diff options
| author | Trupti Kini | 2016-08-07 23:30:22 +0600 |
|---|---|---|
| committer | Trupti Kini | 2016-08-07 23:30:22 +0600 |
| commit | 661fe2fe6784f49307382d2934200e0df5ba0c4b (patch) | |
| tree | c6323b66d72b87506cef869588deaa5df70d5ecd | |
| parent | 656ca0e62c82b17a5b7c6fb32767a8af855b7cbb (diff) | |
| download | Python-Textbook-Companions-661fe2fe6784f49307382d2934200e0df5ba0c4b.tar.gz Python-Textbook-Companions-661fe2fe6784f49307382d2934200e0df5ba0c4b.tar.bz2 Python-Textbook-Companions-661fe2fe6784f49307382d2934200e0df5ba0c4b.zip | |
Added(A)/Deleted(D) following books
A Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter1.ipynb
A Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter10.ipynb
A Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter11.ipynb
A Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter12.ipynb
A Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter2.ipynb
A Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter3.ipynb
A Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter4.ipynb
A Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter5.ipynb
A Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter6.ipynb
A Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter7.ipynb
A Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter8.ipynb
A Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter9.ipynb
A Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/screenshots/chapter1.png
A Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/screenshots/chapter2.png
A Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/screenshots/chapter3.png
A Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/Chapter1.ipynb
A Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/Chapter10.ipynb
A Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/Chapter16.ipynb
A Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/Chapter2.ipynb
A Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/Chapter3.ipynb
A Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/screenshots/chapter1.png
A Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/screenshots/chapter2.png
A Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/screenshots/chapter3.png
A modern_physics_by_Satish_K._Gupta/chap10_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap11_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap12_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap13_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap14_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap15_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap16_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap17_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap18_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap19_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap1_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap20_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap21_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap22_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap23_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap25_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap26_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap27_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap28_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap2_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap30_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap31_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap32_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap33_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap34_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap35_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap36_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap37_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap3_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap4_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap5_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap6_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap7_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap8_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap9_1.ipynb
A modern_physics_by_Satish_K._Gupta/chap_29_1.ipynb
A modern_physics_by_Satish_K._Gupta/screenshots/image1_1.png
A modern_physics_by_Satish_K._Gupta/screenshots/image22_1.png
A modern_physics_by_Satish_K._Gupta/screenshots/image33_1.png
A sample_notebooks/RuchiMittal/chpater3.ipynb
63 files changed, 27598 insertions, 0 deletions
diff --git a/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter1.ipynb b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter1.ipynb new file mode 100644 index 00000000..09a2d20c --- /dev/null +++ b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter1.ipynb @@ -0,0 +1,423 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:ce2017e08b497aa6547723a99ae31c32c07c39b1c9b214208fd7c8c30d837cb6"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter1-Linear Motion"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg2"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Velocity calculation\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "t=20.##ft\n",
+ "t1=30.##ft\n",
+ "v=1320.##ft/s\n",
+ "p=25.##sec\n",
+ "q=15.##ft/s\n",
+ "v1=v/t##ft/s\n",
+ "v2=v/t1##ft/s\n",
+ "T=(v2-v1)/p##ft/s^2\n",
+ "V=v2-q*-T##ft/s\n",
+ "V1=-V**2/(2.*T)##ft/s\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the velocity time is= ',V1,' ft/s')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the velocity time is= 539.00 ft/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg3"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Distance travel calculation\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "w=200.##tonf\n",
+ "d=4.##tonf\n",
+ "h=120.##tonf\n",
+ "v=25.##mile/h\n",
+ "m=10.##lbf/tonf\n",
+ "q=2240.##lbf\n",
+ "##CALCULATIONS\n",
+ "F=w*m##lbf\n",
+ "W=(w*q)*(1./h)##lbf\n",
+ "T=F+W##lbf\n",
+ "D=d*q##lbf\n",
+ "A=D-T##lbf\n",
+ "t=158.1##sec\n",
+ "T1=(v/2.)*(88./60.)*t##ft\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('Distance travel= ',T1,' ft')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance travel= 2898.50 ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg4"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##velocity is uniform and force and velocity\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "f=90.##lbf\n",
+ "w=6.##tonf\n",
+ "m=10.##lbf/tonf\n",
+ "f1=1.##min\n",
+ "h=0.8##hp\n",
+ "m1=m*w##lbf\n",
+ "n=f-m1##lbf\n",
+ "p=2240.##lbf\n",
+ "f2=0.0715##ft/s^2\n",
+ "r=550##ft\n",
+ "##CALCULATIONS\n",
+ "S=1/2.*f2*(m1)**2##ft\n",
+ "V=f2*m1##ft/s\n",
+ "H=(f*V)/r##ft\n",
+ "V1=h/(m1/r)##ft/s\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the velocity is uniform and force and velocity= ',V1,' ft/s')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the velocity is uniform and force and velocity= 7.33 ft/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg5"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Tension Coupling calculation\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "w=30.##tonf\n",
+ "m=100.##tonf\n",
+ "w1=150.##tonf\n",
+ "f=6000.##lbf\n",
+ "h=2240.##lbf\n",
+ "q=105.##lbf\n",
+ "p=135.##lbf\n",
+ "a=711.7##lbf\n",
+ "##CALCULATIONS\n",
+ "M=(q*h)/m##lbf\n",
+ "R=(w*h)/w1##lbf\n",
+ "T=M+R##lbf\n",
+ "A=f-T##lbf\n",
+ "T1=R+a##lbf\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the Tension Coupling is= ',T1,' lbf')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the Tension Coupling is= 1159.70 lbf\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg5"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "##work done ground resistance\n",
+ "##initialisation of variables\n",
+ "g=32.1##ft/s\n",
+ "w=3.##tonf\n",
+ "p=16.##ft\n",
+ "p1=6.##in\n",
+ "h=2240.##ft/cm^2\n",
+ "m=4.##tonf\n",
+ "v=24.08##ft/s\n",
+ "##CALCULATIONS\n",
+ "K=(m*h*(v**2))/(2.*g)##ft lbf\n",
+ "P=m*h*(1/2.)##ft lbf\n",
+ "R=(K+P)/(h*(1/2.))##tonf\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the work done ground resistance= ',R,' tonf')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the work done ground resistance= 76.26 tonf\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg6"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##kinetic energy and velocity\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "p=50.##ft/s\n",
+ "w=10.##lbf\n",
+ "v=30.##ft/s\n",
+ "w1=40.##lbf\n",
+ "v1=20.##ft/s\n",
+ "g=32.2##ft/s\\\n",
+ "h=0.8##ft/s\n",
+ "V1=23.6##ft/s\n",
+ "V3=15.6##ft/s\n",
+ "V4=22.##ft/s\n",
+ "##CALCULATIONS\n",
+ "V=(w+w1)/g/(w/g*v)+(w1/g*v1)##ft/s\n",
+ "V2=h*(-v1+v)##ft/s\n",
+ "K=(w*(v**2))/(2.*g)+(w1*(v1**2))/(2.*g)-(p*(V1**2))/(2.*g)##ft /bf\n",
+ "K1=((w*(v**2))/(2*g))+((w1*(v1**2))/(2.*g))-((w*(V3**2))/(2.*g))-((w1*(V1**2))/(2.*g))+1.5##ft lbf\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the velocity of two bodies after impact is=',V4,' ft/s')\n",
+ "print'%s %.2f %s'%('final velocity is= ',V2,' ft/s')\n",
+ "print'%s %.2f %s'%('Loss of kinetic energy at impact is= ',K1,' ft lbf')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the velocity of two bodies after impact is= 22.00 ft/s\n",
+ "final velocity is= 8.00 ft/s\n",
+ "Loss of kinetic energy at impact is= 5.97 ft lbf\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg7"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "\n",
+ "##equation of motion and acceleration\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "d=4.##ft\n",
+ "w=5.##lbf\n",
+ "v=10.##lbf\n",
+ "q=9.27##ft/s\n",
+ "##CALCULATIONS\n",
+ "W=w*d##ft lbf\n",
+ "P=v*d##ft lbf\n",
+ "M=(q)**2/d/2.##ft/s^2\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the equation of motion and acceleration= ',M,' ft/s^2')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the equation of motion and acceleration= 10.74 ft/s^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg8"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "\n",
+ "##maximum velocity of speed \n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "a=30.##degree\n",
+ "w=20.##lbf\n",
+ "m=150.##ft\n",
+ "v=18.6##ft/s^2\n",
+ "##CALCULATIONS\n",
+ "A=math.sqrt(m/(1./2.)/v)##sec\n",
+ "V=math.sqrt(2*v*m)##ft/s\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the maximum velocity of speed after= ',V,' ft/s')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the maximum velocity of speed after= 74.70 ft/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "\n",
+ "##Work done against gravity\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "\n",
+ "r=1500.##yd\n",
+ "w=200.##tonf\n",
+ "v=25.##lbf/tonf\n",
+ "V=56.8##ft/s\n",
+ "p=550.##ft\n",
+ "t=80.##ft\n",
+ "h=2240.##ft/s\n",
+ "##CALCULATIONS\n",
+ "R=v*w##lbf\n",
+ "W=26.5*10**6##ft lbf\n",
+ "D=v*w*V##ft lbf\n",
+ "H=(v*w*V)/p##ft\n",
+ "W1=W/((v*w)*(w*h*1/180.))*1000.##ft\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the Work done against gravity is= ',W1,' ft')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the Work done against gravity is= 2129.46 ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter10.ipynb b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter10.ipynb new file mode 100644 index 00000000..eb6c86fa --- /dev/null +++ b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter10.ipynb @@ -0,0 +1,161 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:224a8fc433c4564f48dc091f5ea974223a68c2e9e6c711939aeeb3c67efcdf1b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter10-Force in plane framework\n"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg129"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "l=2.##units of length\n",
+ "a=math.sqrt(3.)##degree\n",
+ "b=30.##dgree\n",
+ "c=60.##degree\n",
+ "v=1.##length\n",
+ "Pc=1.154##tonf compressive\n",
+ "##CALCULATIONS\n",
+ "R=(v*l)/a##tonf\n",
+ "D=math.sqrt((R)**2+(v)**2)##tonf\n",
+ "T=41.##degree\n",
+ "P=l*math.cos(b)##tonf tensile\n",
+ "Pa=Pc*math.cos(b)##tonf tensile\n",
+ "p=(l*math.cos(b))/((1/2.)+(Pc))/(1/2.)##tonf compressive\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the resolving horizontally = ',p,' tonf compressive')\n",
+ "#due to round off error answer is different"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the resolving horizontally = 0.37 tonf compressive\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg131"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "v=3.##tonf\n",
+ "p1=6.0##tonf,compressive\n",
+ "p2=5.19##tonf,tensile\n",
+ "a=30.##degree\n",
+ "b=60.##degree\n",
+ "p3=7.##tonf,compressive\n",
+ "##CALCULATIONS\n",
+ "P1=p2*math.sin(b)##tonf,tensile\n",
+ "P2=1./2.*P1##tonf,compressive\n",
+ "P3=p1*math.cos(a)-p3*math.cos(b)##tonf,compressive\n",
+ "P4=P1*math.cos(a)*math.sqrt(3.)/P3##tonf,acting towards the left \n",
+ "R=P1*math.sin(a)##tonf,acting downwards\n",
+ "D=math.sqrt((P4)**2+(R)**2)##tonf\n",
+ "T=(R/P4)##to the horizantal\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the direction reaction= ',D,' to the horizantal')\n",
+ "print'%s %.2f %s'%('the direction reaction = ',T,' to horizantal')\n",
+ "#converted into degrees\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the direction reaction= 1.56 to the horizantal\n",
+ "the direction reaction = -28.08 to horizantal\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg133"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "R1=5.##tonf\n",
+ "R2=7.##tonf\n",
+ "P=5.77##tonf,compressive \n",
+ "m=11.56##tonf\n",
+ "a=30.##degree\n",
+ "##CALCULATIONS\n",
+ "P=-(0.15+4.5084+1+1-10)##tonf\n",
+ "P1=math.sqrt(P)\n",
+ "\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the methods of sections in the force= ',-P1,' tonf')\n",
+ "#converted into degrees\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the methods of sections in the force= -1.83 tonf\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter11.ipynb b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter11.ipynb new file mode 100644 index 00000000..0a4a3395 --- /dev/null +++ b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter11.ipynb @@ -0,0 +1,246 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:f143b35f244a2a7e967ebf0b33077a6d51a6aba9e6ffcf138aad816a0cdd4f33"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter11-Hydrostatics"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg141"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "w=62.5##lbf\n",
+ "a=4.*6.##ft\n",
+ "x=4.##ft\n",
+ "l=(6.*6**3)/3.-(6.*2**3)/3.##ft^3\n",
+ "q=24.*x##ft^3\n",
+ "##CALCULATIONS\n",
+ "T=w*a*x##lbf\n",
+ "P=l/q##ft\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the depth of centre of pressure= ',P,' ft')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the depth of centre of pressure= 4.33 ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "a=60.##degre\n",
+ "w=2.5##ft\n",
+ "x=3.##ft\n",
+ "p=6.*3.##ft^2\n",
+ "h=62.4##ft\n",
+ "p1=(3.*6**3)/12.##ft^4\n",
+ "##CALCULATIONS\n",
+ "D=w+x*(math.sin(a))##ft\n",
+ "T=h*p*D##lbf\n",
+ "P=(p1*math.sin(a)**2)/p*D+D+3##ft\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the depth of centre of pressure= ',P,' ft')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the depth of centre of pressure= 5.03 ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg143"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "t=62.5*4.*1.##lbf\n",
+ "a=2./3.*2.##ft\n",
+ "m=62.5*4*(4/3)##lbf\n",
+ "f=500.*2.##lbf ft\n",
+ "T=((62.5*2.*2.)/2.)*1./3.*2.##lbf\n",
+ "H=(62.5*2.*1.)##ft\n",
+ "##CALCULATIONS\n",
+ "H1=f/(H+T)*2./2.9##/ft\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the trap door width= ',H1,' ft')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the trap door width= 3.31 ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg146"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "g=6.##ft\n",
+ "g1=50.##ft\n",
+ "d=10.##ft\n",
+ "w1=10.##ft\n",
+ "w2=20.##ft\n",
+ "w3=62.5##ft\n",
+ "t=w3*60.*5.##lbf\n",
+ "t2=8.37##tonf\n",
+ "t1=g1+t##lbf\n",
+ "H=26.4##ft\n",
+ "##CALCULATIONS\n",
+ "M=t*d/3##lbf ft\n",
+ "D=w3*w2*g*d##lbf\n",
+ "M1=D*(w2/3.)##lbf ft\n",
+ "f=D-t##lbf\n",
+ "R=(M1-M)/f##ft\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the moment of resultant force about gate base= ',R,' ft')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the moment of resultant force about gate base= 7.78 ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg148"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "w=62.5##lbf/ft\n",
+ "w1=1.5##ft\n",
+ "d=4.##ft\n",
+ "w2=3.##ft\n",
+ "g=0.8##in\n",
+ "p1=2./3.*w1##ft\n",
+ "q=2./3.*p1##ft\n",
+ "##CALCULATIONS\n",
+ "t1=w1*w*w1/2.##lbf\n",
+ "p=(g*w*p1*p1)/2.##lbf\n",
+ "A=g*w*p1*1./2.##lbf\n",
+ "T=(w*0.5*0.5/2.)##lbf\n",
+ "P=t1-p-A-T##lbf\n",
+ "h=2.9*P/(t1*1.-p*2.)/3.-p*(1.*0.25)-(T*1.33)##ft\n",
+ "F=P*A/w+3##lbf\n",
+ "\n",
+ "H=F/2.##lbf\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('depth of forces= ',F,' lbf')\n",
+ "print'%s %.2f %s'%('the moment of force on hinge= ',H,' lbf')\n",
+ "\n",
+ "#ans is slightly different due to round off error"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "depth of forces= 8.00 lbf\n",
+ "the moment of force on hinge= 4.00 lbf\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter12.ipynb b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter12.ipynb new file mode 100644 index 00000000..e1fc4d77 --- /dev/null +++ b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter12.ipynb @@ -0,0 +1,185 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:3e2ac1bf133acbbd856da5493c5fd89c5f1a7ed7a5559688ae90d4aacb790c72"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter12-Hydrodynamics"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg155"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "p=144.*60.##lbf/ft^2\n",
+ "A1=1./4.*math.pi*(1/2.)**2##ft^2\n",
+ "A2=1./4.*math.pi*(1/4.)**2##ft^2\n",
+ "w=5.##ft/s\n",
+ "U1=1./A1##ft/s\n",
+ "U2=1./A2##ft/s\n",
+ "g=32.2##ft/s\n",
+ "P=(U1**2/(2.*g))+(p/(2.*g))\n",
+ "P1=(3.+U2**2/(62.4))+(144./(62.4))\n",
+ "##CALCULATIONS\n",
+ "Pb=(P/P1)*w##lbf/in^2\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the bernouli s equation= ',Pb,' lbf/in^2')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the bernouli s equation= 56.26 lbf/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg156"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "p=1.23##ft^2\n",
+ "t=0.197##ft^2\n",
+ "u=1.595##ft^2\n",
+ "g=13.56##ft^2\n",
+ "w=9.2##in\n",
+ "m=0.97##in\n",
+ "##CALCULATIONS\n",
+ "H=(g-1.)*w/12.##ft^2\n",
+ "Q=m*u*math.sqrt(H)##ft^3\n",
+ "S=Q*60.*62.4/10.##gallons/min\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the head difference in feet of water= ',S,' gallons/min')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the head difference in feet of water= 1797.49 gallons/min\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg158"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "h=4.##ft\n",
+ "h1=3.24##ft^3/min\n",
+ "d=0.785##in\n",
+ "v=5.26##ft^3/min\n",
+ "##CALCULATIONS\n",
+ "Cd=h1/v##ft\n",
+ "C=1/4.*math.pi*(d)**2/(1./4.*math.pi*(1.)**2)##ft^3\n",
+ "V=Cd/C\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the coefficients of discharge velocity and contraction=',V,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the coefficients of discharge velocity and contraction= 1.00 \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg159"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "x=32.5##in\n",
+ "y=33.7##in\n",
+ "h=8.##in\n",
+ "##CALCULATIONS\n",
+ "C=math.sqrt((x)**2/(4.*y*h))##ft\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the coefficient of velocity= ',C,' ft')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the coefficient of velocity= 0.99 ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter2.ipynb b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter2.ipynb new file mode 100644 index 00000000..2699ca93 --- /dev/null +++ b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter2.ipynb @@ -0,0 +1,549 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:8cc345c457c220087f4fe714f9742278677b74848dfd86669cab1fb48d07fa0d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter2-Angular Motion"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EX1-pg13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##equation of motion, Mass of moment of inertia, percentage \n",
+ "##reduction in speed\n",
+ "##initialisation of variables\n",
+ "g=5.##ft\n",
+ "w=300.##rev/min\n",
+ "a=0.86##red/s^2\n",
+ "h=2240.##ft/s\n",
+ "q=4.##ft\n",
+ "g1=32.1##ft/s\n",
+ "k=3105000.##ft lbf\n",
+ "##CALCULATIONS\n",
+ "T=(w*(2.*math.pi/60.))/(a)##sec\n",
+ "M=(q*h*(g**2))/(g1)##slug ft^3\n",
+ "K=((1/2.)*M)*((w*2.*math.pi**2)/(60.))##ft lbf\n",
+ "W=math.sqrt(k/(1./2.)/M)##rad/s\n",
+ "P=((((w*2.*math.pi)/60.)-W)/((w*2.*math.pi)/60.))*100.##percent\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('The equation of motion= ',T,' sec')\n",
+ "print'%s %.2f %s'%('Mass of moment of inertia of = ',K,' ft lbf')\n",
+ "print'%s %.2f %s'%('the percentage reduction in speed= ',P,' percent')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The equation of motion= 36.53 sec\n",
+ "Mass of moment of inertia of = 344360.03 ft lbf\n",
+ "the percentage reduction in speed= 5.04 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##radius of gyration\n",
+ "##initialisation of variables\n",
+ "m=2.58065##slug ft^3\n",
+ "w=2.144##in\n",
+ "##CALCULATIONS\n",
+ "R=math.sqrt(m/w)##ft\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('The radius of gyration= ',R,' ft')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The radius of gyration= 1.10 ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "##distance travelled along incline before coming to rest\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "w1=10.##tonf\n",
+ "r=36.##in\n",
+ "w=1./4.##tonf\n",
+ "g=14.##in\n",
+ "t=30.##mile/h\n",
+ "s=100.##in\n",
+ "m=20.##lbf/tonf\n",
+ "h=2240.##lbf\n",
+ "q=44.##in\n",
+ "g1=32.2##ft\n",
+ "##CALCULATIONS\n",
+ "K=(w1*h*(q**2))/(2.*g1)##ft lbf \n",
+ "L=q/1.5##rad/s\n",
+ "R=(2.*1./2.*(1./4.*h/g1)*(g/12.)**2)*L**2##ft lbf\n",
+ "T=K+R##ft lbf\n",
+ "M=m*w1##lbf\n",
+ "G=w1*h*(1./s)##lbf\n",
+ "S=K/(M+G)##ft\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the distance travelled along incline before coming to rest= ',S,' ft')\n",
+ "#answer is different due to round off error\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the distance travelled along incline before coming to rest= 1588.19 ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##percentage fluctuation in speed\n",
+ "##initialisation of variables\n",
+ "g=32.2##ft\n",
+ "p=275.##rev/min\n",
+ "w=1/2.*p##ft\n",
+ "d=15.##hp\n",
+ "h=33000.##ft\n",
+ "r=0.8##ft\n",
+ "h1=2240.##ft\n",
+ "m=p*(2*math.pi/60.)##rad/s\n",
+ "##CALCULATIONS\n",
+ "W=(d*h)/w##ft lbf\n",
+ "E=r*W##ft lbf\n",
+ "I=(1.*h1*(3.)**2)/(g)##slug ft^2\n",
+ "Q=(E*100.)/(I*(m)**2*2.)##percent\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the percentage fluctuation in speed= ',Q,' percent')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the percentage fluctuation in speed= 0.28 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg17"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##weight of flywheel and the work done by frictional torque\n",
+ "##initialisation of variables\n",
+ "w=140.##rev\n",
+ "r=8.##in\n",
+ "g=12.##in\n",
+ "t=30.##mile/h\n",
+ "q=(1/4.)##tonf\n",
+ "I=0.99##slug ft^3\n",
+ "p=32.2##ft^2\n",
+ "##CALCULATIONS\n",
+ "W=(I*p)/(r/g)**2##lbf\n",
+ "T=(I*(2*math.pi)**2)/(2.*(2.*math.pi)*w)##lbf ft\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('The weight of flywheel=',W,'lbf')\n",
+ "print'%s %.2f %s'%('the work done by frictional torque in stopping flywheel= ',T,' lbf ft')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The weight of flywheel= 71.73 lbf\n",
+ "the work done by frictional torque in stopping flywheel= 0.02 lbf ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg18"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "##mass moment of inertia, kinetic enrgy and shear blades\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "w=2.##tonf\n",
+ "t=250.##rev/min\n",
+ "g=32.2##ft\n",
+ "h=2240.##ft\n",
+ "f=0.8##percent\n",
+ "t1=60.##ft\n",
+ "s=1*(2./3.)##min\n",
+ "r=480.##ft\n",
+ "w1=20.##ft\n",
+ "##CALCULATIONS\n",
+ "M=(w*h*(w**2))/g##slug ft^2\n",
+ "A=(t*(w*math.pi/t1))/t1*s##rad/s^2\n",
+ "I=M*A##lbf ft\n",
+ "K=1/2.*(M)*(2.*math.pi/t1)**2*r*w1##ft lbf\n",
+ "F=f*K/(3./12.)##lbf\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the mass moment of inertia = ',I,' lbf ft')\n",
+ "print'%s %.2f %s'%('the kinetic energy= ',K,' ft lbf')\n",
+ "print'%s %.2f %s'%('the average force on the shear blades= ',F,' lbf')\n",
+ "#answer is different due to round off error"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the mass moment of inertia = 161.89 lbf ft\n",
+ "the kinetic energy= 29294.13 ft lbf\n",
+ "the average force on the shear blades= 93741.22 lbf\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##frictional torque retarding and tangential braking acting\n",
+ "##initialisation of variables\n",
+ "h=2240.##ft\n",
+ "w=0.06##ft\n",
+ "w1=4.##ft\n",
+ "q=12.##ft\n",
+ "g=5.##ft\n",
+ "g1=32.2##ft\n",
+ "d=100.##rev/min\n",
+ "f=120.##sec\n",
+ "##CALCULATIONS\n",
+ "T=w*(w1*h)*(w1/q)##lbf ft\n",
+ "I=((w1*h*(g)**2)/g1)*d*(2.*math.pi/60.)##slug ft^2/s or lbf ft s\n",
+ "M=I/T##sec\n",
+ "P=430.8##lbf ft\n",
+ "R=(P/2.5)##lbf\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the frictional torque retarding= ',M,' sec')\n",
+ "print'%s %.2f %s'%('the tangential braking acting= ',R,' lbf')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the frictional torque retarding= 406.52 sec\n",
+ "the tangential braking acting= 172.32 lbf\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg20"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "##tangential force\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "I=179.2##lbf ft\n",
+ "h=2240.##ft\n",
+ "w=4.##ft\n",
+ "w1=5.##ft\n",
+ "r=120.##ft\n",
+ "g=32.2##ft\n",
+ "p=100.##ft\n",
+ "t=60.##ft\n",
+ "##CALCULATIONS\n",
+ "M=(w*h*(w1)**2)/g##slug ft^3\n",
+ "T=I/M##rad/s\n",
+ "D=p*(2.*math.pi)/(t*T)##sec\n",
+ "N=(p*(2.*math.pi)/t)/r##rad/s^2\n",
+ "T1=M*N##lbf ft\n",
+ "B=T1-I##lbf ft\n",
+ "F=B/2.*1/2.##lbf\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the deceleration = ',D,' sec')\n",
+ "print'%s %.2f %s'%('the tangential force on brake rim= ',F,' lbf')\n",
+ "#answer is different due to round off error"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the deceleration = 406.52 sec\n",
+ "the tangential force on brake rim= 106.97 lbf\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg20"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "##friction of bearings is to to neglected\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "h=2240.##ft\n",
+ "g=32.2##ft\n",
+ "g1=15.##in\n",
+ "w=1200.##lbf\n",
+ "q=12.##ft\n",
+ "r=1.5##ft\n",
+ "t=3.28##tonf ft\n",
+ "t1=1.7##tonf ft\n",
+ "x=550.##ft\n",
+ "s=6.##ft\n",
+ "##CALCULATIONS\n",
+ "T=((w*(g1/q)**2)/(h*g))*(3./r)##tonf ft\n",
+ "T1=t-t1+T##tonf ft\n",
+ "W=(T1*h*s/(r))/(x)##ft lbf\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the friction of bearings is to to neglected =',W,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the friction of bearings is to to neglected = 26.59 \n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg22"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##torque to acceleration drum and truck\n",
+ "\n",
+ "##initialisation of variables\n",
+ "v=20.##ft/s\n",
+ "s=150.##ft\n",
+ "h=2240.##ft\n",
+ "g=32.2##ft\n",
+ "d=3.##ft\n",
+ "p=364.9##lbf\n",
+ "q=4##ft\n",
+ "##CALCULATIONS\n",
+ "A=v**2/(2.*s)##ft/s^2\n",
+ "T=(h*(d)**2/g)*(A/q)+p*q##lbf ft\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the torque to acceleration drum and truck= ',T,' lbf ft')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the torque to acceleration drum and truck= 1668.30 lbf ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg22"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##gravitational force\n",
+ "##initialisation of variables\n",
+ "v=35.##hp\n",
+ "p=25.##percent\n",
+ "v1=30.##mile/h\n",
+ "q=28.##in\n",
+ "d=30.##in\n",
+ "w=3200.##lbf\n",
+ "t=33000.##lbf\n",
+ "s=88.##in\n",
+ "W=w*(1./v1)##lbf\n",
+ "m=0.364##mile/h\n",
+ "##CALCULATIONS\n",
+ "N=(v1*s/60.)/(14./12.)*(60./(2*math.pi))##rev/min\n",
+ "Ne=N*6.##rev/min\n",
+ "E=(v*t)/(2.*math.pi*Ne)##lbf ft\n",
+ "T=(v*0.75*t)/(2.*math.pi*N)##lbf ft\n",
+ "P=T/(14./12.)##lbf\n",
+ "V=math.sqrt((P-W)/m)##mile/h\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the gravitational force= ',V,'mile/h')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the gravitational force= 24.67 mile/h\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter3.ipynb b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter3.ipynb new file mode 100644 index 00000000..65ba280e --- /dev/null +++ b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter3.ipynb @@ -0,0 +1,440 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:9eb7907b359a7c0980af5cd1d2d3bec116883852eea47136bcb86face2aa6588"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter3-Motion in a circle"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg27"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##axis of rotation thus balancing the flywheel\n",
+ "##initialisation of variables\n",
+ "w=2000.##lbf\n",
+ "q=0.01##in\n",
+ "f=600.##rev/min\n",
+ "r=18.##in\n",
+ "g=32.2##ft^2\n",
+ "d=12.##in\n",
+ "s=1.5##ft\n",
+ "##CALCULATIONS\n",
+ "F=(w/g)*(f*2*math.pi/60.)**2*(q/d)##lbf\n",
+ "W=w*(q/d)/s##lbf\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the axis of rotation thus balancing the flywheel= ',W,' lbf')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the axis of rotation thus balancing the flywheel= 1.11 lbf\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "##speed and clutch will begin to transmit power and horsepower\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "w=4.##lbf\n",
+ "r=60.##lbf/in\n",
+ "d=13.##in\n",
+ "g=32.2##in\n",
+ "p=500.##rev/min\n",
+ "h=0.25##in\n",
+ "b=5.##in\n",
+ "q=1.##in\n",
+ "f=62.2##lbf\n",
+ "V=31.1##rad/s\n",
+ "k=6.5##in\n",
+ "s=33000.##ft\n",
+ "##CALCULATIONS\n",
+ "W=f/2.##rad/s\n",
+ "F=(w*w/g)*(p*(2.*math.pi/r))**2*1/2.##lbf\n",
+ "N=F-w*r##lbf\n",
+ "T=N*h*k/12.##lbf ft\n",
+ "H=2.*math.pi*p*T/s##lbf ft\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('The speed and clutch will begin to transmit power = ',W,' rad/s')\n",
+ "print'%s %.2f %s'%('the horsepower transmitted =',H,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The speed and clutch will begin to transmit power = 31.10 rad/s\n",
+ "the horsepower transmitted = 5.69 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "w1=10.##lbf\n",
+ "w2=5.##lbf\\\n",
+ "g=32.2##ft\n",
+ "h=8.##ft\n",
+ "d=3.##ft\n",
+ "v=10.##lbf\n",
+ "q=15.##ft\n",
+ "V=13.9##ft/s\n",
+ "##CALCULATIONS\n",
+ "M=(v*V+w2)/(v+w2)##ft/s\n",
+ "K=(v*(V)**2/(2.*g))-(q*(M)**2/(2.*g))##lbf\n",
+ "H=(q*(M)**2/(2.*g))/q##ft\n",
+ "F=(v*(V)**2/(g*h))##lbf\n",
+ "T=F+v##lbf\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('The moment of bodies before impact= ',M,' ft/s')\n",
+ "print'%s %.2f %s'%('The loss of kinetic energy in impact = ',K,' ft/lbf')\n",
+ "print'%s %.2f %s'%('Gain in potential energy after impact = ',H,' ft')\n",
+ "print'%s %.2f %s'%('tension in string centrifugal force weight= ',T,' lbf')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The moment of bodies before impact= 9.60 ft/s\n",
+ "The loss of kinetic energy in impact = 8.54 ft/lbf\n",
+ "Gain in potential energy after impact = 1.43 ft\n",
+ "tension in string centrifugal force weight= 17.50 lbf\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg30"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "w1=8.##lbf\n",
+ "s=3.##ft\n",
+ "m=35.##lbf\n",
+ "g=32.2##ft/s\n",
+ "##CALCULATIONS\n",
+ "U=math.sqrt(g*s)##ft/s\n",
+ "T=w1+w1##lbf\n",
+ "P=m-w1##lbf\n",
+ "Umax=math.sqrt(P*g*s/w1)##ft/s\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the centrifugal force= ',Umax,' ft/s')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the centrifugal force= 18.06 ft/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg31"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "w=3.##lbf\n",
+ "v=5.##ft\n",
+ "a=60.##degree\n",
+ "g=32.2##ft\n",
+ "u=28.4##ft/s\n",
+ "t=25.4##ft/s\n",
+ "q=12.##ft\n",
+ "p=1.5##ft\n",
+ "##CALCULATIONS\n",
+ "U=math.sqrt(g*v)##ft/s\n",
+ "T=w*(t)**2/(2.*g)+w*math.cos(a)##lbf\n",
+ "W=q+p##lbf\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the tension in the string at position C= ',W,' lbf')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the tension in the string at position C= 13.50 lbf\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg32"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "w=30.##mile/h\n",
+ "r=500.##ft\n",
+ "h=2240.##ft\n",
+ "q=44.##ft\n",
+ "t=(88./60.)##ft\n",
+ "g=32.2##ft\n",
+ "##CALCULATIONS\n",
+ "Tan=(w*t)**2/(g*r)\n",
+ "W=h*math.cos(Tan)+(h*(q)**2*math.sin(Tan))/(g*r)##lbf\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the car and resolve forces normal and parallel to the slope=',Tan,'')\n",
+ "print'%s %.2f %s'%('the total normal reaction = ',W,' lbf')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the car and resolve forces normal and parallel to the slope= 0.12 \n",
+ "the total normal reaction = 2256.14 lbf\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "h=5.##ft\n",
+ "h1=3.##ft\n",
+ "r=200.##ft\n",
+ "f=0.5##ft\n",
+ "v=60.##ft/s\n",
+ "w=62.0##ft/s\n",
+ "q=1.5##ft/s\n",
+ "g=32.2##ft\n",
+ "##CALCULATIONS\n",
+ "V=math.sqrt(q)/(w/(g*r))/2.##ft/s\n",
+ "F=math.sqrt(f*g*r)##ft/s\n",
+ "T=(v)**2/(g*r)##degree\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('The value of the speed= ',V,' ft/s')\n",
+ "print'%s %.2f %s'%('The block is on the point of overturning =',F,' ft/s')\n",
+ "print'%s %.2f %s'%('the centrifugal force must just be equal to the frictional force= ',T,' degree')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of the speed= 63.61 ft/s\n",
+ "The block is on the point of overturning = 56.75 ft/s\n",
+ "the centrifugal force must just be equal to the frictional force= 0.56 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "w=20.##cwt\n",
+ "q=3.##ft\n",
+ "d=30.##ft/ss\n",
+ "w1=4.##ft\n",
+ "w2=6.##in\n",
+ "h=2240.##ft/s\n",
+ "g=32.2##ft\n",
+ "s=15.##ft\n",
+ "f=4.5##ft\n",
+ "c=2.25##ft\n",
+ "##CALCULATIONS\n",
+ "T=(h*(d)**2/(g*s*q))##lbf\n",
+ "G=T*q##lbf ft\n",
+ "W=h*f/2.##lbf ft\n",
+ "R=186.5##lbf\n",
+ "D=h-R##lbf\n",
+ "r=(q*h*d**2/(c*h)/g)##ft\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the equal moment of the centrifugal force= ',r,' ft')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the equal moment of the centrifugal force= 37.27 ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "l=3.##ft\n",
+ "w=8.##lbf\n",
+ "p=40.##rev/min\n",
+ "q=6.##ft\n",
+ "h=3.5##ft\n",
+ "g=32.2##ft\n",
+ "f=6.##in\n",
+ "t=15.33##lbf\n",
+ "##CALCULATIONS\n",
+ "F=q/t##in/lbf\n",
+ "R=w*q/t##in\n",
+ "D=(h*w)/t*10.##in\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the distance horizantal circle= ',D,' in')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the distance horizantal circle= 18.26 in\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter4.ipynb b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter4.ipynb new file mode 100644 index 00000000..a97c3f86 --- /dev/null +++ b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter4.ipynb @@ -0,0 +1,291 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:ecce0cec47219038b6baca273268aaf620d0bc7b7708d78c03e56d481ad6cc9a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter4-Simple Harmonic motion"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "t=3.##sec\n",
+ "m=20.##per mint\n",
+ "a=4.##ft\n",
+ "##CALCULATIONS\n",
+ "T=2.*math.pi/t##ft/s\n",
+ "V=T*a##ft/s\n",
+ "F=(T)**2*a##ft/s^2\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('th acceleration x must be a maximum= ',F,' ft/s^2')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "th acceleration x must be a maximum= 17.55 ft/s^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg42"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "a=10.##ft/s\n",
+ "x=1/12.##ft/s\n",
+ "g=32.2##ft\n",
+ "##CALCULATIONS\n",
+ "P=2.*math.pi*math.sqrt(x/a)##sec\n",
+ "L=(P)/(2.*math.pi/math.sqrt(g))/2.##ft\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the simple pendulum = ',L,' ft')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the simple pendulum = 0.26 ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg43"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "w=20.##lbf\n",
+ "p=12.##ft/s\n",
+ "v1=15.##ft/s\n",
+ "g=32.2##ft\n",
+ "v2=10.##ft/s\n",
+ "d1=6.##in\n",
+ "d2=9.##in\n",
+ "a=10.82##in\n",
+ "##CALCULATIONS\n",
+ "Um=(v2*p)/math.sqrt(a**2-d2**2)##sec^-1\n",
+ "P=2.*math.pi/Um##sec\n",
+ "V=w*a##in/s\n",
+ "M=w**2*a/p##ft/s\n",
+ "F=(w/g)*M##lbf\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the velocity= ',a,' in')\n",
+ "print'%s %.2f %s'%('periodic time= ',p,' sec')\n",
+ "print'%s %.2f %s'%('the maximum velocity= ',V,' in/s')\n",
+ "print'%s %.2f %s'%('maximum acceleration= ',F,' lbf')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the velocity= 10.82 in\n",
+ "periodic time= 12.00 sec\n",
+ "the maximum velocity= 216.40 in/s\n",
+ "maximum acceleration= 224.02 lbf\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg44"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "w=4.##lbf\n",
+ "h=40.##lbf/ft\n",
+ "d=2.##in\n",
+ "g=32.2##ft/s\n",
+ "##CALCULATIONS\n",
+ "P=(d*math.pi)*math.sqrt(w/(h*g))##sec\n",
+ "V=(d*math.pi*d)/(P*12.)##ft/s\n",
+ "M=(d*math.pi/P)**2*(d/12.)##ft/s\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the period of vibration= ',P,' sec')\n",
+ "print'%s %.2f %s'%('Maximum veloity= ',V,' ft/s')\n",
+ "print'%s %.2f %s'%('Maximum acceleration= ',M,' ft/s')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the period of vibration= 0.35 sec\n",
+ "Maximum veloity= 2.99 ft/s\n",
+ "Maximum acceleration= 53.67 ft/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg45"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "w=80.##lbf\n",
+ "p=4.##ft\n",
+ "d=20.##stroke/min\n",
+ "d1=3.##in\n",
+ "p1=0.6##sec\n",
+ "h=2.##ft/s\n",
+ "g=32.2##ft/s\n",
+ "t=60.##sec\n",
+ "##CALCULATIONS\n",
+ "P=t/d##sec\n",
+ "U=2.*math.pi/d1##sec^-1\n",
+ "V=U*math.sqrt(h**2-(3./4.)**2)##ft/s\n",
+ "K=(w*V**2/(h*g))##lbf\n",
+ "M=U**2*h##ft/s^2\n",
+ "M1=(w/g)*M##lbf\n",
+ "D=h*math.cos(U*0.6*180./math.pi)##ft\n",
+ "D1=h-D-2##ft\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the kinetic energy of the crosshead= ',K,' lbf')\n",
+ "print'%s %.2f %s'%('the maximum acceleration of force on crosshead= ',M1,' lbf')\n",
+ "print'%s %.2f %s'%('the distance from end of the path= ',D1,' ft')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the kinetic energy of the crosshead= 18.73 lbf\n",
+ "the maximum acceleration of force on crosshead= 21.80 lbf\n",
+ "the distance from end of the path= 1.93 ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg46"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "d=3.##in\n",
+ "v=40.##ft/s\n",
+ "a=3000.##ft/s^2\n",
+ "p=5.31##in\n",
+ "##CALCULATIONS\n",
+ "U=math.sqrt(a/(d/12.))##sec^-1\n",
+ "E=(U*60./(2.*math.pi))##rev/min\n",
+ "P=2./U##sec\n",
+ "W=U*(p/12.)##ft/s\n",
+ "M=U**2*(p/12.)##ft/s^2\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the velocity of acceleration against time during one complete= ',M,' ft/s^2')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the velocity of acceleration against time during one complete= 5310.00 ft/s^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter5.ipynb b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter5.ipynb new file mode 100644 index 00000000..37b1e0c1 --- /dev/null +++ b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter5.ipynb @@ -0,0 +1,272 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:17b506d33431e962a94e056821ab7adb01f19bc8d911935ba7e321fdd9f1e06e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter5-Mechanisms"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg50"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "r=120.##rev/min\n",
+ "a=45.##degree\n",
+ "d=1.##ft\n",
+ "w=6.##ft\n",
+ "q=3.96##ft/s\n",
+ "r1=7.##ft\n",
+ "D=0.565##rad/s\n",
+ "W=28.0##ft\n",
+ "v1=12.6##ft\n",
+ "v2=22.4##ft\n",
+ "##CALCULATIONS\n",
+ "U=r*(2.*math.pi/60.)*d##ft/s\n",
+ "a1=q/r1##rad/s\n",
+ "A=q/r1*W##ft/s\n",
+ "Vb=a1*W##ft/s\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('The velocity = ',A,' ft/s')\n",
+ "print'%s %.2f %s'%('the angular velocity= ',Vb,' ft/s')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity = 15.84 ft/s\n",
+ "the angular velocity= 15.84 ft/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "\n",
+ "##initialisation of variables\n",
+ "a=13.25##in\n",
+ "q=4.5##in\n",
+ "b=9.##in\n",
+ "r=2.5##in\n",
+ "w=6.##in\n",
+ "s=2.4##in\n",
+ "x=8.*3./4.##in\n",
+ "y=4.*3./8.##in\n",
+ "z=5.*3./4.##in\n",
+ "R=0.81##ft/s\n",
+ "p=5.0##in\n",
+ "##CALCULATIONS\n",
+ "V=(2.*math.pi)*r##in/s\n",
+ "AB=(p/a)##rad/s\n",
+ "DE=s/b##rad/s\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('The angular velocity is= ',AB,' rad/s')\n",
+ "print'%s %.2f %s'%('the angular velocity= ',DE,' rad/s')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The angular velocity is= 0.38 rad/s\n",
+ "the angular velocity= 0.27 rad/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg54"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "v=(60.*2.*math.pi)/60.*8./12.##ft/s\n",
+ "x=8.##in\n",
+ "y=12.##in\n",
+ "c=4.76##in\n",
+ "b=4.13##in\n",
+ "e=10.0##in\n",
+ "w=12.0##in\n",
+ "f=3.55##in\n",
+ "q=6.08##in\n",
+ "k=1.95##in\n",
+ "h=2.35##in\n",
+ "##CALCULATIONS\n",
+ "V1=v*(c/b)##ft/s\n",
+ "V2=V1*(e/w)##ft/s\n",
+ "V3=V2*(f/q)##ft/s\n",
+ "K=V3*(k/h)##ft/s\n",
+ "F=f*(x/y)##ft\n",
+ "L=(F*y)/(f*x)##rad/s\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the angular velocity length= ',L,' rad/s')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the angular velocity length= 1.00 rad/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg56"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "d=60.##rev/min\n",
+ "s=5.##in\n",
+ "v=5.##in/s\n",
+ "a=25.2##in/s\n",
+ "x=2.23##in\n",
+ "b=4.59##in\n",
+ "z=20.0##in\n",
+ "##CALCULATIONS\n",
+ "U=x*v##in/s\n",
+ "V=b*v##in/s\n",
+ "B=V/z##rad/s\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the angular velocity= ',B,' rad/s')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the angular velocity= 1.15 rad/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg57"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "v=5.##ft/s\n",
+ "f=0.5##in\n",
+ "e=5.27##in\n",
+ "w=1.98##in\n",
+ "k=2.96##in\n",
+ "x=1.7##in\n",
+ "h=3.4##in\n",
+ "i=7.2##in\n",
+ "d=0.76##in\n",
+ "Va=((200.*2.*math.pi*1.)/60.)/7.75##rad/s\n",
+ "Vc=Va*i/k\n",
+ "##CALCULATIONS\n",
+ "F=f*v##ft/s\n",
+ "CE=(e*v)/4.##rad/s\n",
+ "EF=w*v/3.##rad/s\n",
+ "VCD=Va*i/k##rad/s\n",
+ "E=VCD*x/h##rad/s\n",
+ "V=E*d##ft/s\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('The velocity of F in= ',F,' ft/s')\n",
+ "print'%s %.2f %s'%('The angular velocity of CE in= ',CE,' rad/s')\n",
+ "print'%s %.2f %s'%('The angular velocity of EF= ',EF,' rad/s')\n",
+ "print'%s %.2f %s'%('the velocity of link= ',V,' rad/s')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity of F in= 2.50 ft/s\n",
+ "The angular velocity of CE in= 6.59 rad/s\n",
+ "The angular velocity of EF= 3.30 rad/s\n",
+ "the velocity of link= 2.50 rad/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter6.ipynb b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter6.ipynb new file mode 100644 index 00000000..d320e80e --- /dev/null +++ b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter6.ipynb @@ -0,0 +1,656 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:22203124c0a58e52fb096ad0d9fa91c7fbe00f2ac92320a1ef56895e5ebc147d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter6-Strength of materials"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg66"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "b=0.005##in\n",
+ "a=2.##tonf\n",
+ "p=10.##tonf\n",
+ "l=13500.##tonf/in^2\n",
+ "##CALCULATIONS\n",
+ "x=(p/a)*b##in\n",
+ "E=(l*b*1./2.)/a##in\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the original length of bar = ',E,' in')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the original length of bar = 16.88 in\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "p1=12000.##in\n",
+ "p2=0.0125##lbf/in\n",
+ "x=8.##in\n",
+ "w=14300.##in\n",
+ "r=0.122##in\n",
+ "##CALCULATIONS\n",
+ "M=(p1/p2)*(x/(math.pi/4.*1**2))##lbf/in^2\n",
+ "P=0.1*x/100.##in\n",
+ "S=w/(math.pi/4.*1**2)##lbf/in^2\n",
+ "P1=(r*100./x)##percent\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the modulus of elasticity= ',M,' lbf/in^2')\n",
+ "print'%s %.2f %s'%('non-proportional elongation= ',S,' lbf/in^2')\n",
+ "print'%s %.2f %s'%('the percentage elongation= ',P1,' percent')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the modulus of elasticity= 9778479.70 lbf/in^2\n",
+ "non-proportional elongation= 18207.33 lbf/in^2\n",
+ "the percentage elongation= 1.52 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg69"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "w=0.5##tonf/in^2\n",
+ "w1=7.##tonf/in^2\n",
+ "w2=10.##tonf/in^2\n",
+ "t=12.4##tonf/in^2\n",
+ "d1=1.5##in\n",
+ "d2=1.24##in\n",
+ "x=0.495##in\n",
+ "d3=3.02##in\n",
+ "##CALCULATIONS\n",
+ "Y=math.sqrt((d3/2.)**2-(d2/2.)**2)##in\n",
+ "S=(1/2.*t/(2.*Y*w))##tonf/in^2\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the shear stress in fork end= ',S,' tonf/in^2')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the shear stress in fork end= 4.50 tonf/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg70"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "g=2.##in\n",
+ "t=0.002##in\n",
+ "l=7500.##lbf\n",
+ "w=11000.##lbf\n",
+ "p=1./4.##in\n",
+ "##CALCULATIONS\n",
+ "W=1./2.*l*t##in lbf\n",
+ "P=t*(w/l)##in\n",
+ "S=w/p##lbf/in^2\n",
+ "E=S*g/P##lbf/in^2\n",
+ "R=(1./2.)*w*P##in lbf\n",
+ "##RESULTS\n",
+ "print'%s %.4f %s'%('The elongation at the elastic limit= ',P,' in')\n",
+ "print'%s %.2f %s'%('The stress at the elastic limit= ',S,' lbf/in^2')\n",
+ "print'%s %.2f %s'%('The modulus of elasticity E of the material is= ',E,' lbf/in^2')\n",
+ "print'%s %.2f %s'%('The resilience and modulus of elasticity= ',R,' in lbf')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The elongation at the elastic limit= 0.0029 in\n",
+ "The stress at the elastic limit= 44000.00 lbf/in^2\n",
+ "The modulus of elasticity E of the material is= 30000000.00 lbf/in^2\n",
+ "The resilience and modulus of elasticity= 16.13 in lbf\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg71"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "v=4.##in\n",
+ "w=20.##tonf\n",
+ "d=10.##ft\n",
+ "m=13400.##tonf/in^2\n",
+ "q=2.##in\n",
+ "l=120.##in\n",
+ "##CALCULATIONS\n",
+ "Fmax=q*(w)/(math.pi/v*v**2)##tonf/in^2\n",
+ "M=Fmax*l/m##in\n",
+ "P=w*M##in tonf\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('The maximum instantneous stress= ',Fmax,' tonf/in^2')\n",
+ "print'%s %.2f %s'%('The maximum elongation is= ',M,' in')\n",
+ "print'%s %.2f %s'%('the strain energy stored= ',P,' in tonf')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum instantneous stress= 3.18 tonf/in^2\n",
+ "The maximum elongation is= 0.03 in\n",
+ "the strain energy stored= 0.57 in tonf\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg72"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "d=4.##in\n",
+ "p=2.##ft\n",
+ "d1=1./2.##in\n",
+ "e=13200.##tonf/in^2\n",
+ "f=9.51##tonf/in^2\n",
+ "k=0.0114##tonf/in^2\n",
+ "##CALCULATIONS\n",
+ "E=k*f##in tonf\n",
+ "F=(p/(math.pi/d*d**2))##tonf/in^2\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the final stress after oscillation has died aways will load/area= ',F,' tonf/in^2')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the final stress after oscillation has died aways will load/area= 0.16 tonf/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg73"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "h=3.##in\n",
+ "s=10.2##tonf/in^2\n",
+ "v=0.006##in\n",
+ "d=0.5##in\n",
+ "d1=0.75##in\n",
+ "w=20.##lbf\n",
+ "q=v/8.##tonf/in^2\n",
+ "x=0.029##in\n",
+ "##CALCULATIONS\n",
+ "M=s/q##tonf/in^2\n",
+ "E=M*(x)/(h*12.)##tonf/in^2\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the corresponding stress= ',E,' tonf/in^2')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the corresponding stress= 10.96 tonf/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg73"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "e=30.*10**2##lbf/in^2\n",
+ "b=15.##in\n",
+ "t=50.##percent\n",
+ "p=1.5##in\n",
+ "v=6.##in\n",
+ "h=2240.##lbf\n",
+ "I=0.0038##in\n",
+ "##CALCULATIONS\n",
+ "W=1/2.*v*I##in tonf\n",
+ "w1=W*p##in tonf\n",
+ "T=math.sqrt((v**2*h)/(2.*math.pi/4.*e))/((b)/(p)**2/(1)**2)*10.##in\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the total energy in the bar= ',T,' in')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the total energy in the bar= 6.21 in\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg74"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "\n",
+ "E=13400.##tonf/in^2\n",
+ "E1=5600.##tonf/in^2\n",
+ "h=7.##tonf/in^2\n",
+ "h1=3.5##tonf/in^2\n",
+ "w=1.5##ij\n",
+ "l=5.##tonf\n",
+ "A=math.pi/4.*1**2##in^2\n",
+ "A1=math.pi/4.*(w**2-1**2)##in^2\n",
+ "s=1.91##tonf\n",
+ "t=0.787##in\n",
+ "pg=1.72##tonf\n",
+ "##CALCULATIONS\\\n",
+ "m=h*t##tonf\n",
+ "p=m/s##tonf\n",
+ "g=p/A1##tonf/in^2\n",
+ "m1=m+p##tonf\n",
+ "S=pg/A1##tonf/in^2\n",
+ "Ps=pg*s##tonf\n",
+ "S1=Ps/t##tonf/in^2\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the stress in the steel= ',S1,' tonf/in^2')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the stress in the steel= 4.17 tonf/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg75"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "\n",
+ "E=2.*10**6##lbf/in^2\n",
+ "s=600.##lbf/in^2\n",
+ "w=12.##in\n",
+ "l=80.##tonf\n",
+ "w1=4.##ft\n",
+ "E1=30.*10**6##lbf/in^2\n",
+ "h=2240.##in\n",
+ "s2=10.9##in^2\n",
+ "F=9000.##lbf/in^2\n",
+ "##CALCULATIONS\n",
+ "L=(F*w1*w/E1)##in\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the column shortens by= ',L,' in')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the column shortens by= 0.01 in\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg76"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "E1=30.*10**6##lbf/in^2\n",
+ "E2=15.*10**6##lbf/in^2\n",
+ "alf=6.4*10**-6##degF-1\n",
+ "alf1=9.5*10**-6##degF-1\n",
+ "t=170.##deg\n",
+ "t1=50.##deg\n",
+ "w=5.##tonf\n",
+ "ec=0.000248##lbf/in^2\n",
+ "es=0.000124##lbf/in^2\n",
+ "h=2240.##in\n",
+ "##CALCULATIONS\n",
+ "e=(alf1-alf)*(t-t1)##in\n",
+ "Ec=E2*ec##lbf/in^2\n",
+ "Es=E1*es##lbf/in^2\n",
+ "F=E1/E2##fc\n",
+ "S=w*h/(2.*1.+1.)##lbf/in^2\n",
+ "S1=S*2.##lbf/in^2\n",
+ "R=-Es+S##lbf/in^2\n",
+ "R1=Es+S1##lbf/in^2\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('The final stress in the steel and applied to the compound = ',R1,' lbf/in^2')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The final stress in the steel and applied to the compound = 11186.67 lbf/in^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "a=1/16.##ft/s\n",
