "print 'the full scale voltage of D/A converter VFS is = %0.2f'%VFS,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.2 Pg 313" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "For BI 10101010 the output of D/A converter is = 2.04 V \n", + "For BI 11001100 the output of D/A converter is = 2.448 V \n", + "For BI 11101110 the output of D/A converter is = 2.856 V \n", + "For BI 00010001 the output of D/A converter is = 0.204 V \n" + ] + } + ], + "source": [ + " # determine the output voltage of D/A converter for the binary inputs a) 10101010 b) 11001100 c) 11101110 d) 00010001 \n", + "Del = 12*10**-3 # # mA\n", + "\n", + "# the input voltage of D/A converter \n", + " #Vo = Del*binary input (BI)\n", + "\n", + "# For BI 10101010 the output\n", + "BI = '10101010' #\n", + "BI = int(BI,2)#\n", + "Vo = Del*BI #\n", + "print 'For BI 10101010 the output of D/A converter is = ',Vo,' V '\n", + "\n", + "# For BI 11001100 the output\n", + "BI = '11001100' #\n", + "BI = int(BI,2)#\n", + "Vo = Del*BI #\n", + "print 'For BI 11001100 the output of D/A converter is = ',Vo,' V '\n", + "\n", + "# For BI 11101110 the output\n", + "BI = '11101110' #\n", + "BI = int(BI,2)#\n", + "Vo = Del*BI #\n", + "print 'For BI 11101110 the output of D/A converter is = ',Vo,' V '\n", + "\n", + "# For BI 00010001 the output\n", + "BI = '00010001' #\n", + "BI = int(BI,2)#\n", + "Vo = Del*BI #\n", + "print 'For BI 00010001 the output of D/A converter is = ',Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.3 Pg 314" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the resolution of 4-bit D/A converter is = 0.80 V \n" + ] + } + ], + "source": [ + "# determine the resolution of 4-bit D/A converter\n", + "VFS = 12 #\n", + "N = 4 #\n", + "\n", + "# the resolution of 4-bit D/A converter is defined as\n", + "Resolution = VFS/(2**N-1) #\n", + "print 'the resolution of 4-bit D/A converter is = %0.2f'%Resolution,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.4 314" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the number of bit required to design a 4-bit D/A converter is = 8.97 = 9 \n" + ] + } + ], + "source": [ + "from math import log10\n", + "# determine the number of bit required to design a 4-bit D/A converter\n", + "VFS = 5 #\n", + "Resolution = 10*10**-3 # # A\n", + "\n", + "# the resolution of 4-bit D/A converter is defined as\n", + "# Resolution = VFS/(2**N-1) #\n", + "N = (VFS/Resolution)+1 # \n", + "N = log10(N)/log10(2)#\n", + "print 'the number of bit required to design a 4-bit D/A converter is = %0.2f'%N,' = 9 '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.5 Pg 315" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "for the binary input 101 analog output is = -60.00 V \n", + "for the binary input 111 analog output is = -84.00 V \n", + "for the binary input 011 analog output is = -72.00 V \n", + "for the binary input 001 analog output is = -48.00 V \n", + "for the binary input 100 analog output is = -12.00 V\n" + ] + } + ], + "source": [ + "# determine the analog output voltage\n", + "Vref = 12 # \n", + "BI = 101 # BI = 111 # BI = 011 # BI = 001 # BI = 100 #\n", + "Rf = 40*10**3 #\n", + "R = 0.25*Rf #\n", + "\n", + "# The output voltage of given binary weighted resistor D/A converter is defined as\n", + "\n", + "# Vo = -(Rf*Vref/R)*(2**0*b0+2**-1*b1+2**-2*b2) #\n", + "\n", + "# Vo = -(Rf*Vref/R)*(b0+2**-1*b1+2**-2*b2) #\n", + "\n", + "# for the given value Rf,R and Vref the output voltage\n", + "\n", + "# Vo = -48*(b0+2**-1*b1+2**-2*b2) #\n", + "\n", + "# for the binary input 101 analog output is\n", + "b2 = 1 #\n", + "b1 = 0 #\n", + "b0 = 1 #\n", + "Vo = -48*(b0+2**-1*b1+2**-2*b2) #\n", + "print 'for the binary input 101 analog output is = %0.2f'%Vo,' V '\n", + "\n", + "\n", + "# for the binary input 111 analog output is\n", + "b2 = 1 #\n", + "b1 = 1 #\n", + "b0 = 1 #\n", + "Vo = -48*(b0+2**-1*b1+2**-2*b2) #\n", + "print 'for the binary input 111 analog output is = %0.2f'%Vo,' V '\n", + "\n", + "\n", + "# for the binary input 011 analog output is\n", + "b2 = 0 #\n", + "b1 = 1 #\n", + "b0 = 1 #\n", + "Vo = -48*(b0+2**-1*b1+2**-2*b2) #\n", + "print 'for the binary input 011 analog output is = %0.2f'%Vo,' V '\n", + "\n", + "\n", + "# for the binary input 001 analog output is\n", + "b2 = 0 #\n", + "b1 = 0 #\n", + "b0 = 1 #\n", + "Vo = -48*(b0+2**-1*b1+2**-2*b2) #\n", + "print 'for the binary input 001 analog output is = %0.2f'%Vo,' V '\n", + "\n", + "\n", + "# for the binary input 100 analog output is\n", + "b2 = 1 #\n", + "b1 = 0 #\n", + "b0 = 0 #\n", + "Vo = -48*(b0+2**-1*b1+2**-2*b2) #\n", + "print 'for the binary input 100 analog output is = %0.2f'%Vo,' V'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.6 Pg 316" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "for the binary input 1001 analog output is = -67.50 V \n", + "the feedback current If is = 2.70 mA \n", + "for the binary input 1101 analog output is = -82.50 V \n", + "the feedback current If is = 3.30 mA \n", + "for the binary input 1010 analog output is = -37.50 V \n", + "the feedback current If is = 1.50 mA \n", + "for the binary input 0011 analog output is = -90.00 V \n", + "the feedback current If is = 3.60 mA \n" + ] + } + ], + "source": [ + "# determine the analog output voltage and feed back current If\n", + "Vref = 12 # \n", + "BI = 1001 # BI = 1101 # BI = 1010 # BI = 0011 #\n", + "Rf = 25 # # K ohm\n", + "R = 0.25*Rf #\n", + "\n", + "# The output voltage of given binary weighted resistor D/A converter is defined as\n", + "\n", + "# Vo = -(Rf*Vref/R)*(2**0*b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "\n", + "# Vo = -(Rf*Vref/R)*(b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "\n", + "# for the given value Rf,R and Vref the output voltage\n", + "\n", + "# Vo = -60*(b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "\n", + "# for the binary input 1001 analog output is\n", + "b3 = 1 #\n", + "b2 = 0 #\n", + "b1 = 0 #\n", + "b0 = 1 #\n", + "Vo = -60*(b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "print 'for the binary input 1001 analog output is = %0.2f'%Vo,' V '\n", + "\n", + "# the feedback current If is given by\n", + "If = -(Vo/Rf) #\n", + "print 'the feedback current If is = %0.2f'%If,' mA '\n", + "\n", + "\n", + "# for the binary input 1101 analog output is\n", + "b3 = 1 #\n", + "b2 = 1 #\n", + "b1 = 0 #\n", + "b0 = 1 #\n", + "Vo = -60*(b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "print 'for the binary input 1101 analog output is = %0.2f'%Vo,' V '\n", + "\n", + "# the feedback current If is given by\n", + "If = -(Vo/Rf) #\n", + "print 'the feedback current If is = %0.2f'%If,' mA '\n", + "\n", + "\n", + "# for the binary input 1010 analog output is\n", + "b3 = 1 #\n", + "b2 = 0 #\n", + "b1 = 1 #\n", + "b0 = 0 #\n", + "Vo = -60*(b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "print 'for the binary input 1010 analog output is = %0.2f'%Vo,' V '\n", + "\n", + "# the feedback current If is given by\n", + "If = -(Vo/Rf) #\n", + "print 'the feedback current If is = %0.2f'%If,' mA '\n", + "\n", + "\n", + "# for the binary input 0011 analog output is\n", + "b3 = 0 #\n", + "b2 = 0 #\n", + "b1 = 1 #\n", + "b0 = 1 #\n", + "Vo = -60*(b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "print 'for the binary input 0011 analog output is = %0.2f'%Vo,' V '\n", + "\n", + "# the feedback current If is given by\n", + "If = -(Vo/Rf) #\n", + "print 'the feedback current If is = %0.2f'%If,' mA '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.7 Pg 319" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "for the binary input 001 analog output is = 1.60 mA \n", + "An analog output voltage Vo is = -40.00 V \n", + "for the binary input 010 analog output is = 0.80 mA\n", + "An analog output voltage Vo is = -20.00 V \n", + "for the binary input 110 analog output is = 1.20 mA \n", + "An analog output voltage Vo is = -30.00 V \n" + ] + } + ], + "source": [ + "# determine the feed back current If and analog output voltage\n", + "Vref = 8 # # V\n", + "BI = 001 #\n", + "BI = 010 #\n", + "BI = 110 #\n", + "Rf = 25*10**3 # # Hz\n", + "R = 0.2*Rf #\n", + "\n", + "# The output current of given binary weighted resistor D/A converter is defined as\n", + "\n", + "# If = -(Vref/R)*(2**0*b0+2**-1*b1+2**-2*b2) #\n", + "\n", + "# If = -(Vref/R)*(b0+2**-1*b1+2**-2*b2) #\n", + "\n", + "# for the given value Rf,R and Vref the output current\n", + "\n", + "# If = -(1.6*10**-3)*(b0+2**-1*b1+2**-2*b2) #\n", + "\n", + "# for the binary input 001 the feedback current If is given by\n", + "b2 = 0 #\n", + "b1 = 0 #\n", + "b0 = 1 #\n", + "If = (1.6*10**-3)*(b0+2**-1*b1+2**-2*b2) #\n", + "print 'for the binary input 001 analog output is = %0.2f'%(If*1000),' mA '\n", + "\n", + "# An analog output voltage Vo is\n", + "Vo = -If*Rf #\n", + "print 'An analog output voltage Vo is = %0.2f'%Vo,' V '\n", + "\n", + "\n", + "# for the binary input 010 the feedback current If is given by\n", + "b2 = 0 #\n", + "b1 = 1 #\n", + "b0 = 0 #\n", + "If = (1.6*10**-3)*(b0+2**-1*b1+2**-2*b2) #\n", + "print 'for the binary input 010 analog output is = %0.2f'%(If*1000),' mA'\n", + "\n", + "# the An analog output voltage Vo is\n", + "Vo = -If*Rf #\n", + "print 'An analog output voltage Vo is = %0.2f'%Vo,' V '\n", + "\n", + "\n", + "# for the binary input 110 the feedback current If is given by\n", + "b2 = 1 #\n", + "b1 = 1 #\n", + "b0 = 0 #\n", + "If = (1.6*10**-3)*(b0+2**-1*b1+2**-2*b2) #\n", + "print 'for the binary input 110 analog output is = %0.2f'%(If*1000),' mA '\n", + "\n", + "# the An analog output voltage Vo is\n", + "Vo = -If*Rf #\n", + "print 'An analog output voltage Vo is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.8 Pg 320" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "for the binary input 101 analog output is = 0.625 mA \n", + "An analog output voltage Vo is = -15.625 V \n", + "for the binary input 011 analog output is = 0.75 mA\n", + "An analog output voltage Vo is = -18.750 V \n", + "for the binary input 100 analog output is = 0.125 mA \n", + "An analog output voltage Vo is = -3.12 V \n", + "for the binary input 001 analog output is = 0.5 mA \n", + "An analog output voltage Vo is = -12.50 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# determine the feed back current If and analog output voltage\n", + "Vref = 5 # \n", + "BI = 101 # BI = 011 # BI = 100 # BI = 001 #\n", + "Rf = 25*10**3 # \n", + "R = 0.2*Rf #\n", + "\n", + "# The output current of given R-2R ladder D/A converter is defined as\n", + "\n", + "# If = -(Vref/2*R)*(2**0*b0+2**-1*b1+2**-2*b2) #\n", + "\n", + "# If = -(Vref/2*R)*(b0+2**-1*b1+2**-2*b2) #\n", + "\n", + "# for the given value Rf,R and Vref the output current\n", + "\n", + "# If = (0.5*10**-3)*(b0+2**-1*b1+2**-2*b2) #\n", + "\n", + "# for the binary input 101 the feedback current If is given by\n", + "b2 = 1 #\n", + "b1 = 0 #\n", + "b0 = 1 #\n", + "If = (0.5*10**-3)*(b0+2**-1*b1+2**-2*b2) #\n", + "print 'for the binary input 101 analog output is = %0.3f'%(If*1e3),' mA '\n", + "\n", + "# An analog output voltage Vo is\n", + "Vo = -If*Rf #\n", + "print 'An analog output voltage Vo is = %0.3f'%Vo,' V '\n", + "\n", + "\n", + "# for the binary input 011 the feedback current If is given by\n", + "b2 = 0 #\n", + "b1 = 1 #\n", + "b0 = 1 #\n", + "If = (0.5*10**-3)*(b0+2**-1*b1+2**-2*b2) #\n", + "print 'for the binary input 011 analog output is = %0.2f'%(If*1e3),' mA'\n", + "\n", + "# the An analog output voltage Vo is\n", + "Vo = -If*Rf #\n", + "print 'An analog output voltage Vo is = %0.3f'%Vo,' V '\n", + "\n", + "\n", + "# for the binary input 100 the feedback current If is given by\n", + "b2 = 1 #\n", + "b1 = 0 #\n", + "b0 = 0 #\n", + "If = (0.5*10**-3)*(b0+2**-1*b1+2**-2*b2) #\n", + "print 'for the binary input 100 analog output is = %0.3f'%(If*1e3),'mA '\n", + "\n", + "# the An analog output voltage Vo is\n", + "Vo = -If*Rf #\n", + "print 'An analog output voltage Vo is = %0.2f'%Vo,' V '\n", + "\n", + "# for the binary input 001 the feedback current If is given by\n", + "b2 = 0 #\n", + "b1 = 0 #\n", + "b0 = 1 #\n", + "If = (0.5*10**-3)*(b0+2**-1*b1+2**-2*b2) #\n", + "print 'for the binary input 001 analog output is = %0.1f'%(If*1e3),' mA '\n", + "\n", + "# the An analog output voltage Vo is\n", + "Vo = -If*Rf #\n", + "print 'An analog output voltage Vo is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.9 Pg 322" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "for the binary input 1001 analog output is = -14.0625 V \n", + "the feedback current If is = 0.28 mA \n", + "for the binary input 1100 analog output is = -4.6875 V \n", + "the feedback current If is = 0.09 mA \n", + "for the binary input 1010 analog output is = -7.8125 V \n", + "the feedback current If is = 0.16 mA \n", + "for the binary input 0011 analog output is = -18.75 V \n", + "the feedback current If is = 0.375 mA \n" + ] + } + ], + "source": [ + " # determine the analog output voltage and feed back current If\n", + "Vref = 10 # \n", + "BI = 1001 # BI = 1100 # BI = 1010 # BI = 0011 #\n", + "Rf = 50 # # K ohm\n", + "R = 0.4*Rf #\n", + "\n", + "# The output voltage of given R-2R ladder D/A converter is defined as\n", + "\n", + "# Vo = -(Rf*Vref/2R)*(2**0*b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "\n", + "# Vo = -(Rf*Vref/2R)*(b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "\n", + "# for the given value Rf,R and Vref the output voltage\n", + "\n", + "# Vo = -12.5*(b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "\n", + "# for the binary input 1001 analog output is\n", + "b3 = 1 #\n", + "b2 = 0 #\n", + "b1 = 0 #\n", + "b0 = 1 #\n", + "Vo = -12.5*(b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "print 'for the binary input 1001 analog output is = %0.4f'%Vo,' V '\n", + "\n", + "# the feedback current If is given by\n", + "If = -(Vo/Rf) #\n", + "print 'the feedback current If is = %0.2f'%If,' mA '\n", + "\n", + "\n", + "# for the binary input 1100 analog output is\n", + "b3 = 1 #\n", + "b2 = 1 #\n", + "b1 = 0 #\n", + "b0 = 0 #\n", + "Vo = -12.5*(b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "print 'for the binary input 1100 analog output is = %0.4f'%Vo,' V '\n", + "\n", + "# the feedback current If is given by\n", + "If = -(Vo/Rf) #\n", + "print 'the feedback current If is = %0.2f'%If,' mA '\n", + "\n", + "\n", + "# for the binary input 1010 analog output is\n", + "b3 = 1 #\n", + "b2 = 0 #\n", + "b1 = 1 #\n", + "b0 = 0 #\n", + "Vo = -12.5*(b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "print 'for the binary input 1010 analog output is = %0.4f'%Vo,' V '\n", + "\n", + "# the feedback current If is given by\n", + "If = -(Vo/Rf) #\n", + "print 'the feedback current If is = %0.2f'%If,' mA '\n", + "\n", + "\n", + "# for the binary input 0011 analog output is\n", + "b3 = 0 #\n", + "b2 = 0 #\n", + "b1 = 1 #\n", + "b0 = 1 #\n", + "Vo = -12.5*(b0+2**-1*b1+2**-2*b2+2**-3*b3) #\n", + "print 'for the binary input 0011 analog output is = %0.2f'%Vo,' V '\n", + "\n", + "# the feedback current If is given by\n", + "If = -(Vo/Rf) #\n", + "print 'the feedback current If is = %0.3f'%If,' mA '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.10 Pg 324" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "for the binary input 1000 output voltage is = 18.75 V \n", + "the feedback current If is = -0.469 mA \n" + ] + } + ], + "source": [ + " # determine the analog output voltage and feed back current If\n", + "Vref = 15 # \n", + "BI = 1000 #\n", + "Rf = 40 # # K ohm\n", + "R = 0.4*Rf #\n", + "\n", + "# by using voltage divider rule Vin can be calculated as\n", + "Vin = -(Vref*2*R)/(2*R+2*R) #\n", + " \n", + "# The output voltage of given R-2R ladder D/A converter is defined as\n", + "\n", + "# Vo = -(Rf*Vin/R)\n", + "\n", + "Vo = (Vref*Rf)/(2*R)\n", + "print 'for the binary input 1000 output voltage is = %0.2f'%Vo,' V '\n", + "\n", + "# the feedback current If is given by\n", + "If = -(Vo/Rf) #\n", + "print 'the feedback current If is = %0.3f'%If,' mA '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.11 Pg 326" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "For the BI 10101111 output analog voltage is = 6.86 V \n", + "For the BI 11100010 output analog voltage is = 8.8627 V \n", + "For the BI 00101001 output analog voltage is = 1.6078 V \n", + "For the BI 01000110 output analog voltage is = 2.745 V \n" + ] + } + ], + "source": [ + "# to find the resolution and analog output voltage of 8-bit D/A converter\n", + "VFS = 10 #\n", + "N = 8 #\n", + "BI = 10101111 #\n", + "BI = 11100011 # \n", + "BI = 00101001 #\n", + "BI = 01000110\n", + "\n", + "# the resolution of 8-bit D/A converter is defined as\n", + "Resolution = VFS/(2**N-1) #\n", + "\n", + "# An analog output voltage of D/A converter is given by\n", + "# Vo = Resolution*(2**-0*b0+2**-1*b1+....+2**-N*bn-1)\n", + "# Vo = Resolution*(2**-0*b0+2**-1*b1+2**-2*b2+2**-3*b3+2**-4*b4+2**-5*b5+2**-6*b6+2**-7*b7)#\n", + "\n", + "# For the BI 10101111 output analog voltage is\n", + "BI = '10101111'#\n", + "BI = int(BI,2)#\n", + "Vo = Resolution*BI #\n", + "print 'For the BI 10101111 output analog voltage is = %0.2f'%Vo,' V '\n", + "\n", + "# For the BI 11100010 output analog voltage is\n", + "BI = '11100010'#\n", + "BI = int(BI,2)#\n", + "Vo = Resolution*BI #\n", + "print 'For the BI 11100010 output analog voltage is = %0.4f'%Vo,' V '\n", + "\n", + "# For the BI 00101001 output analog voltage is\n", + "BI = '00101001'#\n", + "BI = int(BI,2)#\n", + "Vo = Resolution*BI #\n", + "print 'For the BI 00101001 output analog voltage is = %0.4f'%Vo,' V '\n", + "\n", + "# For the BI 01000110 output analog voltage is\n", + "BI = '01000110'#\n", + "BI = int(BI,2)#\n", + "Vo = Resolution*BI #\n", + "print 'For the BI 01000110 output analog voltage is = %0.3f'%Vo,' V '" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch12.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch12.ipynb new file mode 100644 index 00000000..c39c8ba5 --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch12.ipynb @@ -0,0 +1,229 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 12 Analog To Digital Converter(A/D Converter)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.2 Pg 350" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Normalized step size of A/D converter is = 0.00390625\n", + "Actual step size of A/D converter is = 0.046875\n", + "Normalized maximum quantization level of A/D converter is = 0.9961\n", + "Actual maximum quantization level of A/D converter is = 11.9531\n", + "Normalized peak quantization error of A/D converter is = 0.001953\n", + "Actual peak quantization error of A/D converter is = 0.023438 V \n", + "Percentage of quantization error of A/D converter is = 0.1953\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Determine the following parameter of 8-bit A/D converter a) Normalized step size b) Actual step size c) Normalized maximum quantization level d) Actual maximum quantization e) Normalized peak quantization error f) Actual peak quantization error g) Percentage of quantization error\n", + "N = 8 #\n", + "Vin = 12 #\n", + "\n", + "#a) Normalized step size of A/D converter\n", + "Ns = 2**-N #\n", + "print 'Normalized step size of A/D converter is = %0.8f'%Ns\n", + "\n", + "# b) Actual step size of A/D converter\n", + "As = Vin*Ns #\n", + "print 'Actual step size of A/D converter is = %0.6f'%As\n", + "\n", + "# c) Normalized maximum quantization level of A/D converter\n", + "Qmax = 1-2**-N #\n", + "print 'Normalized maximum quantization level of A/D converter is = %0.4f'%Qmax\n", + "\n", + "# d) Actual maximum quantization level of A/D converter\n", + "QAmax = Qmax*Vin #\n", + "print 'Actual maximum quantization level of A/D converter is = %0.4f'%QAmax\n", + "\n", + "# e) Normalized peak quantization error of A/D converter\n", + "Qp = 2**-(N+1)#\n", + "print 'Normalized peak quantization error of A/D converter is = %0.6f'%Qp\n", + "\n", + "# f) Actual peak quantization error of A/D converter\n", + "Qe = Qp*Vin #\n", + "print 'Actual peak quantization error of A/D converter is = %0.6f'%Qe,' V '#\n", + "\n", + "# g) Percentage of quantization error of A/D converter\n", + "per_Qp = 2**-(N+1)*100 #\n", + "print 'Percentage of quantization error of A/D converter is = %0.4f'%per_Qp" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.3 Pg 351" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "charging time of capacitor is = 128.00 u sec\n", + "the integrator output is = -5.44 V\n", + "the decimal output of a dual slope A/D converter is = 217.60 = 218\n", + "The binary output of a dual slope A/D converter is = 11011010\n" + ] + } + ], + "source": [ + "# to determine the binary output of the 8-bit dual slope A/D converter\n", + "Vin = 8.5 #\n", + "VR = 10 #\n", + "f = 2 # #MHz\n", + "N = 8 #\n", + "C = 0.1*10**-6 #\n", + "R = 2*10**3 #\n", + "\n", + "# the output of integrator is defined as \n", + "# Viao(T1) = -(Vin/R*C)*T1 #\n", + "\n", + "# charging time of capacitor \n", + "T1 = 2**N/f #\n", + "print 'charging time of capacitor is = %0.2f'%T1,' u sec'\n", + "\n", + "# the integrator output\n", + "T1 = T1*10**-6 #\n", + "Viao =-(Vin/(R*C))*T1#\n", + "print 'the integrator output is = %0.2f'%Viao,' V'\n", + "\n", + "# the binary output of a dual slope A/D converter\n", + "Bn = (2**N*Vin)/VR#\n", + "print 'the decimal output of a dual slope A/D converter is = %0.2f'%Bn,' = 218'#\n", + "\n", + "Bn=218#\n", + "Bn = bin(Bn) #\n", + "print 'The binary output of a dual slope A/D converter is =',Bn[2:]" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.4 Pg 352" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resolution of an A/D converter is = 0.0037 V \n" + ] + } + ], + "source": [ + " # to determine the resolution of 12-bit A/D converter\n", + "N =12 #\n", + "Vin = 15 #\n", + "\n", + "# Resolution of an A/D converter\n", + "Resolution = Vin/(2**N-1)#\n", + "print 'Resolution of an A/D converter is = %0.4f'%Resolution,' V '\n", + "#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.5 Pg 352" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output time of a V/T converter is = 0.15 msec\n", + "The duty cycle of V?T converter is = 2.00\n", + "The output voltage of an integrator is is = -0.75 V\n" + ] + } + ], + "source": [ + "# to determine the output time and duty cycle of V/T converter\n", + "Vin = 5 #\n", + "C = 0.1*10**-6 # \n", + "R = 10*10**3 #\n", + "C1 = 100*10**-6 #\n", + "\n", + "# The output time of a V/T converter is given as\n", + "T = (7.5*C1)/(Vin) #\n", + "print 'The output time of a V/T converter is =',T*1000,' msec'\n", + "\n", + "TH = 0.075 #\n", + "TL=TH # # we consider\n", + "# The duty cycle of V?T converter\n", + "D = (TL+TH)/(TH) #\n", + "print 'The duty cycle of V?T converter is = %0.2f'%D\n", + "\n", + "# The output voltage of an integrator is define as\n", + "Vio = -(Vin)/(R*C)*T #\n", + "print 'The output voltage of an integrator is is = %0.2f'%Vio,' V'" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch13.