diff options
Diffstat (limited to 'Principles_of_Physics_by_F.J.Bueche/Chapter17.ipynb')
| -rw-r--r-- | Principles_of_Physics_by_F.J.Bueche/Chapter17.ipynb | 641 |
1 files changed, 641 insertions, 0 deletions
diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter17.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter17.ipynb new file mode 100644 index 00000000..c718557a --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter17.ipynb @@ -0,0 +1,641 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 17:DC Circuits" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.1:pg-866" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The number of electrons that pass through bulb is=\n", + "9.375e+17\n", + "electrons\n" + ] + } + ], + "source": [ + "#Example 17_1\n", + " \n", + " \n", + "#To find number of electrons flow through bulb\n", + "current=0.15 #Units in C\n", + "q=1.6*10**-19 #Units in C/electron\n", + "noe=current/q #Units in number of Electrons\n", + "print \"The number of electrons that pass through bulb is=\"\n", + "print noe\n", + "print \"electrons\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.2:pg-867" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance in bulb is= 19.4 Ohms\n" + ] + } + ], + "source": [ + "#Example 17_2\n", + " \n", + " \n", + "#To find the resistance in bulb\n", + "v=1.55 #Units in V\n", + "i=0.08 #Units in A\n", + "r=v/i #Units in Ohms\n", + "print \"The resistance in bulb is=\",round(r,1),\" Ohms\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.3:pg-867" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance in wire is= 0.205 Ohms\n" + ] + } + ], + "source": [ + "#Example 17_3\n", + " \n", + " \n", + "#To find the resistance in wire\n", + "row=1.7*10**-8 #Units in Ohm meter\n", + "l=40 #Units in meters\n", + "a=0.0331*10**-4 #Units in meters**2\n", + "r=(row*l)/a #Units in Ohms\n", + "print \"The resistance in wire is=\",round(r,3),\" Ohms\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.4:pg-868" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The appropriate resistance in wire is Ro= 26.6 Ohms\n" + ] + } + ], + "source": [ + "#Example 17_4\n", + " \n", + " \n", + "#To find the appropriate resistance of the wire\n", + "alpha=0.0045 #Units in Centigrade**-1\n", + "t=1780 #Units in Centigrade\n", + "deltaR=240 #Units in Ohms\n", + "ro=deltaR/(1+(alpha*t)) #Units in ohms\n", + "print \"The appropriate resistance in wire is Ro=\",round(ro,1),\" Ohms\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.5:pg-868" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat generated in bulb is= 48000.0 J\n" + ] + } + ], + "source": [ + "#Example 17_5\n", + " \n", + " \n", + "#To find out the amount of heat developed in bulb\n", + "t=20.0*60 #Units in sec\n", + "pow=40.0 #Units in W\n", + "heat=t*pow #Units in J\n", + "print \"Heat generated in bulb is=\",round(heat),\" J\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.6:pg-869" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Cost needed to operate is= 0.035 Dollars\n" + ] + } + ], + "source": [ + "#Example 17_6\n", + " \n", + " \n", + "#To calculate the cost needed to operate\n", + "power=0.7 #Units in KW\n", + "time=0.5 #Units in h\n", + "heat=power*time #Units in K Wh\n", + "cost=0.10 #Units in Dollars\n", + "tcost=cost*heat #Units in Dollars\n", + "print \"Cost needed to operate is=\",round(tcost,4),\" Dollars\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.7:pg-869" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current in circuit is I= -0.82 A\n" + ] + } + ], + "source": [ + "#Example 