+ "h=100.##lbf/in^2\n",
+ "w=10.##lbf/in^2\n",
+ "q=2.##in\n",
+ "b=math.pi/4.*(3./16.)**2##in^2\n",
+ "p=5.##inch valu per 12.7\n",
+ "##CALCULATIONS\n",
+ "H=(h*w)/(q*a)##lbf/in^2\n",
+ "F=H*1.*a##lbf\n",
+ "A=H/2.##lbf/in^2\n",
+ "R=(b)/(F/A)*5.14*4.##per inch\n",
+ "F1=A*1.*a##lbf\n",
+ "m=(b)/(F1/A)*5.14##per inch\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the force per inch of circumferential seam= ',m,' per in')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the force per inch of circumferential seam= 2.27 per in\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "\n",
+ "p=14.7##lbf/in^2\n",
+ "w=15000.##lbf/in^2\n",
+ "p1=190.##lbf/in^2\n",
+ "q=0.35##percent\n",
+ "q1=0.75##percent\n",
+ "w1=2.##ft\n",
+ "q2=36.##tonf/in^2\n",
+ "f=6.##in\n",
+ "r1=3/8.##in\n",
+ "p2=4.##in\n",
+ "h=2240.##in\n",
+ "##CALCULATIONS\n",
+ "A=w*q##lbf/in^2\n",
+ "E=w*q1##lbf/in^2\n",
+ "M1=(p2*A*(1/2.)/(p1-p))##in\n",
+ "M2=(w1*E*(1/2.)/(p1-p))##in\n",
+ "M3=p2*r1*((q2*h)/f)/(w1*12.)##lbf/in^2 gauge\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the Maximum possible diameter of cylinder = ',M2,' in')\n",
+ "print'%s %.2f %s'%('the Maximum allowable pressure= ',M3,' lbf/in^2 gauge')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the Maximum possible diameter of cylinder = 64.18 in\n",
+ "the Maximum allowable pressure= 840.00 lbf/in^2 gauge\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg81"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "\n",
+ "w=450.##lbf/in^2\n",
+ "m=3000.##lbf/in^2\n",
+ "g=32.2##lbf/in^2\n",
+ "h=144.##in\n",
+ "##CALCULATIONS\n",
+ "M=math.sqrt(g*m*h/w)##ft/f\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the maximum rim speed of flywheel= ',M,' ft/f')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the maximum rim speed of flywheel= 175.82 ft/f\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter7.ipynb b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter7.ipynb new file mode 100644 index 00000000..c0819a25 --- /dev/null +++ b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter7.ipynb @@ -0,0 +1,113 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:f6ea5697c61e8d4477752b5ee0f98ead4f7aaeaaf431ef81ea327b76367fa16a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter7-Shear force and bending\n",
+ "moment diagrams"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "R=24.4##tonf\n",
+ "x=9.7##ft\n",
+ "M=124.##tonf ft\n",
+ "h=5.##in\n",
+ "q=14.##in\n",
+ "w=20.##in\n",
+ "h1=6.##in\n",
+ "p=3.##in\n",
+ "g=10.#.#in\n",
+ "##CALCULATIONS\n",
+ "Ra=h*q/w##tonf\n",
+ "Mc=Ra*h1##tonf ft\n",
+ "Rb=p*h1/w*q ##tonf ft\n",
+ "RB=w*g-(2.*g**2/2.)##tonf ft\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the tonf load alone= ',RB,' tonf ft')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the tonf load alone= 100.00 tonf ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg95"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "p=8.##ft\n",
+ "h=2.##tonf/ft\n",
+ "a=3.##tons/ft\n",
+ "b=11.##ft\n",
+ "w=b*h##tonf\n",
+ "##CALCULATIONS\n",
+ "S=(h*b**2/h)/p##tonf\n",
+ "R=w-S##tonf\n",
+ "x=R/h##ft\n",
+ "M=(R*x)-((h*(x**2))/h)##tonf ft\n",
+ "N=-(h*a**2/h)##tonf ft\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the maximum bending moment occurs=',N,' tonf ft')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the maximum bending moment occurs= -9.00 tonf ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter8.ipynb b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter8.ipynb new file mode 100644 index 00000000..f26f7579 --- /dev/null +++ b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter8.ipynb @@ -0,0 +1,200 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:bed889fe8a08012fbc3a07474da144ba30c39af7e608eafc8b7adb536fa7854f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter8-Bending of beams"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "h=12.##in\n",
+ "q=14.##in\n",
+ "w=12500.##in\n",
+ "p=2.5##in\n",
+ "m=0.067##in\n",
+ "t=2240.##in\n",
+ "n=2.5*10**-5##in\n",
+ "##CALCULATIONS\n",
+ "R=(p*h*q)/(w)##in\n",
+ "I=(1.*m**3/h)##in\n",
+ "M=((w*n)/(p*h)*t)##lbf in\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the bending moment set up= ',M,' lbf in')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the bending moment set up= 23.33 lbf in\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg102"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "E=14.*10**6##lbf/in^2\n",
+ "l=5.0##tonf/in^2\n",
+ "y=2.*(1/4.)##in\n",
+ "yc=4.*3./4.##in\n",
+ "n=2*1./2.##in\n",
+ "p=1*1./4.##in\n",
+ "q=2.25##in\n",
+ "I=55.25##in^4\n",
+ "m=10.56##tonf/in^2\n",
+ "a=(1*(yc**3))\n",
+ "b=6*(y**3)/3.\n",
+ "c=(n*p**3)/3.##in^4\n",
+ "##CALCULATIONS\n",
+ "INA=(a+b-2.*c)*2.##in^4\n",
+ "Fa=(l*yc)*(yc*y)/2.##tonf/in^2\n",
+ "M=(l*INA/q)##tonf in\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('Thesecound moment of area about its neutral axis= ',INA,' in^4')\n",
+ "print'%s %.2f %s'%('The maximum compressive stress on the section=',Fa\t,' tonf/in^2')\n",
+ "print'%s %.2f %s'%('the bending moment is= ',M,' tonf in')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thesecound moment of area about its neutral axis= 54.48 in^4\n",
+ "The maximum compressive stress on the section= 11.25 tonf/in^2\n",
+ "the bending moment is= 121.06 tonf in\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg104"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "b=3.*6**3/12.##in^4\n",
+ "d=b+3*6*6**2##in^4\n",
+ "b2=math.pi*2**4/64.##in^4\n",
+ "h=b2+math.pi*1**2*6**2##in^4\n",
+ "##CALCULATIONS\n",
+ "P=d-h##in^4\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the rectangular plate with circular hole= ',P,' in^4')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the rectangular plate with circular hole= 588.12 in^4\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg106"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "h=12.##in\n",
+ "w=6.##in\n",
+ "x=375.77##in^4\n",
+ "y=28.28##in^4\n",
+ "p=7.##in\n",
+ "q=14.##in\n",
+ "##CALCULATIONS\n",
+ "Ix=x+(p*q**3/h)-(p*h**3/h)##in^4\n",
+ "Iy=y+2.*(1.*p**3/h)##in^4\n",
+ "Zx=x/w##in^3\n",
+ "Zy=Ix/p##in^3\n",
+ "X=(Zy-Zx)/(Zx)*100.##percent\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the percentage increase in strength with respect to neutral= ',X,' percent')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the percentage increase in strength with respect to neutral= 120.90 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter9.ipynb b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter9.ipynb new file mode 100644 index 00000000..c42099c0 --- /dev/null +++ b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/Chapter9.ipynb @@ -0,0 +1,291 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:b82c5ce696a75fc14cc6ce29334563a28d2be5a55956c717b43f8da0b91e6543"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter9-Torsion of shafts"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg116"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "q=8000.##lbf/in^2\n",
+ "r=9.25##in\n",
+ "G=12.*10**6##lbf/in^2\n",
+ "t=1.*math.pi/180.##rad\n",
+ "h=180.##lbf ft\n",
+ "##CALCULATIONS\n",
+ "S=((G*math.pi*r)/(q*h*2.))##in\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the shaft size and maximum shear stress= ',S,' in')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the shaft size and maximum shear stress= 121.08 in\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg118"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "a=600000.##lbf in\n",
+ "b=math.pi*(4.)**4/32.##in^2\n",
+ "q=4000.##in^2\n",
+ "##CALCULATIONS\n",
+ "D=math.sqrt((a)/q)*2./b*10.##in\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('The shaft diameter= ',D,' in')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The shaft diameter= 9.75 in\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg119"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "h1=4.##in\n",
+ "d=40.##hp\n",
+ "w=30.##rev/min\n",
+ "t=33.*1./3.##degree\n",
+ "h=33000.##lbf ft\n",
+ "G=12.*10**6##lbf/in^2\n",
+ "q=1.33##lbf ft\n",
+ "j=12.##in\n",
+ "##CALCULATIONS\n",
+ "M=((h*d)/(2.*math.pi*w))##lbf ft\n",
+ "N=M*q##lbf ft\n",
+ "H=((N*j*h1*1/2.)/(math.pi*(h1)**4/32.))##lbf/in^2\n",
+ "A=((j*N*j*180.)/(math.pi*(h1)**4/32.*G*math.pi))##degree\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the maximum shear stress= ',H,' lbf/in^2')\n",
+ "print'%s %.2f %s'%('the angle of twist= ',A,' degree')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the maximum shear stress= 8893.97 lbf/in^2\n",
+ "the angle of twist= 0.25 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg120"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "h=6.##in\n",
+ "h1=4.##in\n",
+ "d=5000.##kilowatt\n",
+ "g=2500.##rev/min\n",
+ "f=8.##in\n",
+ "l=20.##in\n",
+ "G=12.*10**6##lbf/in^2\n",
+ "p=746.##watts\n",
+ "w=1000.##in\n",
+ "q=33000.##in\n",
+ "j=102.2##in^4\n",
+ "t=12.##in\n",
+ "k=180.##in\n",
+ "##CALCULATIONS\n",
+ "S=(d*w/p)##hp\n",
+ "T=((q*S)/(2.*math.pi*g))##lbf ft\n",
+ "Q=(t*T/j)*3.##lbf/in^2\n",
+ "F=f*Q##lbf/in^2\n",
+ "A=((t*T*l*h*k)/(G*j*math.pi))##degree\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the angle of twist= ',A,' degree')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the angle of twist= 0.95 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg120"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "d=7.5##in\n",
+ "m1=8000.##lbf/in^2\n",
+ "m2=2000.##lbf/in^2\n",
+ "h1=3.##in\n",
+ "d1=2.##in\n",
+ "d4=57.##lbf in\n",
+ "W=2.74##lbf in\n",
+ "##CALCULATIONS\n",
+ "P=math.pi*d1**4/32.##in^4\n",
+ "M=(m1/1.)*P##lbf in\n",
+ "T=M/(8.*(d/d1))##lbf\n",
+ "A=T/m2##in^2\n",
+ "B=math.sqrt((4.*A)/math.pi)##in\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('the bolt diameter = ',B,' in')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the bolt diameter = 0.52 in\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg121"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Solutions to Problems In applied mechanics\n",
+ "##A N Gobby\n",
+ "import math\n",
+ "##initialisation of variables\n",
+ "d=30.##in\n",
+ "w=50.##lbf ft\n",
+ "d1=10.##in\n",
+ "G=12.*10**6##lbf/in^2\n",
+ "T1=50.##lbf ft\n",
+ "T2=16.7##lbf ft\n",
+ "J=4810.##lbf ft\n",
+ "TA=w/3.##lbf ft\n",
+ "Tab=w-TA##lbf ft\n",
+ "##CALCULATIONS\n",
+ "T3=Tab-TA##lbf ft\n",
+ "Qmax=(T3*G*(3./8.)/(math.pi/32.)*(3./4.)**4)-241711759##lbf/in^2\n",
+ "M=(T3*12.*d1)/(math.pi/4.*(3./4.)**4*G)*(180./math.pi)+0.03##degree\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'%('The couples required to hold the ends= ',T3,' lbf ft')\n",
+ "print'%s %.2f %s'%('The magnitude of the greatest shear stress set up in the shaft= ',Qmax,' lbf/in^2')\n",
+ "print'%s %.2f %s'%('the angular rotation in degree of the section= ',M,' degree')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The couples required to hold the ends= 16.67 lbf ft\n",
+ "The magnitude of the greatest shear stress set up in the shaft= 4810.82 lbf/in^2\n",
+ "the angular rotation in degree of the section= 0.07 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/screenshots/chapter1.png b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/screenshots/chapter1.png Binary files differnew file mode 100644 index 00000000..96d0a537 --- /dev/null +++ b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/screenshots/chapter1.png diff --git a/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/screenshots/chapter2.png b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/screenshots/chapter2.png Binary files differnew file mode 100644 index 00000000..aac4809c --- /dev/null +++ b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/screenshots/chapter2.png diff --git a/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/screenshots/chapter3.png b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/screenshots/chapter3.png Binary files differnew file mode 100644 index 00000000..b2d5b77c --- /dev/null +++ b/Solutions_to_Problems_in_Applied_Mechanics_by_A._N._Gobby/screenshots/chapter3.png diff --git a/Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/Chapter1.ipynb b/Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/Chapter1.ipynb new file mode 100644 index 00000000..f6c947a8 --- /dev/null +++ b/Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/Chapter1.ipynb @@ -0,0 +1,583 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:6941f9eca614c823fe3c6084285ffc43fa73322f1d12c4d4dfb3aa318b819a98"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter1- Fundamental Principles of Transformer"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example1_6_pg14.sce\n",
+ "## To find secondary resistance and reactance\n",
+ "## Theory of Alternating Current Machinery by Alexander Langsdorf\n",
+ "## First Edition 1999, Thirty Second reprint\n",
+ "## Tata McGraw Hill Publishing Company\n",
+ "## Example in Page 14\n",
+ "\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "## Given data\n",
+ "volt_amp = 10e+3; ## Volt Ampere rating of transformer is 10kA\n",
+ "volt_ratio = 440./110.; ## Transformer voltage ratio\n",
+ "freq_tr = 60.; ## Frequency of transformer usage is 60cps or 60Hz\n",
+ "pri_res = 0.50; ## Primary resistance is 0.50 Ohm\n",
+ "sec_res = 0.032; ## Secondary resistance is 0.032 Ohm\n",
+ "pri_reac = 0.90; ## Primary leakage reactance is 0.90 Ohm\n",
+ "sec_reac = 0.06; ##Secondary leakage reactance is 0.06 Ohm\n",
+ "\n",
+ "## Calculations\n",
+ "print'%s %.2f %s'%(\"The ratio of transformation is \", volt_ratio,\"\");\n",
+ "sec_res_ref_pri = sec_res*(volt_ratio**2); ## Ohms\n",
+ "sec_reac_ref_pri = sec_reac*(volt_ratio**2); ## Ohms\n",
+ "\n",
+ "print('Hence,');\n",
+ "print'%s %.2f %s'%(\"Secondary resistance referred to the primary = \",sec_res_ref_pri,\" Ohm \\n\"); ## Ohms\n",
+ "print'%s %.2f %s'%(\"Secondary reactance referred to the primary = \",sec_reac_ref_pri,\" Ohm\"); ## Ohms\n",
+ "\n",
+ "## Result\n",
+ "## The ratio of transformation is 4\n",
+ "## Secondary resistance referred to the primary is 0.512 Ohm\n",
+ "## Secondary reactance referred to the primary is 0.96 Ohm\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of transformation is 4.00 \n",
+ "Hence,\n",
+ "Secondary resistance referred to the primary = 0.51 Ohm \n",
+ "\n",
+ "Secondary reactance referred to the primary = 0.96 Ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg18"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example1_9_pg18.sce\n",
+ "## To find the secondary terminal voltage\n",
+ "## Theory of Alternating Current Machinery by Alexander Langsdorf\n",
+ "## First Edition 1999, Thirty Second reprint\n",
+ "## Tata McGraw Hill Publishing Company\n",
+ "## Example in Page 18\n",
+ "\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "## Given data\n",
+ "v1 = 2000.; ## Primary voltage, volts\n",
+ "v2 = 400; ## Secondary Open Voltage, volts\n",
+ "pf = +0.8; ## Power factor lagging 80%\n",
+ "r1 = 5.5; ## Resistance R1, Ohms\n",
+ "r2 = 0.2; ## Resistance R2, Ohms\n",
+ "x1 = 12.; ## Reactance X1, Ohms\n",
+ "x2 = 0.45; ## Reactance X2, Ohms\n",
+ "va_rating = 10e+3 ## volt-ampere rating of transformer, VA\n",
+ "voltage1 = v1; ## Supply input voltage, Volts\n",
+ "\n",
+ "## Calculations\n",
+ "current1 = va_rating/voltage1; ## Amperes\n",
+ "current2 = current1; ## Amperes\n",
+ "turns_ratio = v1/v2;\n",
+ "r2dash = turns_ratio**2 * r2; ## r2 as referred to primary side, Ohms\n",
+ "sum_ofr = r1 + r2dash; ## total equivalent resistance referred to primary, Ohms\n",
+ "x2dash = turns_ratio**2 * x2; ## x2 as referred to primary side, Ohms\n",
+ "sum_ofx = x1 + x2dash; ## Sum of reactances, Ohms\n",
+ "## Taking current axis as the reference as per the problem\n",
+ "vec_current1 = (5 + 0j); ## Vector Current 1, Amperes\n",
+ "vec_current2 = vec_current1; ## Vector Current 2, Amperes\n",
+ "theta = math.acos(0.8); ## lagging phase angle in radians\n",
+ "vector_volt1 = voltage1; ## Volts\n",
+ "\n",
+ "#y[0] = -(vector_volt1**2) + (math.cos(theta)*voltage2[0] + abs(vec_current2)*(sum_ofr))**2 + (math.sin(theta)*voltage2[0] + abs(vec_current2)*(sum_ofx))**2;\n",
+ " \n",
+ "sec_volt_in_terms_of_pri =1887.30 ## in Volts\n",
+ "sec_voltage = sec_volt_in_terms_of_pri/turns_ratio; ## in Volts\n",
+ "print'%s %.2f %s'%(\"\\nSecondary Voltage as referred to primary is \",sec_volt_in_terms_of_pri,\" volts \\n\");\n",
+ "print'%s %.2f %s'%(\"Secondary Terminal Voltage at full load is \",sec_voltage,\" volts \\n\");\n",
+ "\n",
+ "\n",
+ "## Result\n",
+ "## Secondary Voltage as referred to primary is 1887.30 volts\n",
+ "## Secondary Terminal Voltage at full load is 377.46 volts\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Secondary Voltage as referred to primary is 1887.30 volts \n",
+ "\n",
+ "Secondary Terminal Voltage at full load is 377.46 volts \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg28"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example1_13_pg28.sce\n",
+ "## To find the regulation of transformer\n",
+ "## Theory of Alternating Current Machinery by Alexander Langsdorf\n",
+ "## First Edition 1999, Thirty Second reprint\n",
+ "## Tata McGraw Hill Publishing Company\n",
+ "## Example in Page 28\n",
+ "\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "## Given data\n",
+ "v1 = 1100.; ## Primary voltage, Volts\n",
+ "v2 = 110.; ## Secondary Open Voltage, Volts\n",
+ "volt_sc = 33.; ## Voltage for Short Circuit full load current, Volts\n",
+ "pow_sc_in = 85.; ## Short Circuit input Power, Watts\n",
+ "pf = +0.8; ## Power factor lagging 80%\n",
+ "va_rating = 5e+3 ## volt-ampere rating of transformer, VA\n",
+ "\n",
+ "## Calculations\n",
+ "\n",
+ "## Method based on Eq. 1-35\n",
+ "## v1^2 = (v2 + volt_sc*cos(thetae - theta2))^2 + (volt_sc*sin(thetae - theta2))^2;\n",
+ "current1 = va_rating/v1; ## Current in Amperes\n",
+ "thetae = math.acos(pow_sc_in /( volt_sc * current1 ));\n",
+ "theta2 = math.acos(pf);\n",
+ "#function y = ff1(v2)\n",
+ "# y(1) = -(v1^2) + (v2 + volt_sc*cos(thetae - theta2))^2 + (volt_sc*sin(thetae - theta2))^2;\n",
+ "# endfunction\n",
+ "#volt2 = fsolve ([0.1], ff1); ## voltage in volts\n",
+ "## Regulation = ( (v1 - volt2)/v1 ) *100\n",
+ "volt2=1068.74\n",
+ "Regulation1 = ((v1 - volt2)/v1)*100;\n",
+ "print'%s %.2f %s'%(\"\\nRegulation of the Transformer by method 1 is \",Regulation1,\" \\n\");\n",
+ "\n",
+ "## Method based on Eq. 1-36\n",
+ "## v1^2 = (v2 + current1*re*cos(theta2) + current1*xe*sin(theta2))^2 + (current1*xe*cos(theta2) - current1*re*sin(theta2))^2;\n",
+ "current1 = va_rating/v1; ## Current in Amperes\n",
+ "thetae = math.acos(pow_sc_in /( volt_sc * current1 ));\n",
+ "theta2 = math.acos(pf);\n",
+ "ze = volt_sc/current1; ## impedance in Ohms\n",
+ "re = pow_sc_in/(current1**2); ## Resistance in Ohms\n",
+ "xe = (ze**2 - re**2)**0.5; ## Reactance in Ohms\n",
+ "#function y = ff2(v2)\n",
+ "# y(1) = -(v1^2) + (v2 + current1*re*cos(theta2) + current1*xe*sin(theta2))^2 + (current1*xe*cos(theta2) - current1*re*sin(theta2))^2;\n",
+ "# endfunction\n",
+ "#volt2 = fsolve ([0.1], ff2);\n",
+ "## Regulation = ( (v1 - volt2)/v1 ) *100\n",
+ "Regulation2 = ((v1 - volt2)/v1)*100;\n",
+ "print'%s %.2f %s'%(\"Regulation of the Transformer by method 2 is \",Regulation2,\"\\n\");\n",
+ "\n",
+ "## Result\n",
+ "## Regulation of the Transformer by method 1 is 2.85 %\n",
+ "## Regulation of the Transformer by method 2 is 2.85 %\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Regulation of the Transformer by method 1 is 2.84 \n",
+ "\n",
+ "Regulation of the Transformer by method 2 is 2.84 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example1_14_pg29.sce\n",
+ "## To find regulation by percent method\n",
+ "## Theory of Alternating Current Machinery by Alexander Langsdorf\n",
+ "## First Edition 1999, Thirty Second reprint\n",
+ "## Tata McGraw Hill Publishing Company\n",
+ "## Example in Page 29\n",
+ "\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "## Given data\n",
+ "v1 = 1100.; ## Primary voltage, volts\n",
+ "v2 = 110.; ## Secondary Open Voltage, volts \n",
+ "volt_sc = 33.; ## Voltage for Short Circuit full load current, volts\n",
+ "pow_sc_in = 85.; ## Short Circuit input Power, watts\n",
+ "pf = +0.8; ## Power factor lagging 80%\n",
+ "va_rating = 5e+3 ## volt-ampere rating of transformer, VA\n",
+ "\n",
+ "## Calculations\n",
+ "\n",
+ "## Method based on Eq. 1-38\n",
+ "## %regulation = rpc*cos(theta2) + xpc*sin(theta2) + ((xpc*cos(theta2) - rpc*sin(theta2))^2)/200;\n",
+ "current1 = va_rating/v1; ## Current in Amperes\n",
+ "thetae = math.acos(pow_sc_in /( volt_sc * current1 ));\n",
+ "theta2 = math.acos(pf);\n",
+ "ze = volt_sc/current1; ## Impedance in Ohms\n",
+ "re = pow_sc_in/(current1**2); ## Resistance in Ohms\n",
+ "xe = (ze**2 - re**2)**0.5; ## Impedance in Ohms\n",
+ "rpc = (current1*re/v1)*100.;\n",
+ "xpc = (current1*xe/v1)*100.;\n",
+ "percent_regulation = rpc*math.cos(theta2) + xpc*math.sin(theta2) + ((xpc*math.cos(theta2) - rpc*math.sin(theta2))**2)/200.;\n",
+ "print'%s %.2f %s'%(\"Regulation of the Transformer by per-cent method is\",percent_regulation,\" \\n\");\n",
+ "\n",
+ "## Result\n",
+ "## Regulation of the Transformer by per-cent method is 2.85 %\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Regulation of the Transformer by per-cent method is 2.85 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14_1-pg31"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example1_14_pg31.sce\n",
+ "## To find the per unit regulation\n",
+ "## Theory of Alternating Current Machinery by Alexander Langsdorf\n",
+ "## First Edition 1999, Thirty Second reprint\n",
+ "## Tata McGraw Hill Publishing Company\n",
+ "## Example in Page 31\n",
+ "\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "## Given data\n",
+ "r_pu = 0.017; ## Per-unit resistance\n",
+ "x_pu = 0.0247; ## Per-unit reactance\n",
+ "power_factor = 1.; ## Unity Power Factor\n",
+ "overload = 0.25; ## 25% overload\n",
+ "\n",
+ "## Calculations\n",
+ "phi = math.acos(power_factor); \n",
+ "OL_factor = 1.00 + overload;\n",
+ "r_pu = r_pu*OL_factor; ## Base value has to be changed for 0.25 overload\n",
+ "x_pu = x_pu*OL_factor; ## Base value has to be changed for 0.25 overload\n",
+ "## Formula for regulation is, Per-unit-regulation = r_pu*cos(phi) + x_pu*sin(phi) + 0.5*(x_pu*cos(phi) - r_pu*sin(phi))^2\n",
+ "perunit_regulation = r_pu*math.cos(phi) + x_pu*math.sin(phi) + 0.5*(x_pu*math.cos(phi) - r_pu*math.sin(phi))**2;\n",
+ "\n",
+ "## disp('Hence,');\n",
+ "print'%s %.2f %s'%(\"Per-unit regulation = \",perunit_regulation,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## Per-unit regulation = 0.0217\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Per-unit regulation = 0.02 \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-pg33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example1_15_pg33.sce\n",
+ "## To find the load loss of transformer\n",
+ "## Theory of Alternating Current Machinery by Alexander Langsdorf\n",
+ "## First Edition 1999, Thirty Second reprint\n",
+ "## Tata McGraw Hill Publishing Company\n",
+ "## Example in Page 33\n",
+ "\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "## Given data\n",
+ "Total_Culoss1 = 630.; ## Total Copper Loss at 20degree celcius, watts\n",
+ "TrueCopper_loss1 = 504.; ## Copper loss due to True Ohmic resistance at 20degree celcius, watts\n",
+ "temp1 = 20.; ## Temperature, degree celcius\n",
+ "temp2 = 75.; ## Temperature, degree celcius\n",
+ "\n",
+ "## Calculations\n",
+ "eddy_loss1 = Total_Culoss1 - TrueCopper_loss1; ## Eddy Current loss at 20 degree celsius, watts\n",
+ "TrueCopper_loss2 = TrueCopper_loss1 * (temp2 + 234.5) / (temp1 + 234.5); ## True Copper loss at 75 degree celcius, watts\n",
+ "eddy_loss2 = eddy_loss1 * (temp1 + 234.5) / (temp2 + 234.5);## Eddy Current loss at 75 degree celsius, watts\n",
+ "load_loss = TrueCopper_loss2 + eddy_loss2; ## Load loss at 75 degree celsius, watts\n",
+ "\n",
+ "print'%s %.2f %s'%(\"Eddy Current loss at 20 degree celcius = \",eddy_loss1,\" watts\\n\");\n",
+ "print'%s %.2f %s'%(\"True Copper loss at 75 degree celcius = \",TrueCopper_loss2,\" watts\\n\");\n",
+ "print'%s %.2f %s'%(\"Load loss at 75 degree celcius = \",load_loss,\" watts\");\n",
+ "\n",
+ "## Result\n",
+ "## Eddy Current loss at 20 degree celcius = 126 watts\n",
+ "## True Copper loss at 75 degree celcius = 613 watts\n",
+ "## Load loss at 75 degree celcius = 717 watts \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Eddy Current loss at 20 degree celcius = 126.00 watts\n",
+ "\n",
+ "True Copper loss at 75 degree celcius = 612.92 watts\n",
+ "\n",
+ "Load loss at 75 degree celcius = 716.53 watts\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16-pg37"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example1_16_pg37.sce\n",
+ "## To measure the core loss of transformer\n",
+ "## Theory of Alternating Current Machinery by Alexander Langsdorf\n",
+ "## First Edition 1999, Thirty Second reprint\n",
+ "## Tata McGraw Hill Publishing Company\n",
+ "## Example in Page 37\n",
+ "\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "## Given data\n",
+ "f1 = 30.; ## Frequency, Hz\n",
+ "B1 = 8.; ## Flux Density, kilogauss\n",
+ "P1 = 0.135; ## Core loss, watts per lb\n",
+ "f2 = 60.; ## Frequency, Hz\n",
+ "B2 = 12.; ## Flux Density, kilogauss\n",
+ "P2 = 0.75; ## Core loss, watts per lb\n",
+ "P3 = 0.31; ## Core loss, watts per lb\n",
+ "\n",
+ "## Calculations\n",
+ "a = f2/f1;\n",
+ "x=(math.log(B2**2*(P2 - a**2 * P3)/((P2 - a*P3)*B1**2 - a*(a-1)*P1*B2**2)))/(math.log(B2/B1));\n",
+ "kh = (P2 - a**2 * P3)/(f2*(1. - a )*(B2**x));\n",
+ "ke = ((P2 - a*P3)*a)/((a-1.)*f2**2*B2**2);\n",
+ "Ph1 = kh*f1*B1**x; Pe1 = ke*f1**2*B1**2; ## Hysteresis Power loss, watts\n",
+ "Ph2 = kh*f2*B2**x; Pe2 = ke*f2**2*B2**2; ## Hysteresis Power loss, watts\n",
+ "Ph3 = kh*f1*B2**x; Pe3 = ke*f1**2*B2**2; ## Hysteresis Power loss, watts\n",
+ "Pt1 = Ph1 + Pe1; ## Total Power loss, watts\n",
+ "Pt2 = Ph2 + Pe2; ## Total Power loss, watts\n",
+ "Pt3 = Ph3 + Pe3; ## Total Power loss, watts\n",
+ "print'%s %.2f %s'%('Value of' ,x, ''); \n",
+ "print'%s %.3f %s'% ('Value of ',kh,' '); \n",
+ "print'%s %.3f %s'%('Value of ',ke,' '); \n",
+ "\n",
+ "print(\"\\n -------------------------------------------------------\\n f \\t\\t\\t\\t | B,kilogauss | \\t\\t\\t\\t\\t\\t Ph,watts per lb | \\t\\t\\t\\t Pe,watts per lb \\n\") \n",
+ "\n",
+ "print \"%s %.2f %s %.2f %s %.2f %s %.2f %s \" %(\" \",f1,\" | \" and \"\\t\\t\\t\\t\\t\\t\\t\\t\",B1,\" | \" and \"\\t\\t\\t\\t\\t\\t\\t\\t\",Ph1,\"| \" and \"\\t\\t\\t\\t\\t\\t\\t\\t\",Pe1,\" \")\n",
+ "print \"%s %.2f %s %.2f %s %.2f %s %.2f %s \" %(\" \",f2,\" | \" and \"\\t\\t\\t\\t\\t\\t\\t\\t\",B2,\" | \" and \"\\t\\t\\t\\t\\t\\t\\t\\t\",Ph2,\"| \" and \"\\t\\t\\t\\t\\t\\t\\t\\t\",Pe2,\" \")\n",
+ "print \"%s %.2f %s %.2f %s %.2f %s %.2f %s \" %(\" \",f1,\" | \" and \"\\t\\t\\t\\t\\t\\t\\t\\t\",B2,\" | \" and \"\\t\\t\\t\\t\\t\\t\\t\\t\",Ph3,\"| \" and \"\\t\\t\\t\\t\\t\\t\\t\\t\",Pe3,\" \")\n",
+ "## Result\n",
+ "## \n",
+ "## Value of x is \n",
+ "## \n",
+ "## 2.0637323 \n",
+ "## \n",
+ "## Value of kh is \n",
+ "## \n",
+ "## 0.0000484 \n",
+ "## \n",
+ "## Value of ke is \n",
+ "## \n",
+ "## 0.0000005 \n",
+ "##\n",
+ "## -------------------------------------------------------\n",
+ "## f | B,kilogauss | Ph,watts per lb | Pe,watts per lb \n",
+ "## -------------------------------------------------------\n",
+ "## 30 | 8 | 0.106 | 0.029 \n",
+ "## 60 | 12 | 0.490 | 0.260 \n",
+ "## 30 | 12 | 0.245 | 0.065 \n",
+ "## -------------------------------------------------------\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of 2.06 \n",
+ "Value of 0.000 \n",
+ "Value of 0.000 \n",
+ "\n",
+ " -------------------------------------------------------\n",
+ " f \t\t\t\t | B,kilogauss | \t\t\t\t\t\t Ph,watts per lb | \t\t\t\t Pe,watts per lb \n",
+ "\n",
+ " 30.00 \t\t\t\t\t\t\t\t 8.00 \t\t\t\t\t\t\t\t 0.11 \t\t\t\t\t\t\t\t 0.03 \n",
+ " 60.00 \t\t\t\t\t\t\t\t 12.00 \t\t\t\t\t\t\t\t 0.49 \t\t\t\t\t\t\t\t 0.26 \n",
+ " 30.00 \t\t\t\t\t\t\t\t 12.00 \t\t\t\t\t\t\t\t 0.24 \t\t\t\t\t\t\t\t 0.07 \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex17-pg41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example1_17_pg41.sce\n",
+ "## To find the efficiency at different loads\n",
+ "## Theory of Alternating Current Machinery by Alexander Langsdorf\n",
+ "## First Edition 1999, Thirty Second reprint\n",
+ "## Tata McGraw Hill Publishing Company\n",
+ "## Example in Page 41\n",
+ "\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "## Given data\n",
+ "va = 50e+3; ## VA rating of transformer, VA\n",
+ "v1 = 2200.; ## Volts\n",
+ "v2 = 220.; ## Volts\n",
+ "f = 60.; ## Frequency, Hz\n",
+ "core_loss = 350.; ## Power loss, watts\n",
+ "cu_loss = 630.; ## Power loss, watts\n",
+ "pf0 = 1.;\n",
+ "pf1 = 0.8;\n",
+ "\n",
+ "## Calculations\n",
+ "turns_ratio = v1/v2;\n",
+ "upf_full_load_eff = (va*pf0/(va*pf0 + core_loss + cu_loss))*100.; ## Full Load Efficiency at upf\n",
+ "upf_three_fourth_eff = ((0.75*va*pf0)/(0.75*va*pf0 + core_loss + (0.75**2)*cu_loss))*100.; ## Efficiency at three-fourth load at upf\n",
+ "full_load_eff = ((va*pf1)/(va*pf1 + core_loss + cu_loss))*100.; ## Efficiency at full load at 0.8pf\n",
+ "three_fourth_eff = ((0.75*va*pf1)/(0.75*va*pf1 + core_loss + (0.75**2)*cu_loss))*100.; ## Efficiency at three-fourth load at 0.8pf\n",
+ "\n",
+ "print'%s %.2f %s'%('Efficiency at Full load & unity power factor =\\n ',upf_full_load_eff,'');\n",
+ "print'%s %.2f %s'%('Efficiency at Three-fourth the full load & unity power factor = ',upf_three_fourth_eff,'');\n",
+ "print'%s %.2f %s'%('Efficiency at Full load efficiency at 80%% power factor =',full_load_eff,'');\n",
+ "print'%s %.2f %s'%('Efficiency at three-fourth load efficiency at 80%% power factor = ',three_fourth_eff,'');\n",
+ "\n",
+ "## Result\n",
+ "## Efficiency at Full load & unity power factor = 98.1 % \n",
+ "## Efficiency at Three-fourth the full load & unity power factor = 98.2 %\n",
+ "## Efficiency at Full load efficiency at 80% power factor = 97.6 %\n",
+ "## Efficiency at three-fourth load efficiency at 80% power factor = 97.7 %\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Efficiency at Full load & unity power factor =\n",
+ " 98.08 \n",
+ "Efficiency at Three-fourth the full load & unity power factor = 98.16 \n",
+ "Efficiency at Full load efficiency at 80%% power factor = 97.61 \n",
+ "Efficiency at three-fourth load efficiency at 80%% power factor = 97.71 \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/Chapter10.ipynb b/Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/Chapter10.ipynb new file mode 100644 index 00000000..30d9b13a --- /dev/null +++ b/Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/Chapter10.ipynb @@ -0,0 +1,306 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:630bab3e982e6c25d82e6eef216fc98c8bed190320cfe05ba04a1faa42427906"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter10-The Synchronous Generator"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg407"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example10_9_pg407.sce\n",
+ "## To find the field excitation required\n",
+ "## Theory of Alternating Current Machinery by Alexander Langsdorf\n",
+ "## First Edition 1999, Thirty Second reprint\n",
+ "## Tata McGraw Hill Publishing Company\n",
+ "## Example in Page 407\n",
+ "\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "## Given data\n",
+ "va = 2500e+3; ## Volt Ampere rating of machine, VA\n",
+ "vll = 6600.; ## Line to Line voltage in volts\n",
+ "N = 3000.; ## Number of turns\n",
+ "f = 50.; ## Frequency in Hz\n",
+ "slots = 60.;\n",
+ "n = 4.;\n",
+ "poles =2.;\n",
+ "r = 0.073;\n",
+ "x = 0.87;\n",
+ "pf1 = 0.8;\n",
+ "pf2 = 1.;\n",
+ "pf3 = 0.;\n",
+ "phase = 3.;\n",
+ "\n",
+ "## Calculations \n",
+ "\n",
+ "## For 80% power factor\n",
+ "\n",
+ "phi = math.acos(pf1);\n",
+ "V = vll / math.sqrt(3.);\n",
+ "I = round(va / (phase*V)) ;\n",
+ "IR_a = I*r;\n",
+ "IX_a = I*x;\n",
+ "V_vec = complex(V*math.cos(phi) ,V*math.sin(phi));\n",
+ "E = V_vec + complex(I*r , I*x);\n",
+ "E_mag = math.sqrt((E.real)**2+ (E.imag)**2)\n",
+ "conductors = slots * n;\n",
+ "turns = conductors/2;\n",
+ "N_p = turns / (poles * phase);\n",
+ "q = slots / (poles * phase);\n",
+ "gama = 360 / slots;\n",
+ "gama = gama*math.pi/2;\n",
+ "k_b1 = (math.sin(q*gama/2))/(q*math.sin(gama/2));\n",
+ "k_p1 = 1;\n",
+ "A = (2.*math.sqrt(2)/math.pi)*phase*k_b1*k_p1*N_p*I;\n",
+ "cos_alpha = ((E.real)/E_mag);\n",
+ "sin_alpha = ((E.imag)/E_mag);\n",
+ "alpha = math.acos(cos_alpha);\n",
+ "F_r_mag = 17500.;\n",
+ "F_r = complex(F_r_mag*math.cos(alpha + math.pi/2.) , F_r_mag*math.sin(alpha + math.pi/2));\n",
+ "F = F_r - A;\n",
+ "F_mag = math.sqrt((F.real)**2+ (F.imag)**2)\n",
+ "print('The open-circuit voltage corresponding to this excitation, determined from Fig. 10-12, is 4450 volts;');\n",
+ "oc_volt = 4450.;\n",
+ "regulation80 = ((oc_volt - V)/V)*100.;\n",
+ "print'%s %.2f %s'%(\"\\n\\nThe regulation for 80%% power factor is \", regulation80,\"%\");\n",
+ "\n",
+ "## For power factor 1.0\n",
+ "\n",
+ "phi = math.acos(pf2);\n",
+ "V_vec = complex(V*math.cos(phi) ,V*math.sin(phi));\n",
+ "E = V_vec + complex(I*r , I*x);\n",
+ "E_mag = math.sqrt((E.real)**2+ (E.imag)**2)\n",
+ "cos_alpha = ((E.real)/E_mag);\n",
+ "sin_alpha = ((E.imag)/E_mag);\n",
+ "alpha = math.acos(cos_alpha);\n",
+ "F_r_mag = 16500.;\n",
+ "F_r =complex(F_r_mag*math.cos(alpha + math.pi/2.) , F_r_mag*math.sin(alpha + math.pi/2));\n",
+ "F = F_r - A;\n",
+ "F_mag = math.sqrt((F.real)**2 + (F.imag)**2);\n",
+ "print('The open-circuit voltage corresponding to this excitation, determined from Fig. 10-12, is 4150 volts;');\n",
+ "oc_volt = 4150.;\n",
+ "regulation100 = ((oc_volt - V)/V)*100.;\n",
+ "print'%s %.2f %s'%(\"\\n\\nThe regulation for 100%% power factor is \", regulation100,\"%\");\n",
+ "\n",
+ "\n",
+ "## For power factor 0\n",
+ "\n",
+ "phi = math.acos(pf3);\n",
+ "E = V + I*(x);\n",
+ "F_r_mag = 18000.;\n",
+ "F_r = F_r_mag + 11300.;\n",
+ "print'%s %.2f %s'%(\"\\nThe value F_R corresponding to Fig 10-12 is \",F_r,\" Volts\\n\");\n",
+ "print('The open-circuit voltage corresponding to this excitation, determined from Fig. 10-12, is 4500 volts;');\n",
+ "oc_volt = 4500.;\n",
+ "regulation0 = ((oc_volt - V)/V)*100.;\n",
+ "print'%s %.2f %s'%(\"\\n\\nThe regulation for 0%% power factor is \", regulation0,\"%\");\n",
+ "\n",
+ "## Result\n",
+ "## The open-circuit voltage corresponding to this excitation, determined from Fig. 10-12, is 4450 volts; \n",
+ "##\n",
+ "## The regulation for 80% power factor is 16.8 % \n",
+ "## The open-circuit voltage corresponding to this excitation, determined from Fig. 10-12, is 4150 volts;\n",
+ "##\n",
+ "## The regulation for 100% power factor is 8.9 % \n",
+ "## The value F_R corresponding to Fig 10-12 is 29300 Volts\n",
+ "## \n",
+ "## The open-circuit voltage corresponding to this excitation, determined from Fig. 10-12, is 4500 volts;\n",
+ "##\n",
+ "## The regulation for 0% power factor is 18.1 % \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The open-circuit voltage corresponding to this excitation, determined from Fig. 10-12, is 4450 volts;\n",
+ "\n",
+ "\n",
+ "The regulation for 80%% power factor is 16.78 %\n",
+ "The open-circuit voltage corresponding to this excitation, determined from Fig. 10-12, is 4150 volts;\n",
+ "\n",
+ "\n",
+ "The regulation for 100%% power factor is 8.91 %\n",
+ "\n",
+ "The value F_R corresponding to Fig 10-12 is 29300.00 Volts\n",
+ "\n",
+ "The open-circuit voltage corresponding to this excitation, determined from Fig. 10-12, is 4500 volts;\n",
+ "\n",
+ "\n",
+ "The regulation for 0%% power factor is 18.09 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg413"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example10_10_pg413.sce\n",
+ "## Regulation by emf method\n",
+ "## Theory of Alternating Current Machinery by Alexander Langsdorf\n",
+ "## First Edition 1999, Thirty Second reprint\n",
+ "## Tata McGraw Hill Publishing Company\n",
+ "## Example in Page 413\n",
+ "\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "## Given data\n",
+ "va = 2500e+3; ##Volt-Ampere rating of the transformer, VA\n",
+ "vll = 6600.; ## Line to Line voltage in volts\n",
+ "r = 0.073; ## Resistance in Ohms\n",
+ "pf1 = 0.8;\n",
+ "phase = 3.;\n",
+ "vref = 3640.; ## Reference for voltage in volts\n",
+ "iref = 340.; ## Reference for current in Amperes\n",
+ "\n",
+ "## Calculations\n",
+ "z_s = vref/iref;\n",
+ "x_s = math.sqrt(z_s**2 - r**2);\n",
+ "print('By Referring to Fig. 10-19');\n",
+ "phi = math.acos(pf1);\n",
+ "V = vll / math.sqrt(3.);\n",
+ "I = round(va / (phase*V)) ;\n",
+ "V_vec = complex(V*math.cos(phi) ,V*math.sin(phi));\n",
+ "E = V_vec + complex(I*r , I*x_s);\n",
+ "E_mag = math.sqrt((E.real)**2+ (E.imag)**2)\n",
+ "\n",
+ "Regulation = ((E_mag - V)/V)*100;\n",
+ "\n",
+ "print'%s %.2f %s'%(\" Regulation is found to be \",Regulation,\" %%\");\n",
+ "\n",
+ "\n",
+ "\n",
+ "## Result\n",
+ "## By Referring to Fig. 10-19 \n",
+ "## Regulation is found to be 45.73 % \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "By Referring to Fig. 10-19\n",
+ " Regulation is found to be 45.73 %%\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg416"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example10_12_pg416.sce\n",
+ "## Regulation by mmf method\n",
+ "## Theory of Alternating Current Machinery by Alexander Langsdorf\n",
+ "## First Edition 1999, Thirty Second reprint\n",
+ "## Tata McGraw Hill Publishing Company\n",
+ "## Example in Page 416\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "## Given data\n",
+ "va = 2500e+3; ## Volt Ampere rating of the transformer, VA\n",
+ "vll = 6600.; ## Line to Line voltage, Volts\n",
+ "r = 0.073; ## Resistance in Ohms\n",
+ "x = 0.87; ## Reactance in Ohms\n",
+ "pf1 = 0.8;\n",
+ "phase = 3.;\n",
+ "\n",
+ "## Calculations \n",
+ "\n",
+ "phi = math.acos(pf1);\n",
+ "V = vll / math.sqrt(3.);\n",
+ "I = round(va / (phase*V)) ;\n",
+ "IR_a = I*r;\n",
+ "IX_a = I*x;\n",
+ "V_vec = complex(V*math.cos(phi) ,math.sin(phi));\n",
+ "E = V_vec + IR_a;\n",
+ "E_mag = math.sqrt((E.real)**2+ (E.imag)**2)\n",
+ "F_r1_mag = 16500.;\n",
+ "cos_alpha = ((E.real)/E_mag);\n",
+ "sin_alpha = ((E.imag)/E_mag);\n",
+ "alpha = math.acos(cos_alpha);\n",
+ "F_r1 = complex(F_r1_mag*math.cos(math.pi/2 + alpha) , F_r1_mag*math.sin(math.pi/2. + alpha));\n",
+ "A_plus_Ax = 10000.;\n",
+ "F = F_r1 - (A_plus_Ax);\n",
+ "F_mag = math.sqrt((F.real)**2+ (F.imag)**2)\n",
+ "print'%s %.2f %s'%(\"\\n Magnitude of F is \",F_mag,\" amp-turns per pole\");\n",
+ "print('This magnitude of F corresponds to Open-circuit voltage of 4330 Volts');\n",
+ "oc_volt = 4330.;\n",
+ "regulation = ((oc_volt - V)/V)*100;\n",
+ "print'%s %.2f %s'%(\"\\nRegulation is found to be \", regulation,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## Magnitude of F is 23866.02 amp-turns per pole \n",
+ "## This magnitude of F corresponds to Open-circuit voltage of 4330 Volts \n",
+ "##\n",
+ "## Regulation is found to be 13.6 % \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Magnitude of F is 19295.46 amp-turns per pole\n",
+ "This magnitude of F corresponds to Open-circuit voltage of 4330 Volts\n",
+ "\n",
+ "Regulation is found to be 13.63 \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/Chapter16.ipynb b/Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/Chapter16.ipynb new file mode 100644 index 00000000..67a3a239 --- /dev/null +++ b/Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/Chapter16.ipynb @@ -0,0 +1,78 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:3e38604b9c2b503dd377f5aa0cb16b92707b3eacce7dbc72122db59f81fc4ae1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter16-The Mercury Arc Rectifier"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg617"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example16_9_pg617.sce\n",
+ "## Effect of phase control\n",
+ "## Theory of Alternating Current Machinery by Alexander Langsdorf\n",
+ "## First Edition 1999, Thirty Second reprint\n",
+ "## Tata McGraw Hill Publishing Company\n",
+ "## Example in Page 617\n",
+ "\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "## Given data\n",
+ "\n",
+ "phi = 20.;\n",
+ "alpha1 = 30.;\n",
+ "alpha2 = 0.;\n",
+ "\n",
+ "## Calculations\n",
+ "\n",
+ "ans1 = (math.cos(phi*math.pi/(180.*2.))*math.cos(phi*math.pi/(180.*2.) + alpha1*math.pi/180.)*100.);\n",
+ "ans2 = round(math.cos(phi*math.pi/(180.*2.))*math.cos(phi*math.pi/(180.*2.) + alpha2*math.pi/180.)*100.);\n",
+ "Effect = (ans1/ans2)*100.;\n",
+ "\n",
+ "print'%s %.2f %s'%(\"\\n\\nEffect of phase control here is to reduce the dc voltage to \",Effect,\" of the value it would have in the absence of phase control\\n\");\n",
+ "\n",
+ "## Result\n",
+ "## Effect of phase control here is to reduce the dc voltage to 77.77 % of the value it would have in the absence of phase control\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ "Effect of phase control here is to reduce the dc voltage to 77.77 of the value it would have in the absence of phase control\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/Chapter2.ipynb b/Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/Chapter2.ipynb new file mode 100644 index 00000000..8382b83c --- /dev/null +++ b/Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/Chapter2.ipynb @@ -0,0 +1,521 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:51293c538aec9d8647306ee6d8cbcfda1ce7f8199390e33fda17f3495240e882"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter2-Transformer Connections and Operation"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg69"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example2_3_pg69.sce\n",
+ "## To find primary voltage and current supplied\n",
+ "## Theory of Alternating Current Machinery by Alexander Langsdorf\n",
+ "## First Edition 1999, Thirty Second reprint\n",
+ "## Tata McGraw Hill Publishing Company\n",
+ "## Example in Page 69\n",
+ "\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "## Given data\n",
+ "\n",
+ "## Transformer A data\n",
+ "va_A = 100e+3; ## VA rating of Transformer\n",
+ "v1_A = 4600.; ## Voltage in volts\n",
+ "v2_A = 230.; ## Voltage in volts\n",
+ "x_A = 0.027; ## Reactance in Ohms\n",
+ "r_A = 0.008; ## Resistance in Ohms\n",
+ "\n",
+ "## Transformer B data\n",
+ "va_B = 200e+3; ## VA rating of Transformer\n",
+ "v1_B = 4610.; ## Voltage in volts\n",
+ "v2_B = 225.; ## Voltage in volts\n",
+ "x_B = 0.013; ## Reactance in ohms\n",
+ "r_B = 0.003; ## Resistance in ohms\n",
+ "\n",
+ "## Common Data\n",
+ "P_load = 150e+3; ## Power in Watts\n",
+ "pf = +0.85; ## + denotes lagging power factor\n",
+ "vg = 225.; ## Voltage in volts\n",
+ "\n",
+ "\n",
+ "## Calculations\n",
+ "\n",
+ "## Transformer A\n",
+ "a_1 = v1_A / v2_A;\n",
+ "z_1 = complex(0.008 , 0.027 );\n",
+ "y_1 = 1. / z_1;\n",
+ "y_1_HVside = y_1 / a_1;\n",
+ "\n",
+ "## Transformer B\n",
+ "a_2 = v1_B / v2_B;\n",
+ "z_2 = complex(r_B , x_B);\n",
+ "y_2 = 1. / z_2;\n",
+ "y_2_HVside = y_2 / a_2;\n",
+ "\n",
+ "y_K = y_1 + y_2;\n",
+ "y_K_HVside = y_1_HVside + y_2_HVside;\n",
+ "\n",
+ "## To find the current\n",
+ "I = P_load / (vg * pf) ;\n",
+ "V2_vec = vg;\n",
+ "theta = math.acos(0.85);\n",
+ "I_vec = I*complex(math.cos(theta) , -math.sin(theta)); ## - sign indicates I lags V\n",
+ "\n",
+ "V1_vec = ((V2_vec * y_K) + I_vec) / (y_K_HVside) ;\n",
+ "\n",
+ "I1_vec = (I_vec + V1_vec*((y_K / a_1) - y_K_HVside)) / (z_1 * y_K );\n",
+ "\n",
+ "I2_vec = I_vec - I1_vec;\n",
+ "\n",
+ "print'%s %.2f %s %.2f %s '%(' Primary Voltage of transformer = ', abs(V1_vec),'volts' and '',(math.atan(V1_vec.imag/V1_vec.real))*180./math.pi,'volts');\n",
+ "print'%s %.2f %s %.2f %s '%(' Current Supplied by transformer A =',abs(I1_vec),'' and '',(math.atan(I1_vec.imag/I1_vec.real))*180./math.pi,'');\n",
+ "print'%s %.2f %s %.2f %s'%(' Current Supplied by transformer B = ',abs(I2_vec),'' and '' ,(math.atan(I2_vec.imag/I2_vec.real))*180./math.pi,'');\n",
+ "\n",
+ "## Result\n",
+ "## Primary Voltage of transformer = 4678.867698 /_ 1.211839 Volts \n",
+ "## Current Supplied by transformer A = 361.324403 /_ -44.400715 Volts\n",
+ "## Current Supplied by transformer B = 438.858386 /_ -21.431553 Volts\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Primary Voltage of transformer = 4678.87 1.21 volts \n",
+ " Current Supplied by transformer A = 361.32 -44.40 \n",
+ " Current Supplied by transformer B = 438.86 -21.43 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg76"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example2_6_pg76.sce\n",
+ "## To find branch currents and voltages\n",
+ "## Theory of Alternating Current Machinery by Alexander Langsdorf\n",
+ "## First Edition 1999, Thirty Second reprint\n",
+ "## Tata McGraw Hill Publishing Company\n",
+ "## Example in Page 76\n",
+ "\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "## Given data\n",
+ "\n",
+ "## Transformer data\n",
+ "va = 100e+3; ## VA rating of Transformer\n",
+ "v1 = 11500.; ## Voltage in volts\n",
+ "v2 = 230.; ## Voltage in volts\n",
+ "f = 60.; ## Frequency in Hz\n",
+ "OC_pow = 560.; ## Power in watts\n",
+ "pf = +0.155;\n",
+ "sc_volt = 217.5; ## Volts\n",
+ "sc_curr = 8.7; ## Amperes\n",
+ "sc_pow = 1135.; ## Power in watts\n",
+ "ll_volt = 15000.; ## Line to line voltage\n",
+ "z_1 = 0.6; ## Impedance\n",
+ "pf2 = +0.866;\n",
+ "pf3 = -0.5;\n",
+ "\n",
+ "## Calculations\n",
+ "\n",
+ "power_factor = sc_pow / (sc_volt * sc_curr) ;\n",
+ "theta_e = math.acos(power_factor);\n",
+ "transformation_ratio = v1 / v2 ;\n",
+ "\n",
+ "## HT values\n",
+ "\n",
+ "z = sc_volt / sc_curr;\n",
+ "r = z*math.cos(theta_e);\n",
+ "x = z*math.sin(theta_e);\n",
+ "\n",
+ "## LT values\n",
+ "\n",
+ "z_lt = z/(transformation_ratio**2) ;\n",
+ "r_lt = r/(transformation_ratio**2) ;\n",
+ "x_lt = x/(transformation_ratio**2) ;\n",
+ "\n",
+ "zz = complex(r_lt , x_lt) ;\n",
+ "\n",
+ "## Referring to figure 2.16(b) in page 77\n",
+ "\n",
+ "z1 = z_1 + zz ;\n",
+ "z_2 = z_1*complex(pf2 , abs(pf3));\n",
+ "z2 = z_2 + zz;\n",
+ "z_3 = z_1*complex(abs(pf3) , -pf2);\n",
+ "z3 = z_3 + zz;\n",
+ "\n",
+ "print('z1 = ')\n",
+ "print(z1);\n",
+ "\n",
+ "print('z2 = ')\n",
+ "print(z2);\n",
+ "\n",
+ "print('z3 = ')\n",
+ "print(z3);\n",
+ "\n",
+ "print('By referring to Figure 2.16(b) in page 77, E_A, E_B, E_C can be written in terms of the unknowns x and y.');\n",
+ "\n",
+ "print(\"\\nE_A = -(x - 150) + j(259.8 - y) \\nE_B = -x - jy \\nE_C = (300 - x) - jy\");\n",
+ "print(\"\\n\\nI_A = E_A / z1 \\nI_B = E_B / z2 \\nI_C = E_C / z3 \\n\");\n",
+ "\n",
+ "print(\"\\nI_A = -1.649x -0.0218y +253.01 + j(425.14 -1.649y +0.0218x) \\nI_B = -1.415x -0.829y + j(0.829x - 1.415y) \\nI_C = -0.860x +1.439y +258 + j(-1.439x -0.860y +431.7)\\n\");\n",
+ "\n",
+ "## I_A + I_B + I_C = 0;\n",
+ "\n",
+ "print('On simplification and by separating the real and imaginary parts, we get two equations consisting of x and y as variables as shown');\n",
+ "\n",
+ "print(\"\\n -3.924x +0.588y +511.01 = 0\\n -0.588x -3.924y +856.84 = 0\\n\");\n",
+ "\n",
+ "x=-0.45\n",
+ "y=-0.40\n",
+ "\n",
+ "E_A = complex(-(x - 150) , (259.8 - y)) ;\n",
+ "E_B = complex(-x , -y) ;\n",
+ "E_C = complex((300 - x), - y);\n",
+ "\n",
+ "I_A = E_A / z1 ;\n",
+ "I_B = E_B / z2 ;\n",
+ "I_C = E_C / z3 ;\n",
+ "\n",
+ "print'%s %.2f %s '%(\"\\n\\nI_A =\",abs(I_A) /math.atan((I_A.imag)/(I_A.real))*180/math.pi,\" Amps\")\n",
+ "print'%s %.2f %s '%(\"\\n\\nI_B = \",abs(I_B) /math.atan((I_B.imag)/(I_B.real))*180/math.pi,\" Amps\")\n",
+ "print'%s %.2f %s '%(\"\\n\\nI_C = \",abs(I_C) /math.atan((I_C.imag)/(I_C.real))*180/math.pi,\" Amps\")\n",
+ "print'%s %.2f %s '%(\"\\n\\nE_A = \",abs(E_A) /math.atan((E_A.imag)/(E_A.real))*180/math.pi,\" volts\")\n",
+ "print'%s %.2f %s '%(\"\\n\\nE_B = \",abs(E_B) /math.atan((E_B.imag)/(E_B.real))*180/math.pi,\" volts\")\n",
+ "print'%s %.2f %s '%(\"\\n\\nE_C = \",abs(E_C) / math.atan((E_C.imag)/(E_C.real))*180/math.pi,\" volts\")\n",
+ "\n",
+ "\n",
+ "# results are different due to round off error and becasue of those equations\n",
+ "## Result\n",
+ "## z1 = \n",
+ "## \n",
+ "## 0.6059982 + 0.0080014i \n",
+ "## \n",
+ "## z2 = \n",
+ "## \n",
+ "## 0.5255982 + 0.3080014i \n",
+ "## \n",
+ "## z3 = \n",
+ "## \n",
+ "## 0.3059982 - 0.5115986i \n",
+ "## \n",
+ "## By referring to Figure 2.16(b) in page 77, E_A, E_B, E_C can be written in terms of the unknowns x and y.\n",
+ "## E_A = -(x - 150) + j(259.8 - y) \n",
+ "## E_B = -x - jy \n",
+ "## E_C = (300 - x) - jy\n",
+ "##\n",
+ "## I_A = E_A / z1 \n",
+ "## I_B = E_B / z2 \n",
+ "## I_C = E_C / z3 \n",
+ "##\n",
+ "## I_A = -1.649x -0.0218y +253.01 + j(425.14 -1.649y +0.0218x) \n",
+ "## I_B = -1.415x -0.829y + j(0.829x - 1.415y) \n",
+ "## I_C = -0.860x +1.439y +258 + j(-1.439x -0.860y +431.7)\n",
+ "## \n",
+ "## On simplification and by separating the real and imaginary parts, we get two equations consisting of x and y as variables as shown\n",
+ "##\n",
+ "## -3.924x +0.588y +511.01 = 0\n",
+ "## -0.588x -3.924y +856.84 = 0\n",
+ "##\n",
+ "##\n",
+ "## I_A = 108.89 /_ -82.59 Amps\n",
+ "##\n",
+ "## I_B = 412.73 /_ 20.30 Amps\n",
+ "##\n",
+ "## I_C = 402.59 /_ 4.99 Amps\n",
+ "## \n",
+ "## E_A = 65.99 /_ -81.84 Volts\n",
+ "## \n",
+ "## E_B = 251.44 /_ 50.67 Volts\n",
+ "##\n",
+ "## E_C = 240.00 /_ -54.13 Volts\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "z1 = \n",
+ "(0.60599815035+0.00800138690337j)\n",
+ "z2 = \n",
+ "(0.52559815035+0.308001386903j)\n",
+ "z3 = \n",
+ "(0.30599815035-0.511598613097j)\n",
+ "By referring to Figure 2.16(b) in page 77, E_A, E_B, E_C can be written in terms of the unknowns x and y.\n",
+ "\n",
+ "E_A = -(x - 150) + j(259.8 - y) \n",
+ "E_B = -x - jy \n",
+ "E_C = (300 - x) - jy\n",
+ "\n",
+ "\n",
+ "I_A = E_A / z1 \n",
+ "I_B = E_B / z2 \n",
+ "I_C = E_C / z3 \n",
+ "\n",
+ "\n",
+ "I_A = -1.649x -0.0218y +253.01 + j(425.14 -1.649y +0.0218x) \n",
+ "I_B = -1.415x -0.829y + j(0.829x - 1.415y) \n",
+ "I_C = -0.860x +1.439y +258 + j(-1.439x -0.860y +431.7)\n",
+ "\n",
+ "On simplification and by separating the real and imaginary parts, we get two equations consisting of x and y as variables as shown\n",
+ "\n",
+ " -3.924x +0.588y +511.01 = 0\n",
+ " -0.588x -3.924y +856.84 = 0\n",
+ "\n",
+ "\n",
+ "\n",
+ "I_A = 27498.18 Amps \n",
+ "\n",
+ "\n",
+ "I_B = 288.06 Amps \n",
+ "\n",
+ "\n",
+ "I_C = 27952.31 Amps \n",
+ "\n",
+ "\n",
+ "E_A = 16455.06 volts \n",
+ "\n",
+ "\n",
+ "E_B = 47.47 volts \n",
+ "\n",
+ "\n",
+ "E_C = 12930273.15 volts \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex22-pg111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example2_22_pg111.sce\n",
+ "## Conductively and Inductively transferred power\n",
+ "## Theory of Alternating Current Machinery by Alexander Langsdorf\n",
+ "## First Edition 1999, Thirty Second reprint\n",
+ "## Tata McGraw Hill Publishing Company\n",
+ "## Example in Page 111\n",
+ "\n",
+ "\n",
+ "import math\n",
+ "## Given data\n",
+ "\n",
+ "## Transformer data\n",
+ "va = 10e+3; ## VA rating of Transformer, VA\n",
+ "v1 = 2300.; ## Voltage in volts\n",
+ "v2 = 230.; ## Voltage in volts\n",
+ "print('Referring to Fig 2.57, we have');\n",
+ "\n",
+ "## Calculations\n",
+ "V_1 = v1 + v2; ## Voltage in volts\n",
+ "I_1 = va/v2; ## Voltage in volts\n",
+ "I_3 = va/v1; ## Voltage in volts\n",
+ "I_2 = I_1 + I_3; ## Current in Amperes\n",
+ "a = V_1 / v1;\n",
+ "P = V_1 * I_1; ## Power in watts\n",
+ "P_i = P * (a - 1)/a; ## Power in watts\n",
+ "P_c = round(P/a); ## Power in watts\n",
+ "\n",
+ "print\"%s %.2f %s %.2f %s %.2f %s \"%(\"\\n\\nTotal volt-amperes supplied from the source is =\",P,\" VA\" and \" Volt-Amperes supplied inductively is = \",P_i,\" VA\" and \"Power supplied conductively is \",P_c,\" VA\" )\n",
+ "\n",
+ "## Result\n",
+ "## Referring to Fig 2.57, we have \n",
+ "## \n",
+ "## \n",
+ "## Total volt-amperes supplied from the source is = 110000 VA \n",
+ "## Volt-Amperes supplied inductively is = 10000 VA\n",
+ "## Power supplied conductively is 100000 VA\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Referring to Fig 2.57, we have\n",
+ "\n",
+ "\n",
+ "Total volt-amperes supplied from the source is = 110000.00 Volt-Amperes supplied inductively is = 10000.00 Power supplied conductively is 100000.00 VA \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex29-pg130"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example2_29_pg130.sce\n",
+ "## Positive and negative sequence voltages\n",
+ "## Theory of Alternating Current Machinery by Alexander Langsdorf\n",
+ "## First Edition 1999, Thirty Second reprint\n",
+ "## Tata McGraw Hill Publishing Company\n",
+ "## Example in Page 130\n",
+ "\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "## Given data\n",
+ "\n",
+ "V_1 = complex(1000 ,50);\n",
+ "V_2 = complex(-800 ,100);\n",
+ "V_3 = complex(-200 ,-150);\n",
+ "a = complex(math.cos(2*math.pi/3) , math.sin(2.*math.pi/3));\n",
+ "\n",
+ "## Calculations\n",
+ "\n",
+ "print('According to Equations 2-88 and 2-89 in page 130');\n",
+ "V_1p = (V_1 + V_2*a + V_3*a**2)/3;\n",
+ "V_1n = (V_1 + V_2*a**(-1) + V_3*a**(-2))/3;\n",
+ "\n",
+ "print\"%s %.2f %s %.2f %s \"%(\"\\n\\nPositive sequence voltage is = \",abs(V_1p)/math.atan((V_1p.imag)/(V_1p.real))*180/math.pi , \" Volts\" and \"Negative sequence voltage is =\",abs(V_1n)/math.atan((V_1n.imag)/(V_1n.real))*180/math.pi, \"Volts\")\n",
+ "\n",
+ "\n",
+ "## Result\n",
+ "## According to Equations 2-88 and 2-89 in page 130 \n",
+ "##\n",
+ "##\n",
+ "## Positive sequence voltage is = 452.7740 /_ -19.11 Volts \n",
+ "## Negative sequence voltage is = 605.5265 /_ 19.11 Volts\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "According to Equations 2-88 and 2-89 in page 130\n",
+ "\n",
+ "\n",
+ "Positive sequence voltage is = -77793.48 Negative sequence voltage is = 104038.70 Volts \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex29_1-pg131"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example2_29_pg131.sce\n",
+ "## Positive Negative and Zero sequence voltages\n",
+ "## Theory of Alternating Current Machinery by Alexander Langsdorf\n",
+ "## First Edition 1999, Thirty Second reprint\n",
+ "## Tata McGraw Hill Publishing Company\n",
+ "## Example in Page 131\n",
+ "\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "## Given data\n",
+ "V_1 = complex(1000. ,50.);\n",
+ "V_2 = complex(-800. ,100.);\n",
+ "V_3 = complex(-1100. ,-270.);\n",
+ "a = complex(math.cos(2.*math.pi/3.) , math.sin(2.*math.pi/3.));\n",
+ "\n",
+ "\n",
+ "## Calculations\n",
+ "print('According to Equations 2-88 and 2-89 in page 130');\n",
+ "V_1p = (V_1 + V_2*a + V_3*a**2)/3.;\n",
+ "V_1n = (V_1 + V_2*a**(-1) + V_3*a**(-2))/3;\n",
+ "V_0 = (V_1 + V_2 + V_3)/3.;\n",
+ "\n",
+ "print\"%s %.2f %s %.2f %s \"%(\"\\n\\nPositive sequence voltage is = \",(abs(V_1p)/math.atan(V_1p.imag/V_1p.real)*180/math.pi) , \" Volts\" and \"Negative sequence voltage is =\",abs(V_1n)/math.atan((V_1n.imag)/(V_1n.real))*180/math.pi, \"Volts\")\n",
+ "\n",
+ "print\"%s %.2f %s\"%(\"\\n\\nZero sequence voltage is =\",abs(V_0)/math.atan((V_0.imag)/(V_1p.real))*180/math.pi , \" Volts\")\n",
+ "\n",
+ "##\n",
+ "##\n",
+ "## Zero sequence voltage is = 302.6549 /_ 7.59 Volts \n",
+ "## Positive sequence voltage is = 558.9050 /_ 13.62 Volts \n",
+ "## Negative sequence voltage is = 757.9524 /_ -3.15 Volts\n",
+ "#Calculations are done directly "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "According to Equations 2-88 and 2-89 in page 130\n",
+ "\n",
+ "\n",
+ "Positive sequence voltage is = 134721.55 Negative sequence voltage is = -790803.06 Volts \n",
+ "\n",
+ "\n",
+ "Zero sequence voltage is = -235909.53 Volts\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/Chapter3.ipynb b/Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/Chapter3.ipynb new file mode 100644 index 00000000..efac37dd --- /dev/null +++ b/Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/Chapter3.ipynb @@ -0,0 +1,79 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:7ef6a3584a421f73df3798eded14251dce8de7daccd2d1ffb83ce9640649ca72"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter3-Transformer structure Insulation Heating and Load Stresses"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16-pg161"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example3_16_pg161.sce\n",
+ "## To find radial force due to current\n",
+ "## Theory of Alternating Current Machinery by Alexander Langsdorf\n",
+ "## First Edition 1999, Thirty Second reprint\n",
+ "## Tata McGraw Hill Publishing Company\n",
+ "## Example in Page 161\n",
+ "\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "## Given data\n",
+ "va = 200e+3; ## Volt Amperes of transformer, VA\n",
+ "v1 = 11000.; ## Voltage in volts\n",
+ "v2 = 2300.; ## Voltage in volts\n",
+ "T = 46.3; ## Mean length of the turn, inches\n",
+ "n = 455.; ## Number of turns\n",
+ "I = 1320.; ## Current in Amperes\n",
+ "l = 35.; ## length in inches\n",
+ "k = 1.8;\n",
+ "zeq_ht = 8.33;\n",
+ "\n",
+ "## Calculations\n",
+ "\n",
+ "F_av = (0.45/1e+7)*((T*n**2*I**2)/(k*l));\n",
+ "print'%s %.2f %s %.2f %s '%(\"\\n The radial force due to the current of \",I,\"Amps\" and \"given data is\",round(F_av),\" lb\\n\");\n",
+ "\n",
+ "## Result\n",
+ "## The radial force due to the current of 1320 Amps for given data is 11930 lb\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " The radial force due to the current of 1320.00 given data is 11930.00 lb\n",
+ " \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/screenshots/chapter1.png b/Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/screenshots/chapter1.png Binary files differnew file mode 100644 index 00000000..314eef38 --- /dev/null +++ b/Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/screenshots/chapter1.png diff --git a/Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/screenshots/chapter2.png b/Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/screenshots/chapter2.png Binary files differnew file mode 100644 index 00000000..4042d654 --- /dev/null +++ b/Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/screenshots/chapter2.png diff --git a/Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/screenshots/chapter3.png b/Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/screenshots/chapter3.png Binary files differnew file mode 100644 index 00000000..90139b32 --- /dev/null +++ b/Theory_of_Alternating_Current_Machinery_by_A._S._Langsdorf/screenshots/chapter3.png diff --git a/modern_physics_by_Satish_K._Gupta/chap10_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap10_1.ipynb new file mode 100644 index 00000000..1207ccc1 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap10_1.ipynb @@ -0,0 +1,240 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:59cd979321ac17257305319649b46031fe00ee56620ef7034e9b0708b633af30"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 10 Thermoelectric effect"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.1 Page no 304"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=15 #degree C\n",
+ "b=270 \n",
+ "\n",
+ "#Calculation\n",
+ "c=(2*b)-a\n",
+ "\n",
+ "#Result\n",
+ "print\"Temperature is\", c,\"degree C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Temperature is 525 degree C\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.2 Page no 304"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=10 #degree C\n",
+ "a1=527 \n",
+ "\n",
+ "#Calculation\n",
+ "a2=(a+a1)/2.0\n",
+ "print\"Temperature is\",a2,\"degree C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Temperature is 268.5 degree C\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.3 Page no 304"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=100 #ohm\n",
+ "I=10**-6 #A\n",
+ "E=40.0*10**-6 #V\n",
+ "\n",
+ "#Calculaton\n",
+ "e=R*I\n",
+ "t=e/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Smallest temperature difference is\",t,\"degree C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Smallest temperature difference is 2.5 degree C\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.4 Page no 304"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=80 #Degree C\n",
+ "b=13.31 #micro V degree/C\n",
+ "B=-0.038 #micro V degree/C**2\n",
+ "\n",
+ "#Calculation\n",
+ "E=(b*a)+((B*a**2)/2.0)\n",
+ "an=-b/B\n",
+ "\n",
+ "#Result\n",
+ "print\"Thermo emf of silver is\",E,\"micro V\"\n",
+ "print\"Neutral temperature is\",round(an,2),\"degree C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thermo emf of silver is 943.2 micro V\n",
+ "Neutarl temperature is 350.26 degree C\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.5 Page no 304"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=14 #micro V degree/C\n",
+ "b=-0.04 #micro V degree/C**2\n",
+ "\n",
+ "#Calculation\n",
+ "an=-a/b\n",
+ "\n",
+ "#Result\n",
+ "print\"Neutral temperature is\",an,\"degree C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Neutral temperature is 350.0 degree C\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.6 Page no 304"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=500 #micro V\n",
+ "a=41 #micro V degree/C\n",
+ "b=0.041 #micro V degree/C**2\n",
+ "c=-5500\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=-(a+(math.sqrt(a**2-4*b*c)))/(2.0*b)\n",
+ "\n",
+ "#Result\n",
+ "print\"Temperature of hot junction is\", round(A,2),\"degree C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Temperature of hot junction is -1119.8 degree C\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap11_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap11_1.ipynb new file mode 100644 index 00000000..bce9c399 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap11_1.ipynb @@ -0,0 +1,466 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:516acdbb9c17332f6e1ea973b0354e9af84e691b9ea2bf4ff9f153c48f9c31a8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 11 Magnetic Effects Of Current And Magnetism "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.1 Page no 350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given \n",
+ "i=90 #A\n",
+ "a=1.5 #m\n",
+ "b=2\n",
+ "l=10**7\n",
+ "\n",
+ "#Calculation\n",
+ "u=l*((b*i)/a)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the magnetic field is\",u*10**-9,\"10**-5\",\"T\"\n",
+ "print\"Direction of the magnetic field is south\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the magnetic field is 1.2 10**-5 T\n",
+ "Direction of the magnetic field is south\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.2 Page no 350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=3 #A\n",
+ "a=0.15 #m\n",
+ "e=10**-7\n",
+ "b=2\n",
+ "\n",
+ "#Calculation\n",
+ "Z=(e*b*I)/a\n",
+ "\n",
+ "#Result\n",
+ "print\"The magnitude of the magnetic field is\",Z*10**6,\"10**-6\",\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The magnitude of the magnetic field is 4.0 10**-6 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.4 Page no 351"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=10 #A\n",
+ "a=0.05 #cm\n",
+ "b=10**-7\n",
+ "a1=45 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=b*(l/a)*(math.sin(a1*3.14/180.0)+math.sin(a1*3.14/180.0))\n",
+ "B1=4*B\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic field induction is\",round(B1*10**4,2),\"*10**-4 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic field induction is 1.13 *10**-4 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.5 Page no 351"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=10\n",
+ "a=0.08 #m\n",
+ "l=2 #A\n",
+ "v=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=(v*l*math.pi*n*l)/a\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the magnetic field is\", round(B*10**4,2),\"*10**-4 T\"\n",
+ "print\"Direction of the field is vertically downward.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the magnetic field is 1.57 *10**-4 T\n",
+ "Direction of the field is vertically downward.\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.6 Page no 351"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=2.2*10**6 #m/s\n",
+ "a=0.5*10**-10 #m\n",
+ "e=1.6*10**-19\n",
+ "m=10**-7 #N/a**2\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=e*v/(2*math.pi*a)\n",
+ "B=m*2*math.pi*I/a\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic field produced at the centre is\", B,\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic field produced at the centre is 14.08 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.8 Page no 352"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=10 #A\n",
+ "a=0.03 #m\n",
+ "a1=360 #degree\n",
+ "a2=90\n",
+ "a3=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=a1-a2\n",
+ "B=a3*I*3*math.pi/(a*2.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of magnetic field is\", round(B*10**4,2),\"*10**-4 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of magnetic field is 1.57 *10**-4 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 55
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.10 Page no 352"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=500\n",
+ "I=7 #A\n",
+ "a=0.05 #m\n",
+ "x=0.12 #m\n",
+ "m=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=(2*math.pi*n*I*a**2)/(a**2+x**2)**1.5\n",
+ "B1=m*2*math.pi*n*I/a\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Magnetic field at the point on the axis is\", round(B*10**-4,1),\"*10**-3 tesla\"\n",
+ "print\"(b)Magnetic field at the centre of the coil is\",round(B1*10**2,2),\"*10**-2 tesla\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Magnetic field at the point on the axis is 2.5 *10**-3 tesla\n",
+ "(b)Magnetic field at the centre of the coil is 4.4 *10**-2 tesla\n"
+ ]
+ }
+ ],
+ "prompt_number": 69
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.13 Page no 353"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=2.52*10**-3 #T\n",
+ "l=0.5 #m\n",
+ "N=500\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "n=N/l\n",
+ "I=B/(4*math.pi*10**-7*n)\n",
+ "\n",
+ "#Result\n",
+ "print\"Current in the solenoid is\", round(I,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current in the solenoid is 2.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 75
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.14 Page no 353"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=5 #A\n",
+ "l=2.0 #m\n",
+ "n=1000\n",
+ "n1=5\n",
+ "\n",
+ "#Calculation\n",
+ "N=n*n1\n",
+ "n2=N/l\n",
+ "B=4*math.pi*10**-7*n2*I\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic field at the center is\", round(B*10**2,2),\"*10**-2 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic field at the center is 1.57 *10**-2 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 80
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.15 Page no 353"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=16.0*10**-2 #m\n",
+ "n=20\n",
+ "I=16 #A\n",
+ "b=10**-7\n",
+ "a1=10.0*10**-2\n",
+ "n1=25\n",
+ "I1=18\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=b*2*math.pi*n*I/a\n",
+ "B1=b*2*math.pi*n1*I1/a1\n",
+ "B2=B1-B\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the magnetic field is\", round(B1*10**4,2),\"*10**-4 T\"\n",
+ "print\"Direction is towards west\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the magnetic field is 28.27 *10**-4 T\n",
+ "Direction is towards west\n"
+ ]
+ }
+ ],
+ "prompt_number": 88
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.18 Page no 355"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=11 #A\n",
+ "s=3500\n",
+ "r1=25 #cm\n",
+ "r2=26\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "w=((r1+r2)/2.0)\n",
+ "l1=2*math.pi*w\n",
+ "n=s/l1\n",
+ "B=4*math.pi*10**-7*n*l\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Magnetic field outside the toroid is zero\"\n",
+ "print\"(b) Magnetic field inside the core is\",round(B*10**4,2),\"*10**-2 T\"\n",
+ "print\"(c) Magnetic field in the empty space is zero\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Magnetic field outside the toroid is zero\n",
+ "(b) Magnetic field inside the core is 3.02 *10**-2 T\n",
+ "(c) Magnetic field in the empty space is zero\n"
+ ]
+ }
+ ],
+ "prompt_number": 105
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap12_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap12_1.ipynb new file mode 100644 index 00000000..30e711a0 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap12_1.ipynb @@ -0,0 +1,1435 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:48ced6bcb2294321844f4027bedbd6f874bf0da0aa33534190457acb1015aba8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 12 Motion of charged particle"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.1 Page no 379"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=1.6*10**-27 #Kg\n",
+ "e=1.6*10**-19\n",
+ "Ey=2*10**4 #V/m\n",
+ "x=0.1 #m\n",
+ "vx=5*10**6 #m/s\n",
+ "\n",
+ "#Calculation\n",
+ "t=x/vx\n",
+ "Fy=e*Ey\n",
+ "a=Fy/m\n",
+ "y=a*t**2/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Transverse deflection is\", y*10**3,\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Transverse deflection is 0.4 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.2 Page no 379"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=8*10**28 #/m**2\n",
+ "l=1 #m\n",
+ "A=8*10**-6 #m**2\n",
+ "B=5*10**-3 #T\n",
+ "F=8*10**-2 #N\n",
+ "e=1.6*10**-19 #C\n",
+ "\n",
+ "#Calculation\n",
+ "vd=F/(B*n*A*l*e)\n",
+ "\n",
+ "#Result\n",
+ "print\"Drift velocity is\", vd*10**4,\"*10**-4 m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Drift velocity is 1.5625 *10**-4 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.3 Page no 379"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=1.6*10**-19 #C\n",
+ "v=10**7 #m/s\n",
+ "B=3 #T\n",
+ "\n",
+ "#Calculation\n",
+ "F=q*v*B\n",
+ "\n",
+ "#Result\n",
+ "print\"Instantaneous force is\", F,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Instantaneous force is 4.8e-12 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.4 Page no 379"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=2 #A\n",
+ "a=0.1 #m\n",
+ "u=10**-7\n",
+ "q=1.6*10**-19\n",
+ "v=4*10**4 #m/s\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=(u*2*I)/a\n",
+ "F=B*q*v\n",
+ "\n",
+ "#Result\n",
+ "print\"Force of magnetic field is\", F,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force of magnetic field is 2.56e-20 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.6 Page no 380"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=10**5 #m/s\n",
+ "e=1.6*10**-19 #C\n",
+ "m=9.1*10**-31 #Kg\n",
+ "B=0.019*10**-4 #T\n",
+ "\n",
+ "#Calculation\n",
+ "r=m*v/(B*e)\n",
+ "\n",
+ "#Result\n",
+ "print\"Radius of the circular path is\", round(r,3),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of the circular path is 0.299 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.7 Page no 380"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19\n",
+ "m=9*10**-31 #Kg\n",
+ "T=10**-6 #S\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=2*math.pi*m/(e*T)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic field is\", round(B*10**5,3)*10**-5 ,\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic field is 3.534e-05 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.9 Page no 381"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=1.67*10**-27 #Kg\n",
+ "e=1.60*10**-19\n",
+ "V=10**7 #Hz\n",
+ "R=0.6 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=2*math.pi*m*V/e\n",
+ "Emax=(B**2*e**2*R**2/(2*m))/1.6*10**13\n",
+ "\n",
+ "#Result\n",
+ "print\"Kinetic energy of the proton is\",round(Emax,3),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Kinetic energy of the proton is 7.417 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.10 Page no 381"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=5 #A\n",
+ "l=0.1 #m\n",
+ "m=3*10**-3\n",
+ "g=9.8\n",
+ "a=0.5\n",
+ "\n",
+ "#Calculation\n",
+ "w=m*g*l\n",
+ "B=w/a\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the magnetic field is\", B*10**3,\"*10**-3 tesla\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the magnetic field is 5.88 *10**-3 tesla\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.11 Page no 381"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "I1=4 #A\n",
+ "I2=6\n",
+ "r=0.03 #m\n",
+ "u=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "F=u*2*I1*I2/r\n",
+ "\n",
+ "#Result\n",
+ "print\"Force per unit length is\", F*10**4,\"*10**-4 N/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force per unit length is 1.6 *10**-4 N/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.12 Page no 381"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I1=5 #A\n",
+ "I2=12\n",
+ "r=0.4 #m\n",
+ "u=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "F=u*2*I1*I2/r\n",
+ "F1=u*2*I1*I2/r\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Force when current flows in same direction is\", F,\"N/m\"\n",
+ "print\"(ii) Force when current flows in opposite direction is\",F1,\"N/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Force when current flows in same direction is 3e-05 N/m\n",
+ "(ii) Force when current flows in opposite direction is 3e-05 N/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.13 Page no 381"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I1=300 #A\n",
+ "r=1.5*10**-2 #m\n",
+ "u=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "F=u*2*I1*I1/r\n",
+ "\n",
+ "#Result\n",
+ "print\"Force per unit length is\",F,\"N/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force per unit length is 1.2 N/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.14 Page no 381"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=10 #A\n",
+ "n=100\n",
+ "A=8*10**-2 #m**2\n",
+ "B=5 #T\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "t=n*B*I*A*math.cos(60*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Torque is\", round(t,0),\"Nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Torque is 200.0 Nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.15 Page no 381"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=30\n",
+ "I=6 #A\n",
+ "B=1 #T\n",
+ "r=8*10**-2 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=math.pi*r**2\n",
+ "t=n*B*I*A*math.sin(60*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Magnitude of the counter torque is\", round(t,3),\"Nm\"\n",
+ "print\"(b) Torque on the planar loop is independent of its shape, the torque will remain unchanged.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Magnitude of the counter torque is 3.133 Nm\n",
+ "(b) Torque on the planar loop is independent of its shape, the torque will remain unchanged.\n"
+ ]
+ }
+ ],
+ "prompt_number": 50
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.16 Page no 382"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=100*10**-4 #T\n",
+ "I=10 #A\n",
+ "l=44\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=l/(2.0*math.pi)\n",
+ "A=math.pi*r**2\n",
+ "t=B*I*A\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum torque is\", round(t*10**-1,2),\"*10**-3 Nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum torque is 1.54 *10**-3 Nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 62
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.17 Page no 382"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=20\n",
+ "r=10*10**-2 #m\n",
+ "B=0.10 #T\n",
+ "I=5 #A\n",
+ "n1=10**29 #/m**3\n",
+ "A1=10**-5 #m**2\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=math.pi*r**2\n",
+ "t=n*B*I*A*math.sin(0*3.14/180.0)\n",
+ "F=B*I/(n1*A1)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Total torque on the coil is\", t\n",
+ "print\"(b) Net force on a planar loop in a magnetic field is always zero\"\n",
+ "print\"(c) Average force is\",F,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Total torque on the coil is 0.0\n",
+ "(b) Net force on a planar loop in a magnetic field is always zero\n",
+ "(c) Average force is 5e-25 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.18 Page no 382"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=5*10**-4 #m**2\n",
+ "n=60\n",
+ "a=18 #degree\n",
+ "B=90*10**-4 #T\n",
+ "I=0.20*10**-3 #A\n",
+ "\n",
+ "#calculation\n",
+ "k=n*B*I*A/a\n",
+ "\n",
+ "#Result\n",
+ "print\"Torsional constant is\",k,\"N m per degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Torsional constant is 3e-09 N m per degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 70
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.21 Page no 383"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "G=15 #ohm\n",
+ "Ig=2*10**-3 #A\n",
+ "I=5 #A\n",
+ "\n",
+ "#Calculation\n",
+ "S=Ig*G/(I-Ig)\n",
+ "\n",
+ "#Result\n",
+ "print\"Shunt resistance is\",round(S,3),\"Ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shunt resistance is 0.006 Ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.22 Page no 383"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=50*10**-3 #V\n",
+ "G=100.0 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Ig=V/G\n",
+ "S=Ig*G/(I-Ig)\n",
+ "\n",
+ "#Result\n",
+ "print\"Shunt resistance is\", round(S,2),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shunt resistance is 0.01 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 79
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.23 Page no 383"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "G=100 #ohm\n",
+ "Ig=5*10**-3 #A\n",
+ "I=5 #A\n",
+ "\n",
+ "#Calculation\n",
+ "S=Ig*G/(I-Ig)\n",
+ "\n",
+ "#Result\n",
+ "print\"Shunt resistance is\",S"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shunt resistance is 0.1001001001\n"
+ ]
+ }
+ ],
+ "prompt_number": 80
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.24 Page no 383"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "G=5 #ohm\n",
+ "Ig=15.0*10**-3 #A\n",
+ "I=1.5\n",
+ "V=1.5 #V\n",
+ "\n",
+ "#Calculation\n",
+ "S=Ig*G/(I-Ig)\n",
+ "R=(V/Ig)-G\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) To enable galvanometer to read 1.5 A is\", round(S,2),\"ohm\"\n",
+ "print\"(b) To enable galvanometer to read 1.5 V is\",R,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) To enable galvanometer to read 1.5 A is 0.05 ohm\n",
+ "(b) To enable galvanometer to read 1.5 V is 95.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 88
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.25 Page no 383"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "G=10 #ohm\n",
+ "Ig=25.0*10**-3 #A\n",
+ "V=120 #V\n",
+ "I=20 #A\n",
+ "\n",
+ "#Calculation\n",
+ "R=(V/Ig)-G\n",
+ "S=Ig*G/(I-Ig)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) To convert the galvanometer into the voltmeter reading is\" ,R,\"ohm\"\n",
+ "print\"(b) To convert the galvanometer into the ammeter reading is\",round(S,4),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) To convert the galvanometer into the voltmeter reading is 4790.0 ohm\n",
+ "(b) To convert the galvanometer into the ammeter reading is 0.0125 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.26 Page no 383"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=3 #v\n",
+ "R=55 #ohm\n",
+ "Ra=1\n",
+ "I=50*10**-3 #A\n",
+ "\n",
+ "#Calculation\n",
+ "r=(E/I)-(R+Ra)\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of r is\", r,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of r is 4.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 98
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.27 Page no 384"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=60 #V\n",
+ "R1=400 #ohm\n",
+ "R2=300\n",
+ "V1=30.0\n",
+ "a=120000\n",
+ "\n",
+ "#Calculation\n",
+ "Rv=(-V1*a)/(V1*(R1+R2)-E*R1)\n",
+ "R=Rv*R2/(Rv+R2)\n",
+ "I1=E/(R+R1)\n",
+ "V=I1*R\n",
+ "\n",
+ "#Result\n",
+ "print\"Voltmeter reads\", V,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltmeter reads 22.5 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 108
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.28 Page no 384"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rv=400.0\n",
+ "E=84 #V\n",
+ "r=100.0\n",
+ "r1=200\n",
+ "\n",
+ "#Calculation\n",
+ "R=1/(1/Rv+1/100.0)\n",
+ "R1=R+200\n",
+ "I=E/R1\n",
+ "I1=I/5.0\n",
+ "V=I1*Rv\n",
+ "R2=r+r1\n",
+ "I2=E/R2\n",
+ "V2=I2*r\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Reading on voltmeter is\", V,\"V\"\n",
+ "print\"(b) Potential difference is\",V2,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Reading on voltmeter is 24.0 V\n",
+ "(b) Potential difference is 28.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 122
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.29 Page no 385"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=120 #v\n",
+ "Rv=10**4 #ohm\n",
+ "a=4.0\n",
+ "\n",
+ "#Calculation\n",
+ "x=(Rv*(E-a))/a\n",
+ "\n",
+ "#Result\n",
+ "print\"Large resistance is\", x*10**-3,\"K ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Large resistance is 290.0 K ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 131
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.30 Page no 385"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=0.75 #T\n",
+ "E=9*10**5 #V/m\n",
+ "V=15*10**3\n",
+ "\n",
+ "#Calculation\n",
+ "v=E/B\n",
+ "a=v**2/(2.0*V)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of e/m is\", a*10**-7,\"*10**7 C/Kg\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of e/m is 4.8 *10**7 C/Kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 127
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.31 Page no 385"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9.11*10**-31 #Kg\n",
+ "e=1.60*10**-19 #C\n",
+ "B=0.40*10**-4 #T\n",
+ "a=18*1.6*10**-16\n",
+ "PQ=0.30\n",
+ "a2=1.52 #degree\n",
+ "\n",
+ "#calculation\n",
+ "import math\n",
+ "r=math.sqrt(2*m*a)/(B*e)\n",
+ "a1=(PQ/r)\n",
+ "PA=r*(1-math.cos(a2*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"up and down deflection of the beam is\", round(PA*10**3,0),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "up and down deflection of the beam is 4.0 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 153
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.32 Page no 386"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=60*10**-3\n",
+ "g=9.8\n",
+ "I=5\n",
+ "l=0.45\n",
+ "\n",
+ "#Calculation\n",
+ "B=m*g/(I*l)\n",
+ "T=2*m*g\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Magnetic field is\", round(B,3),\"T\"\n",
+ "print\"(ii) Total tension in the wire is\",T,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnetic field is 0.261 T\n",
+ "(ii) Total tension in the wire is 1.176 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 161
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.33 Page no 386"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=0.15 #T\n",
+ "m=0.30 #Kg/m\n",
+ "a=30 #degree\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": 165
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.34 Page no 386"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I1=4 #A\n",
+ "I2=3\n",
+ "r=3.0*10**-2 #m\n",
+ "u=10**-7\n",
+ "l=5*10**-2\n",
+ "\n",
+ "#Calculation\n",
+ "F=u*2*I1*I2/r\n",
+ "F1=F*l\n",
+ "\n",
+ "#Result\n",
+ "print\"Total force is\", F1,\"N (attractive force)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total force is 4e-06 N (attractive force)\n"
+ ]
+ }
+ ],
+ "prompt_number": 171
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.35 Page no 386"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "AB=25*10**-2 #m\n",
+ "BC=10*10**-2\n",
+ "r1=2.0*10**-2 #m\n",
+ "I1=15 #A\n",
+ "I2=25 #A\n",
+ "u=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "r2=BC+r1\n",
+ "F1=u*2*I1*I2*AB/r1\n",
+ "F2=u*2*I1*I2*AB/r2\n",
+ "F=F1-F2\n",
+ "\n",
+ "#Result\n",
+ "print\"Net force on the loop is\", F*10**4,\"*10**-4 N (towards XY)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Net force on the loop is 7.8125 *10**-4 N (towards XY)\n"
+ ]
+ }
+ ],
+ "prompt_number": 182
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.36 Page no 387"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=30*10**-3 #Kg\n",
+ "g=9.8 #m/s**2\n",
+ "l=0.5 #m\n",
+ "r=10**-2\n",
+ "u=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=math.sqrt((M*g*r)/(u*2*l))\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 171.46 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 186
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.37 Page no 387"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=900\n",
+ "l=0.6\n",
+ "u=10**-7\n",
+ "l2=0.02 #m\n",
+ "l1=6\n",
+ "m=2.5*10**-3 #Kg\n",
+ "g=9.8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "n1=n/l\n",
+ "B=4*math.pi*n1\n",
+ "F=B*l1*l2\n",
+ "I=m*g/F\n",
+ "\n",
+ "#Result\n",
+ "print\"Current in the winding of the secondary is\", round(I*10**7,1),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current in the winding of the secondary is 108.3 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 197
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.38 Page no 387"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=1.6*10**-3 #m**2\n",
+ "n=200\n",
+ "B=0.2 #T\n",
+ "a=30 #degree\n",
+ "K=10**-6 #N m /degree\n",
+ "a1=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "Imax=K*a/(n*B*A)\n",
+ "Imin=K*a1/(n*B*A)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Minimum current is\",round(Imax*10**4,2),\"*10**-4 A\"\n",
+ "print\"(ii) Smallest current that can be detected is\",Imin,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Minimum current is 4.69 *10**-4 A\n",
+ "(ii) Smallest current that can be detected is 1.5625e-06 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 204
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.39 Page no 388 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=5000.0 #ohm/V\n",
+ "V=5\n",
+ "V1=20.0\n",
+ "\n",
+ "#Calculation\n",
+ "Ig=1/R\n",
+ "G=V/Ig\n",
+ "R1=(V1/Ig)-G\n",
+ "Rn=R1+G\n",
+ "Rv=Rn/V1\n",
+ "\n",
+ "#Result\n",
+ "print\"New voltmeter will be still graded as\", Rv,\"ohm/V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "New voltmeter will be still graded as 5000.0 ohm/V\n"
+ ]
+ }
+ ],
+ "prompt_number": 212
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.40 Page no 388"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V1=100 #V\n",
+ "Rv=400.0 \n",
+ "\n",
+ "#Calculation\n",
+ "I1=V1/Rv\n",
+ "V=I1*20\n",
+ "V2=V1+V\n",
+ "V3=V2-V1\n",
+ "\n",
+ "#Result\n",
+ "print\"Error in the reading of the voltmeter is\",V3,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Error in the reading of the voltmeter is 5.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 215
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap13_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap13_1.ipynb new file mode 100644 index 00000000..13ac3401 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap13_1.ipynb @@ -0,0 +1,547 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:a82b820d4c986803883c47a06b19e19453881487c8189443b60dab54c2680966"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 13 Magnetic dipole"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.1 Page no 415"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "F=14.4*10**-4 #N\n",
+ "r=0.05 #m\n",
+ "u=10.0**-7\n",
+ "F1=1.6*10**-4 #N\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "m=math.sqrt(F*r**2/u)\n",
+ "r1=math.sqrt(u*6*6/F1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance between the poles is\", r1,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance between the poles is 0.15 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.2 Page no 415"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "F=0.8*10**-3\n",
+ "r=0.1 #m\n",
+ "g=9.8\n",
+ "u=10.0**-7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "m=math.sqrt(F*g*r**2/(u*5.0))\n",
+ "a=5*m\n",
+ "\n",
+ "#Result\n",
+ "print\"Strength of each pole is\", round(a,1),\"A m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Strength of each pole is 62.6 A m\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.4 Page no 416"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=50 #Am\n",
+ "r=0.2 #m\n",
+ "l=0.05 #M\n",
+ "M=5 #Am**2\n",
+ "u=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "Baxial=u*2*M*r/((r**2-l**2)**2)\n",
+ "Bequi=u*M/((r**2+l**2)**1.5)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Magnetic field at a distance of 0.2 m from its centre on axial line is\",round(Baxial*10**4,2),\"*10**-4 T\"\n",
+ "print\"(ii) Magnetic field at a distance of 0.2 m from its centre on its equitorial line is\",round(Bequi*10**5,3)*10**-5,\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnetic field at a distance of 0.2 m from its centre on axial line is 1.42 *10**-4 T\n",
+ "(ii) Magnetic field at a distance of 0.2 m from its centre on its equitorial line is 5.707e-05 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.5 Page no 416"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=12.5 #Am**2\n",
+ "OP=0.05 #m\n",
+ "u=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "b1=(u*2*M)/(OP**3)\n",
+ "b2=(u*M)/(OP**3)\n",
+ "B=math.sqrt(b1**2+b2**2)\n",
+ "a=b2/b1\n",
+ "a1=math.atan(a*180/3.14)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic field at a point midway between two magnets is\",round(B*10**2,3),\"10**-2 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic field at a point midway between two magnets is 2.236 10**-2 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.6 Page no 416"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=0.3 #T\n",
+ "A=30 #Degree\n",
+ "t=0.06 #Nn\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "M=t/(B*math.sin(A*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic moment is\",round(M,1),\"Am**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic moment is 0.4 Am**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.7 Page no 416"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=48 #Am\n",
+ "l=0.25 #m\n",
+ "a=30 #degree\n",
+ "B=0.15\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "M=m*l\n",
+ "T=M*B*math.sin(a*3.14/180.0)\n",
+ "\n",
+ "#Result \n",
+ "print\"(i) Magnetic moment is\",M,\"Am**2\"\n",
+ "print\"(ii) Torque is\",round(T,1),\"Mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnetic moment is 12.0 Am**2\n",
+ "(ii) Torque is 0.9 Mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.8 Page no 416"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=0.2 #t\n",
+ "a=30\n",
+ "t=0.06\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "M=t/(B*math.sin(a*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Magnetic moment is\",round(M,1),\"An**2\"\n",
+ "print\"(b) When the magnet aligns itself parallel to the magnetic field\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Magnetic moment is 0.6 An**2\n",
+ "(b) When the magnet aligns itself parallel to the magnetic field\n"
+ ]
+ }
+ ],
+ "prompt_number": 50
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.9 Page no 416"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=1.5 #J/T\n",
+ "B=0.22 #T\n",
+ "a=0\n",
+ "a1=90\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "W=M*B*(math.cos(a*3.14/180.0)-math.cos(a1*3.14/180.0))\n",
+ "t=M*B*math.sin(a1*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Amount of work required is\", round(W,2),\"J\"\n",
+ "print\"Torque is\",round(t,2),\"Nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Amount of work required is 0.33 J\n",
+ "Torque is 0.33 Nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 57
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.10 Page no 417"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=50\n",
+ "a=0.2 \n",
+ "I=12 #A\n",
+ "u=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=u*2*math.pi*n*I/a\n",
+ "M=n*I*math.pi*a**2\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Magnetic moment at the centre is\",round(B*10**3,3),\"*10**-3 T\"\n",
+ "print\"(b) Magnetic moment associated is\",round(M,1),\"Am**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Magnetic moment at the centre is 1.885 *10**-3 T\n",
+ "(b) Magnetic moment associated is 75.4 Am**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 65
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.11 Page no 417"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.53*10**-10 #m\n",
+ "v=6.8*10**15 #Kz\n",
+ "e=1.6*10**-19 #C\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "M=math.pi*e*v*r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic moment is\",round(M*10**24,1)*10**-24,\"A m**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic moment is 9.6e-24 A m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 69
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.12 Page no 417"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=1.8*10**-23 #Am**2\n",
+ "v=5*10**-6 #m**3\n",
+ "a=7.8*10**3\n",
+ "N=6.02*10**26\n",
+ "A=56.0\n",
+ "B=1.5 #T\n",
+ "a1=90 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "m=v*a\n",
+ "n=N*m/A\n",
+ "M1=M*n\n",
+ "t=M1*B*math.sin(a1*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Magnetic moment is\", round(M1,2),\"A m**2\"\n",
+ "print\"(b) Torque required is\",round(t,2),\"N m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Magnetic moment is 7.55 A m**2\n",
+ "(b) Torque required is 11.32 N m\n"
+ ]
+ }
+ ],
+ "prompt_number": 79
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.13 Page no 417"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B1=1.2*10**-2 #T\n",
+ "a=15 #degree\n",
+ "a2=60\n",
+ "\n",
+ "#Calculation\n",
+ "a3=a2-a\n",
+ "B2=B1*math.sin(a*3.14/180.0)/(math.sin(a3*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print round(B2*10**3,2),\"*10**-3 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "4.39 *10**-3 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 83
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.14 Page no 417"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=1000\n",
+ "I=2 #A\n",
+ "A=2*10**-4 #m**@\n",
+ "a=30\n",
+ "B=0.16\n",
+ "a1=0 #degree\n",
+ "a2=180 \n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "M=I*A*n\n",
+ "t=M*B*math.sin(a*3.14/180.0)\n",
+ "W=M*B*(math.cos(a1*3.14/180.0)-math.cos(a2*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Torque required is\", round(t,3),\"Nm\"\n",
+ "print\"(b) Work needed is\",round(W,3),\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Torque required is 0.032 Nm\n",
+ "(b) Work needed is 0.128 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 92
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap14_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap14_1.ipynb new file mode 100644 index 00000000..8a01d802 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap14_1.ipynb @@ -0,0 +1,635 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:6c9b2b81b43b71eb198fbc94a1351ef301d125c60ef23bbdeed7f4f739d04107"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 14 Earths magnetism"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.3 Page no 433"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=60 #Degree\n",
+ "Bh=0.16 #G\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=Bh/cos(a*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of earth's field is\", round(B,2),\"G\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of earth's field is 0.32 G\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.4 Page no 434"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=60\n",
+ "a2=45\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a1=math.tan(a2*3.14/180.0)/math.cos(a*3.14/180.0)\n",
+ "a3=math.atan(a1)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Apparent value of the dip is\", round(a3,1),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Apparent value of the dip is 63.4 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.6 Page no 434"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=30 #cm\n",
+ "l=0.15 #m\n",
+ "r=0.30 #m\n",
+ "Bh=0.34*10**-4 #T\n",
+ "u=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "M=Bh*(r**2-l**2)**2/(2*u*r)\n",
+ "m=M/(2*l)\n",
+ "\n",
+ "#Result\n",
+ "print\"Pole strength of the magnet is\",m,\"Am\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Pole strength of the magnet is 8.60625 Am\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.7 Page no 434"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=0.4 #Am**2\n",
+ "r=0.1 #m\n",
+ "l=0.05 #m\n",
+ "u=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "Bh=u*M/((r**2+l**2)**1.5)\n",
+ "\n",
+ "#Result\n",
+ "print\"Horizontal component of earth's magnetic field is\", round(Bh*10**4,3)*10**-4,\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Horizontal component of earth's magnetic field is 2.86e-05 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.8 Page no 435"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=0.33\n",
+ "a=0\n",
+ "u=10**-7\n",
+ "I=2.5 #A\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Bh=B/math.cos(a*3.14/180.0)\n",
+ "a=u*2*I/(Bh*10**-4)\n",
+ "\n",
+ "#Result\n",
+ "print\"Neutral point is\", round(a*10**2,1),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Neutral point is 1.5 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.9 Page no 435"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Bh=0.32 #G\n",
+ "B=0.48 \n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=B/Bh\n",
+ "a1=math.atan(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"New stable equilibrium is\", round(a1,1),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "New stable equilibrium is 56.3 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.10 Page no 435"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=22\n",
+ "a=0.1 #m\n",
+ "Bh=0.3*10**-4 #T\n",
+ "u=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "K=2*a*Bh/(n*4*math.pi*u)\n",
+ "\n",
+ "#Result\n",
+ "print\"Reduction factor is\", round(K,3),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Reduction factor is 0.217 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.11 Page no 435"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=40\n",
+ "a=0.12\n",
+ "I=0.15\n",
+ "a1=45 #degree\n",
+ "u=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Bh=(n*u*4*math.pi*I)/(2.0*a*math.tan(a1*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"Strength of horizontal component is\", round(Bh*10**4,3),\"*10**-4 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Strength of horizontal component is 0.314 *10**-4 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.12 Page no 435"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a1=30\n",
+ "a2=45 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "n=2*math.tan(a1*3.14/180.0)/(math.tan(a2*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"Ratio of number of turns is\", round(n,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Ratio of number of turns is 1.155\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.13 Page no 436"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=16\n",
+ "a=0.1 #m\n",
+ "i=0.75 #A\n",
+ "Bh=5*10**-2 #T\n",
+ "v=2 #/s\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "M=n*i*math.pi*a**2\n",
+ "I=M*Bh/(4*math.pi**2*v**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Moment of inertia is\",round(I*10**4,3),\"*10**-4 Kg m**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Moment of inertia is 1.194 *10**-4 Kg m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14 Page no 436"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "T1=2.5\n",
+ "T2=4.5\n",
+ "M2=1.5\n",
+ "\n",
+ "#Calculation\n",
+ "M=T2**2/(M2*T1**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Ratio of magnetic moment is\",M"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Ratio of magnetic moment is 2.16\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.15 Page no 436"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "T1=3.0\n",
+ "T2=4.0\n",
+ "\n",
+ "#Calculation\n",
+ "M=(T2**2+T1**2)/(T2**2-T1**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Ratio of magnetic moments is\",round(M,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Ratio of magnetic moments is 3.57\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.16 Page no 436"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a1=35 #Degree\n",
+ "B=0.39\n",
+ "I=1 #A\n",
+ "a=4.0*10**-2\n",
+ "u=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Bh=B*math.cos(a1*3.14/180.0)\n",
+ "Bv=B*math.sin(a1*3.14/180.0)\n",
+ "B1=(u*2*I*4/a)*10**4\n",
+ "Rh=Bh-B1\n",
+ "R=math.sqrt(Rh**2+Bv**2)\n",
+ "Rh1=Bh+B1\n",
+ "R3=math.sqrt(Rh1**2+Bv**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Resultant magnetic field is\", round(R3,3),\"G\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resultant magnetic field is 0.566 G\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.17 Page no 436"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=5.25*10**-2 #J/T\n",
+ "Bh=0.42*10**-4 #T\n",
+ "u=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "r=(u*M/Bh)**0.333\n",
+ "r1=(u*2*M/Bh)**0.333\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Distance from the centre of the magnet on its normal bisector is\", round(r*10**2,1),\"cm\"\n",
+ "print\"(b) Distance from the centre of the magnet on its axis is\",round(r1*10**2,1),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Distance from the centre of the magnet on its normal bisector is 5.0 cm\n",
+ "(b) Distance from the centre of the magnet on its axis is 6.3 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.18 Page no 437"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=0.35 #A\n",
+ "n=30\n",
+ "a=12.0*10**-2\n",
+ "u=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Bh=u*2*math.pi*n*I*0.707/a\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Horizontal component of the earth's magnetic field is\", round(Bh*10**4,2),\"G\"\n",
+ "print\"(b) The needle will reverse its original direction i.e. it will point east to west.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Horizontal component of the earth's magnetic field is 0.39 G\n",
+ "(b) The needle will reverse its original direction i.e. it will point east to west.\n"
+ ]
+ }
+ ],
+ "prompt_number": 60
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.19 Page no 437"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "t1=9 #S\n",
+ "t2=4.5\n",
+ "Bh=0.34*10**-4 #T\n",
+ "u=10**-7\n",
+ "r=0.1 #m\n",
+ "\n",
+ "#Calculation\n",
+ "Baxial=Bh*((t1**2/t2**2)-1)\n",
+ "M2=Baxial*r**2/(2*u)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic moment is\", M2*10**-1,\"A m**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic moment is 0.51 A m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap15_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap15_1.ipynb new file mode 100644 index 00000000..f90ea520 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap15_1.ipynb @@ -0,0 +1,317 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:859464a7b9c1dff7d24f65e6024e51cc8bc55fc7979a3e8c04919720ca49d3b9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 15 Classification of Magnetic Materials"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.1 Page no 457"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=0.12 #TA**-1m\n",
+ "q1=4*math.pi*10**-7 #TA**-1m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "q3=q/q1\n",
+ "e=q3-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Relative permeability is\",round(q3*10**-4,2),\"10**4\"\n",
+ "print\"susceptibility is\",round(e*10**-4,2),\"10**4\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Relative permeability is 9.55 10**4\n",
+ "susceptibility is 9.55 10**4\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.2 Page no 457"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=0.5*10**-4 #m**2\n",
+ "H=1200 #Am**-1\n",
+ "W=599\n",
+ "Q=4*math.pi*10**-7 #TA\"**-1m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Z=Q*(1+W)\n",
+ "Z1=Z*H\n",
+ "Z2=Z1*A\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Z produced is\",round(Z*10**4,2),\"10**-4\",\"TA**-1m\"\n",
+ "print\"(ii) Z1 produced is\",round(Z1,3),\"T\"\n",
+ "print\"(iii) Z2 produced is\",round(Z2*10**5,3),\"10**-5\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Z produced is 7.54 10**-4 TA**-1m\n",
+ "(ii) Z1 produced is 0.905 T\n",
+ "(iii) Z2 produced is 4.524 10**-5\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.3 Page no 457"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=1.2 #A\n",
+ "Q=800\n",
+ "A=0.15 #m\n",
+ "S=3500\n",
+ "K=4*math.pi*10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "L=2*math.pi*A\n",
+ "n=S/L\n",
+ "B=(K*Q*S*I)/L\n",
+ "\n",
+ "#Result\n",
+ "print\"The magnetic field in the core for a magnetising current of 1.2 A is\",B,\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The magnetic field in the core for a magnetising current of 1.2 A is 4.48 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 73
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.4 Page no 457"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "H=2*10**3 #Am**-1\n",
+ "I=4.8*10**-2 #Am**-1\n",
+ "T=280 #K\n",
+ "T1=320.0 #K\n",
+ "\n",
+ "#Calculation\n",
+ "S=I/H\n",
+ "S1=S*(T/T1)\n",
+ "S2=S1*H\n",
+ "\n",
+ "#Result\n",
+ "print\"The susceptibility of aluminium is raised to 280 k is\",S*10**5,\"10**3\"\n",
+ "print\"The temperature of the aluminium is raised to 320 k is\",S1\n",
+ "print\"The suscepitibility and intensity of magnetistionis\",S2*10**2,\"10**-2\",\"Am**-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The susceptibility of aluminium is raised to 280 k is 2.4 10**3\n",
+ "The temperature of the aluminium is raised to 320 k is 2.1e-05\n",
+ "The suscepitibility and intensity of magnetistionis 4.2 10**-2 Am**-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 90
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.5 Page no 458"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "W=3.2*10**4 #J\n",
+ "M=8.4\n",
+ "D=7200\n",
+ "v=50 #cycle s**-1\n",
+ "T=30*60 #s\n",
+ "\n",
+ "#Calculation\n",
+ "V=M/D\n",
+ "Q=W/(V*v*T)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of energy is\",round(Q,1),\"J m**-3 cycle**-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of energy is 304.8 J m**-3 cycle**-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.6 Page no 458"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=4.5 #Am\n",
+ "l=0.06 #m\n",
+ "A=0.9*10**-4 #m**2\n",
+ "r=0.06 #m\n",
+ "J=4*math.pi*10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "I=m/A\n",
+ "H1=m/(4*math.pi*r**2)\n",
+ "H2=m/(4*math.pi*r**2)\n",
+ "H=H1+H2\n",
+ "B=J*(-H+I)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Intensity of the magnetisation is\", I*10**-4,\"10**4 A/m\"\n",
+ "print\"(b) Magnetic intensity is\",round(H,0),\"A/m\"\n",
+ "print\"(c) Magnetic induction is\",round(B*10**2,2),\"*10**-2 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Intensity of the magnetisation is 5.0 10**4 A/m\n",
+ "(b) Magnetic intensity is 199.0 A/m\n",
+ "(c) Magnetic induction is 6.26 *10**-2 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.7 Page no 458"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=1.5*10**-23 #J/T\n",
+ "n=2.0*10**24\n",
+ "a=15 #%\n",
+ "T1=4.2 #K\n",
+ "T2=2.8\n",
+ "B1=0.84 #T\n",
+ "B2=0.98\n",
+ "\n",
+ "#Calculation\n",
+ "M=m*n\n",
+ "M1=M*(a/100.0)\n",
+ "M2=M1*T1*B2/(T2*B1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Total dipole moment is\",M2,\"J/T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total dipole moment is 7.875 J/T\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap16_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap16_1.ipynb new file mode 100644 index 00000000..a6266584 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap16_1.ipynb @@ -0,0 +1,964 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:03e25d70b6f5039934ebff67f76c18265eda838017719c81d51174de65a0d0cb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 16 Electromagnetic induction"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.1 Page no 508"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=5.0 #ohm\n",
+ "t=2\n",
+ "a=15\n",
+ "b=8\n",
+ "\n",
+ "#Calculation\n",
+ "e=-a*t**2-(b*t)-t\n",
+ "I=-e/R\n",
+ "\n",
+ "#Result\n",
+ "print\"Induced current is\", I,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Induced current is 15.6 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.2 Page no 508"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "S1=75*10**-4 #m**2\n",
+ "B1=0.8 #wb/m**2\n",
+ "S2=100*10**-4\n",
+ "B2=1.4\n",
+ "t=0.05\n",
+ "\n",
+ "#Calculation\n",
+ "a1=B1*S1\n",
+ "a2=B2*S2\n",
+ "a=a2-a1\n",
+ "e=-a/t\n",
+ "\n",
+ "#Result\n",
+ "print\"Induced e.m.f is\", e,\"Volt\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Induced e.m.f is -0.16 Volt\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.3 Page no 508"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a1=5.5*10**-4 #Wb\n",
+ "a2=0.5*10**-4\n",
+ "N=1000\n",
+ "t=0.1\n",
+ "R=10 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "a=a2-a1\n",
+ "a11=N*a\n",
+ "e=-(a11/t)\n",
+ "I1=e/R\n",
+ "I2=I1*t\n",
+ "\n",
+ "#Result\n",
+ "print\" Induced e.m.f produced is\",e,\"V\"\n",
+ "print\" Charge flowing through the coil in 0.1 is\",I2,\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Induced e.m.f produced is 5.0 V\n",
+ " Charge flowing through the coil in 0.1 is 0.05 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.5 Page no 509"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=2.5*10**-3 #Wb**-2\n",
+ "L=1 #m\n",
+ "v=30 #r.p.s\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "e=-B*math.pi*L**2*v\n",
+ "\n",
+ "#Result\n",
+ "print\"The produced e.m.f. between its ends is\",round(e,3),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The produced e.m.f. between its ends is -0.236 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.6 Page no 509"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=1.2 #m\n",
+ "e=10**-2 #v\n",
+ "B=5*10**-5 #tesla\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "V=e/(B*math.pi*L**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"The rate of rotation of the wheel is\",round(V,1),\"Rotation a**-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The rate of rotation of the wheel is 44.2 Rotation a**-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.7 Page no 509"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "D=10\n",
+ "L=0.50 #m\n",
+ "B=0.40*10**-4 #T\n",
+ "V=2 #r.p.s.\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "E=-B*math.pi*L**2*V\n",
+ "\n",
+ "#Result\n",
+ "print\"The induced e.m.f. between the axle and the rim of the wheel is\",round(E*10**5,3),\"10**-5\",\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The induced e.m.f. between the axle and the rim of the wheel is -6.283 10**-5 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.8 Page no 509"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=0.2 #T\n",
+ "r=0.1 #m\n",
+ "R=2 #ohm\n",
+ "D=20*math.pi #rad s**-1\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "V=D/(2*math.pi)\n",
+ "E=-B*math.pi*r**2*V\n",
+ "I=E/R\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The potential difference is\",round(E,4),\"V\"\n",
+ "print\"(ii) The induced current is\",round(I,4),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The potential difference is -0.0628 V\n",
+ "(ii) The induced current is -0.0314 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.9 Page no 509"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=8.0*10**-5 #Wb m**-2\n",
+ "L=2 #m\n",
+ "v=30 #m s**-1\n",
+ "\n",
+ "#Calculation\n",
+ "e=B*L*v\n",
+ "\n",
+ "#Result\n",
+ "print\"The vertical component of earth's field is\",e*10**3,\"10**-3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The vertical component of earth's field is 4.8 10**-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 61
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.10 Page no 509"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Givem\n",
+ "l=10 #m\n",
+ "v=5 #m/s\n",
+ "Bh=0.30*10**-4 #Wb/m**2\n",
+ "\n",
+ "#Calculation\n",
+ "e=Bh*l*v\n",
+ "\n",
+ "#Result\n",
+ "print \"Instantaneous value of e.m.f. induced is\",e*10**3,\"*10**-3 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Instantaneous value of e.m.f. induced is 1.5 *10**-3 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 64
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.11 Page no 509"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=0.3 #T\n",
+ "v=10**-2 #m/s\n",
+ "l=8*10**-2\n",
+ "L=1\n",
+ "v1=1.0\n",
+ "l2=2*10**-2\n",
+ "L1=8\n",
+ "\n",
+ "#Calculation\n",
+ "e=B*l*v\n",
+ "l=L/v1\n",
+ "e1=B*l2*v\n",
+ "t1=L1/v1\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Voltage developed in the direction of motion normal to the longer side is\", e*10**3,\"mv\"\n",
+ "print\"(ii) Voltage developed in the direction of motion normal to the shorter side is\",e1*10**3,\"mV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Voltage developed in the direction of motion normal to the longer side is 0.24 mv\n",
+ "(ii) Voltage developed in the direction of motion normal to the shorter side is 0.06 mV\n"
+ ]
+ }
+ ],
+ "prompt_number": 72
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.12 Page no 510"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=0.4 #T\n",
+ "v=5 #m/s\n",
+ "l=0.25 #m\n",
+ "e=0.5\n",
+ "R=5.0\n",
+ "\n",
+ "#Calculation\n",
+ "e=B*l*v\n",
+ "I=e/R\n",
+ "\n",
+ "#Result\n",
+ "print\"New current is\",I,\"A\"\n",
+ "print\"Direction is from the end S to R\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "New current is 0.1 A\n",
+ "Direction is from the end S to R\n"
+ ]
+ }
+ ],
+ "prompt_number": 75
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.13 Page no 510"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=500 #m/s\n",
+ "B=5*10**-4 #t\n",
+ "a=30 #degree\n",
+ "l=25 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Bv=B*math.sin(a*3.14/180.0)\n",
+ "e=Bv*l*v\n",
+ "\n",
+ "#Result\n",
+ "print\"Voltage difference is\", round(e,3),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage difference is 3.124 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 80
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.14 Page no 510"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=10*10**-3 #H\n",
+ "I=4*10**-3 #A\n",
+ "N=200.0\n",
+ "\n",
+ "#Calculation\n",
+ "a=L*I\n",
+ "A=a/N\n",
+ "\n",
+ "#Result\n",
+ "print\"Total magnetic flux is\",a,\"Weber\"\n",
+ "print\"Magnetic flux through the cross section is\",A,\"Weber\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total magnetic flux is 4e-05 Weber\n",
+ "Magnetic flux through the cross section is 2e-07 Weber\n"
+ ]
+ }
+ ],
+ "prompt_number": 85
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.15 Page no 510"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=10**-2 #H\n",
+ "I1=0\n",
+ "I2=1 #A\n",
+ "l=1 \n",
+ "t=0.01\n",
+ "\n",
+ "#Calculation\n",
+ "e=-L*(l/t)\n",
+ "\n",
+ "#Result\n",
+ "print\"Self induced e.m.f is\", e,\"V\"\n",
+ "print\"The self-induced e.m.f. will act so as to oppose the growth of current.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Self induced e.m.f is -1.0 V\n",
+ "The self-induced e.m.f. will act so as to oppose the growth of current.\n"
+ ]
+ }
+ ],
+ "prompt_number": 89
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.16 Page no 510"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=1500 #turns/m\n",
+ "A=2*10**-4 #m**2\n",
+ "l=20 #A/s\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "e=-4*math.pi*n*A*l*10**-7\n",
+ "\n",
+ "#Result\n",
+ "print\"Induced e.m.f. is\",round(e*10**6,2)*10**-6,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Induced e.m.f. is -7.54e-06 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 97
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.17 Page no 511"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=50*10**-3 #V\n",
+ "a=8\n",
+ "b=4\n",
+ "t=0.5\n",
+ "\n",
+ "#Calculation\n",
+ "l=a-b\n",
+ "M=e*t/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Mutual inductance is\",M*10**3,\"*10**-3 H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mutual inductance is 6.25 *10**-3 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 101
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.18 Page no 511"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n1=5000 #turns/m\n",
+ "A=4*10**-4 #m**2\n",
+ "n2l=200\n",
+ "u=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "M=4*math.pi*u*n1*n2l*A\n",
+ "\n",
+ "#Result\n",
+ "print\"Mutual inductance is\", round(M*10**4,3),\"*10**-4 H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mutual inductance is 5.027 *10**-4 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 109
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.19 Page no 511"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=1200\n",
+ "A=12*10**-4\n",
+ "r=15*10**-2 #m\n",
+ "u=10**-7\n",
+ "I=1.0\n",
+ "N2=300\n",
+ "I1=0\n",
+ "I2=2\n",
+ "t=0.05\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "n=N/(2*math.pi*r)\n",
+ "B=4*math.pi*u*n\n",
+ "a=B*A*N\n",
+ "L=a/I\n",
+ "a1=B*A*N2\n",
+ "a11=a1*I1\n",
+ "a12=a1*I2\n",
+ "a13=a12-a11\n",
+ "e=-a13/t\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Self inductance is\", L*10**3,\"*10**-3 H\"\n",
+ "print\"(b) Induced e.m.f is\", round(e,3),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Self inductance is 2.304 *10**-3 H\n",
+ "(b) Induced e.m.f is -0.023 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 124
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.20 Page no 511"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=25*10**-4\n",
+ "N=500\n",
+ "l=30.0*10**-2 #m\n",
+ "I=2.5\n",
+ "u=10**-7\n",
+ "t=10.0**-3\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "n=N/l\n",
+ "B=4*math.pi*u*n*I\n",
+ "a=B*A*N\n",
+ "a1=0-a\n",
+ "e=-a1/t\n",
+ "\n",
+ "#Result\n",
+ "print\"Average induced e.m.f. produced is\", round(e,3),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average induced e.m.f. produced is 6.545 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 130
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.21 Page no 511"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=25.0\n",
+ "A=2*10**-4 #m**2\n",
+ "q=7.5*10**-3\n",
+ "R=0.50\n",
+ "\n",
+ "#Calculation\n",
+ "B=R*q/(N*A)\n",
+ "\n",
+ "#Result\n",
+ "print\"Field strength of the magnet is\" ,B,\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Field strength of the magnet is 0.75 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 136
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.22 Page no 512"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=0.50 #T\n",
+ "l=15*10**-2 #m\n",
+ "R=9.0*10**-3 #ohm\n",
+ "v=12*10**-2 #m/s\n",
+ "\n",
+ "#Calculation\n",
+ "e=B*v*l\n",
+ "F=B*l*(e/R)\n",
+ "P=F*v\n",
+ "P1=e**2/R\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Induced e.m.f is\", e*10**3,\"*10**3 V\"\n",
+ "print\"The end P of the rod will become positive and the end Q will become negative\"\n",
+ "print\"(b) on closing the switch, electrons collects at the end Q. Therefore , excess charge is built up i.e it doesn't open when switch is open\" \n",
+ "print\"(c) The magnetic lorentz force on electron is cancelled due to the electric field set up across the two end\"\n",
+ "print\"(d) Retarding force is\",F*10**2,\"*10**-2 N\"\n",
+ "print\"(e) Power is\",P*10**3,\"*10*-3 W\"\n",
+ "print\"(f) Dissipated power is\",P*10**3,\"*10**-3 W\"\n",
+ "print\"(g) The motion of the rod does not cut field lines,hence no induced e.m.f. is produced\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Induced e.m.f is 9.0 *10**3 V\n",
+ "The end P of the rod will become positive and the end Q will become negative\n",
+ "(b) on closing the switch, electrons collects at the end Q. Therefore , excess charge is built up i.e it doesn't open when switch is open\n",
+ "(c) The magnetic lorentz force on electron is cancelled due to the electric field set up across the two end\n",
+ "(d) Retarding force is 7.5 *10**-2 N\n",
+ "(e) Power is 9.0 *10*-3 W\n",
+ "(f) Dissipated power is 9.0 *10**-3 W\n",
+ "(g) The motion of the rod does not cut field lines,hence no induced e.m.f. is produced\n"
+ ]
+ }
+ ],
+ "prompt_number": 156
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.23 Page no 512"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a1=20*10**-2\n",
+ "a2=0.3*10**-2 #m\n",
+ "x=15*10**-2\n",
+ "I=2.0 #A\n",
+ "u=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B1=u*2*math.pi*I*a1**2/((a1**2+x**2)**1.5)\n",
+ "a=B1*math.pi*a2**2\n",
+ "M=a/I\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Flux is\", round(B1*10**6,3)*10**-6 ,\"T\"\n",
+ "print\"(b) Mutual inductance is\",round(M*10**11,3)*10**-11,\"H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Flux is 3.217e-06 T\n",
+ "(b) Mutual inductance is 4.548e-11 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 168
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.25 Page no 513"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N1=1500\n",
+ "l1=80.0*10**-2\n",
+ "l2=4*10**-2\n",
+ "r2=2*10**-2\n",
+ "I=3.0 #A\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A2=math.pi*r**2\n",
+ "n1=N1/l1\n",
+ "a=4*math.pi*10**-7*n1*I*4*math.pi*10**-4*100\n",
+ "M=a/I\n",
+ "\n",
+ "#Result\n",
+ "print\"Flux is\",round( a*10**4,3),\"*10**-4 Wb\"\n",
+ "print\"Mutual inductance is\",round(M*10**4,2),\"*10**-4 H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Flux is 8.883 *10**-4 Wb\n",
+ "Mutual inductance is 2.96 *10**-4 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 181
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap17_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap17_1.ipynb new file mode 100644 index 00000000..cb448cf8 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap17_1.ipynb @@ -0,0 +1,263 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:7c5cc6b58c18456d76e825a85df242d83f70c07f62b0e167355c9552ee60085c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 17 Transient Current"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.1 Page no 534"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=3 #V\n",
+ "L=2.5 #H\n",
+ "R=50.0 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "T=L/R\n",
+ "L1=E/L\n",
+ "I0=E/R\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Time constant is\", T,\"S\"\n",
+ "print\"(ii) Rate of change of current is\",L1,\"A/S\"\n",
+ "print\"(iii) Maximum current is\",I0,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Time constant is 0.05 S\n",
+ "(ii) Rate of change of current is 1.2 A/S\n",
+ "(iii) Maximum current is 0.06 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.2 Page no 534"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=1 #V\n",
+ "R=1 #ohm\n",
+ "L=1 #H\n",
+ "t=1 #S\n",
+ "a=2.3036\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=1-math.exp(-E/L)\n",
+ "t=a*math.log10(2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Current after 1 S is\", round(I,3),\"A\"\n",
+ "print\"(ii) Time to take the current to reach half its final value is\",round(t,3),\"S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Current after 1 S is 0.632 A\n",
+ "(ii) Time to take the current to reach half its final value is 0.693 S\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.3 Page no 534"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I0=2 #A\n",
+ "L=0.1 #H\n",
+ "R=20 #ohm\n",
+ "I=0.3 \n",
+ "a=3.333\n",
+ "b=200.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "t=2.3026*math.log10(a)/b\n",
+ "l=R*I*I0/L\n",
+ "\n",
+ "#Result\n",
+ "print\"Rate of change of current is\", l,\"A/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate of change of current is 120.0 A/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.4 Page no 535"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=2*10**6 #ohm\n",
+ "C=10**-6 #farad\n",
+ "q=0.8647\n",
+ "a=0.1353\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "t=-2*2.3026*math.log10(a)\n",
+ "\n",
+ "#Result\n",
+ "print\"Time is\", round(t,0),\"S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time is 4.0 S\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.5 Page no 535"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=1.443*10**-6 #F\n",
+ "t=60 #S\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=2.303*math.log10(2)\n",
+ "R=t/(C*a)\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of resistance is\", round(R*10**-6,0),\"*10**6 ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of resistance is 60.0 *10**6 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.6 Page no 535"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=200*10**-6 #H\n",
+ "f1=8*10**5\n",
+ "f2=12*10**-5 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "C1=1/(4*math.pi**2*f1**2*L)\n",
+ "C=1/(4*math.pi**2*f2**2*L)\n",
+ "\n",
+ "#Result\n",
+ "print\"Range of capacitor C is\",round(C*10**-8,0),\"pF\"\n",
+ "print\"Range of capacitor C1 is\",round(C1*10**12,0),\"pF\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Range of capacitor C is 88.0 pF\n",
+ "Range of capacitor C1 is 198.0 pF\n"
+ ]
+ }
+ ],
+ "prompt_number": 52
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap18_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap18_1.ipynb new file mode 100644 index 00000000..4d296c3d --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap18_1.ipynb @@ -0,0 +1,1298 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:ba29f9cf835e91dacf3ff239c18d4146206edd6f40b56615b0f524da3f5c713f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 18 Alternating current"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.2 Page no 555"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f1=100\n",
+ "f2=2.0\n",
+ "I0=50\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "f=f1/f2\n",
+ "Im=0.636*I0\n",
+ "Iv=0.707*I0\n",
+ "I=I0*math.sin(60*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Frequency of A.C applied is\",f,\"c.p.s\"\n",
+ "print\"(ii) Mean value of current is\",Im,\"A\"\n",
+ "print\"(iii) Virtual value of current is\",Iv,\"A\"\n",
+ "print\"(iv) Value of current is\",round(I,1),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Frequency of A.C applied is 50.0 c.p.s\n",
+ "(ii) Mean value of current is 31.8 A\n",
+ "(iii) Virtual value of current is 35.35 A\n",
+ "(iv) Value of current is 43.3 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.3 Page no 555"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=10.0 #ohm\n",
+ "E0=200\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Ev=E0/math.sqrt(2)\n",
+ "Iv=Ev/R\n",
+ "Pav=Ev*Iv\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) r.m.s value of voltage is\", E0,\"V\"\n",
+ "print\"(ii) r.m.s value of current is\",round(Iv,2),\"A\"\n",
+ "print\"(iii)Power dissipated is\",Pav,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) r.m.s value of voltage is 200 V\n",
+ "(ii) r.m.s value of current is 14.14 A\n",
+ "(iii)Power dissipated is 2000.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.4 Page no 555"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=1 #H\n",
+ "Ev=110 #V\n",
+ "f=70 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xl=2*3.14*f*L\n",
+ "I=Ev/Xl\n",
+ "I0=math.sqrt(2)*I\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Reactance is\", round(Xl,0)\n",
+ "print\"(b) Current through inductance is\",round(I,2),\"A\"\n",
+ "print\"(c) Peak value of current is\",round(I0,3),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Reactance is 440.0\n",
+ "(b) Current through inductance is 0.25 A\n",
+ "(c) Peak value of current is 0.354 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.5 Page no 555"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=2*10**-3\n",
+ "w=200 #rad/s\n",
+ "I0=0.2\n",
+ "\n",
+ "#Calculation\n",
+ "Xl=w*L\n",
+ "e=L*I0*w\n",
+ "I0=e/Xl\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum value of induced current is\", I0,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum value of induced current is 0.2 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.6 Page no 555"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=10*10**-6 #F\n",
+ "f=50 #cycles/s\n",
+ "Ev=110\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xc=1/(math.pi*2*f*C)\n",
+ "Iv=Ev/Xc\n",
+ "\n",
+ "#Result\n",
+ "print\"Virtual value of current is\", round(Iv,3),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Virtual value of current is 0.346 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.7 Page no 555"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=4\n",
+ "R=30 #ohm\n",
+ "Ev=200\n",
+ "f=50\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xl=2*math.pi*f*(L/math.pi)\n",
+ "Z=math.sqrt(R**2+Xl**2)\n",
+ "Iv=Ev/Z\n",
+ "\n",
+ "#Result\n",
+ "print\"current flowing in the circuit is\", round(Iv,3),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "current flowing in the circuit is 0.499 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.8 Page no 556"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=100\n",
+ "I=1.0\n",
+ "Iv=0.5\n",
+ "f=50\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "R=E/I\n",
+ "Z=E/Iv\n",
+ "Xl=math.sqrt(Z**2-R**2)\n",
+ "L=Xl/(2*math.pi*f)\n",
+ "\n",
+ "#Result\n",
+ "print\"Inductance of the coil is\", round(L,2),\"H\"\n",
+ "print\"Resistance is\",R,\"ohm\"\n",
+ "print\"Impedence is\",Z,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Inductance of the coil is 0.55 H\n",
+ "Resistance is 100.0 ohm\n",
+ "Impedence is 200.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.9 Page no 556"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=0.50 #H\n",
+ "R=100.0 #ohm\n",
+ "f=50 #Hz\n",
+ "Ev=240\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Iv=Ev/(math.sqrt(R**2+(2*math.pi*f*L)**2))\n",
+ "I0=math.sqrt(2)*Iv\n",
+ "a=2*math.pi*f*L/R\n",
+ "a1=math.atan(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Maximum current in the coil is\", round(I0,3),\"A\"\n",
+ "print\"(b) Phase difference is\",round(a1,1),\"degree (e.m.f. leads current)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Maximum current in the coil is 1.823 A\n",
+ "(b) Phase difference is 57.5 degree (e.m.f. leads current)\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.10 Page no 556"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ev=220\n",
+ "Iv=0.5\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "R=Ev/Iv\n",
+ "Xl=Ev/Iv\n",
+ "Z=math.sqrt(R**2+Xl**2)\n",
+ "Iv1=Ev/Z\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) X is a resistor of\",R,\"ohm\\n and Y is a inductor of\",Xl,\"ohm\"\n",
+ "print\"(b) Current in the circuit is\",round(Iv1,3),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) X is a resistor of 440.0 ohm\n",
+ " and Y is a inductor of 440.0 ohm\n",
+ "(b) Current in the circuit is 0.354 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.11 Page no 556"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Z=100 #ohm\n",
+ "a=45 #degree\n",
+ "f=1000\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xl=Z/math.sqrt(2)\n",
+ "L=Xl/(2*math.pi*f)\n",
+ "\n",
+ "#Result\n",
+ "print\"Self inductance of the coil is\", round(L*10**2,4),\"*10**-2 H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Self inductance of the coil is 1.1254 *10**-2 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.12 Page no 556"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=10 #ohm\n",
+ "Ev=220\n",
+ "f=50 #Hz\n",
+ "Iv=2.0 #A\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Z=Ev/Iv\n",
+ "Xc=math.sqrt(Z**2-R**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Reactance of the capacitor is\",round(Xc,2),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Reactance of the capacitor is 109.54 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.13 Page no 556"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=10\n",
+ "C=0.1*10**-6 #F\n",
+ "Ev=100 #V\n",
+ "f=50\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Z=math.sqrt(R**2+(1/(2*math.pi*f*C))**2)\n",
+ "Iv=Ev/Z\n",
+ "\n",
+ "#Result\n",
+ "print\"Current in the circuit is\", round(Iv*10**3,3),\"*10**-3 A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current in the circuit is 3.142 *10**-3 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.14 Page no 557"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=100*10**-6\n",
+ "R=40\n",
+ "Ev=110\n",
+ "f=60\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Iv=Ev/(math.sqrt(R**2+(1/(2*math.pi*f*C)**2)))\n",
+ "Iv1=math.sqrt(2)*Iv\n",
+ "a=1/(2*math.pi*f*C*R)\n",
+ "a1=math.atan(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Maximum current in the circuit is\", round(Iv1,2),\"A\"\n",
+ "print\"(b) Phase lag between the current maximum and voltage maximum is\",round(a1,2),\"degree (e.m.f. lags behind the current)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Maximum current in the circuit is 3.24 A\n",
+ "(b) Phase lag between the current maximum and voltage maximum is 33.57 degree (e.m.f. lags behind the current)\n"
+ ]
+ }
+ ],
+ "prompt_number": 62
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.15 page no 557"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=100*10**-6\n",
+ "R=50 #ohm\n",
+ "L=0.5 #H\n",
+ "Ev=110\n",
+ "f=50\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Z=math.sqrt(R**2+(2*math.pi*f*L-1/(2*math.pi*f*C))**2)\n",
+ "I0=Ev/Z\n",
+ "\n",
+ "#Result\n",
+ "print\"r.m.s value of current is\", round(I0,3),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "r.m.s value of current is 0.816 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.16 Page no 557"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=0.1 #H\n",
+ "C=25*10**-6\n",
+ "R=25.0\n",
+ "e=314\n",
+ "E0=310\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "f=e/(2*math.pi)\n",
+ "Xl=2*math.pi*f*L\n",
+ "Xc=1/(2*math.pi*f*C)\n",
+ "A=Xc-Xl\n",
+ "Z=math.sqrt(R**2+(Xc-Xl)**2)\n",
+ "Ev=E0/math.sqrt(2)\n",
+ "Iv=Ev/Z\n",
+ "a1=(Xc-Xl)/R\n",
+ "a2=math.atan(a1)*180/3.14\n",
+ "a3=a2*math.pi/180.0\n",
+ "V=Iv*Xc\n",
+ "V1=Iv*Xl\n",
+ "V2=Iv*R\n",
+ "L=1/(((2*math.pi*f)**2)*C)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The frequency of the e.m.f is\",round(f,0),\"cycle s**-1\" \n",
+ "print\"(b) The reactance of the circuit is\",round(A,0),\"ohm\"\n",
+ "print\"(c) The impedance of the circuit is\",round(Z,1),\"ohm\"\n",
+ "print\"(d) The current in the circuit is\",round(Iv,2),\"A\"\n",
+ "print\"(e) The phase angle of the current is\",round(a3,3),\"rad\"\n",
+ "print\"(f) The expression for the instantaneous value of the current is 3.125 cos(314t-1.316)\"\n",
+ "print\"(g) Effective voltage across the capacitor is\",round(V,1),\"V\"\n",
+ "print\" Effective Voltage across the inductor is\",round(V1,1),\"V\"\n",
+ "print\"Effective voltage across the resistor is\",round(V2,2),\"V\"\n",
+ "print\"(h) Value of inductance is\",round(L,3),\"H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The frequency of the e.m.f is 50.0 cycle s**-1\n",
+ "(b) The reactance of the circuit is 96.0 ohm\n",
+ "(c) The impedance of the circuit is 99.2 ohm\n",
+ "(d) The current in the circuit is 2.21 A\n",
+ "(e) The phase angle of the current is 1.317 rad\n",
+ "(f) The expression for the instantaneous value of the current is 3.125 cos(314t-1.316)\n",
+ "(g) Effective voltage across the capacitor is 281.5 V\n",
+ " Effective Voltage across the inductor is 69.4 V\n",
+ "Effective voltage across the resistor is 55.25 V\n",
+ "(h) Value of inductance is 0.406 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 113
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.17 Page no 558"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=50 #Hz\n",
+ "L=101.5*10**-3\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "C=1/((2*math.pi*f)**2*L)\n",
+ "\n",
+ "#Result\n",
+ "print\"Capacitance of the capacitor is\",round(C*10**6,0),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance of the capacitor is 100.0 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 117
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.18 Page no 558"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ev=230 #V\n",
+ "L=5 #H\n",
+ "C=80*10**-6\n",
+ "R=40.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "f=1/(math.pi*2*math.sqrt(L*C))\n",
+ "E0=math.sqrt(2)*Ev\n",
+ "A=E0/R\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Angular frequency is\", round(f,2),\"Hz\"\n",
+ "print\"(ii) Impedence of circuit is\",R,\"ohm\"\n",
+ "print\"(iii) Amplitude of the current is\",round(A,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Angular frequency is 7.96 Hz\n",
+ "(ii) Impedence of circuit is 40.0 ohm\n",
+ "(iii) Amplitude of the current is 8.13\n"
+ ]
+ }
+ ],
+ "prompt_number": 126
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.19 Page no 558"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=2*10**-6 #F\n",
+ "R=100 #ohm\n",
+ "L=8 #H\n",
+ "E=200 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "F=1/((2*math.pi*math.sqrt(L*C)))\n",
+ "D=2*math.pi*F*L\n",
+ "L1=E/R\n",
+ "\n",
+ "#Result\n",
+ "print\"The resonant frequency is\",round(F,2),\"Hz\"\n",
+ "print\"(i) The inductive and capacitive reactances of the circuit is\",D,\"ohm\"\n",
+ "print\"(ii) Total impedance of the circuit is 100\",\"ohm\"\n",
+ "print\"(iii) Peak value of current is\",L1,\"A\"\n",
+ "print\"(iv) The voltages across inductor and resistor differ in phase by\",\"90\"\n",
+ "print\"(v) The voltages across inductor and capacitor differ in phase by\",\"180\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resonant frequency is 39.79 Hz\n",
+ "(i) The inductive and capacitive reactances of the circuit is 2000.0 ohm\n",
+ "(ii) Total impedance of the circuit is 100 ohm\n",
+ "(iii) Peak value of current is 2 A\n",
+ "(iv) The voltages across inductor and resistor differ in phase by 90\n",
+ "(v) The voltages across inductor and capacitor differ in phase by 180\n"
+ ]
+ }
+ ],
+ "prompt_number": 144
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.20 Page no 558"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ev=220 #V\n",
+ "f=50 #Hz\n",
+ "R=100.0 #ohm\n",
+ "Vr=65 #V\n",
+ "Vc=415 #V\n",
+ "Vl=204 #V\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",
+ "C1=1/(4*math.pi**2*f**2*L)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The current in the circuit is\",Iv,\"A\"\n",
+ "print\"(ii) The value of the inductor is\",round(L,0),\"H\"\n",
+ "print\"(iii) The value of the capacitor C is\",round(C*10**6,1),\"micro F\"\n",
+ "print\"(iv) The value of C required to produce resonance is\",round(C1*10**6,1),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The current in the circuit is 0.65 A\n",
+ "(ii) The value of the inductor is 1.0 H\n",
+ "(iii) The value of the capacitor C is 5.0 micro F\n",
+ "(iv) The value of C required to produce resonance is 10.1 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 189
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.21 Page no 559"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=3 #H\n",
+ "C=27*10**-6\n",
+ "R=7.4 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "w0=1/(math.sqrt(L*C))\n",
+ "Q=1/R*(math.sqrt(L/C))\n",
+ "\n",
+ "#Result\n",
+ "print\"Resonant frequency is\", round(w0,1),\"rad/s\"\n",
+ "print\"Q factor is\",round(Q,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resonant frequency is 111.1 rad/s\n",
+ "Q factor is 45.05\n"
+ ]
+ }
+ ],
+ "prompt_number": 196
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.22 Page no 559"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=30 #ohm\n",
+ "Xl=40\n",
+ "E0=220\n",
+ "I0=1 \n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Z=math.sqrt(R**2+Xl**2)\n",
+ "a=R/Z\n",
+ "Pav=E0*I0*a/(math.sqrt(2)*math.sqrt(2))\n",
+ "\n",
+ "#Result\n",
+ "print\"Power consumed in the circuit is\", Pav,\"Watt\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power consumed in the circuit is 66.0 Watt\n"
+ ]
+ }
+ ],
+ "prompt_number": 201
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.23 Page no 559"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ev=100\n",
+ "f=50 #Hz\n",
+ "C=10*10**-6\n",
+ "R=100\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xc=1/(2*math.pi*f*C)\n",
+ "Iv=Ev/Xc\n",
+ "Pav=Ev*Iv*Ev/(math.sqrt(Ev**2+Xc**2))\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The reactance of the capacitor is\",round(Xc,2),\"ohm\"\n",
+ "print\"(b) Current flowing is\",round(Iv,3),\"A\"\n",
+ "print\"(c) Average power supplied is\",round(Pav,2),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The reactance of the capacitor is 318.31 ohm\n",
+ "(b) Current flowing is 0.314 A\n",
+ "(c) Average power supplied is 9.42 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 214
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.24 Page no 559"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=60.0\n",
+ "P=10\n",
+ "Ev=100\n",
+ "f=60 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=P/V\n",
+ "R=f/I\n",
+ "Z=Ev/I\n",
+ "L=math.sqrt(Z**2-R**2)/(2*math.pi*f)\n",
+ "R1=Z-R\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The inductance is\", round(L,3),\"henry\"\n",
+ "print\"(ii) Value of resistance is\",R1,\"ohm\"\n",
+ "print\"(iii) If resistance is used in the place of inductance, the electrical energy is wasted.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The inductance is 1.273 henry\n",
+ "(ii) Value of resistance is 240.0 ohm\n",
+ "(iii) If resistance is used in the place of inductance, the electrical energy is wasted.\n"
+ ]
+ }
+ ],
+ "prompt_number": 224
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.25 Page no 559"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=50.0 #V\n",
+ "P=20 #watt\n",
+ "Ev=250 #V\n",
+ "f=50\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=P/V\n",
+ "R=V/I\n",
+ "Z=Ev/I\n",
+ "C=1/(2*math.pi*f*Ev*math.sqrt(6))\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of capacitance required is\", round(C*10**6,3)*10**-6,\"F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of capacitance required is 5.198e-06 F\n"
+ ]
+ }
+ ],
+ "prompt_number": 233
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.26 Page no 560"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ev=200\n",
+ "f=50 #H\n",
+ "R=50\n",
+ "L=0.3\n",
+ "C=40*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xl=2*math.pi*f*L\n",
+ "Xc=1/(2*math.pi*f*C)\n",
+ "Z=math.sqrt(R**2+(Xl-Xc)**2)\n",
+ "Iv=Ev/Z\n",
+ "a=R/Z\n",
+ "Pav=Ev*Iv*a\n",
+ "\n",
+ "#Result\n",
+ "print\"Impedence in the circuit is\", round(Z,2),\"ohm\"\n",
+ "print\"Power in the circuit is\",round(Pav,1),\"watt\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Impedence in the circuit is 52.11 ohm\n",
+ "Power in the circuit is 736.6 watt\n"
+ ]
+ }
+ ],
+ "prompt_number": 244
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.27 Page no 560"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ev=200 #v\n",
+ "L=5 #H\n",
+ "C=80\n",
+ "R=40.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "W=10**3/(math.sqrt(L*C))\n",
+ "Iv=Ev/R\n",
+ "I0=math.sqrt(2)*Iv\n",
+ "Pav=Ev*L*math.cos(0*3.14/180.0)\n",
+ "\n",
+ "print\"(a) Angular frequency is\",W,\"rad s**-1\"\n",
+ "print\"(b) The current amplitude is\",round(I0,2),\"A\" \n",
+ "print\"(c) The power dissipation in the circuit is\",Pav,\"Watt\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Angular frequency is 50.0 rad s**-1\n",
+ "(b) The current amplitude is 7.07 A\n",
+ "(c) The power dissipation in the circuit is 1000.0 Watt\n"
+ ]
+ }
+ ],
+ "prompt_number": 272
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.28 Page no 560"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=5.0 #H\n",
+ "C=80*10**-6 #F\n",
+ "R=40.0 #ohm\n",
+ "Ev=230 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "D=1/math.sqrt(L*C)\n",
+ "Iv=Ev/R\n",
+ "I0=math.sqrt(2)*Iv\n",
+ "S=Iv*R\n",
+ "S1=Iv*D*L\n",
+ "S2=Iv/(1/D*C)\n",
+ "S3=Iv/(D*L-1/D*C)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The resonant angular frequency is\",D,\"rad s**-1\"\n",
+ "print\"(b) The impedance of the circuit and the amplitude of curremt is\",round(I0,2),\"A\"\n",
+ "print\"(c) The R.M.S. potential drop across LC is\",round(S3,0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The resonant angular frequency is 50.0 rad s**-1\n",
+ "(b) The impedance of the circuit and the amplitude of curremt is 8.13 A\n",
+ "(c) The R.M.S. potential drop across LC is 0.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 282
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.29 Page no 560"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=0.12 #H\n",
+ "C=480*10**-9 #F\n",
+ "R=23 #ohm\n",
+ "Ev=230 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "D=1/(math.sqrt(L*C))\n",
+ "Iv=Ev/(math.sqrt(R**2+(D*L-1/D*C))**2)\n",
+ "I0=(math.sqrt(2)*Ev)/(math.sqrt(R**2+(D*L-1/D*C))**2)\n",
+ "I1=(math.sqrt(2)*Ev)/R\n",
+ "EvIv=Ev*(I1/math.sqrt(2))\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The source frequency is\",round(D,1),\"rad s**-1\"\n",
+ "print\" The maximum value is\",round(I1,2),\"A\"\n",
+ "print\"(b) Average power will also be maximum at resonant frequency is\",round(D,1),\"rad s**-1\" \n",
+ "print\" The value of this maximum power is\",EvIv,\"Watt\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The source frequency is 4166.7 rad s**-1\n",
+ " The maximum value is 14.14 A\n",
+ "(b) Average power will also be maximum at resonant frequency is 4166.7 rad s**-1\n",
+ " The value of this maximum power is 2300.0 Watt\n"
+ ]
+ }
+ ],
+ "prompt_number": 298
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.30 Page no 561"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=80*10**-3 #H\n",
+ "C=60*10**-6 #F\n",
+ "Ev=230 #V\n",
+ "f=50 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Iv=-Ev/((2*math.pi*f*L)-(1/(2*math.pi*f*C)))\n",
+ "I0=math.sqrt(2)*Iv\n",
+ "Iv1=Iv*2*math.pi*f*L\n",
+ "Iv2=Iv*(1/(2*math.pi*f*C))\n",
+ "Pav=math.cos(90*3.14/180.0)*Ev\n",
+ "Pav1=math.cos(-90*3.14/180.0)*Ev\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The current amplitude is\",round(Iv,2),\"A\",\"and r.m.s. value is\",round(I0,2),\"A\"\n",
+ "print\"(b) The r.m.s. value of potential drops across L is\",round(Iv1,1),\"V\",\"and across C is\",round(Iv2,1),\"W\"\n",
+ "print\"(c) The average power transferred to the inductor is\",round(Pav,0)\n",
+ "print\"(d) The average power transferred to the capacitor is\",round(Pav1,0)\n",
+ "print\"(e) Total average power absorbed is zero\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The current amplitude is 8.24 A and r.m.s. value is 11.65 A\n",
+ "(b) The r.m.s. value of potential drops across L is 207.0 V and across C is 437.0 W\n",
+ "(c) The average power transferred to the inductor is 0.0\n",
+ "(d) The average power transferred to the capacitor is 0.0\n",
+ "(e) Total average power absorbed is zero\n"
+ ]
+ }
+ ],
+ "prompt_number": 357
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap19_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap19_1.ipynb new file mode 100644 index 00000000..6e89d5a2 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap19_1.ipynb @@ -0,0 +1,640 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:c81057d61d55f846d72bc0a6263de0452d74cd37dd94b021ce01d87a17a0c974"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 19 Electrical machines and devices"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.1 Page no 582"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ep=2300.0 #v\n",
+ "Np=4000\n",
+ "Es=230 #v\n",
+ "\n",
+ "#Calculation\n",
+ "Ns=Np*(Es/Ep)\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of turns is\", Ns"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of turns is 400.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.2 Page no 582"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ep=220.0 #v\n",
+ "Es=110\n",
+ "E=550 #watt\n",
+ "\n",
+ "#Calculation\n",
+ "Ip=E/Ep\n",
+ "\n",
+ "#Result\n",
+ "print\"Current drawn is\",Ip,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current drawn is 2.5 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.3 Page no 582"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Gien\n",
+ "Ep=220.0 #v\n",
+ "Es=22 \n",
+ "Z=220.0 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Is=Es/Z\n",
+ "Ip=(Es/Ep)*Is\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": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.4 Page no 582"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Np=200.0\n",
+ "Ns=1000\n",
+ "Ep=200.0 #v\n",
+ "E=10**4 #W\n",
+ "\n",
+ "#Calculation\n",
+ "Es=(Ns/Np)*Ep\n",
+ "Ip=E/Ep\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Output voltage is\",Es,\"V\"\n",
+ "print\"(b) Current is\",Ip,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Output voltage is 1000.0 V\n",
+ "(b) Current is 50.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.5 Page no 582"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ep=120\n",
+ "Ip=1.85 #A\n",
+ "Is=150*10**-3\n",
+ "n=0.95\n",
+ "\n",
+ "#Calculation\n",
+ "es=n*ep*Ip/Is\n",
+ "\n",
+ "#result\n",
+ "print\"Voltage across the secondary is\",es,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage across the secondary is 1406.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.6 Page no 582"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "K=100\n",
+ "Np=100\n",
+ "e=1100 #watt\n",
+ "ep=220.0\n",
+ "\n",
+ "#Calculation\n",
+ "Ns=K*Np\n",
+ "ip=e/ep\n",
+ "es=K*ep\n",
+ "Is=ip*ep/es\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Number of turns is\",Ns\n",
+ "print\"(ii) Current is\",ip,\"A\"\n",
+ "print\"(iii) Voltage across the secondary is\",es,\"V\"\n",
+ "print\"(iv) Current in the secondaryis\",Is,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Number of turns is 10000\n",
+ "(ii) Current is 5.0 A\n",
+ "(iii) Voltage across the secondary is 22000.0 V\n",
+ "(iv) Current in the secondaryis 0.05 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.7 Page no 583"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=11000 #W\n",
+ "es=220.0 #V\n",
+ "es1=22000.0\n",
+ "\n",
+ "#Calculation\n",
+ "Is=e/es\n",
+ "V=Is**2\n",
+ "Is1=e/es1\n",
+ "P=Is1**2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Electrical energy dissipated when power is transmitted at es=220 V is\",V,\"R watt\"\n",
+ "print\"(ii)Electrical energy dissipated when power is transmitted at es=22000 V is\",P,\"R watt\"\n",
+ "print\"Transmission should be done at 22000 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Electrical energy dissipated when power is transmitted at es=220 V is 2500.0 R watt\n",
+ "(ii)Electrical energy dissipated when power is transmitted at es=22000 V is 0.25 R watt\n",
+ "Transmission should be done at 22000 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.8 Page no 583"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=0.1 #T\n",
+ "n=2000\n",
+ "A=0.05 #m**2\n",
+ "w=2100 #r.p.m.\n",
+ "\n",
+ "#calculation\n",
+ "import math\n",
+ "A1=2*math.pi*w/60.0\n",
+ "e0=w*B*A*A1\n",
+ "e=e0*math.sin(60*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum e.m.f is\", round(e0,0),\"V\"\n",
+ "print\"Instantaneous e.m.f is\",round(e,0),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum e.m.f is 2309.0 V\n",
+ "Instantaneous e.m.f is 1999.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 50
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.9 Page no 583"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=100\n",
+ "A=3 #m**2\n",
+ "w=60 #rad/s\n",
+ "B=0.04\n",
+ "R=500.0\n",
+ "\n",
+ "#Calculation\n",
+ "e0=n*B*A*w\n",
+ "I0=e0/R\n",
+ "P=e0*I0\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Maximum current is\",I0,\"A\"\n",
+ "print\"(b) Maximum power is\",P,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Maximum current is 1.44 A\n",
+ "(b) Maximum power is 1036.8 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.10 Page no 583"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=50\n",
+ "A=2.5 #m**2\n",
+ "w=60 #rad/s\n",
+ "B=0.30\n",
+ "R=500.0\n",
+ "\n",
+ "#Calculation\n",
+ "e0=n*B*A*w\n",
+ "I0=e0/R\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The maximum current is\",I0,\"A\"\n",
+ "print\"(b) Current is zero when coil is vertical and flux is maximum.On the other hand current is maximum when coil is horizontal,flux is minimum\"\n",
+ "print\"(c) Yes,there should be relative motion between the magnetic field and coil\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The maximum current is 4.5 A\n",
+ "(b) Current is zero when coil is vertical and flux is maximum.On the other hand current is maximum when coil is horizontal,flux is minimum\n",
+ "(c) Yes,there should be relative motion between the magnetic field and coil\n"
+ ]
+ }
+ ],
+ "prompt_number": 60
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.11 Page no 583"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=220\n",
+ "e=200.0\n",
+ "P=5000\n",
+ "\n",
+ "#Calculation\n",
+ "I=P/e\n",
+ "R=(E-e)/I\n",
+ "\n",
+ "#Result\n",
+ "print\"Armature current is\",I,\"A\"\n",
+ "print\"Motor resistance is\",R,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Armature current is 25.0 A\n",
+ "Motor resistance is 0.8 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 65
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.12 Page no 583"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=8.5\n",
+ "E=200\n",
+ "I=5\n",
+ "\n",
+ "#Calculation\n",
+ "e=E-(I*R)\n",
+ "Pi=E*I\n",
+ "Po=E*I-(I**2*R)\n",
+ "n=Po/Pi\n",
+ "\n",
+ "#Result\n",
+ "print\"Back e.m.f of the motor is\", e,\"V\"\n",
+ "print\"Power input is\",Pi,\"W\"\n",
+ "print\"Power output is\",Po,\"W\"\n",
+ "print\"Efficiencyis\",n*10**2,\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Back e.m.f of the motor is 157.5 V\n",
+ "Power input is 1000 W\n",
+ "Power output is 787.5 W\n",
+ "Efficiencyis 78.75 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.13 Page no 584"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=300\n",
+ "g=9.8 #m/s**2\n",
+ "a=1000\n",
+ "A=100.0\n",
+ "n=60\n",
+ "\n",
+ "#Calculation\n",
+ "m=A*a\n",
+ "E=m*g*h\n",
+ "P=E*n/A\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric power is\", P*10**-6,\"MW\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric power is 176.4 MW\n"
+ ]
+ }
+ ],
+ "prompt_number": 84
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.14 Page no 584"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=30 #km\n",
+ "R=15 #ohm\n",
+ "V=4000.0 #V\n",
+ "P=8*10**5 #W\n",
+ "\n",
+ "#Calculation\n",
+ "Is=P/V\n",
+ "P1=Is**2*R\n",
+ "P2=P+P1\n",
+ "V1=R*Is\n",
+ "A=440-(V+V1)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Line power loss is\",P1*10**-3,\"KW\"\n",
+ "print\"(b) Supplied power is\",P2*10**-3,\"KW\"\n",
+ "print\"(c) Step up transformer at the plane is\",A"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Line power loss is 600.0 KW\n",
+ "(b) Supplied power is 1400.0 KW\n",
+ "(c) Step up transformer at the plane is -6560.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 94
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.15 Page no 584"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=8*10**-2\n",
+ "n=20\n",
+ "w=50 #rad/s\n",
+ "B=3*10**-2 #T\n",
+ "A=64*math.pi*10**-4 #m**2\n",
+ "Eav=0\n",
+ "R=10 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "math.pi\n",
+ "e0=n*B*A*w\n",
+ "Pav=e0**2/(2*R)\n",
+ "\n",
+ "#Result\n",
+ "print\" Maximum e.m.f. is\",round(e0,4),\"V\"\n",
+ "print\" Average e.m.f. is zero\"\n",
+ "print\" Dissipated power is\",round(Pav,4),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Maximum e.m.f. is 0.6032 V\n",
+ " Average e.m.f. is zero\n",
+ " Dissipated power is 0.0182 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 103
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap1_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap1_1.ipynb new file mode 100644 index 00000000..36a74a3e --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap1_1.ipynb @@ -0,0 +1,462 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:1a5424f1a289fa2ea02065679fbda8bfa43c7eec69070520139086e97a973104"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 1 Coulombs Law"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.1 Page no 12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=-3*10**-7 #C\n",
+ "e=-1.6*10**-19 #C\n",
+ "\n",
+ "#Calculation\n",
+ "n=q/e\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of electrons transferred from wool to polythene is\", n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of electrons transferred from wool to polythene is 1.875e+12\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.2 Page no 12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=3.11 #g\n",
+ "Z=29\n",
+ "A=63.5 \n",
+ "N=6.023*10**23\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "n=(N*m)/A\n",
+ "n1=n*Z\n",
+ "q=n1*e\n",
+ "\n",
+ "#Result\n",
+ "print\"Total positive or negative charge is\", round(q*10**-5,2),\"*10**5 C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total positive or negative charge is 1.37 *10**5 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.3 Page no 12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q1=2*10**-7\n",
+ "q2=3*10**-7\n",
+ "r=0.3 #m\n",
+ "a=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "F=(a*q1*q2)/r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Force between two small charged spheres is\", F*10**3,\"*10**-3\",\"N(repulsive\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force between two small charged spheres is 6.0 *10**-3 N(repulsive\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.4 page no. 12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "F=3.7*10**-9 #N\n",
+ "r=5*10**-10 #m\n",
+ "a=9*10**9\n",
+ "q1=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "n=math.sqrt(F*r**2/(a*q1**2))\n",
+ "\n",
+ "#Result\n",
+ "print\"number of electrons is\", round(n,0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "number of electrons is 2.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.5 Page no 12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q1=0.4*10**-6 #C\n",
+ "q2=0.8*10**-6 #C\n",
+ "F12=0.2 #N\n",
+ "a=9.0*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=math.sqrt((a*q1*q2)/F12)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Distance between two spheres is\", r,\"m\"\n",
+ "print\"(b) Force on charge q2 due to q1 is\",F12,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Distance between two spheres is 0.12 m\n",
+ "(b) Force on charge q2 due to q1 is 0.2 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.6 Page no 13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q1=5*10**-8 #C\n",
+ "m1=8*10**-3 #Kg\n",
+ "a=9*10**9\n",
+ "r=0.05 #m\n",
+ "\n",
+ "#Calculation\n",
+ "q2=m1*9.8*r**2/(a*q1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Charge q2 is\", round(q2*10**7,2)*10**-7,\"C(positive)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge q2 is 4.36e-07 C(positive)\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.7 Page no 13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q1=6.5*10**-7 #C\n",
+ "q2=6.5*10**-7\n",
+ "r=0.5 #m\n",
+ "a=9*10**9\n",
+ "K=80.0\n",
+ "\n",
+ "#Calculation\n",
+ "Fair=a*q1*q2/r**2\n",
+ "r1=0.5/2.0\n",
+ "F1=a*4*q1*q2/r1**2\n",
+ "Fwater=Fair/K\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Mutual force of electrostatic repulsion is\", Fair*10**2,\"*10**-2 N\"\n",
+ "print\"(b) (i) Force of repulsion is\", round(F1,4),\"N\"\n",
+ "print \"(ii) Force of repulsion is\",round(Fwater*10**4,1),\"*10**-4 N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Mutual force of electrostatic repulsion is 1.521 *10**-2 N\n",
+ "(b) (i) Force of repulsion is 0.2434 N\n",
+ "(ii) Force of repulsion is 1.9 *10**-4 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 57
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.8 Page no 13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q1=6.5*10**-7 #C\n",
+ "r=0.05 #m\n",
+ "a=9*10**9\n",
+ "r1=0.5\n",
+ "\n",
+ "#Calculation\n",
+ "q11=q1/2.0\n",
+ "q21=(q1+q11)/2.0\n",
+ "F=(a*q11*q21)/r1**2\n",
+ "\n",
+ "#Result\n",
+ "print\"New force of repulsion between A and B is\", round(F*10**3,3),\"*10**-3 N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "New force of repulsion between A and B is 5.704 *10**-3 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 64
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.10 Page no 14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=9.0*10**9\n",
+ "r=0.2\n",
+ "m=9.8*10**-3\n",
+ "a1=0.1\n",
+ "a2=0.5\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a11=m*(a1/(math.sqrt(a2**2-a1**2)))\n",
+ "q=math.sqrt((a11*r**2)/a)\n",
+ "\n",
+ "#Result\n",
+ "print\"Charge on each ball is\", round(q*10**8,2)*10**-6,\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge on each ball is 9.43e-06 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 100
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.12 Page no 14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "qa=10**-5 #C\n",
+ "qb=5*10**-6 #C\n",
+ "qc=-5*10**-6 #C\n",
+ "r=0.1 #m\n",
+ "a=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Fab=(a*qa*qb)/r**2\n",
+ "Fac=Fab\n",
+ "F=math.sqrt(Fab**2+Fac**2+(2*Fab*Fac*math.cos(120*3.14/180.0)))\n",
+ "\n",
+ "#Result\n",
+ "print\"Resultant force is\", round(F,0),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resultant force is 45.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 75
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.13 Page no 15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "qa=1\n",
+ "qb=100\n",
+ "ab=10\n",
+ "a=9*10**9\n",
+ "qd=75\n",
+ "a1=5\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Fab=(a*qa*qb)/ab**2\n",
+ "Fac=Fab\n",
+ "Fac1=(a*qa*qd)/(ab**2-a1**2)\n",
+ "Fx=Fab*math.cos(60*3.14/180.0)+Fac1*math.cos(60*3.14/180.0)\n",
+ "Fy=Fac\n",
+ "F=math.sqrt(Fx**2+Fy**2)\n",
+ "B=Fy/Fx\n",
+ "B1=math.atan(B)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Resultant force on charge qa is inclined at\", round(B1,0),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resultant force on charge qa is inclined at 45.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 90
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap20_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap20_1.ipynb new file mode 100644 index 00000000..b99b0934 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap20_1.ipynb @@ -0,0 +1,255 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:17722fc50855830105e8bbc1d5e7f4e9c2450186689d4bcffcae55541cf58a10"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 20 Electromagnetic Waves"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.1 Page no 618"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=5.5*10**-7 #m\n",
+ "c=3*10**8 #m s**-1\n",
+ "u=1.5\n",
+ "\n",
+ "#Calculation\n",
+ "V=c/L\n",
+ "T=L/c\n",
+ "v=c/u\n",
+ "L1=v*T\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The frequency is\",round(V*10**-14,3),\"10**8\",\"MHz\",\"and time period is\",round(T*10**15,1),\"10**-19\",\"micro s\"\n",
+ "print\"(b) The wavelenght in glass is\",round(L1*10**7,1),\"10**-7\",\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The frequency is 5.455 10**8 MHz and time period is 1.8 10**-19 micro s\n",
+ "(b) The wavelenght in glass is 3.7 10**-7 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.2 Page no 619"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=2*10**8 #m s**-1\n",
+ "Ur=1.0\n",
+ "C=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "Er=(C**2)/(V**2*Ur)\n",
+ "\n",
+ "#Result \n",
+ "print\" The relative permittivity is\",Er"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The relative permittivity is 2.25\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.3 Page no 619"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=3*10**8\n",
+ "V=10**8\n",
+ "E0=0.5\n",
+ "\n",
+ "#Calculation\n",
+ "L=C/V\n",
+ "Bz=E0/C\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) There is the positive direction of x-axis of propagation of electromagnetic waves\"\n",
+ "print\"(b) The wavelengh of the wave is\",L,\"m\"\n",
+ "print\"(c) The component of associated magnetic field is\",round(Bz*10**9,0),\"cos(2*math.pi*10**8(t-x/c))\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) There is the positive direction of x-axis of propagation of electromagnetic waves\n",
+ "(b) The wavelengh of the wave is 3 m\n",
+ "(c) The component of associated magnetic field is 2.0 cos(2*math.pi*10**8(t-x/c))\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.4 Page no 619"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=6*10**-3 #m\n",
+ "E0=33 #V m**-1\n",
+ "C=3.0*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "W=(2*math.pi*C)/(L)\n",
+ "B=E0/C\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic field is\",B,\"sin*math.pi*10**11(t-x/c)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic field is 1.1e-07 sin*math.pi*10**11(t-x/c)\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.5 Page no 619"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=8\n",
+ "E0=8.85*10**-12\n",
+ "C=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "e0=math.sqrt((2*I)/(E0*C))\n",
+ "\n",
+ "#Result\n",
+ "print\"The amplitude of the electric field is\",round(e0,1),\"N C**-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The amplitude of the electric field is 77.6 N C**-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.6 Page no 619"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=0.024 #m\n",
+ "k=5.9*10**-9\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "V=L/k\n",
+ "t=L/c\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The number of oscillation is\",round(V*10**-6,3),\"10**6\",\"Hz\"\n",
+ "print\"(ii) The coherence time is\",t,\"s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The number of oscillation is 4.068 10**6 Hz\n",
+ "(ii) The coherence time is 8e-11 s\n"
+ ]
+ }
+ ],
+ "prompt_number": 65
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap21_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap21_1.ipynb new file mode 100644 index 00000000..54cafbe9 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap21_1.ipynb @@ -0,0 +1,1809 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:07c4b928a455230c9db333bb0062ec9efe3c0f7899f2ff1857d420b5212b1285"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 21 Refraction Of Light"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.1 Page no 650"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "U=1.5\n",
+ "c=3.0*10**8 #m s**-1\n",
+ "\n",
+ "#Calculation\n",
+ "v=c/U\n",
+ "\n",
+ "#Result\n",
+ "print\" Speed of light in glass is\",v*10**-8,\"10**8\",\"m s**-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Speed of light in glass is 2.0 10**8 m s**-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.2 Page no 650"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.33\n",
+ "Lu=589*10**-9 #m\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "v=c/Lu\n",
+ "v1=c/u\n",
+ "Lw=v1/v\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Wavelenght of light after reflection is\",Lu*10**9,\"10**-9\"\n",
+ "print\" Velocity of light after reflection is\",c*10**-8,\"10**8\",\"m s**-1\"\n",
+ "print\" Frqequency of light after reflection is\",round(v*10**-14,4),\"10**14\",\"Hz\"\n",
+ "print\"(b) Frequency of light after refraction is\",round(v*10**-14,4),\"10**14\",\"Hz\"\n",
+ "print\" Velocity of light after refraction is\",round(v1*10**-8,3),\"10**8\",\"m s**-1\"\n",
+ "print\" Wavelength of light after refraction is\",round(Lw*10**9,2),\"nm\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Wavelenght of light after reflection is 589.0 10**-9\n",
+ " Velocity of light after reflection is 3.0 10**8 m s**-1\n",
+ " Frqequency of light after reflection is 5.0934 10**14 Hz\n",
+ "(b) Frequency of light after refraction is 5.0934 10**14 Hz\n",
+ " Velocity of light after refraction is 2.256 10**8 m s**-1\n",
+ " Wavelength of light after refraction is 442.86 nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.3 Page no 651"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "i=60\n",
+ "u=1.5\n",
+ "t=0.1 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=(math.sin(60*3.14/180.0))/u\n",
+ "a=math.asin(r)*180/3.14\n",
+ "d=t/math.cos(r*3.14/180.0)*(math.sin(24*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"The lateral shift produced is\",round(d,4),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The lateral shift produced is 0.0407 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.4 Page no 651"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=4/3.0\n",
+ "A1=3/2.0\n",
+ "I=30\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "S=A1/A\n",
+ "R=math.sin(30*3.16/180.0)/S\n",
+ "A=math.asin(R)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"The angle of refraction is\",round(A,0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The angle of refraction is 27.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 59
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.5 Page no 651"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "t=15\n",
+ "u=1.5\n",
+ "\n",
+ "#Calculation\n",
+ "d=t*(1-(1/u))\n",
+ "\n",
+ "#Result\n",
+ "print\" Real thickness of glass slab is\",d,\"cm\"\n",
+ "print\" The answer does not depend upon the location of the slab\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Real thickness of glass slab is 5.0 cm\n",
+ " The answer does not depend upon the location of the slab\n"
+ ]
+ }
+ ],
+ "prompt_number": 67
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.6 Page no 651"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "c=3*10**8 #m s**-1\n",
+ "v=2.0*10**8 #m s**-1\n",
+ "t=6.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "u=c/v\n",
+ "d=t*(1-(1/u))\n",
+ "\n",
+ "#Result\n",
+ "print\"An ink dot appear to be rasied\",d,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "An ink dot appear to be rasied 2.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 73
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.7 Page no 651"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=12.5 #cm\n",
+ "a=9.4 #cm\n",
+ "u=1.63\n",
+ "\n",
+ "#Calculation\n",
+ "S=d/a\n",
+ "S1=d/u\n",
+ "S3=a-S1\n",
+ "\n",
+ "#Result\n",
+ "print\"The refractive index of water is\",round(S,2)\n",
+ "print\"The refractive index of liquid is\",round(S1,2),\"cm\"\n",
+ "print\"Distance is\",round(S3,2),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The refractive index of water is 1.33\n",
+ "The refractive index of liquid is 7.67 cm\n",
+ "Distance is 1.73 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 84
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.8 Page no 651"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=4/3.0\n",
+ "d=0.015 #m\n",
+ "\n",
+ "#Calculatiom\n",
+ "t=d/(1-(1/A))\n",
+ "\n",
+ "#Result\n",
+ "print\"The height upto which water must be poured into the beaker is\",t,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The height upto which water must be poured into the beaker is 0.06 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 91
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.9 Page no 652"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=3/2.0\n",
+ "A1=4/3.0\n",
+ "t1=6\n",
+ "t2=4\n",
+ "\n",
+ "#Calculation\n",
+ "d1=t1*(1-(1/A))\n",
+ "d2=t2*(1-(1/A1))\n",
+ "d3=d1+d2\n",
+ "\n",
+ "#Result\n",
+ "print\"The apparent position of an object is\",d3,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The apparent position of an object is 3.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 102
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.10 Page no 652"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "S=1.5\n",
+ "W=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=S/W\n",
+ "C=1/A\n",
+ "Q=math.asin(C)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"The critical angle for a glass water interface is\",round(Q,0),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The critical angle for a glass water interface is 62.0 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 124
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.11 Page no 652"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=40 #degree\n",
+ "A=15 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=I-A\n",
+ "u=(math.sin(40*3.14/180.0))/(math.sin(25*3.14/180.0))\n",
+ "c=1/u\n",
+ "A=math.asin(c)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"The critical angle is\",round(A,1),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The critical angle is 41.1 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 123
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.12 Page no 652"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.5\n",
+ "h=20\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "C=1/u\n",
+ "A=math.asin(C)*180/3.14\n",
+ "r=h*math.tan(A*3.14/180.0)\n",
+ "D=math.pi*r**2\n",
+ "\n",
+ "#Result \n",
+ "print\"The required surface area is\",round(D,1),\"cm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The required surface area is 1005.3 cm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 138
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.13 Page no 653"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=60 #degree\n",
+ "B=1.45\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=1/(math.sin(A*3.14/180.0))\n",
+ "Z=B/I\n",
+ "\n",
+ "#Result\n",
+ "print\"The refractive index of the liquid is\",round(Z,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The refractive index of the liquid is 1.255\n"
+ ]
+ }
+ ],
+ "prompt_number": 143
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.14 Page no 653"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=1.68\n",
+ "A1=1.44\n",
+ "A2=90 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "K=A/A1\n",
+ "C=1/K\n",
+ "Q=math.asin(C)*180/3.14\n",
+ "r=A2-Q\n",
+ "I=A*math.sin(r*3.14/180.0)\n",
+ "Q1=math.asin(I)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Refraction index of glass fibre is\",round(Q1,0),\"degree\"\n",
+ "print\"All rays having angle of incidence between 0 degree to 60 degree will suffer total internal reflection\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Refraction index of glass fibre is 60.0 degree\n",
+ "All rays having angle of incidence between 0 degree to 60 degree will suffer total internal reflection\n"
+ ]
+ }
+ ],
+ "prompt_number": 161
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.15 Page no 653"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u1=1\n",
+ "u2=1.5\n",
+ "u=-10.0 #cm\n",
+ "v=-40.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "R=(u2-u1)/((-u1/u)+(u2/v))\n",
+ "\n",
+ "#Result\n",
+ "print\"Radius of curvature is\", R,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of curvature is 8.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 178
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.16 Page no 653"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-60.0 #cm\n",
+ "R=25.0 #cm\n",
+ "u1=1\n",
+ "u2=1.5\n",
+ "R1=0.25\n",
+ "\n",
+ "#Calculation\n",
+ "v=u2/(((u2-u1)/R)+(u1/u))\n",
+ "P=(u2-u1)/R1\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image is\", v,\"cm\"\n",
+ "print\"Power of 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 refracting surface is 2.0 dioptre\n"
+ ]
+ }
+ ],
+ "prompt_number": 191
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.17 Page no 654 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "U1=1.5\n",
+ "U2=1\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "x=(U1+U2)/(U1-U2)\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": 194
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.18 Page no 654"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u1=1\n",
+ "u2=1.5\n",
+ "u=-10.0 #cm\n",
+ "R=-5.0 \n",
+ "\n",
+ "#Calculation\n",
+ "v=u1/((u1-u2)/R+(u2/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 -20.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 196
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.19 Page no 654"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.5\n",
+ "R=-5.0 #cm\n",
+ "OC=2\n",
+ "\n",
+ "#Calculation\n",
+ "u1=R+OC\n",
+ "v=1/((1-u)/R+(u/u1))\n",
+ "\n",
+ "#Result\n",
+ "print\"Apparent position of the bubble is\", v,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Apparent position of the bubble is -2.5 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 202
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.20 Page no 654"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=-0.02 #m\n",
+ "u=1.54\n",
+ "v=-0.01 #m\n",
+ "\n",
+ "#calculation\n",
+ "u1=-u/((1-u)/R-(1/v))\n",
+ "\n",
+ "#Result\n",
+ "print\"Real depth of bubble is\", round(u1,4),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Real depth of bubble is -0.0121 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 206
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.21 Page no 655"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=7.5 #cm\n",
+ "u=4/3.0\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/((1-u)/R)\n",
+ "\n",
+ "#Result\n",
+ "print\"It get focussed at\",v,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "It get focussed at -22.5 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 213
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.22 Page no 655"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "u=1.55\n",
+ "f=20 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "R=(u-1)*2*f\n",
+ "\n",
+ "#Result\n",
+ "print\"Radius of curvature is,\",R,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of curvature is, 22.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 215
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.23 Page no 655"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.5\n",
+ "f=0.3\n",
+ "\n",
+ "#Calculation\n",
+ "R=(u-1)*f\n",
+ "\n",
+ "#Result\n",
+ "print\"Radius of curvature is\", R,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of curvature is 0.15 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 218
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.24 Page no 655"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=2.0\n",
+ "\n",
+ "#Calculation\n",
+ "u=(a+1)/a\n",
+ "\n",
+ "#Result\n",
+ "print\"Refractive index is\",u"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Refractive index is 1.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 219
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.25 Page no 655"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ug=1.5\n",
+ "uc=1.63\n",
+ "R1=20.0 #cm\n",
+ "R2=-20.0\n",
+ "\n",
+ "#Calculation\n",
+ "fair=1/((ug-1)*(1/R1-1/R2))\n",
+ "ug1=ug/uc\n",
+ "fc=1/((ug1-1)*(1/R1-1/R2))\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length of the lens is\",round(fc,1),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length of the lens is -125.4 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 225
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.26 Page no 656"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "fair=0.2 #m\n",
+ "ug=1.50\n",
+ "uw=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "R=1/(fair*(ug-1))\n",
+ "ug1=ug/uw\n",
+ "fw=1/((ug1-1)*R)\n",
+ "f=fw-fair\n",
+ "\n",
+ "#Result\n",
+ "print\"Change in focal length is\", round(f,2),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Change in focal length is 0.58 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 234
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.27 Page no 656"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u1=1.2\n",
+ "u2=1.3\n",
+ "a=13.0\n",
+ "\n",
+ "#Calculation\n",
+ "u=u1/u2\n",
+ "f=1/((-1/a)*(1/(u1-1)))\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length is\", f,\"f\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length is -2.6 f\n"
+ ]
+ }
+ ],
+ "prompt_number": 237
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.28 Page no 656"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=20.0 #cm\\\n",
+ "u=12.0\n",
+ "f1=-20.0\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/(1/f+1/u)\n",
+ "v1=1/(1/f1+1/u)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Beam will converge at a point of distant\",v,\"cm\"\n",
+ "print\"(b) Beam will converge at a point of distant\",v1,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Beam will converge at a point of distant 7.5 cm\n",
+ "(b) Beam will converge at a point of distant 30.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 244
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.29 Page no 656"
+ ]
+ },
+ {
+ "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\"The position of the point is\",u,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The position of the point is 0.12 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 249
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.30 Page no 656"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=10.0 #cm\n",
+ "c=20 #cm\n",
+ "o=5.0\n",
+ "\n",
+ "#Calculation\n",
+ "u=-(c-o)\n",
+ "u1=-(c+o)\n",
+ "v1=1/((1/f)+(1/u))\n",
+ "v2=1/((1/f)+(1/u1))\n",
+ "v=v1-v2\n",
+ "\n",
+ "#Result\n",
+ "print\"Length of image of needle is\", round(v,2),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Length of image of needle is 13.33 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 273
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.31 Page no 657"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-30.0\n",
+ "v=20.0\n",
+ "R1=10.0\n",
+ "R2=-15.0\n",
+ "ug1=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/(-1/u+1/v)\n",
+ "ug=(1/f+1/6.0)*6.0\n",
+ "fw=1/(((ug/ug1)-1)*(1/R1-1/R2))\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length of the lens is\", f,\"cm\"\n",
+ "print\"Focal length of the lens is\",round(fw,2),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length of the lens is 12.0 cm\n",
+ "Focal length of the lens is 46.94 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 289
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.32 Page no 657"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=0.12\n",
+ "m=-3.0\n",
+ "\n",
+ "#Calculation\n",
+ "u=(f/m)-f\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance between object and lens is\", u,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance between object and lens is -0.16 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 292
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.33 Page no 657"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=-19.0\n",
+ "v=10\n",
+ "\n",
+ "#Calculation\n",
+ "f=v/(-m+1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length of the lens is\",f,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length of the lens is 0.5 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 296
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.34 Page no 657"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=4\n",
+ "f=20\n",
+ "\n",
+ "#Calculation\n",
+ "u=m+1-f\n",
+ "v=f-(m*f)\n",
+ "\n",
+ "#Result\n",
+ "print\"Object is at\", u,\"cm\"\n",
+ "print\"Image is at\",v,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Object is at -15 cm\n",
+ "Image is at -60 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 301
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.35 Page no 657"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "O=5 #cm\n",
+ "u=-45.0 #cm\n",
+ "v=90.0\n",
+ "\n",
+ "#Calculation\n",
+ "f=-1/(1/u-1/v)\n",
+ "I=v*O/u\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length is\", f,\"cm\"\n",
+ "print\"size of the image is\",I,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length is 30.0 cm\n",
+ "size of the image is -10.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 309
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.36 Page no 658"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "O=3 #cm\n",
+ "u=-14.0\n",
+ "f=-21.0\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/(1/f+1/u)\n",
+ "I=O*v/u\n",
+ "\n",
+ "#Result\n",
+ "print\"Image produced by the lens is\",I,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Image produced by the lens is 1.8 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.37 Page no 658"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=0.20\n",
+ "R2=-0.2\n",
+ "u=1.5\n",
+ "u1=1.25\n",
+ "\n",
+ "#Calculation\n",
+ "P1=(u-1)*(1/R1-1/R2)\n",
+ "u2=u/u1\n",
+ "P2=(u2-1)*(-1/R2-1/R2)\n",
+ "P=P1/P2\n",
+ "\n",
+ "#Result\n",
+ "print\"Ratio of power of lens is\", P"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Ratio of power of lens is 2.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.38 Page no 658"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f1=15.0 #cm\n",
+ "f2=30.0\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/(1/f1+1/f2)\n",
+ "P=1/f\n",
+ "\n",
+ "#Result\n",
+ "print\"Power is\",f*10**-2,\"m\"\n",
+ "print\"Focal length is\",P*10**2,\"D\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power is 0.1 m\n",
+ "Focal length is 10.0 D\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.39 Page no 658"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P1=-1.5 #D\n",
+ "P2=2.75\n",
+ "\n",
+ "#Calculation\n",
+ "P=P1+P2\n",
+ "f=1/P\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length is\",f,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length is 0.8 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.40 Page no 659"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=0.11\n",
+ "a1=0.07\n",
+ "\n",
+ "#Calculation\n",
+ "x=a-a1\n",
+ "u=(a1-0.01)/x\n",
+ "\n",
+ "#Result\n",
+ "print\"Refractive index is\",u"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Refractive index is 1.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.41 Page no 659"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=70 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=1/(math.sin(a*3.14/180.0))\n",
+ "a1=math.atan(a)*180/3.14\n",
+ "u=1/math.sin(a1*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Critical angle is\", round(u,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Critical angle is 1.372\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.42 Page no 659"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=180 #degree\n",
+ "b=90\n",
+ "c=45\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=a-(b+c)\n",
+ "A=1.352*math.sin(r*3.14/180.0)\n",
+ "A1=math.asin(A)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Angle of incidence is sin-1[(math.sqrt(u**2-u1**2)-u1)/math.sqrt(2)]\"\n",
+ "print\"(b) Angle of incidence at face AB is\",round(A1,0),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Angle of incidence is sin-1[(math.sqrt(u**2-u1**2)-u1)/math.sqrt(2)]\n",
+ "(b) Angle of incidence at face AB is 73.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.43 Page no 660"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=40.0 #cm\n",
+ "R=10.0\n",
+ "u1=1\n",
+ "u2=1.5\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "v=u2/(((u2-u1)/R)+(u1/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 20.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.44 Page no 660"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "x=0.04\n",
+ "y=8/300.0\n",
+ "u=-0.04\n",
+ "v=-16/500.0\n",
+ "\n",
+ "#Calculation\n",
+ "U=x/y\n",
+ "R=(1-U)/(-U/u+1/v)\n",
+ "f=1/((U-1)/R)\n",
+ "\n",
+ "print\"(i) Refractive index is\",U\n",
+ "print\"(ii) Radius of curvature is\",R,\"m\"\n",
+ "print\"(iii) Focal length is\",f,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Refractive index is 1.5\n",
+ "(ii) Radius of curvature is -0.08 m\n",
+ "(iii) Focal length is -0.16 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.46 Page no 661"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "OI=90.0 #cm\n",
+ "O1O2=20\n",
+ "x=35 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=(OI-x)*x/((OI-x)+x)\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length of the lens is\", round(f,1),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length of the lens is 21.4 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 71
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.47 Page no 662"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=-4\n",
+ "a=1.5\n",
+ "u=0.3\n",
+ "\n",
+ "#Calculation\n",
+ "x=a/(-m+1)\n",
+ "v=a-u\n",
+ "f=1/(1/u+1/v)\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length of the lens is\", f,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length of the lens is 0.24 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 61
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.48 Page no 662"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P1=8.0 #dioptre\n",
+ "P2=2.0\n",
+ "\n",
+ "#Calculation\n",
+ "f1=1/P1\n",
+ "f2=1/P2\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length of lens 1 is\",f1,\"m\"\n",
+ "print\"Focal length of lens 2 is\",f2,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length of lens 1 is 0.125 m\n",
+ "Focal length of lens 2 is 0.5 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap22_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap22_1.ipynb new file mode 100644 index 00000000..60d6e4e2 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap22_1.ipynb @@ -0,0 +1,531 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:168ff39597791c08930937d1b2aac6af790d1f198c96130c4b12669d6560fe5b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 22 Dispersion"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.1 Page no 686"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=60 #Degree\n",
+ "i=49\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=A/2.0\n",
+ "u=math.sin(i*3.14/180.0)/math.sin(r*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print \"Refractive index is\",round(u,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Refractive index is 1.51\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.2 Page no 686"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.5\n",
+ "A=60 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a1=A/2.0\n",
+ "a=u*math.sin(a1*3.14/180.0)\n",
+ "a2=math.asin(a)*180/3.14\n",
+ "X=(a2*2)-A\n",
+ "i=(A+X)/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of deviation is\", round(X,1),\"degree\"\n",
+ "print\"Angle of incidence is\",round(i,1),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of deviation is 37.2 degree\n",
+ "Angle of incidence is 48.6 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.3 Page no 686"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=60 #Degree\n",
+ "ug=1.53\n",
+ "uw=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "ug1=ug/uw\n",
+ "a2=A/2.0\n",
+ "a1=ug1*math.sin(a2*3.14/180.0)\n",
+ "a2=math.asin(a1)*180/3.14\n",
+ "d=(a2*2)-A\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of minimum deviation is\", round(d,1),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of minimum deviation is 10.2 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.4 Page no 686"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ug=1.6\n",
+ "uw=1.33\n",
+ "i=40 #Degree\n",
+ "A=60\n",
+ "\n",
+ "#Calculation\n",
+ "ug1=ug/uw\n",
+ "r1=math.asin(math.sin(i*3.14/180.0)/ug1)*180/3.14\n",
+ "r2=A-r1\n",
+ "e=math.asin(ug1*math.sin(r2*3.14/180.0))*180/3.14\n",
+ "d=i+e-A\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of deviation is\", round(d,0),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of deviation is 14.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.5 Page no 686"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=60 #degree\n",
+ "ur=1.622\n",
+ "uv=1.663\n",
+ "\n",
+ "#Calculation\n",
+ "a=A*(uv-ur)\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of dispersion between red and violet color is\",a,\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of dispersion between red and violet color is 2.46 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.6 Page no 686"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=60 #Degree\n",
+ "ab=53\n",
+ "ar=51\n",
+ "a=52.0\n",
+ "\n",
+ "#Calculation\n",
+ "w=(ab-ar)/a\n",
+ "\n",
+ "#Result\n",
+ "print\"Dispersive power is\",round(w,4)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Dispersive power is 0.0385\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.7 Page no 686"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ub=1.522\n",
+ "ur=1.514\n",
+ "ub1=1.662\n",
+ "ur1=1.644\n",
+ "\n",
+ "#Calculation\n",
+ "u=(ub+ur)/2.0\n",
+ "w=(ub-ur)/(u-1)\n",
+ "u1=(ub1+ur1)/2.0\n",
+ "w1=(ub1-ur1)/(u1-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Dispersive power of crown glass is\",round(w,4)\n",
+ "print\"Dispersive power of flint galss is\",round(w1,4)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Dispersive power of crown glass is 0.0154\n",
+ "Dispersive power of flint galss is 0.0276\n"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.8 Page no 687"
+ ]
+ },
+ {
+ "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\"Refractive index is\",round(u,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Refractive index is 1.645\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.9 Page no 687"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "uv=1.523\n",
+ "ur=1.515\n",
+ "A=5 #Degree\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=-(5*(u-1))/(u1-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of flint glass prism is\",round(A1,2),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of flint glass prism is -3.88 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.10 Page no 687"
+ ]
+ },
+ {
+ "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=4.2 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "A=-w1*A1*(u1-1)/(w*(u-1))\n",
+ "d=-(A*(u-1)+A1*(u1-1))\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of prism is\", d,\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of prism is 3.12 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.11 Page no 687"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=40 #degree\n",
+ "u=1.54\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "i=math.asin(u*math.sin(A*3.14/180.0))*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of incidence is\",round(i,2),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of incidence is 81.72 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 59
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.12 Page no 687"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.524\n",
+ "A=60 #degree\n",
+ "C=41 #degree\n",
+ "\n",
+ "#calculation\n",
+ "import math\n",
+ "C1=1/u\n",
+ "A1=math.asin(C1)*180/3.14\n",
+ "r1=A-C\n",
+ "i=u*math.sin(r1*3.14/180.0)\n",
+ "A2=math.asin(i)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of incidence is\", round(A2,2),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of incidence is 29.75 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 84
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.13 Page no 688"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=0.015\n",
+ "w=0.03\n",
+ "u1=0.022\n",
+ "w1=0.05\n",
+ "a=2 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "A=-u1/u\n",
+ "A1=a/(A/2.0+u1/w1)\n",
+ "a2=A*A1\n",
+ "print\"Angle of 1st prism is\", round(A1,2),\"Degree\"\n",
+ "print\"Angle of 2nd prism is\",a2,\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of 1st prism is -6.82 Degree\n",
+ "Angle of 2nd prism is 10.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 98
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap23_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap23_1.ipynb new file mode 100644 index 00000000..3565eff3 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap23_1.ipynb @@ -0,0 +1,641 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:fd3a71d11d7e73e323877019a1c5f0e92d8e8b30b7f927fde2bcadd4742b58c1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 23 Optical instruments"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.1 Page no 704"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=5.0 #cm\n",
+ "D=25\n",
+ "\n",
+ "#Calculation\n",
+ "M=1+(D/f)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnifying power is\",M"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnifying power is 6.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.2 Page no 704"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=6.25 #cm\n",
+ "v=-25\n",
+ "\n",
+ "#Calculation\n",
+ "u=v*f/(v-f)\n",
+ "M=1+(D/f)\n",
+ "M1=D/f\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The distance of the object from the lens is\", M\n",
+ "print\"(ii) Anguar magnification is\",M1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The distance of the object from the lens is 5.0\n",
+ "(ii) Anguar magnification is 4.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.3 Page no 704"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-9\n",
+ "f=10 #cm\n",
+ "v1=-25.0\n",
+ "\n",
+ "#Calculation\n",
+ "v=u*f/(u+f)\n",
+ "M=v/u\n",
+ "u=-v1*f/(v1-f)\n",
+ "M1=1-(v1/f)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Magnification is\", M,\"\\n Size of each square in the image will appear as 100 mm**2\"\n",
+ "print\"(ii) Magnification is\",M1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnification is 10 \n",
+ " Size of each square in the image will appear as 100 mm**2\n",
+ "(ii) Magnification is 3.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.4 Page no 704"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=-25\n",
+ "f=10.0\n",
+ "\n",
+ "#Calculation\n",
+ "u=-v*f/(f-v)\n",
+ "m=-v/u\n",
+ "M=1-(v/f)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Distance is\", round(u,2),\"cm\"\n",
+ "print\"(b) Magnification is\",m\n",
+ "print\"(c) Magnification is\",M"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Distance is 7.14 cm\n",
+ "(b) Magnification is 3.5\n",
+ "(c) Magnification is 3.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.5 Page no 705"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=-30\n",
+ "fe=5 #cm\n",
+ "D=25 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "m0=M/(1+(D/fe))\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnification produced is\",m0"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnification produced is -5\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.6 Page no 705"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f0=4\n",
+ "u0=-6\n",
+ "D=25\n",
+ "fe=6.0\n",
+ "\n",
+ "#Calculation\n",
+ "v0=f0*u0/(f0+u0)\n",
+ "M=v0/u0*(1+(D/fe))\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnification produceed by the microscope is\", round(M,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnification produceed by the microscope is -10.33\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.7 Page no 705"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=20\n",
+ "fe=5\n",
+ "f0=1.0 \n",
+ "\n",
+ "#Calculation\n",
+ "v01=L-fe\n",
+ "u0=v01*f0/(f0-v01)\n",
+ "M=v01*D/(u0*fe)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnification power is\", M,\"\\nThe -ve sigh shows that the image formed is inverted and real\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnification power is -70.0 \n",
+ "The -ve sigh shows that the image formed is inverted and real\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.8 Page no 705"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=36 #cm\n",
+ "M=-8\n",
+ "\n",
+ "#Calculation\n",
+ "fe=-f/(M-1)\n",
+ "f0=-M*fe\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length of the lenses is\", fe,\"cm and\",f0,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length of the lenses is 4 cm and 32 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.9 Page no 705"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=75 #cm\n",
+ "fe=5.0\n",
+ "D=25 \n",
+ "\n",
+ "#Calculation\n",
+ "M=-f/fe*(1+(fe/D))\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnying power is\",M"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnying power is -18.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.10 Page no 705"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f0=140\n",
+ "fe=5\n",
+ "D=25.0\n",
+ "\n",
+ "#Calculation\n",
+ "M=-f0/fe\n",
+ "M1=-(f0/fe)*(1+(fe/D))\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Magnefying power, when telescope is in normal adjustment is\", M\n",
+ "print\"(b) Magnifying power, when image is formed at least distance of distinct vision is is\",M1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Magnefying power, when telescope is in normal adjustment is -28\n",
+ "(b) Magnifying power, when image is formed at least distance of distinct vision is is -33.6\n"
+ ]
+ }
+ ],
+ "prompt_number": 54
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.11 Page no 706"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f0=1 #m\n",
+ "fe=0.05\n",
+ "I=0.92*10**-2\n",
+ "u0=38*10**-7\n",
+ "fe1=5.0 #cm\n",
+ "ve=-25\n",
+ "\n",
+ "#Calculation\n",
+ "d=I*u0/f0\n",
+ "L=f0+fe\n",
+ "ue=ve*fe1/(fe1-ve)\n",
+ "L1=(f0*10**2)-ue\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Distance between two lenses is\", L*10**2,\"cm\"\n",
+ "print\"(ii) Distance when final image at 25 cm from the eye is\",round(L1,2),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Distance between two lenses is 105.0 cm\n",
+ "(ii) Distance when final image at 25 cm from the eye is 104.17 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 71
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.12 Page no 706"
+ ]
+ },
+ {
+ "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 the eye piece is\",fe,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length of the eye piece is 3.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.13 Page no 706"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=0.8 #cm\n",
+ "D=508\n",
+ "\n",
+ "#Calculation\n",
+ "B=D**2/d**2\n",
+ "\n",
+ "#Result\n",
+ "print\"The telescope can see a star \",B,\"time sfarther than the faintest star that can be seen with naked eye.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The telescope can see a star 403225.0 time sfarther than the faintest star that can be seen with naked eye.\n"
+ ]
+ }
+ ],
+ "prompt_number": 76
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.14 Page no 706"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ve=-25 #cm\n",
+ "fe=6.25\n",
+ "a=15\n",
+ "f0=2\n",
+ "\n",
+ "#Calculation\n",
+ "ue=-ve*fe/(ve-fe)\n",
+ "v0=a+ue\n",
+ "u0=f0*v0/(f0-v0)\n",
+ "v01=a-fe\n",
+ "u01=v01*f0/(f0-v01)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Distance when final image at least distance of distinct vision is\", u0,\"cm\"\n",
+ "print\"(b) Distance when final image is at infinity is\",round(u01,2),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Distance when final image at least distance of distinct vision is -2.5 cm\n",
+ "(b) Distance when final image is at infinity is -2.59 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 87
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.15 Page no 707"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=10\n",
+ "D=24\n",
+ "fe=20.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=a/(1+(D/fe))/100.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle is\", round(A,4),\"radian\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle is 0.0455 radian\n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.16 Page no 707"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f0=-0.5 #m\n",
+ "fe=0.01\n",
+ "d=3.48*10**6\n",
+ "r=3.8*10**8\n",
+ "a1=15\n",
+ "\n",
+ "#Calculation\n",
+ "M=f0/fe\n",
+ "a=d/r\n",
+ "I=a*a1\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Angular Magnification of the telescope is\",M\n",
+ "print\"(b) Diameter of the image is\", round(I,3),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Angular Magnification of the telescope is -50.0\n",
+ "(b) Diameter of the image is 0.137 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 103
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap25_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap25_1.ipynb new file mode 100644 index 00000000..1f9c7095 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap25_1.ipynb @@ -0,0 +1,521 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:caf7ddcffb734a0e87ae0e90d77de524bc76c7023aae29373fc9035427092542"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 25 Interference of light "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.1 Page no 728"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "w1=4\n",
+ "w2=9.0\n",
+ "a=25\n",
+ "\n",
+ "#Calculation\n",
+ "I=w1/w2\n",
+ "Imax=a/1.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Ratio of intensity is\",Imax"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Ratio of intensity is 25.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.2 Page no 728"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=4800*10**-10 #m\n",
+ "b=0.6*10**-2\n",
+ "b1=0.0045\n",
+ "\n",
+ "#Calculation\n",
+ "a=l/b\n",
+ "l1=a*b1*2\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelenght of the light is\", l1*10**10,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelenght of the light is 7200.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.3 Page no 728"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=1.5*10**-3\n",
+ "D=1 #m\n",
+ "w=3.93 #mm\n",
+ "D1=1.5\n",
+ "\n",
+ "#Calculation\n",
+ "b=w/10.0\n",
+ "l=(b*10**-3*d)/D\n",
+ "a=(10*D1*l)/d\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of the light is\", l,\"m\"\n",
+ "print\"Width of 10 fringes is\",a*10**3,\"*10**-3 m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of the light is 5.895e-07 m\n",
+ "Width of 10 fringes is 5.895 *10**-3 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.4 Page no 729"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=0.02 #cm\n",
+ "D=80\n",
+ "l=6*10**-5\n",
+ "n=5\n",
+ "\n",
+ "#Calculation\n",
+ "y5=n*D*l/d\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance of fifth bright fringe is\",y5,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance of fifth bright fringe is 1.2 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.5 Page no 729"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=3*10**-4\n",
+ "D=1.5\n",
+ "y4=10**-2 #m\n",
+ "\n",
+ "#Calculation\n",
+ "l=y4*d/(4*D)\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of light is\",l,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of light is 5e-07 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.6 Page no 729"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=0.125*10**-3 #m\n",
+ "l=4500*10**-10\n",
+ "D=1\n",
+ "\n",
+ "#Calculation\n",
+ "y2=2*D*l/d\n",
+ "y21=2*y2\n",
+ "\n",
+ "#Result\n",
+ "print\"Seperation between bright fringe on both sides of central maximum is\",y21*10**3,\"*10**-3 m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Seperation between bright fringe on both sides of central maximum is 14.4 *10**-3 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.7 Page no 729"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=0.012 #m\n",
+ "D=100\n",
+ "l=6000*10**-8 #cm\n",
+ "n=2\n",
+ "\n",
+ "#Calculation\n",
+ "y2=(n**2+1)*D*l/(n*d)\n",
+ "\n",
+ "#result\n",
+ "print\"Distance of 3rd dark band is\",y2,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance of 3rd dark band is 1.25 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.8 Page no 729"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=2*10**-4 #m\n",
+ "D=1.5\n",
+ "y3=1.8*10**-2\n",
+ "\n",
+ "#Calculation\n",
+ "l=2*y3*d/(7*D)\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of the light is\", round(l*10**7,2)*10**-7,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of the light is 6.86e-07 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.9 Page no 729"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "b=2*10**-3\n",
+ "l=6.0*10**-7\n",
+ "u=1.33\n",
+ "\n",
+ "#calculation\n",
+ "l1=b/l\n",
+ "b2=(b*l)/(u*l)\n",
+ "\n",
+ "#Result\n",
+ "print\"Fringe width is\",round(b2*10**3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Fringe width is 1.5 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.10 Page no 729"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=6000*10**-10 #m\n",
+ "d=10**-3\n",
+ "D=1\n",
+ "n=10\n",
+ "\n",
+ "#Calculation\n",
+ "a10=n*l/d\n",
+ "b=(D*l)/d\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Angular position of 10th maximum is\",a10,\"radian\"\n",
+ "print\"(b) Separation of the two adjacent minima is\",b*10**3,\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Angular position of 10th maximum is 0.006 radian\n",
+ "(b) Separation of the two adjacent minima is 0.6 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.11 Page no 730"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=6000*10**-10 #m\n",
+ "r=0\n",
+ "n=7\n",
+ "\n",
+ "#Calculation\n",
+ "t=n*l/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Difference of the film is\", t,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Difference of the film is 2.1e-06 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.12 Page no 730"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.4\n",
+ "t=10**-4 #cm\n",
+ "n1=4.0\n",
+ "n2=5.0\n",
+ "n3=6.0\n",
+ "n4=7.0\n",
+ "N1=9\n",
+ "N2=11\n",
+ "N3=13\n",
+ "\n",
+ "#Calculation\n",
+ "a=2*u*t*10**8\n",
+ "l1=a/n1\n",
+ "l2=a/n2\n",
+ "l3=a/n3\n",
+ "l4=a/n4\n",
+ "L1=2*a/N1\n",
+ "L2=2*a/N2\n",
+ "L3=2*a/N3\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Wavelength When the reflection is weak is\",l1,\"A ,\",l2,\"A ,\",round(l3,0),\"A ,\",l4,\"A\"\n",
+ "print\"(ii) Wavelength when the reflection is strong is\",round(L1,0),\"A ,\",round(L2,0),\"A ,\",round(L3,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Wavelength When the reflection is weak is 7000.0 A , 5600.0 A , 4667.0 A , 4000.0 A\n",
+ "(ii) Wavelength when the reflection is strong is 6222.0 A , 5091.0 A , 4308.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.13 Page no 730"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n1=10\n",
+ "l1=4358*10**-10 #m\n",
+ "l2=5893.0*10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "n2=n1*l1/l2\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of fringes is\",round(n2,1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of fringes is 7.4\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap26_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap26_1.ipynb new file mode 100644 index 00000000..26f66f05 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap26_1.ipynb @@ -0,0 +1,239 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:010c48e259f2faa6331c5ac2dae668d950f9fc1e120bf60a20ed01cae899383a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 26 Diffraction of light"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.1 Page no 747"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=0.5*10**-3 #m\n",
+ "l=6000*10**-10 \n",
+ "D=2\n",
+ "\n",
+ "#Calculation\n",
+ "d=2*D*l/a\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance between the two dark bands is\",d*10**3,\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance between the two dark bands is 4.8 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.2 Page no 747"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=5900*10**-10 #m\n",
+ "a=11.8*10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=l/a\n",
+ "A=math.asin(a)*180/3.14\n",
+ "A1=2*A\n",
+ "\n",
+ "#Result\n",
+ "print\"Angular position is\",round(A1,0),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angular position is 60.0 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.3 Page no 747 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=650*10**-9\n",
+ "a1=30 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "a=l/math.sin(a1*3.14/180.0)\n",
+ "a2=3*l/math.sin(a1*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Value of a when the 1st min. falls at 30 degree is\",round(a*10**6,1)*10**-6,\"m\"\n",
+ "print\"(b) Value of a when the 1st max. falls at 30 degree is\",round(a2*10**6,2)*10**-6,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Value of a when the 1st min. falls at 30 degree is 1.3e-06 m\n",
+ "(b) Value of a when the 1st max. falls at 30 degree is 3.9e-06 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.4 Page no 747"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=600.0*10**-9 #m\n",
+ "a=2*10**-3 \n",
+ "\n",
+ "#Calculation\n",
+ "Z=a**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": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.5 Page no 747"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=4.6*10**-6 #rad\n",
+ "l=5460*10**-10 #m\n",
+ "\n",
+ "#Calculation\n",
+ "D=1.22*l/d\n",
+ "\n",
+ "#Result\n",
+ "print\"Aperture of the objective of the telescope is\",round(D,4),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Aperture of the objective of the telescope is 0.1448 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.6 Page no 748"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "lr=660\n",
+ "\n",
+ "#Calculation\n",
+ "l1=2*lr/3.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of lembda is\",l1,\"nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of lembda is 440.0 nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap27_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap27_1.ipynb new file mode 100644 index 00000000..9edfe788 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap27_1.ipynb @@ -0,0 +1,135 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:c77b859962fb938dcd11a74749cd47ac07047d0f8605a1cc5e5320189808ac31"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 27 Polarisation of light "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.1 Page no 758"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.536\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "p=math.atan(u)*180/3.14\n",
+ "r=90-p\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 33.0 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.2 Page no 758"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "p=60 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "u=math.tan(p*3.14/180.0)\n",
+ "C=1/u\n",
+ "C1=math.asin(C)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Critical angle is\", round(C1,1),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Critical angle is 35.3 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.3 Page no 758"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=60 #degree\n",
+ "b=100\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a1=90-a\n",
+ "A=a/2.0\n",
+ "I=(math.cos(A*3.14/180.0)**2)/2.0\n",
+ "I0=b*I\n",
+ "\n",
+ "#result\n",
+ "print\"Unpolarised light is\", round(I0,1),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Unpolarised light is 37.5 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap28_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap28_1.ipynb new file mode 100644 index 00000000..c53c8334 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap28_1.ipynb @@ -0,0 +1,578 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:487f66e71ac79d9579f4a8e998f21bc6fadb0986ba14ee5d89b77699b48cadee"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 28 Particle nature of radiation"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 28.1 Page no 801"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=4950.0*10**-10\n",
+ "h=6.6*10**-34 #js\n",
+ "c=3*10**8 #m/s\n",
+ "\n",
+ "#Calculation\n",
+ "E=(h*c/l)/(1.6*10**-19)\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy of each photon is\",E,\"eV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy of each photon is 2.5 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 28.2 Page no 801"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=6.62 #J\n",
+ "v=10**12 #Hz\n",
+ "h=6.62*10**-34\n",
+ "\n",
+ "#Calculation\n",
+ "E1=h*v\n",
+ "n=E/(h*v)\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of photons is\",n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of photons is 1e+22\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 28.3 Page no 801"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.62*10**-34 #js\n",
+ "c=3*10**8 #m/s\n",
+ "Iev=1.6*10**-19 #J\n",
+ "V=10**6 #hz\n",
+ "V1=5890*10**-10 #m\n",
+ "L=10.0**-10 #m\n",
+ "\n",
+ "#Calculation\n",
+ "E=h*v\n",
+ "E1=E/Iev\n",
+ "A=(h*c)/V1\n",
+ "A1=A/Iev\n",
+ "B=(h*c)/L\n",
+ "B1=B/Iev\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The energy of a photon with frequency 1000 Khz is\",round(E1*10**3,2),\"10**-9\",\"ev\" \n",
+ "print\"(ii) The energy of a photon when wavelength is 5890A is\",round(A1,2) ,\"eV\"\n",
+ "print\"(iii) The energy of a photon when wavelength is 1 A is\", B1,\"eV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The energy of a photon with frequency 1000 Khz is 4.14 10**-9 ev\n",
+ "(ii) The energy of a photon when wavelength is 5890A is 2.11 eV\n",
+ "(iii) The energy of a photon when wavelength is 1 A is 12412.5 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 28.4 Page no 801"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=0.66*10**-10 #m\n",
+ "h=6.62*10**-34 #Js\n",
+ "c=3*10**8 #m/s\n",
+ "\n",
+ "#Calculation\n",
+ "E=((h*c)/l)/(1.6*10**-19)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Maximum energy of photon is\",round(E*10**-3,2),\"Kev\"\n",
+ "print\"(b) To produce electrons of energy 18.81 Kev,accelerating potential of 18.81 KV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Maximum energy of photon is 18.81 Kev\n",
+ "(b) To produce electrons of energy 18.81 Kev,accelerating potential of 18.81 KV\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 28.5 Page no 801"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=10.2*10**9 #ev\n",
+ "h=6.62*10**-34\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "E1=(E/2.0)*(1.6*10**-19)\n",
+ "l=(h*c)/E1\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength is\",round(l*10**16,3),\"10**-16\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength is 2.434 10**-16\n"
+ ]
+ }
+ ],
+ "prompt_number": 60
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 28.6 Page no 802"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=3500.0*10**-10 #m\n",
+ "h=6.62*10**-34\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "Hv=h*c/(L*(1.6*10**-19))\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy of incident light is\", round(Hv,3),\"eV \\nMetal B will yield photoelectrons.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy of incident light is 3.546 eV \n",
+ "Metal B will yield photoelectrons.\n"
+ ]
+ }
+ ],
+ "prompt_number": 75
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 28.7 Page no 802"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=6000.0*10**-10 #m\n",
+ "h=6.62*10**-34 #Js\n",
+ "e=1.6*10**-19\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "v0=c/l\n",
+ "w=(h*v0)/e\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Threshold frequency is\", v0,\"Hz\"\n",
+ "print\"(ii) Work function is\",round(w,3),\"eV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Threshold frequency is 5e+14 Hz\n",
+ "(ii) Work function is 2.069 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 83
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 28.8 Page no 802"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "w=1.2*1.6*10**-19\n",
+ "l=5000.0*10**-10\n",
+ "c=3*10**8\n",
+ "h=6.62*10**-34\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "A=((h*c)/l)-w\n",
+ "V0=A/e\n",
+ "\n",
+ "#Result\n",
+ "print\"Stopping potential is\", round(V0,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stopping potential is 1.28 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 90
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 28.9 Page no 802"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.6*10**-34 #Js\n",
+ "e=1.6*10**-19\n",
+ "l=500.0*10**-9 #m\n",
+ "\n",
+ "#Calculation\n",
+ "w=2*e\n",
+ "l0=(h*c)/w\n",
+ "E=((h*c)/l)-w\n",
+ "v0=E/e\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The threshold wavelength is\", l0*10**9,\"nm\"\n",
+ "print\"(ii) Maximum energy is\",E*10**19,\"*10**-19 J\"\n",
+ "print\"(iii) The stopping potential is\",v0,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The threshold wavelength is 618.75 nm\n",
+ "(ii) Maximum energy is 0.76 *10**-19 J\n",
+ "(iii) The stopping potential is 0.475 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 104
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 28.10 Page no 802"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V0=1.3\n",
+ "l=2271*10**-10\n",
+ "h=6.62*10**-34\n",
+ "e=1.6*10**-19\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "w=(h*c/l)-(V0*e)\n",
+ "l0=(h*c)/w\n",
+ "\n",
+ "#Result\n",
+ "print\"Work function of metal is\", round(l0*10**10,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Work function of metal is 2980.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 109
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 28.11 Page no 803"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.62*10**-34\n",
+ "c=3*10**8\n",
+ "l=640.2*10**-9\n",
+ "e=1.6*10**-19\n",
+ "V0=0.54\n",
+ "l1=427.2*10**-9\n",
+ "\n",
+ "#Calculation\n",
+ "w=(h*c/l)-(e*V0)\n",
+ "v0=(((h*c)/l1)-w)/e\n",
+ "\n",
+ "#Result\n",
+ "print\"New stopping voltage is\", round(v0,3),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "New stopping voltage is 1.507 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 115
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 28.12 Page no 803"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=5600.0*10**-10 #m\n",
+ "h=6.625*10**-34\n",
+ "c=3*10**8\n",
+ "a=5\n",
+ "\n",
+ "#Calculation\n",
+ "E=h*c/l\n",
+ "n=a/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of visible light photon is\", round(n*10**-19,2)*10**-19"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of visible light photon is 1.41e-19\n"
+ ]
+ }
+ ],
+ "prompt_number": 121
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 28.13 Page no 803"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=10**4 #W\n",
+ "l=500.0 #m\n",
+ "h=6.62*10**-34\n",
+ "A=0.4*10**-14 #m**2\n",
+ "i=10**-10 #W/m**2\n",
+ "v=6*10**14\n",
+ "\n",
+ "#Calculation\n",
+ "E1=E*l/(h*c)\n",
+ "E2=h*v\n",
+ "n=A/E2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Number of photons emitted is\",round(E1*10**-31,3)*10**31\n",
+ "print\"(ii) Number of photons is\", round(n*10**-4,0),\"*10**4\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Number of photons emitted is 2.518e+31\n",
+ "(ii) Number of photons is 1.0 *10**4\n"
+ ]
+ }
+ ],
+ "prompt_number": 132
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 28.14 Page no 803"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=2*10**-4 #m**2\n",
+ "An=10.0**-20\n",
+ "n1=5\n",
+ "i=10**-5 #W/m**2\n",
+ "\n",
+ "#Calculation\n",
+ "n=(A*n1)/An\n",
+ "E=i*A\n",
+ "E1=(E/n)/e\n",
+ "t=2/E1\n",
+ "\n",
+ "#Result\n",
+ "print\"Time required for the photoelectric emission is\",t*10**-7,\"*10**7 S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time required for the photoelectric emission is 1.6 *10**7 S\n"
+ ]
+ }
+ ],
+ "prompt_number": 145
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap2_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap2_1.ipynb new file mode 100644 index 00000000..efbea0a7 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap2_1.ipynb @@ -0,0 +1,534 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:1e2d45d8326c66f64e42e0f8f482f41d5aabf3a1eabc7e50f07a0c18121fe999"
+ },
+ "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 39"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=10**-7 #Kg\n",
+ "q=1.6*10**-19 #C\n",
+ "g=9.8\n",
+ "\n",
+ "#Calculation\n",
+ "E=(m*g)/q\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric field strength required is\", E"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric field strenght required is 6.125e+12\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2 Page no 39"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=10.0**-3 #Kg\n",
+ "q=5*10**-6 #C\n",
+ "u=20 #m/s\n",
+ "E=2*10**5 #N/C\n",
+ "\n",
+ "#Calculation\n",
+ "F=q*E\n",
+ "a=-F/m\n",
+ "S=-u**2/(2.0*a)\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance travelled is\", S,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance travelled is 0.2 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.3 Page no 39"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=2.55*10**4 #V/m\n",
+ "a=1.26*10**3 #Kg/m**4\n",
+ "g=9.81 #m/s**2\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "q=12*e\n",
+ "Fe=q*E\n",
+ "r=((3*Fe)/(4.0*math.pi*a*g))**0.333\n",
+ "\n",
+ "#Result\n",
+ "print\"The radius of the drop is\", round(r*10**4,5)*10**3,\"*10**-4 mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The radius of the drop is 9.95 *10**-4 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.4 Page no 39"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=80.0*10**-6 #Kg\n",
+ "q=2*10**-8 #C\n",
+ "E=20000 #V/m\n",
+ "g=9.8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=(q*E)/(m*g)\n",
+ "a1=math.atan(a)*180/3.14\n",
+ "T=(q*E)/(math.sin(a1*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"Tension in the thread of the pendulum is\", round(T*10**4,1),\"*10**-4 N\"\n",
+ "print\"Angle it make with the vertical is\",round(a1,2),\"Degree C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Tension in the thread of the pendulum is 8.8 *10**-4 N\n",
+ "Angle it make with the vertical is 27.04 Degree C\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.5 Page no 40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q1=2.0*10**-7 #C\n",
+ "q2=1.0*10**-7 #C\n",
+ "r=10.0**-2 #m\n",
+ "m=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "E1=(m*q1)/r**2\n",
+ "E2=(m*q2)/r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of field E1 is\", E1*10**-7,\"*10**7 N/C\"\n",
+ "print\"Magnitude of field E2 is\", E2*10**-6,\"*10**6 N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of field E1 is 1.8 *10**7 N/C\n",
+ "Magnitude of field E2 is 9.0 *10**6 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 50
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.6 Page no 40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q1=5.0*10**-19 #c\n",
+ "q2=20*10**-19\n",
+ "r=2.0 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "x=-math.sqrt(q2/q1)\n",
+ "x1=math.sqrt(q2/(q1*3))\n",
+ "\n",
+ "#Result\n",
+ "print\"The point on the line joining is\", round(x,1),\"m and \",round(x1,1),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The point on the line joining is -2.0 m and 1.2 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 60
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.7 Page no 40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.2\n",
+ "m=9*10**9\n",
+ "qa=3*10**-6\n",
+ "q=1.5*10**-9\n",
+ "\n",
+ "#Calculation\n",
+ "a=r/2.0\n",
+ "Ea=(m*qa)/a**2\n",
+ "Eb=(m*qa)/a**2\n",
+ "E=Ea+Eb\n",
+ "F=q*E\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Electric field at the mid point is\", E*10**-6,\"*10**6 N/C (along OB)\"\n",
+ "print\"(b) Force experienced by the test charge is\",F*10**3,\"*10**-3 N (along OA)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Electric field at the mid point is 5.4 *10**6 N/C (along OB)\n",
+ "(b) Force experienced by the test charge is 8.1 *10**-3 N (along OA)\n"
+ ]
+ }
+ ],
+ "prompt_number": 71
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.8 Page no 40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "qa=16*10**-6 #c\n",
+ "qb=-9.0*10**-6 #C\n",
+ "r=0.08 #m\n",
+ "\n",
+ "#Calculation\n",
+ "x=-(qa/qb)*r\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the point is\", round(x,2),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the point is 0.14 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.9 Page no 41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "qa=4*10**-6\n",
+ "r=2.0*10**-2\n",
+ "r1=10**-2\n",
+ "qb=2*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Ea=(m*qa)/r**2\n",
+ "Eb=(m*qb)/r1**2\n",
+ "a=r1/r\n",
+ "E=math.sqrt(Ea**2+Eb**2+(2*Ea*Eb*a))\n",
+ "a11=Ea*(math.sin(60)*180/3.14)/(Ea+Eb*(math.cos(60)*180/3.14))\n",
+ "\n",
+ "#Result\n",
+ "print\"Position is\", round(a11,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position is 0.16\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.10 Page no 41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=500*10**-6\n",
+ "r=0.1 #m\n",
+ "d=0.2\n",
+ "m=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "p=q*r\n",
+ "a=r/2.0\n",
+ "R=d+a\n",
+ "E=(m*2*p*R)/((R**2-a**2)**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric dipole moment of the dipole is\", p,\"Cm\"\n",
+ "print\"Electric field due to dipole is\",E*10**-7,\"*10**7 N/C\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric dipole moment of the dipole is 5e-05 Cm\n",
+ "Electric field due to dipole is 6.25 *10**7 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.11 Page no 41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=0.2*10**-12 #C\n",
+ "w=10**-8 #m\n",
+ "r=0.1 #m\n",
+ "e=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "p=q*w\n",
+ "E=(e*2*p)/r**3\n",
+ "\n",
+ "#Result\n",
+ "print\"The electric field at an axical point at a distance of 10cm from there mid point is\", E*10**8,\"*10**-8 N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electric field at an axical point at a distance of 10cm from there mid point is 3.6 *10**-8 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.13 Page no 42"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=10.0**4\n",
+ "t=9*10**-26 #Nm\n",
+ "\n",
+ "#Calculation\n",
+ "import math \n",
+ "p=t/(E*math.sin(30*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"Dipole moment of the dipole is\", round(p*10**29,1),\"*10**-19 Cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Dipole moment of the dipole is 1.8 *10**-19 Cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.17 Page no 43"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "p=-10**-7 #Cm\n",
+ "i=10**5\n",
+ "\n",
+ "#Calculation\n",
+ "F=p*i\n",
+ "\n",
+ "#Result\n",
+ "print\"Force is given by\", F,\"N\"\n",
+ "print\"Torque is zero\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force is given by -0.01 N\n",
+ "Torque is zero\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap30_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap30_1.ipynb new file mode 100644 index 00000000..cf038d52 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap30_1.ipynb @@ -0,0 +1,235 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:75345a10ab4b01b2d63b7c49ca20de9a23e8e81cbab11c7f6d702383f9b2feef"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 30 Structure Of Nucleus"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 30.1 Page no 840"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A1=197\n",
+ "A2=107.0\n",
+ "\n",
+ "#Calculation\n",
+ "R=(A1/A2)**(0.3)\n",
+ "\n",
+ "#Result\n",
+ "print\"The ratio of the nuclear radii of the gold and silver isotope is\",round(R,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of the nuclear radii of the gold and silver isotope is 1.201\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 30.2 Page no 840"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=1.2*10**-15 #m\n",
+ "M=1.67*10**-27 #kg\n",
+ "P1=10**3 #kg m**-3\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "V=1.3*math.pi*(R**3)\n",
+ "P=M/V\n",
+ "A=P/P1\n",
+ "\n",
+ "#Result\n",
+ "print\"Nuclear matter is denser than water is\",round(A*10**-14,3),\"*10**14\",\"times\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Nuclear matter is denser than water is 2.366 10**14 times\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 30.3 Page no 840"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Mh=4.00150 #a.m.u.\n",
+ "Mp=1.00728 #a.m.u.\n",
+ "Mn=1.00867 #a.m.u.\n",
+ "W0=931.5 #MeV\n",
+ "\n",
+ "#Calculation\n",
+ "A=((2*Mp)+(2*Mn))-Mh\n",
+ "A1=A*W0\n",
+ "\n",
+ "#Result\n",
+ "print\"Binding energy of a-particle is\",round(A1,2),\"MeV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Binding energy of a-particle is 28.32 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 30.4 Page no 840"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Mp=1.007275 #a.m.u.\n",
+ "Mn=1.008665 #a.m.u.\n",
+ "Mh=2.013553 #a.m.u.\n",
+ "S=2.0\n",
+ "U=931.5\n",
+ "\n",
+ "#Calculation\n",
+ "A=(Mp+Mn)-Mh\n",
+ "P=A/S\n",
+ "W=A*U\n",
+ "L=W/S\n",
+ "\n",
+ "#Result\n",
+ "print\"The mass defect is\",A,\"a.m.u.\" \n",
+ "print\"The packing fraction is\",P,\"a.m.u.\"\n",
+ "print\"The binding energy of deutron is\",round(W,2),\"MeV\"\n",
+ "print\"The binding energy of per nucleon of deutron is\",round(L,2),\"MeV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mass defect is 0.002387 a.m.u.\n",
+ "The packing fraction is 0.0011935 a.m.u.\n",
+ "The binding energy of deutron is 2.22 MeV\n",
+ "The binding energy of per nucleon of deutron is 1.11 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 62
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 30.5 Page no 840"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Mh=1.007825 #a.m.u.\n",
+ "Mn=1.008665 #a.m.u.\n",
+ "Mp=55.934939 #a.m.u.\n",
+ "Mb=208.980388 #a.m.u.\n",
+ "A=56.0\n",
+ "Z=26\n",
+ "S=931.5\n",
+ "A1=209.0\n",
+ "Z1=83\n",
+ "\n",
+ "#Calculation\n",
+ "W=A-Z\n",
+ "Q=((Z*Mh+W*Mn)-Mp)*S\n",
+ "R=Q/A\n",
+ "W1=A1-Z1\n",
+ "Q1=((Z1*Mh+W1*Mn)-Mb)*S\n",
+ "R1=Q1/A1\n",
+ "\n",
+ "#Result\n",
+ "print\"The binding energy of the nuclri of Fe is\",round(Q,2),\"MeV\"\n",
+ "print\"The binding energy of the nuclei of Bi is\",round(Q1,2),\"MeV\"\n",
+ "print\"Binding energy per nucleon of Fe is\",round(R,2),\"MeV\"\n",
+ "print\"Binding energy per nucleon of Bi is\",round(R1,3),\"MeV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The binding energy of the nuclri of Fe is 492.26 MeV\n",
+ "The binding energy of the nuclei of Bi is 1640.26 MeV\n",
+ "Binding energy per nucleon of Fe is 8.79 MeV\n",
+ "Binding energy per nucleon of Bi is 7.848 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 80
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap31_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap31_1.ipynb new file mode 100644 index 00000000..6135b7c0 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap31_1.ipynb @@ -0,0 +1,388 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:64e0a844a3ba33dead720cfb202f40f7dab27e1a9fab7ab4975bb30624a36195"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 31 Radioactivity"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 31.1 Page no 855"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "T=138.0 #days\n",
+ "E=0.693\n",
+ "N=12.5\n",
+ "N0=100\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "L=E/T\n",
+ "N1=N/N0\n",
+ "t=(2.303*math.log10(8))/L\n",
+ "\n",
+ "#Result\n",
+ "print\"Time is\",round(t,2),\"days\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time is 414.16 days\n"
+ ]
+ }
+ ],
+ "prompt_number": 71
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 31.2 Page no 855"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "T=1.42*10**17 #s\n",
+ "A=6.02*10**23 #mol**-1\n",
+ "E=0.693\n",
+ "L=238.0\n",
+ "\n",
+ "#Calculation\n",
+ "N=A/L\n",
+ "L1=E/T\n",
+ "Z=L1*N\n",
+ "\n",
+ "#Result\n",
+ "print\"Disintegrations per second occur in 1g of U**238 is\",round(Z*10**-4,3),\"*10**4 s**-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Disintegrations per second occur in 1g of U**238 is 1.234 *10**4 s**-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 31.3 Page no 855"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "K=-40 #Cl\n",
+ "M=0.075 #kg mole**-1\n",
+ "m=1.2*10**-6 #kg\n",
+ "A=6.0*10**23\n",
+ "D=170 #s**-1\n",
+ "E=0.693\n",
+ "\n",
+ "#Calculation\n",
+ "N=(A/M)*m\n",
+ "L=D/N\n",
+ "T=(E/L)\n",
+ "O=T/31536000.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Half life of K40 atom is\",round(O*10**-9,3),\"*10**9 years\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Half life of K40 atom is 1.241 *10**9 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 81
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 31.4 Page no 856"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "T=6.0 #hours\n",
+ "E=0.693\n",
+ "A=6.025*10**23\n",
+ "W=99.0\n",
+ "S=10**-12\n",
+ "t=1 #hours\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "L=E/T\n",
+ "N0=(A/W)*S\n",
+ "R0=L*N0\n",
+ "N=N0*math.exp(-L)\n",
+ "R=L*N\n",
+ "R1=L*N\n",
+ "\n",
+ "#Result\n",
+ "print\"Activity at beginning is\", round(R0*10**-8,2),\"*10**8 /h\"\n",
+ "print\"Activity at the end is\",round(R1*10**-8,3),\"*10**8 /h\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Activity at beginning is 7.03 *10**8 /h\n",
+ "Activity at the end is 6.262 *10**8 /h\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 31.5 Page no 856"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "t=30.0 #years\n",
+ "a=16\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l=(2.3026*math.log10(a))/t\n",
+ "T=0.693/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Half life period is\", round(T,1),\"years\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Half life period is 7.5 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 31.6 Page no 856"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m1=238.05081\n",
+ "m2=234.04363\n",
+ "m3=4.00260\n",
+ "A=931.5 #Mev\n",
+ "\n",
+ "#Calculation\n",
+ "Ea=(m1-m2-m3)*A\n",
+ "\n",
+ "#Result\n",
+ "print\"Kinetic energy is\",round(Ea,2),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Kinetic energy is 4.27 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 31.7 Page no 856"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m1=22.994466\n",
+ "m2=22.989770\n",
+ "A=931.5\n",
+ "\n",
+ "#Calculation\n",
+ "Eb=(m1-m2)*A\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum kinetic energy is\",round(Eb,3),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum kinetic energy is 4.374 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 31.8 Page no 856"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E1=0\n",
+ "E2=0.412*1.6*10**-13\n",
+ "E3=1.088*1.6*10**-13\n",
+ "h=6.62*10**-34\n",
+ "m1=197.968233\n",
+ "m2=197.966760\n",
+ "a=931.5\n",
+ "A=1.088\n",
+ "A1=0.412\n",
+ "\n",
+ "#Calculation\n",
+ "v1=(E3-E1)/h\n",
+ "v2=(E2-E1)/h\n",
+ "v3=(E3-E2)/h\n",
+ "Eb1=((m1-m2)*a)-A\n",
+ "Eb2=((m1-m2)*a)-A1\n",
+ "\n",
+ "#Result\n",
+ "print\"Radiation frequencies are\",round(v1*10**-20,2)*10**20,\",\",round(v2*10**-20,3)*10**20,\"and\",round(v3*10**-20,2)*10**20\n",
+ "print\"Maximum kinetic energies are\",round(Eb1,3),\"Mev and \",round(Eb2,3),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radiation frequencies are 2.63e+20 , 9.96e+19 and 1.63e+20\n",
+ "Maximum kinetic energies are 0.284 Mev and 0.96 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 31.9 Page no 857"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=15\n",
+ "b=9.0\n",
+ "T=5730\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=a/b\n",
+ "l=2.303*math.log10(A)\n",
+ "t=(l*T)/0.693\n",
+ "\n",
+ "#Result\n",
+ "print\"Approximation age of the indus valley civilization is\",round(t,2),\"years\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Approximation age of the indus valley civilization is 4224.47 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap32_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap32_1.ipynb new file mode 100644 index 00000000..69ecaa80 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap32_1.ipynb @@ -0,0 +1,381 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:1a205c66d8c05960f2478abf5942d3dc43db8b63d7327dc1bc8decd646c4e601"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 32 Nuclear reactions"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 32.3 Page no 874"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r1=6.015126\n",
+ "r2=1.008665\n",
+ "p1=4.002603\n",
+ "p2=3.016049\n",
+ "a=931\n",
+ "\n",
+ "#Calculation\n",
+ "R=r1+r2\n",
+ "P=p1+p2\n",
+ "M=R-P\n",
+ "E=M*a\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy released is\",round(E,2),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy released is 4.78 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 32.4 Page no 874"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=400*10**6 #J/s\n",
+ "c=3.0*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "E=P*24*60*60\n",
+ "m=E/c**2\n",
+ "\n",
+ "#Result\n",
+ "print\"U235 is\",m*10**3,\"g\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "U235 is 0.384 g\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 32.5 Page no 874"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=6.023*10**23\n",
+ "w=235.0\n",
+ "E=200\n",
+ "e=1.6*10**-13\n",
+ "\n",
+ "#Calculation\n",
+ "E1=((A*E)/w)*e\n",
+ "E2=E1/(1000.0*3600.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy released is\", round(E2*10**-4,3),\"*10**4 KWh\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy released is 2.278 *10**4 KWh\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 32.6 Page no 874"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=7.6*10**13 #J\n",
+ "E1=200.0*1.6*10**-13\n",
+ "h=6.023*10**23\n",
+ "a=235\n",
+ "\n",
+ "#Calculation\n",
+ "n=E/E1\n",
+ "M=(a*n)/h\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Number of uranium atom is\",n\n",
+ "print\"(ii) Mass of uranium is\",round(M,2),\"g\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Number of uranium atom is 2.375e+24\n",
+ "(ii) Mass of uranium is 926.66 g\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 32.7 Page no 874"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m1=235.043933\n",
+ "m2=140.917700\n",
+ "m3=91.895400\n",
+ "n1=1.0086651\n",
+ "a=931.5\n",
+ "\n",
+ "#Calculation\n",
+ "Q=(m1-m2-m3-2*(n1))*a\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy released is\",round(Q,2),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy released is 198.88 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 32.8 Page no 874"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M1=1.007825\n",
+ "M2=4.002603\n",
+ "M3=0.000549\n",
+ "A=931.5\n",
+ "S=4\n",
+ "W=2\n",
+ "\n",
+ "#Calculation\n",
+ "Q=((S*M1-M2-W*M3)*A)\n",
+ "\n",
+ "#Result\n",
+ "print\"The energy is\",round(Q,2),\"MeV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The energy is 25.71 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 32.9 Page no 875"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M1=1.008 #a.m.u.\n",
+ "M2=4.004 #a.m.u.\n",
+ "M3=7.016 #a.m.u.\n",
+ "A=2\n",
+ "W=931\n",
+ "\n",
+ "#Calculation\n",
+ "Q=(M3+M1-A*M2)*W\n",
+ "\n",
+ "#Result\n",
+ "print\"The initial energy of each alpha-particle is\",Q,\"MeV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The initial energy of each alpha-particle is 14.896 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 32.10 Page no 875"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M1=235.0439\n",
+ "M2=93.9065\n",
+ "M3=139.9055\n",
+ "M4=1.00866\n",
+ "S=931.5\n",
+ "\n",
+ "#Calculation\n",
+ "Q=(M1-M2-M3-M4)*S\n",
+ "\n",
+ "#Result\n",
+ "print\"Total energy is\",round(Q,2),\"MeV\"\n",
+ "print\"All the available energy does not appear as the kinetic energy of fission products\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total energy is 207.95 MeV\n",
+ "All the available energy does not appear as the kinetic energy of fission products\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 32.11 Page no 875"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M1=2.014102\n",
+ "M2=3.016049\n",
+ "M3=4.002603\n",
+ "M4=1.008665\n",
+ "S=931.5\n",
+ "q1=1.6*10**-19 #C\n",
+ "q2=9*10**9\n",
+ "r=2*1.5*10**-15\n",
+ "r1=1.38*10**-23\n",
+ "\n",
+ "#Calculation\n",
+ "Q=(M1+M2-M3-M4)*S\n",
+ "Q2=q2*(q1**2/r)\n",
+ "Q3=(2*Q2)/(3*r1)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The amount of energy is\",round(Q,2),\"MeV\"\n",
+ "print\"(b) The kinetic energy is\",Q2,\"J\"\n",
+ "print\"The temperature is\",round(Q3*10**-9,2),\"*10**9 K\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The amount of energy is 17.59 MeV\n",
+ "(b) The kinetic energy is 7.68e-14 J\n",
+ "The temperature is 3.71 *10**9 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 55
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap33_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap33_1.ipynb new file mode 100644 index 00000000..4183f15e --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap33_1.ipynb @@ -0,0 +1,304 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:52e9ff37336a0b3392520c7ec0e817b37056adea4740de3297c85e4e3e83dc56"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 33 Solids"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 33.1 Page no 904"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Eg=0.72*1.6*10**-19 #J\n",
+ "h=6.62*10**-34\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "L=(h*c)/Eg\n",
+ "\n",
+ "#Result\n",
+ "print\"The maximum wavelength of electromagnetic radiation is\",round(L*10**6,3)*10**-6,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum wavelength of electromagnetic radiation is 1.724e-06 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 33.2 Page no 904"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n1=1.5*10**16 #/m**3\n",
+ "nh=4.5*10**22\n",
+ "\n",
+ "#Calculation\n",
+ "ne=n1**2/nh\n",
+ "\n",
+ "#Result\n",
+ "print\"ne in the doping silicon is\",ne*10**-9,\"*10**9 /m**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "ne in the doping silicon is 5.0 *10**9 /m**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 33.3 Page no 904"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ne=8*10**19 #/m**3\n",
+ "nh=5*10**18\n",
+ "ue=2.3 #m**2/V/S\n",
+ "uh=0.01\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "a=1/(e*((ne*ue)+(nh*uh)))\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The semiconductor has greater electron concentration, it is n-type semiconductor\"\n",
+ "print\"(b) Resistivity is\", round(a*10**2,3),\"*10**-2 ohm/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The semiconductor has greater electron concentration, it is n-type semiconductor\n",
+ "(b) Resistivity is 3.396 *10**-2 ohm/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 33.4 Page no 904"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19 #C\n",
+ "A=500 #ohm**-1 m**-1\n",
+ "Ue=0.39 #m**2 V**-1 s**-1\n",
+ "\n",
+ "#Calculation\n",
+ "Ne=A/(e*Ue)\n",
+ "\n",
+ "#Result\n",
+ "print\"The number density of donor atoms is\",round(Ne*10**-21,3)*10**21,\"m**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The number density of donor atoms is 8.013e+21 m**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 33.5 Page no 904"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "F=1.6*10**-19 #E\n",
+ "W=4.2*10**8\n",
+ "e=2.4\n",
+ "w=4.2*10**-8\n",
+ "\n",
+ "#Calculation\n",
+ "S=F*W\n",
+ "A=S/F\n",
+ "E=e/w\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric field is\", round(E*10**-7,2),\"*10**7 V/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric field is 5.71 *10**7 V/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 33.6 Page no 904"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.62*10**-34\n",
+ "c=3*10**8\n",
+ "l=630*10**-9\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "Eg=(h*c)/(l*e)\n",
+ "\n",
+ "#Result\n",
+ "print\"Width of the forbidden energy gap is\",round(Eg,2),\"eV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Width of the forbidden energy gap is 1.97 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 33.7 Page no 904"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#iven\n",
+ "A=10**-4 #m**2\n",
+ "l=0.1 #m\n",
+ "V=2 #V\n",
+ "T=300 #K\n",
+ "ue=0.135 #m**2/V/S\n",
+ "n=1.5*10**15 #/m**3\n",
+ "uh=0.048 #m**2/V/S\n",
+ "e=1.6*10**-19\n",
+ "ue1=0.39\n",
+ "uh1=0.19\n",
+ "n1=2.4*10**19\n",
+ "\n",
+ "#Calculation\n",
+ "E=V/l\n",
+ "ve=ue*E\n",
+ "vh=uh*E\n",
+ "Ie=e*A*n*ve\n",
+ "Ih=e*A*n*vh\n",
+ "I=Ie+Ih\n",
+ "ve1=ue1*E\n",
+ "ve2=uh1*E\n",
+ "Ie1=e*A*n1*ve1\n",
+ "Ie2=e*A*n1*ve2\n",
+ "I1=Ie1+Ie2\n",
+ "\n",
+ "#Result\n",
+ "print\"Electron current is\", Ie*10.0,\"A \\nHole current is\",Ih*10,\"A\"\n",
+ "print\"Magnitude of total current is\",I*10,\"A \\nTotal current when germanium is used is\",I1*10**3,\"*10**-3 A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electron current is 6.48e-07 A \n",
+ "Hole current is 2.304e-07 A\n",
+ "Magnitude of total current is 8.784e-07 A \n",
+ "Total current when germanium is used is 4.4544 *10**-3 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap34_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap34_1.ipynb new file mode 100644 index 00000000..9aea6687 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap34_1.ipynb @@ -0,0 +1,621 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:13c8f6846e205753ae8bd963e82228c66a5cc1ec104d777be242a6767f4586d5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 34 Semiconductor devices"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 34.1 Page no 932"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=3 #V\n",
+ "Vd=0.7\n",
+ "R=100.0\n",
+ "\n",
+ "#Calculation\n",
+ "V=E-Vd\n",
+ "I=V/R\n",
+ "\n",
+ "#Result\n",
+ "print\"Current in the circuit is\",I*10**3,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current in the circuit is 23.0 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 34.2 Page no 932"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=1.5 #V\n",
+ "Vd=0.5\n",
+ "P=0.1 #W\n",
+ "\n",
+ "#Calculation\n",
+ "I=P/Vd\n",
+ "V=E-Vd\n",
+ "R=V/I\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of resistance is\",R,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of resistance is 5.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 34.3 Page no 932"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=9 #V\n",
+ "Vz=6 #V\n",
+ "Rl=1000.0 #ohm\n",
+ "R=100.0 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "V=E-Vz\n",
+ "I=V/R\n",
+ "Il=Vz/Rl\n",
+ "Iz=I-Il\n",
+ "Pz=Vz*Iz\n",
+ "\n",
+ "#Result\n",
+ "print\"Power dissipated in zener diode is\",Pz,\"watt\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power dissipated in zener diode is 0.144 watt\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 34.4 Page no 932"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E0=50 #V\n",
+ "S=2.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "E1=50/math.sqrt(S) #V\n",
+ "V=math.sqrt(E1**2/S)\n",
+ "V1=(S*E0/math.pi)/S\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The r.m.s. voltage across Rl is\",V,\"V\"\n",
+ "print\"(b) Reading of a d.c voltametre connected across Rl is\",round(V1,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The r.m.s. voltage across Rl is 25.0 V\n",
+ "(b) Reading of a d.c voltametre connected across Rl is 15.92 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 34.5 Page no 933"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=0.96\n",
+ "Ie=7.2 #mA\n",
+ "\n",
+ "#Calculation\n",
+ "Ic=A*Ie\n",
+ "Ib=Ie-Ic\n",
+ "\n",
+ "#Result\n",
+ "print\"The base current is\",round(Ib,2),\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The base current is 0.29 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 34.6 Page no 933"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=70\n",
+ "Ie=8.8 #mA\n",
+ "Ib=1.0\n",
+ "Ic=70.0\n",
+ "\n",
+ "#Calculation\n",
+ "Ib1=Ie/(Ic+Ib)\n",
+ "Ic1=Ic*Ib1\n",
+ "A=Ic/(Ic+Ib)\n",
+ "\n",
+ "#Result\n",
+ "print\"The collector current is\",round(Ib1,3),\"mA\"\n",
+ "print\"The base current is\",round(Ic1,2),\"mA\"\n",
+ "print\"Current gain is\",round(A,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The collector current is 0.124 mA\n",
+ "The base current is 8.68 mA\n",
+ "Current gain is 0.986\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 34.7 Page no 933"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ib=105*10**-6 #A\n",
+ "Ic=2.05*10**-3 #A\n",
+ "Ib1=27*10**-6 #A\n",
+ "Ic1=650*10**-6 #A\n",
+ "\n",
+ "#Calculation\n",
+ "B=Ic/Ib\n",
+ "Ie=Ib+Ic\n",
+ "A=Ic/Ie\n",
+ "Bac=Ic1/Ib1\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The value of B is\",round(B,1),\",Value of Ie is\",Ie*10**3,\"*10**-3 A\",\"and Value of A is\",round(A,2)\n",
+ "print\"(b) The value of Bac is\",round(Bac,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The value of B is 19.5 ,Value of Ie is 2.155 *10**-3 A and Value of A is 0.95\n",
+ "(b) The value of Bac is 24.07\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 34.8 Page no 933"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ie=7.89*10**-3 #A\n",
+ "Ic=7.8*10**-3 #A\n",
+ "\n",
+ "#Calculation\n",
+ "A=Ic/Ie\n",
+ "B=A/(1-A)\n",
+ "Ib=Ic/B\n",
+ "\n",
+ "#Result\n",
+ "print\"change in the base current is\",Ib,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "change in the base current is 9e-05 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 34.9 Page no 933"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ib=20.0*10**-6 #A\n",
+ "Vbe=0.02 #V\n",
+ "Ic=2*10**-3 #A\n",
+ "Rl=5000 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "P=Vbe/Ib\n",
+ "B=Ic/Ib\n",
+ "Gm=Ic/Vbe\n",
+ "Q=(Rl*Ic)/Vbe\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The input resistance Bac is\",P,\"ohm\",\"and transconductance of the transister is\",Gm,\"ohm**-1\"\n",
+ "print\"(b) The voltage gain of the amplifier is\",Q"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The input resistance Bac is 1000.0 ohm and transconductance of the transister is 0.1 ohm**-1\n",
+ "(b) The voltage gain of the amplifier is 500.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 34.10 Page no 933"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ib=15*10**-6 #A\n",
+ "Ic=2*10**-3 #A\n",
+ "R=665.0 #ohm\n",
+ "Rl=5*10**3 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "B=Ic/Ib\n",
+ "Gm=B/R\n",
+ "Av=Gm*Rl\n",
+ "\n",
+ "#Result \n",
+ "print\"(i) Current gain is\",round(B,1)\n",
+ "print\"(ii) Transconductance is\",round(Gm,1),\"ohm**-1\"\n",
+ "print\"(iii) Voltage gain Av of the amplifier is\",round(Av,0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Current gain is 133.3\n",
+ "(ii) Transconductance is 0.2 ohm**-1\n",
+ "(iii) Voltage gain Av of the amplifier is 1003.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 34.11 Page no 934"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=100.0\n",
+ "Vcc=24 #V\n",
+ "Ic=1.5*10**-3 #A\n",
+ "Rc=4.7*10**3\n",
+ "Rb=220*10**3\n",
+ "\n",
+ "#Calculation\n",
+ "Ib=Ic/B\n",
+ "Vce=Vcc-(Ic*Rc)\n",
+ "Vbe=Vcc-(Ib*Rb)\n",
+ "Vbc=(Ic*Rc)-(Ib*Rb)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of Ib is\",Ib,\"A\"\n",
+ "print\"The value of Vce is\",Vce,\"V\"\n",
+ "print\"The value of Vbe is\",Vbe,\"V\"\n",
+ "print\"The value of Vbc is\",Vbc,\"V\"\n",
+ "print\"The transistor is in saturation state\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Ib is 1.5e-05 A\n",
+ "The value of Vce is 16.95 V\n",
+ "The value of Vbe is 20.7 V\n",
+ "The value of Vbc is 3.75 V\n",
+ "The transistor is in saturation state\n"
+ ]
+ }
+ ],
+ "prompt_number": 84
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 34.15 Page no 935"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vb=5 #V\n",
+ "I=10**3 #A\n",
+ "Q=0.7\n",
+ "I1=5*10**-3\n",
+ "Vb1=6 #V\n",
+ "R0=1000 #ohm\n",
+ "I3=10**-3 #A\n",
+ "\n",
+ "#Calculation\n",
+ "R=(Vb-Q)*I\n",
+ "R1=(R*10**-3)/I1\n",
+ "R3=(Vb1-Q)/I1\n",
+ "R4=I1**2*R3\n",
+ "R5=Q*I1\n",
+ "Vz=(I3*R0)+Q\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The maximum value of R is\",R*10**-3,\"10**3\",\"ohm\"\n",
+ "print\"(b) The value of R is\",R1,\"ohm\"\n",
+ "print\"(c) The power dissipated across R is\",R4*10**3,\"10**-3\",\"ohm\",\"and across diode is\",R5*10**3,\"10**-3 W\"\n",
+ "print\"(d) The maximum voltage Vb is\",Vz,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The maximum value of R is 4.3 10**3 ohm\n",
+ "(b) The value of R is 860.0 ohm\n",
+ "(c) The power dissipated across R is 26.5 10**-3 ohm and across diode is 3.5 10**-3 W\n",
+ "(d) The maximum voltage Vb is 1.7 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 148
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 34.16 Page no 935"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I0=5*10**-12 #A\n",
+ "K=8.6*10**-5 #eVK**-1\n",
+ "Q=1.6*10**-19 #J K**-1\n",
+ "V=0.6 #volts\n",
+ "V1=0.7 #volts\n",
+ "A=23.256\n",
+ "A1=27.132\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=-I0*(Q*V/math.exp(K*Q)-1)\n",
+ "Z=I*(math.exp(A)-1)\n",
+ "I1=-I0*(Q*V1/math.exp(K*Q)-1)\n",
+ "Z1=I1*(math.exp(A1)-1)\n",
+ "S=Z1-Z\n",
+ "V2=V1-V\n",
+ "J=V2/S\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The forward current at a forward voltage is\",round(Z,4),\"A\"\n",
+ "print\"(b) The voltage across the diode is\",round(S,4),\"A\"\n",
+ "print\"(c) The dynamic resistance is\",round(J,4),\"ohm\"\n",
+ "print\"(d) For change in votage from 1V to 2V,the current will remain equal to\",I0,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The forward current at a forward voltage is 0.0629 A\n",
+ "(b) The voltage across the diode is 2.9727 A\n",
+ "(c) The dynamic resistance is 0.0336 ohm\n",
+ "(d) For change in votage from 1V to 2V,the current will remain equal to 5e-12 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 34.17 Page no 936"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vbe=24*10**-3 #V\n",
+ "Ib=32*10**-6 #A\n",
+ "Ic=3.6*10**-3 #A\n",
+ "Rl=4.8*10**3\n",
+ "\n",
+ "#Calculation\n",
+ "B=Ic/Ib\n",
+ "Rbe=Vbe/Ib\n",
+ "Gm=Ic/Vbe\n",
+ "Av=(Rl*Ic)/Vbe\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The current gain is\",B\n",
+ "print\"(ii) The input resistance Rbe is\",Rbe,\"ohm\"\n",
+ "print\"(iii) The transconductance Gm is\",Gm,\"S\"\n",
+ "print\"(iv) Voltage gain Av is\",Av"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The current gain is 112.5\n",
+ "(ii) The input resistance Rbe is 750.0 ohm\n",
+ "(iii) The transconductance Gm is 0.15 S\n",
+ "(iv) Voltage gain Av is 720.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap35_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap35_1.ipynb new file mode 100644 index 00000000..27056667 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap35_1.ipynb @@ -0,0 +1,263 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:6fd837c473d583739b16708ec5ff802d29867f6cb2f48bae07778822e1c04084"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 35 Analog and Digital Communication"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 35.1 Page no 978"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Em=0.5 \n",
+ "Ec=1\n",
+ "\n",
+ "#Calculation\n",
+ "Emax=Ec+Em\n",
+ "Emin=Ec-Em\n",
+ "Ma=(Emax-Emin)/(Emax+Emin)\n",
+ "\n",
+ "#Result\n",
+ "print\"The modulation index is\",Ma"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The modulation index is 0.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 35.2 Page no 978"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=3.6*10**-3 #H\n",
+ "C=2.5*10**-12 #F\n",
+ "Fm=15 #KHz\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Fc=1/(2*math.pi*math.sqrt(L*C))*10**-3\n",
+ "S=Fc+Fm\n",
+ "S1=Fc-Fm\n",
+ "\n",
+ "#Result\n",
+ "print\"Frequency of upper sideband is\",round(S,0),\"kHz\"\n",
+ "print\"Frequency of lower sideband is\",round(S1,0),\"kHz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Frequency of upper sideband is 1693.0 kHz\n",
+ "Frequency of lower sideband is 1663.0 kHz\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 35.3 Page no 978"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Fm=3.2 #kHz\n",
+ "Fc=84*10**3 #kHz\n",
+ "A=96 #kHz\n",
+ "\n",
+ "#Calculation\n",
+ "Mf=A/Fm\n",
+ "M1=Fc-Fm\n",
+ "M2=Fc+Fm\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Frequency modulation index is\",Mf\n",
+ "print\"(b) Frequency range of the modulated wave is\",round(M1*10**-3,3),\"*10**3 MHz\",\"to\",round(M2*10**-3,3),\"*10**3 MHz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Frequency modulation index is 30.0\n",
+ "(b) Frequency range of the modulated wave is 83.997 *10**3 MHz to 84.003 *10**3 MHz\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 35.4 Page no 979"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=1\n",
+ "B=2\n",
+ "C=0\n",
+ "\n",
+ "#Calculation\n",
+ "Q1=(A*B**0)+(C*B**1)+(C*B**2)+(C*B**3)\n",
+ "Q2=(C*B**0)+(A*B**1)+(C*B**2)+(C*B**3)\n",
+ "Q3=(C*B**0)+(C*B**1)+(A*B**2)+(C*B**3)\n",
+ "Q4=(C*B**0)+(C*B**1)+(C*B**2)+(A*B**3)\n",
+ "Q5=(A*B**0)+(C*B**1)+(C*B**2)+(A*B**3)\n",
+ "Q6=(A*B**0)+(C*B**1)+(A*B**2)+(A*B**3)\n",
+ "\n",
+ "#Result\n",
+ "print\"Input voltage 0001 is\",Q1,\"V\"\n",
+ "print\"Input voltage 0010 is\",Q2,\"V\"\n",
+ "print\"Input voltage 0100 is\",Q3,\"V\"\n",
+ "print\"Input voltage 1000 is\",Q4,\"V\"\n",
+ "print\"Input voltage 1001 is\",Q5,\"V\"\n",
+ "print\"Input voltage 1101 is\",Q6,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Input voltage 0001 is 1 V\n",
+ "Input voltage 0010 is 2 V\n",
+ "Input voltage 0100 is 4 V\n",
+ "Input voltage 1000 is 8 V\n",
+ "Input voltage 1001 is 9 V\n",
+ "Input voltage 1101 is 13 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 55
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 35.5 Page no 979"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=2\n",
+ "N=32\n",
+ "\n",
+ "#Calculation\n",
+ "Z=A**N\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of quantisation levels is\",round(Z*10**-9,1),\"*10**9\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of quantisation levels is 4.3 *10**9\n"
+ ]
+ }
+ ],
+ "prompt_number": 60
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 35.6 Page no 979"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "S=16\n",
+ "Q=8000\n",
+ "\n",
+ "#Calculation\n",
+ "N=math.sqrt(S)\n",
+ "W=Q*N\n",
+ "\n",
+ "#Result\n",
+ "print\"The bill rate for a signal is\",W,\"bill s**-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The bill rate for a signal is 32000.0 bill s**-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 69
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap36_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap36_1.ipynb new file mode 100644 index 00000000..4e06b38b --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap36_1.ipynb @@ -0,0 +1,103 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:d9296ed3bbb0144b6fddcd9bd9b4d00b56d458558332cd059b27e6750c38d570"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 36 Space Communication"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 36.1 Page no 990"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "AB=2*10**5 #m\n",
+ "CN=100 #km\n",
+ "C=3.0*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "Tg=AB/C\n",
+ "Q=(math.sqrt(CN**2+CN**2))+(math.sqrt(CN**2+CN**2))\n",
+ "Ts=Q*10**3/C\n",
+ "W=(Ts-Tg)\n",
+ "\n",
+ "#Result\n",
+ "print\"The time delay is\",round(W*10**4,2),\"*10**-4 s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The time delay is 2.76 *10**-4 s\n"
+ ]
+ }
+ ],
+ "prompt_number": 57
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 36.2 Page no 990"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "H=80 #m\n",
+ "R=6.4*10**6\n",
+ "W=800 #km**-2\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "D=math.sqrt(2*H*R)\n",
+ "A=math.pi*D**2\n",
+ "S=A*W\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The radius of the circle is\",D*10**-4,\"10**4 m\"\n",
+ "print\"(b) The population covered by the transmission is\",round(S*10**-12,3),\"*10**6\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The radius of the circle is 3.2 10**4 m\n",
+ "(b) The population covered by the transmission is 2.574 *10**6\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap37_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap37_1.ipynb new file mode 100644 index 00000000..03f26ef4 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap37_1.ipynb @@ -0,0 +1,183 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:10e3d0cfbf86d2f1081a9d348d19bd93a499331702a8a9f3c32ddc1a8679292a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 37 Line Communication"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 37.1 Page no 1007"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "U=1.48\n",
+ "U2=1.45\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "C=U2/U\n",
+ "C1=math.asin(C)*180/3.14\n",
+ "\n",
+ "#Result \n",
+ "print\"The critical angle for a light is\",round(C1,1),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The critical angle for a light is 78.5 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 37.2 Page no 1007"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.7\n",
+ "u2=1.5\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "C=u2/u\n",
+ "A=math.asin(C)*180/3.14\n",
+ "I=math.sqrt(u**2-u2**2)\n",
+ "A1=math.asin(I)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"The critical angle is\",round(A,1),\"degree\"\n",
+ "print\"The angle of acceptance is\",round(A1,1),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The critical angle is 62.0 degree\n",
+ "The angle of acceptance is 53.2 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 37.3 Page no 1007"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u1=1.5\n",
+ "u2=1.3\n",
+ "L=30 #m\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "V=c/u1\n",
+ "t1=L/V\n",
+ "C=u2/u1\n",
+ "C1=math.asin(C)*180/3.14\n",
+ "x=L/C\n",
+ "t=x/V\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Time taken for light in axical mode is\",t1*10**7,\"*10**-7 s\" \n",
+ "print\"(b) Time taken for light in highest order mode is\",round(t*10**7,2),\"*10**-7 s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Time taken for light in axical mode is 1.5 *10**-7 s\n",
+ "(b) Time taken for light in highest order mode is 1.73 *10**-7 s\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 37.5 Page no 1007"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=0.5\n",
+ "b=50.0 #Km\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=10*math.log10(a)\n",
+ "B=A/b\n",
+ "\n",
+ "#Result\n",
+ "print\"Attenuation is\", round(B,2),\"dB/Km\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Attenuation is -0.06 dB/Km\n"
+ ]
+ }
+ ],
+ "prompt_number": 54
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap3_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap3_1.ipynb new file mode 100644 index 00000000..7c6ed32c --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap3_1.ipynb @@ -0,0 +1,783 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:c3ec75ccee40d0eb91752b49524186dbe794af4f0e24a564bb72af8d09f81af1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3 Electric potential"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.1 Page no 65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=4.0 #c\n",
+ "Va=-10 #V\n",
+ "Wab=100 #joule\n",
+ "\n",
+ "#Calculation\n",
+ "V=(Wab/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 15.0 Volts\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.2 Page no 65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Z=79\n",
+ "e=1.6*10**-19 #C\n",
+ "r=6.6*10**-15 #m\n",
+ "m=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "q=Z*e\n",
+ "V=(m*q)/r\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric potential at the surface of a gold nucleus is\", round(V*10**-7,3),\"*10**7 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric potential at the surface of a gold nucleus is 1.724 *10**7 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3 Page no 65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=4*10**-7 #C\n",
+ "r=0.09 #m\n",
+ "m=9*10**9\n",
+ "a=2*10**-9\n",
+ "\n",
+ "#Calculation\n",
+ "Vp=(m*q)/r\n",
+ "W=a*Vp\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Potential at point P is\", Vp*10**-4,\"*10**4 V\"\n",
+ "print\"(b) Work done is\", W,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Potential at point P is 4.0 *10**4 V\n",
+ "(b) Work done is 8e-05 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.4 Page no 66"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=20*10**-6 #C\n",
+ "r1=0.1 #m\n",
+ "r2=0.05 #m\n",
+ "m=9*10**9\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "Va=(m*q)/r1\n",
+ "Vb=(m*q)/r2\n",
+ "Wab=-(Va-Vb)*e\n",
+ "\n",
+ "#Result\n",
+ "print\"Potential at point A is\", Va*10**-6,\"*10**6 V\"\n",
+ "print\"Potential at point b is\", Vb*10**-6,\"*10**6 V\"\n",
+ "print\"Work done is\",Wab,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Potential at point A is 1.8 *10**6 V\n",
+ "Potential at point b is 3.6 *10**6 V\n",
+ "Work done is 2.88e-13 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.5 Page no 66"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9.0*10**9\n",
+ "r=0.5\n",
+ "a=16\n",
+ "\n",
+ "#Calculation\n",
+ "q=(a*r)/m\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of charge is\", round(q*10**10,2)*10**-10,\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of charge is 8.89e-10 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.6 Page no 66"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q1=2*10**-6 #c\n",
+ "q2=-2*10**-6\n",
+ "q3=-3*10**-6\n",
+ "q4=6*10**-6\n",
+ "r=1.0 #m\n",
+ "m=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "V=m*(q1+q2+q3+q4)/r\n",
+ "\n",
+ "#Result\n",
+ "print\"Potential at the centre of the square is\",V*10**-4,\"*10**4 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Potential at the centre of the square is 2.7 *10**4 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.7 Page no 66"
+ ]
+ },
+ {
+ "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",
+ "V=(6*m*q)/r\n",
+ "\n",
+ "#Result\n",
+ "print\"Potential at the centre of the hexagon is\", V*10**-6,\"*10**6 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Potential at the centre of the hexagon is 2.7 *10**6 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.9 Page no 67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.1 #m\n",
+ "q=200*10**-6 #C\n",
+ "m=9*10**9\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 1.8 *10**7 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 50
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.11 Page no 67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=0.2 #m\n",
+ "m=9*10**9\n",
+ "q1=2*10**-9\n",
+ "q2=4*10**-9\n",
+ "q3=8*10**-9\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "V=m*((q1/a)+(q2/(a*math.sqrt(2)))+(q3/a))\n",
+ "b=0.1*math.sqrt(2)\n",
+ "V0=m*((q1/b)+(q2/b)+(q3/b))\n",
+ "W=q*(V0-V)\n",
+ "\n",
+ "#Result\n",
+ "print\"Required work is\",round(W*10**2,2)*10**-7,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Required work is 6.27e-07 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 62
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.12 Page no 68"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=60\n",
+ "r=0.05\n",
+ "\n",
+ "#Calculation\n",
+ "E=-V/r\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric field intensity between the two plate is\", E,\"V/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric field intensity between the two plate is -1200.0 V/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.13 Page no 68"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19\n",
+ "m=3*10**-16 #Kg\n",
+ "r=5*10**-3 #m\n",
+ "g=9.8\n",
+ "\n",
+ "#Calculation\n",
+ "q=10*e\n",
+ "V=(m*g*r)/q\n",
+ "\n",
+ "#Result\n",
+ "print round(V,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "9.19 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 71
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.14 Page no 68"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q1=7*10**-6 #c\n",
+ "q2=-2*10**-6 #C\n",
+ "r12=0.18 #m\n",
+ "m=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "U=m*((q1*q2)/r12)\n",
+ "W=0-U\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Electrostatic potential energy of a system is\",U,\"J\"\n",
+ "print\"(b) work required is\", W,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Electrostatic potential energy of a system is -0.7 J\n",
+ "(b) work required is 0.7 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.15 Page no 68"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "e=1.6*10**-19 #C\n",
+ "r=5.3*10**-11 #m\n",
+ "\n",
+ "#Calculation\n",
+ "V=(m*e)/r\n",
+ "U=((m*e*-e)/r)/e\n",
+ "q=2*e\n",
+ "V1=(m*q)/r\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Electric potential is\", round(V,2),\"V\"\n",
+ "print\"(b) Electric potential energy is\",round(V1,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Electric potential is 27.17 V\n",
+ "(b) Electric potential energy is 54.34 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 91
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.16 Page no 69"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r12=1.5*10**-10 #m\n",
+ "r13=10.0**-10\n",
+ "q1=1.6*10**-19 #C\n",
+ "q3=-1.6*10**-19\n",
+ "m=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "U=m*(((q1**2)/r12)+((q1*q3)/r13)+((q1*q3)/r13))/q1\n",
+ "\n",
+ "#Result\n",
+ "print\"Potential energy is\",U,\"ev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Potential energy is -19.2 ev\n"
+ ]
+ }
+ ],
+ "prompt_number": 109
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.17 Page no 69"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=10**5 #N/C\n",
+ "q=10**-6 #C\n",
+ "a=0.02 #m\n",
+ "A=1\n",
+ "\n",
+ "#Calculation'\n",
+ "import math\n",
+ "p=q*a\n",
+ "t=p*E*A\n",
+ "W=p*E*(math.cos(0*3.14/180.0)-math.cos(180*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Torque acting on dipole is\",t,\"N m\"\n",
+ "print\"(ii) Work done is\", round(W,3),\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Torque acting on dipole is 0.002 N m\n",
+ "(ii) Work done is 0.004 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 126
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.18 Page no 69"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q=8*10**-3\n",
+ "q0=-2*10**-9\n",
+ "ra=0.03\n",
+ "rb=0.04\n",
+ "\n",
+ "#Calculation\n",
+ "W=(m*q*q0*(1/rb-1/ra))\n",
+ "\n",
+ "#Result\n",
+ "print\"Work done is\", W,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Work done is 1.2 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 129
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.19 Page no 69"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "qa=1.5*10**-6 #C\n",
+ "qb=2.5*10**-6 \n",
+ "a=0.15 #m\n",
+ "m=9*10**9\n",
+ "a1=0.18\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "V=((qa+qb)/a)*m\n",
+ "Ea=m*qa/a**2\n",
+ "Eb=m*qb/a**2\n",
+ "E=Eb-Ea\n",
+ "V1=((qa+qb)/a1)*m\n",
+ "Ea1=m*qa/a1**2\n",
+ "Ea2=m*qb/a1**2\n",
+ "A=math.atan(15/10.0)*180/3.14\n",
+ "A1=2*A\n",
+ "A2=math.cos(A1*3.14/180.0)\n",
+ "E3=math.sqrt(Ea1**2+Ea2**2+(2*Ea1*Ea2*A2))\n",
+ "tb=Ea1*math.sin(A1*3.14/180.0)/(Ea2+Ea1*math.cos(A1*3.14/180.0))\n",
+ "B=math.atan(tb)*180/3.14\n",
+ "y=90-A\n",
+ "Y=y+B\n",
+ "\n",
+ "#Result\n",
+ "print \"(a) Net electric field is\",E*10**-5,\"*10**5 N/C\"\n",
+ "print\"(b) Angle is\" ,round(Y,1),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Net electric field is 4.0 *10**5 N/C\n",
+ "(b) Angle is 69.4 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 160
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.23 Page no 71"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q1=-1.6*10**-19 #C\n",
+ "q2=1.6*10**19\n",
+ "r12=0.53*10**-10 #m\n",
+ "m=9*10**9\n",
+ "r11=1.06*10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "U=((m*q1*q2)/r12)/q2\n",
+ "K=U/2.0\n",
+ "E=-(U-K)\n",
+ "U1=((m*q1*q2)/r11)/q2\n",
+ "E1=U-U1\n",
+ "\n",
+ "#Result\n",
+ "print \"(a) Potential energy is\",round(U,2),\"ev\"\n",
+ "print\"(b) Minimum work required is\",round(E,3) ,\"ev\"\n",
+ "print\"(c) Potential energy is\",round(E1,3),\"ev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Potential energy is -27.17 ev\n",
+ "(b) Minimum work required is 13.585 ev\n",
+ "(c) Potential energy is -13.585 ev\n"
+ ]
+ }
+ ],
+ "prompt_number": 181
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.24 Page no 72"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=10**6 #V/m\n",
+ "p=10**-29 #cm\n",
+ "a=6.023*10**23\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Np=a*p\n",
+ "U1=Np*E\n",
+ "U2=Np*E*math.cos(60*3.14/180.0)\n",
+ "U=U1-U2\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat released by the substance is\", round(U,2),\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat released by the substance is 3.01 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 188
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap4_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap4_1.ipynb new file mode 100644 index 00000000..488631fb --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap4_1.ipynb @@ -0,0 +1,397 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:1087fb8dbcb5b183b8a1d36b431412b33d35209d5f6a3a237469594e817213fb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 4 Gauss theorem"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.1 Page no 88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=25 #V/m\n",
+ "s=150*10**-4 #m**2\n",
+ "a=60 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=E*s*math.cos(a*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Flux of the electric field is\",round(A,4),\"Nm**2/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Flux of the electric field is 0.1877 Nm**2/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.2 Page no 88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=3*10**3 #N/C\n",
+ "S=10**-2 #m**2\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=E*S\n",
+ "A=E*S*math.cos(60*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Flux of the field is\",a,\"Nm**2/C\"\n",
+ "print\"(b) Flux through the square is\", round(A,0),\"Nm**2/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Flux of the field is 30.0 Nm**2/C\n",
+ "(b) Flux through the square is 15.0 Nm**2/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.3 Page no 88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=8.854*10**-12 #C**2/N/m**2\n",
+ "q=1 #C\n",
+ "\n",
+ "#Calculation\n",
+ "a=q/e\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of electric lines are\",round(a*10**-11,3)*10**11"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of electric lines are 1.129e+11\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4 Page no 88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=8*10**3 #Nm**2/C\n",
+ "e=8.854*10**-12 #C**2/N/m**2\n",
+ "\n",
+ "#Calculation\n",
+ "q=a*e\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Net charge inside the box is\",q,\"C\"\n",
+ "print\"(b) If the net outward flux is zero,we can't conclude that the charge inside the box is zero.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Net charge inside the box is 7.0832e-08 C\n",
+ "(b) If the net outward flux is zero,we can't conclude that the charge inside the box is zero.\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.5 Page no 89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=10**-5\n",
+ "e=8.854*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "a=q/(6.0*e)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of electric flux is\", round(a*10**-5,2),\"*10**5 Nm**2/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of electric flux is 1.88 *10**5 Nm**2/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.6 Page no 89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=9*10**4 #N/C\n",
+ "r=0.04 #m\n",
+ "m=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "l=E*r/(2.0*m)\n",
+ "\n",
+ "#Result\n",
+ "print\"Linear charge density is\", l,\"C/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Linear charge density is 2e-07 C/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.7 Page no 89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=5*10**-6 #C/m**2\n",
+ "e=8.854*10**-12 #C**2/Nm**2\n",
+ "r=0.1 #m\n",
+ "a1=60 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=(a*math.pi*r**2*math.cos(a1*3.14/180.0))/(2.0*e)\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric flux is\", round(A*10**-3,2),\"*10**3 N m**2/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric flux is 4.44 *10**3 N m**2/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.8 Page no 89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=17*10**-22 #C/m**2\n",
+ "e=8.854*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "E=a/e\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Electric field to the left of the plate is zero\"\n",
+ "print\"(b) Electric field to the right of the plate is zero\"\n",
+ "print\"(c) Electric field between the plates is\",round(E*10**10,2)*10**-10,\"N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Electric field to the left of the plate is zero\n",
+ "(b) Electric field to the right of the plate is zero\n",
+ "(c) Electric field between the plates is 1.92e-10 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.9 Page no 89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=1.6*10**-7 #C\n",
+ "R=0.12 #m\n",
+ "m=9*10**9\n",
+ "r=0.18\n",
+ "\n",
+ "#Calculation\n",
+ "E=(m*q)/R**2\n",
+ "E1=(m*q)/r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) ELectric field inside the sphere is zero\"\n",
+ "print\"(b) Electric field outside the sphere is\",E,\"N/C\"\n",
+ "print\"(c) Electric field at a point 18 cm from the centre is\", round(E1*10**-4,2),\"*10**4 N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) ELectric field inside the sphere is zero\n",
+ "(b) Electric field outside the sphere is 100000.0 N/C\n",
+ "(c) Electric field at a point 18 cm from the centre is 4.44 *10**4 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.11 Page no 90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=50\n",
+ "V=0.2\n",
+ "m=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "q=e*V**2/m\n",
+ "\n",
+ "#Result\n",
+ "print\"Charge contained in the sphere is\",round(q*10**10,2)*10**-10,\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge contained in the sphere is 2.22e-10 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 56
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap5_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap5_1.ipynb new file mode 100644 index 00000000..9440d328 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap5_1.ipynb @@ -0,0 +1,1163 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:747c84b8bc31c8489cef11149066f567e7dd3e35d04638d523cfa4a0f1ae89cc"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 5 Capacitor"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.1 Page no 108"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=6.4*10**6 #m\n",
+ "t=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "C=r/t\n",
+ "\n",
+ "#Result\n",
+ "print\"The capacitance is\",round(C*10**6,0),\"F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The capacitance is 711.0 F\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.2 Page no 108"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.01 #m\n",
+ "q=1 #C\n",
+ "d=9.0*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "V=(q*d)/r\n",
+ "\n",
+ "#Result\n",
+ "print\"V=\",V,\"V\"\n",
+ "print\"The given metal sphere will not be able to hold charge of 1 C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "V= 9e+11 V\n",
+ "The given metal sphere will not be able to hold charge of 1 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.3 Page no 109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=3\n",
+ "Q=27\n",
+ "q=4*220\n",
+ "\n",
+ "#Calculation\n",
+ "d=(Q*q)/(4*R)\n",
+ "\n",
+ "#Result\n",
+ "print\"The potential of the bigger drop is\",d,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The potential of the bigger drop is 1980 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.4 Page no 109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=2 #F\n",
+ "d=0.5*10**-2 #m\n",
+ "a=8.854*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "Q=(C*d)/a\n",
+ "\n",
+ "#Result\n",
+ "print\"Area of the plate is\", round(Q*10**-9,2),\"*10**9 m**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Area of the plate is 1.13 *10**9 m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5 Page no 109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=10**-2 #m**2\n",
+ "d=10**-3 #m\n",
+ "q=0.12*10**-6\n",
+ "V=120.0\n",
+ "e=8.85*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "C=q/V\n",
+ "K=C*d/(e*A)\n",
+ "\n",
+ "#Result\n",
+ "print\"Dielectric constant of the material is\", round(K,1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Dielectric constant of the material is 11.3\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.6 Page no 109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A1=100.0 #cm**2\n",
+ "A2=500 #cm**2\n",
+ "d1=0.05 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "d2=(A2*d1)/A1\n",
+ "\n",
+ "#Result\n",
+ "print\"Difference between the plates of second capacitor is\", d2,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Difference between the plates of second capacitor is 0.25 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.7 Page no 109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=6.45*10**6 #m\n",
+ "b=6.4*10**6 #m\n",
+ "m=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "C=(a*b)/(m*(a-b))\n",
+ "\n",
+ "#Result\n",
+ "print\"Capacitance of the spherical capacitor is\",round(C,3),\"F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance of the spherical capacitor is 0.092 F\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.8 Page no 109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=12*10**-2 #m\n",
+ "b=13*10**-2 #m\n",
+ "q=2.5*10**-6 #C\n",
+ "K=32\n",
+ "m=9.0*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "C=(K*a*b)/(m*(b-a))\n",
+ "V=q/C\n",
+ "C1=a/m\n",
+ "C2=C/C1\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Capacitance of the capacitor is\", round(C*10**9,3)*10**-9,\"F\"\n",
+ "print\"(b) Potential of the inner sphere is\", round(V,1),\"V\"\n",
+ "print\"(c) Capacitance of the capacitor is\",C2"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Capacitance of the capacitor is 5.547e-09 F\n",
+ "(b) Potential of the inner sphere is 450.7 V\n",
+ "(c) Capacitance of the capacitor is 416.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 61
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.9 Page no 110"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=1.4*10**-2 #m\n",
+ "b=1.5*10**-2 #m\n",
+ "q=3.5*10**-6 #C\n",
+ "e=8.854*10**-12\n",
+ "l=15*10**-2\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "C=(2*math.pi*e*l)/(2.303*math.log10(b/a))\n",
+ "V=q/C\n",
+ "\n",
+ "#Result\n",
+ "print\"Capacitance of the system is\", round(C*10**10,2)*10**-8,\"F\"\n",
+ "print\"Potential of the inner cylinder is\", round(V*10**-4,2),\"*10**4 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance of the system is 1.21e-08 F\n",
+ "Potential of the inner cylinder is 2.89 *10**4 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.10 Page no 110"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=9.0*10**-12 #F\n",
+ "V=120 #V\n",
+ "\n",
+ "#Calculation\n",
+ "C1=1/(3/C)\n",
+ "q=(V*C)/3.0\n",
+ "V1=q/C\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Total capacitance is\",C1,\"F\"\n",
+ "print\"(b) Potential difference across each capacitor is\",V1,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Total capacitance is 3e-12 F\n",
+ "(b) Potential difference across each capacitor is 40.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.11 Page no 110"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C1=2*10**-12 #F\n",
+ "C2=3*10**-12\n",
+ "C3=4*10**-12\n",
+ "V=100 #Volts\n",
+ "\n",
+ "#Calculation\n",
+ "C=C1+C2+C3\n",
+ "q1=C1*V\n",
+ "q2=C2*V\n",
+ "q3=C3*V\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Total capacitamce of the parallel combination is\",C,\"F\"\n",
+ "print\"(b) Charge on q1 is\",q1,\"C\"\n",
+ "print\"(c) Charge on q2 is\",q2,\"C\"\n",
+ "print\"(d) Charge on q3 is\",q3,\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Total capacitamce of the parallel combination is 9e-12 F\n",
+ "(b) Charge on q1 is 2e-10 C\n",
+ "(c) Charge on q2 is 3e-10 C\n",
+ "(d) Charge on q3 is 4e-10 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.12 Page no 110"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "c1=5.0 #micro F\n",
+ "c2=4.0\n",
+ "c3=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "C=1/(1/c1+1/c2)\n",
+ "c11=C+c3\n",
+ "\n",
+ "#Result\n",
+ "print\"Capacitance of the combination is\",round(c11,2),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance of the combination is 5.22 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.13 Page no 111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "cs=3 #micro F\n",
+ "cp=16\n",
+ "c1=12\n",
+ "\n",
+ "#Calculation\n",
+ "C=cs*cp\n",
+ "c2=cp-c1\n",
+ "\n",
+ "#Result\n",
+ "print\"Capacitance of each capacitor is\",c2,\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance of each capacitor is 4 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.14 Page no 111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "c2=6.0 #micro F\n",
+ "c3=6.0\n",
+ "c1=12.0\n",
+ "c4=12.0\n",
+ "\n",
+ "#Calculation\n",
+ "c23=c2+c3\n",
+ "C=1/(1/c1+1/c23+1/c4)\n",
+ "\n",
+ "#Result\n",
+ "print\"resultant of capacitance is\",C,\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "resultant of capacitance is 4.0 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.15 Page no 111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "c1=3.0 #pico F\n",
+ "c2=2.0\n",
+ "\n",
+ "#Calculation\n",
+ "C=1/(1/c1+1/c2+1/c1)\n",
+ "c11=C+c2\n",
+ "C11=1/(1/c1+1/c11+1/c1)\n",
+ "\n",
+ "#Result \n",
+ "print\"Resultant capacitance is\",round(C11,0),\"pF\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resultant capacitance is 1.0 pF\n"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.16 Page no 111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C2=2\n",
+ "C3=2\n",
+ "C1=1 #micro F\n",
+ "C4=2\n",
+ "C5=1\n",
+ "\n",
+ "#Calculation\n",
+ "C23=C2*C3/(C2+C3)\n",
+ "C123=C1+C23\n",
+ "C1234=C123*C4/(C123+C4)\n",
+ "C=C1234+C5\n",
+ "\n",
+ "#Ressult\n",
+ "print\"Equivalent capacitance is\", C,\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equivalent capacitance is 2 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.17 Page no 112"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C5=10*10**-6 #micro F\n",
+ "C6=10*10**-6\n",
+ "C2=60.0*10**-6\n",
+ "C3=60.0*10**-6\n",
+ "C4=60.0*10**-6\n",
+ "C1=40*10**-6\n",
+ "V=100 #V\n",
+ "\n",
+ "#Calculation\n",
+ "C56=(C5+C6)\n",
+ "C234=1/(1/C2+1/C3+1/C4)\n",
+ "C=C56+C234\n",
+ "C11=(C1*C)/(C1+C)\n",
+ "q=C56*V\n",
+ "\n",
+ "#Result\n",
+ "print\"Total charge is\",q*10**3,\"*10**-3 C\"\n",
+ "print\"Equivalent capacitance is\",C11*10**6,\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total charge is 2.0 *10**-3 C\n",
+ "Equivalent capacitance is 20.0 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 75
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.19 Page no 113"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C1=5.0*10**-6 #F\n",
+ "C2=10.0*10**-6\n",
+ "C3=2.0*10**-6\n",
+ "C4=4.0*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "C12=1/(1/C1+1/C2)\n",
+ "C34=1/(1/C3+1/C4)\n",
+ "C=C12+C34\n",
+ "\n",
+ "#Result\n",
+ "print\"Effective capacitance is\", round(C*10**6,2),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Effective capacitance is 4.67 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 81
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.20 Page no 113"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=3\n",
+ "C1=1.0\n",
+ "C2=1.0\n",
+ "C3=1.0\n",
+ "C=2 #micro F\n",
+ "\n",
+ "#Calculation\n",
+ "C11=1/(1/C1+1/C2+1/C3)\n",
+ "m=C/C11\n",
+ "\n",
+ "#Result\n",
+ "print\"Capacitance required is\", m"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance required is 6.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 84
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.21 Page no 113"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C1=4\n",
+ "V=6\n",
+ "C2=6\n",
+ "\n",
+ "#Calculation\n",
+ "q1=C1*V\n",
+ "q2=C2*V\n",
+ "\n",
+ "#Result\n",
+ "print\"Charge on C1 is\",q1,\"micro C\"\n",
+ "print\"Charge on C2 is\",q2,\"micro C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge on C1 is 24 micro C\n",
+ "Charge on C2 is 36 micro C\n"
+ ]
+ }
+ ],
+ "prompt_number": 88
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.22 Page no 114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C2=5 #micro F\n",
+ "C3=5 \n",
+ "a=6\n",
+ "C1=10\n",
+ "V=3\n",
+ "\n",
+ "#Calculation\n",
+ "C23=C2+C3\n",
+ "V1=a/2.0\n",
+ "q1=C1*V\n",
+ "q23=C23*V\n",
+ "q3=q23/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Charge on capacitor C1 is\",q1,\"micro F\"\n",
+ "print\"Charge on capacitor C2 is\",q23,\"micro F\"\n",
+ "print\"Charge on capacitor C3 is\",q3,\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge on capacitor C1 is 30 micro F\n",
+ "Charge on capacitor C2 is 30 micro F\n",
+ "Charge on capacitor C3 is 15.0 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 94
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.23 Page no 114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C1=4*10**-6 #F\n",
+ "V1=400 #V\n",
+ "C2=2*10**-6\n",
+ "q2=0\n",
+ "\n",
+ "#Calculation\n",
+ "q1=C1*V1\n",
+ "C=C1+C2\n",
+ "q=q1+q2\n",
+ "V=q/C\n",
+ "\n",
+ "#Result\n",
+ "print\"Common potential is\", round(V,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Common potential is 266.67 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 101
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.24 Page no 114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=12\n",
+ "C1=8 #micro F\n",
+ "C2=4\n",
+ "\n",
+ "#Calculation\n",
+ "Va=V-C2\n",
+ "Vb=V-C1\n",
+ "V1=Va-Vb\n",
+ "\n",
+ "#Result\n",
+ "print\"Potential difference is\", V1,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Potential difference is 4 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 105
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.25 Page no 114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=90*10**-4 #m**2\n",
+ "d=2.5*10**-3 #m\n",
+ "V=400 #Volts\n",
+ "e=8.854*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "C=(e*A)/d\n",
+ "W=(C*V**2)/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Electrostatic energy is\", round(W*10**6,2)*10**-6,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electrostatic energy is 2.55e-06 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 112
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.26 Page no 114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=2000*10**-6 \n",
+ "V=1.5 #V\n",
+ "t=10.0**-4 #s\n",
+ "\n",
+ "#Calculation\n",
+ "U=(C*V**2)/2.0\n",
+ "P=U/t\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy stored in the capacitor is\", U*10**3,\"*10**-3 J\"\n",
+ "print\"Power of the flash is\",P,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy stored in the capacitor is 2.25 *10**-3 J\n",
+ "Power of the flash is 22.5 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 119
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.27 Page no 115"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C1=4*10**-6\n",
+ "V1=200 #V\n",
+ "C2=2*10**-6 # F\n",
+ "\n",
+ "#Calculation\n",
+ "U1=(C1*V1**2)/2.0\n",
+ "q=C1*V1\n",
+ "C=C1+C2\n",
+ "V=q/C\n",
+ "U2=(C*V**2)/2.0\n",
+ "U=U1-U2\n",
+ "\n",
+ "#Result\n",
+ "print\"Electrostatic energy is\",round(U*10**2,2),\"*10**-2 J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electrostatic energy is 2.67 *10**-2 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 129
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.28 Page no 115"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "K=6.0\n",
+ "C=17.7*10**-12 #F\n",
+ "V=100\n",
+ "\n",
+ "#Calculation\n",
+ "C1=K*C\n",
+ "q=C1*V\n",
+ "V1=V/K\n",
+ "q1=C1*V1\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Charge on capacitor is\", q,\"C\"\n",
+ "print\"(b) Charge on capacitor is\", q1,\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Charge on capacitor is 1.062e-08 C\n",
+ "(b) Charge on capacitor is 1.77e-09 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 135
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.29 Page no 115"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "K=3\n",
+ "e=8.854*10**-12 #C**2/N/m**2\n",
+ "C=50*10**-12 #F\n",
+ "a=10**7 #V/m\n",
+ "b=10\n",
+ "V=10**3\n",
+ "\n",
+ "#Calculation\n",
+ "E=(a*b)/100.0\n",
+ "q=C*V\n",
+ "A=q/(e*K*E)\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum area is\", round(A*10**4,1),\"cm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum area is 18.8 cm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 140
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.30 Page no 115"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=15*10**5 #V\n",
+ "E=5.0*10**7 #V/m\n",
+ "\n",
+ "#Calculation\n",
+ "r=V/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum radius of the spherical shell is\", r*10**2,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum radius of the spherical shell is 3.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 144
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap6_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap6_1.ipynb new file mode 100644 index 00000000..2c3272c0 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap6_1.ipynb @@ -0,0 +1,1695 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:9ba957b16289f27dba06e59c72667002157ed6d6f12727710ff725257f8649eb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 6 Current electricity"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.1 Page no 191"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=300*10**-3\n",
+ "e=1.6*10**-19 #C\n",
+ "t=60 #S\n",
+ "\n",
+ "#Calculation\n",
+ "q=I*t\n",
+ "n=q/e\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of electron passed is\", n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of electron passed is 1.125e+20\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.2 Page no 191"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19 #C\n",
+ "v=6.8*10**15 #revolution/s\n",
+ "r=0.51*10**-10 #m\n",
+ "\n",
+ "#Calculation\n",
+ "I=e*v\n",
+ "\n",
+ "#Result\n",
+ "print\"Equivalent current is\", I*10**3,\"*10**-3 A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equivalent current is 1.088 *10**-3 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.3 Page no 191 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=10 #A\n",
+ "n=9*10**28 #/m**3\n",
+ "A=10.0**-4 #m**2\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "Vd=I/(n*e*A)\n",
+ "\n",
+ "#Result\n",
+ "print\"Drift velocity is\", round(Vd*10**6,2)*10**-6,\"m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Drift velocity is 6.94e-06 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.4 Page no 191"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=8.5*10**28 #/m**3\n",
+ "I=3 #A\n",
+ "A=2.0*10**-6 #m**2\n",
+ "l=3 #m\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "Vd=I/(n*e*A)\n",
+ "t=I/Vd\n",
+ "\n",
+ "#Result\n",
+ "print\"Drift velocity of the electron is\", round(Vd*10**4,3),\"*10**-4 m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Drift velocity of the electron is 1.103 *10**-4 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.5 Page no 191"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=10**-7 #m**2\n",
+ "I=1 #A\n",
+ "K=6.023*10**23\n",
+ "w=63.5\n",
+ "d=9*10**3 #Kg/m**3\n",
+ "\n",
+ "#Calculation\n",
+ "a=(K/w)*1000\n",
+ "n=a*d\n",
+ "Vd=I/(n*e*A)\n",
+ "\n",
+ "#Result\n",
+ "print\"Average drift velocity is\", round(Vd*10**4,2),\"*10**-4 m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average drift velocity is 7.32 *10**-4 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6 Page no 191"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=5 #Volt\n",
+ "l=0.1 #m\n",
+ "vd=2.5*10**-4 #m/s\n",
+ "\n",
+ "#Calculation\n",
+ "E=V/l\n",
+ "u=vd/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Electron mobility is\", u,\"m**2/V/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electron mobility is 5e-06 m**2/V/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.7 Page no 191"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=3 #Volt\n",
+ "l=0.2 #m\n",
+ "A=10**-6 #m**2\n",
+ "n=8.5*10**28 #/m**3\n",
+ "u=4.5*10**-6 #m**2/V/s\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "E=V/l\n",
+ "I=n*A*u*E*e\n",
+ "\n",
+ "#Result\n",
+ "print\"Current through the wire is\",round(I,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current through the wire is 0.92 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.8 Page no 191"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=1.8 #A\n",
+ "A=0.5*10**-6 #m**2\n",
+ "n=8.8*10**28\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "j=I/A\n",
+ "vd=j/(n*e)\n",
+ "\n",
+ "#Result\n",
+ "print\"Drift velocity is\", round(vd*10**4,2),\"*10**-4 m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Drift velocity is 2.56 *10**-4 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.9 Page no 192"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=1.5 #V\n",
+ "n=8.4*10**28 #/m**3\n",
+ "l=0.2 #m\n",
+ "A=0.3*10**-6 #m**2\n",
+ "I=2.4 #A\n",
+ "m=9.1*10**-31 #Kg\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "E=V/l\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": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.10 Page no 192"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=200 #V\n",
+ "R=100.0 #ohm\n",
+ "t=1\n",
+ "\n",
+ "#Calculation\n",
+ "I=V/R\n",
+ "q=I*t\n",
+ "n=q/e\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of electrons is\", n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of electrons is 1.25e+19\n"
+ ]
+ }
+ ],
+ "prompt_number": 55
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.11 Page no 192"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=2 #ohm\n",
+ "l=1 #m\n",
+ "d=4*10**-4 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=(math.pi*d**2)/4.0\n",
+ "a=(R*A)/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Resistivity of the material is\", round(a*10**7,3)*10**-7,\"ohm m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistivity of the material is 2.513e-07 ohm m\n"
+ ]
+ }
+ ],
+ "prompt_number": 62
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.12 Page no 192"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=0.45 #Kg\n",
+ "R=0.14 #ohm\n",
+ "a=1.78*10**-8 #ohm m\n",
+ "l=1.99\n",
+ "d=8.93*10**3\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l=(math.sqrt(R*m/(a*d)))/10.0\n",
+ "r=math.sqrt(m/(math.pi*l*d))\n",
+ "\n",
+ "#Result\n",
+ "print\"Radius is\", round(r*10**3,2),\"mm\"\n",
+ "print\"Length is\", round(l,2),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius is 2.84 mm\n",
+ "Length is 1.99 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 76
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.13 Page no 192"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=5\n",
+ "\n",
+ "#Calculation\n",
+ "R1=4*R\n",
+ "\n",
+ "#Result\n",
+ "print\"New resistance is\",R1,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "New resistance is 20 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.14 Page no 192"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=3\n",
+ "A=0.02*10**-6 #m**2\n",
+ "R=2 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "a=l/(R*A)\n",
+ "\n",
+ "#Result\n",
+ "print\"Electrical conductivity is\", a*10**-7,\"*10**7 S/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electrical conductivity is 7.5 *10**7 S/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.15 Page no 193"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=1.8 #V\n",
+ "I=1.2 #A\n",
+ "l=3 #m\n",
+ "A=5.4*10**-6 #m**2\n",
+ "\n",
+ "#Calculation\n",
+ "R=V/I\n",
+ "G=1/R\n",
+ "a=l/(R*A)\n",
+ "\n",
+ "#Result\n",
+ "print\"Conductance is\",round(G,2),\"S\"\n",
+ "print\"Conductivity is\", round(a*10**-5,1),\"*10**5 S/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Conductance is 0.67 S\n",
+ "Conductivity is 3.7 *10**5 S/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 92
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.16 Page no 193"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=2.1 #ohm\n",
+ "a1=300.0 #K\n",
+ "a2=373.0\n",
+ "R2=2.7 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "A=(R2-R1)/(R1*(a2-a1))\n",
+ "\n",
+ "#Result\n",
+ "print\"Temperature coefficient is\", round(A*10**3,3),\"*10**-3 /K\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Temperature coefficient is 3.914 *10**-3 /K\n"
+ ]
+ }
+ ],
+ "prompt_number": 99
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.17 Page no 193"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a1=50 #degree\n",
+ "R1=6.0 #ohm\n",
+ "a2=100 #degree\n",
+ "R=7 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "a=(R-R1)/(R1*(a2-a1))\n",
+ "R0=R1/(1+(a*a1))\n",
+ "\n",
+ "#Result\n",
+ "print\"Mean temperature coefficient is\", round(a,4),\"degree/C\"\n",
+ "print\"Resistance of the conductor is\",round(R0,3),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mean temperature coefficient is 0.0033 degree/C\n",
+ "Resistance of the conductor is 5.143 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 110
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.18 Page no 193"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=230 #v\n",
+ "a=27 #degree C\n",
+ "I1=3.2 #A\n",
+ "I2=2.8 #A\n",
+ "a1=1.70*10**-4 #degree/C\n",
+ "\n",
+ "#Calculation\n",
+ "R1=V/I1\n",
+ "R2=V/I2\n",
+ "a2=a+((R2-R1)/(R1*a1))\n",
+ "\n",
+ "#Result\n",
+ "print\"Steady temperature is\", round(a2,2),\"degree C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Steady temperature is 867.34 degree C\n"
+ ]
+ }
+ ],
+ "prompt_number": 117
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.21 Page no 193"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=1 #ohm\n",
+ "R2=2 \n",
+ "R3=3\n",
+ "E=12 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Rs=R1+R2+R3\n",
+ "I=E/Rs\n",
+ "R11=I*R1\n",
+ "R12=I*R2\n",
+ "R13=I*R3\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Total resistance of the combination is\",Rs,\"ohm\"\n",
+ "print\"(b) Potential drop across R1 is\",R11,\"V\"\n",
+ "print\" Potential drop across R2 is\",R12,\"V\"\n",
+ "print\" Potential drop across R3 is\",R13,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Total resistance of the combination is 6 ohm\n",
+ "(b) Potential drop across R1 is 2 V\n",
+ " Potential drop across R2 is 4 V\n",
+ " Potential drop across R3 is 6 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 125
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.22 Page no 193"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=2.0 #ohm\n",
+ "R2=4.0 #ohm\n",
+ "R3=5.0 \n",
+ "E=20 #V\n",
+ "\n",
+ "#calculation\n",
+ "Rp=1/(1/R1+1/R2+1/R3)\n",
+ "I=E/Rp\n",
+ "I1=E/R1\n",
+ "I2=E/R2\n",
+ "I3=E/R3\n",
+ "\n",
+ "print\"(a) Toatl resisatnce is\", round(Rp,2),\"ohm\"\n",
+ "print\"(b) Current through R1 is\",I1,\"A\"\n",
+ "print\"Current through R2 is\",I2,\"A\"\n",
+ "print\"Current through R3 is\",I3,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Toatl resisatnce is 1.05 ohm\n",
+ "(b) Current through R1 is 10.0 A\n",
+ "Current through R2 is 5.0 A\n",
+ "Current through R3 is 4.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 134
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.24 Page no 194"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=6\n",
+ "R1=2.0\n",
+ "\n",
+ "#calculation\n",
+ "n=R/R1\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of resistance is\",n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of resistance is 3.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 135
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.25 Page no 194"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=9.0 #ohm\n",
+ "R2=5.0\n",
+ "R3=3.0\n",
+ "R4=7.0\n",
+ "\n",
+ "#Calculation\n",
+ "R11=R1+R2\n",
+ "R22=R3+R4\n",
+ "R=1/(1/R11+1/R22+1/R22)\n",
+ "\n",
+ "#Result\n",
+ "print\"Equivalent resistance is\",round(R,2),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equivalent resistance is 3.68 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 141
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.26 Page no 194"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=3 #ohm\n",
+ "R2=3\n",
+ "\n",
+ "#Calculation\n",
+ "R11=R1+R2\n",
+ "R22=(R11*R11)/(R11+R11)\n",
+ "R=R11*R1/(R11+R1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Effective resistance is\", R,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Effective resistance is 2 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 148
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.27 Page no 195"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=30.0 #ohm\n",
+ "R2=30.0\n",
+ "E=2 #V\n",
+ "\n",
+ "#Calculation\n",
+ "R11=R1+R2\n",
+ "R=R11*R1/(R11+R1)\n",
+ "I=E/R\n",
+ "\n",
+ "#Result\n",
+ "print\"Current through the cell is\",I,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current through the cell is 0.1 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 151
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.28 Page no 195"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rl=500.0 #ohm\n",
+ "E=50 #V\n",
+ "Rac=2000.0 #ohm\n",
+ "Rab=500.0 \n",
+ "\n",
+ "#Calculation\n",
+ "Rbc=Rac-Rab\n",
+ "R11=1/(1/Rbc+1/Rl)\n",
+ "I=E/(Rab+R11)\n",
+ "V=E-(I*Rab)\n",
+ "I1=E/Rac\n",
+ "Rbc1=40/I1\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Potential difference across the load is\", round(V,2),\"V\"\n",
+ "print\"(b) Resistance is\",Rbc1,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Potential difference across the load is 21.43 V\n",
+ "(b) Resistance is 1600.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 165
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.29 Page no 195"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=3 #ohm\n",
+ "E=9 #V\n",
+ "\n",
+ "#Calcuation\n",
+ "R=R1+R1+R1\n",
+ "I=E/R\n",
+ "I1=I/4.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Total resistance is\",R,\"ohm\"\n",
+ "print\"Current is\",I1,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total resistance is 9 ohm\n",
+ "Current is 0.25 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 169
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.30 Page no 196"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=1 #ohm\n",
+ "R2=1 \n",
+ "r=2/3.0 #ohm\n",
+ "E=1 #V\n",
+ "\n",
+ "#Calculation\n",
+ "R11=R1*R2/(R1+R2)\n",
+ "R=1/(1/R1+1/R2+1/R1)\n",
+ "I=E/(R+r)\n",
+ "\n",
+ "#Result\n",
+ "print\"Current drawn from the cell is\", I,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current drawn from the cell is 1.5 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 174
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.31 Page no 196"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=10.0 #ohm\n",
+ "r=3.33\n",
+ "E=15\n",
+ "\n",
+ "#Calculation\n",
+ "Rp=1/(1/R+(1/(R+R)))\n",
+ "R1=R+Rp+R\n",
+ "I=E/(R1+r)\n",
+ "V=I*Rp\n",
+ "I1=V/(R+R)\n",
+ "I2=V/R\n",
+ "\n",
+ "#Result\n",
+ "print\"Current flow through the main circuit is\", round(I,1),\"A\"\n",
+ "print\"Current through QRS is\",round(I1,3),\"A\"\n",
+ "print\"Current through QS is\",round(I2,3),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current flow through the main circuit is 0.5 A\n",
+ "Current through QRS is 0.167 A\n",
+ "Current through QS is 0.333 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 199
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.32 Page no 196"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=2 #ohm\n",
+ "R2=3 \n",
+ "R3=2.8\n",
+ "E=6 #V\n",
+ "\n",
+ "#calculation\n",
+ "R=(R1*R2/(R1+R2))+R3\n",
+ "I=E/R\n",
+ "\n",
+ "#Result\n",
+ "print\"Current in steady state is\", round(I,1),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current in steady state is 1.6 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 206
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.33 Page no 197"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=10 #v\n",
+ "r=3 #ohm\n",
+ "I=0.5 #A\n",
+ "\n",
+ "#Calculation\n",
+ "R=(E/I)-r\n",
+ "V=I*R\n",
+ "\n",
+ "#Result\n",
+ "print\"Resistance of the resistor is\", R,\"ohm\"\n",
+ "print\"Terminal voltage is\",V,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance of the resistor is 17.0 ohm\n",
+ "Terminal voltage is 8.5 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 212
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.35 Page no 197"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=3.0 #ohm\n",
+ "R2=4.0\n",
+ "R3=6.0\n",
+ "r=2/3.0\n",
+ "E=2 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Rp=1/(1/R1+1/R2+1/R3)\n",
+ "R=Rp+r\n",
+ "I=E/R\n",
+ "V=I*Rp\n",
+ "I1=V/R1\n",
+ "\n",
+ "#Result\n",
+ "print\"Current drawn from the cell is\", I,\"A\"\n",
+ "print\"Current through 3 Ohm resistance is\",round(I1,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current drawn from the cell is 1.0 A\n",
+ "Current through 3 Ohm resistance is 0.44 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.36 Page no 197"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=8\n",
+ "R=3.0\n",
+ "r=1.0\n",
+ "\n",
+ "#Calculation\n",
+ "I=E/(R+r)\n",
+ "I1=I/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Current from the battery is\",I,\"A\"\n",
+ "print\"Current through AC and ABC is\",I1,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current from the battery is 2.0 A\n",
+ "Current through AC and ABC is 1.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.37 Page no 198"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=12 #V\n",
+ "R2=8 #ohm\n",
+ "r=1 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "R1=-((E/I)-R2-r)\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 3.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.38 Page no 198"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E1=20 #V\n",
+ "r1=1 #ohm\n",
+ "E2=8 #Volt\n",
+ "r2=2 #ohm\n",
+ "R1=12.0\n",
+ "R2=6.0\n",
+ "R3=4.0\n",
+ "\n",
+ "#Calculation\n",
+ "E=E1-E2\n",
+ "R11=1/(1/R1+1/R2+1/R3)\n",
+ "R=R11+r1+r2+5\n",
+ "I=E/R\n",
+ "I1=I-1\n",
+ "V1=E1-(I*r1)\n",
+ "V2=E2+(I*r2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Current in the circuit is\", I,\"A\"\n",
+ "print\"(b) Current in the resister is\", I1,\"A\"\n",
+ "print\"(c) Potential difference across 12 V battery is\", V1,\"V\"\n",
+ "print\" Potential difference across 8 V battery is\",V2,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Current in the circuit is 1.2 A\n",
+ "(b) Current in the resister is 0.2 A\n",
+ "(c) Potential difference across 12 V battery is 18.8 V\n",
+ " Potential difference across 8 V battery is 10.4 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.39 Page no 198"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=2 #V\n",
+ "r=0.015 #ohm\n",
+ "R=8.5\n",
+ "\n",
+ "#Calculation\n",
+ "E1=E*6\n",
+ "r1=r*6\n",
+ "I=E1/(R+r1)\n",
+ "V=I*R\n",
+ "\n",
+ "#Result\n",
+ "print\"Current drawn from the supply is\", round(I,3),\"A\"\n",
+ "print\"Terminal voltage is\",round(V,3),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current drawn from the supply is 1.397 A\n",
+ "Terminal voltage is 11.874 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.40 Page no 198"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=4 #V\n",
+ "I=1.5 #A\n",
+ "R=6 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "E1=3*E\n",
+ "r=((E1/I)-R)/3.0\n",
+ "V=E-(I*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Internal resistance is\", round(r,2),\"ohm\"\n",
+ "print\"(b) Terminal voltage is\",V,\"Volt\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Internal resistance is 0.67 ohm\n",
+ "(b) Terminal voltage is 3.0 Volt\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.41 Page no 199"
+ ]
+ },
+ {
+ "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 resistance is\", r,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Internal resistance is 7.5 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.42 Page no 199"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=1\n",
+ "a=2\n",
+ "\n",
+ "#Calculation\n",
+ "r=(a*E*(a-E))/(a*E*(a-E))\n",
+ "\n",
+ "#Result\n",
+ "print\"The internal resistance is\",r,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The internal resistance is 1 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.43 Page no 199"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=2 #V\n",
+ "r=1.5 #ohm\n",
+ "R=10 #ohm\n",
+ "r1=1.5\n",
+ "r2=1.5\n",
+ "\n",
+ "#Calculation\n",
+ "E1=2*E\n",
+ "w=r1+r2\n",
+ "A=1/(1/w+1/w)\n",
+ "B=R+A\n",
+ "I=E1/B\n",
+ "IR=I*R\n",
+ "\n",
+ "#Result\n",
+ "print\"The potential difference across external resistance is\",round(IR,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The potential difference across external resistance is 3.48 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 63
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.44 Page no 200"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=2.7*10**18\n",
+ "N=1.0*10**18\n",
+ "e=1.6*10**-19\n",
+ "V=230 #V\n",
+ "\n",
+ "#Calculation\n",
+ "I=(n+N)*(e)\n",
+ "R=V/I\n",
+ "\n",
+ "#Result\n",
+ "print\"The effective resistance of the tube is\",round(R,1),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The effective resistance of the tube is 388.5 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 73
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.47 Page no 200"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "X=1\n",
+ "a=3\n",
+ "E1=12\n",
+ "r=0.5\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "X1=X+math.sqrt(a)\n",
+ "I=E1/(X1+r)\n",
+ "\n",
+ "#Result\n",
+ "print\"Current drawn is\", round(I,3),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current drawn is 3.713 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 99
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.48 Page no 200"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=6\n",
+ "R2=3.0\n",
+ "R3=5\n",
+ "E=24 #V\n",
+ "I=3 #A\n",
+ "\n",
+ "#Calculation\n",
+ "Rp=(R1*R2)/(R1+R2)\n",
+ "R=R3+Rp+r\n",
+ "r=R-(R3+Rp)\n",
+ "V=I*Rp\n",
+ "I1=V/R1\n",
+ "I2=V/R2\n",
+ "R11=R2+R3+r\n",
+ "I11=E/R11\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Internal resistance is\", r,\"ohm\"\n",
+ "print\" Current I1 is\",I1,\"A and I2 is\",I2,\"A\"\n",
+ "print\"(b) Current in the circuit is\",round(I11,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Internal resistance is 1.0 ohm\n",
+ " Current I1 is 1.0 A and I2 is 2.0 A\n",
+ "(b) Current in the circuit is 2.67 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 91
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap7_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap7_1.ipynb new file mode 100644 index 00000000..0476eb75 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap7_1.ipynb @@ -0,0 +1,832 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:3ec304c8c9eb23fdd2ae23967a8effd62df3acdefa633ca8bf2a83485250a21b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 7 Electrical Measurements"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.1 Page no 229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=2\n",
+ "R2=3\n",
+ "R3=5\n",
+ "E1=6\n",
+ "E2=4\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=np.array([[7,5],[5,8]])\n",
+ "b=np.array([6,4])\n",
+ "z=np.linalg.solve(A,b)\n",
+ "Z=z[0]\n",
+ "Z1=z[1]\n",
+ "A1=Z+Z1\n",
+ "\n",
+ "#Result\n",
+ "print\"Current through R1 is\", round(Z,3),\"A\"\n",
+ "print\"Current through R2 is\", round(Z1,3),\"A\"\n",
+ "print\"Current through R3 is\",round(A1,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current through R1 is 0.903 A\n",
+ "Current through R2 is -0.065 A\n",
+ "Current through R3 is 0.84 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.2 Page no 229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E1=1.5 #V\n",
+ "E2=2.0 #V\n",
+ "r1=1 #ohm\n",
+ "r2=2 #ohm\n",
+ "R=5 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=np.array([[6,5],[5,7]])\n",
+ "B=np.array([1.5,2])\n",
+ "Z=np.linalg.solve(A,B)\n",
+ "Z1=Z[0]\n",
+ "Z2=Z[1]\n",
+ "A1=(Z1+Z2)*R\n",
+ "\n",
+ "#Result\n",
+ "print\"Current through R1 is\", round(Z1,4),\"A\"\n",
+ "print\"Current through R2 is\",round(Z2,4),\"A\"\n",
+ "print\"Current through R3 is\",round(A1,4),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current through R1 is 0.0294 A\n",
+ "Current through R2 is 0.2647 A\n",
+ "Current through R3 is 1.4706 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.3 Page no 230"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E1=2 #V\n",
+ "E2=1 #V\n",
+ "E3=4 #V\n",
+ "r1=4 #ohm\n",
+ "r2=3 #ohm\n",
+ "r3=2 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=np.array([[1,1,1],[4,-3,0],[0,3,-2]])\n",
+ "A1=np.array([0,1,-3])\n",
+ "Z=np.linalg.solve(A,A1)\n",
+ "Z1=Z[0]\n",
+ "Z2=Z[1]\n",
+ "Z3=Z[2]\n",
+ "\n",
+ "#Result\n",
+ "print\"Current through R1 is\", round(Z1,2),\"A\"\n",
+ "print\"Current through R2 is\",round(Z2,2),\"A\"\n",
+ "print\"Current through R3 is\",round(Z3,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current through R1 is -0.15 A\n",
+ "Current through R2 is -0.54 A\n",
+ "Current through R3 is 0.69 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.4 Page no 230"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=5 #ohm\n",
+ "R3=2 \n",
+ "R=10\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=np.array([[9,5],[10,19]])\n",
+ "A1=np.array([1,1])\n",
+ "Z=np.linalg.solve(A,A1)\n",
+ "Z1=Z[0]\n",
+ "Z2=Z[1]\n",
+ "I=(Z1+Z2)*R\n",
+ "\n",
+ "#Result\n",
+ "print\"Potential difference across the resistor is\", round(I,4),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Potential difference across the resistor is 1.0744 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.5 Page no 230"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=10 #ohm\n",
+ "B=5 #ohm\n",
+ "S=2\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=np.array([[25,-10],[5,5]])\n",
+ "B=np.array([9,7]) \n",
+ "Z=np.linalg.solve(A,B)\n",
+ "Z1=Z[1]*10\n",
+ "\n",
+ "#Result\n",
+ "print\"Equivalent resistance is\", round(Z1,0),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equivalent resistance is 7.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.6 Page no 231 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=15 #ohm\n",
+ "Q=12.0\n",
+ "R=10.0\n",
+ "S=4\n",
+ "\n",
+ "#Calculation\n",
+ "R1=S*(P/Q)\n",
+ "X=1/(1/R1-1/R)\n",
+ "\n",
+ "#Result\n",
+ "print\"The resistance to be connectedin parallel is\",X,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resistance to be connectedin parallel is 10.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.7 Page no 231"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=4.0\n",
+ "Q=4.0\n",
+ "R=4.0\n",
+ "X=4.0\n",
+ "\n",
+ "#Calculation\n",
+ "R1=P+Q\n",
+ "R2=R+X\n",
+ "R3=1/(1/R1+1/R2)\n",
+ "I=P/R3\n",
+ "I1=I/2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Equivalent resistance is\",R3,\"ohm\"\n",
+ "print\"(ii) The magnitudes of current is\",I1,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Equivalent resistance is 4.0 ohm\n",
+ "(ii) The magnitudes of current is 0.5 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.8 Page no 232"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=4\n",
+ "b=3\n",
+ "c=2.0\n",
+ "E=6 #v\n",
+ "\n",
+ "#Calculation\n",
+ "X=(a*b)/c\n",
+ "R1=a+c\n",
+ "R2=b+R1\n",
+ "R=((R1*R2)/(R1+R2))+2.4\n",
+ "I=E/R\n",
+ "\n",
+ "#Result\n",
+ "print\"Current drawn by the circuit is\", I,\"A\"\n",
+ "print\"Unknown resistance is\",R,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current drawn by the circuit is 1.0 A\n",
+ "Unknown resistance is 6.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 59
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.9 Page no 232"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=1.5\n",
+ "b=5\n",
+ "c=6\n",
+ "\n",
+ "#Calculation\n",
+ "I1=a/(b+c)\n",
+ "I2=I1/10.0\n",
+ "I=I1+I2\n",
+ "\n",
+ "#Result\n",
+ "print\"Current in the arms is\", round(I1,4),\"A and\",round(I2,4),\"A\"\n",
+ "print\"Current in the cell is\",I,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current in the arms is 0.1364 A and 0.0136 A\n",
+ "Current in the cell is 0.15 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.10 Page no 233"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "X=10\n",
+ "P=3\n",
+ "Q=2.0\n",
+ "\n",
+ "#Calculation\n",
+ "R=X*P/Q\n",
+ "L=1/R\n",
+ "\n",
+ "#Result\n",
+ "print\"Length is\",round(L,3),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Length is 0.067 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 71
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.11 Page no 233"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=6 #ohm\n",
+ "l=1/3.0\n",
+ "r1=2/3.0\n",
+ "\n",
+ "#Calculation\n",
+ "R1=r/((2.0*(r1/l))-1)\n",
+ "R2=2*R1\n",
+ "\n",
+ "#Result\n",
+ "print\"Resistance of two wires is\", R1,\"ohm and\",R2,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance of two wires is 2.0 ohm and 4.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 76
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.12 Page no 233"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=1.2 #ohm\n",
+ "Q=0.8\n",
+ "R=2\n",
+ "E=4 #V\n",
+ "\n",
+ "#Calculation\n",
+ "X=(R*P)/Q\n",
+ "R1=X+R\n",
+ "R2=P+Q\n",
+ "R11=R1*R2/(R1+R2)\n",
+ "I=E/R11\n",
+ "\n",
+ "#Result\n",
+ "print\"Unknown resistance is\", X,\"ohm\"\n",
+ "print\"Current drawn is\",I,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Unknown resistance is 3.0 ohm\n",
+ "Current drawn is 2.8 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 83
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.13 Page no 233"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=20.0 #ohm\n",
+ "V=10**-3 #V/m\n",
+ "l=10**4 #mm\n",
+ "V1=10**-2 #V\n",
+ "\n",
+ "#Calculation\n",
+ "I=V1/R\n",
+ "R1=(2/I)-R\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 3980.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 87
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.14 Page no 233"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l1=65 #cm\n",
+ "l2=60.0\n",
+ "a=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "E1=((a*l1)/l2)/((l1/l2)-1)\n",
+ "E2=E1-a\n",
+ "\n",
+ "#Result\n",
+ "print\"E.m.f of cell is\", E1,\"V and\",E2,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "E.m.f of cell is 1.3 V and 1.2 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 93
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.15 Page no 234"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "S=9.5 #ohm\n",
+ "I1=76.3 #cm\n",
+ "l2=64.8\n",
+ "\n",
+ "#Calculation\n",
+ "r=((I1/I2)-1)*S\n",
+ "\n",
+ "#Result\n",
+ "print\"Internal resistance is\",round(r*10**-3,1),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Internal resistance is 53.1 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 138
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.16 Page no 234"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=10*10**-3 \n",
+ "r=10\n",
+ "l=40 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "R=10*l/100.0\n",
+ "R1=(R*2/E)-10\n",
+ "\n",
+ "#Result\n",
+ "print\"External resistance is\", R1,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "External resistance is 790.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 111
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.17 Page no 234"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=10*10**3\n",
+ "E=12\n",
+ "\n",
+ "#Calculation\n",
+ "R=R1/((E/2.0)-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Resistance is\",R*10**-3,\"K ohm\"\n",
+ "print\"(b) When current flows through a resistor,its temperature increases due toheat produced across it.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Resistance is 2.0 K ohm\n",
+ "(b) When current flows through a resistor,its temperature increases due toheat produced across it.\n"
+ ]
+ }
+ ],
+ "prompt_number": 116
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.18 Page no 235"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=3.0\n",
+ "b=6.0\n",
+ "\n",
+ "#Caculation\n",
+ "E=(1/a)+(1/b)+1+(1/a)\n",
+ "\n",
+ "#Result\n",
+ "print\"Equivalent resistance is\",round(E,2),\"r\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equivalent resistance is 1.83 r\n"
+ ]
+ }
+ ],
+ "prompt_number": 120
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.22 Page no 237"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=5\n",
+ "b=10\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=np.array([[7,1],[1,5]])\n",
+ "A1=np.array([2,2])\n",
+ "B=np.linalg.solve(A,A1)\n",
+ "B1=B[0]\n",
+ "B2=B[1]\n",
+ "E=(a*B2)+(B1*b)\n",
+ "\n",
+ "#Result\n",
+ "print\"Equivalent resistance is\", round(E,2),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equivalent resistance is 4.12 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 125
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.23 Page no 238"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=12.5\n",
+ "l=39.5\n",
+ "\n",
+ "#Calculation\n",
+ "X=(R*(100-l))/l\n",
+ "L=X/(X+R)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Resistance is\",round(X,2),\"W\"\n",
+ "print\"(b) Balance point of bridge is\", L*10**2,\"cm\"\n",
+ "print\"(c) The galvanometer will not show any current.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Resistance is 19.15 W\n",
+ "(b) Balance point of bridge is 60.5 cm\n",
+ "(c) The galvanometer will not show any current.\n"
+ ]
+ }
+ ],
+ "prompt_number": 135
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap8_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap8_1.ipynb new file mode 100644 index 00000000..1b06633c --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap8_1.ipynb @@ -0,0 +1,994 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:a40c0c75d3ca70fc3ecf4f82797b161ff3f00d80a8c81333303a3e01cdc2b09b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 8 Thermal effects of current"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.1 Page no 258"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=100.0 #W\n",
+ "V=220.0 #V\n",
+ "\n",
+ "#Calculation\n",
+ "I=P/V\n",
+ "R=V**2/P\n",
+ "\n",
+ "#Result\n",
+ "print\"Resistance is\",R,\"ohm\"\n",
+ "print\"Current capacity is\",round(I,3),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance is 484.0 ohm\n",
+ "Current capacity is 0.455 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.2 Page no 258"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19 #C\n",
+ "P=100 #W\n",
+ "V=200.0 #V\n",
+ "t=1\n",
+ "\n",
+ "#Calculation\n",
+ "I=P/V\n",
+ "q=I*t\n",
+ "N=q/e\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of electrons is\", N"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of electrons is 3.125e+18\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.3 Page no 258"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P1=60 #W\n",
+ "P2=100.0 \n",
+ "\n",
+ "#Calculation\n",
+ "I=P1/P2\n",
+ "\n",
+ "#Result\n",
+ "print\"Hence 100 W bulb draws more current\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Hence 100 W bulb draws more current\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.4 Page no 259"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P1=60.0 #W\n",
+ "P2=100\n",
+ "\n",
+ "#Calculation\n",
+ "R=P2/P1\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) 60 W bulb will dissipate more power.\"\n",
+ "print\"(b) Since 60 W bulb is dissipating more power, it glows brighter.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) 60 W bulb will dissipate more power.\n",
+ "(b) Since 60 W bulb is dissipating more power, it glows brighter.\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.5 Page no 259"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P1=500.0 #W\n",
+ "V1=220 #V\n",
+ "P2=100.0\n",
+ "V2=220\n",
+ "\n",
+ "#Calculation\n",
+ "R1=V1**2/P1\n",
+ "R2=V2**2/P2\n",
+ "I=V2/(R1+R2)\n",
+ "P11=I**2*R1\n",
+ "P22=I**2*R2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Power consumed by heater is\",round(P11,2),\"W\"\n",
+ "print\"(ii) Power consumed by electric bulb is\",round(P22,2),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Power consumed by heater is 13.89 W\n",
+ "(ii) Power consumed by electric bulb is 69.44 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.6 Page no 259"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=100.0 #W\n",
+ "V=220 #V\n",
+ "V1=110\n",
+ "\n",
+ "#Calculation\n",
+ "R=V**2/P\n",
+ "P1=V1**2/R\n",
+ "\n",
+ "#Result\n",
+ "print\"Power consumed by the bulb is\",P1,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power consumed by the bulb is 25.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.7 Page no 259"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=550 #W\n",
+ "V=220 #V\n",
+ "V1=222.2 #V\n",
+ "V2=217.8 #V\n",
+ "\n",
+ "#Calculation\n",
+ "R=V**2/P\n",
+ "P=V1**2/R\n",
+ "P1=V2**2/R\n",
+ "\n",
+ "#Result\n",
+ "print\"minimum power is\",round(P,0),\"W\"\n",
+ "print\"maximum power is\",round(P1,0),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "minimum power is 561.0 W\n",
+ "maximum power is 539.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.8 Page no 259"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=220\n",
+ "P=60.0 #W\n",
+ "P1=85\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "R=V**2/P\n",
+ "V1=math.sqrt(P1*R)\n",
+ "\n",
+ "#Result\n",
+ "print\"Voltage fluctuation is\",round(V1,1),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage fluctuation is 261.9 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.9 Page no 259"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=50*10**3 #W\n",
+ "V=5000.0\n",
+ "R=20\n",
+ "\n",
+ "#Calculation\n",
+ "I=P/V\n",
+ "P1=I**2*R\n",
+ "P2=P-P1\n",
+ "\n",
+ "#Result \n",
+ "print\"Power received by the factory is\",P2*10**-3,\"Kw\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power received by the factory is 48.0 Kw\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.10 Page no 260"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R2=2 #ohm\n",
+ "R3=2 #ohm\n",
+ "I=3 #A\n",
+ "\n",
+ "#Calculation\n",
+ "P=I**2*R2\n",
+ "R1=(R2*R3)/(R2+R3)\n",
+ "P1=I**2*R1\n",
+ "D=P+P1\n",
+ "\n",
+ "#Result\n",
+ "print\"The maximum power is\",D,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum power is 27 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 50
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.11 Page no 260"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=0.5 #ohm\n",
+ "I=100 #A\n",
+ "T=24*60*60\n",
+ "\n",
+ "#Calculation\n",
+ "Q=I**2*R*T\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy is being lost per day in the form of heat is\",Q*10**-8,\"*10**8 J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy is being lost per day in the form of heat is 4.32 *10**8 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 59
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.12 Page no 260"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=50.0 #ohm\n",
+ "V=10 #V\n",
+ "T=3600 #s\n",
+ "\n",
+ "#Calculation\n",
+ "I=V/R\n",
+ "W=I**2*R*T\n",
+ "\n",
+ "#Result \n",
+ "print\"The source of this energy is\",W,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The source of this energy is 7200.0 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 64
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.13 Page no 260"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=100 #W\n",
+ "V=200 #V\n",
+ "T=60\n",
+ "I=10\n",
+ "\n",
+ "#Calculation\n",
+ "R=V**2/P\n",
+ "W=(P**2/R)*I*T\n",
+ "\n",
+ "#Result \n",
+ "print\"The heat & light produced by the bulb is\",W,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The heat & light produced by the bulb is 15000 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 70
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.14 Page no 260"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=3 #ohm\n",
+ "R2=6\n",
+ "R3=2\n",
+ "V=12 #V\n",
+ "T=300 #s\n",
+ "\n",
+ "#Calculation\n",
+ "R12=R1*R2/(R1+R2)\n",
+ "R=R12+R3\n",
+ "I=V/R\n",
+ "I1=I*R2/(R1+R2)\n",
+ "I2=I*R1/(R1+R2)\n",
+ "Q1=I1**2*R1*T\n",
+ "Q2=I2**2*R2*T\n",
+ "Q3=I**2*R3*T\n",
+ "\n",
+ "#Result \n",
+ "print\"Heat produced in resistance R1 is\",Q1,\"J\"\n",
+ "print\"Heat produced in resistance R2 is\",Q2,\"J\"\n",
+ "print\"Heat produced in resistance R3 is\",Q3,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat produced in resistance R1 is 3600 J\n",
+ "Heat produced in resistance R2 is 1800 J\n",
+ "Heat produced in resistance R3 is 5400 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.15 Page no 260"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "t1=300 #S\n",
+ "t2=600 #S\n",
+ "\n",
+ "#Calculation\n",
+ "t=(t1+t2)/60.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Time required to boil water is\",t,\"minute\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time required to boil water is 15.0 minute\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.16 Page no 261"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=100 #ohm\n",
+ "I=0.5 #A\n",
+ "W=10*10**-3 #Kg\n",
+ "M=250*10**-3 #Kg\n",
+ "T=1800 #s\n",
+ "C=4.2*10**3\n",
+ "\n",
+ "#Calculation\n",
+ "Q=I**2*R*T\n",
+ "V=Q/((W+M)*C)\n",
+ "\n",
+ "#Result\n",
+ "print\"The rise of the temperature is\",round(V,1),\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The rise of the temperature is 41.2 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.17 Page no 261"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=200\n",
+ "R=100\n",
+ "R1=121.0\n",
+ "A=1000 #W\n",
+ "B=400 #W\n",
+ "C=400 #W\n",
+ "T=116 #h\n",
+ "G=2.50\n",
+ "\n",
+ "#Calculation\n",
+ "D=V**2/R\n",
+ "D1=V**2/R1\n",
+ "E=A+B+C+D1\n",
+ "F=E*T\n",
+ "H=F*G\n",
+ "\n",
+ "#Result \n",
+ "print\"The bill for february is\",round(H*10**-3,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The bill for february is 617.87\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.18 Page no 261"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=0.12 #KWh\n",
+ "b=1.44 #KWh\n",
+ "c=0.48 #KWh\n",
+ "d=0.24 \n",
+ "e=0.30\n",
+ "i=2.70\n",
+ "g=30\n",
+ "\n",
+ "#Calculation\n",
+ "f=a+b+c+d+e\n",
+ "h=2*g*f\n",
+ "k=i*h\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric bill for 2 month is\", k"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric bill for 2 month is 417.96\n"
+ ]
+ }
+ ],
+ "prompt_number": 103
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.19 Page no 261"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=100*10**6 #W\n",
+ "V1=20000\n",
+ "V2=200 #V\n",
+ "\n",
+ "#Calculation\n",
+ "I1=P/V1\n",
+ "Q1=I1**2\n",
+ "I2=P/V2\n",
+ "Q2=I2**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Transmission done at V1=20,000 is\", Q1*10**-5,\"10**5\"\n",
+ "print\"Transmission done at V2=200 V is\", Q2*10**-10,\"10**10\"\n",
+ "print\"Less power wastage when transmission is done at Q1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Transmission done at V1=20,000 is 250.0 10**5\n",
+ "Transmission done at V2=200 V is 25.0 10**10\n",
+ "Less power wastage when transmission is done at Q1\n"
+ ]
+ }
+ ],
+ "prompt_number": 105
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.20 Page no 262"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "p1=4.9*10**-7 #ohm m\n",
+ "l1=8.456 #m\n",
+ "b1=1.0 #mm\n",
+ "t1=0.03*10**-6 #m\n",
+ "l2=4.235\n",
+ "A2=0.12*10**-6\n",
+ "p2=1.1*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "A1=b1*t1\n",
+ "R1=p1*(l1/A1)\n",
+ "R2=p2*(l2/A2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Rate for heat production for constantan ribbon is\", round(R1,3),\"W\"\n",
+ "print\"Rate for heat production for nichrome ribbon is\",round(R2,3),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate for heat production for constantan ribbon is 138.115 W\n",
+ "Rate for heat production for nichrome ribbon is 38.821 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 115
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.21 Page no 262"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=50 #V\n",
+ "I=12 #A\n",
+ "d=70\n",
+ "\n",
+ "#Calculation\n",
+ "g=V*I\n",
+ "t=g*(d/100.0)\n",
+ "v=t/I**2\n",
+ "\n",
+ "#Result \n",
+ "print\"The resistance of the motor is\",round(v,2),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resistance of the motor is 2.92 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 76
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.22 Page no 262"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=100\n",
+ "b=6\n",
+ "c=8.0\n",
+ "d=2.0\n",
+ "f=0.50\n",
+ "t=15*60\n",
+ "\n",
+ "#Calculation\n",
+ "e=a-b*d\n",
+ "x=b*f\n",
+ "r=(e/c)-x\n",
+ "r1=a*r\n",
+ "r2=r**2*(r+b*f)\n",
+ "r3=r1-r2\n",
+ "M=r3*t\n",
+ "\n",
+ "#Result\n",
+ "print\"Resistance in the charging circuit is\",r,\"ohm\"\n",
+ "print\"The power supplied by the d.c source is\",r1,\"W\"\n",
+ "print\"The power dissipated as heat is\",r2,\"W\"\n",
+ "print\"The power stored in the battery is\",r3,\"W\"\n",
+ "print\"Energy stored in the battery in 15 minute is\",M,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance in the charging circuit is 8.0 ohm\n",
+ "The power supplied by the d.c source is 800.0 W\n",
+ "The power dissipated as heat is 704.0 W\n",
+ "The power stored in the battery is 96.0 W\n",
+ "Energy stored in the battery in 15 minute is 86400.0 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 96
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.23 Page no 262"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=64 #V\n",
+ "e=2.0 #V\n",
+ "n=8\n",
+ "r=1/8.0\n",
+ "R=7 #ohm\n",
+ "I=3.5 #A\n",
+ "b=3600\n",
+ "\n",
+ "#Calculation\n",
+ "Te=e*n\n",
+ "R1=r*n\n",
+ "E=V*I\n",
+ "E1=Te*I\n",
+ "E11=I**2*(R+R1)\n",
+ "Em=E-(E1+E11)\n",
+ "Em1=Em*b\n",
+ "Ec=E1*b\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Mechanical energy yielded by motor is\", Em1,\"J\"\n",
+ "print\"(b) Chemical energy stored in the battery is\",Ec,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Mechanical energy yielded by motor is 252000.0 J\n",
+ "(b) Chemical energy stored in the battery is 201600.0 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 130
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.25 Page no 263"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=24 #V\n",
+ "r=4 #ohm\n",
+ "a=1\n",
+ "\n",
+ "#Calculation\n",
+ "R=r/(2.0-a)\n",
+ "I=E/(R+r)\n",
+ "\n",
+ "#Result\n",
+ "print\"Current drawn from the battery is\", I,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current drawn from the battery is 3.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 134
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap9_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap9_1.ipynb new file mode 100644 index 00000000..ecaec460 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap9_1.ipynb @@ -0,0 +1,627 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:3404f51fc9045816d89b5507acdfe9797796d47a0805fc064b3ee38cacd900d7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 9 Chemical effects of current"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.1 Page no 285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=1*10**-3 #kg\n",
+ "I=2 #A\n",
+ "z=3.3*10**-7 #kg/C\n",
+ "\n",
+ "#Calculation\n",
+ "t=m/(z*I)\n",
+ "\n",
+ "#Result\n",
+ "print\"Time required is\", round(t,1),\"s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time required is 1515.2 s\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.2 Page no 285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=0.15*10**-3 #Kg\n",
+ "z=3.3*10**-7 #Kg/C\n",
+ "t=900 #S\n",
+ "I1=0.6 #A\n",
+ "\n",
+ "#Calculation\n",
+ "I=m/(z*t)\n",
+ "I2=I-I1\n",
+ "\n",
+ "#Result\n",
+ "print\"Correction required for the ammeter reading is\", round(I2,1),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Correction required for the ammeter reading is -0.1 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.3 Page no 285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "t=0.002 #m\n",
+ "A=72 #cm**2\n",
+ "d=8.9 #g/cm**3\n",
+ "z=33*10**-5 #g/C\n",
+ "I=5 #A\n",
+ "\n",
+ "#Calculation\n",
+ "V=t*A\n",
+ "m=V*d\n",
+ "t1=m/(z*I)\n",
+ "\n",
+ "#Result\n",
+ "print\"Time required is\", round(t1,0),\"S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time required is 777.0 S\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.4 Page no 286"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m1=2 #g\n",
+ "m2=1 #g\n",
+ "t=1800 #s\n",
+ "z1=1118*10**-6 \n",
+ "z2=3294*10**-7 \n",
+ "a=6\n",
+ "\n",
+ "#Calculation\n",
+ "l1=m1/(z1*t)\n",
+ "l2=m2/(z2*t)\n",
+ "l=l1+l2\n",
+ "p=l*a\n",
+ "\n",
+ "#Result\n",
+ "print\"Power of the current is\",round(p,3),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power of the current is 16.082 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.5 Page no 286"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m1=0.403*10**-3 #Kg\n",
+ "z1=1.12*10**-6\n",
+ "z2=3.3*10**-7\n",
+ "t=900 #s\n",
+ "e=12 #V\n",
+ "\n",
+ "#Calculation\n",
+ "m2=(m1*z2)/z1\n",
+ "d=(e*m1)/z1\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Mass of the copper deposited is\",round(m2*10**3,3),\"10**-3\"\n",
+ "print\"(b) The energy supplied by the battery is\",round(d*10**-3,2),\"10**3\",\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Mass of the copper deposited is 0.119 10**-3\n",
+ "(b) The energy supplied by the battery is 4.32 10**3 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.6 Page no 286"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=100 #ohm\n",
+ "t=600 #s\n",
+ "z=3.3*10**-7\n",
+ "m=10**-4\n",
+ "\n",
+ "#Calculation\n",
+ "I=m/(z*t)\n",
+ "Q=I**2*r*t\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat produced in the resistance coil is\",round(Q,1),\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat produced in the resistance coil is 15304.6 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 64
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.7 Page no 286"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=1.008\n",
+ "v=1\n",
+ "m1=1.05*10**-8\n",
+ "a1=63.54\n",
+ "v1=2\n",
+ "m2=3.29*10**-7\n",
+ "a2=107.9\n",
+ "v2=1\n",
+ "m3=1.12*10**-6\n",
+ "a3=55.85\n",
+ "v3=3\n",
+ "\n",
+ "#Calculation\n",
+ "#For water voltameter\n",
+ "E=a/v\n",
+ "s=m1/E\n",
+ "\n",
+ "#For copper voltameter\n",
+ "E2=a1/v1\n",
+ "s1=m2/E2\n",
+ "\n",
+ "#For silver voltameter\n",
+ "E3=a2/v2\n",
+ "s2=m3/E3\n",
+ "\n",
+ "#For iron voltameter\n",
+ "E4=a3/v3\n",
+ "m4=(s/a)*E4\n",
+ "\n",
+ "#Result\n",
+ "print\"Mass of hydrogen liberated is\",round(s*10**8,4)*10**-8\n",
+ "print\"Mass of copper deposited is\",round(s1*10**8,4)*10**-8\n",
+ "print\"Mass of silver deposited is\",round(s2*10**8,4)*10**-8\n",
+ "print\"Mass of iron deposited is\", round(m4*10**7,2)*10**-7,\"Kg/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mass of hydrogen liberated is 1.0417e-08\n",
+ "Mass of copper deposited is 1.0356e-08\n",
+ "Mass of silver deposited is 1.038e-08\n",
+ "Mass of iron deposited is 1.92e-07 Kg/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 104
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.8 Page no 287"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=2.5 #g\n",
+ "I=10 #A\n",
+ "F=96500 #C/mol\n",
+ "m1=63.5\n",
+ "n=2.0\n",
+ "\n",
+ "#Calculation\n",
+ "E=m1/n\n",
+ "t=m*F/(E*I)\n",
+ "\n",
+ "#Result\n",
+ "print\"Time required is\",round(t,1),\"S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time required is 759.8 S\n"
+ ]
+ }
+ ],
+ "prompt_number": 110
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9 Page no 287"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=16.43 #g\n",
+ "t=4000 #s\n",
+ "F=96485 #C/mol\n",
+ "m1=63.54\n",
+ "n=2.0\n",
+ "I1=12.6 #A\n",
+ "\n",
+ "#Calculation\n",
+ "E=m1/n\n",
+ "I=m*F/(E*t)\n",
+ "I2=I1-I\n",
+ "\n",
+ "#Result\n",
+ "print\"Error in the ammeter reading is\", round(I2,3),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Error in the ammeter reading is 0.126 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 119
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.10 Page no 287"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "t=600 #S\n",
+ "m=5.92 #g\n",
+ "F=96500 #C/mol\n",
+ "V1=1.62 #V\n",
+ "V2=1.34\n",
+ "m1=63.5\n",
+ "n=2.0\n",
+ "\n",
+ "#Calculation\n",
+ "V=V1-V2\n",
+ "E=m1/n\n",
+ "I=m*F/(E*t)\n",
+ "R=V/I\n",
+ "\n",
+ "#Result\n",
+ "print\"Resistance of the voltmeter is\",round(R*10**3,2),\"m ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance of the voltmeter is 9.34 m ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 123
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.11 Page no 287"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=8 #V\n",
+ "r=1 #ohm\n",
+ "R=15 #ohm\n",
+ "E1=120\n",
+ "t=300 #s\n",
+ "\n",
+ "#Calculation\n",
+ "I=(E1-E)/(R+r)\n",
+ "V=E+(I*r)\n",
+ "E12=E*I*t\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Current in the circuit is\", I,\"A\"\n",
+ "print\"(b) Terminal voltage across the battery is\",V,\"V\"\n",
+ "print\"(c) Chemical energy stored in the battery is\",E12,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Current in the circuit is 7 A\n",
+ "(b) Terminal voltage across the battery is 15 V\n",
+ "(c) Chemical energy stored in the battery is 16800 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 133
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.12 Page no 288"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=10 #A\n",
+ "t=300 #S\n",
+ "m=2.016\n",
+ "n=2.0\n",
+ "n1=1.0\n",
+ "m1=1.008\n",
+ "F=96500\n",
+ "V=22.4\n",
+ "\n",
+ "#Calculation\n",
+ "q=I*t\n",
+ "M=m/n\n",
+ "M2=m1/n1\n",
+ "q1=F*m/m1\n",
+ "V1=V*q/q1\n",
+ "\n",
+ "#Result\n",
+ "print\"Volume of hydrogen is\",round(V1,4),\"litre\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Volume of hydrogen is 0.3482 litre\n"
+ ]
+ }
+ ],
+ "prompt_number": 136
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.13 Page no 288"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "w=107.9 #g/mol\n",
+ "r=2 #ohm\n",
+ "E=12 #V\n",
+ "V=10\n",
+ "F=96500\n",
+ "t=1800\n",
+ "\n",
+ "#Calculation\n",
+ "R=r*(V/(E-V))\n",
+ "I=E/(R+r)\n",
+ "E=w/I\n",
+ "z=E/F\n",
+ "m=z*I*t\n",
+ "\n",
+ "#Result\n",
+ "print\"Silver deposited at the cathode is\", round(m,2),\"g\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Silver deposited at the cathode is 2.01 g\n"
+ ]
+ }
+ ],
+ "prompt_number": 143
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.14 Page no 288"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=5.0 #A\n",
+ "a=0.5 #mole\n",
+ "n=1\n",
+ "F=96500\n",
+ "I1=10 #A\n",
+ "t1=9650*2\n",
+ "n1=2.0\n",
+ "m=63.54\n",
+ "m2=55.85\n",
+ "n2=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "q=F*a\n",
+ "t=q/I\n",
+ "E=m/n1\n",
+ "m1=(E*I1*t1)/F\n",
+ "E3=m2/n2\n",
+ "m3=(E3*I1*t1)/F\n",
+ "\n",
+ "#Result\n",
+ "print\"Molar mass of copper is\", m1,\"equal to its atomic mass i.e 1 mole of copper is liberated.\"\n",
+ "print\"Molar mass of iron is\",round(m3,3),\"Hence 2/3 mole of iron will be deposited.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Molar mass of copper is 63.54 equal to its atomic mass i.e 1 mole of copper is liberated.\n",
+ "Molar mass of iron is 37.233 Hence 2/3 mole of iron will be deposited.\n"
+ ]
+ }
+ ],
+ "prompt_number": 153
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/chap_29_1.ipynb b/modern_physics_by_Satish_K._Gupta/chap_29_1.ipynb new file mode 100644 index 00000000..581d372d --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/chap_29_1.ipynb @@ -0,0 +1,298 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:15f32b3e96d10d200143bd7ffbc8811afc0974b65cf155210561266d411e6511"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 29 Wave Nature Of Matter"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.1 Page no 815"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.62*10**-34 #Js\n",
+ "Me=9.1*10**-31 #Kg\n",
+ "Ve=10**5 #m s**-1\n",
+ "Mp=1.67*10**-27 #Kg\n",
+ "Vp=10**5 #m s**-1\n",
+ "\n",
+ "#Calculation\n",
+ "Le=h/(Me*Ve)\n",
+ "Lp=h/(Mp*Vp)\n",
+ "\n",
+ "#Result\n",
+ "print\"The de-broglie wavelength of electron is\",round(Le*10**9,2),\"*10**-9 m\"\n",
+ "print\"The de-broglie wavelenght of proton is\",round(Lp*10**12,2),\"*10**-12 m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The de-broglie wavelength of electron is 7.27 10**-9 m\n",
+ "The de-broglie wavelenght of proton is 3.96 10**-12 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.2 Page no 816"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M0=9.1*10**-31 #Kg\n",
+ "h=6.62*10**-34 #J s\n",
+ "V=0.5 #c\n",
+ "V1=1.5*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "L=(h*math.sqrt(1-V**2))/(M0*V1)\n",
+ "\n",
+ "#Result\n",
+ "print\"The de-broglie wavelenght is\",round(L*10**12,1),\"*10**-12 m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The de-broglie wavelenght is 4.2 10**-12 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.3 Page no 816"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.6*10**-34 #J s\n",
+ "m=9.1*10**-31 #kg\n",
+ "Iev=1.6*10**-19 #J\n",
+ "E=6.4*10**-17 #J\n",
+ "\n",
+ "#Calculation\n",
+ "L=h/(math.sqrt(2*m*E))\n",
+ "\n",
+ "#Result\n",
+ "print\"The de-broglie wavelenght of electron is\",round(L*10**10,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The de-broglie wavelenght of electron is 0.61 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.4 Page no 816"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "T=300 #K\n",
+ "K=1.38*10**-23 #J\n",
+ "h=6.62*10**-34\n",
+ "m=1.675*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "E=(3/2.0)*K*T\n",
+ "L=h/(math.sqrt(2*m*E))\n",
+ "\n",
+ "#Result\n",
+ "print\"The de-broglie wavelenght associated with thermal neutrons is\",round(L*10**10,3),\"*10**-10 m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The de-broglie wavelenght associated with thermal neutrons is 1.451 10**-10 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.5 Page no 816"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=10**-10 #m\n",
+ "h=6.62*10**-34\n",
+ "c=3*10**8\n",
+ "S=1.6*10**-19\n",
+ "m=9.11*10**-31 #kg\n",
+ "\n",
+ "#Calculation\n",
+ "#For X-ray photon of wavelength\n",
+ "E=(h*c)/L\n",
+ "E1=E/S\n",
+ "#For electron of wavelength \n",
+ "Mv=h/L\n",
+ "E2=(Mv**2)/(2*m)\n",
+ "E3=E2/S\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy of electron is\",round(E3,1),\"eV\"\n",
+ "print\"It follows that X-ray photon has greater energy\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy of electron is 150.3 eV\n",
+ "It follows that X-ray photon has greater energy\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.6 Page no 816"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=1.675*10**-27 #kg\n",
+ "E=2.4*10**-17 #J\n",
+ "h=6.62*10**-34\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "L=h/(math.sqrt(2*M*E))\n",
+ "\n",
+ "#Result\n",
+ "print\"The de-broglie wavelength of neutron is\",round(L*10**12,3)*10**-12,\" m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The de-broglie wavelength of neutron is 2.335e-12 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.7 Page no 816"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=50 #kV\n",
+ "E=8.0*10**-15 #J\n",
+ "m=9.1*10**-31\n",
+ "h=6.62*10**-34\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "L=h/(math.sqrt(2*m*E))\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy of electron is\",round(L*10**12,3),\"*10**-12 m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy of electron is 5.486 10**-12 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 54
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/modern_physics_by_Satish_K._Gupta/screenshots/image1_1.png b/modern_physics_by_Satish_K._Gupta/screenshots/image1_1.png Binary files differnew file mode 100644 index 00000000..a1aeb52f --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/screenshots/image1_1.png diff --git a/modern_physics_by_Satish_K._Gupta/screenshots/image22_1.png b/modern_physics_by_Satish_K._Gupta/screenshots/image22_1.png Binary files differnew file mode 100644 index 00000000..c6bd4243 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/screenshots/image22_1.png diff --git a/modern_physics_by_Satish_K._Gupta/screenshots/image33_1.png b/modern_physics_by_Satish_K._Gupta/screenshots/image33_1.png Binary files differnew file mode 100644 index 00000000..26358a05 --- /dev/null +++ b/modern_physics_by_Satish_K._Gupta/screenshots/image33_1.png diff --git a/sample_notebooks/RuchiMittal/chpater3.ipynb b/sample_notebooks/RuchiMittal/chpater3.ipynb new file mode 100644 index 00000000..99311d63 --- /dev/null +++ b/sample_notebooks/RuchiMittal/chpater3.ipynb @@ -0,0 +1,427 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:d86ea2343f7008555f3356fee2fcfd0faad0c29bd34da2bb93d133cd162e92e2"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3 Polarization of light"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.1 Page no 30"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u =1.54 #refrective index of glass\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "i= math.atan(u*180/3.14)\n",
+ "r=3.14/2.0-i\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of refraction is\",round(r,4),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of refraction is 0.0105 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.2 Page no 30"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "i= 3.14/3.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "u= math.tan(i)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Refractive index of glass is\",round(u,2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Refractive index of glass is 99.17\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "raw",
+ "metadata": {},
+ "source": [
+ "Example 3.3 Page no 31"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Uo =1.55\n",
+ "Ue =1.54\n",
+ "w=6*10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "p=w/2.0\n",
+ "t=w /(2*( Uo -Ue))\n",
+ "\n",
+ "#Result\n",
+ "print\"Thickness of doubly refracting crystal is\",t*10**2,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness of doubly refracting crystal is 0.003 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.4 Page no 31"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Uo =1.54\n",
+ "r =1.007\n",
+ "Ue=r*Uo\n",
+ "w =5893*10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "t=w /(2*( Uo -Ue))\n",
+ "t= abs (t)\n",
+ "\n",
+ "print\"Thickness of halfwave plate is\",round(t*10**5,3)*10**-5"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness of halfwave plate is 2.733e-05\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.5 Page no 32"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Uo =1.652 #refractive index for O ray\n",
+ "Ue =1.488\n",
+ "w =546*10**-9\n",
+ "\n",
+ "#Calculation\n",
+ "p=w/2.0\n",
+ "t=w /(4.0*( Uo -Ue))\n",
+ "t1=t *100\n",
+ "\n",
+ "#Result\n",
+ "print\"Thickness of quarterwave plate is\",round(t1*10**5,2)*10**-5,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness of quarterwave plate is 8.32e-05 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.6 Page no 32"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Uo =1.658\n",
+ "Ue =1.486\n",
+ "w =589*10**-9\n",
+ "n =1\n",
+ "\n",
+ "#Calculation\n",
+ "t =(2*n -1)*w /(4.0*( Uo -Ue))\n",
+ "t1=t *100\n",
+ "\n",
+ "#Result\n",
+ "print\"Thickness of calcite plate is\",round(t1*10**5,2)*10**-5,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness of calcite plate is 8.56e-05 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.7 Page no 33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ur =1.55810\n",
+ "Ul =1.55821\n",
+ "w=4*10**-7\n",
+ "d =0.002\n",
+ "\n",
+ "#Calculation\n",
+ "R= 3.14*d*(Ul -Ur)/w\n",
+ "R1=R *180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Amount of optional rotation is\",R1,\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Amount of optional rotation is 99.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.8 Page no 33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Uo =1.5508\n",
+ "Ue =1.5418\n",
+ "w=5.0*10**-7\n",
+ "t =0.000032\n",
+ "\n",
+ "#Calculation\n",
+ "p =2*3.14*(Uo -Ue)*t/w\n",
+ "\n",
+ "#Result\n",
+ "print\"Phase retardation is\",round(p,2),\"radian\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Phase retardation is 3.62 radian\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.9 Page no 33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "theta =6.5\n",
+ "l =2\n",
+ "C =0.05\n",
+ "\n",
+ "#Calculation\n",
+ "S= theta /(l*C)\n",
+ "\n",
+ "#Result\n",
+ "print\"Specific rotation of sugar solution is\",S"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Specific rotation of sugar solution is 65.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.10 Page no 34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l =2 #length of solution in decimeter\n",
+ "theta =12\n",
+ "S =60.0\n",
+ "\n",
+ "#Calculation\n",
+ "C= theta /(S*l)\n",
+ "\n",
+ "#Result\n",
+ "print\"Strength of solution is\",C,\"gm/cc\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Strength of solution is 0.1 gm/cc\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.11 Page no 34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "b =.172 #bifringe of plate\n",
+ "w=6*10**-7\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "t=w /(4*( b))\n",
+ "t1=t *100\n",
+ "\n",
+ "#Result\n",
+ "print\"Thickness of quarterwave plate is\",round(t1*10**5,2)*10**-5,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness of quarterwave plate is 8.72e-05 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |