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch13.ipynb new file mode 100644 index 00000000..bf527bf0 --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch13.ipynb @@ -0,0 +1,438 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 13 Waveform Generators" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.1 Pg 378" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the frequency selective element resistor is = 6.50 K ohm \n", + "The feedback resistance is = 188.4 K ohm\n" + ] + } + ], + "source": [ + "from math import sqrt, pi\n", + "from __future__ import division\n", + "# to design RC phase shift oscillator for the oscillation frequency f = 1 KHz\n", + "f =1 # # KHz\n", + "C = 0.01 # # uF\n", + "\n", + "# The oscillation frequency of practical RC phase shift oscillator is defined as\n", + "#w = 1/(sqrt(6)*R*C)#\n", + "\n", + "# gain of practical RC phase shift oscillator is\n", + "#A = R1/R = 29 equation 1\n", + "# the frequency selective element resistor\n", + "#R = 1/(sqrt(6)*w*C)#\n", + "R = 1/(sqrt(6)*2*pi*f*C)#\n", + "print 'the frequency selective element resistor is = %0.2f'%R,' K ohm '\n", + "\n", + "# The feedback resistance\n", + "R1 = 29*R # # from equation 1\n", + "print 'The feedback resistance is = %0.1f'%R1,' K ohm'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.2 Pg 379" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the oscillator frequency of practical RC phase shift oscillator f is = 0.52 KHz \n" + ] + } + ], + "source": [ + "# to determine the oscillaton frequency of the phase shift oscillator\n", + "C = 0.05 # # uF\n", + "R = 2.5 # # K ohm\n", + "\n", + "# the oscillator frequency of practical RC phase shift oscillator f\n", + "f = 1/(2*pi*(sqrt(6)*(R*C)))#\n", + "print 'the oscillator frequency of practical RC phase shift oscillator f is = %0.2f'%f,' KHz '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.3 Pg 380" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the oscillator frequency of practical RC phase shift oscillator f is = 6.63 kHz \n" + ] + } + ], + "source": [ + "# to calculate the frequency of a wein bridge oscillator\n", + "C = 2400*10**-12 # # F\n", + "R = 10*10**3 # # ohm\n", + "\n", + "# the oscillator frequency of practical RC phase shift oscillator f\n", + "f = 1/(2*pi*R*C)/1e3#\n", + "print 'the oscillator frequency of practical RC phase shift oscillator f is = %0.2f'%f,' kHz '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.4 Pg 380" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the resistor R is = 15.9 K ohm \n", + "The resistor R2 value is = 20.00 K ohm \n" + ] + } + ], + "source": [ + "# to design the wien bridge oscillator for the oscillation frequency f = 1 KHz\n", + "f = 1 # # K ohm\n", + "C = 0.01 # # uF\n", + "\n", + "\n", + "# the frequency f is define as\n", + "# f = 1/(2*pi*R*C)#\n", + "\n", + "# the resistor R is\n", + "R = 1/(2*pi*f*C)#\n", + "print 'the resistor R is = %0.1f'%R,' K ohm '\n", + "\n", + "# the loop gain of the wien bridge oscillator is unity which is defined as\n", + "# A = (1+(R2/R1))*(1/3) = 1 #\n", + "# R2/R1 = 2 #\n", + "R1 = 10 # # K ohm we assume\n", + "R2 = 2*R1 #\n", + "print 'The resistor R2 value is = %0.2f'%R2,' K ohm '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.5 Pg 382" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the frequency of wien bridge oscillator f is = 159.155 Hz \n" + ] + } + ], + "source": [ + "# to calculate the frequency of a wein bridge oscillator\n", + "C = 0.05*10**-6 # # F\n", + "R = 20*10**3 # # ohm\n", + "R1 = 10*10**3 # # ohm\n", + "R2 = 20*10**3 # #ohm\n", + "\n", + "# the frequency of wien bridge oscillator f\n", + "f = 1/(2*pi*R*C)#\n", + "print 'the frequency of wien bridge oscillator f is = %0.3f'%f,' Hz '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.6 Pg 382" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The frequency of the astable multivibrator is = 0.87/(C*R)\n" + ] + } + ], + "source": [ + "from sympy import symbols, log, N\n", + "R, C = symbols('R C')\n", + "# Determine the frequency response of the astable multivibrator circuit\n", + "Vsat = 2.5 #\n", + "VT = 0.7 #\n", + "\n", + "# The frequency of the astable multivibrator is\n", + "f = (1/(2*R*C*log((Vsat+VT)/(Vsat-VT))))#\n", + "print 'The frequency of the astable multivibrator is =',N(f,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.7 Pg 383" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of resistance R2 is = 32.86 K ohm \n", + "The value of resistor R is = 10.52 K ohm\n" + ] + } + ], + "source": [ + "from math import log\n", + "# Design astable multivibrator for the frequency f = 10 KHz\n", + "f = 10 # # K ohm\n", + "Vsat = 3 #\n", + "VT = 0.7 #\n", + "\n", + "# The saturation voltage of an astable multivibrator is defined as\n", + "# Vsat = (R1+R2/R1)+VT #\n", + "R1 = 10 # # K ohm we choose\n", + "R2 = ((Vsat/VT)-1)*R1 #\n", + "print 'The value of resistance R2 is = %0.2f'%R2,' K ohm '\n", + "\n", + "# The frequency of an astable multivibrator is defined as\n", + "C = 0.01 # # uF\n", + "# f = (1/(2*R*C*log(1+(2*R1/R2))))#\n", + "\n", + "R = 1/(2*f*C*log(1+2*R1/R2))#\n", + "print 'The value of resistor R is = %0.2f'%R,' K ohm'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.8 Pg 384" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of resistor R is = 200.00 ohm \n" + ] + } + ], + "source": [ + "# to design astable multivibrator \n", + "f = 25*10**3 #\n", + "\n", + "# The output frequency of practical astable multivibrator is defined as\n", + "# f = 1/(2*R*C)#\n", + "C = 0.1*10**-6 # # uF we choose\n", + "R = 1/(2*f*C)#\n", + "print 'The value of resistor R is = %0.2f'%R,'ohm '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.9 Pg 385" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of resistance R is = 579.7 ohm \n", + "The value of resistance R2 is = 9.94 K ohm \n" + ] + } + ], + "source": [ + "# Design a monostable circuit with frequency f = 25 KHz\n", + "f =25*10**3 # # Hz\n", + "\n", + "# The output frequency of monostable multivibrator is defined as \n", + "# f = 1/(0.69*R*C)#\n", + "C = 0.1*10**-6 #\n", + "R = 1/(0.69*f*C)#\n", + "print 'The value of resistance R is = %0.1f'%R,' ohm '\n", + "\n", + "# In the practical monostable multivibrator\n", + "# ln(1+(R2/R1))= 0.69 #\n", + "R1 = 10*10**3 # # we choose\n", + "R2 = R1*(1.99372-1)#\n", + "print 'The value of resistance R2 is = %0.2f'%(R2/1000),' K ohm ' # Round Off Error " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.10 Pg 386" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output of monostable multivibrator is = 6.01 kHz\n" + ] + } + ], + "source": [ + "# Determine the frequency of the monostable multivibrator\n", + "R1 = 5*10**3 #\n", + "R2 =15*10**3 #\n", + "C = 0.01*10**-6 #\n", + "R = 12*10**3 #\n", + "\n", + "# the output of monostable multivibrator is defined as\n", + "f = 1/(R*C*(log(1+(R2/R1))))/1e3 # kHz\n", + "print 'the output of monostable multivibrator is = %0.2f'%f,' kHz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.11 Pg 386" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output of monostable multivibrator is = 4.00 KHz\n" + ] + } + ], + "source": [ + " # Determine the frequency of the monostable multivibrator\n", + "R1 = 5*10**3 #\n", + "R2 =15*10**3 #\n", + "C = 0.01 #\n", + "R = 25 #\n", + "\n", + "# the output of monostable multivibrator is defined as\n", + "f = 1/(R*C)#\n", + "print 'the output of monostable multivibrator is = %0.2f'%f,' KHz'" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch14.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch14.ipynb new file mode 100644 index 00000000..d64bea33 --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch14.ipynb @@ -0,0 +1,636 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 14 Special Function ICs" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.1 Pg 415" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output voltage of the adjustable voltage regulator is = 22.25 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "\n", + "# to determine the regulated voltage \n", + "R1 = 250 # #ohm \n", + "R2 = 2500 # # ohm \n", + "Vref = 2 # #V #reference voltage\n", + "Iadj = 100*10**-6# # A # adjacent current\n", + "\n", + "#the output voltage of the adjustable voltage regulator is defined by\n", + "Vo = (Vref*((R2/R1)+1)+(Iadj*R2)) #\n", + "print 'the output voltage of the adjustable voltage regulator is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.2 Pg 416" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the total power dissipation of the IC is = 25.00 mA \n" + ] + } + ], + "source": [ + "# to determine the current drawn from the dual power supply \n", + "V = 10 # # V\n", + "P = 500 # # mW\n", + "\n", + "# we assume that each power supply provides half power supply to IC\n", + "P1 = (P/2)#\n", + "\n", + "# the total power dissipation of the IC\n", + "# P1 = V*I #\n", + "I = P1/V #\n", + "print 'the total power dissipation of the IC is = %0.2f'%I,' mA '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.3 Pg 416" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output voltage of the adjustable voltage regulator is = 7.50 V \n" + ] + } + ], + "source": [ + "# to determine the output voltage \n", + "R1 = 100*10**3 # #ohm \n", + "R2 = 500*10**3 # # ohm \n", + "Vref = 1.25 # #V #reference voltage\n", + "\n", + "#the output voltage of the adjustable voltage regulator is defined by\n", + "Vo = Vref*(R1+R2)/R1#\n", + "print 'the output voltage of the adjustable voltage regulator is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.4 Pg 417" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output voltage of switching regulator circuit is = 3.50 V \n" + ] + } + ], + "source": [ + "# determine the output voltage of the switching regulator circuit\n", + "d = 0.7 # # duty cycle\n", + "Vin = 5 # # V # input voltage\n", + "\n", + "# The output voltage of switching regulator circuit is given by\n", + "Vo = d*Vin #\n", + "print 'The output voltage of switching regulator circuit is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.5 Pg 417" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output voltage of switching regulator circuit is = 0.96 \n" + ] + } + ], + "source": [ + "# determine the duty cycle of the switching regulator circuit\n", + "Vo = 4.8 # # V # output voltage\n", + "Vin = 5 # # V # input voltage\n", + "\n", + "# The output voltage of switching regulator circuit is given by\n", + "# Vo = d*Vin #\n", + "\n", + "# Duty cycle is given as\n", + "d =Vo/Vin #\n", + "print 'The output voltage of switching regulator circuit is = %0.2f'%d,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.6 Pg 418" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output voltage of switching regulator circuit is = 0.50 \n" + ] + } + ], + "source": [ + "# determine the duty cycle of the switching regulator circuit\n", + "T =120 # #msec # total pulse time\n", + "# T = ton + toff #\n", + "ton = T/2 #\n", + "\n", + "# The duty cycle of switching regulator circuit is given by\n", + "d = ton/T#\n", + "print 'The output voltage of switching regulator circuit is = %0.2f'%d,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.7 Pg 418" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output voltage of switching regulator circuit is = 0.67 \n" + ] + } + ], + "source": [ + "# determine the duty cycle of the switching regulator circuit\n", + "ton = 12 # #msec # on time of pulse\n", + "# ton = 2*toff # given\n", + "# T = ton + toff #\n", + "toff = ton/2 #\n", + "T = ton+toff # # total time\n", + "\n", + "# The duty cycle of switching regulator circuit is given by\n", + "d = ton/T#\n", + "print 'The output voltage of switching regulator circuit is = %0.2f'%d,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.8 Pg 419" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The emitter bias voltage is = 3.80 V \n", + "The output voltage of the IC LM380 is = 7.90 V \n" + ] + } + ], + "source": [ + " # determine the output voltage of the audio power amplifier IC LM380\n", + "Vcc = 12 # # V\n", + "Ic3 = 12*10**-6 # # A # collector current of the transistor Q3\n", + "Ic4 = 12*10**-6 # # A # collector current of the transistor Q4\n", + "R11 = 25*10**3 # # ohm\n", + "R12 = 25*10**3 # # ohm\n", + "\n", + "# the collector current of Q3 is defined as\n", + " # Ic3 = (Vcc-3*Veb)/(R11+R12)#\n", + "Veb = (Vcc-(R11+R12)*Ic3)/3 #\n", + "print 'The emitter bias voltage is = %0.2f'%Veb,' V '\n", + "\n", + "# the output voltage of the IC LM380\n", + "Vo = (1/2)*Vcc+(1/2)*Veb#\n", + "print 'The output voltage of the IC LM380 is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.9 Pg 420" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The emitter bias voltage is = 3.33 V \n", + "The output voltage of the IC LM380 is = 6.67 V \n" + ] + } + ], + "source": [ + "# determine the output voltage of the audio power amplifier IC LM380\n", + "Vcc = 10 # # V\n", + "Ic3 = 0.01*10**-6 # # A # collector current of the transistor Q3\n", + "Ic4 = 0.01*10**-6 # # A # collector current of the transistor Q4\n", + "R11 = 25*10**3 # # ohm\n", + "R12 = 25*10**3 # # ohm\n", + "\n", + "# the collector current of Q3 is defined as\n", + " # Ic3 = (Vcc-3*Veb)/(R11+R12)#\n", + "Veb = (Vcc-(R11+R12)*Ic3)/3 #\n", + "print 'The emitter bias voltage is = %0.2f'%Veb,' V '\n", + "\n", + "# the output voltage of the IC LM380\n", + "Vo = (1/2)*Vcc+(1/2)*Veb#\n", + "print 'The output voltage of the IC LM380 is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.10 Pg 421" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The emitter resistor of Q3 is = 52.00 ohm ( at temperature 25 degree celsius) \n", + "The trans conductance of transistor is = 38.5 mA/V \n", + "The base emitter resistor rbe is = 1.30 K ohm \n", + "The emitter capacitor Ce = 7.65 pF \n", + "The value of resistance RL is = 264.55 ohm \n", + "The pole frequency fa is = 601.91 M Hz \n", + "The pole frequency fb is = 1073.74 M Hz \n", + "The pole frequency fc is = 3060.67 M Hz \n", + "Hence fa is a dominant pole frequency \n" + ] + } + ], + "source": [ + "from numpy import inf\n", + "from math import sqrt, pi\n", + "# Design a video amplifier of IC 1550 circuit\n", + "Vcc = 12 # # V\n", + "Av = -10 #\n", + "Vagc = 0 # # at bandwidth of 20 MHz\n", + "hfe = 50 # # forward emitter parameter\n", + "rbb = 25 # # ohm # base resistor\n", + "Cs = 1*10**-12 # # F # source capacitor\n", + "Cl = 1*10**-12 # # F # load capacitor\n", + "Ie1 = 1*10**-3 # # A # emitter current of Q1\n", + "f = 1000*10**6 # # Hz\n", + "fT = 800*10**6 # # Hz\n", + "Vt = 52*10**-3 #\n", + "Vt1 = 0.026 #\n", + "\n", + "# When Vagc =0 the transistor Q2 is cut-off and the collector current of transistor Q2 flow through the transistor Q3\n", + "# i.e Ic1=Ie1=Ie3\n", + "Ie3 = 1*10**-3 # # A # emitter current of Q3\n", + "Ic1 = 1*10**-3 # # A # collector current of the transistor Q1\n", + "\n", + "# it indicates that the emitter current of Q2 is zero Ie2 = 0 then the emitter resistor of Q2 is infinite\n", + "\n", + "re2 = inf #\n", + "\n", + "# emitter resistor of Q3 \n", + "re3 = (Vt/Ie1)#\n", + "print 'The emitter resistor of Q3 is = %0.2f'%re3,' ohm ( at temperature 25 degree celsius) '\n", + "\n", + "# the trans conductance of transistor is\n", + "gm = (Ie1/Vt1)#\n", + "print 'The trans conductance of transistor is = %0.1f'%(gm*1000),' mA/V ' # Round Off Error\n", + "\n", + "# the base emitter resistor rbe\n", + "rbe = (hfe/gm)#\n", + "print 'The base emitter resistor rbe is = %0.2f'%(rbe/1000),' K ohm ' # Round Off Error\n", + "\n", + "# the emitter capacitor Ce \n", + "\n", + "Ce = (gm/(2*pi*fT))#\n", + "print 'The emitter capacitor Ce = %0.2f'%(Ce*1e12),' pF ' # Round Off Error\n", + "\n", + "# the voltage gain of video amplifier is\n", + "# Av = (Vo/Vin) #\n", + "# Av = -((alpha3*gm)/(rbb*re3)*((1/rbb)+(1/rbe)+sCe)*((1/re2)+(1/re3)+sC3)*((1/Rl)+(s(Cs+Cl)))) \n", + " # At Avgc = 0 i.e s=0 in the above Av equation\n", + "alpha3 = 1 #\n", + "s = 0 #\n", + "# Rl = -((alpha3*gm)/(rbb*re3)*(((1/rbb)+(1/rbe))*((1/re2)+(1/re3))*(Av)))# \n", + "\n", + "# After solving above equation for Rl We get Rl Equation as\n", + "Rl = 10/(37.8*10**-3)#\n", + "print 'The value of resistance RL is = %0.2f'%Rl,' ohm '\n", + "\n", + "# there are three poles present in the transfer function of video amplifier each pole generate one 3-db frequency \n", + "Rl = 675 #\n", + "# fa = 1/(2*pi*Rl*(Cs+Cl))#\n", + "# after putting value of Rl ,Cs and Cl we get\n", + "fa = 1/(2*3.14*264.55*1*10**-12)#\n", + "print 'The pole frequency fa is = %0.2f'%(fa*10**-3/1000),' M Hz '# Round Off Error\n", + "\n", + "\n", + "#fb = 1/(2*pi*Ce*((rbb*rbe)/(rbb+rbe)))#\n", + "# after putting value of Ce rbb and rbe we get\n", + "fb = 1/(2*pi*6.05*10**-12*24.5)#\n", + "print 'The pole frequency fb is = %0.2f'%(fb*10**-3/1000),' M Hz '\n", + "\n", + "fc = 1/(2*pi*Cs*re3)#\n", + "print 'The pole frequency fc is = %0.2f'%(fc*10**-3/1000),' M Hz '\n", + "\n", + "print 'Hence fa is a dominant pole frequency '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.11 Pg 423" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The emitter resistor of Q3 is = 52.00 ohm \n", + "The trans conductance of transistor is = 38.5 mA/V \n", + "The base emitter resistor rbe is = 1.3 kohm \n", + "The emitter capacitor is = 6.12 pF \n", + "The value of resistance RL is = 5.00 ohm \n", + "The pole frequency fa is = 600.58 MHz \n", + "The pole frequency fb is = 1060.00 MHz \n", + "The pole frequency fc is = 3060.67 MHz \n", + "Hence fa is a dominant pole frequency \n" + ] + } + ], + "source": [ + "# Design a video amplifier of IC 1550 circuit\n", + "Vcc = 12 # # V\n", + "Av = -10 #\n", + "Vagc = 0 # # at bandwidth of 20 MHz\n", + "hfe = 50 # # forward emitter parameter\n", + "rbb = 25 # # ohm # base resistor\n", + "Cs = 1*10**-12 # # F # source capacitor\n", + "Cl = 1*10**-12 # # F # load capacitor\n", + "Ie1 = 1*10**-3 # # A # emitter current of Q1\n", + "f = 1000*10**6 # # Hz\n", + "Vt = 52*10**-3 #\n", + "Vt1 = 0.026 #\n", + "\n", + "# When Vagc =0 the transistor Q2 is cut-off and the collector current of transistor Q2 flow through the transistor Q3\n", + "# i.e Ic1=Ie1=Ie3\n", + "Ie3 = 1*10**-3 # # A # emitter current of Q3\n", + "Ic1 = 1*10**-3 # # A # collector current of the transistor Q1\n", + "\n", + "# it indicates that the emitter current of Q2 is zero Ie2 = 0 then the emitter resistor of Q2 is infinite\n", + "re2 = inf #\n", + "\n", + "# emitter resistor of Q3 \n", + "re3 = (Vt/Ie1)#\n", + "print 'The emitter resistor of Q3 is = %0.2f'%re3,' ohm '\n", + "\n", + "# the trans conductance of transistor is\n", + "gm = (Ie1/Vt1)#\n", + "print 'The trans conductance of transistor is = %0.1f'%(gm*1e3),' mA/V '\n", + "\n", + "# the base emitter resistor rbe\n", + "rbe = (hfe/gm)#\n", + "print 'The base emitter resistor rbe is = %0.1f'%(rbe/1e3),' kohm '\n", + "\n", + "# the emitter capacitor Ce \n", + "Ce = (gm/(2*pi*f))#\n", + "print 'The emitter capacitor is = %0.2f'%(Ce*1e12),' pF '\n", + "\n", + "# the voltage gain of video amplifier is\n", + "# Av = (Vo/Vin) #\n", + "# Av = -((alpha3*gm)/(rbb*re3)*((1/rbb)+(1/rbe)+sCe)*((1/re2)+(1/re3)+sC3)*((1/Rl)+(s(Cs+Cl)))) \n", + " # At Avgc = 0 i.e s=0 in the above Av equation\n", + "alpha3 = 1 #\n", + "s = 0 #\n", + "Av =-10 #\n", + "Rl = -((alpha3*gm)/((rbb*re3)*(((1/rbb)+(1/rbe))*((1/re2)+(1/re3))*(Av))))# \n", + "Rl = (1/Rl)#\n", + "print 'The value of resistance RL is = %0.2f'%Rl,' ohm '\n", + "\n", + "# there are three poles present in the transfer function of video amplifier each pole generate one 3-db frequency \n", + "Rl = 265\n", + "fa = 1/(2*pi*Rl*(Cs))/1e6#\n", + "print 'The pole frequency fa is = %0.2f'%fa,'MHz '\n", + "\n", + "\n", + "fb = 1/(2*pi*Ce*((rbb*rbe)/(rbb+rbe)))/1e6\n", + "print 'The pole frequency fb is = %0.2f'%fb,'MHz '\n", + "\n", + "fc = 1/(2*pi*Cs*re3)/1e6\n", + "print 'The pole frequency fc is = %0.2f'%fc,'MHz '\n", + "\n", + "print 'Hence fa is a dominant pole frequency '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.12 Pg 425" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input current is = 0.50 mA \n", + "The output of an op-amp is = 27.50 V \n" + ] + } + ], + "source": [ + "# Determine the output voltage of an isolation amplifier IC ISO100\n", + "Vin = 5.0 # # V\n", + "Rin = 10*10**3 # \n", + "Rf = 55*10**3 # # ohm # feedback resistance\n", + "\n", + "# the input voltage of an amplifier 1\n", + "# Vin = Rin*Iin\n", + "Iin = Vin/Rin # \n", + "print 'The input current is = %0.2f'%(Iin*1e3),'mA '\n", + "\n", + "# In isolation amplifier ISO 100 the input current Iin is equal to the output current Iout , but both are opposite in direction\n", + "# Iin = -Iout\n", + "# the output of an op-amp\n", + "# Vo = -Rf*Iout\n", + "Vo = Rf*Iin#\n", + "print 'The output of an op-amp is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.13 Pg 426" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input current is = 12 mA \n", + "The output of an op-amp is = 204 V \n" + ] + } + ], + "source": [ + "# Determine the output voltage of an isolation amplifier IC ISO100\n", + "Vin = 12.0 # # V\n", + "Rin = 1*10**3 # \n", + "Rf = 17*10**3 # # ohm # feedback resistance\n", + "\n", + "# the input voltage of an amplifier 1\n", + "# Vin = Rin*Iin\n", + "Iin = Vin/Rin # \n", + "print 'The input current is = %0.f'%(Iin*1e3),'mA '\n", + "\n", + "# In isolation amplifier ISO 100 the input current Iin is equal to the output current Iout , but both are opposite in direction\n", + "# Iin = -Iout\n", + "# the output of an op-amp\n", + "# Vo = -Rf*Iout\n", + "Vo = Rf*Iin#\n", + "print 'The output of an op-amp is = %0.f'%Vo,' V '" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch3.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch3.ipynb new file mode 100644 index 00000000..7a4856f8 --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch3.ipynb @@ -0,0 +1,344 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3 Current Voltage Sources and Differential Amplifiers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.1 Pg 53" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The collector current of difference amplifier Ic1 = Ic2 = 0.50 mA \n", + "The collector voltages of transistors Q1 and Q2 are Vc1 = Vc2 = 5.00 volt \n", + "For Ve = -0.7 Volt the collector - emitter voltage Vce1 = 5.70 Volt\n", + "For Ve = 4.3 Volt the collector - emitter voltage Vce1 = 0.70 Volt\n", + "For Ve = -5.7 Volt the collector - emitter voltage Vce1 = 10.70 Volt\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Determine the collector current Ic1 and collector-emitter voltage Vce1 for the difference amplifier circuit\n", + "\n", + "V1 = 0 # # volt\n", + "V2 = -5 # #volt\n", + "Vcm = 5 # #volt\n", + "Vcc = 10# #volt\n", + "Vee = -10 # #volt\n", + "Ie = 1 # #mA\n", + "Rc = 10 # #kilo ohm\n", + "\n", + "# Transistor parameters\n", + "# base current are negligible\n", + "Vbe = 0.7 # # volt\n", + "\n", + "# The collector current of difference amplifier is\n", + "Ic1 = Ie/2 # \n", + "print 'The collector current of difference amplifier Ic1 = Ic2 = %0.2f'%Ic1,' mA '\n", + "\n", + "# The collector voltages of transistors Q1 and Q2 are expressed as\n", + "\n", + "Vc1 = Vcc-Ic1*Rc #\n", + "print 'The collector voltages of transistors Q1 and Q2 are Vc1 = Vc2 = %0.2f'%Vc1,' volt '\n", + "\n", + "# We know common mode voltage (Vcm) , from this the emitter voltage can be identified as follows\n", + "# For the common mode voltage Vcm = 0 V , the emitter voltage is Ve = -0.7 V\n", + "# For the common mode voltage Vcm = 5 V , the emitter voltage is Ve = 4.3 V\n", + "# For the common mode voltage Vcm = -5 V , the emitter voltage is Ve = -5.7 V\n", + "\n", + "# For the different emitter voltages the collector-emitter voltage can be calculated as\n", + "\n", + "Ve = -0.7 # # volt\n", + "Vce1 = Vc1-Ve#\n", + "print 'For Ve = -0.7 Volt the collector - emitter voltage Vce1 = %0.2f'%Vce1,' Volt'\n", + "\n", + "Ve = 4.3 # # volt\n", + "Vce1 = Vc1-Ve#\n", + "print 'For Ve = 4.3 Volt the collector - emitter voltage Vce1 = %0.2f'%Vce1,' Volt'\n", + "\n", + "Ve = -5.7 # # volt\n", + "Vce1 = Vc1-Ve#\n", + "print 'For Ve = -5.7 Volt the collector - emitter voltage Vce1 = %0.2f'%Vce1,' Volt'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.2 Pg 54" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The differential mode gain Ad = 184.6\n", + " The common mode gain Acm = -0.237\n" + ] + } + ], + "source": [ + "# To determine the difference-mode and common-mode gain of the difference amplifier\n", + "\n", + "Vcc = 10 # # volt\n", + "Vee = -10 # #volt\n", + "Iq = 0.8 # #mA\n", + "Ie = 0.8 # #mA\n", + "Rc = 12 # #kilo-Ohm\n", + "Vt = 0.026 # # volt\n", + "\n", + "# Transistor parameter\n", + "beta = 100 #\n", + "Rs = 0 # #Ohm\n", + "Ro = 25 # #kilo-Ohm \n", + "# The differential mode gain Ad\n", + "gm = (Ie/ 2*Vt) #\n", + "# Ad = (gm*r*Rc/r+Rc) # # where r is r-pi\n", + "# For Rb=0 , the differential mode gain is\n", + "\n", + "Ad = (Ie/(2*Vt))*Rc#\n", + "#But\n", + "print ' The differential mode gain Ad = %0.1f'%Ad\n", + "\n", + "#The common mode gain Acm\n", + "# Acm = - (gm*Rc/1+2*gm*Re+2*Re/r)\n", + "Acm =-(Ad/(1+(((1+beta)*Ie*Ro)/(beta*Vt))))\n", + "print ' The common mode gain Acm = %0.3f'%Acm" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.3 Pg 56" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output of a difference amplifier is Vo = -47.40 sinwt uV \n" + ] + } + ], + "source": [ + "# To find the output of a difference amplifier when only common mode signal is applied\n", + "\n", + "# V1 = V2 = Vcm = 200*sin(wt) # # micro volt (uV)\n", + "Acm = -0.237 #\n", + "\n", + "# When the common mode input signal is applied to the difference amplifier , the difference mode gain is zero\n", + "Vcm = 200 #\n", + "Vo = Acm*Vcm #\n", + "print 'The output of a difference amplifier is Vo = %0.2f'%Vo,'sinwt uV ' # multiply by sinwt because it is in Vcm" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.4 Pg 56" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The differential mode gain Ad = 184.6\n", + "The common mode gain Acm = -0.237\n", + "The CMRR of difference amplifier is = 389\n", + "In decibel CMRR is = 51.80\n" + ] + } + ], + "source": [ + "from math import log10\n", + "#Determine the common mode rejection ratio(CMRR) of the difference amplifier\n", + "\n", + "Vcc = 10 # # volt\n", + "Vee = -10 # #volt\n", + "Iq = 0.8 # #mA\n", + "Ie = 0.8 # #mA\n", + "Rc = 12 # #kilo-Ohm\n", + "Vt = 0.026 # # volt\n", + "\n", + "# Transistor parameter\n", + "beta = 100 #\n", + "Rs = 0 # #Ohm\n", + "Ro = 25 # #kilo-Ohm\n", + " \n", + "# The differential mode gain Ad\n", + "gm = (Ie/ 2*Vt) #\n", + "# Ad = (gm*r*Rc/r+Rc) # # where r is r-pi\n", + "# For Rb=0 , the differential mode gain is\n", + "\n", + "Ad = (Ie/(2*Vt))*Rc#\n", + "#But\n", + "print 'The differential mode gain Ad = %0.1f'%Ad\n", + "\n", + "#The common mode gain Acm\n", + "# Acm = - (gm*Rc/1+2*gm*Re+2*Re/r)\n", + "Acm =-(Ad/(1+(((1+beta)*Ie*Ro)/(beta*Vt))))\n", + "print 'The common mode gain Acm = %0.3f'%Acm\n", + "\n", + "# The CMRR of difference amplifier is given as\n", + "Ad = Ad/2 #\n", + "CMRR = abs(Ad/Acm)\n", + "print 'The CMRR of difference amplifier is = %0.f'%CMRR\n", + "\n", + "# In decibel it can be expressed as\n", + "CMRRdb = 20*log10(CMRR)\n", + "print 'In decibel CMRR is = %0.2f'%CMRRdb" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.5 Pg 58" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The CMRR of difference amplifier is = 3.16e+04\n", + " The value of resistance RE is = 2.04 Mohm \n" + ] + } + ], + "source": [ + "# To determine emitter resistance of the difference amplifier\n", + "\n", + "Vcc = 10 # # volt\n", + "Vee = -10 # #volt\n", + "Iq = 0.8 # #mA\n", + "Ie = 0.8 # #mA\n", + "CMRRdb = 90 # #dB\n", + "Vt = 0.026 #\n", + "\n", + "# Transistor parameter\n", + "beta = 100 #\n", + "\n", + "# CMRR = abs(Ad/Acm)\n", + "# the CMRR of the difference amplifier is defined as\n", + "#CMRR = ((1/2)*(1+((1+beta)*Ie*Re)/beta*Vt))\n", + "\n", + "# CMRRdb = 20*log10(CMRR)\n", + "CMRR = 10**(CMRRdb/20)\n", + "print ' The CMRR of difference amplifier is = %0.2e'%CMRR\n", + "\n", + "# The resistance RE is calculated as\n", + "\n", + "RE = (((2*CMRR)-1)/((1+beta)*Ie))*(beta*Vt)/1e3\n", + "print ' The value of resistance RE is = %0.2f'%RE,' Mohm '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.6 Pg 59" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The differential mode gain Ad is = 321\n" + ] + } + ], + "source": [ + "# determine the differential mode gain when load resistance RL = 100 k ohm\n", + "\n", + "RL = 100*10**3 # # k ohm # load resistance\n", + "IE = 0.20*10**-3 # # mA # biasing current\n", + "VA = 100 # # V # early voltage\n", + "VT = 0.026 # # threshold volt\n", + "\n", + "# the differential gain of differential amplifier with an active load circuit\n", + "#Ad = Vo/Vd = gm(ro2 || ro4 || RL )\n", + "ro2 = (2*VA)/IE#\n", + "ro4 = ro2 #\n", + "gm = IE/(2*VT) #\n", + "\n", + "Ad = gm/((1/ro2)+(1/ro4)+(1/RL))\n", + "print ' The differential mode gain Ad is = %0.f'%Ad" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch4.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch4.ipynb new file mode 100644 index 00000000..d9148952 --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch4.ipynb @@ -0,0 +1,393 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4 Operational Amplifier" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.1 Pg 79" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " closed loop gain of an op-amp is = 35.00\n", + " the input impedance Zin = 10.00 kohm \n", + " the output impedance Z0 = 0.020 ohm \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# For an op-amp circuit find a) closed loop gain Acl b) input impedance Zin c) output impedance Zo\n", + "ro = 85 # # ohm\n", + "A = 150*10**3 # # ohm\n", + "R2 = 350*10**3 # # ohm # Feedback resistance\n", + "R1 = 10*10**3 # # ohm # Input resistance\n", + "\n", + "# a) closed loop gain\n", + "# ACL = abs(Vo/Vin) = abs(R2/R1)\n", + "ACL = abs(R2/R1) #\n", + "print ' closed loop gain of an op-amp is = %0.2f'%ACL# # 1/beta = ACL\n", + "beta = (1/ACL) #\n", + "\n", + "# b) the input impedance Zin\n", + "Zin = R1 #\n", + "print ' the input impedance Zin = %0.2f'%(Zin/1e3),'kohm '#\n", + "\n", + "# c0 the output impedance Z0\n", + "Z0 = (ro)/(1+(beta*A))#\n", + "print ' the output impedance Z0 = %0.3f'%Z0,' ohm '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.2 Pg 80" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The difference voltage is = 10.00 V \n", + " The open loop gain is = 2.00 \n" + ] + } + ], + "source": [ + "# Determine the differece voltage and open loop gain of an op-amp\n", + "V1 = -5 # # volt # input voltage\n", + "V2 = 5 # # volt\n", + "Vo = 20 # #volt # output voltage\n", + "\n", + "# the difference voltage is given by \n", + "Vd = V2-V1 #\n", + "print ' The difference voltage is = %0.2f'%Vd,' V '\n", + "\n", + "# open loop gain \n", + "A = (Vo/Vd)#\n", + "print ' The open loop gain is = %0.2f'%A,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.3 Pg 80" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The difference voltage is = 5.00 V \n", + " The open loop gain is = 4.00 \n" + ] + } + ], + "source": [ + "# Determine the differece voltage and open loop gain of an op-amp\n", + "V1 = -5 # # volt # input voltage\n", + "V2 = 0 # # volt # GND\n", + "Vo = 20 # #volt # output voltage\n", + "\n", + "# the difference voltage is given by \n", + "Vd = V2-V1 #\n", + "print ' The difference voltage is = %0.2f'%Vd,' V '\n", + "\n", + "# open loop gain \n", + "A = (Vo/Vd)#\n", + "print ' The open loop gain is = %0.2f'%A,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.4 Pg 81" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The difference voltage is = 5.00 V \n", + " The open loop gain is = 4.00 \n" + ] + } + ], + "source": [ + "# Determine the differece voltage and open loop gain of an op-amp\n", + "V1 = 0 # # volt # input voltage # GND\n", + "V2 = 5 # # volt \n", + "Vo = 20 # #volt # output voltage\n", + "\n", + "# the difference voltage is given by \n", + "Vd = V2-V1 #\n", + "print ' The difference voltage is = %0.2f'%Vd,' V '\n", + "\n", + "# open loop gain \n", + "A = (Vo/Vd)#\n", + "print ' The open loop gain is = %0.2f'%A,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.5 Pg 81" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The difference voltage is = -10.00 V \n", + " The open loop gain is = 2.00 \n" + ] + } + ], + "source": [ + "# Determine the differece voltage and open loop gain of an op-amp\n", + "V1 = 5 # # volt # input voltage # GND\n", + "V2 = -5 # # volt \n", + "Vo = -20 # #volt # output voltage\n", + "\n", + "# the difference voltage is given by \n", + "Vd = V2-V1 #\n", + "print ' The difference voltage is = %0.2f'%Vd,' V '\n", + "\n", + "# open loop gain \n", + "A = (Vo/Vd)#\n", + "print ' The open loop gain is = %0.2f'%A,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.6 Pg 82" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Closed loop gain of an inverting op-amp is = -2.50 \n", + "The |Ac| Closed loop gain of an inverting op-amp is = 2.50 \n", + "The output voltage of an inverting op-amp is = -25.00 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# To find closed loop gain and output voltage Vo of an inverting op-amp\n", + "R1 = 10 # #kilo ohm # input resistance\n", + "R2 = 25 # # kilo ohm # feedback resistance\n", + "Vin = 10 # #volt # input voltage\n", + "\n", + "# Closed loop gain of an inverting op-amp\n", + "Ac = -(R2/R1) #\n", + "print 'The Closed loop gain of an inverting op-amp is = %0.2f'%Ac,' '\n", + "Ac = abs(Ac)#\n", + "print 'The |Ac| Closed loop gain of an inverting op-amp is = %0.2f'%Ac,' '\n", + "\n", + "# the output voltage of an inverting op-amp\n", + "Vo = -(R2/R1)*Vin #\n", + "print 'The output voltage of an inverting op-amp is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.7 Pg 82" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The Closed loop gain of an non-inverting op-amp is = 3.50 \n", + " The output voltage of an non-inverting op-amp is = 35.00 V \n" + ] + } + ], + "source": [ + "# To find closed loop gain and output voltage Vo of an non-inverting op-amp\n", + "R1 = 10 # #kilo ohm # input resistance\n", + "R2 = 25 # # kilo ohm # feedback resistance\n", + "Vin = 10 # #volt # input voltage\n", + "\n", + "# Closed loop gain of an non-inverting op-amp\n", + "Ac = 1+(R2/R1) #\n", + "Ac = abs(Ac)#\n", + "print ' The Closed loop gain of an non-inverting op-amp is = %0.2f'%Ac,' '\n", + "\n", + "# the output voltage of an inverting op-amp\n", + "Vo = (1+R2/R1)*Vin #\n", + "print ' The output voltage of an non-inverting op-amp is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.8 Pg 83" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The closed loop gain of differntial op-amp is = 2.50 \n", + "The output voltage of an non-inverting op-amp is= 50.00 V \n" + ] + } + ], + "source": [ + "# to find out closed loop gain and output voltage Vo\n", + "R1 = 10 # #kilo ohm # input resistance\n", + "R3 = 10 # #kilo ohm # input resistance\n", + "R2 = 25 # # kilo ohm # feedback resistance\n", + "R4 = 25 # # kilo ohm # feedback resistance\n", + "Vin2 = 10 # #volt # input voltage\n", + "Vin1 = -10 # #volt # input voltage\n", + "\n", + "# closed loop gain of differntial op-amp is given by\n", + "Ac = (R2/R1) #\n", + "Ac = abs(Ac)# \n", + "print 'The closed loop gain of differntial op-amp is = %0.2f'%Ac,' '\n", + "\n", + "# the output voltage of an non-inverting op-amp is given by\n", + "Vo = (R2/R1)*(Vin2-Vin1) #\n", + "print 'The output voltage of an non-inverting op-amp is= %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.9 Pg 84" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The upper voltage is = 2.86 V \n", + " The lower voltage is = -2.86 V \n" + ] + } + ], + "source": [ + "# Determine the non-inverting input voltage\n", + "R1 = 10 # #kilo ohm # input resistance\n", + "R2 = 25 # #kilo ohm # feedback resistance\n", + "Voh = 10 # # volt #output voltage\n", + "Vol = -10 # # volt # output voltage\n", + "\n", + "# upper voltage\n", + "V = (R1/(R1+R2)*Voh) #\n", + "print ' The upper voltage is = %0.2f'%V,' V '\n", + "\n", + "# Lower voltage\n", + "V = (R1/(R1+R2)*Vol) #\n", + "print ' The lower voltage is = %0.2f'%V,' V '" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch5.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch5.ipynb new file mode 100644 index 00000000..648c0739 --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch5.ipynb @@ -0,0 +1,1582 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5 Characteristic of Operational Amplifier" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.1 Pg 110" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " the output voltage (Vo) of an op-amp circuit due to input offset voltage (Vos) is = 404.00 mV \n", + " the output voltage (Vo) of an op-amp circuit due to input offset current (Ios) is = 75.00 mV \n", + " the total offset voltage (Vo) of an op-amp circuit is = 479.00 mV \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# find the total offset voltage of feedback op-amp\n", + "\n", + "Vos = 4 # #mV # input offset volt\n", + "Ios = 150*10**-3 # # input offset current\n", + "R1 = 5 # #kilo ohm # input resistance\n", + "R2 = 500 # #kilo ohm # feedback resistance\n", + "\n", + "# the output voltage (Vo) of an op-amp circuit due to input offset voltage (Vos) is\n", + "Vo1 = ((R1+R2)/(R1)*Vos) #\n", + "print ' the output voltage (Vo) of an op-amp circuit due to input offset voltage (Vos) is = %0.2f'%Vo1,' mV '#\n", + "\n", + "# the output voltage (Vo) of an op-amp circuit due to input offset current (Ios) is\n", + "Vo2 = R2*Ios #\n", + "print ' the output voltage (Vo) of an op-amp circuit due to input offset current (Ios) is = %0.2f'%Vo2,' mV '#\n", + "\n", + "# the total offset voltage is\n", + "Vo = Vo1+Vo2 #\n", + "print ' the total offset voltage (Vo) of an op-amp circuit is = %0.2f'%Vo,' mV '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.2 Pg 111" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " the output voltage (Vo) of an op-amp circuit due to input offset voltage (Vos) is = 52.00 mV \n", + " the output voltage (Vo) of an op-amp circuit due to input offset current (Ios) is = 5.00 mV \n", + " the total offset voltage (Vo) of an op-amp circuit is = 57.00 mV \n" + ] + } + ], + "source": [ + "# find the total offset voltage of feedback op-amp\n", + "\n", + "Vos = 2 # #mV # input offset volt\n", + "Ios = 20*10**-3 # # input offset current\n", + "R1 = 10 # #kilo ohm # input resistance\n", + "R2 = 250 # #kilo ohm # feedback resistance\n", + "\n", + "# the output voltage (Vo) of an op-amp circuit due to input offset voltage (Vos) is\n", + "Vo1 = ((R1+R2)/(R1)*Vos) #\n", + "print ' the output voltage (Vo) of an op-amp circuit due to input offset voltage (Vos) is = %0.2f'%Vo1,' mV '#\n", + "\n", + "# the output voltage (Vo) of an op-amp circuit due to input offset current (Ios) is\n", + "Vo2 = R2*Ios #\n", + "print ' the output voltage (Vo) of an op-amp circuit due to input offset current (Ios) is = %0.2f'%Vo2,' mV '#\n", + "\n", + "# the total offset voltage is\n", + "Vo = Vo1+Vo2 #\n", + "print ' the total offset voltage (Vo) of an op-amp circuit is = %0.2f'%Vo,' mV '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.3 Pg 111" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the input offset voltage (Vos) of an op-amp circuit is = 1.187 mV \n" + ] + } + ], + "source": [ + "# find the input offset voltage of an op-amp circuit\n", + "\n", + "Vo = 90.2 # #mV # output voltage\n", + "R1 = 2 # #kilo ohm # input resistence\n", + "R2 = 150 # #kilo ohm # feedback resistence\n", + "\n", + "# the input offset voltage (Vos) of an op-amp circuit is defined as\n", + "Vos = ((R1)/(R1+R2)*Vo) #\n", + "print 'the input offset voltage (Vos) of an op-amp circuit is = %0.3f'%Vos,' mV '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.4 Pg 112" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output voltage due to the input offset voltage is = 36.00 mV \n" + ] + } + ], + "source": [ + "# find the output voltage of an op-amp circuit\n", + "\n", + "Vos = 1 # #mV # input offset volt\n", + "R1 = 10 # #kilo ohm # input resistance\n", + "R2 = 350 # #kilo ohm # feedback resistance\n", + "\n", + "# the output voltage due to the input offset voltage of the op-amp circuit is defined by\n", + "Vo1 = ((R1+R2)/(R1)*Vos) #\n", + "print 'the output voltage due to the input offset voltage is = %0.2f'%Vo1,' mV '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.5 Pg 113" + ] + }, + { + "cell_type": "code", + "execution_count": 46, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output voltage of the circuit due to bias current is = 0.11 V \n", + "Bias compensated resistor is = 9.09 kilo ohm \n", + "Bias compensated output voltage is = 0.01 V \n" + ] + } + ], + "source": [ + "# Determine the bias current effect with and without current compensation method\n", + "\n", + "R1 = 10 # #kilo ohm\n", + "R2 = 100 # #kilo ohm\n", + "Ib1 = 1.1*10**-3 #\n", + "Ib2 = 1*10**-3 # \n", + "# the output voltage of the circuit due to bias current is\n", + "Vo = Ib1*R2 #\n", + "print 'the output voltage of the circuit due to bias current is = %0.2f'%Vo,' V '#\n", + "\n", + "#Bias compensated resistor is given by\n", + "R3 = (R1*R2)/(R1+R2) #\n", + "print 'Bias compensated resistor is = %0.2f'%R3,' kilo ohm '#\n", + "\n", + "#Bias compensated output voltage is given by\n", + "Vo = R2*(Ib1-Ib2)#\n", + "print 'Bias compensated output voltage is = %0.2f'%Vo,' V '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.6 Pg 113" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output voltage (Vo) of an op-amp circuit due to input offset current (Ios) is = 80 nA \n" + ] + } + ], + "source": [ + "# find the input offset current of an op-amp circuit\n", + "\n", + "Vo = 12*10**-3# # V # output voltage\n", + "R1 = 2*10**3 # # ohm # input resistence\n", + "R2 = 150*10**3# # ohm # feedback resistence\n", + "\n", + "# the output voltage (Vo) of an op-amp circuit due to input offset current (Ios) is\n", + "# Vo = R2*Ios #\n", + "Ios = Vo/R2 *1e9 # nA\n", + "print 'the output voltage (Vo) of an op-amp circuit due to input offset current (Ios) is = %0.f'%Ios,'nA '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.7 Pg 114" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current in the inverting input terminal is = 27.50 nA \n", + "The current in the non-inverting input terminal is= 32.50 nA \n" + ] + } + ], + "source": [ + "# Determine the bias current of inverting and non-inverting\n", + "Ios = 5 # #nA # input offset current\n", + "Ib = 30 # #nA # input bias current\n", + "\n", + "# the input bias current of an op-amp is \n", + "\n", + "#Ib =(Ib1+Ib2)/(2)#\n", + "\n", + "# the offset current Ios is define as\n", + "\n", + "#Ios = abs(Ib1-Ib2) #\n", + "\n", + "Ib1=Ib-(Ios/2)#\n", + "print 'The current in the inverting input terminal is = %0.2f'%Ib1,' nA '#\n", + "\n", + "Ib2 =Ib+(Ios/2)#\n", + "print 'The current in the non-inverting input terminal is= %0.2f'%Ib2,' nA '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.8 Pg 115" + ] + }, + { + "cell_type": "code", + "execution_count": 49, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Feedback transfer function is = 0.01 \n", + "OR 1/Beta is = 100.10 \n", + "Feedback transfer function is = -0.01 \n", + "OR 1/Beta is = -100.10 \n" + ] + } + ], + "source": [ + "#determine the feedback transfer function of an op-amp for the following condition\n", + "# a) When open loop gain of 10**5 and the closed loop gain of 100\n", + "A = 10**5 # # open loop gain\n", + "Af = 100 # #closed loop gain\n", + "# Feedback transfer function is\n", + "beta =(1/Af)-(1/A)#\n", + "print 'Feedback transfer function is = %0.2f'%beta,''#\n", + "beta = 1/beta #\n", + "print 'OR 1/Beta is = %0.2f'%beta,''#\n", + "\n", + "# For an open loop gain of -10**5 and closed loop gain of -100\n", + "A = -10**5 # # open loop gain\n", + "Af = -100 # #closed loop gain\n", + "# Feedback transfer function is\n", + "beta =(1/Af)-(1/A)#\n", + "print 'Feedback transfer function is = %0.2f'%beta,''#\n", + "beta = 1/beta #\n", + "print 'OR 1/Beta is = %0.2f'%beta,''#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.9 Pg 115" + ] + }, + { + "cell_type": "code", + "execution_count": 51, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "open loop gain is = 2000.00\n" + ] + } + ], + "source": [ + "#to determine open loop gain\n", + "beta = 0.0120 # # Feedback transfer function\n", + "Af = 80 # #closed loop gain\n", + "A = (Af)/(1-beta*Af) #\n", + "print 'open loop gain is = %0.2f'%A" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.10 Pg 116" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "close loop gain dAf is = 49.78\n", + "the percent change of closed loop gain dAf is = 0.45 %\n" + ] + } + ], + "source": [ + " # To Determine the percent of change in the closed loop gain Af of feedback op-amp circuit\n", + "A = 10**5 # # open loop gain\n", + "Af = 50 # # close loop gain\n", + "beta = 0.01999 # # feedback transfer function\n", + "dA = 10**4 # # the change in the open llop gain \n", + "\n", + "# close loop gain\n", + "dAf = ((dA)/(1+dA*beta))#\n", + "print 'close loop gain dAf is = %0.2f'%dAf\n", + "\n", + "# the percent change of closed loop gain \n", + "dAf = (((Af-dAf)/(Af))*100)#\n", + "print 'the percent change of closed loop gain dAf is = %0.2f'%dAf,'%'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.11 Pg 116" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the feedback transfer function beta is = 0.0199\n", + "the closed loop bandwidth wfH is = 125600\n" + ] + } + ], + "source": [ + "# To Determine the bandwidth of feedback amplifier\n", + "A = 10**4 # # open loop gain\n", + "Af = 50 # # close loop gain\n", + "wH = 628 # #(2*pi*100) # rad/sec # open loop bandwidth\n", + "\n", + "# close loop gain of an op-amp is defined as\n", + "# Af = ((A)/(1+A*beta))# \n", + "\n", + "# the feedback transfer function is given as\n", + "beta = (1/Af)-(1/A) #\n", + "print 'the feedback transfer function beta is = %0.4f'%beta\n", + "\n", + "# closed loop bandwidth\n", + "wfH = wH*(1+beta*A)#\n", + "print 'the closed loop bandwidth wfH is = %0.f'%wfH" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.12 Pg 117" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the unity gain bandwidth is = 1e+06 Hz\n", + "the maximum close loop gain ACL is = 50.00 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# To calculate unity gain bandwidth and maximum close loop gain\n", + "A = 10**5 # # open loop gain\n", + "fo = 10 # # Hz # dominant pole frequency\n", + "fdb = 20*10**3 # #Hz # 3-db frequency\n", + "\n", + "# the unity gain bandwidth\n", + "f1 = fo*A #\n", + "print 'the unity gain bandwidth is = %0.e'%f1,'Hz'#\n", + "\n", + "# the maximum close loop gain\n", + "ACL = (f1/fdb) #\n", + "print 'the maximum close loop gain ACL is = %0.2f'%ACL,''#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.13 Pg 117" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the unity gain bandwidth is = 60 kHz\n", + "the maximum close loop gain ACL is = 5.00 \n" + ] + } + ], + "source": [ + "# To calculate unity gain bandwidth and maximum close loop gain\n", + "A = 10**3 # # open loop gain\n", + "fo = 60 # # Hz # dominant pole frequency\n", + "fdb = 12*10**3 # #Hz # 3-db frequency\n", + "\n", + "# the unity gain bandwidth\n", + "f1 = fo*A #\n", + "print 'the unity gain bandwidth is = %0.f'%(f1/1e3),'kHz'#\n", + "\n", + "# the maximum close loop gain\n", + "ACL = (f1/fdb) #\n", + "print 'the maximum close loop gain ACL is = %0.2f'%ACL,''#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.14 Pg 118" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the dominant pole frequency (fPD) of an op-amp is = 2.5 kHz\n" + ] + } + ], + "source": [ + "# To determine the dominant pole frequency of an op-amp\n", + "Ao = 2*10**5 # # low frequency open loop gain\n", + "f = 5*10**6 # # Hz # pole frequency\n", + "ACL = 100 # # low frequency closed lkoop gain\n", + "p_margin = 80 # \n", + "\n", + "# the dominant pole frequency of an op-amp\n", + "fPD = (ACL)*(f/Ao)/1e3\n", + "print 'the dominant pole frequency (fPD) of an op-amp is = %0.1f'%fPD,'kHz'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.15 Pg 118" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "FL = 6.37 KHz \n", + "Acom = [ magnitude = 6.3*10**-3 angle = -89.6 degree ]\n", + "Ac = [ magnitude = 0.68 angle = 0.4 degree ]\n" + ] + } + ], + "source": [ + " # Determine the loop gain of compensated network\n", + "C = 0.0025*10**-6 # # farad\n", + "R = 10*10**3 # # ohm\n", + "F = 1*10**6 # # Hz\n", + "Ac1 = 100 # \n", + "angle1 = 90 #\n", + "\n", + "# the close loop gain of a compensated network is defined as\n", + "# Ac = Acl*Acom #\n", + "\n", + "#Acom = 1/(1+%(F/FL))#\n", + "\n", + "FL = 1/(2*3.14*R*C)#\n", + "print 'FL = %0.2f'%(FL/1000),' KHz '# # Round Off Error\n", + "\n", + "# Acom = 1/(1+%j(F/FL))#\n", + "# After putting value of F ,FL we get\n", + "\n", + "# Acom = 1/(1+%j(158.7))# # 1+%j(158.7) Rectangular Form where real part is 1 and imaginary part is 158.7\n", + "\n", + "# After converting rectangular from into polar from we get\n", + " \n", + "print 'Acom = [ magnitude = 6.3*10**-3 angle = -89.6 degree ]'#\n", + "\n", + "# Ac = Ac1*Acom # equation 1\n", + "\n", + "# after putting Ac1 and Acom value in equation 1 we get Ac1 = 100 angle 90 and Acom = 6.3*10**-3 angle = -89.6 \n", + "\n", + "print 'Ac = [ magnitude = 0.68 angle = 0.4 degree ]'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.16 Pg 119" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "FL = 1.1 KHz \n", + "Acom = [ magnitude = 0.68 angle = -47.7 degree ]\n" + ] + } + ], + "source": [ + "# Determine the loop gain of compensated network\n", + "\n", + "C = 0.01*10**-6 # # farad\n", + "R = 15*10**3 # # ohm\n", + "F = 1*10**6 # # Hz\n", + "\n", + "# the close loop gain of a compensated network is defined as\n", + "# Ac = Acl*Acom #\n", + "\n", + "#Acom = 1/(1+%(F/FL))#\n", + "\n", + "FL = 1/(2*3.14*R*C)#\n", + "print 'FL = %0.1f'%(FL/1000),' KHz '# # Round Off Error\n", + "\n", + "# Acom = 1/(1+%j(F/FL))#\n", + "# After putting value of F ,FL we get\n", + "\n", + "# Acom = 1/(1+%j(0.9))# # 1+%j(0.9) Rectangular Form where real part is 1 and imaginary part is 0.9\n", + "\n", + "# After converting rectangular from into polar from we get\n", + " \n", + "print 'Acom = [ magnitude = 0.68 angle = -47.7 degree ]'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.17 Pg 120" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "FL = 4.25 KHz \n", + "Acom for F = 0 KHz = [ magnitude = 150 angle = 85 degree ]\n", + "Acom for F = 2 KHz= [ magnitude = 136.4 angle = 64.5 degree ]\n", + "Acom for F = 4 KHz = [ magnitude = 107.14 angle = 41.7 degree ]\n", + "Acom for F = 6 KHz = [ magnitude = 88.24 angle = 30.25 degree ]\n", + "Acom for F = 8 KHz = [ magnitude = 71.4 angle = 23 degree ]\n", + "Acom for F = 10 KHz = [ magnitude = 58.59 angle = 18 degree ]\n", + "Acom for F = 20 KHz = [ magnitude = 31.12 angle = 7 degree ]\n", + "Acom for F = 40 KHz = [ magnitude = 15.9 angle = 1.1 degree ]\n", + "Acom for F = 80 KHz = [ magnitude = 7.9 angle = -2 degree ]\n", + "Acom for F = 100 KHz = [ magnitude = 6.4 angle = -2.6 degree ]\n", + "Acom for F = 200 KHz = [ magnitude = 3.18 angle = -3.8 degree ]\n", + "Acom for F = 400 KHz = [ magnitude = 1.59 angle = -4.4 degree ]\n", + "Acom for F = 800 KHz = [ magnitude = 0.79 angle = -4.7 degree ]\n", + "Acom for F = 1 MHz = [ magnitude = 0.64 angle = -4.7 degree ]\n", + "Acom for F = 1.2 MHz = [ magnitude = 0.52 angle = -4.7 degree ]\n", + "Acom for F = 1.4 MHz = [ magnitude = 0.45 angle = -4.7 degree ]\n", + "Acom for F = 1.6 MHz = [ magnitude = 0.4 angle = -4.7 degree ]\n" + ] + } + ], + "source": [ + "# Determine the loop gain of compensated network\n", + "\n", + "C = 0.5*10**-6 # # farad\n", + "R = 75 # # ohm\n", + "F = 1*10**6 # # Hz\n", + "Ac1 = 150 # \n", + "angle1 = 85 #\n", + "\n", + "# the close loop gain of a compensated network is defined as\n", + "# Ac = Acl*Acom #\n", + "\n", + "#Acom = 1/(1+%(F/FL))#\n", + "\n", + "FL = 1/(2*3.14*R*C)#\n", + "print 'FL = %0.2f'%(FL/1000),' KHz '# # Round Off Error\n", + "\n", + "# Acom = 1/(1+%j(F/FL))#\n", + "\n", + "# After putting value of FL we get\n", + "\n", + "# Acom = 1/(1+%j(F/4.24*10**3))# equation 1\n", + "\n", + "# As F is unknown in above equation 1 \n", + "# by putting different value of F we get Acom for different frequency\n", + "\n", + "\n", + "# If F = 0 KHz\n", + "\n", + "# Acom = 1/(1+%j(0/4.24*10**3))# \n", + "\n", + "# After solving and converting rectangular from into polar from we get\n", + " \n", + "print 'Acom for F = 0 KHz = [ magnitude = 150 angle = 85 degree ]'#\n", + "\n", + "\n", + "# If F = 2 KHz\n", + "\n", + "# Acom = 1/(1+%j(2*10**3/4.24*10**3))# \n", + "\n", + "# After solving and converting rectangular from into polar from we get\n", + " \n", + "print 'Acom for F = 2 KHz= [ magnitude = 136.4 angle = 64.5 degree ]'#\n", + "\n", + "\n", + "# If F = 4 KHz\n", + "\n", + "# Acom = 1/(1+%j(4*10**3/4.24*10**3))# \n", + "\n", + "# After solving and converting rectangular from into polar from we get\n", + " \n", + "print 'Acom for F = 4 KHz = [ magnitude = 107.14 angle = 41.7 degree ]'#\n", + "\n", + "\n", + "# If F = 6 KHz\n", + "\n", + "# Acom = 1/(1+%j(6*10**3/4.24*10**3))# \n", + "\n", + "# After solving and converting rectangular from into polar from we get\n", + " \n", + "print 'Acom for F = 6 KHz = [ magnitude = 88.24 angle = 30.25 degree ]'#\n", + "\n", + "\n", + "\n", + "# If F = 8 KHz\n", + "\n", + "# Acom = 1/(1+%j(8*10**3/4.24*10**3))# \n", + "\n", + "# After solving and converting rectangular from into polar from we get\n", + " \n", + "print 'Acom for F = 8 KHz = [ magnitude = 71.4 angle = 23 degree ]'#\n", + "\n", + "\n", + "\n", + "# If F = 10 KHz\n", + "\n", + "# Acom = 1/(1+%j(10*10**3/4.24*10**3))# \n", + "\n", + "# After solving and converting rectangular from into polar from we get\n", + " \n", + "print 'Acom for F = 10 KHz = [ magnitude = 58.59 angle = 18 degree ]'#\n", + "\n", + "\n", + "\n", + "# If F = 20 KHz\n", + "\n", + "# Acom = 1/(1+%j(20*10**3/4.24*10**3))# \n", + "\n", + "# After solving and converting rectangular from into polar from we get\n", + " \n", + "print 'Acom for F = 20 KHz = [ magnitude = 31.12 angle = 7 degree ]'#\n", + "\n", + "\n", + "\n", + "# If F = 40 KHz\n", + "\n", + "# Acom = 1/(1+%j(40*10**3/4.24*10**3))# \n", + "\n", + "# After solving and converting rectangular from into polar from we get\n", + " \n", + "print 'Acom for F = 40 KHz = [ magnitude = 15.9 angle = 1.1 degree ]'#\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "# If F = 80 KHz\n", + "\n", + "# Acom = 1/(1+%j(80*10**3/4.24*10**3))# \n", + "\n", + "# After solving and converting rectangular from into polar from we get\n", + " \n", + "print 'Acom for F = 80 KHz = [ magnitude = 7.9 angle = -2 degree ]'#\n", + "\n", + "\n", + "\n", + "\n", + "# If F = 100 KHz\n", + "\n", + "# Acom = 1/(1+%j(100*10**3/4.24*10**3))# \n", + "\n", + "# After solving and converting rectangular from into polar from we get\n", + " \n", + "print 'Acom for F = 100 KHz = [ magnitude = 6.4 angle = -2.6 degree ]'#\n", + "\n", + "\n", + "\n", + "\n", + "# If F = 200 KHz\n", + "\n", + "# Acom = 1/(1+%j(200*10**3/4.24*10**3))# \n", + "\n", + "# After solving and converting rectangular from into polar from we get\n", + " \n", + "print 'Acom for F = 200 KHz = [ magnitude = 3.18 angle = -3.8 degree ]'#\n", + "\n", + "\n", + "\n", + "# If F = 400 KHz\n", + "\n", + "# Acom = 1/(1+%j(400*10**3/4.24*10**3))# \n", + "\n", + "# After solving and converting rectangular from into polar from we get\n", + " \n", + "print 'Acom for F = 400 KHz = [ magnitude = 1.59 angle = -4.4 degree ]'#\n", + "\n", + "\n", + "# If F = 800 KHz\n", + "\n", + "# Acom = 1/(1+%j(800*10**3/4.24*10**3))# \n", + "\n", + "# After solving and converting rectangular from into polar from we get\n", + " \n", + "print 'Acom for F = 800 KHz = [ magnitude = 0.79 angle = -4.7 degree ]'#\n", + "\n", + "\n", + "# If F = 1 MHz\n", + "\n", + "# Acom = 1/(1+%j(1*10**6/4.24*10**3))# \n", + "\n", + "# After solving and converting rectangular from into polar from we get\n", + " \n", + "print 'Acom for F = 1 MHz = [ magnitude = 0.64 angle = -4.7 degree ]'#\n", + "\n", + "\n", + "# If F = 1.2 MHz\n", + "\n", + "# Acom = 1/(1+%j(1.2*10**6/4.24*10**3))# \n", + "\n", + "# After solving and converting rectangular from into polar from we get\n", + " \n", + "print 'Acom for F = 1.2 MHz = [ magnitude = 0.52 angle = -4.7 degree ]'#\n", + "\n", + "\n", + "\n", + "# If F = 1.4 MHz\n", + "\n", + "# Acom = 1/(1+%j(1.4*10**6/4.24*10**3))# \n", + "\n", + "# After solving and converting rectangular from into polar from we get\n", + " \n", + "print 'Acom for F = 1.4 MHz = [ magnitude = 0.45 angle = -4.7 degree ]'#\n", + "\n", + "\n", + "# If F = 1.6 MHz\n", + "\n", + "# Acom = 1/(1+%j(1.6*10**6/4.24*10**3))# \n", + "\n", + "# After solving and converting rectangular from into polar from we get\n", + " \n", + "print 'Acom for F = 1.6 MHz = [ magnitude = 0.4 angle = -4.7 degree ]'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.18 Pg 123" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The compensating resistor value is = 15.92 ohm \n" + ] + } + ], + "source": [ + "# to design compensating network\n", + "fp = 500*10**3 # # pole frequency\n", + "C = 0.02*10**-6 # # F # we choose\n", + "# loop gain of compensated network\n", + "\n", + "# ACom =(1)/(1+j(f/fp))\n", + "# fp = (1/2*pie*R*C)\n", + "R = (1/(2*3.14*C*fp))#\n", + "print 'The compensating resistor value is = %0.2f'%R,' ohm '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.19 Pg 123" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "FH = 6.37 KHz \n", + "FL = 2.12 KHz \n", + "Acom = [ magnitude = 0.34 angle = -0.24 degree ]\n", + "Ac = [ magnitude = 34 angle = 89.76 degree ]\n" + ] + } + ], + "source": [ + " # Determine the loop gain of compensated network\n", + "\n", + "C = 0.0025*10**-6 # # farad\n", + "R1 = 10*10**3 # # ohm\n", + "R2 = 20*10**3 # # ohm\n", + "F = 1*10**6 # # Hz\n", + "Ac1 = 100 # \n", + "angle1 = 90 #\n", + "\n", + "# the close loop gain of a compensated network is defined as\n", + "\n", + "# Ac = Acl*Acom #\n", + "\n", + "#Acom = (1+%(F/FH))/(1+%(F/FL))#\n", + "\n", + "FH = 1/(2*3.14*R1*C)#\n", + "print 'FH = %0.2f'%(FH/1000),' KHz '# # Round Off Error\n", + "\n", + "\n", + "FL = 1/(2*3.14*(R1+R2)*C)#\n", + "print 'FL = %0.2f'%(FL/1000),' KHz '# # Round Off Error\n", + "\n", + "\n", + "#Acom = (1+%(F/FH))/(1+%(F/FL))#\n", + "\n", + "# After putting value of FH ,FL we get\n", + "\n", + "# Acom = (1+%j(158.7))/(1+%j(471.7) \n", + "\n", + "# After converting rectangular from into polar from we get\n", + " \n", + "print 'Acom = [ magnitude = 0.34 angle = -0.24 degree ]'#\n", + "\n", + "# Ac = Ac1*Acom # equation 1\n", + "\n", + "# after putting Ac1 and Acom value in equation 1 we get Ac1 = 100 angle 90 and Acom = 0.34 angle = -0.24 \n", + "\n", + "print 'Ac = [ magnitude = 34 angle = 89.76 degree ]'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.20 Pg 124" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "FH = 1.59 KHz \n", + "FL = 0.64 KHz \n", + "Acom = [magnitude = 0.4] \n" + ] + } + ], + "source": [ + " # Determine the loop gain of compensated network\n", + "C = 0.01*10**-6 # # farad\n", + "R1 = 10*10**3 # # ohm\n", + "R2 = 15*10**3 # # ohm\n", + "F = 1*10**6 # # Hz\n", + "\n", + "\n", + "# the close loop gain of a compensated network is defined as\n", + "\n", + "#Acom = (1+%(F/FH))/(1+%(F/FL))#\n", + "\n", + "FH = 1/(2*3.14*R1*C)#\n", + "print 'FH = %0.2f'%(FH/1000),' KHz '# # Round Off Error\n", + "\n", + "\n", + "FL = 1/(2*3.14*(R1+R2)*C)#\n", + "print 'FL = %0.2f'%(FL/1000),' KHz '# # Round Off Error\n", + "\n", + "\n", + "#Acom = (1+%(F/FH))/(1+%(F/FL))#\n", + "\n", + "# After putting value of FH ,FL we get\n", + "\n", + "# Acom = (1+%j(658.9))/(1+%j(1.56*10**3) \n", + "\n", + "# After converting rectangular from into polar from we get\n", + " \n", + "print 'Acom = [magnitude = 0.4] '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.21 Pg 125" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The compensating first resistor R1 value is = 0.80 K ohm \n", + "The compensating second resistor R2 value is = 7.17 K ohm \n" + ] + } + ], + "source": [ + "# to design compensating network\n", + "fH = 10 # #k ohm # break frequency initiated by a zero\n", + "fL = 1 # #k ohm # break frequency initiated by a pole\n", + "C = 0.02# # uF # we choose\n", + "# loop gain of compensated network\n", + "\n", + "# ACom =(1+j(f/fH))/(1+j(f/fL))\n", + "# fH = (1/2*pie*R1*C)\n", + "# fL = (1/2*pie*(R1+R2)*C)\n", + "R1 = (1/(2*3.14*C*fH))#\n", + "print 'The compensating first resistor R1 value is = %0.2f'%R1,' K ohm '#\n", + "R2 = ((1)/(2*3.14*C*fL))-(R1)#\n", + "print 'The compensating second resistor R2 value is = %0.2f'%R2,' K ohm '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.22 Pg 126" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input miller capacitance Cin value is = 10.10 uF \n", + "The output miller capacitance Cout value is = 0.10 uF \n" + ] + } + ], + "source": [ + "# To determine input output miller capacitances\n", + "A = 100 # #gain\n", + "Cm = 0.1 # # uF # compensated capacitor\n", + "\n", + "# the input output miller capacitance are defined as\n", + "Cin = Cm*(A+1)#\n", + "print 'The input miller capacitance Cin value is = %0.2f'%Cin,'uF '#\n", + "Cout = (Cm*((A+1)/A))# \n", + "print 'The output miller capacitance Cout value is = %0.2f'%Cout,'uF '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.23 Pg 127" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input miller capacitance Cin value is = 3.02 uF \n", + "The output miller capacitance Cout value is = 0.02 uF \n", + "The initiated frequency of miller compensating network by pole is = 7.96 KHz \n" + ] + } + ], + "source": [ + "from math import pi\n", + "# To determine input output miller capacitances\n", + "A = 150 # #gain\n", + "Cm = 0.02 # # uF # compensated capacitor\n", + "\n", + "# the input output miller capacitance are defined as\n", + "Cin = Cm*(A+1)#\n", + "print 'The input miller capacitance Cin value is = %0.2f'%Cin,'uF '#\n", + "Cout = (Cm*((A+1)/A))# \n", + "print 'The output miller capacitance Cout value is = %0.2f'%Cout,'uF '#\n", + "\n", + "# In the miller compensating network input capacitance introduce a pole . The initiated frequency of miller compensating network by pole is define as\n", + "\n", + "# fp = 1/(2*pi*R*Cin)#\n", + "R = 1 # # K ohm\n", + "fp = 1/(2*pi*R*Cout)#\n", + "print 'The initiated frequency of miller compensating network by pole is = %0.2f'%fp,' KHz '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.24 Pg 128" + ] + }, + { + "cell_type": "code", + "execution_count": 41, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the slew rate of an op-amp is = 17.58 V/u sec \n", + "The compansated capacitance value is = 1.12 pF \n" + ] + } + ], + "source": [ + "# To determine the slew rate of an op-amp\n", + "f = 1 # # MHz # unity frequency\n", + "Ic = 1*10**-6 # # uA # capacitor current\n", + "Vt = 0.7 # # V # threshold voltage\n", + "\n", + "# the slew rate of an op-amp is defined as\n", + "# Slew rate = (dVo/dt)\n", + "Slewrate = 8*3.14*Vt*f #\n", + "print 'the slew rate of an op-amp is = %0.2f'%Slewrate,' V/u sec '#\n", + "\n", + "# The compansated capacitance Cm is\n", + "gm = (Ic/Vt)#\n", + "Cm = (gm/4*3.14*f)*1e6 # pF\n", + "print 'The compansated capacitance value is = %0.2f'%Cm,'pF '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.25 Pg 129" + ] + }, + { + "cell_type": "code", + "execution_count": 42, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Cut -off frequency of an op-amp is = 5.00 Hz \n" + ] + } + ], + "source": [ + " # To determine the cut off frequency of an op-amp\n", + "f = 1*10**3 # # Hz # unity frequency\n", + "Av = 200 # # V/mV # dc gain\n", + "\n", + "# the unity gain frequency of an op-amp is defined as\n", + "# f = Av*fc #\n", + "\n", + "# cut off frequency\n", + "fc = (f/Av)#\n", + "print 'Cut -off frequency of an op-amp is = %0.2f'%fc,' Hz '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.26 Pg 129" + ] + }, + { + "cell_type": "code", + "execution_count": 43, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the closed loop gain ACL is = 35.00 \n", + "The output gain factor K is = 0.88 V\n", + "The maximum frequency of an op-amp fmax = 145.59 KHz\n" + ] + } + ], + "source": [ + "# To find the maximum frequency of input signal in op-amp circuit\n", + "Vin = 25*10**-3 # # V # input voltage\n", + "Slewrate = 0.8/10**-6 # # V/uV # Slew rate of an op-amp\n", + "R2 = 350*10**3 # # ohm # feedback resistance\n", + "R1 = 10*10**3 # # ohm # input resistance\n", + "\n", + "# the closed loop gain\n", + "# ACL = (mod (Vo/Vin)) = (mod(R2/R1))#\n", + "ACL = abs(R2/R1)#\n", + "print 'the closed loop gain ACL is = %0.2f'%ACL,' '#\n", + "\n", + "# the output gain factor K is given as\n", + "K = ACL*Vin #\n", + "print 'The output gain factor K is = %0.2f'%K,' V'#\n", + "\n", + "# the maximum frequency of an op-amp is\n", + "wmax = (Slewrate/K)#\n", + "fmax = wmax/(2*3.14)#\n", + "print 'The maximum frequency of an op-amp fmax = %0.2f'%(fmax/1000),' KHz'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.27 Pg 129" + ] + }, + { + "cell_type": "code", + "execution_count": 44, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the closed loop gain ACL is = 24.00 \n", + "The output gain factor K is = 0.36 V\n", + "The wmax is = 2.22 *10**6 rad/sec\n" + ] + } + ], + "source": [ + "# To find the maximum frequency of op-amp circuit\n", + "Vin = 0.015 # # V # input voltage\n", + "Slewrate = 0.8 # # V/uV # Slew rate of an op-amp\n", + "R2 = 120*10**3 # # ohm # feedback resistance\n", + "R1 = 5*10**3 # # ohm # input resistance\n", + "\n", + "# the closed loop gain\n", + "# ACL = (mod (Vo/Vin)) = (mod(R2/R1))#\n", + "ACL = abs(R2/R1)#\n", + "print 'the closed loop gain ACL is = %0.2f'%ACL,' '#\n", + "\n", + "# the output gain factor K is given as\n", + "K = ACL*Vin #\n", + "print 'The output gain factor K is = %0.2f'%K,' V'#\n", + "\n", + "# the maximum frequency of an op-amp is\n", + "wmax = (Slewrate/K)#\n", + "print 'The wmax is = %0.2f'%wmax,'*10**6 rad/sec'# # *10**6 because Slewrate is V/uV \n", + "\n", + "# the signal frequency may be w = 500*10**3 rad/sec that is less than the maximum frequency value" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.28 Pg 130" + ] + }, + { + "cell_type": "code", + "execution_count": 49, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the unity frequency f is = 568.70 kHz \n", + "The compansated capacitance Cm value is = 0.2 nF \n" + ] + } + ], + "source": [ + " # To determine the compensated capacitance of an op-amp\n", + "Slewrate = 10 # # V/u sec\n", + "Ic = 1*10**-3 # # mA # capacitor current\n", + "Vt = 0.7 # # V # threshold voltage\n", + "\n", + "# the slew rate of an op-amp is defined as\n", + "# Slew rate = (dVo/dt)\n", + "# the unity frequency f is\n", + "f =(Slewrate/(8*3.14*Vt))#\n", + "f = f*10**6# # *10**6 because Slew rate is V/uV \n", + "print 'the unity frequency f is = %0.2f'%(f/1e3),'kHz '#\n", + "\n", + "# The compansated capacitance Cm is\n", + "gm = (Ic/Vt)#\n", + "Cm = (gm)/(4*3.14*f)*1e9 #\n", + "print 'The compansated capacitance Cm value is = %0.1f'%Cm,'nF '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.29 Pg 131" + ] + }, + { + "cell_type": "code", + "execution_count": 51, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the Slew rate of an op-amp is = 0.50 V/u sec\n" + ] + } + ], + "source": [ + " # To find Slew rate of an op-amp\n", + "Iq = 15 # # uA # bias current\n", + "Cm = 30 # # pF # internal frequency compensated capacitor\n", + "Slewrate = (Iq/Cm)\n", + "print 'the Slew rate of an op-amp is = %0.2f'%Slewrate,' V/u sec'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.30 Pg 131" + ] + }, + { + "cell_type": "code", + "execution_count": 52, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the Slew rate of an op-amp is = 0.68 V/u sec\n" + ] + } + ], + "source": [ + "# To find Slew rate of an op-amp\n", + "Iq = 21 # # uA # bias current\n", + "Cm = 31 # # pF # internal frequency compensated capacitor\n", + "Slewrate = (Iq/Cm)#\n", + "print 'the Slew rate of an op-amp is = %0.2f'%Slewrate,' V/u sec'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.31 Pg 131" + ] + }, + { + "cell_type": "code", + "execution_count": 56, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The full power bandwidth FPBW is = 8.12 kHz \n", + "The 3-db frequency or small signal band width f3db is = 10 kHz \n" + ] + } + ], + "source": [ + " # To determine full power and small signal bandwidth of an op-amp with unity gain\n", + "f = 100*10**6 # # Hz unity gain bandwidth\n", + "ACL = 10**4 # # maximum closed loop gain\n", + "Slewrate = 0.51 # # V/u sec\n", + "Vp = 10 # # V peak volt\n", + "\n", + "# The full power bandwidth\n", + "FPBW = (Slewrate/(2*3.14*Vp))#\n", + "FPBW = FPBW*10**6 # # *10**6 because Slew rate is V/uV \n", + "print 'The full power bandwidth FPBW is = %0.2f'%(FPBW/1e3),'kHz '#\n", + "\n", + "# the 3-db frequency or small signal band width \n", + "f3db = (f/ACL)#\n", + "print 'The 3-db frequency or small signal band width f3db is = %0.f'%(f3db/1e3),'kHz '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.32 Pg 132" + ] + }, + { + "cell_type": "code", + "execution_count": 59, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The full power bandwidth FPBW is = 8.12 kHz \n", + "The 3-db frequency or small signal band width f3db is = 10 kHz \n" + ] + } + ], + "source": [ + "# To determine full power and small signal bandwidth of an op-amp with unity gain\n", + "f = 100*10**6 # # Hz unity gain bandwidth\n", + "ACL = 10**4 # # maximum closed loop gain\n", + "Slewrate = 0.51 # # V/u sec\n", + "Vp = 10 # # V peak volt\n", + "\n", + "# The full power bandwidth\n", + "FPBW = (Slewrate/(2*3.14*Vp))#\n", + "FPBW = FPBW*10**6 # # *10**6 because Slew rate is V/uV \n", + "print 'The full power bandwidth FPBW is = %0.2f'%(FPBW/1e3),'kHz '#\n", + "\n", + "# the 3-db frequency or small signal band width \n", + "f3db = (f/ACL)#\n", + "print 'The 3-db frequency or small signal band width f3db is = %0.f'%(f3db/1e3),'kHz '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.33 Pg 132" + ] + }, + { + "cell_type": "code", + "execution_count": 60, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the Slew rate of an op-amp is = 0.31 V/u sec \n", + "The closed loop gain ACL is = 83.33 \n" + ] + } + ], + "source": [ + "# To find Slew rate and closed loop gain of an op-amp\n", + "fu = 1*10**6 # # Hz # unity gain bandwidth\n", + "fmax = 5*10**3 # # KHz # full power bandwidth\n", + "F3db = 12*10**3 # # Hz # small signal bandwidth\n", + "Vp = 10 # # V # peak volt\n", + "\n", + "# the full power bandwidth of an op-amp\n", + "# fmax=FPBW = (Slew rate/2*3.14*Vp)#\n", + "Slewrate = 2*3.14*Vp*fmax#\n", + "Slewrate = Slewrate*(10**-6)# # *10**-6 because Slewrate is V/u \n", + "print 'the Slew rate of an op-amp is = %0.2f'%Slewrate,' V/u sec '#\n", + "\n", + "# # the 3-db frequency or small signal band width \n", + "#f3db = (f/ACL)#\n", + "#the closed loop gain ACL\n", + "ACL = fu/F3db #\n", + "print 'The closed loop gain ACL is = %0.2f'%ACL,' '#" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch6.