17_7\n", + " \n", + " \n", + "#To find the current in circuit\n", + "v1=3 #Units in V\n", + "v2=12.0#Units in V\n", + "r1=5.0 #Units in Ohms\n", + "r2=6 #Units in Ohms\n", + "i=(v1-v2)/(r1+r2) #Units in A\n", + "print \"The current in circuit is I=\",round(i,2),\" A\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.8:pg-870" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current in wire 1 is I1= 1.75 A\n", + "Current in wire 2 is I2= -0.5 A\n", + "Current in wire 3 is I3= 1.25 A\n", + "\n" + ] + } + ], + "source": [ + "#Example 17_8\n", + " \n", + " \n", + "#To find the current in all wires\n", + "v=9 #Units in V\n", + "r1=18.0 #Units in Ohms\n", + "i2=-v/r1 #Units in A\n", + "v1=6.0 #Units in V\n", + "r2=12.0 #Units in Ohms\n", + "i3=(v+v1)/r2 #Units in A\n", + "i1=i3-i2 #Units in A\n", + "print \"Current in wire 1 is I1=\",round(i1,2),\" A\\nCurrent in wire 2 is I2=\",round(i2,2),\" A\\nCurrent in wire 3 is I3=\",round(i3,2),\" A\\n\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.9:pg-871" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current I= 2.0 A\n" + ] + } + ], + "source": [ + "#Example 17_9\n", + " \n", + " \n", + "#To find the current I in the battery\n", + "r1=3 #Units in Ohms\n", + "r2=6.0 #Units in Ohms\n", + "rbc=(r1*r2)/(r1+r2)#Units in Ohms\n", + "r3=4#Units in Ohms\n", + "rac=r3+rbc #Units in Ohms\n", + "v=12.0 #Units in V\n", + "i=v/rac #Units in A\n", + "print \"The current I=\",round(i),\" A\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.10:pg-872" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "ename": "SyntaxError", + "evalue": "invalid syntax (<ipython-input-10-e97062f38a8b>, line 17)", + "output_type": "error", + "traceback": [ + "\u001b[1;36m File \u001b[1;32m\"<ipython-input-10-e97062f38a8b>\"\u001b[1;36m, line \u001b[1;32m17\u001b[0m\n\u001b[1;33m print \"The current in battery is I=\",round( ,2),\" A\",i)\u001b[0m\n\u001b[1;37m ^\u001b[0m\n\u001b[1;31mSyntaxError\u001b[0m\u001b[1;31m:\u001b[0m invalid syntax\n" + ] + } + ], + "source": [ + "#Example 17_10\n", + " \n", + " \n", + "#To find the current in battery\n", + "r1=3 #Units in Ohms\n", + "r2=6 #Units in Ohms\n", + "ra=(r1*r2)/(r1+r2)#Units in Ohms\n", + "r3=2 #Units in Ohms\n", + "r4=4 #Units in Ohms\n", + "rb=r3+r4 #Units in Ohms\n", + "r5=6 #Units in Ohms\n", + "rc=(r5*rb)/(r5+rb) #Units in Ohms\n", + "r6=9 #Units in Ohms\n", + "r=r6+rc #Units in Ohms\n", + "v=6 #Units in V\n", + "i=v/r #Units in Ohms\n", + "print \"The current in battery is I=\",round( ,2),\" A\",i)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.11:pg-872" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current in wire 1 is I1= 1.2 A\n", + "Current in wire 2 is I2= 0.6 A\n", + "Current in wire 3 is I3= 0.6 A\n", + "\n" + ] + } + ], + "source": [ + "#Example 17_11\n", + " \n", + " \n", + "#To find the current in the wires\n", + "v1=12.0 #Units in V\n", + "r3=20.0 #Units in Ohms\n", + "v2=6 #Units in V\n", + "r2=10.0 #Units in Ohms\n", + "r1=5 #Units in Ohms\n", + "i3=((v1*r3)-(v2*r1))/((r2*r3)+(r1*r3)+(r1*r2)) #Units in A\n", + "i2=((r2*i3)+v2)/r3 #Units in A\n", + "i1=i3+i2 #Units in A\n", + "print \"Current in wire 1 is I1=\",round(i1,1),\" A\\nCurrent in wire 2 is I2=\",round(i2,1),\" A\\nCurrent in wire 3 is I3=\",round(i3,1),\" A\\n\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.12:pg-873" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current in wire 1 is I1= 0.667 A\n", + "Current in wire 2 is I2= 0.444 A\n", + "Current in wire 3 is I3= -0.222 A\n", + "\n" + ] + } + ], + "source": [ + "#Example 17_12\n", + " \n", + " \n", + "#To find I1, I2 and I3 in the circuit\n", + "v1=40 #Units in V\n", + "r1=10.0 #Units in Ohms\n", + "r2=30.0 #Units in Ohms\n", + "v2=60.0 #Units in V\n", + "r3=15.0 #Units in Ohms\n", + "v3=50 #Units in V\n", + "i1=((-v1*r2)+(-r3*v1)+(60*r3)+(v3*r2))/((r1*r2)+(r2*r3)+(r3*r1)) #Units in A\n", + "i=2 #Units in A\n", + "i2=(i-i1)/3 #Units in A\n", + "i3=i2-i1 #Units in A\n", + "print \"Current in wire 1 is I1=\",round(i1,3),\" A\\nCurrent in wire 2 is I2=\",round(i2,3),\" A\\nCurrent in wire 3 is I3=\",round(i3,3),\" A\\n\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.13:pg-874" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The current I= 2.5 A\n", + " Resistance is R= 20.0 Ohms\n", + " The value E is= 41.0 V\n" + ] + } + ], + "source": [ + "#Example 17_13\n", + " \n", + " \n", + "#To find the values of e, R and I\n", + "i1=2 #Units in A\n", + "i2=0.5 #Units in A\n", + "i=i1+i2 #Units in A\n", + "v1=6 #Units in V\n", + "v2=16.0 #Units in V\n", + "r=-(v1-v2)/0.5 #Units in Ohms\n", + "v3=25.0 #Units in V\n", + "e=v2+v3 #Units in V\n", + "print \" The current I=\",round(i,1),\" A\\n Resistance is R=\",round(r),\" Ohms\\n The value E is=\",round(e),\" V\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.14:pg-874" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current in wire 1 is I1= 2.0 A\n", + "Current in wire 2 is I2= 4.0 A\n", + "Current in wire 3 is I3= 2.0 A\n", + "\n", + "The charge on the capacitor is q= 1e-05 C\n" + ] + } + ], + "source": [ + "#Example 17_14\n", + " \n", + " \n", + "#To find the I1,I2,I3 values and charge on the capacitor\n", + "v1=12.0 #Units in V\n", + "r1=6.0 #Units in Ohms\n", + "i1=v1/r1 #Units in A\n", + "v2=4.0 #Units in V\n", + "r2=8.0 #Units in Ohms\n", + "i3=(v1+v2)/r2 #Units in A\n", + "i2=i1+i3 #Units in A\n", + "print \"Current in wire 1 is I1=\",round(i1),\" A\\nCurrent in wire 2 is I2=\",round(i2),\" A\\nCurrent in wire 3 is I3=\",round(i3),\" A\\n\"\n", + "v3=10.0 #Units in V\n", + "vfg=-v3+(r1*i1) #Units in V\n", + "c=5*10**-6 #Units in F\n", + "q=c*vfg #Units in C\n", + "print \"The charge on the capacitor is q=\",round(q,5),\" C\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.15:pg-874" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Potential difference between d to c is= 23.8 V\n", + "\n", + "The potential difference between b to a is= 7.8 V\n" + ] + } + ], + "source": [ + "#Example 17_15\n", + " \n", + " \n", + "#To find the terminal potential of each battery\n", + "v=18 #Units in V\n", + "r=9 #Units in Ohms\n", + "i=v/r #Units in A\n", + "r1=0.1 #Units in Ohms\n", + "v1=-i*r1 #Units in V\n", + "v2=24 #Units in V\n", + "v11=v1+v2 #Units in V\n", + "r2=0.9 #Units in Ohms\n", + "v3=i*r2 #Units in V\n", + "v4=6 #Units in V\n", + "v22=v3+v4 #Units in V \n", + "print \"The Potential difference between d to c is=\",round(v11,1),\" V\"\n", + "print \"\\nThe potential difference between b to a is=\",round(v22,1),\" V\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.16:pg-875" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance of the recording device is= 1000000.0 Ohms\n" + ] + } + ], + "source": [ + "#Example 17_16\n", + " \n", + " \n", + "#To findout how large a a resistance must the recording device must have\n", + "r1=10000.0 #Units in Ohms\n", + "percent=1.0 #Units in Percentage \n", + "vo=1/(r1*(percent*100)) #Units In terms of Ro\n", + "Ro=1/vo #Units in Ohms\n", + "print \"The resistance of the recording device is=\",round(Ro),\" Ohms\"\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": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |