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author | Trupti Kini | 2016-02-27 23:30:09 +0600 |
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committer | Trupti Kini | 2016-02-27 23:30:09 +0600 |
commit | 5e73d776147364bfb09a8d64c59508ed7dfb4f49 (patch) | |
tree | 479b08ccfeec5d75827a2180d8938575f8137a72 /sample_notebooks/AnkitKumar | |
parent | bda5fec689914d5a882b64f1bbc5d8c1a05ddb3b (diff) | |
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Added(A)/Deleted(D) following books
A sample_notebooks/AnkitKumar/Ch16.ipynb
A sample_notebooks/GauravMittal/Ch3.ipynb
A sample_notebooks/Mohdarif/Ch21.ipynb
Diffstat (limited to 'sample_notebooks/AnkitKumar')
-rw-r--r-- | sample_notebooks/AnkitKumar/Ch16.ipynb | 364 |
1 files changed, 364 insertions, 0 deletions
diff --git a/sample_notebooks/AnkitKumar/Ch16.ipynb b/sample_notebooks/AnkitKumar/Ch16.ipynb new file mode 100644 index 00000000..86539c04 --- /dev/null +++ b/sample_notebooks/AnkitKumar/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 +} |