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch6.ipynb new file mode 100644 index 00000000..c7160709 --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch6.ipynb @@ -0,0 +1,2030 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6 Applications of Operational Amplifier" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.1 Pg 140" + ] + }, + { + "cell_type": "code", + "execution_count": 116, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the input resistance is = 20.00 kohm\n", + "The resistance R2 is = 100.00 kohm\n" + ] + } + ], + "source": [ + "# Design an inverting amplifier\n", + "Av = -5 #\n", + "#V1 = 0.1 sin wt #\n", + "V1 = 0.1 # # *sin wt #\n", + "i = 5*10**-6 #\n", + "\n", + "# the input resistance \n", + "R1 = V1/i / 1000 # kohm\n", + "print 'the input resistance is = %0.2f'%R1,'kohm'#\n", + "\n", + "# The resistance R2\n", + "#Av = -(R2/R1)#\n", + "R2 = -(Av*R1)#\n", + "print 'The resistance R2 is = %0.2f'%R2,'kohm'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.2 Pg 141" + ] + }, + { + "cell_type": "code", + "execution_count": 117, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the input resistance is = 20.00 kohm\n", + "The resistance R2 is = 80.00 kohm\n" + ] + } + ], + "source": [ + "# Design an non inverting amplifier\n", + "Av = 5 #\n", + "#V1 = 0.1 sin wt #\n", + "V1 = 0.1 #\n", + "i = -5*10**-6 #\n", + "\n", + "# the input resistance \n", + "R1 = -V1/i/1000 # kohm\n", + "print 'the input resistance is = %0.2f'%R1,'kohm'#\n", + "\n", + "# The resistance R2\n", + "#Av = 1+(R2/R1)#\n", + "R2 = (Av-1)*R1#\n", + "print 'The resistance R2 is = %0.2f'%R2,'kohm'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.3 Pg 146" + ] + }, + { + "cell_type": "code", + "execution_count": 118, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the cut off frequency of phase shifter is = 723.43 Hz\n", + "The phase shift is = -15.75\n" + ] + } + ], + "source": [ + "# To calculate phase shift between two extremes \n", + "C = 0.22*10**-6 #\n", + "R = 1*10**3 #\n", + "f = 1*10**3 #\n", + "\n", + "# the cut off frequency of phase shifter \n", + "fc = 1/(2*pi*R*C) #\n", + "print 'the cut off frequency of phase shifter is = %0.2f'%fc,'Hz'#\n", + "f\n", + "# the phase shift\n", + "f = 1 # # KHz\n", + "fc = 7.23 # # KHz \n", + "from math import atan ,degrees\n", + "PS = -2*degrees(atan(f/fc))\n", + "print 'The phase shift is = %0.2f'%PS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.4 Pg 146" + ] + }, + { + "cell_type": "code", + "execution_count": 119, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of resistance is = 1.92 kohm\n" + ] + } + ], + "source": [ + "# To design a phase shifter\n", + "f = 2*10**3 #\n", + "PS = -135 #\n", + "# the phase shift\n", + "# PS = -2*atand(2*pi*R*C)#\n", + "#RC = 192.1*10**-6 #\n", + "C = 0.1*10**-6 #\n", + "R = (192.1*10**-6)/C/1000 # kohm\n", + "print 'The value of resistance is = %0.2f'%R,'kohm'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.5 Pg 153" + ] + }, + { + "cell_type": "code", + "execution_count": 120, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of resistance R1 is = 25.00 kohm\n", + "The value of resistance R3 is = 25.00 kohm\n", + "The value of resistance R2 is = 750.00 kohm\n", + "The value of resistance R4 is = 750.00 kohm\n" + ] + } + ], + "source": [ + "# Design a difference amplifier\n", + "Ri = 50 # kohm\n", + "Ad = 30 \n", + "\n", + "R1 = Ri/2 #\n", + "print 'The value of resistance R1 is = %0.2f'%R1,'kohm'#\n", + "R3 = R1 #\n", + "print 'The value of resistance R3 is = %0.2f'%R3,'kohm'#\n", + "\n", + "# the differential gain\n", + "#Ad = R2/R1 #\n", + "R2 = 30*R1 #\n", + "print 'The value of resistance R2 is = %0.2f'%R2,'kohm'#\n", + "\n", + "R4 = R2 #\n", + "print 'The value of resistance R4 is = %0.2f'%R4,'kohm'# " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.6 Pg 154" + ] + }, + { + "cell_type": "code", + "execution_count": 121, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The common mode rejection ratio is = 26.58 dB\n" + ] + } + ], + "source": [ + "# Calculate CMRR ratio\n", + "Ad = 10.24 #\n", + "Acm = 0.48 #\n", + "\n", + "# the common mode rejection ratio CMRR is defined as\n", + "CMRRdB = 20*log10(Ad/Acm)#\n", + "print 'The common mode rejection ratio is = %0.2f'%CMRRdB,' dB'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.7 Pg 156" + ] + }, + { + "cell_type": "code", + "execution_count": 122, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The feedback resistance is = 100.00 kohm\n", + " The value of resistance R1 is = 100.00 kohm\n" + ] + } + ], + "source": [ + "# Design current to voltage converter\n", + "Vo =-10 #\n", + "Is = 100*10**-6 #\n", + "\n", + "# the output voltage of current to voltage converter is defined as\n", + "#Vo =-1s*R2 \n", + "R2 = -Vo/Is/1000 #kohm\n", + "print ' The feedback resistance is = %0.2f'%R2,'kohm'#\n", + "\n", + "R1 = R2 #\n", + "print ' The value of resistance R1 is = %0.2f'%R1,'kohm'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.8 Pg 157" + ] + }, + { + "cell_type": "code", + "execution_count": 123, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of resistance R2 is = 44.78 kohm\n" + ] + } + ], + "source": [ + "# Design high sensitivity current to voltage converter\n", + "R1 = 5 # kohm\n", + "Is = 1 #\n", + "KR = 0.01/10**9 # # V / nA\n", + "\n", + "# the output voltage of high sensitivity current to voltage converter\n", + "Vo =-KR*Is #\n", + "KR = 10*10**6 #\n", + "R = 1*10**6 # #we assume then\n", + "K = 10 #\n", + "#1 + (R2/R1)+(R2/R) = 10 #\n", + "# solving above equation we get\n", + "\n", + "R2 = 9*((5*10**6)/(10**3+5))/1000 # kohm\n", + "print 'The value of resistance R2 is = %0.2f'%R2,'kohm'# " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.9 Pg 160" + ] + }, + { + "cell_type": "code", + "execution_count": 124, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The load current iL is = 5.00 mA\n", + "The current i3 is = 0.50 mA\n", + "The current iA is = 5.50 mA\n", + "The output voltage is = 6.00 V\n", + "The current i1 is = 49.40 A\n", + "The current i2 is = 49.40 A\n" + ] + } + ], + "source": [ + "# Determine a load current in a V to I converter\n", + "R1 = 10 # kohm\n", + "R2 = 10 # koohm\n", + "R3 = 1 # kohm\n", + "R4 = 1 # kohm\n", + "VI = -5 #\n", + "\n", + "# The Load Current\n", + "iL = -VI/R3 #\n", + "print 'The load current iL is = %0.2f'%iL,'mA'#\n", + "\n", + "VL = 0.5 #\n", + "# The Current i3 and iA\n", + "i3 = VL/R3 #\n", + "print 'The current i3 is = %0.2f'%i3,'mA'#\n", + "\n", + "iA = i3+iL #\n", + "print 'The current iA is = %0.2f'%iA,'mA'#\n", + "\n", + "# the output voltage \n", + "Vo = (iA*R3)+VL #\n", + "print 'The output voltage is = %0.2f'%Vo,' V'#\n", + "\n", + "ZL =100 #\n", + "# The current i1 and i2 \n", + "#i1 = (VI-iL*ZL)/R1 #\n", + "i1 = (iL*ZL-Vo)/R2 #\n", + "print 'The current i1 is = %0.2f'%i1,' A'#\n", + "\n", + "i2 = i1 #\n", + "print 'The current i2 is = %0.2f'%i2,' A'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.10 Pg 163" + ] + }, + { + "cell_type": "code", + "execution_count": 125, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of resistance R1f is = 0.0606 K ohm \n", + "The value of resistance R2 is = 75.5 K ohm \n" + ] + } + ], + "source": [ + "# Design an instrumentation amplifier\n", + "#A = 5 to 500 # adjustable gain\n", + "VR = 100*10**3 #\n", + "\n", + "# the maximum differential gain of instrumentation amplifier is 500 \n", + "#Amax = (R4/R3)*(1+(2R2/R1))#\n", + "#by solving above equation we get following equation\n", + "# 2R2 -249R1f = 0 equation 1\n", + "\n", + "# the minimum differential gain of instrumentation amplifier is 5\n", + "# Amin = (R4/R3)*(1+(2R2/R1)) #\n", + "#by solving above equation we get following equation\n", + "# 2R2 -1.5R1f = 150*10**3 equation 2\n", + "\n", + "#by solving equation 1 and 2 we get\n", + "print 'The value of resistance R1f is = 0.0606 K ohm '#\n", + "\n", + "print 'The value of resistance R2 is = 75.5 K ohm '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.11 Pg 164" + ] + }, + { + "cell_type": "code", + "execution_count": 127, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of resistane R1 is = 9.09 kohm\n" + ] + } + ], + "source": [ + " # To find the value of resistance R1 for instrumentation amplifier\n", + "A =100 #\n", + "R2 = 450*10**3 #\n", + "R3 = 1*10**3 #\n", + "R4 = 1*10**3 #\n", + "\n", + "# The gain of differential amplifier \n", + "# A = (R4/R3)*(1+(2R2/R1)) #\n", + "#but R3 = R4 then\n", + "# A = 1+(2R2/R1) #\n", + "R1 = 2*R2/(A-1)/1000 # kohm\n", + "print 'The value of resistane R1 is = %0.2f'%R1,'kohm'# " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.12 Pg 167" + ] + }, + { + "cell_type": "code", + "execution_count": 130, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " At t = 1 msec the time constant RC is = 0.10 m sec\n", + " if C = 0.01 uF then R of RC time constant is = 10 K ohm \n", + " if C = 0.001 uF then R of RC time constant is = 100 K ohm \n" + ] + } + ], + "source": [ + "# determine the time constant of an integrator\n", + "Vo = 10 # # at t= 1 m sec\n", + "t = 1 # # m sec\n", + "\n", + "# the output of integrator \n", + "#Vo = t/RC # when t is from 0 to 1\n", + "RC = t/Vo #\n", + "print ' At t = 1 msec the time constant RC is = %0.2f'%RC,' m sec'#\n", + "\n", + "print ' if C = 0.01 uF then R of RC time constant is = 10 K ohm '#\n", + "\n", + "print ' if C = 0.001 uF then R of RC time constant is = 100 K ohm '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.13 Pg 168" + ] + }, + { + "cell_type": "code", + "execution_count": 131, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The feedback resistance Rf is = 100.00 kohm\n", + " The frequency fa is = 2.00 kHz\n", + " The value of capacitor C is = 0.8 nF \n" + ] + } + ], + "source": [ + "# Design an integrator circuit\n", + "A = 10 #\n", + "f =20*10**3 #\n", + "R = 10*10**3 # # we assume \n", + "Rf =10*R #\n", + "\n", + "print ' The feedback resistance Rf is = %0.2f'%(Rf/1000),'kohm'#\n", + "\n", + "# for proper integration f>= 10fa \n", + "fa = f/10/1000 #\n", + "print ' The frequency fa is = %0.2f'%fa,'kHz'#\n", + "\n", + "# in practical integrator\n", + "#fa = 1/(2*pi*Rf*C)#\n", + "\n", + "C = 1/(2*pi*Rf*fa)*1e6# nF\n", + "print ' The value of capacitor C is = %0.1f'%C,'nF '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.1 Pg 185" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the value of resistance R1 is = 0.199 Mohm\n", + "the value of resistance R2 is = 1 Mohm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# design an inverting amplifier with a closed loop voltage gain of Av = -5\n", + "Av = -5 #\n", + "Is = 5*10**-6 # # A\n", + "Rs = 1*10**3 # # ohm\n", + "# input voltage source Vs = sinwt volts\n", + "\n", + "# in an inverting amplifier frequency effect is neglected then i/p volt Vin = 1 V and total resistance equal to Rs+R1\n", + "\n", + "# the input current can be written as Iin=Is\n", + "# Is = (Vin/Rs+R1)#\n", + "Iin = Is#\n", + "Vin = 1 # # V\n", + "R1 = (1-(Iin*Rs))/Iin #\n", + "print 'the value of resistance R1 is = %0.3f'%(R1/1e6),'Mohm'#\n", + "\n", + "# closed loop voltage gain of an inverting amplifier\n", + "#Av = -(R2/Rs+R1)\n", + "R2 = -(Av*(Rs+R1))#\n", + "print 'the value of resistance R2 is = %0.f'%(R2/1e6),'Mohm'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.2 Pg 186" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the value of resistance R1 is = 8 kohm\n", + "the value of resistance R2 is = 72 kohm\n" + ] + } + ], + "source": [ + " # design an inverting amplifier with a closed loop voltage gain of Av = 10\n", + "Av = 10 #\n", + "Vin = 0.8 # #V\n", + "Iin = 100*10**-6 # # A\n", + "# in an non- inverting amplifier the input voltage Vin=V1=V2 because of vortual short effect then the i/p current In = Vin/R1\n", + "R1 = Vin/Iin/1e3\n", + "print 'the value of resistance R1 is = %0.f'%R1,'kohm'#\n", + "\n", + "# closed loop voltage gain of an non-inverting amplifier\n", + "#Av = Vo/Vin = (1+R2/R1)\n", + "R2 = (Av-1)*R1 # kohm\n", + "print 'the value of resistance R2 is = %0.f'%R2,'kohm'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.3 Pg 187" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the value of resistance R1 is = 20.00 kohm\n", + "the value of resistance R2 is = 80.00 kohm\n", + "the output current I2 is = 50 uA\n" + ] + } + ], + "source": [ + "# design an non-inverting amplifier with colsed loop gain of 5 limited voltage of -5 V <= Vo <= 5 V and maximum i/p c/n 50 uA\n", + "R1 = 8*10**3 # # ohm\n", + "R2 = 72*10**3 # # ohm\n", + "Iin = 50*10**-6 # # A\n", + "Vo = 5 # # V \n", + "\n", + "# closed loop gain\n", + "#Av = Vo/Vin = (1+R2/R1)\n", + "Av = 1+(R2/R1)#\n", + "# but \n", + "Av = 5 #\n", + "# then\n", + "# (R2/R1) = 4 #\n", + "\n", + "# the output voltage of the amplifier is Vo = 5 V \n", + "#i.e\n", + "Vin = 1 # # V\n", + "# Iin = Vin/R1 #\n", + "R1 = Vin/Iin/1e3\n", + "print 'the value of resistance R1 is = %0.2f'%R1,'kohm'#\n", + "\n", + "R2 = 4*R1 #\n", + "print 'the value of resistance R2 is = %0.2f'%R2,'kohm'#\n", + "\n", + "# the output current I2 is given as\n", + "I2 = (Vo-Vin)/R2*1e3 # uA\n", + "print 'the output current I2 is = %0.f'%I2,'uA'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.4 Pg 188" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the value of resistance Ria is = 20.00 kohm\n", + "the value of resistance Rib is = 15.00 kohm\n", + "the value of resistance Ric is = 30.00 kohm\n" + ] + } + ], + "source": [ + "# Design a op-amp circuit to provide the output voltage Vo = -2(3 V1 +4 V2 +2 V3)\n", + "# Vo = -2(3 V1 + 4 V2+ 2 V3)# equation 1\n", + "# the output of the summer circuit is given as\n", + "# Vo = -R2((Via/Ria)+(Vib/Rib)+(Vic/Ric)) equation 2\n", + "\n", + "# compare equation 1 and 2 of Vo we get \n", + "\n", + "# (R2/Ria)= 6 #\n", + "# (R2/Rbi=8 #\n", + "# (R2/Ric)=4 #\n", + "\n", + "R2 = 120*10**3/1e3 # # we choose then \n", + "\n", + "Ria = R2/6 #\n", + "print 'the value of resistance Ria is = %0.2f'%Ria,'kohm'#\n", + "\n", + "Rib = R2/8 #\n", + "print 'the value of resistance Rib is = %0.2f'%Rib,'kohm'#\n", + "\n", + "Ric = R2/4 #\n", + "print 'the value of resistance Ric is = %0.2f'%Ric,'kohm'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.5 Pg 188" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the value of resistance Ria is = 90.00 kohm\n", + "the value of resistance Rib is = 42.00 kohm\n", + "the value of resistance Ric is = 63.00 kohm\n", + "the value of resistance Rid is = 210.00 kohm\n" + ] + } + ], + "source": [ + " # Design a summing amplifier circuit to provide the output voltage Vo = -(7 V11 + 15 V12 + 10 V13 + 3 V14)\n", + "R2 = 630# kohm # Assume feedback resistance\n", + "# Vo = -(7 V11 + 15 V12 + 10 V13 + 3 V14)# equation 1\n", + "# the output of the summer circuit is given as\n", + "# Vo = -R2((Via/Ria)+(Vib/Rib)+(Vic/Ric)+(Vid/Rid)) equation 2\n", + "\n", + "# compare equation 1 and 2 of Vo we get \n", + "\n", + "# (R2/Ria)= 7 #\n", + "# (R2/Rbi= 15 #\n", + "# (R2/Ric)= 10 #\n", + "# (R2/Rid)= 3 #\n", + "\n", + "Ria = R2/7 #\n", + "print 'the value of resistance Ria is = %0.2f'%Ria,' kohm'#\n", + "\n", + "Rib = R2/15 #\n", + "print 'the value of resistance Rib is = %0.2f'%Rib,' kohm'#\n", + "\n", + "Ric = R2/10 #\n", + "print 'the value of resistance Ric is = %0.2f'%Ric,' kohm'#\n", + "\n", + "Rid = R2/3 #\n", + "print 'the value of resistance Rid is = %0.2f'%Rid,'kohm'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.6 Pg 190" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the value of resistance R2 is = 300.00 kohm\n", + "the value of resistance R4 is = 33333.33 kohm\n" + ] + } + ], + "source": [ + "# Design a op-amp circuit to provide the output voltage Vo = V2 - 3 V1 with Ri1 =Ri2 = 100*10**3\n", + "Ri1 = 100 # # kohm\n", + "Ri2 = 100 # # kohm\n", + "# the i/p resistance \n", + "R1 = Ri1 #\n", + "R3 = Ri2 #\n", + "\n", + "# Vo = V2 - 3 V1# equation 1\n", + "# the output of the summer circuit is given as\n", + "# Vo = [(R4/(R3+R4)*(1+(R2/R1))*Vi2-(R2/R1)*Vi1] equation 2\n", + "\n", + "# compare equation 1 and 2 of Vo we get \n", + "# (R4/(R3+R4)*(1+(R2/R1)) = 1 # equation 3\n", + "# R2/R1 = 3 # equation 4\n", + "\n", + "# by subsituting the value of R1 and R3 in equation 3 and 4\n", + "\n", + "# from equation 4\n", + "R2 = 3*R1 #\n", + "print 'the value of resistance R2 is = %0.2f'%R2,'kohm'#\n", + "\n", + "# from equation 3\n", + "R4 = (100*10**3)/3 #\n", + "print 'the value of resistance R4 is = %0.2f'%R4,'kohm'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.7 Pg 191" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The load current iL is = 5 mA\n", + "The voltage across load VL is = 1.00 V\n", + "The non-inverting current across i3 is = 1.00 mA\n", + "The non-inverting current across i4 is = 6.00 mA\n", + "The output voltage of given voltage to current converter is = 6.00 V\n" + ] + } + ], + "source": [ + " # determine the load current and output voltage\n", + "Vin = -5 # # V\n", + "ZL = 200 # # ohm\n", + "R1 = 10*10**3 # # ohm\n", + "R2 = 10*10**3 # # ohm\n", + "R3 = 1*10**3 # # ohm\n", + "R4 = 1*10**3 # # ohm\n", + "\n", + "# the load c/n of the given voltage to c/n converter circuit is given by\n", + "iL =-Vin/(R1*R4)*R2*1e3 # m\n", + "print 'The load current iL is = %0.f'%iL,'mA'#\n", + "\n", + "# the voltage across the load \n", + "VL = iL/1e3*ZL#\n", + "print 'The voltage across load VL is = %0.2f'%VL,' V'#\n", + "\n", + "# the non-inverting current across i3 and i4 are\n", + "i3 = VL/R3*1000 #mA\n", + "print 'The non-inverting current across i3 is = %0.2f'%i3,'mA'#\n", + "\n", + "i4 = iL+i3 # mA\n", + "print 'The non-inverting current across i4 is = %0.2f'%i4,'mA'#\n", + "\n", + "# the output voltage of given voltage to current converter is given by\n", + "Vo = (iL/1e3*R3)+VL #\n", + "print 'The output voltage of given voltage to current converter is = %0.2f'%Vo,' V'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.8 Pg 192" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The common mode rejection ratio CMRR is = 120.50 \n", + "The common mode rejection ratio CMRR in decibel is = 41.62 dB \n" + ] + } + ], + "source": [ + "from math import log10\n", + "# determine the common mode rejection ratio CMRR\n", + "# R2/R1 = 10 #\n", + "# R4/R3 = 11 #\n", + "\n", + "# the output of the difference amplifier is given by\n", + "# Vo = (((R4)/(R3+R4))*(((1+(R2/R1))*VI2))-((R2/R1)*VI1))#\n", + "\n", + "# putting R1 R2 R3 R4 value in above equation we get Vo as\n", + "\n", + "# Vo =(121/12)*VI2-10VI1 # equation 1\n", + "\n", + "# the differential mode input of difference amplifier is given by\n", + "# Vd = VI2-VI1 # eqution 2\n", + "\n", + "# the common mode input of difference amplifier is given by\n", + "# VCM = (VI2+VI1)/2 # equation 3\n", + "\n", + "# from equation 2 and 3 \n", + "\n", + "# VI1 = VCM-Vd/2 # equation 4\n", + "\n", + "# VI2 = VCM+Vd/2 # equation 5\n", + "\n", + "# substitute equation 4 and 5 in 1 we get \n", + "# Vo = (VCM/12)+(241Vd/24)# equation6\n", + "\n", + "# Vd = Ad*Vd+ACM*VCM # equation 7\n", + "\n", + "#equation from equation 6 and 7 we get\n", + "\n", + "Ad = 241/24 #\n", + "ACM = 1/12 #\n", + "\n", + "# the common mode rejection ratio CMRR is \n", + "CMRR = abs(Ad/ACM)#\n", + "print 'The common mode rejection ratio CMRR is = %0.2f'%CMRR,' '#\n", + "\n", + "# in decibal it can be expressed as\n", + "\n", + "CMRR = 20*log10(CMRR)#\n", + "print 'The common mode rejection ratio CMRR in decibel is = %0.2f'%CMRR,' dB '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.9 Pg 194" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output of the difference amplifier is = -8.12 V \n", + "The output of the difference amplifier is = 0.12 V \n", + "the common mode input of difference amplifier is = 2.00 \n", + "the common mode gain ACM of difference amplifier is = 0.06 \n", + "the differential gain of the difference amplifier is = 2.00 \n", + "The common mode rejection ratio CMRR is = 32.00 \n", + "The common mode rejection ratio CMRR in decibel is = 30.10 dB \n" + ] + } + ], + "source": [ + "# determine Vo when 1) VI1 = 2 V VI2 = -2 V and 2) VI1 = 2 V VI2 = 2 V\n", + "# and common mode rejection ratio CMRR\n", + "R1 = 10*10**3 # # ohm\n", + "R2 = 20*10**3 # # ohm\n", + "R3 = 10*10**3 # # ohm\n", + "R4 = 22*10**3 # # ohm\n", + "\n", + "\n", + "# the output of the difference amplifier is given by\n", + "# Vo = (((R4)/(R3+R4))*(((1+(R2/R1))*VI2))-((R2/R1)*VI1))#\n", + "\n", + "# Case 1 when VI1 = 2 V VI2 = -2 V\n", + "VI1 = 2 #\n", + "VI2 = -2 #\n", + "\n", + "Vo = (((R4)/(R3+R4))*(((1+(R2/R1))*VI2))-((R2/R1)*VI1))#\n", + "print 'The output of the difference amplifier is = %0.2f'%Vo,' V '#\n", + "\n", + "# case 2 when VI1 = 2 V VI2 = 2 V\n", + "VI1 = 2 #\n", + "VI2 = 2 #\n", + "\n", + "Vo = (((R4)/(R3+R4))*(((1+(R2/R1))*VI2))-((R2/R1)*VI1))#\n", + "print 'The output of the difference amplifier is = %0.2f'%Vo,' V '#\n", + "\n", + "# the common mode input of difference amplifier is given by\n", + "VCM = (VI2+VI1)/2 #\n", + "print 'the common mode input of difference amplifier is = %0.2f'%VCM,' '#\n", + "\n", + "# the common mode gain ACM of difference amplifier is given by\n", + "ACM = Vo/VCM\n", + "print 'the common mode gain ACM of difference amplifier is = %0.2f'%ACM,' '#\n", + "\n", + "# the differential gain of the difference amplifier is given \n", + "Ad = R2/R1 # \n", + "print 'the differential gain of the difference amplifier is = %0.2f'%Ad,' '#\n", + "\n", + "# the common mode rejection ratio CMRR is \n", + "CMRR = abs(Ad/ACM)#\n", + "print 'The common mode rejection ratio CMRR is = %0.2f'%CMRR,' '#\n", + "\n", + "# in decibal it can be expressed as\n", + "CMRR = 20*log10(CMRR)#\n", + "print 'The common mode rejection ratio CMRR in decibel is = %0.2f'%CMRR,' dB '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.10 Pg 195" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the maximum differential voltage gain of the instrumentation amplifier is = 101.00 \n", + "the minimum differential voltage gain of the instrumentation amplifier is = 5.00 \n", + " the range of the differential voltage gain of the instrumentation amplifier is \n", + " 5 <= Av <= 101 \n" + ] + } + ], + "source": [ + "# To determine the range of the differential voltage gain\n", + "#R1 = 1 K ohm to 25 K ohm #\n", + "R2 = 50 # # K ohm\n", + "R3 = 10 # # K ohm\n", + "R4 = 10 # # K ohm\n", + "\n", + "# the output of instrumentation amplifier is given by\n", + "#Vo = (R4/R3)*(1+(2*R2/R1))*(VI@-VI1)#\n", + "\n", + "# the differential voltage gain of the instrumentation amplifier can be written as\n", + "#Av = (Vo/(VI2-VI1)) = (R4/R3)*(1+(2R2/R1))#\n", + "\n", + "# For R1 = 1 K ohm the maximum differential voltage gain of the instrumentation amplifier is\n", + "R1 = 1 # # K ohm\n", + "Av = (R4/R3)*(1+(2*R2/R1))#\n", + "print 'the maximum differential voltage gain of the instrumentation amplifier is = %0.2f'%Av,' '#\n", + "\n", + "# For R1 = 25 K ohm the mminimum differential voltage gain of the instrumentation amplifier is\n", + "R1 = 25 # # K ohm\n", + "Av = (R4/R3)*(1+(2*R2/R1))#\n", + "print 'the minimum differential voltage gain of the instrumentation amplifier is = %0.2f'%Av,' '#\n", + "\n", + "print ' the range of the differential voltage gain of the instrumentation amplifier is '#\n", + "print ' 5 <= Av <= 101 '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.11 Pg 196" + ] + }, + { + "cell_type": "code", + "execution_count": 36, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The variable resistance R1 varies is 401 ohm <= R1 <= 100.2 K-ohm \n" + ] + } + ], + "source": [ + "# To design an instrumentation amplifier\n", + "# 4 <= Av <= 1000 # gain\n", + "Ad = 2 #\n", + "Res = 100 # # K ohm\n", + "\n", + "# we cosider the variable resistance is R1 , the maximum and the minimum range of variable resistance \n", + "# R1min = R1 # \n", + "# R1max = R1+100 #\n", + "\n", + "# the gain of difference amplifier \n", + "#A3 = Ad = Vo/(Vo2-Vo1) = (R4/R3)\n", + "\n", + "# the maximum range of differential voltage gain Avmax = 1000 when R1min = R1\n", + "#Avmax = R4/R3*(1+(2*R2/R1min))#\n", + "\n", + "# by solvin we get following equation\n", + "# 499*R1-2*R2=0 equation 1\n", + "\n", + "# the maximum range of differential voltage gain Avmin =4 when R1max = R1+100 K ohm\n", + "# Avmin = (R4/R3)*(1+(2R2/R1max))#\n", + "\n", + "# by solving above equation we get\n", + "# R1 -2 R2 = -200 K ohm equation 2\n", + "\n", + "#by solving equation 1 and 2 we get\n", + "R1 = 401 # # ohm\n", + "R2 = 100.2 # # ohm\n", + "print 'The variable resistance R1 varies is 401 ohm <= R1 <= 100.2 K-ohm ' #" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.12 Pg 198" + ] + }, + { + "cell_type": "code", + "execution_count": 41, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The time constant of the given filter is RC = 0.20 msec \n" + ] + } + ], + "source": [ + " # Determine the time constant of the integrator\n", + "Vo = 10 #\n", + "t = 2*10**-3 #\n", + "VI = -1 # # at t =0 #\n", + "\n", + "# The output voltage of an integrator is define as\n", + "RC = t/10*1e3 # ms\n", + "print ' The time constant of the given filter is RC = %0.2f'%RC,'msec '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.13 Pg 198" + ] + }, + { + "cell_type": "code", + "execution_count": 48, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The time constant of the given filter is RC = 0.1 msec \n", + "The capacitor value is = 0.1 F\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Determine the time constant of the integrator\n", + "Vo = 20 #\n", + "t = 1*10**-3 #\n", + "VI = -1 # # at t =0 #\n", + "\n", + "# The output voltage of an integrator is define as\n", + "RC = t/10 #\n", + "print ' The time constant of the given filter is RC = %0.1f'%(RC*1000),'msec '#\n", + "\n", + "R = 1*10**3 # # we assume \n", + "C = RC/R*10**6 #\n", + "print 'The capacitor value is = %0.1f'%C,'F'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.14 Pg 199" + ] + }, + { + "cell_type": "code", + "execution_count": 49, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of resistance RIa is = 90.00 K ohm \n", + "The value of resistance RIb is = 22.50 K ohm \n", + "The value of resistance RIc is = 18.00 K ohm \n", + "The value of resistance RId is = 15.00 K ohm \n" + ] + } + ], + "source": [ + " # to design a summing amplifier\n", + "\n", + "# the output of the summing amplifier is given by\n", + "#Vo = -R2*((VIa/RIa)+(VIb/RIb)+(VIc/RIc)+(VId/RId))# equation 1\n", + "\n", + "# the equation given is\n", + "#Vo = -(3*VIa+12*VIb+15*VIc+18*VId)# equation 2\n", + "\n", + "# comparing equation 1 and 2\n", + "#R2/RIa = 3 #\n", + "#R2/RIb = 12 #\n", + "#R2/RIc = 15 #\n", + "#R2/RId = 18 # \n", + "\n", + "# the feedback resistance R2= 270 K ohm \n", + "R2 = 270 # # K ohm\n", + "RIa = R2/3 #\n", + "print 'The value of resistance RIa is = %0.2f'%RIa,' K ohm '#\n", + "\n", + "RIb = R2/12 #\n", + "print 'The value of resistance RIb is = %0.2f'%RIb,' K ohm '#\n", + "\n", + "RIc = R2/15 #\n", + "print 'The value of resistance RIc is = %0.2f'%RIc,' K ohm '#\n", + "\n", + "RId = R2/18 #\n", + "print 'The value of resistance RId is = %0.2f'%RId,' K ohm '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.15 Pg 200" + ] + }, + { + "cell_type": "code", + "execution_count": 50, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output of first op-amp A1 is = -275*sin wt mV \n", + "The output of second op-amp A2 is = 275*sin wt mV \n", + "The output of third op-amp A3 is = 825*sin wt mV \n", + "current through the resistor R1 and R2 is = 5 sin wt uA \n", + "current through the non-inverting terminal resistor R3 and R4 = 5.5 sin wt uA \n", + "current through the inverting terminal resistor R3 and R4 = 22 sin wt uA \n" + ] + } + ], + "source": [ + "# for the instrumentation amplifier find Vo1 , Vo2 , Vo \n", + "# Vi1 = -25 sin wt # # mV\n", + "# Vi2 = 25 sin wt # # mV\n", + "R1 = 10*10**3 #\n", + "R2 = 20*10**3 #\n", + "R3 = 20*10**3 #\n", + "R4 = 10*10**3 #\n", + "\n", + "# the output of first op-amp A1 is given by\n", + "# Vo1 = (1+(R2/R1))*Vi1-(R2/R1)*Vi2 #\n", + "#by solving above equation we get\n", + "print 'The output of first op-amp A1 is = -275*sin wt mV '#\n", + "\n", + "# the output of second op-amp A2 is given by\n", + "# Vo2 = (1+(R2/R1))*Vi2-(R2/R1)*Vi1 #\n", + "#by solving above equation we get\n", + "print 'The output of second op-amp A2 is = 275*sin wt mV '#\n", + "\n", + "# the output of third op-amp A3 is given by\n", + "# Vo = (R4/R3)-(1+(2R2/R1)*(Vi2-Vi1) #\n", + "#by solving above equation we get\n", + "print 'The output of third op-amp A3 is = 825*sin wt mV '#\n", + "\n", + "# current through the resistor R1 and R2 is\n", + "#i = (Vi1-Vi2)/R1 #\n", + "print 'current through the resistor R1 and R2 is = 5 sin wt uA '#\n", + "\n", + "# current through the non-inverting terminal resistor R3 and R4 \n", + "#i3 = Vo2/(R3+R4)#\n", + "print 'current through the non-inverting terminal resistor R3 and R4 = 5.5 sin wt uA '#\n", + "\n", + "# current through the inverting terminal resistor R3 and R4 \n", + "#i2 = Vo1-(R3/(R3+R4))*Vo2/R3 #\n", + "print 'current through the inverting terminal resistor R3 and R4 = 22 sin wt uA '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.16 Pg 202" + ] + }, + { + "cell_type": "code", + "execution_count": 52, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input resistance Rin is = 0.00999 ohm \n", + "The value of Resistance Rs is = 1.0990 K ohm \n" + ] + } + ], + "source": [ + " # for the a current to voltage converter show a) Rin = (Rf/1+Aop) b) Rf = 10 K ohm Aop = 1000 \n", + "\n", + "#a) The input resistance given as\n", + "#Rin = (Rf)/(1+Aop) #\n", + "\n", + "# The input resistance of the circuit can be written as\n", + "#Rin = (V1/i!)#\n", + "\n", + "# the feedback current of the given circuit is defined as\n", + "#i1 =(V1-Vo)/RF #\n", + "\n", + "# the feedback resistance RF is \n", + "#RF =(V1-Vo)/i1 #\n", + "\n", + "# The output voltage Vo is\n", + "#Vo = -Aop*V1 #\n", + "\n", + "#by using this output feedback currenty i1 can be reformed as\n", + "#i1 = (V1-(-Aop*V1))/RF #\n", + "\n", + "#i1 = V1*(1+Aop)/RF #\n", + "\n", + "# Then Rin Becomes \n", + "#Rin =Rf/(1+Aop)#\n", + "\n", + "Rf =10*10**3 #\n", + "Aop = 1000 #\n", + "\n", + "# the input current and output voltage of the circuit are defined as\n", + "#i1 =(Rs)/(Rs+Rin) #\n", + "# Vo = -(Aop*(RF/1+Aop))*i1 #\n", + "\n", + "#the input resistance Rin is \n", + "Rin =(Rf/(1+Aop)) #\n", + "\n", + "# subsituting the value of RF Aop Rin and Vo we get \n", + "RF = 10 #\n", + "Rin = RF/(1+Aop)\n", + "print 'The input resistance Rin is = %0.5f'%Rin,' ohm '#\n", + "\n", + "Aop = 1000 #\n", + "#(1000/1001)*(Rs/(Rs*0.00999))> 0.99 #\n", + "# by solving above equation we get \n", + "Rs = 1.099 # # K ohm \n", + "print 'The value of Resistance Rs is = %0.4f'%Rs,' K ohm '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.17 Pg 204" + ] + }, + { + "cell_type": "code", + "execution_count": 54, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "for Aop = 10**4 closed loop gain is = 0.9999 \n", + "for Aop = 10**3 closed loop gain is = 0.9990 \n", + "for Aop = 10**2 closed loop gain is = 0.9901 \n", + "for Aop = 10**1 closed loop gain is = 0.9091 \n" + ] + } + ], + "source": [ + " # determine the closed loop gain\n", + "\n", + "# the output of the voltage follower is given as\n", + "#Vo = Aop(V1-Vo)#\n", + "\n", + "# the closed loop gain of the voltage follower \n", + "#A = 1/(1+(1/Aop))#\n", + " \n", + "# for Aop = 10**4 closed loop gain\n", + "Aop = 10**4 #\n", + "A = 1/(1+(1/Aop))#\n", + "print 'for Aop = 10**4 closed loop gain is = %0.4f'%A,' '#\n", + "\n", + "# for Aop = 10**3 closed loop gain\n", + "Aop = 10**3 #\n", + "A = 1/(1+(1/Aop))#\n", + "print 'for Aop = 10**3 closed loop gain is = %0.4f'%A,' '#\n", + "\n", + "# for Aop = 10**2 closed loop gain\n", + "Aop = 10**2 #\n", + "A = 1/(1+(1/Aop))#\n", + "print 'for Aop = 10**2 closed loop gain is = %0.4f'%A,' '#\n", + "\n", + "# for Aop = 10**1 closed loop gain\n", + "Aop = 10**1 #\n", + "A = 1/(1+(1/Aop))#\n", + "print 'for Aop = 10**1 closed loop gain is = %0.4f'%A,' '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.18 Pg 205" + ] + }, + { + "cell_type": "code", + "execution_count": 56, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "for the frequency f = 10 Hz the output is = 106.10 V \n", + "for the frequency f = 1000 Hz the output is = 1.06 V \n", + "for the frequency f = 10000 Hz the output is = 0.106 V \n" + ] + } + ], + "source": [ + "from math import pi\n", + "# To determine the output voltage of integrator\n", + "Vin = 1 #\n", + "R = 150*10**3 ## ohm\n", + "C = 1*10**-9 # # F\n", + "\n", + "# the output voltage of an integrator is given as\n", + "#Vo = (fc/f)*Vin #\n", + "\n", + "#fc = 1/(2*pi*R*C)#\n", + "\n", + "#Vo = (1/(2*pi*R*C*f))*Vin#\n", + "\n", + "#for the frequency f = 10 Hz the output is\n", + "f = 10 # # Hz\n", + "Vo = (1/(2*pi*R*C*f))*Vin#\n", + "print 'for the frequency f = 10 Hz the output is = %0.2f'%Vo,' V '#\n", + "\n", + "#for the frequency f = 1000 Hz the output is\n", + "f = 1000 # # Hz\n", + "Vo = (1/(2*pi*R*C*f))*Vin#\n", + "print 'for the frequency f = 1000 Hz the output is = %0.2f'%Vo,' V '#\n", + "\n", + "#for the frequency f = 10000 Hz the output is\n", + "f = 10000 # # Hz\n", + "Vo = (1/(2*pi*R*C*f))*Vin#\n", + "print 'for the frequency f = 10000 Hz the output is = %0.3f'%Vo,' V '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.19 Pg 206" + ] + }, + { + "cell_type": "code", + "execution_count": 63, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The cutoff frequency of the integrator is = 13.263 kHz\n", + "The gain of the integrator is = 3.18\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "# To determine the magnitude gain of the integrator\n", + "Vin = 1 #\n", + "f = 50*10**3 #\n", + "Rf = 120*10**3 #\n", + "R = 10*10**3 #\n", + "C = 0.1*10**-9 #\n", + "\n", + "# the magnitude gain of the integrator is given by\n", + "#A = (Rf/R)/(sqrt(1+(f/fc)**2))#\n", + "\n", + "# the cutoff frequency of the integrator \n", + "fc = 1/(2*pi*Rf*C)/1e3\n", + "print 'The cutoff frequency of the integrator is = %0.3f'%fc,'kHz'#\n", + "\n", + "\n", + "A = (Rf/R)/(sqrt(1+(f/fc)**2))*1e3#\n", + "print 'The gain of the integrator is = %0.2f'%A" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.20 Pg 207" + ] + }, + { + "cell_type": "code", + "execution_count": 67, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the break frequency fa is = 31.83 kHz \n", + "the break frequency fb is = 21.22 kHz \n", + "The gain of the differentiator is = 0.6667 \n" + ] + } + ], + "source": [ + " # To determine the magnitude gain of the differentiator\n", + "Vin = 1 #\n", + "f = 50*10**3 #\n", + "R = 75*10**3 #\n", + "R1 = 50*10**3 #\n", + "C = 0.1*10**-9 #\n", + "\n", + "# the magnitude gain of the differentiator is given by\n", + "#A = (f/fa)/(sqrt(1+(f/fb)**2))#\n", + "\n", + "# the break frequency fa is defined as\n", + "fa = 1/(2*pi*R1*C) / 1e3\n", + "print 'the break frequency fa is = %0.2f'%fa,'kHz '#\n", + "\n", + "# the break frequency fb is defined as\n", + "fb = 1/(2*pi*R*C) /1e3\n", + "print 'the break frequency fb is = %0.2f'%fb,'kHz '#\n", + "\n", + "\n", + "A = (f/fa)/(sqrt(1+(f/fb)**2))#\n", + "print 'The gain of the differentiator is = %0.4f'%A,' '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.21 Pg 209" + ] + }, + { + "cell_type": "code", + "execution_count": 70, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input voltage of an op-amp is = -40 mV\n" + ] + } + ], + "source": [ + " # to determine the input voltage of an op-amp\n", + "Vo = 2 # # V\n", + "R1 = 20*10**3 # # ohm\n", + "R2 = 1*10**6 # # ohm\n", + "\n", + "# the input voltage of an op-amp\n", + "Vin = -(R1/R2)*Vo *1000 # mV\n", + "print 'The input voltage of an op-amp is = %0.f'%Vin,'mV'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.22 Pg 209" + ] + }, + { + "cell_type": "code", + "execution_count": 71, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output voltage of follower Vo1 is = 2.00 V\n", + "The output voltage of an inverting amplifier is = -20.00 V \n" + ] + } + ], + "source": [ + "# To determine the output voltage\n", + "Vin = 2 #\n", + "R2 = 20*10**3 #\n", + "R1 = 2*10**3 #\n", + "\n", + "# the output voltage of follower Vo1 is\n", + "Vo1 = Vin #\n", + "print 'the output voltage of follower Vo1 is = %0.2f'%Vo1,' V'#\n", + "# the output voltage of an inverting amplifier\n", + "Vo = -(R2/R1)*Vo1 #\n", + "print 'The output voltage of an inverting amplifier is = %0.2f'%Vo,' V '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.23 Pg 210" + ] + }, + { + "cell_type": "code", + "execution_count": 72, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output of the inverting amplifier is = -15.00 V\n", + "The output of the non-inverting amplifier is = 20.00 V\n" + ] + } + ], + "source": [ + "# to determine the output voltage of an op-amp\n", + "Vin = 5 # # V\n", + "R1 = 25*10**3 # # ohm\n", + "R2 = 75*10**3 # # ohm\n", + "\n", + "# in this problem op-amp A1 perform the voltage follower and op-amp A2 perform inverting amplifier and op-amp A3 perform non-inverting amplifier\n", + "\n", + "# the output voltage of follower op-amp A1\n", + "Vo1 = Vin #\n", + "\n", + "# the output of the inverting amplifier A2\n", + "Vo2 = -((R2/R1)*Vo1) #\n", + "print 'The output of the inverting amplifier is = %0.2f'%Vo2,' V'#\n", + "\n", + "# the output of the non-inverting amplifier A3\n", + "Vo =(1+(R2/R1))*Vo1 #\n", + "print 'The output of the non-inverting amplifier is = %0.2f'%Vo,' V'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.24 Pg 211" + ] + }, + { + "cell_type": "code", + "execution_count": 73, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output voltage of an inverting amplifier is = 27.50 V \n", + "the output voltage of follower Vo1 is = 2.50 V\n", + "the output of the inverting summing amplifier is = 101.25 V \n" + ] + } + ], + "source": [ + " # To determine the output voltage\n", + "Vin = 2.5 #\n", + "Rf = 100*10**3 #\n", + "R1 = 10*10**3 #\n", + "RI1 = 25*10**3 #\n", + "RI2 = 10*10**3 #\n", + "R2 = 100*10**3 #\n", + "\n", + "# the output voltage of an inverting amplifier\n", + "Vo1 = (1+(R2/R1))*Vin # #\n", + "print 'The output voltage of an inverting amplifier is = %0.2f'%Vo1,' V '#\n", + "\n", + "# the output voltage of follower Vo2 is\n", + "Vo2 = Vin #\n", + "print 'the output voltage of follower Vo1 is = %0.2f'%Vo2,' V'#\n", + "\n", + "# the output of the inverting summing amplifier\n", + "R2 = 75*10**3 #\n", + "Vo = R2*((Vo1/RI1)+(Vo2/RI2))#\n", + "print 'the output of the inverting summing amplifier is = %0.2f'%Vo,' V '#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.25 Pg 212" + ] + }, + { + "cell_type": "code", + "execution_count": 74, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the total gain of the circuit is = 36.00 \n", + "The output voltage of the op-amp is = 90.00 V\n" + ] + } + ], + "source": [ + "# To calculate the output voltage\n", + "Vin = 2.5 # \n", + "R1 = 10*10**3 #\n", + "R2 = 10*10**3 #\n", + "R3 = 10*10**3 #\n", + "Rf = 30*10**3 #\n", + "\n", + "# the total gain of the circuit \n", + "#Av =A1v*A2v*A3v #\n", + "Av = (1+(Rf/R1))*(-Rf/R2)*(-Rf/R3)#\n", + "print 'the total gain of the circuit is = %0.2f'%Av,' '#\n", + "\n", + "# The output voltage of the op-amp \n", + "Vo = Av*Vin #\n", + "print 'The output voltage of the op-amp is = %0.2f'%Vo,' V'#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.26 Pg 212" + ] + }, + { + "cell_type": "code", + "execution_count": 76, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output of op-amp A1 is = -10.00 V1\n", + "The output of op-amp A2 is Vo = 40V1 - 2V2 \n", + "The output is equal to the difference between 40V1 and 2V2 \n" + ] + } + ], + "source": [ + "# to calculate the output voltage of op-amp circuit\n", + "Rf = 100*10**3 # # ohm\n", + "R1 = 10*10**3 # # ohm\n", + "R2 = 25*10**3 # # ohm\n", + "R3 = 50*10**3 # # ohm\n", + "\n", + "# the output of op-amp A1 is\n", + "# VA1 = (-Rf/R1)*V1 #\n", + "VA1 = (-Rf/R1)#\n", + "print 'The output of op-amp A1 is = %0.2f'%VA1,'V1' # # *V1 because the output is come from 1 op-amp\n", + "\n", + "# the output of op-amp A2 is\n", + "# Vo = -Rf*((VA1/R2)+(V2/R3))#\n", + "#Vo = -100*(-0.4*V1+0.02V2)#\n", + "print 'The output of op-amp A2 is Vo = 40V1 - 2V2 '# \n", + "\n", + "print 'The output is equal to the difference between 40V1 and 2V2 '# " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.27 Pg 213" + ] + }, + { + "cell_type": "code", + "execution_count": 77, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the upper crossover voltage of schmitt trigger is = 1.00 V\n", + "the lower crossover voltage of schmitt trigger is = -1.00 V\n", + "the hysteresis width HW of schmitt trigger is = 2.00 V\n" + ] + } + ], + "source": [ + " # to determine the hysteresis width of a schmitt trigger\n", + "R1 = 25*10**3 # # ohm\n", + "R2 = 75*10**3 # # ohm\n", + "VTH = 4 # # V\n", + "VTL = -4 # # V\n", + "\n", + "# the upper crossover voltage of schmitt trigger is defined as\n", + "VU = (R1/(R1+R2))*VTH#\n", + "print 'the upper crossover voltage of schmitt trigger is = %0.2f'%VU,' V'\n", + "\n", + "# the lower crossover voltage of schmitt trigger is defined as\n", + "VL = (R1/(R1+R2))*VTL#\n", + "print 'the lower crossover voltage of schmitt trigger is = %0.2f'%VL,' V'\n", + "\n", + "# the hysteresis width of schmitt trigger is\n", + "HW = VU-VL #\n", + "print 'the hysteresis width HW of schmitt trigger is = %0.2f'%HW,' V'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.28 Pg 214" + ] + }, + { + "cell_type": "code", + "execution_count": 78, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the upper crossover voltage of schmitt trigger is = 1.43 V\n", + "the lower crossover voltage of schmitt trigger is = -1.43 V\n", + "the hysteresis width HW of schmitt trigger is = 2.86 V\n" + ] + } + ], + "source": [ + " # to determine the hysteresis width of a schmitt trigger\n", + "R1 = 15*10**3 # # ohm\n", + "R2 = 90*10**3 # # ohm\n", + "VTH = 10 # # V\n", + "VTL = -10 # # V\n", + "\n", + "# the upper crossover voltage of schmitt trigger is defined as\n", + "VU = (R1/(R1+R2))*VTH#\n", + "print 'the upper crossover voltage of schmitt trigger is = %0.2f'%VU,' V'\n", + "\n", + "# the lower crossover voltage of schmitt trigger is defined as\n", + "VL = (R1/(R1+R2))*VTL#\n", + "print 'the lower crossover voltage of schmitt trigger is = %0.2f'%VL,' V'\n", + "\n", + "# the hysteresis width of schmitt trigger is\n", + "HW = VU-VL #`\n", + "print 'the hysteresis width HW of schmitt trigger is = %0.2f'%HW,' V'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.29 Pg 214" + ] + }, + { + "cell_type": "code", + "execution_count": 80, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the value of resistance R1 is = 5.00 kohm\n" + ] + } + ], + "source": [ + "# to determine the resistance R1 when low and high saturated output states are given\n", + "R2 = 20*10**3 # # ohm\n", + "VH = 2 # # V crossover voltage\n", + "VL = -2 # # V crossover voltage\n", + "VOH = 10 # # V saturated output states\n", + "VOL = -10 # # V saturated output states\n", + "\n", + "# the upper crossover voltage of schmitt trigger is defined as\n", + "# V = (R1/(R1+R2))*VOH#\n", + "# solving above equation we get \n", + "# 2R1+2R2 = 10R1 #\n", + "R1 = (2*R2)/8/1000 # kohm\n", + "print 'the value of resistance R1 is = %0.2f'%R1,'kohm'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.30 Pg 215" + ] + }, + { + "cell_type": "code", + "execution_count": 82, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the value of resistance R2 is = 30.00 kohm\n" + ] + } + ], + "source": [ + "# to determine the value of resistance R1 and R2 when low and high saturated output states are given\n", + "VH = 3 # # V crossover voltage\n", + "VL = -3 # # V crossover voltage\n", + "VOH = 12 # # V saturated output states\n", + "VOL = -12 # # V saturated output states\n", + "\n", + "# the upper crossover voltage of schmitt trigger is defined as\n", + "# V = (R1/(R1+R2))*VOH#\n", + "# solving above equation we get \n", + "# 3R1+3R2 = 12R1 #\n", + "\n", + "# 3*R1 = R2 #\n", + "R1 = 10*10**3 # # ohm we assume\n", + "R2 = 3*R1 / 1e3#\n", + "print 'the value of resistance R2 is = %0.2f'%R2,'kohm'" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch7.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch7.ipynb new file mode 100644 index 00000000..b665ce02 --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch7.ipynb @@ -0,0 +1,722 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 7 Filters and Rectifiers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.1 Pg 232" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistor value is = 1.59 k ohm \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, sqrt\n", + "# Design active low filter with cut-off frequency 10 kHz\n", + "fc = 10 # # kHz\n", + "C = 0.01 # #uF # we assume\n", + "\n", + "# the cut-off frequency of active low pass filter is defined as\n", + "# fc = (1/2*pi*R3*C)#\n", + "\n", + "# R3 can be calculated as\n", + "R3 = (1/(2*pi*fc*C))#\n", + "print 'The resistor value is = %0.2f'%R3,' k ohm '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.2 Pg 233" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistor value is = 106 ohm \n", + "The pass band gain is = 1.50 \n" + ] + } + ], + "source": [ + " # Design active low filter with cut-off frequency 15 kHz\n", + "fc = 15*10**3 # # Hz \n", + "C = 0.1*10**-6 # #F # we assume\n", + "\n", + "# the cut-off frequency of active low pass filter is defined as\n", + "# fc = (1/2*pi*R3*C)#\n", + "\n", + "# R3 can be calculated as\n", + "R3 = (1/(2*pi*fc*C))#\n", + "print 'The resistor value is = %0.f'%R3,' ohm '\n", + "\n", + "# the pass band gain of filter is given by\n", + "# Af = 1+(R2/R1)#\n", + "# assume that the inverting terminal resistor R2=0.5*R1#\n", + "# in Af equation if we put R2=0.5R1 in R1 R1 cancellout each other \n", + "Af = 1+(0.5)\n", + "print 'The pass band gain is = %0.2f'%Af,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.3 Pg 234" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistor value is = 159 Kohm \n", + "The resistor R2 value is = 900.00 k ohm \n", + "The magnitude of an active low pass filter is = 1.96 \n", + "The phase angle of the filter is = -78.69 \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# Design active low filter with cut-off frequency 20 kHz\n", + "fc = 20 # # kHz \n", + "f = 100 # # frequency of filter\n", + "Af = 10 # # desired pass band gain\n", + "C = 0.05 # #nF # we assume\n", + "\n", + "# the cut-off frequency of active low pass filter is defined as\n", + "# fc = (1/2*pi*R3*C)#\n", + "\n", + "# R3 can be calculated as\n", + "R3 = (1/(2*pi*fc*1e3*C*1e-9))/1e3 # Kohm\n", + "print 'The resistor value is = %0.f'%R3,' Kohm '\n", + "\n", + "# the pass band gain of filter is given by\n", + "# Af = 1+(R2/R1)#\n", + "# assume that the inverting terminal resistor R1= 100 k ohm#\n", + "R1 = 100 # # k ohm\n", + "R2 = (Af*R1)-R1#\n", + "print 'The resistor R2 value is = %0.2f'%R2,' k ohm '\n", + "\n", + "# the magnitude of an active low pass filter is given as\n", + "A = Af/(sqrt(1+(f/fc)**2))#\n", + "print 'The magnitude of an active low pass filter is = %0.2f'%A,' '\n", + "\n", + "#the phase angle of the filter\n", + "from math import atan , degrees\n", + "Angle = -degrees(atan(f/fc))#\n", + "print 'The phase angle of the filter is = %0.2f'%Angle,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.4 Pg 236" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The frequency of the first order low pass filter is = 2.65 kHz \n", + "The pass band gain of filter is = 13.00 \n" + ] + } + ], + "source": [ + " # to determine the cut-off frequency and pass band gain Af\n", + "R1 = 1 # # k ohm\n", + "R2 = 12 # # k ohm\n", + "R3 = 1.2 # # k ohm\n", + "C = 0.05 # #uF # we assume\n", + "\n", + "# the frequency of the first order low pass filter is defined as\n", + "fc = (1/(2*pi*R3*C))#\n", + "print 'The frequency of the first order low pass filter is = %0.2f'%fc,' kHz '\n", + "\n", + "# the pass band gain of filter is given by\n", + "Af =(1+R2/R1)#\n", + "print 'The pass band gain of filter is = %0.2f'%Af,''" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.5 Pg 236" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The capacitor of high pass filter is = 25.13 uF \n", + "The second resistor value is = 90.00 K ohm \n" + ] + } + ], + "source": [ + " # to design a first order high pass filter with cut-off frequency 2kHz\n", + "Af = 10 #\n", + "fc = 2 # # kHz \n", + "R3 = 2 # #K ohm # we assume\n", + "R1 = 10 # # k ohm\n", + "# the capacitor of high pass filter is given by\n", + "C = 2*pi*R3*fc#\n", + "print 'The capacitor of high pass filter is = %0.2f'%C,' uF '\n", + "\n", + "# the voltage gain of the high pass filter is\n", + "# Af = 1+(R2/R1)#\n", + "R2 = R1*(Af-1)#\n", + "print 'The second resistor value is = %0.2f'%R2,' K ohm '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.6 Pg 237" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance R3 is = 1.59 K ohm \n" + ] + } + ], + "source": [ + " # to design an active high pass filter with cut-off frequency 10kHz\n", + "fc = 10 # # kHz \n", + "C = 0.01 # #uF # we assume\n", + "# the cut-off frequency of active high pass filter is given by\n", + "# fc = 2*pi*R3*C#\n", + "# R3 can be calculated as\n", + "R3 = (1/(2*pi*fc*C))#\n", + "print 'The resistance R3 is = %0.2f'%R3,' K ohm '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.7 Pg 238" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance R3 is = 64 Kohm \n", + "The pass band gain is = 1.20 \n" + ] + } + ], + "source": [ + "# to design an active high pass filter with cut-off frequency 25kHz\n", + "fc = 25 # # kHz \n", + "C = 0.1 # #nF # we assume\n", + "# the cut-off frequency of active high pass filter is given by\n", + "# fc = 2*pi*R3*C#\n", + "# R3 can be calculated as\n", + "R3 = (1/(2*pi*fc*1e3*C*1e-9)) / 1e3 # Kohm\n", + "print 'The resistance R3 is = %0.f'%R3,' Kohm '\n", + "\n", + "# the desire pass band gain of filter is given by \n", + "#Af = 1+(R2/R1)#\n", + "# assume that the inverting terminal resistor R2=0.2*R1#\n", + "# in Af equation if we put R2=0.2R1 in R1 R1 cancellout each other \n", + "Af = 1+(0.2)\n", + "print 'The pass band gain is = %0.2f'%Af,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.8 Pg 239" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance R3 is = 159 K ohm \n", + "The resistance R2 is = 700.00 K ohm \n", + "The magnitude of an active high pass filter is = 14.55 \n", + "The phase angle of the filter is = 14.04 degree\n" + ] + } + ], + "source": [ + "## # to design an active high pass filter with cut-off frequency 20kHz \n", + "Af = 15 #\n", + "fc = 20 # #kHz\n", + "f = 80 # # kHz the frequency of filter \n", + "C = 0.05 # #nF # we assume\n", + "# the cut-off frequency of active high pass filter is given by\n", + "# fc = 2*pi*R3*C#\n", + "# R3 can be calculated as\n", + "R3 = (1/(2*pi*fc*C))#\n", + "print 'The resistance R3 is = %0.f'%(R3*1000),' K ohm ' # Round Off Error\n", + "\n", + "# the desire pass band gain of filter is given by \n", + "#Af = 1+(R2/R1)#\n", + "# assume that the inverting terminal resistor R1=50 K ohm#\n", + "R1 = 50 # # K ohm\n", + "R2 = (R1*Af)-(R1)\n", + "print 'The resistance R2 is = %0.2f'%R2,' K ohm '\n", + "\n", + "# the magnitude of an active high pass filter is given as\n", + "A = Af*(f/fc)/(sqrt(1+(f/fc)**2))#\n", + "print 'The magnitude of an active high pass filter is = %0.2f'%A,' '\n", + "\n", + "#the phase angle of the filter\n", + "from numpy import inf\n", + "Angle = degrees(-atan(f/fc)+atan(inf))\n", + "print 'The phase angle of the filter is = %0.2f'%Angle,' degree' # Round Off Error" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.9 Pg 241" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The lower cut-off frequency FLC of band pass filter is = 159.2 Hz \n", + "The upper cut-off frequency FUC of band pass filter is = 15.92 kHz \n" + ] + } + ], + "source": [ + "# to calculate upper and lower cut-off frequency of the band pass filter\n", + "R1 = 10*10**3 # #K ohm\n", + "R2 = 10 # #K ohm\n", + "C1 = 0.1*10**-6 # # uF\n", + "C2 = 0.001 # #uF\n", + "\n", + "# the lower cut-off frequency of band pass filter is\n", + "fLC = 1/(2*pi*R1*C1)#\n", + "print 'The lower cut-off frequency FLC of band pass filter is = %0.1f'%fLC,' Hz '\n", + "\n", + "# The upper cut-off frequency of band pass filter is\n", + "fUC = 1/(2*pi*R2*C2)#\n", + "print 'The upper cut-off frequency FUC of band pass filter is = %0.2f'%fUC,' kHz '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.10 Pg 242" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance R3 Value is = 7.96 M ohm \n", + "The resistance R6 value is = 1.59 M ohm \n" + ] + } + ], + "source": [ + "# to design an active band pass filter with lower cut-off frequency 10 kHz an upper 50 kHz\n", + "fL = 10 # # kHz\n", + "fH = 50 # # kHz\n", + "C1 = 0.002 # # nF\n", + "C2 = 0.002 # # nF\n", + "\n", + "# the lower cut-off frequency of band pass filter is\n", + "# fL = 1/(2*pi*R3*C1)#\n", + "R3 = 1/(2*pi*fL*C1)#\n", + "print 'The resistance R3 Value is = %0.2f'%R3,' M ohm '\n", + "\n", + "# The upper cut-off frequency of band pass filter is\n", + "# fH = 1/(2*pi*R6*C2)#\n", + "R6 = 1/(2*pi*fH*C2)#\n", + "print 'The resistance R6 value is = %0.2f'%R6,' M ohm '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.11 Pg 243" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance R3 Value is = 7.96 M ohm \n", + "The resistance R6 value is = 3.98 M ohm \n", + "The desire pass band gain of filter is = 15 \n" + ] + } + ], + "source": [ + "# to design an active band pass filter with lower cut-off frequency 20 kHz an upper 40 kHz\n", + "fL = 20 # # kHz\n", + "fH = 40 # # kHz\n", + "# the inverting terminal resistance 2R1=R2 and 4R4=R5\n", + "C1 = 0.001 # # nF\n", + "C2 = 0.001 # # nF\n", + "\n", + "# the lower cut-off frequency of band pass filter is\n", + "# fL = 1/(2*pi*R3*C1)#\n", + "R3 = 1/(2*pi*fL*C1)#\n", + "print 'The resistance R3 Value is = %0.2f'%R3,' M ohm '\n", + "\n", + "# The upper cut-off frequency of band pass filter is\n", + "# fH = 1/(2*pi*R6*C2)#\n", + "R6 = 1/(2*pi*fH*C2)#\n", + "print 'The resistance R6 value is = %0.2f'%R6,' M ohm '\n", + "\n", + "# the desire pass band gain of filter is defined as\n", + "R1 = 1 # # M ohm we assume\n", + "#we define inverting terminal resistance 2R1=R2\n", + "R2 = 2 # # M ohm\n", + "# then\n", + "R4 = 1 # #M ohm\n", + "R5 = 4 # # M ohm\n", + "Af = (1+(R2/R1))*(1+(R5/R4))#\n", + "print 'The desire pass band gain of filter is = %d'%Af,' '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.12 Pg 244" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance R3 Value is = 7.96 M ohm \n", + "The resistance R6 value is = 1.99 M ohm \n", + "The desire pass band gain of filter is = 15.00 \n", + "The magnitude of gain of band pass filter is = 11.49 \n", + "The phase angle of gain of band pass filter is = 50 degree\n" + ] + } + ], + "source": [ + "# to design an active band pass filter with lower cut-off frequency 20 kHz an upper 80 kHz\n", + "f = 100 # # kHz the frequency of band pass filter\n", + "fL = 20 # # kHz\n", + "fH = 80 # # kHz\n", + "# the inverting terminal resistance R1=0.5*R2 and R4=0.25*R5\n", + "C1 = 0.001 # # nF\n", + "C2 = 0.001 # # nF\n", + "\n", + "# the lower cut-off frequency of band pass filter is\n", + "# fL = 1/(2*pi*R3*C1)#\n", + "R3 = 1/(2*pi*fL*C1)#\n", + "print 'The resistance R3 Value is = %0.2f'%R3,' M ohm '\n", + "\n", + "# The upper cut-off frequency of band pass filter is\n", + "# fH = 1/(2*pi*R6*C2)#\n", + "R6 = 1/(2*pi*fH*C2)#\n", + "print 'The resistance R6 value is = %0.2f'%R6,' M ohm ' # Round Off Error\n", + "\n", + "# the desire pass band gain of filter is defined as\n", + "R1 = 1 # # M ohm we assume\n", + "#we define inverting terminal resistance R1=0.5*R2\n", + "R2 = 2 # # M ohm\n", + "# then\n", + "R4 = 1 # #M ohm\n", + "R5 = 4 # # M ohm\n", + "Af = (1+(R2/R1))*(1+(R5/R4))#\n", + "print 'The desire pass band gain of filter is = %0.2f'%Af,' '\n", + "\n", + "# the magnitude of gain of band pass filter is given as\n", + "A = Af*(f**2/(fL*fH))/((sqrt(1+(f/fL)**2))*(sqrt(1+(f/fH)**2)))#\n", + "print 'The magnitude of gain of band pass filter is = %0.2f'%A,' ' # Round Off Error\n", + "\n", + "#the phase angle of the filter\n", + "from numpy import inf\n", + "Angle = degrees(2*atan(inf)-atan(f/fL)-atan(f/fH))\n", + "print 'The phase angle of gain of band pass filter is = %0.f'%Angle,'degree' # Round Off Error" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.13 Pg 247" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output of the half wave precision rectifier Vo is = -20.00 V \n" + ] + } + ], + "source": [ + " # to determine the output voltage of the precision rectifier circuit\n", + "Vi = 10 # #V i/p volt\n", + "R1 = 20 # # K ohm\n", + "R2 = 40 # # K ohm\n", + "Vd = 0.7 # # V the diode voltage drop\n", + "\n", + "# the output of the half wave precision rectifier is defined as\n", + "# Vo = -(R2/R1)*Vi # for Vi < 0\n", + "# = 0 otherwise\n", + "# i.e for Vi > 0\n", + "# Vo = 0\n", + "# for Vi < 0\n", + "Vo = -(R2/R1)*Vi\n", + "print 'The output of the half wave precision rectifier Vo is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.14 Pg 247" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output of the half wave precision rectifier Vo is = -15.00 V \n", + "The output of the half wave precision rectifier Vo is = 15.00 V \n" + ] + } + ], + "source": [ + "# to determine the output voltage of the precision rectifier circuit for i/p voltage a) Vi = 5 b) Vi = -5\n", + "Vi = 5 # #V i/p volt\n", + "R1 = 5 # # K ohm\n", + "R2 = 15 # # K ohm\n", + "Vd = 0.7 # # V the diode voltage drop\n", + "\n", + "# the output of the half wave precision rectifier is defined as\n", + "# Vo = -(R2/R1)*Vi # for Vi < 0\n", + "# = 0 otherwise\n", + "\n", + "# for Vi = 5 V\n", + "# i.e for Vi > 0\n", + "# Vo = 0\n", + "# for Vi < 0\n", + "Vo = -(R2/R1)*Vi#\n", + "print 'The output of the half wave precision rectifier Vo is = %0.2f'%Vo,' V '\n", + "\n", + "# for Vi = -5 V\n", + "# i.e for Vi > 0\n", + "# Vo = 0\n", + "# for Vi < 0\n", + "Vi =-5 # # V\n", + "Vo = -(R2/R1)*Vi#\n", + "print 'The output of the half wave precision rectifier Vo is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.15 Pg 248" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The gain of precision full wave rectifier A is = 6.00 \n", + "The output voltage Vo is = 42.00 V \n", + "The output voltage Vo is = 42.00 V \n" + ] + } + ], + "source": [ + " # to determine the output voltage of the precision rectifier circuit for i/p voltage a) Vi = 7 b) Vi = -7\n", + "Vi = 7 # #V i/p volt\n", + "R1 = 5 # # K ohm\n", + "R3 = 5 # # K ohm\n", + "R4 = 5 # # K ohm\n", + "R2 = 15 # # K ohm\n", + "R5 = 15 # # K ohm\n", + "Vd = 0.7 # # V the diode voltage drop\n", + "\n", + "# the output of the full wave precision rectifier is defined as\n", + "# Vo = -A*Vi # for Vi < 0 equation 1\n", + "# = A*Vi # otherwise equation 2\n", + "\n", + "# or Vo = abs(A*Vi) #\n", + "\n", + "# The gain of precision full wave rectifier\n", + "A = (((R2*R5)/(R1*R3))-(R5/R4)) #\n", + "print 'The gain of precision full wave rectifier A is = %0.2f'%A,' '\n", + "\n", + "\n", + "# for Vi = 7 V the output voltage is\n", + "Vi = 7 #\n", + "Vo = -A*Vi # # from equation 1\n", + "Vo = A*Vi # # from equation 2\n", + "Vo = abs(A*Vi) #\n", + "print 'The output voltage Vo is = %0.2f'%Vo,' V '\n", + "\n", + "# for Vi = -7 V the output voltage is\n", + "Vi = -7 #\n", + "Vo = -A*Vi # # from equation 1\n", + "Vo = A*Vi # # from equation 2\n", + "Vo = abs(A*Vi) #\n", + "print 'The output voltage Vo is = %0.2f'%Vo,' V '" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch8.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch8.ipynb new file mode 100644 index 00000000..df35b96e --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch8.ipynb @@ -0,0 +1,180 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8 Analog Multiplier" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.1 Pg 267" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output voltage of inverting amplifier (V2) is = 3.00 V\n" + ] + } + ], + "source": [ + "from math import sqrt, pi\n", + "from __future__ import division\n", + "# to determine the output voltage of inverting amplifier (V2)\n", + "Vin = 18 # # V\n", + "V1 = -6 # # V\n", + "\n", + "# in the op-amp due to the infinite i/p resiostance the input current is = 0\n", + "# i1+i2 = 0\n", + "# it gives relation\n", + "Vo = -Vin #\n", + "\n", + "# the output of multiplier is defined as\n", + "#Vo = K*V1*V2\n", + "\n", + "K = 1 # # we assume\n", + "\n", + "V2 = (Vo/(K*V1))#\n", + "print 'the output voltage of inverting amplifier (V2) is = %0.2f'%(V2),'V'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.2 Pg 267" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output voltage of multiplier is = 225.00 V\n" + ] + } + ], + "source": [ + "# to determine the output voltage of multiplier\n", + "Vin = 15 # # V\n", + "\n", + "# the output of multiplier is defined as\n", + "#Vo = K*V1*V2\n", + "# because of i/p terminal the circuit performs mathematical operation squaring\n", + "# i.e V1 = V2 = Vin\n", + "K = 1 # # we assume\n", + "Vo = K*(Vin)**2#\n", + "print 'the output voltage of multiplier is = %0.2f'%(Vo),'V' " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.3 Pg 268" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output voltage of inverting amplifier is = 4.00 V \n", + "the output voltage of multiplier is = 16.00 V \n" + ] + } + ], + "source": [ + "from math import sqrt, pi\n", + "# to determine the output voltage of multiplier and inverting amplifier\n", + "Vin = 16 #\n", + "# the output of the inverting amplifier\n", + "K =1 # # we assume\n", + "Vos = sqrt(abs(Vin)/K) #\n", + "print 'the output voltage of inverting amplifier is = %0.2f'%(Vos),' V '\n", + "\n", + "# the output of the multiplier\n", + "Vo = K*Vos**2 #\n", + "print 'the output voltage of multiplier is = %0.2f'%(Vo),' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.5 Pg 269" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "output voltage of RMS detector is = 10.00 V \n" + ] + } + ], + "source": [ + "# output voltage of of RMS detector\n", + "Vin = 10 # \n", + "T = 1 # # we assume that the charging and discharging period of capacitor\n", + "\n", + "# the output voltage of RMS detector\n", + "# Vo =sqrt(1/T*)integrate(Vin**2*(t),t,0,1 [,atol [,rtol]]) #\n", + "Vo = 10 #\n", + "print 'output voltage of RMS detector is = %0.2f'%(Vo),'V '" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch9.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch9.ipynb new file mode 100644 index 00000000..5ae5cfbe --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch9.ipynb @@ -0,0 +1,404 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9 Phase Locked Loop" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.1 Pg 284" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output voltage of switching regulator circuit is = -0.30 V \n", + "The output voltage of switching regulator circuit is = 1.50 V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt, pi\n", + "# to find output voltage for a constant input signal frequency of 200 KHz\n", + "fo = 2*pi*1*10**3 # # KHz/V # VCO sensitivity range 4.1\n", + "fc = 500 # # Hz a free running frequency\n", + "f1 = 200 # # Hz input frequency\n", + "f2 = 2*10**3 # # Hz input frequency\n", + "\n", + "# the output voltage of PLL is defined as\n", + "#Vo = (wo-wc)/ko\n", + "ko = fo #\n", + "# when i/p locked with o/p wo=wi\n", + "# Vo = (wi-wc)/ko #\n", + "\n", + "#for the i/p frequency fi = 200 Hz\n", + "fi = 200 # # Hz\n", + "Vo = (((2*pi*fi)-(2*pi*fc))/ko)#\n", + "print 'The output voltage of switching regulator circuit is = %0.2f'%Vo,' V '\n", + "\n", + "#for the i/p frequency fi = 200 Hz\n", + "fi = 2*10**3 # # Hz\n", + "Vo = (((2*pi*fi)-(2*pi*fc))/ko)#\n", + "print 'The output voltage of switching regulator circuit is = %0.2f'%Vo,' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.2 Pg 285" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The sum frequency produce by phase detector is = 900.00 KHz \n", + "The difference frequency produce by phase detector is = 100.00 KHz \n", + "The phase detector frequencies are outside of the low pass filter\n", + "The VCO will be in its free running frequency \n" + ] + } + ], + "source": [ + " # to find VCO output frequency\n", + "fc = 400 # # KHz a free running frequency\n", + "f = 10 # # KHz low pass filter bandwidth\n", + "fi = 500 # # KHz input frequency\n", + "\n", + "# In PLL a phase detector produces the sum and difference frequencies are defined as\n", + "\n", + "sum = fi+fc #\n", + "print 'The sum frequency produce by phase detector is = %0.2f'%sum,' KHz '\n", + "\n", + "difference = fi-fc #\n", + "print 'The difference frequency produce by phase detector is = %0.2f'%difference,' KHz '\n", + "\n", + "print 'The phase detector frequencies are outside of the low pass filter'#\n", + "\n", + "print 'The VCO will be in its free running frequency '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.3 Pg 286" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The sensitivity of phase detector Kd is = 0.45 \n", + "The maximum control voltage of VCO Vfmax = 1.40 V\n", + "The maximum frequency swing of VCO = 35.00 KHz\n", + "The maximum range of frequency which lock a PLL is = 15.00 KHz \n", + "The maximum range of frequency which lock a PLL is = 85.00 KHz \n", + "The maximum and minimum rage between 15 KHz to 85 KHZ \n", + "The lock range is = 70.00 KHz \n" + ] + } + ], + "source": [ + "# to determine the lock range of PLL\n", + "Ko = 25 # # KHz\n", + "fo = 50 # # KHz\n", + "A = 2 #\n", + "Vd = 0.7 #\n", + "AL = 1 #\n", + "\n", + "# the amximum output swing of phase detector \n", + "# Vd = Kd*(pi/2) #\n", + "\n", + "# the sensitivity of phase detector Kd is\n", + "Kd = Vd*(2/pi) #\n", + "print 'The sensitivity of phase detector Kd is = %0.2f'%Kd,''\n", + "\n", + "# The maximum control voltage of VCO Vfmax\n", + "Vfmax = (pi/2)*Kd*A #\n", + "print 'The maximum control voltage of VCO Vfmax = %0.2f'%Vfmax,' V'\n", + "\n", + "# the maximum frequency swing of VCO\n", + "fL = (Ko*Vfmax)#\n", + "print 'The maximum frequency swing of VCO = %0.2f'%fL,' KHz'\n", + "\n", + "# The maximum range of frequency which lock a PLL are\n", + "fi = fo-fL #\n", + "print 'The maximum range of frequency which lock a PLL is = %0.2f'%fi,' KHz '\n", + "\n", + "fi = fo+fL #\n", + "print 'The maximum range of frequency which lock a PLL is = %0.2f'%fi,' KHz '\n", + "\n", + "print 'The maximum and minimum rage between 15 KHz to 85 KHZ '\n", + "\n", + "\n", + "# the lock range is\n", + "fLock = 2*fL #\n", + "print 'The lock range is = %0.2f'%fLock,' KHz '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.4 Pg 286" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current through the control resistor R is = 0.60 mA \n", + "The charging time of capacitor is = 5.00 msec \n", + "The total time period of tringular and square wave is = 10.00 msec \n", + "The output frequency of VCO is = 0.10 KHz \n" + ] + } + ], + "source": [ + "# to determine the output frequency capacitor charging time of VCO\n", + "Vcc = 12 #\n", + "Vcs = 6\n", + "R = 10 # # K ohm\n", + "C = 1 # # uF\n", + "\n", + "# the current through the control resistor R\n", + "i =(Vcc-Vcs)/R #\n", + "print 'The current through the control resistor R is = %0.2f'%i, ' mA '\n", + "\n", + "# The charging time of capacitor \n", + "t = (0.25*Vcc*C)/i #\n", + "print 'The charging time of capacitor is = %0.2f'%t, ' msec '\n", + "\n", + "# In VCO the capacitor charging and discharging time period are equal ,so the total time period of tringular and square wave forms can be written as 2*t #\n", + "t = ((0.5*Vcc*C)/i)#\n", + "print 'The total time period of tringular and square wave is = %0.2f'%t, ' msec '\n", + "\n", + "# the output frequency of VCO is\n", + "fo = 1/t #\n", + "print 'The output frequency of VCO is = %0.2f'%fo, ' KHz '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.5 Pg 287" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The charging or discharging time of capacitor is = 25.00 msec \n", + "The output frequency of VCO is is = 20.00 Hz \n", + "The output frequency of VCO is = 625.00 Kohm\n", + "The current through the control resistor R is = 1.60 uA \n", + "The capacitor charging current is = 2000.00 V = 0.33Vcc \n" + ] + } + ], + "source": [ + "# to design VCO with output square wave pulse time of 50 msec\n", + "Vcc =6 #\n", + "Vcs = 5 #\n", + "R = 22 # #K ohm\n", + "C = 0.02 # # uF\n", + "t = 50*10**-3 # # sec output square wave pluse\n", + "\n", + "# In VCO the capacitor charging and discharging time period are equal ,so the total time period of tringular and square wave forms can be written as 2*t #\n", + "\n", + "\n", + "# the charging or discharging time of capacitor \n", + "tcap = t/2*1e3 #\n", + "print 'The charging or discharging time of capacitor is = %0.2f'%tcap, ' msec '\n", + "\n", + "# the output frequency of VCO is\n", + "fo = 1/t #\n", + "print 'The output frequency of VCO is is = %0.2f'%fo, ' Hz '\n", + "\n", + "# the output frequency of VCO\n", + " # fo = (1/4*R*C)#\n", + "R = 1/(4*fo*1e3*C*1e-9)/1e3 # Kohm\n", + "print 'The output frequency of VCO is = %0.2f'%R, ' Kohm'\n", + "\n", + "# the current through the control resistor R\n", + "i =(Vcc-Vcs)/R*1e3 #\n", + "print 'The current through the control resistor R is = %0.2f'%i, ' uA '\n", + "\n", + "# the capacitor charging current \n", + "# (V/t)=(i/C) #\n", + "V = (i/C)*tcap #\n", + "print 'The capacitor charging current is = %0.2f'%V, ' V = 0.33Vcc '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.6 Pg 289" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The center frequency of VCO is is = 0.17 kHz \n", + "The lock range of PLL is = 2.67 KHz/V \n", + "The lock range of PLL is = 25.59 k Hz/V \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# to determine the center frequency of VCO lock and capture range of PLL\n", + "R = 15 # # K ohm\n", + "C = 0.12 # # uF\n", + "Vcc = 12 #\n", + "\n", + "# the center frequency of VCO fo\n", + "fo = (1.2/4/(R*1e3)/(C*1e-6))/1e3#\n", + "print 'The center frequency of VCO is is = %0.2f'%fo, ' kHz '\n", + "\n", + "fo = 4 # # KHz\n", + "# the lock range of PLL\n", + "fL = (8*fo/Vcc) #\n", + "print 'The lock range of PLL is = %0.2f'%fL, ' KHz/V '\n", + "\n", + "# the capture range of PLL\n", + "fc = ((fo-fL)/(2*pi*3.6*10**3*C*1e-6)**(1/2)) #\n", + "print 'The lock range of PLL is = %0.2f'%fc, 'k Hz/V '\n", + "# ans wrong in the textbook." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.7 Pg 290" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total time period of VCO is = 5.00 usec \n", + "The charging or discharging time of capacitor is = 2.50 usec \n", + "The voltage swing of VCO for 12 V supply is = 3.00 V \n", + "The lock range of PLL FL is = 0.955 Hz \n", + "The capture range is = 437.02 Hz \n" + ] + } + ], + "source": [ + "# determine the lock range of the FSK demodulator\n", + "Vcc = 12 #\n", + "Fvco = 0.25*Vcc #\n", + "f = 200*10**3 # # Hz\n", + "\n", + "\n", + "# the total time period of VCO \n", + "t = 1/f*1e6 #\n", + "print 'The total time period of VCO is = %0.2f'%t, ' usec '\n", + "\n", + "# In VCO the capacitor charging and discharging time period are equal ,so the total time period of tringular and square wave forms can be written as 2*t #\n", + "\n", + "\n", + "# the charging or discharging time of capacitor \n", + "tcap = t/2 #\n", + "print 'The charging or discharging time of capacitor is = %0.2f'%tcap, ' usec '\n", + "\n", + "# the voltage swing of VCO for 12 V supply\n", + "Fvco = 0.25*Vcc #\n", + "print 'The voltage swing of VCO for 12 V supply is = %0.2f'%Fvco, ' V '\n", + "\n", + "# The lock range of PLL \n", + "#FL = (1/2*pi*f)*(Fvco/tcap)#\n", + "FL = (3/(2*pi*f*tcap*1e-6))#\n", + "print 'The lock range of PLL FL is = %0.3f'%FL, ' Hz '\n", + "\n", + "# the capture range \n", + "fcap = sqrt(f*FL)#\n", + "print 'The capture range is = %0.2f'%fcap, ' Hz '" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/BiasingVoltage.png b/Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/BiasingVoltage.png Binary files differnew file mode 100644 index 00000000..3079217b --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/BiasingVoltage.png diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/InOutCrrnt.png b/Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/InOutCrrnt.png Binary files differnew file mode 100644 index 00000000..ebf45d77 --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/InOutCrrnt.png diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/inAndOutCurrnt.png b/Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/inAndOutCurrnt.png Binary files differnew file mode 100644 index 00000000..a6158aa6 --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/inAndOutCurrnt.png diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter10.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter10.ipynb new file mode 100644 index 00000000..c3f165d1 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter10.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.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter28.ipynb new file mode 100644 index 00000000..afb57de5 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter28.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.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter29.ipynb new file mode 100644 index 00000000..9f1a9134 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter29.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_/chapter1.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter1.ipynb new file mode 100644 index 00000000..f6180b5a --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter1.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_/chapter11.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter11.ipynb new file mode 100644 index 00000000..11322719 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter11.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.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter12.ipynb new file mode 100644 index 00000000..c7cfe98a --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter12.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.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter13.ipynb new file mode 100644 index 00000000..cf5ab315 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter13.ipynb @@ -0,0 +1,1642 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:9195a4838870eb225ab235ab7007db341156360ed8273d6ec6c10a406508614b"
+ },
+ "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 tiem 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 tiem is 0.00246 S\n"
+ ]
+ }
+ ],
+ "prompt_number": 65
+ },
+ {
+ "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.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter14.ipynb new file mode 100644 index 00000000..7bb51839 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter14.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.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter15.ipynb new file mode 100644 index 00000000..111cd390 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter15.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.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter16.ipynb new file mode 100644 index 00000000..f74b4c1b --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter16.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.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter17.ipynb new file mode 100644 index 00000000..6ddbfa96 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter17.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.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter18.ipynb new file mode 100644 index 00000000..46d58979 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter18.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.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter19.ipynb new file mode 100644 index 00000000..3a7eac21 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter19.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_/chapter2.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter2.ipynb new file mode 100644 index 00000000..e9ce1a12 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter2.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_/chapter20.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter20.ipynb new file mode 100644 index 00000000..e96a7bf3 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter20.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.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter21.ipynb new file mode 100644 index 00000000..5e2ded02 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter21.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.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter22.ipynb new file mode 100644 index 00000000..e91e3a27 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter22.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.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter23.ipynb new file mode 100644 index 00000000..7d0fa841 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter23.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.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter24.ipynb new file mode 100644 index 00000000..b7102dac --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter24.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.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter25.ipynb new file mode 100644 index 00000000..95708b68 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter25.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.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter26.ipynb new file mode 100644 index 00000000..98e47e10 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter26.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.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter27.ipynb new file mode 100644 index 00000000..01e54494 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter27.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_/chapter3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter3.ipynb new file mode 100644 index 00000000..8aff7044 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter3.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.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter4.ipynb new file mode 100644 index 00000000..5987ff7f --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter4.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.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter5.ipynb new file mode 100644 index 00000000..639e5486 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter5.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.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter6.ipynb new file mode 100644 index 00000000..f1c6f4b7 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter6.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.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter7.ipynb new file mode 100644 index 00000000..74f3f441 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter7.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.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter8.ipynb new file mode 100644 index 00000000..baadbaea --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter8.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.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter9.ipynb new file mode 100644 index 00000000..7669d0e6 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter9.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": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/screenshots/image_1.png b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/screenshots/image_1.png Binary files differnew file mode 100644 index 00000000..4fac604e --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/screenshots/image_1.png diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/screenshots/image_2.png b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/screenshots/image_2.png Binary files differnew file mode 100644 index 00000000..36961025 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/screenshots/image_2.png diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/screenshots/image_3.png b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/screenshots/image_3.png Binary files differnew file mode 100644 index 00000000..dcb82ffc --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/screenshots/image_3.png diff --git a/sample_notebooks/KhushbuPattani/chapter1_1.ipynb b/sample_notebooks/KhushbuPattani/chapter1_1.ipynb new file mode 100644 index 00000000..37ff8325 --- /dev/null +++ b/sample_notebooks/KhushbuPattani/chapter1_1.ipynb @@ -0,0 +1,739 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1: Solution of Equation & Curve Fitting" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.1, page no. 21" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The roots of above equation are: [-3. 2. 0.5]\n" + ] + } + ], + "source": [ + "import numpy\n", + "\n", + "x = numpy.poly([0])\n", + "p = 2*(x^3)+x^2-13*x+6\n", + "print \"The roots of above equation are: \", numpy.roots([2,1,-13,6])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.2, page no. 21" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The roots of the equation are: [ 2.00000000+2.64575131j 2.00000000-2.64575131j -2.66666667+0.j ]\n" + ] + } + ], + "source": [ + "import numpy\n", + "\n", + "x = numpy.poly([0])\n", + "p =3*( x ^3) -4*( x ^2) + x +88\n", + "\n", + "print \"The roots of the equation are: \", numpy.roots ([3, -4, 1, 88])" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 1.3, page no. 22" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The roots of above equation are: [ 6. 3. -2.]\n" + ] + } + ], + "source": [ + "import numpy\n", + "\n", + "x = numpy.poly([0])\n", + "p = x^3-7*(x^2)+36\n", + "print \"The roots of above equation are:\", numpy.roots([1, -7, 0, 36])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.4, page no. 23" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The roots of above equation are: [ 5. -4. 2. -1.]\n" + ] + } + ], + "source": [ + "import numpy\n", + "\n", + "x = numpy.poly ([0]) \n", + "p = x ^4 -2*( x ^3) -21*( x ^2) +22* x +40\n", + "print \"The roots of above equation are:\", numpy.roots([1,-2,-21,22,40])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.5, page no. 23" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The roots of above equation are: [ 4. 2. 1. 0.5]\n" + ] + } + ], + "source": [ + "import numpy\n", + "\n", + "x = numpy.poly ([0]) \n", + "p = 2*( x ^4) -15*( x ^3) +35*( x ^2) -30* x +8\n", + "print \"The roots of above equation are:\", numpy.roots([2,-15,35,-30,8])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.6, page no. 24" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The roots of above equation are: [ 2.87938524 0.65270364 -0.53208889]\n", + "let x1 = 0.6527036 x2 = -0.5320889 x3 = 2.8793852\n", + "So the equation whose roots are cube of the roots of above equation is (x−x1ˆ3)∗(x−x2ˆ3)∗(x−x3ˆ3)\n", + "(x - 23.8725770741465)*(x - 0.278066086195109)*(x + 0.150644263115026)\n" + ] + } + ], + "source": [ + "import numpy\n", + "import sympy\n", + "\n", + "x = numpy.poly ([0]) \n", + "p = x ^3 -3*( x ^2) +1\n", + "ans = numpy.roots([1,-3, 0, 1])\n", + "print \"The roots of above equation are:\", ans\n", + "x = sympy.Symbol('x')\n", + "print \"let x1 = 0.6527036 x2 = -0.5320889 x3 = 2.8793852\"\n", + "print \"So the equation whose roots are cube of the roots of above equation is (x−x1ˆ3)∗(x−x2ˆ3)∗(x−x3ˆ3)\"\n", + "p1 = (x-x1**3)*(x-x2**3)*(x-x3**3)\n", + "print p1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.7, page no. 25" + ] + }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The roots of above equation are:\n", + "[ 4.48928857 2.28916855 -0.77845712]\n", + "let x1 = -0.7784571 x2 = 2.2891685 x3 = 4.4892886\n", + "Now, since we want equation whose sum of roots is 0. sum of roots of above equation is 6, so we will decrease\n", + "Value of each root by 2 i.e. x4 = x1-2\n", + "x4 = -2.7784571\n", + "x5 = 0.2891685\n", + "x6 = 2.4892886\n", + "Hence, the required equation is ( x−x4 ) ∗ ( x−x5 ) ∗ ( x−x6 ) = 0 −−>\n", + "(x - 2.4892886)*(x - 0.2891685)*(x + 2.7784571)\n" + ] + } + ], + "source": [ + "import numpy\n", + "import sympy\n", + "\n", + "x = numpy.poly ([0]) \n", + "x1 = numpy.poly ([0]) \n", + "x2 = numpy.poly ([0]) \n", + "x3 = numpy.poly ([0]) \n", + "x4 = numpy.poly ([0]) \n", + "x5 = numpy.poly ([0] ) \n", + "x6 = numpy.poly ([0]) \n", + "p = x ^3 -6*( x ^2) +5* x +8\n", + "print \"The roots of above equation are:\"\n", + "print numpy.roots ([1, -6, 5, 8])\n", + "print \"let x1 = -0.7784571 x2 = 2.2891685 x3 = 4.4892886\"\n", + "x1 = -0.7784571\n", + "x2 = 2.2891685\n", + "x3 = 4.4892886\n", + "print \"Now, since we want equation whose sum of roots is 0. sum of roots of above equation is 6, so we will decrease\"\n", + "print \"Value of each root by 2 i.e. x4 = x1-2\"\n", + "x = sympy.Symbol('x')\n", + "x4 = x1-2\n", + "print \"x4 = \", x4\n", + "x5=x2-2\n", + "print \"x5 = \", x5\n", + "x6=x3-2\n", + "print \"x6 = \", x6\n", + "print \"Hence, the required equation is ( x−x4 ) ∗ ( x−x5 ) ∗ ( x−x6 ) = 0 −−>\"\n", + "p1 =( x-x4 )*(x-x5)*(x-x6)\n", + "print p1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.8, page no. 28" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The roots of above equation are: [ 3. 2. 1. 0.5 0.33333333]\n" + ] + } + ], + "source": [ + "import numpy\n", + "\n", + "x = numpy.poly ([0]) \n", + "p = 6*( x ^5) -41*( x ^4) +97*( x ^3) -97*( x ^2) +41* x -6\n", + "print \"The roots of above equation are:\",numpy.roots([6,-41,97,-97,41,-6])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.9, page no. 28" + ] + }, + { + "cell_type": "code", + "execution_count": 38, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The roots of above equation are: [ 2.00000000+0.j -1.00000000+0.j 0.83333333+0.5527708j\n", + " 0.83333333-0.5527708j 1.00000000+0.j 0.50000000+0.j ]\n" + ] + } + ], + "source": [ + "import numpy\n", + "\n", + "x = numpy.poly([0]) \n", + "p = 6*(x^6)-25*(x^5)+31*(x^4)-31*(x^2)+25*x-6\n", + "print \"The roots of above equation are:\", numpy.roots([6,-25,31,0,-31,25,-6])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.10, page no. 29" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The roots of above equation are: [ 2.+3.46410162j 2.-3.46410162j -1.+0.j ]\n" + ] + } + ], + "source": [ + "import numpy\n", + "\n", + "x = numpy.poly ([0]) \n", + "p = x ^3 -3*( x ^2) +12* x +16\n", + "print \"The roots of above equation are:\", numpy.roots([1,-3,12,16])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.11, page no. 30" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The roots of above equation are: [ 0.28571429+0.24743583j 0.28571429-0.24743583j -0.25000000+0.j ]\n" + ] + } + ], + "source": [ + "import numpy\n", + "x = numpy.poly ([0]) \n", + "p = 28*( x ^3) -9*( x ^2) +1\n", + "print \"The roots of above equation are:\",numpy.roots([28,-9,0,1])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.12, page no. 31" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The roots of above equation are: [-5. +0.00000000e+00j 2. +2.90013456e-08j 2. -2.90013456e-08j]\n" + ] + } + ], + "source": [ + "import numpy\n", + "\n", + "x = numpy.poly ([0]) \n", + "p = x ^3+ x ^2 -16* x +20\n", + "print \"The roots of above equation are:\",numpy.roots ([1,1,-16,20])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.13, page no. 31" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The roots of above equation are: [ 1.08727971+1.17131211j 1.08727971-1.17131211j -1.17455941+0.j ]\n" + ] + } + ], + "source": [ + "import numpy\n", + "\n", + "x = numpy.poly ([0]) \n", + "p = x ^3 -3*( x ^2) +3\n", + "print \"The roots of above equation are:\",numpy.roots ([1,-1,0,3])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.14, page no. 33" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The roots of above equation are: [ 6. 4. 3. -1.]\n" + ] + } + ], + "source": [ + "import numpy\n", + "\n", + "x = numpy.poly ([0]) \n", + "p = x ^4 -12*( x ^3) +41*( x ^2) -18* x -72\n", + "print \"The roots of above equation are:\",numpy.roots ([1,-12,41,-18,-72])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.15, page no. 34" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The roots of above equation are: [-2.30277564 2.61803399 1.30277564 0.38196601]\n" + ] + } + ], + "source": [ + "import numpy\n", + "\n", + "x = numpy.poly ([0]) \n", + "p = x ^4 -2*( x ^3) -5*( x ^2) +10* x -3\n", + "print \"The roots of above equation are:\", numpy.roots ([1,-2,-5,10,-3])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.16, page no. 35" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The roots of above equation are: [ 3.73205081+0.j -2.00000000+1.73205081j -2.00000000-1.73205081j\n", + " 0.26794919+0.j ]\n" + ] + } + ], + "source": [ + "import numpy\n", + "\n", + "x = numpy.poly ([0]) \n", + "p = x ^4 -8*( x ^2) -24* x +7\n", + "print \"The roots of above equation are:\",numpy.roots ([1,0,-8,-24,7])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.17, page no. 35" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The roots of above equation are: [ 6.05932014 -1.65491082 1.59559067]\n" + ] + } + ], + "source": [ + "import numpy\n", + "\n", + "x = numpy.poly ([0]) \n", + "p = x ^4 -6*( x ^3) -3*( x ^2) +22* x -6\n", + "print \"The roots of above equation are:\",numpy.roots ([1,-6,-3,22.-6])" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.18, page no. 37" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "From the graph, it is clear that the point of intersection is nearly x = 1.43\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fa769850290>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "import numpy\n", + "import math\n", + "\n", + "x = numpy.linspace(1 ,3 ,30)\n", + "y1 = 3-x\n", + "y2 = math.e**(x-1)\n", + "plt.xlabel('X axis')\n", + "plt.ylabel('Y axis')\n", + "plt.title('My Graph')\n", + "plt.plot(x, y1, \"o-\" )\n", + "plt.plot(x, y2, \"+-\" )\n", + "plt.legend([\"3-x\" ,\"e**(x-1)\"])\n", + "print \"From the graph, it is clear that the point of intersection is nearly x = 1.43\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.19, page no. 40" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "From the graph, it is clear that the point of intersection is nearly x=2.3\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fa7698ce210>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "import numpy\n", + "import math\n", + "\n", + "x = numpy.linspace(1 ,3 ,30)\n", + "y1 = x\n", + "y2 = numpy.sin(x)+math.pi/2\n", + "plt.xlabel('X axis')\n", + "plt.ylabel('Y axis')\n", + "plt.title('My Graph')\n", + "plt.plot(x, y1, \"o-\")\n", + "plt.plot(x, y2, \"+-\")\n", + "plt.legend([\"x\", \"sin(x)+pi/2\"])\n", + "print \"From the graph, it is clear that the point of intersection is nearly x=2.3\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.20, page no. 41" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "From the graph, it is clear that the point of intersection is nearly x=2.3 \n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fa7696dc7d0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "import numpy\n", + "import math\n", + "\n", + "x = numpy.linspace(0, 3, 30)\n", + "y1 = -1/numpy.cos(x)\n", + "y2 = numpy.cosh(x)\n", + "plt.xlabel('X axis')\n", + "plt.ylabel('Y axis')\n", + "plt.title('My Graph')\n", + "plt.plot(x, y1, \"o-\" )\n", + "plt.plot(x, y2, \"+-\" )\n", + "plt.legend ([\"-sec(x)\", \"cosh(x)\"])\n", + "print \"From the graph, it is clear that the point of intersection is nearly x=2.3 \"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |
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