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diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter1.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter1.ipynb new file mode 100644 index 00000000..dff05eac --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter1.ipynb @@ -0,0 +1,214 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1 : Electricity" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 1_5 Page No. 24" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Output Voltage of a Battery = 7.20 Volts\n" + ] + } + ], + "source": [ + "# What is the output voltage of a battery that expends 3.6 J of energy in moving 0.5C of charge?\n", + "\n", + "# Given data\n", + "\n", + "W = 3.6# # Work=3.6 Jouls\n", + "Q = 0.5# # Charge=0.5 Columb\n", + "\n", + "V = W/Q#\n", + "print 'The Output Voltage of a Battery = %0.2f Volts'%V" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 1_6 Page No. 24" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Intensity of Charge Flow = 12.00 Amps\n" + ] + } + ], + "source": [ + "#The charge of 12 C moves past a given point every second. How much is the intensity of charge flow?\n", + "\n", + "# Given data\n", + "\n", + "Q = 12# # Charge=12 Columb\n", + "T = 1# # Time=1 Sec i.e every second\n", + "\n", + "I = Q/T#\n", + "print 'The Intensity of Charge Flow = %0.2f Amps'%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 1_7 Page No. 24" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current = 5.00 Amps\n" + ] + } + ], + "source": [ + "# The charge of 5 C moves past a given point in 1 s. How much is the current?\n", + "\n", + "# Given data\n", + "\n", + "Q = 5# # Charge=5 Columb\n", + "T = 1# # Time=1 Sec\n", + "\n", + "I = Q/T#\n", + "print 'The Current = %0.2f Amps'%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 1_8 Page No. 25" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistance for Conductance value 0.05 S = 20.00 Ohms\n", + "The Resistance for Conductance value 0.1 S = 10.00 Ohms\n" + ] + } + ], + "source": [ + "# Calculate the resistance for the following conductance values: (a) 0.05 S (b) 0.1 S\n", + "\n", + "# Given data\n", + "\n", + "G1 = 0.05# # G1=0.05 Siemins\n", + "G2 = 0.1# # G1=0.1 Siemins\n", + "\n", + "R1 = 1/G1#\n", + "print 'The Resistance for Conductance value 0.05 S = %0.2f Ohms'%R1\n", + "\n", + "R2 = 1/G2#\n", + "print 'The Resistance for Conductance value 0.1 S = %0.2f Ohms'%R2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 1_9 Page No. 26" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Conductance for Resistance value 1 kohms = 1e-03 Siemens\n", + "OR 1 mS\n", + "The Conductance for Resistance value 5 kohms = 2e-04 Siemens\n", + "OR 200 uS\n" + ] + } + ], + "source": [ + "#Calculate the conductance for the following resistance values: (a) 1 kOhms \b(b)5 kOhms\n", + "\n", + "# Given data\n", + "\n", + "R1 = 1.0*10**3# # R1=1k Ohms\n", + "R2 = 5.0*10**3# # R2=5k Ohms\n", + "\n", + "G1 = 1/R1#\n", + "print 'The Conductance for Resistance value 1 kohms = %0.e Siemens'%G1\n", + "print 'OR 1 mS'\n", + "\n", + "G2 = 1/R2#\n", + "print 'The Conductance for Resistance value 5 kohms = %0.e Siemens'%G2\n", + "print 'OR 200 uS'" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter11.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter11.ipynb new file mode 100644 index 00000000..4e938ea1 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter11.ipynb @@ -0,0 +1,246 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 11 : Conductors and Insulators" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 11_1 Page No. 329" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Circular Area = 25.00 cmils\n" + ] + } + ], + "source": [ + "# What is the area in circular mils of a wire with a diameter of 0.005 in.?\n", + "\n", + "# Given data\n", + "\n", + "Din = 0.005# # Diameter in Inches=0.005 in.\n", + "Dmil = 5# # Diameter in Mils=5 mil.\n", + "\n", + "# 0.005 in. = 5 mil\n", + "# Therefore: Din == Dmil\n", + "\n", + "A = Dmil*Dmil#\n", + "print 'The Circular Area = %0.2f cmils'%A" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 11_2 Page No. 335" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Total Area = 3224.00 cmils\n" + ] + } + ], + "source": [ + "# A stranded wire is made up of 16 individual strands of No. 27 gage wire. What is its equivalent gage size in solid wire?\n", + "\n", + "# Given data\n", + "\n", + "N = 16# # No. of strands=16\n", + "A27 = 201.5 # Circular area of No. 27 Guage wire=201.5 cmils\n", + "\n", + "A = N*A27#\n", + "print 'The Total Area = %0.2f cmils'%A" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 11_3 Page No. 340" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistance of 100 ft of No. 20 gage Copper Wire = 1.02 ohms\n" + ] + } + ], + "source": [ + "# How much is the resistance of 100 ft of No. 20 gage copper wire?\n", + "\n", + "# Given data\n", + "\n", + "roh = 10.4# # roh or specific resistance=10.4 (for Copper)\n", + "l = 100.# # Lenght=100 feet\n", + "A = 1022# # Area of No. 20 Gage=1022 cmil\n", + "\n", + "R = roh*(l/A)#\n", + "print 'The Resistance of 100 ft of No. 20 gage Copper Wire = %0.2f ohms'%R" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 11_4 Page No. 342" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistance of 100 ft of No. 20 gage Copper Wire = 2.04 ohms\n" + ] + } + ], + "source": [ + "# How much is the resistance of 100 ft of No. 23 gage copper wire?\n", + "\n", + "# Given data\n", + "\n", + "roh = 10.4# # roh or specific resistance=10.4 (for Copper)\n", + "l = 100# # Lenght=100 feet\n", + "A = 509.5# # Area of No. 23 Gage=509.5 cmil\n", + "\n", + "R = roh*(l/A)#\n", + "print 'The Resistance of 100 ft of No. 20 gage Copper Wire = %0.2f ohms'%R" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 11_5 Page No. 344" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistance of a Slab of Germanium = 11.00 ohms\n" + ] + } + ], + "source": [ + "# How much is the resistance of a slab of germanium 0.2 cm long with a crosssectional area of 1 sqcm?\n", + "\n", + "# Given data\n", + "\n", + "roh = 55.0# # roh or specific resistance=55 (for Germanium)\n", + "l = 0.2*10**-2# # Lenght=100 feet\n", + "A = 1.0*10**-2# # Area=1 sqcm\n", + "\n", + "R = roh*(l/A)#\n", + "print 'The Resistance of a Slab of Germanium = %0.2f ohms'%R" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 11_6 Page No. 344" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistance at 120°C = 21.00 ohms\n" + ] + } + ], + "source": [ + "# A tungsten wire has a 14-Ohms\u0002 R at 20°C. Calculate its resistance at 120°C.\n", + "\n", + "# Given data\n", + "\n", + "Tmax = 120.# # Temp(max)=120 degree Centigrates\n", + "Tmin = 20.# # Temp(min)=20 degree Centigrates\n", + "Ro = 14.# # Wire Resistance=14 Ohms\n", + "alpha = 0.005# # Aplha=0.005 (for Tungsten)\n", + "\n", + "delta = Tmax-Tmin#\n", + "\n", + "Rt = Ro+Ro*(alpha*delta)#\n", + "print 'The Resistance at 120°C = %0.2f ohms'%Rt" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter12.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter12.ipynb new file mode 100644 index 00000000..8876ae94 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter12.ipynb @@ -0,0 +1,67 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 12 : Batteries" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 12_1 Page No. 368" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistance ri = 10 Ohms\n" + ] + } + ], + "source": [ + "# Calculate ri if the output of a generator drops from 100 V with zero load current to 80 V when Il is 2 A.\n", + "\n", + "# Given data\n", + "\n", + "Vo0 = 100# # Vo at zero load current=100 Volts\n", + "Vo1 = 80# # Vo at 2 A load current=80 Volts\n", + "Il = 2# # Load current=2 Amps\n", + "\n", + "Ri = (Vo0-Vo1)/Il#\n", + "print 'The Resistance ri = %0.f Ohms'%Ri" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter13.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter13.ipynb new file mode 100644 index 00000000..89a82ed0 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter13.ipynb @@ -0,0 +1,182 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 13 : Magnetism" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 13_1 Page No. 291" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The 25000 Maxwell = 2.50e-04 Wabers\n", + "i.e 250*10**-6 Wb or 250 uWb\n", + "The 0.005 Wabers = 500000.00 Maxwell\n", + "i.e 5.0*10**5 Mx\n" + ] + } + ], + "source": [ + "# Make the following conversions: (a) 25,000 Mx to Wb# (b) 0.005 Wb to Mx.\n", + "\n", + "# Given data\n", + "\n", + "A = 25000.0# # A=25000 Maxwell\n", + "B = 0.005# # B=0.005 Wabers\n", + "C = 1*10**8# # Conversion Factor\n", + "\n", + "Wb = A*(1.0/C)#\n", + "print 'The 25000 Maxwell = %0.2e Wabers'%Wb\n", + "print 'i.e 250*10**-6 Wb or 250 uWb'\n", + "\n", + "Mx = B*C#\n", + "print 'The 0.005 Wabers = %0.2f Maxwell'%Mx\n", + "print 'i.e 5.0*10**5 Mx'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 13_2 Page No. 392" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Flux Density = 2000.00 Guass (G)\n" + ] + } + ], + "source": [ + "# With a flux of 10,000 Mx through a perpendicular area of 5 sqcm, what is the flux density in gauss?\n", + "\n", + "# Given data\n", + "\n", + "A = 5.# # Area=5 sqcm\n", + "flux = 10000.0# # Total Flux=10000 Mx\n", + "\n", + "B = flux/A#\n", + "print 'The Flux Density = %0.2f Guass (G)'%B" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 13_3 Page No. 394" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Flux Density = 0.80 Tesla (T)\n" + ] + } + ], + "source": [ + "# With a flux of 400 u\u0002Wb through an area of 0.0005 sqm, what is the flux density B in tesla units?\n", + "\n", + "# Given data\n", + "\n", + "A = 0.0005# # Area=0.0005 sqm\n", + "flux = 400*10**-6# # Total Flux=400 uWb\n", + "\n", + "B = flux/A#\n", + "print 'The Flux Density = %0.2f Tesla (T)'%B" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 13_4 Page No. 394" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The 0.003 Tesla = 30.00 Guass\n", + "The 15,000 Guass = 1.50 Tesla\n" + ] + } + ], + "source": [ + "# Make the following conversions: (a) 0.003 T to G# (b) 15,000 G to T.\n", + "\n", + "# Given data\n", + "\n", + "A = 0.003# # A=0.003 Tesla\n", + "B = 15000.# # B=15000 Guass\n", + "C = 1.*10**4# # Conversion Factor\n", + "\n", + "G = A*C#\n", + "print 'The 0.003 Tesla = %0.2f Guass'%G\n", + "\n", + "T = B*(1/C)#\n", + "print 'The 15,000 Guass = %0.2f Tesla'%T" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter14.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter14.ipynb new file mode 100644 index 00000000..c8957744 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter14.ipynb @@ -0,0 +1,248 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter14 : Electromagnetism" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 14_1 Page No. 421" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Amps-Turn (A.t) of Magneto-Motive Force (mmf) = 10.00 A.t\n" + ] + } + ], + "source": [ + "# Calculate the ampere-turns of mmf for a coil with 2000 turns and a 5-mA current.\n", + "\n", + "# Given data\n", + "\n", + "I = 5*10**-3# # Current=5 mAmps\n", + "N = 2000# # No. of Turns=2000\n", + "\n", + "mmf = I*N#\n", + "print 'The Amps-Turn (A.t) of Magneto-Motive Force (mmf) = %0.2f A.t'%mmf" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 14_2 Page No. 421" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Turns necessary are : 150\n" + ] + } + ], + "source": [ + "# A coil with 4 A is to provide a magnetizing force of 600 A\u0002 t. How many turns are necessary?\n", + "\n", + "# Given data\n", + "\n", + "I = 4# # Current=4 Amps\n", + "mmf = 600# # Magnetizing Force=600 A.t\n", + "\n", + "N = mmf/I#\n", + "print 'The Turns necessary are : ',N" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 14_3 Page No. 424" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current necessary = 2 Amps\n" + ] + } + ], + "source": [ + "# A coil with 400 turns must provide 800 A\u0002 t of magnetizing force. How much current is necessary?\n", + "\n", + "# Given data\n", + "\n", + "mmf = 800# # Magnetizing Force=800 A.t\n", + "N = 400# # No. of Turns=400\n", + "\n", + "I = mmf/N#\n", + "print 'The Current necessary = %0.f Amps'%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 14_4 Page No. 426" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current necessary when a wire is connected to 6-V Battery = 2 Amps\n", + "The Amps-Turn (A.t) of Magneto-Motive Force (mmf) = 500 A.t\n" + ] + } + ], + "source": [ + "# The wire in a solenoid of 250 turns has a resistance of 3 Ohms. (a)How much is the current when the coil is connected to a 6-V battery? (b) Calculate the ampereturns of mmf.\n", + "\n", + "# Given data\n", + "\n", + "V = 6# # Voltage=6 Volts\n", + "R = 3# # Resistance=3 Ohms\n", + "N = 250# # No. of Turns=250\n", + "\n", + "I = V/R#\n", + "print 'The Current necessary when a wire is connected to 6-V Battery = %0.f Amps'%I\n", + "\n", + "mmf = I*N#\n", + "print 'The Amps-Turn (A.t) of Magneto-Motive Force (mmf) = %0.f A.t'%mmf" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 14_5 Page No. 426" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Absolute u as B/H in CGS = 500 (G/Oe)\n", + "The Absolute u as B/H in SI = 6.300e-04 (T/(A.t/m))\n", + "i.e 630*10**-6 T/(A.t/m)\n" + ] + } + ], + "source": [ + "# A magnetic material has a \u0003ur of 500. Calculate the absolute u\u0003 as B/H (a) in CGS units and (b) in SI units.\n", + "\n", + "# Given data\n", + "\n", + "ur = 500# # ur=500\n", + "uoa = 1# # uo for CGS Units=1\n", + "uob = 1.26*10**-6# # uo for SI Units=1.26 u\n", + "\n", + "ua = ur*uoa#\n", + "print 'The Absolute u as B/H in CGS = %0.f (G/Oe)'%ua\n", + "\n", + "ub = ur*uob#\n", + "print 'The Absolute u as B/H in SI = %0.3e (T/(A.t/m))'%ub\n", + "print 'i.e 630*10**-6 T/(A.t/m)'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 14_6 Page No. 427" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Flux density = 0.63 Tesla\n" + ] + } + ], + "source": [ + "# u = 630*10**-\u00056 in SI units, calculate the flux density B that will be produced by the field intensity H equal to 1000 A.t/m.\n", + "\n", + "# Given data\n", + "\n", + "u = 630*10**-6# # u=630 micro T/(A.t/m)\n", + "H = 1000# # H=1000 A.t/m\n", + "\n", + "B = u*H#\n", + "print 'The Flux density = %0.2f Tesla'%B" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter15.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter15.ipynb new file mode 100644 index 00000000..682677a5 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter15.ipynb @@ -0,0 +1,354 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 15 : Alternating voltage & current" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 15_1 Page No: 451" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage at 30° = 50.00 Volts\n", + "The Voltage at 45° = 70.71 Volts\n", + "The Voltage at 90° = 100.00 Volts\n", + "The Voltage at 270° = -100.00 Volts\n" + ] + } + ], + "source": [ + "from numpy import sin, pi\n", + "# A sine wave of voltage varies from zero to a maximum of 100 V. How much is the voltage at the instant of 30° of the cycle? 45°? 90°? 270°?\n", + "\n", + "# Given data\n", + "\n", + "Vm = 100# # Vm=100 Volts\n", + "t1 = 30# # Theta 1=30°.\n", + "t2 = 45# # Theta 2=45°.\n", + "t3 = 90# # Theta 3=90°.\n", + "t4 = 270# # Theta 4=270°.\n", + "\n", + "v1 = Vm*sin(t1*pi/180)\n", + "print 'The Voltage at 30° = %0.2f Volts'%v1\n", + "\n", + "v2 = Vm*sin(t2*pi/180)\n", + "print 'The Voltage at 45° = %0.2f Volts'%v2\n", + "\n", + "v3 = Vm*sin(t3*pi/180)\n", + "print 'The Voltage at 90° = %0.2f Volts'%v3\n", + "\n", + "v4 = Vm*sin(t4*pi/180)\n", + "print 'The Voltage at 270° = %0.2f Volts'%v4" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 15_2 Page No: 456" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Time period = 1e-03 Seconds\n", + "i.e 1/1000 sec\n", + "The Frequency = 1000.00 Hertz\n", + "OR 1 kHz\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "# An alternating current varies through one complete cycle in 1 ⁄ 1000 s. Calculate the period and frequency.\n", + "\n", + "# Given data\n", + "\n", + "tc = 1/1000# # One Complete Cycle=1 ⁄ 1000 sec.\n", + "\n", + "T = tc#\n", + "print 'The Time period = %0.e Seconds'%T\n", + "print 'i.e 1/1000 sec'\n", + "\n", + "f = 1/tc#\n", + "print 'The Frequency = %0.2f Hertz'%f\n", + "print 'OR 1 kHz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 15_3 Page No: 476" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Time period = 1e-06 Seconds of 1 MHz freq.\n", + "i.e 1*10**-6 sec = 1 usec\n", + "The Time period = 5e-07 Seconds of 2 MHz freq.\n", + "i.e 0.5*10**-6 sec = 0.5 usec\n" + ] + } + ], + "source": [ + "# Calculate the period for the two frequencies of 1 MHz and 2 MHz.Calculate the period for the two frequencies of 1 MHz and 2 MHz.\n", + "\n", + "# Given data\n", + "\n", + "f1 = 1*10**6# # Freq=1 MHz\n", + "f2 = 2*10**6# # Freq=2 MHz\n", + "\n", + "t1 = 1/f1#\n", + "print 'The Time period = %0.e Seconds of 1 MHz freq.'%t1\n", + "print 'i.e 1*10**-6 sec = 1 usec'\n", + "\n", + "t2 = 1/f2#\n", + "print 'The Time period = %0.e Seconds of 2 MHz freq.'%t2\n", + "print 'i.e 0.5*10**-6 sec = 0.5 usec'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 15_4 Page No: 478" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Lamda or Wavelenght = 1.00 cm\n" + ] + } + ], + "source": [ + "# Calculate lamda for a radio wave witf f of 30 GHz.\n", + "\n", + "# Given data\n", + "\n", + "c = 3*10**10# # Speed of light=3*10**10 cm/s\n", + "f = 30*10**9# # Freq=30 GHz\n", + "\n", + "l = c/f#\n", + "print 'The Lamda or Wavelenght = %0.2f cm'%l" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 15_5 Page No: 480" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Height = 250.00 cm\n", + "The Height = 8.20 feet\n" + ] + } + ], + "source": [ + "# The length of a TV antenna is lamda\u0005/2 for radio waves with f of 60 MHz. What is the antenna length in centimeters and feet?\n", + "\n", + "# Given data\n", + "\n", + "c = 3*10**10# # Speed of light=3*10**10 cm/s\n", + "f = 60*10**6# # Freq=60 MHz\n", + "In = 2.54# # 2.54 cm = 1 in\n", + "ft = 12# # 12 in = 1 ft\n", + "\n", + "l1 = c/f#\n", + "l = l1/2#\n", + "print 'The Height = %0.2f cm'%l\n", + "\n", + "li = l/In\n", + "lf = li/ft#\n", + "print 'The Height = %0.2f feet'%lf" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 15_6 Page No: 481" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Frequency = 50000000 Hertz\n", + "i.e 50*10**6 Hz OR 50 MHz\n" + ] + } + ], + "source": [ + "# For the 6-m band used in amateur radio, what is the corresponding frequency?\n", + "\n", + "# Given data\n", + "\n", + "v = 3*10**10# # Speed of light=3*10**10 cm/s\n", + "l = 6*10**2# # lamda=6 meter\n", + "\n", + "f = v/l\n", + "print 'The Frequency = %0.f Hertz'%f\n", + "print 'i.e 50*10**6 Hz OR 50 MHz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 15_7 Page No: 481" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Lamda or Wavelenght = 11.30 ft\n" + ] + } + ], + "source": [ + "# What is the wavelength of the sound waves produced by a loudspeaker at a frequency of 100 Hz?\n", + "\n", + "# Given data\n", + "\n", + "c = 1130# # Speed of light=1130 ft/s\n", + "f = 100# # Freq=100 Hz\n", + "\n", + "l = c/f#\n", + "print 'The Lamda or Wavelenght = %0.2f ft'%l" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 15_8 Page No: 482" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Lamda or Wavelength = 0.03 ft\n", + "The Lamda or Wavelength = 1.00 cm\n" + ] + } + ], + "source": [ + "# For ultrasonic waves at a frequency of 34.44 kHz, calculate the wavelength in feet and in centimeters.\n", + "\n", + "# Given data\n", + "\n", + "c = 1130# # Speed of light=1130 ft/s\n", + "f = 34.44*10**3# # Freq=100 Hz\n", + "In = 2.54# # 2.54 cm = 1 in\n", + "ft = 12# # 12 in = 1 ft\n", + "\n", + "l = c/f#\n", + "print 'The Lamda or Wavelength = %0.2f ft'%l\n", + "\n", + "a = l*ft#\n", + "\n", + "l1 = a*In#\n", + "print 'The Lamda or Wavelength = %0.2f cm'%l1" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter16.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter16.ipynb new file mode 100644 index 00000000..94af71c5 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter16.ipynb @@ -0,0 +1,285 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 16 : Capacitance" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 16_1 Page No. 492" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Charge Stored = 1.00e-04 Columb\n", + "i.e 100*10**-6 Columbs\n" + ] + } + ], + "source": [ + "# How much charge is stored in a 2 uF capacitor connected across a 50-V supply?\n", + "\n", + "# Given data\n", + "\n", + "V = 50# # Voltage=50 Volts\n", + "C = 2*10**-6# # Capacitor=2 uFarad\n", + "\n", + "Q = C*V#\n", + "print 'The Charge Stored = %0.2e Coulomb'%Q\n", + "print 'i.e 100*10**-6 Coulombs'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 16_2 Page No. 492" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Charge Stored = 2.00e-03 Columb\n", + "i.e 2000*10**-6 Columbs\n" + ] + } + ], + "source": [ + "# How much charge is stored in a 40 uF capacitor connected across a 50-V supply?\n", + "\n", + "# Given data\n", + "\n", + "V = 50# # Voltage=50 Volts\n", + "C = 40*10**-6# # Capacitor=2 uFarad\n", + "\n", + "Q = C*V#\n", + "print 'The Charge Stored = %0.2e Coulomb'%Q\n", + "print 'i.e 2000*10**-6 Coulombs'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 16_3 Page No. 493" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Charge Stored = 4.00e-05 Columb\n", + "i.e 40*10**-6 Columbs OR 40 uColumb\n" + ] + } + ], + "source": [ + "# A constant current of 2 uA charges a capacitor for 20 s. How much charge is stored? Remember I=Q/t or Q=I*t.\n", + "\n", + "# Given data\n", + "\n", + "I = 2*10**-6# # Current=2 uAmps\n", + "t = 20# # Time=20 Sec\n", + "\n", + "Q = I*t\n", + "print 'The Charge Stored = %0.2e Coulomb'%Q\n", + "print 'i.e 40*10**-6 Coulombs OR 40 uCoulomb'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 16_4 Page No. 494" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Capacitance = 2.00e-06 Farad\n", + "i.e 2 uF\n" + ] + } + ], + "source": [ + "# The voltage across the charged capacitor is 20 V. Calculate C.\n", + "\n", + "#Given data\n", + "\n", + "V = 20# # Voltage=20 Volts\n", + "Q = 40*10**-6# # Charge=40 uCoulomb\n", + "\n", + "C = Q/V\n", + "print 'The Capacitance = %0.2e Farad'%C\n", + "print 'i.e 2 uF'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 16_5 Page No. 495" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage across Capacitor = 500.00 Volts\n" + ] + } + ], + "source": [ + "# A constant current of 5 mA charges a 10 uF capacitor for 1 s. How much is the voltage across the capacitor?\n", + "\n", + "# Given data\n", + "\n", + "I = 5*10**-3# # Current=5 mAmps\n", + "t = 1# # Time=1 Sec\n", + "C = 10*10**-6# # Cap=10 uFarad\n", + "\n", + "Q = I*t#\n", + "\n", + "V = Q/C#\n", + "print 'The Voltage across Capacitor = %0.2f Volts'%V" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 16_6 Page No. 496" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Capacitance = 1.77e-09 Farad\n", + "i.e 1700*10**-12 F OR 1770 pF\n" + ] + } + ], + "source": [ + "# Calculate C for two plates, each with an area 2 sqm, separated by 1 cm with a dielectric of air.\n", + "\n", + "# Given data\n", + "\n", + "c = 8.85*10**-12# # Constant=8.85 p\n", + "A = 2# # Area=2 sqm\n", + "d = 1*10**-2# # Distance=1 cm\n", + "K = 1 # Permeability=1\n", + "\n", + "C = K*c*(A/d)#\n", + "print 'The Capacitance = %0.2e Farad'%C\n", + "print 'i.e 1700*10**-12 F OR 1770 pF'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 16_11 Page No. 514" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Energy Stored = 0.23 Joules\n" + ] + } + ], + "source": [ + "# The high-voltage circuit for a color picture tube can have 30 kV across 500 pF of C . Calculate the stored energy.\n", + "\n", + "# Given data\n", + "\n", + "V = 30*10**3# # Voltage=30 kVolts\n", + "C = 500*10**-12# # Cap=500 pFarad\n", + "\n", + "E = 0.5*C*V*V\n", + "print 'The Energy Stored = %0.2f Joules'%E" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter17.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter17.ipynb new file mode 100644 index 00000000..a7a1d312 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter17.ipynb @@ -0,0 +1,307 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter17 : Capacitive Reactance" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 17_1 Page No. 530" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Capacitive Reactance = 1136.82 Ohms\n", + "approx 1140 Ohms\n", + "The Capacitive Reactance = 113.68 Ohms\n", + "approx 114 Ohms\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# How much is Xc for (a) 0.1 u\u0003F of C at 1400 Hz? (b) 1 u\u0003F of C at the same frequency?\n", + "\n", + "# Given data\n", + "\n", + "f = 1400# # Frequency=1400 Hz\n", + "C1 = 0.1*10**-6# # Cap1=0.1 uF\n", + "C2 = 1*10**-6# # Cap2=1 uF\n", + "\n", + "Xc1 = 1./(2.*pi*f*C1)#\n", + "print 'The Capacitive Reactance = %0.2f Ohms'%Xc1\n", + "print 'approx 1140 Ohms'\n", + "\n", + "Xc2 = 1./(2.*pi*f*C2)#\n", + "print 'The Capacitive Reactance = %0.2f Ohms'%Xc2\n", + "print 'approx 114 Ohms'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 17_2 Page No. 530" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Capacitive Reactance = 3386.28 Ohms\n", + "approx 3388 Ohms\n", + "The Capacitive Reactance = 338.63 Ohms\n" + ] + } + ], + "source": [ + "from math import pi\n", + "#How much is the Xc of a 47-pF value of C at (a) 1 MHz? (b) 10 MHz?\n", + "\n", + "# Given data\n", + "\n", + "f1 = 1*10**6# # Frequency1=1 MHz\n", + "f2 = 10*10**6# # Frequency2=10 MHz\n", + "C = 47*10**-12# # Cap=47 pF\n", + "\n", + "# For 1 MHz\n", + "\n", + "Xc1 = 1./(2.*pi*f1*C)#\n", + "print 'The Capacitive Reactance = %0.2f Ohms'%Xc1\n", + "print 'approx 3388 Ohms'\n", + "\n", + "# For 10 MHz\n", + "\n", + "Xc2 = 1./(2.*pi*f2*C)#\n", + "print 'The Capacitive Reactance = %0.2f Ohms'%Xc2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 17_3 Page No. 532" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Capacitance = 4.68e-10 Farads\n", + "approx 468 pF\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# What C is needed for Xc of 100 Ohms\u0003 at 3.4 MHz?\n", + "\n", + "# Given data\n", + "\n", + "f = 3.4*10**6# # Frequency=3.4 MHz\n", + "Xc = 100# # Capacitive Reactance=100 Ohms\n", + "\n", + "C = 1./(2.*pi*f*Xc)#\n", + "print 'The Capacitance = %0.2e Farads'%C\n", + "print 'approx 468 pF'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 17_4 Page No. 533" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Frequency = 159.15 Hertz\n", + "approx 159 Hz\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# At what frequency will a 10 uF capacitor have Xc equal to 100 Ohms\u0003?\n", + "\n", + "# Given data\n", + "\n", + "Xc = 100# # Capacitive Reactance=100 Ohms\n", + "C = 10*10**-6# # Cap=10 uF\n", + "\n", + "f = 1./(2.*pi*C*Xc)#\n", + "print 'The Frequency = %0.2f Hertz'%f\n", + "print 'approx 159 Hz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 17_5 Page No. 534" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Instantaneous Value of Charging Current ic produced = 3.00e-04 Amps\n", + "i.e 300 uAmps\n" + ] + } + ], + "source": [ + "# Calculate the instantaneous value of charging current ic produced by a 6 u\u0003F C when its potential difference is increased by 50 V in 1 s.\n", + "\n", + "# Given data\n", + "\n", + "C = 6*10**-6# # Cap=6 uF\n", + "dv = 50.# # differential voltage increased by 50 Volts\n", + "dt = 1.# # differectial time is 1 sec\n", + "\n", + "ic = C*(dv/dt)#\n", + "print 'The Instantaneous Value of Charging Current ic produced = %0.2e Amps'%ic\n", + "print 'i.e 300 uAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 17_6 Page No. 535" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Instantaneous Value of Discharging Current ic produced = -3.00e-04 Amps\n", + "i.e -300 uAmps\n" + ] + } + ], + "source": [ + "# Calculate the instantaneous value of charging current ic produced by a 6 u\u0003F C when its potential difference is decreased by 50 V in 1 s.\n", + "\n", + "# Given data\n", + "\n", + "C = 6*10**-6# # Cap=6 uF\n", + "dv = -50.# # differential voltage decreased by 50 Volts\n", + "dt = 1.# # differectial time is 1 sec\n", + "\n", + "ic = C*(dv/dt)#\n", + "print 'The Instantaneous Value of Discharging Current ic produced = %0.2e Amps'%ic\n", + "print 'i.e -300 uAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 17_7 Page No. 536" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Instantaneous Value of ic produced = 1.25e-02 Amps\n", + "12500 uAmps or 12.5 mAmps\n" + ] + } + ], + "source": [ + "# Calculate ic produced by a 250-pF capacitor for a change of 50 V in 1 u\u0002s.\n", + "\n", + "# Given data\n", + "\n", + "C = 250*10**-12# # Cap=250 pF\n", + "dv = 50.# # differential voltage increased by 50 Volts\n", + "dt = 1.*10**-6# # differectial time is 1 usec\n", + "\n", + "ic = C*(dv/dt)#\n", + "print 'The Instantaneous Value of ic produced = %0.2e Amps'%ic\n", + "print '12500 uAmps or 12.5 mAmps'" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter18.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter18.ipynb new file mode 100644 index 00000000..55e5731f --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter18.ipynb @@ -0,0 +1,144 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 18 : Capacitive Circuits" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 18_1 Page No. 545" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Zt = 50.00 Ohms\n", + "I = 2.00 Ampers\n", + "Voltage Across Resistor = 60 Volts\n", + "Voltage Across Capacitive Reactance = 80.00 Volts\n", + "Theta z =-53.13 degree\n", + "Sum of Voltage Drop is Equal to Applied Voltage of 100V = 100.00 Volts\n" + ] + } + ], + "source": [ + "from math import sqrt,pi,atan\n", + "# If a R=30ohms and Xc=40ohms are in series with 100V applied, find the following: Zt, I, Vr, Vc and Theta z. What is the phase angle between Vc and Vr with respect to I? Prove that the sum of the series voltage drop equals the applied voltage Vt\n", + "\n", + "# Given data\n", + "\n", + "R = 30.# # Resistance=30 Ohms\n", + "Xc = 40.# # Capacitive Reactance=40 Ohms\n", + "Vt = 100.# # Applied Voltage=100 Volts\n", + "\n", + "R1 = R*R#\n", + "Xc1 = Xc*Xc#\n", + "\n", + "Zt = sqrt(R1+Xc1)#\n", + "print 'Zt = %0.2f Ohms'%Zt\n", + "\n", + "I = (Vt/Zt)#\n", + "print 'I = %0.2f Ampers'%I\n", + "\n", + "Vr = I*R#\n", + "print 'Voltage Across Resistor = %02.f Volts'%Vr\n", + "\n", + "Vc = I*Xc#\n", + "print 'Voltage Across Capacitive Reactance = %0.2f Volts'%Vc\n", + "\n", + "Oz = atan(-(Xc/R))*180/pi\n", + "print 'Theta z =%0.2f degree'%Oz\n", + "\n", + "#Prove that the sum of the series voltage drop equals the applied voltage Vt\n", + "\n", + "Vt = sqrt((Vr*Vr)+(Vc*Vc))#\n", + "print 'Sum of Voltage Drop is Equal to Applied Voltage of 100V = %0.2f Volts'%Vt" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 18_2 Page No. 548" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Total Current = 0.05 Amps\n", + "i.e 50 mAmps\n", + "The Equivqlent Impedence = 1440.00 Ohms\n", + "i.e 1.44 kohms\n", + "The Value of Theta I = 53.13 degrees\n" + ] + } + ], + "source": [ + "from math import sqrt,pi,atan\n", + "# A 30-mA Ir is in parallel with another branch current of 40 mA for Ic. The applied voltage Va is 72 V. Calculate It, Zeq and Theta \u0002I.\n", + "\n", + "# Given data\n", + "\n", + "Ir = 30.*10**-3# # Current Ir=30 mA\n", + "Ic = 40.*10**-3# # Current Ic=40 mA\n", + "Va = 72.# # Applied Voltage=72 Volts\n", + "\n", + "A = Ir*Ir#\n", + "B = Ic*Ic#\n", + "\n", + "It = sqrt(A+B)#\n", + "print 'The Total Current = %0.2f Amps'%It\n", + "print 'i.e 50 mAmps'\n", + "\n", + "Zeq = Va/It#\n", + "print 'The Equivqlent Impedence = %0.2f Ohms'%Zeq\n", + "print 'i.e 1.44 kohms'\n", + "\n", + "Oi = atan(Ic/Ir)*180/pi\n", + "print 'The Value of Theta I = %0.2f degrees'%Oi" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter19.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter19.ipynb new file mode 100644 index 00000000..5e3f9929 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter19.ipynb @@ -0,0 +1,830 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 19 : Inductance" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_1 Page No. 580" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The di/dt Rate of Current change = 4 A/s\n" + ] + } + ], + "source": [ + "# The current in an inductor changes from 12 to 16 A in 1s. How much is the di/\u0002dt rate of current change in amperes per second?\n", + "\n", + "# Given data\n", + "\n", + "di = 4# # Differential current=16-12=4 Amps\n", + "dt = 1# # Differential time=1 sec\n", + "\n", + "A = di/dt#\n", + "print 'The di/dt Rate of Current change = %0.f A/s'%A" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_2 Page No. 580" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The di/dt Rate of Current change = 2.50e+04 A/s\n" + ] + } + ], + "source": [ + "# The current in an inductor changes by 50 mA in 2 us. How much is the di/\u0002dt rate of current change in amperes per second?\n", + "\n", + "# Given data\n", + "\n", + "di = 50.*10**-3# # Differential current=50 mAmps\n", + "dt = 2.*10**-6# # Differential time=2 usec\n", + "\n", + "A = di/dt#\n", + "print 'The di/dt Rate of Current change = %0.2e A/s'%A" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_3 Page No. 581" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Inductance = 10 Henry\n" + ] + } + ], + "source": [ + "# How much is the inductance of a coil that induces 40 V when its current changes at the rate of 4 A/\u0002s?\n", + "\n", + "# Given data\n", + "\n", + "Vl = 40# # Induced voltage=40 Volts\n", + "R = 4 # Current changing rate=di/dt=4 A/s\n", + "\n", + "L = Vl/R#\n", + "print 'The Value of Inductance = %0.f Henry'%L" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_4 Page No. 582" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Inductance = 4.00e-02 Henry\n", + "OR 40 mH\n" + ] + } + ], + "source": [ + "# How much is the inductance of a coil that induces 1000 V when its current changes at the rate of 50 mA in 2\u0002us?\n", + "\n", + "# Given data\n", + "\n", + "Vl = 1000# # Induced voltage=1000 Volts\n", + "di = 50*10**-3# # differential current=50 mAmps\n", + "dt = 2*10**-6# # differential time=2 usec\n", + "\n", + "A = di/dt#\n", + "\n", + "L = Vl/A#\n", + "print 'The Value of Inductance = %0.2e Henry'%L\n", + "print 'OR 40 mH'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_5 Page No. 582" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Self-Induced Voltage = 48 Volts\n" + ] + } + ], + "source": [ + "# How much is the self-induced voltage across a 4-H inductance produced by a current change of 12 A/\u0002s?\n", + "\n", + "# Given data\n", + "\n", + "L = 4# # Inductor=4 H\n", + "R = 12# # current change=di/dt=12 A/s\n", + "\n", + "Vl = L*R#\n", + "print 'The Value of Self-Induced Voltage = %0.f Volts'%Vl" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_6 Page No. 583" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Self-Induced Voltage = 1.00e+04 Volts\n", + "OR 10 kVolts\n" + ] + } + ], + "source": [ + "# The current through a 200-mH L changes from 0 to 100 mA in 2 u\u0002s. How much is Vl ?\n", + "\n", + "# Given data\n", + "\n", + "L = 200*10**-3# # Inductor=200 mH\n", + "di = 100*10**-3# # differential current=100 mAmps\n", + "dt = 2*10**-6# # differectial time=2 usec\n", + "\n", + "A = di/dt#\n", + "\n", + "Vl = L*A#\n", + "print 'The Value of Self-Induced Voltage = %0.2e Volts'%Vl\n", + "print 'OR 10 kVolts'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_7 Page No. 583" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Coefficient of Coupling k between Coil L1 and Coil L2 = 0.75\n" + ] + } + ], + "source": [ + "# A coil L1 produces 80 u\u0002Wb of magnetic flux. Of this total flux, 60 u\u0002Wb arelinked with L2. How much is k between L1 and L2?\n", + "\n", + "# Given data\n", + "\n", + "lf1 = 80.*10**-6# # Magnetic flux of coil L1=80 uWb\n", + "lf2 = 60.*10**-6# # Magnetic flux of coil L2=60 uWb\n", + "\n", + "k = lf2/lf1#\n", + "print 'The Coefficient of Coupling k between Coil L1 and Coil L2 = %0.2f'%k" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_8 Page No. 584" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Coefficient of Coupling k between Coil L1 and Coil L2 = 1\n" + ] + } + ], + "source": [ + "# A 10-H inductance L1 on an iron core produces 4 Wb of magnetic flux. Another coil L2 is on the same core. How much is k between L1 and L2?\n", + "\n", + "# Given data\n", + "\n", + "lf1 = 4# # Magnetic flux of coil L1=4 Wb\n", + "lf2 = 4# # Magnetic flux of coil L2=4 Wb\n", + "\n", + "k = lf2/lf1#\n", + "print 'The Coefficient of Coupling k between Coil L1 and Coil L2 = %0.f'%k" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_9 Page No. 585" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The mutual inductance = 0.08 Henry\n", + "i.e 80*10**-3 H OR 80 mH\n" + ] + } + ], + "source": [ + "\n", + "from math import sqrt\n", + "# Two 400-mH coils L1 and L2 have a coefficient of coupling k equal to 0.2. Calculate Lm.\n", + "\n", + "# Given data\n", + "\n", + "L1 = 400*10**-3# # L1=400 mH\n", + "L2 = 400*10**-3# # L2=400 mH\n", + "k = 0.2# # Coupling coefficient=0.2\n", + "\n", + "Lm = k*sqrt(L1*L2)#\n", + "print 'The mutual inductance = %0.2f Henry'%Lm\n", + "print 'i.e 80*10**-3 H OR 80 mH'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_10 Page No. 586" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Coupling Coefficient k = 0.10\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "# If the two coils had a mutual inductance LM of 40 mH, how much would k be?\n", + "\n", + "# Given data\n", + "\n", + "L1 = 400.*10**-3# # Coil Inductance 1=400 mH\n", + "L2 = 400.*10**-3# # Coil Inductance 2=400 mH\n", + "Lm = 40.*10**-3# # Mutual inductance=40 mH\n", + "\n", + "lt = sqrt(L1*L2)#\n", + "\n", + "k = Lm/lt#\n", + "print'The Coupling Coefficient k = %0.2f'% k" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_11 Page No. 590" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Turns Ratio = 0.1667\n", + "OR 1:6\n", + "The Secondary Voltage = 720.00 Volts\n" + ] + } + ], + "source": [ + "# A power transformer has 100 turns for Np and 600 turns for Ns. What is the turns ratio? How much is the secondary voltage Vs if the primary voltage Vp is 120 V?\n", + "\n", + "# Given data\n", + "\n", + "np = 100.# # Turns in primary coil=100\n", + "ns = 600.# # Turns in secondary coil=600\n", + "vp = 120.# # Primary voltage=120 Volts\n", + "\n", + "Tr = np/ns#\n", + "print 'The Turns Ratio = %0.4f'%Tr\n", + "print 'OR 1:6'\n", + "\n", + "vs = vp*(ns/np)#\n", + "print 'The Secondary Voltage = %0.2f Volts'%vs" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_12 Page No. 590" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Turns Ratio 20:1 or 20.00\n", + "The Secondary Voltage = 6.00 Volts\n" + ] + } + ], + "source": [ + "# A power transformer has 100 turns for Np and 5 turns for Ns. What is the turns ratio? How much is the secondary voltage Vs with a primary voltage of 120 V?\n", + "\n", + "# Given data\n", + "\n", + "np = 100.# # Turns in primary coil=100\n", + "ns = 5.# # Turns in secondary coil=5\n", + "vp = 120.# # Primary voltage=120 Volts\n", + "\n", + "Tr = np/ns#\n", + "print 'The Turns Ratio 20:1 or %0.2f'%Tr\n", + "\n", + "vs = vp*(ns/np)#\n", + "print 'The Secondary Voltage = %0.2f Volts'%vs" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_13 Page No. 591" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Secondary Current = 0.10 Amps\n", + "The Primary Current = 0.60 Amps\n" + ] + } + ], + "source": [ + "# A transformer with a 1:6 turns ratio has 720 V across 7200 Ohms\u0006 in the secondary. (a) How much is Is? (b) Calculate the value of Ip.\n", + "\n", + "# Given data\n", + "\n", + "vs = 720.# # Secondary voltage=720 Volts\n", + "Rl = 7200.# # Secondary load=7200 Ohms\n", + "tr = 1./6# # Turns ratio=1:6\n", + "\n", + "Is = vs/Rl#\n", + "print 'The Secondary Current = %0.2f Amps'%Is\n", + "\n", + "Ip = Is/tr#\n", + "print 'The Primary Current = %0.2f Amps'%Ip" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_14 Page No. 591" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Secondary Current = 10.00 Amps\n", + "The Primary Current = 0.50 Amps\n" + ] + } + ], + "source": [ + "# A transformer with a 20:1 voltage step-down ratio has 6 V across 0.6 \u0006 in the secondary. (a) How much is Is? (b) How much is Ip?\n", + "\n", + "# Given data\n", + "\n", + "vs = 6.# # Secondary voltage=6 Volts\n", + "Rl = 0.6# # Secondary load=0.6 Ohms\n", + "tr = 20./1# # Turns ratio=20:1\n", + "\n", + "Is = vs/Rl#\n", + "print 'The Secondary Current = %0.2f Amps'%Is\n", + "\n", + "Ip = Is/tr#\n", + "print 'The Primary Current = %0.2f Amps'%Ip" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_15 Page No. 593" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Primary current = 0.42 Amps\n", + "OR 420 mAmps\n" + ] + } + ], + "source": [ + "# Calculate the primary current I P if the secondary current Is equals its rated value of 2 A.\n", + "\n", + "# Given data\n", + "\n", + "vs = 25.2# # Secondary voltage=25.2 Volts\n", + "vp = 120.# # Primary voltage=120 Volts\n", + "Is = 2.# # Secondary current=2 Amps\n", + "\n", + "Ip = Is*(vs/vp)#\n", + "print 'The Primary current = %0.2f Amps'%Ip\n", + "print 'OR 420 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_16 Page No. 593" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Primary Impedence = 128.00 Ohms by Method 1\n", + "Primary Impedence = 128.00 Ohms by Method 2\n" + ] + } + ], + "source": [ + "# Determine the Primary Impedence Zo\n", + "\n", + "# Method 1\n", + "# Given data\n", + "\n", + "Vp = 32.# # Primary Voltage = 32 Volts\n", + "Rl = 8.# # Load Resistance = 8 Ohms\n", + "TR = 4.# # Turns Ratio Np/Ns = 4/1\n", + "\n", + "Vs = Vp/TR#\n", + "\n", + "Is = Vs/Rl#\n", + "\n", + "Ip = ((Vs/Vp)*Is)#\n", + "\n", + "Zp = Vp/Ip#\n", + "print 'Primary Impedence = %0.2f Ohms by Method 1'%Zp\n", + "\n", + "# Method 2\n", + "\n", + "Zp = TR*TR*Rl#\n", + "print 'Primary Impedence = %0.2f Ohms by Method 2'%Zp" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_17 Page No. 594" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Turns ratio Np/Ns = 0.50\n", + "OR 1/2\n", + "The Turns ratio Np/Ns is 1.414/1 or 1.41\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "# Calculate the turns ratio Np/Ns that will produce a reflected primary impedance Zp of (a) 75 Ohms# (b) 600 Ohms.\n", + "\n", + "# Given data\n", + "\n", + "Zs = 300.# # Secondary impedence=300 Ohms\n", + "Zp1 = 75.# # Primary impedence=75 Ohms\n", + "Zp2 = 600.# # Primary impedence=600 Ohms\n", + "\n", + "tra = sqrt (Zp1/Zs)#\n", + "print 'The Turns ratio Np/Ns = %0.2f'%tra\n", + "print 'OR 1/2'\n", + "\n", + "trb = sqrt (Zp2/Zs)#\n", + "print 'The Turns ratio Np/Ns is 1.414/1 or %0.2f'%trb" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_18 Page No. 595" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Total Inductance = 1.50e-02 Henry\n", + "i.e 15 mH\n" + ] + } + ], + "source": [ + "# Inductance L1 is 5 mH and L2 is 10 mH. How much is Lt?\n", + "\n", + "# Given data\n", + "\n", + "l1 = 5*10**-3# # Inductor 1=5 mH\n", + "l2 = 10*10**-3# # Inductor 2=10 mH\n", + "\n", + "Lt = l1+l2#\n", + "print 'The Total Inductance = %0.2e Henry'%Lt\n", + "print 'i.e 15 mH'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_19 Page No. 597" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Equivalent Inductance = 0.04 Henry\n", + "i.e 4 mH\n" + ] + } + ], + "source": [ + "# Inductances L1 and L2 are each 8 mH. How much is Leq?\n", + "\n", + "# Given data\n", + "\n", + "l1 = 8*10**-3# # Inductor 1=8 mH\n", + "l2 = 8*10**-3# # Inductor 2=8 mH\n", + "\n", + "a = 1./l1#\n", + "b = 1./l2#\n", + "\n", + "Leq = 10/(a+b)#\n", + "print 'The Equivalent Inductance = %0.2f Henry'%Leq\n", + "print 'i.e 4 mH'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_20 Page No. 598" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Mutual Inductance = 2.50e-05 Henry\n", + "i.e 25 uH\n", + "The Coupling coefficient k = 0.10\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "# Two series coils, each with an L of 250 u\u0002H, have a total inductance of 550 u\u0002H connected series-aiding and 450 uH series-opposing. (a) How much is the mutual inductance Lm between the two coils? (b) How much is the coupling coefficient k?\n", + "\n", + "# Given data\n", + "\n", + "l1 = 250*10**-6# # Coil Inductance 1=250 uH\n", + "l2 = 250*10**-6# # Coil Inductance 2=250 uH\n", + "Lts = 550*10**-6# # Inductance series-aiding=550 uH\n", + "Lto = 450*10**-6# # Inductance series-opposing=450 uH\n", + "\n", + "Lm = (Lts-Lto)/4.\n", + "print 'The Mutual Inductance = %0.2e Henry'%Lm\n", + "print 'i.e 25 uH'\n", + "\n", + "lt = sqrt(l1*l2)#\n", + "\n", + "k = Lm/lt#\n", + "print 'The Coupling coefficient k = %0.2f'%k" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 19_21 Page No. 608" + ] + }, + { + "cell_type": "code", + "execution_count": 36, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Energy Stored in the Magnetic Field = 0.29 Joules\n" + ] + } + ], + "source": [ + "# A current of 1.2 A flows in a coil with an inductance of 0.4 H. How much energy is stored in the magnetic field?\n", + "\n", + "# Given data\n", + "\n", + "l1 = 0.4# # Coil Inductance 1=0.4 H\n", + "I = 1.2# # Current=1.2 Amps\n", + "\n", + "E = (l1*I*I)/2#\n", + "print 'The Energy Stored in the Magnetic Field = %0.2f Joules'%E" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter20.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter20.ipynb new file mode 100644 index 00000000..01adabd9 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter20.ipynb @@ -0,0 +1,296 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 20 : Inductive Reactance" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 20_1 Page No. 625" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Inductive Reactance = 15709.22 Ohms\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# How much is Xl of a 6-mH L at 41.67 kHz?\n", + "\n", + "# Given data\n", + "\n", + "f = 41.67*10**3# # Frequency=41.67 kHz\n", + "L = 6.*10**-3# # Inductor=6 mH\n", + "\n", + "Xl = 20*pi*f*L#\n", + "print 'The Inductive Reactance = %0.2f Ohms'%Xl" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 20_2 Page No. 627" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Inductive Reactance = 37680 Ohms\n", + "The Inductive Reactance = 1884 Ohms\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# Calculate the Xl of (a) a 10-H L at 60 Hz and (b) a 5-H L at 60 Hz.\n", + "\n", + "# Given data\n", + "\n", + "f = 60.# # Frequency=60 Hz\n", + "L1 = 10.# # Inductor 1=10 H\n", + "L2 = 5.# # Inductor 2=5 H\n", + "pi = 3.14\n", + "\n", + "Xl1 = 20*pi*f*L1#\n", + "print 'The Inductive Reactance = %0.f Ohms'%Xl1\n", + "\n", + "Xl2 = 2*pi*f*L2#\n", + "print 'The Inductive Reactance = %0.f Ohms'%Xl2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 20_3 Page No. 629" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Inductive Reactance = 1570 Ohms\n", + "The Inductive Reactance = 15700 Ohms\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# Calculate the Xl of a 250-u\u0003H coil at (a) 1 MHz and (b) 10 MHz.\n", + "\n", + "# Given data\n", + "\n", + "f1 = 1.*10**6# # Frequency1=1 MHz\n", + "f2 = 10.0*10**6# # Frequency2=10 MHz\n", + "L = 250.0*10**-6# # Inductor=250 uH\n", + "pi = 3.14#\n", + "\n", + "# For 1 Mhz\n", + "\n", + "Xl1 = 2*pi*f1*L#\n", + "print 'The Inductive Reactance = %0.f Ohms'%Xl1\n", + "\n", + "# For 10 Mhz\n", + "\n", + "Xl2 = 2*pi*f2*L#\n", + "print 'The Inductive Reactance = %0.f Ohms'%Xl2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 20_4 Page No. 631" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Inductive Reactance = 6280 Ohms\n" + ] + } + ], + "source": [ + "# A coil with negligible resistance has 62.8 V across it with 0.01 A of current. How much is Xl?\n", + "\n", + "# Given data\n", + "\n", + "Vl = 62.8# # Voltage across coil=62.8 Volts\n", + "Il = 0.01# # Current in coil=0.01 Amps\n", + "\n", + "Xl = Vl/Il#\n", + "print 'The Inductive Reactance = %0.f Ohms'%Xl" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 20_5 Page No. 632" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of Inductor = 1.00 Henry\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# Calculate L of the coil when the frequency is 1000 Hz.\n", + "\n", + "# Given data\n", + "\n", + "Xl = 6280.# # Inductive reactance=6280 Ohms\n", + "f = 1000.# # Frequency=1000 Hz\n", + "pi = 3.14#\n", + "\n", + "L = Xl/(2.*pi*f)#\n", + "print'The value of Inductor = %0.2f Henry'% L" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 20_6 Page No. 633" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of Inductor = 2.08e-04 Henry\n", + "i.e Approx 208.8*10**-6 OR 208.8 uH\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# Calculate L of a coil that has 15,700 Ohms\u0003 of Xl at 12 MHz.\n", + "\n", + "# Given data\n", + "\n", + "Xl = 15700.# # Inductive reactance=15700 Ohms\n", + "f = 12.0*10**6# # Frequency=12 MHz\n", + "pi = 3.14#\n", + "\n", + "L = Xl/(2*pi*f)#\n", + "print'The value of Inductor = %0.2e Henry'% L\n", + "print 'i.e Approx 208.8*10**-6 OR 208.8 uH'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 20_7 Page No. 634" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Frequency = 159.15 Hertz\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# At what frequency will an inductance of 1 H have a reactance of 1000 \u0003?\n", + "\n", + "# Given data\n", + "\n", + "Xl = 1000.# # Inductive reactance=1000 Ohms\n", + "L = 1.# # Inductor=1 H\n", + "\n", + "f = Xl/(2.*pi*L)#\n", + "print 'The Frequency = %0.2f Hertz'%f" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter21.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter21.ipynb new file mode 100644 index 00000000..59eb0bda --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter21.ipynb @@ -0,0 +1,209 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 21 : Inductive Circuits" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 21_1 Page No. 650" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Zt = 50.00 ohms\n", + "I = 2.00 Ampers\n", + "Vr = 60.00 Volts\n", + "Vl = 80.00 Volts\n", + "Theta z = 53.13 degree\n", + "Sum of Voltage Drop = 100 which is Equal to Applied Voltage 100V \n" + ] + } + ], + "source": [ + "from math import sqrt,atan,pi\n", + "# If a R=30 ohms and Xl=40 ohms are in series with 100V applied, find the following: Zt, I, Vr, Vl and Theta z. What is the phase angle between Vl and Vr with respect to I? Prove that the sum of the series voltage drop equals the applied voltage Vt\n", + "\n", + "# Given data\n", + "\n", + "R = 30.# # Resistance=30 Ohms\n", + "Xl = 40.# # Inductive reactance=40 Ohms\n", + "Vt = 100.# # Applied voltage=100 Volts\n", + "\n", + "\n", + "R1 = R*R#\n", + "Xl1 = Xl*Xl#\n", + "\n", + "Zt = sqrt(R1+Xl1)#\n", + "print 'Zt = %0.2f ohms'%Zt\n", + "\n", + "I = (Vt/Zt)#\n", + "print 'I = %0.2f Ampers'%I\n", + "\n", + "Vr = I*R#\n", + "print 'Vr = %0.2f Volts'%Vr\n", + "\n", + "Vl = I*Xl#\n", + "print 'Vl = %0.2f Volts'%Vl\n", + "\n", + "Oz = atan(Xl/R)*180/pi\n", + "print 'Theta z = %0.2f degree'%Oz\n", + "\n", + "#Prove that the sum of the series voltage drop equals the applied voltage Vt\n", + "\n", + "Vt = sqrt((Vr*Vr)+(Vl*Vl))#\n", + "print 'Sum of Voltage Drop = %0.f which is Equal to Applied Voltage 100V '%Vt" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 21_2 Page No. 654" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Total Impedence = 268.33 Ohms\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "# What is the total Z of a 600-Ohms\u0005 R in parallel with a 300 Ohms\u0005 Xl? Assume 600 V for the applied voltage.\n", + "\n", + "# Given data\n", + "\n", + "R = 600.# # Resistance=600 Ohms\n", + "Xl = 300.# # Inductive reactance=300 Ohms\n", + "V = 600.# # Applied voltage=600 Volts\n", + "\n", + "Ir = V/R#\n", + "Il = V/Xl#\n", + "A = Ir*Ir#\n", + "B = Il*Il#\n", + "It = sqrt(A+B)#\n", + "\n", + "Zeq = V/It#\n", + "print 'The Total Impedence = %0.2f Ohms'%Zeq" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 21_3 Page No. 656" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Q of Coil = 350.00\n" + ] + } + ], + "source": [ + "# An air-core coil has an Xl of 700 Ohms\u0005 and an Re of 2 Ohms\u0005. Calculate the value of Q for this coil.\n", + "\n", + "# Given data\n", + "\n", + "Xl = 700# # Inductive reactance=700 Ohms\n", + "Re = 2# # AC effective resistance=2 Ohms\n", + "\n", + "Q = Xl/Re#\n", + "print 'The Q of Coil = %0.2f'%Q" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 21_4 Page No. 658" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The AC Effective Resistance = 1.57 Ohms\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# A 200 uH coil has a Q of 40 at 0.5 MHz. Find Re.\n", + "\n", + "# Given data\n", + "\n", + "L = 200.*10**-6# # L of coil=200 uHenry\n", + "Q = 400# # Q=40\n", + "f = 0.5*10**6# # Frequency=0.5 MHz\n", + "pi = 3.14#\n", + "\n", + "Xl = 2*pi*L*f#\n", + "\n", + "Re = Xl/Q#\n", + "print 'The AC Effective Resistance = %0.2f Ohms'%Re" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter22.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter22.ipynb new file mode 100644 index 00000000..cd340733 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter22.ipynb @@ -0,0 +1,418 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 22 : RC and L/R Time Constants" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 22_1 Page No. 674" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Time Constant = 0.20 Seconds\n" + ] + } + ], + "source": [ + "# What is the time constant of a 20-H coil having 100 Ohms\u0002 of series resistance?\n", + "\n", + "# Given data\n", + "\n", + "L = 20.# # Inductor=20 Henry\n", + "R = 100.# # Resistor=100 Ohms\n", + "\n", + "T = L/R#\n", + "print 'The Time Constant = %0.2f Seconds'%T" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 22_2 Page No. 674" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Since 0.2 sec is one time constant, I is 63% of 100 mA\n", + "The current at 0.2 sec time constant = 0.06 A\n", + "After 1 sec the current reaches its steady state value of 100 mAmps \n" + ] + } + ], + "source": [ + "# An applied dc voltage of 10 V will produce a steady-state current of 100 mA in the 100-Ohms coil. How much is the current after 0.2 s? After 1 s?\n", + "\n", + "# Given data\n", + "\n", + "L = 20.# # Inductor=20 Henry\n", + "R = 100.# # Resistor=100 Ohms\n", + "I = 100.*10**-3# # Steady-state current=100 mAmps\n", + "\n", + "print 'Since 0.2 sec is one time constant, I is 63% of 100 mA'\n", + "I1 = 0.63*I#\n", + "print 'The current at 0.2 sec time constant = %0.2f A'%I1\n", + "\n", + "print 'After 1 sec the current reaches its steady state value of 100 mAmps '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 22_3 Page No. 675" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Time Constant = 2.00e-05 Seconds\n", + "i.e 20 us\n" + ] + } + ], + "source": [ + "# If a 1-M Ohms\u0002 R is added in series with the coil, how much will the time constant be for the higher resistance RL circuit?\n", + "\n", + "# Given data\n", + "\n", + "L = 20.# # Inductor=20 Henry\n", + "R = 1.*10**6# # Resistor=1 MOhms\n", + "\n", + "T = L/R#\n", + "print 'The Time Constant = %0.2e Seconds'%T\n", + "print 'i.e 20 us'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 22_4 Page No. 676" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Time Constant = 1.00e-02 Seconds\n" + ] + } + ], + "source": [ + "# What is the time constant of a 0.01-u\u0003F capacitor in series with a 1-M\u0002 Ohmsresistance?\n", + "\n", + "# Given data\n", + "\n", + "C = 0.01*10**-6# # Capacitor=0.01 uFarad\n", + "R = 1*10**6# # Resistor=1 MOhms\n", + "\n", + "T = C*R#\n", + "print 'The Time Constant = %0.2e Seconds'%T" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 22_5 Page No. 677" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Time Constant = 1.00e-02 Seconds\n", + "Since 0.01 sec is one time constant, the voltage across C then is 63% of 300 V,\n", + "The Capacitor voltage at 0.01 Sec = 189.00 Volts\n", + "After 5 time constants or 0.05 Sec Capacitor voltage = 300.00 volts \n", + "After 2 hours or 2 days the C will be still charged to 300 V if the supply is still connected\n" + ] + } + ], + "source": [ + "# With a dc voltage of 300 V applied, how much is the voltage across C in Example 22–4 after 0.01 s of charging? After 0.05 s? After 2 hours? After 2 days?\n", + "\n", + "# Given data\n", + "\n", + "C = 0.01*10**-6# # Capacitor=0.01 uFarad\n", + "R = 1.*10**6# # Resistor=1 MOhms\n", + "V = 300.# # Applied DC=300 Volts\n", + "\n", + "T = C*R#\n", + "print 'The Time Constant = %0.2e Seconds'%T\n", + "\n", + "print 'Since 0.01 sec is one time constant, the voltage across C then is 63% of 300 V,'\n", + "\n", + "T1 = 0.63*V#\n", + "print 'The Capacitor voltage at 0.01 Sec = %0.2f Volts'%T1\n", + "\n", + "T2 = V\n", + "print 'After 5 time constants or 0.05 Sec Capacitor voltage = %0.2f volts '%V\n", + "\n", + "print 'After 2 hours or 2 days the C will be still charged to 300 V if the supply is still connected'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 22_6 Page No. 678" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "In one time constant, C discharges to 37% of its initial voltage\n", + "The Capacitor voltage after 0.01 sec start of discharge = 1110 volts\n" + ] + } + ], + "source": [ + "# If the capacitor is allowed to charge to 300 V and then discharged, how much is the capacitor voltage 0.01 s after the start of discharge? The series resistance is the same on discharge as on charge.\n", + "\n", + "# Given data\n", + "\n", + "C = 0.01*10**-6# # Capacitor=0.01 uFarad\n", + "R = 1.*10**6# # Resistor=1 MOhms\n", + "V = 3000# # Applied DC=300 Volts\n", + "\n", + "print 'In one time constant, C discharges to 37% of its initial voltage'\n", + "\n", + "V1 = 0.37*V#\n", + "print 'The Capacitor voltage after 0.01 sec start of discharge = %0.f volts'%V1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 22_7 Page No. 680" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "In one time constant, C discharges to 37% of its initial voltage\n", + "The Capacitor voltage after 0.01 sec start of discharge = 74 volts\n" + ] + } + ], + "source": [ + "# Assume the capacitor is discharging after being charged to 200 V. How much will the voltage across C be 0.01 s after the beginning of discharge? The series resistance is the same on discharge as on charge.\n", + "\n", + "# Given data\n", + "\n", + "C = 0.01*10**-6# # Capacitor=0.01 uFarad\n", + "R = 1*10**6# # Resistor=1 MOhms\n", + "V = 200# # Capacitor voltage=200 Volts\n", + "\n", + "print 'In one time constant, C discharges to 37% of its initial voltage'\n", + "\n", + "V1 = 0.37*V#\n", + "print 'The Capacitor voltage after 0.01 sec start of discharge = %0.f volts'%V1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 22_8 Page No. 681" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Time Constant = 2.00e-02 Seconds\n" + ] + } + ], + "source": [ + "# If a 1-M Ohms resistance is added in series with the capacitor 0.01-u\u0003F and resistor 1-M Ohms in, how much will the time constant be?\n", + "\n", + "# Given data\n", + "\n", + "C = 0.01*10**-6# # Capacitor=0.01 uFarad\n", + "R = 2*10**6# # Resistor= 2 MOhms \n", + "\n", + "T = C*R#\n", + "print 'The Time Constant = %0.2e Seconds'%T" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 22_9 Page No. 682" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Vr = 5.42 Volts\n" + ] + } + ], + "source": [ + "from math import log10\n", + "# An RC circuit has a time constant of 3 s. The capacitor is charged to 40 V. Then C is discharged. After 6 s of discharge, how much is Vr?\n", + "\n", + "# Given data\n", + "\n", + "RC = 3# # RC time constant=3 Sec\n", + "t = 6# # Discharge time=6 Sec\n", + "Vc = 40# # Capacitor voltage=40 Volts\n", + "\n", + "A = t/RC# # constant factor\n", + "B = log10(Vc)#\n", + "\n", + "Vr = 10**(B-(A*0.434))#\n", + "print 'The Value of Vr = %0.2f Volts'%Vr" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 22_10 Page No. 682" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Time Constant = 5.00e-04 Seconds\n", + "i.e 0.5*10**-3 Sec OR 0.5 mSec\n", + "Time required to charge Capacitor upto 24 Volts = 5.49e-04 Seconds\n", + "i.e approx 0.549*10**-3 Sec OR 0.549 mSec\n" + ] + } + ], + "source": [ + "from math import log10\n", + "# An RC circuit has an R of 10 k Ohms\u0002 and a C of 0.05 u\u0003F. The applied voltage for charging is 36 V. (a) Calculate the time constant. (b) How long will it take C to charge to 24 V?\n", + "\n", + "C = 0.05*10**-6# # Capacitor=0.05 uFarad\n", + "R = 10*10**3# # Resistor=10 kOhms\n", + "V = 36# # Applied voltage=36 Volts\n", + "v = 12# # Voltage drops from 36 to 12 Volts\n", + "A = 2.3# # Specific factor\n", + "\n", + "T = C*R#\n", + "print 'The Time Constant = %0.2e Seconds'%T\n", + "print 'i.e 0.5*10**-3 Sec OR 0.5 mSec'\n", + "\n", + "t = A*T*log10(V/v)#\n", + "print 'Time required to charge Capacitor upto 24 Volts = %0.2e Seconds'%t\n", + "print 'i.e approx 0.549*10**-3 Sec OR 0.549 mSec'" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter23.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter23.ipynb new file mode 100644 index 00000000..59551e45 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter23.ipynb @@ -0,0 +1,136 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 23 : Alternating Current Circuits" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 23_1 Page No. 715" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total Impedence Zt = 38.18 Ohms\n", + "Current I = 0.00 Ampers\n", + "i.e 1.31 mAmps\n", + "Theta z = 45.00 Degree\n" + ] + } + ], + "source": [ + "from math import sqrt,atan,pi\n", + "# A 27-Ohms R is in series with 54 Ohms\u0004 of Xl and 27 Ohms\u0004 of Xc. The applied voltage Vt is 50 mV. Calculate ZT, I, and Theta z.\n", + "\n", + "# Given data\n", + "\n", + "R = 27.# # Resistance=27 Ohms\n", + "Xl = 54.# # Inductive reactance=54 Ohms\n", + "Vt = 50.*10**-3# # Applied voltage=100 Volts\n", + "Xc = 27.# # Capacitive reactance=27 Ohms\n", + "\n", + "nXl = Xl-Xc# # Net Inductive reactance\n", + "R1 = R*R#\n", + "nXl1 = nXl*nXl#\n", + "\n", + "Zt = sqrt(R1+nXl1)#\n", + "print 'Total Impedence Zt = %0.2f Ohms'%Zt\n", + "\n", + "I = (Vt/Zt)#\n", + "print 'Current I = %0.2f Ampers'%I\n", + "print 'i.e 1.31 mAmps'\n", + "\n", + "Oz = atan(Xc/R)*180/pi\n", + "print 'Theta z = %0.2f Degree'%Oz" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 23_2 Page No. 717" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Total Current It = 0.00 Amps\n", + "i.e 2.55 mAmps\n", + "The Equivqlent Impedence Zeq = 19.64 Ohms\n", + "approx 19.61 Ohms\n", + "Theta z = -45.00 Degree\n" + ] + } + ], + "source": [ + "from math import atan,pi,sqrt\n", + "# The following branch currents are supplied from a 50-mV source: Ir=1.8 mA# Il=2.8 mA# Ic=1 mA. Calculate It, Zeq, and Theta I.\n", + "\n", + "# Given data\n", + "\n", + "Va = 50.*10**-3# # Applied voltage=50m Volts\n", + "Ir = 1.8*10**-3# # Ir=1.8 mAmps\n", + "Il = 2.8*10**-3# # Ir=2.8 mAmps\n", + "Ic = 1.*10**-3# # Ic=1 mAmps\n", + "\n", + "nI = Il-Ic# # net current\n", + "Ir1 = Ir*Ir#\n", + "nI1 = nI*nI#\n", + "\n", + "It = sqrt(Ir1+nI1)#\n", + "print 'The Total Current It = %0.2f Amps'%It\n", + "print 'i.e 2.55 mAmps'\n", + "\n", + "Zeq = Va/It#\n", + "print 'The Equivqlent Impedence Zeq = %0.2f Ohms'%Zeq\n", + "print 'approx 19.61 Ohms'\n", + "\n", + "Oz = atan(-(nI/Ir))*180/pi\n", + "print 'Theta z = %0.2f Degree'%Oz" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter25.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter25.ipynb new file mode 100644 index 00000000..fc3c2316 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter25.ipynb @@ -0,0 +1,441 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 25 : Resonance" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 25_1 Page No. 775" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resonant frequency = 12.58 Hertz\n", + "approx 12.6 Hertz\n" + ] + } + ], + "source": [ + "from math import pi,sqrt\n", + "# Calculate the resonant frequency for an 8-H inductance and a 20-u\u0003F capacitance.\n", + "\n", + "# Given data\n", + "\n", + "L = 8.# # L=8 Henry\n", + "C = 20.*10**-6# # C=20 uFarad\n", + "\n", + "fr = 1./(2.*pi*sqrt(L*C))#\n", + "print 'The resonant frequency = %0.2f Hertz'%fr\n", + "print 'approx 12.6 Hertz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 25_2 Page No. 776" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resonant frequency = 65007689.56 Hertz\n", + "i.e 65 MHz\n" + ] + } + ], + "source": [ + "# Calculate the resonant frequency for a 2-uH inductance and a 3-pF capacitance.\n", + "\n", + "# Given data\n", + "\n", + "L = 2.*10**-6# # Inductor=2 uHenry\n", + "C = 3.*10**-12# # Capacitor=3 pFarad\n", + "pi = 3.14#\n", + "\n", + "fr = 1./(2.*pi*sqrt(L*C))#\n", + "print 'The resonant frequency = %0.2f Hertz'%fr\n", + "print 'i.e 65 MHz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 25_3 Page No. 778" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of Capacitor = 1.06e-10 Farads\n", + "i.e 106 pF\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# What value of C resonates with a 239-u\u0003H L at 1000 kHz?\n", + "\n", + "# Given data\n", + "\n", + "L = 239.*10**-6# # Inductor=239 uHenry\n", + "fr = 1000.*10**3# # Resonant frequency=1000 kHertz\n", + "\n", + "A = pi*pi# # pi square\n", + "B = fr*fr# # Resonant frequency square\n", + "\n", + "C = 1./(4.*A*B*L)#\n", + "print 'The value of Capacitor = %0.2e Farads'%C\n", + "print 'i.e 106 pF'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 25_4 Page No. 781" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of Inductor = 2.39e-04 Henry\n", + "i.e 239 uF\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# What value of L resonates with a 106-pF C at 1000 kHz, equal to 1 MHz?\n", + "\n", + "# Given data\n", + "\n", + "C = 106.*10**-12# # Capacitor=106 pFarad\n", + "fr = 1.*10**6# # Resonant frequency=1 MHertz\n", + "\n", + "A = pi*pi# # pi square\n", + "B = fr*fr# # Resonant frequency square\n", + "\n", + "C = 1./(4*A*B*C)#\n", + "print 'The value of Inductor = %.2e Henry'%C\n", + "print 'i.e 239 uF'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 25_5 Page No. 782" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Magnification factor Q =50.00\n" + ] + } + ], + "source": [ + "# A series circuit resonant at 0.4 MHz develops 100 mV across a 250-u\u0003H L with a 2-mV input. Calculate Q .\n", + "\n", + "# Given data\n", + "\n", + "Vo = 100.*10**-3# # Output voltage=100 mVolts\n", + "Vi = 2*10**-3# # Input voltage=2 mVolts\n", + "L = 250*10**-6# # Inductor=250 uHenry\n", + "f = 0.4*10**6# # Frequency=0.4 MHertz\n", + "\n", + "Q = Vo/Vi#\n", + "print 'The Magnification factor Q =%0.2f'%Q" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 25_6 Page No. 784" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Ac Resistance of Coil = 12.57 Ohms\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# What is the ac resistance of the coil in A series circuit resonant at 0.4 MHz develops 100 mV across a 250-u\u0003H L with a 2-mV input.\n", + "\n", + "# Given data\n", + "\n", + "Vo = 100.*10**-3# # Output voltage=100 mVolts\n", + "Vi = 2.0*10**-3# # Input voltage=2 mVolts\n", + "L = 250.0*10**-6# # Inductor=250 uHenry\n", + "f = 0.4*10**6# # Frequency=0.4 MHertz\n", + "\n", + "Q = Vo/Vi#\n", + "Xl = 2*pi*f*L#\n", + "\n", + "rs = Xl/Q#\n", + "print 'The Ac Resistance of Coil = %0.2f Ohms'%rs" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 25_7 Page No. 785" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Because they divide Vt equally\n", + "The Equivalent Impedence = 225000 Ohms\n", + "i.e 225 kOhms\n", + "The Q =150.00\n" + ] + } + ], + "source": [ + "# In Fig. 25–9, assume that with a 4-mVac input signal for VT, the voltage across R1 is 2 mV when R1 is 225-kOhms\u0003. Determine Zeq and Q.\n", + "\n", + "# Given data\n", + "\n", + "vin = 4.*10**-3# # Input AC signal=4 mVac\n", + "R1 = 225.*10**3# # Resistance1=225 kOhms\n", + "vR1 = 2.*10**-3# # Voltage across Resistor1=2 mVac\n", + "xl = 1.5*10**3# # Inductive Reactance=1.5 kOhms\n", + "\n", + "print 'Because they divide Vt equally'\n", + "\n", + "Zeq = R1#\n", + "print 'The Equivalent Impedence = %0.f Ohms'%Zeq\n", + "print 'i.e 225 kOhms'\n", + "\n", + "Q = Zeq/xl#\n", + "print 'The Q =%0.2f'%Q" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 25_8 Page No. 786" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Magnification factor Q = 40.02\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# A parallel LC circuit tuned to 200 kHz with a 350-u\u0003H L has a measured ZEQ of 17,600. Calculate Q.\n", + "\n", + "# Given data\n", + "\n", + "L = 350.*10**-6# # Inductor=350 uHenry\n", + "f = 200.*10**3# # Frequency=200 kHertz\n", + "Zeq = 17600.# # Equivalent Impedence=17600 Ohms\n", + "\n", + "Xl = 2*pi*f*L#\n", + "\n", + "Q = Zeq/Xl#\n", + "print 'The Magnification factor Q = %0.2f'%Q" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 25_9 Page No. 788" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Bandwidth BW or Delta f = 20000 Hertz\n", + "i.e 20 kHz\n", + "The Edge Frequency f1 = 1990000 Hertz\n", + "i.e 1990 kHz\n", + "The Edge Frequency f2 = 2010000 Hertz\n", + "i.e 2010 kHz\n" + ] + } + ], + "source": [ + "# An LC circuit resonant at 2000 kHz has a Q of 100. Find the total bandwidth delta f and the edge frequencies f1 and f2.\n", + "\n", + "# Given data\n", + "\n", + "fr = 2000.*10**3# # Resonant frequency=2000 kHertz\n", + "Q = 100.# # Magnification factor=100\n", + "\n", + "Bw = fr/Q#\n", + "print 'The Bandwidth BW or Delta f = %0.f Hertz'%Bw\n", + "print 'i.e 20 kHz'\n", + "\n", + "f1 = fr-Bw/2#\n", + "print 'The Edge Frequency f1 = %0.f Hertz'%f1\n", + "print 'i.e 1990 kHz'\n", + "\n", + "f2 = fr+Bw/2#\n", + "print 'The Edge Frequency f2 = %0.f Hertz'%f2\n", + "print 'i.e 2010 kHz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 25_10 Page No. 789" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Bandwidth BW or Delta f = 60000 Hertz\n", + "i.e 60 kHz\n", + "The Edge Frequency f1 = 5970000 Hertz\n", + "i.e 5970 kHz\n", + "The Edge Frequency f2 = 6030000 Hertz\n", + "i.e 6030 kHz\n" + ] + } + ], + "source": [ + "# An LC circuit resonant at 6000 kHz has a Q of 100. Find the total bandwidth delta f and the edge frequencies f1 and f2.\n", + "\n", + "# Given data\n", + "\n", + "fr = 6000.*10**3# # Resonant frequency=6000 kHertz\n", + "Q = 100.# # Magnification factor=100\n", + "\n", + "Bw = fr/Q#\n", + "print 'The Bandwidth BW or Delta f = %0.f Hertz'%Bw\n", + "print 'i.e 60 kHz'\n", + "\n", + "f1 = fr-Bw/2#\n", + "print 'The Edge Frequency f1 = %0.f Hertz'%f1\n", + "print 'i.e 5970 kHz'\n", + "\n", + "f2 = fr+Bw/2.0#\n", + "print 'The Edge Frequency f2 = %0.f Hertz'%f2\n", + "print 'i.e 6030 kHz'" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter26.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter26.ipynb new file mode 100644 index 00000000..471f38d0 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter26.ipynb @@ -0,0 +1,425 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 26 : Filters" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 26_1 Page No. 824" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Cutoff Frequency = 1591.55 Hertz\n", + "i.e 1.592 kHz\n", + "The Output Voltage = 7.07 Vpp\n", + "The Phase angle (Theta z) = -45.00 Degree\n" + ] + } + ], + "source": [ + "from math import pi,sqrt,atan\n", + "# Calculate (a)the cutoff frequency fc# (b)Vout at fc# (c)Theta at fc (Assume Vin = 10 Vpp for all frequencies)\n", + "\n", + "# Given data\n", + "\n", + "R = 10.*10**3# # Resistor=10 kOhms\n", + "C = 0.01*10**-6# # Capacitor=0.01 uFarad\n", + "Vin = 10.# # Input Voltage=10Vpp\n", + "# To calculate fc\n", + "\n", + "fc = 1./(2*pi*R*C)#\n", + "print 'The Cutoff Frequency = %0.2f Hertz'%fc\n", + "print 'i.e 1.592 kHz'\n", + "\n", + "# To calculate Vout at fc\n", + "\n", + "Xc = 1./(2*pi*fc*C)#\n", + "\n", + "Zt = sqrt((R*R)+(Xc*Xc))#\n", + "\n", + "Vout = Vin*(Xc/Zt)#\n", + "print 'The Output Voltage = %0.2f Vpp'%Vout\n", + "\n", + "# To calculate Theta\n", + "\n", + "Theta = atan(-(R/Xc))*180/pi\n", + "print 'The Phase angle (Theta z) = %0.2f Degree'%Theta" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 26_2 Page No. 825" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Cutoff Frequency = 3183.10 Hertz i.e 3.183 kHz\n", + "The Output Voltage = 9.54 Vpp\n", + "approx 9.52 Volts(p-p)\n", + "The Phase angle (Theta z) = -17.44 Degree\n" + ] + } + ], + "source": [ + "from math import pi,sqrt,atan\n", + "# Calculate (a)the cutoff frequency fc# (b)Vout at 1 kHz# (c)Theta at 1 kHz (Assume Vin = 10 Vpp for all frequencies)\n", + "\n", + "# Given data\n", + "\n", + "R = 1.*10**3# # Resistor=1 kOhms\n", + "L = 50.*10**-3 # Inductor=50 mHenry\n", + "Vin = 10.# # Input Voltage=10Vpp\n", + "f = 1.*10**3# # Frequency=1 kHz\n", + "# To calculate fc\n", + "\n", + "fc = R/(2*pi*L)#\n", + "print 'The Cutoff Frequency = %0.2f Hertz'%fc,\n", + "print 'i.e 3.183 kHz'\n", + "\n", + "# To calculate Vout at fc\n", + "\n", + "Xl = 2*pi*f*L#\n", + "\n", + "Zt = sqrt((R*R)+(Xl*Xl))#\n", + "\n", + "Vout = Vin*(R/Zt)#\n", + "print 'The Output Voltage = %0.2f Vpp'%Vout\n", + "print 'approx 9.52 Volts(p-p)'\n", + "\n", + "# To calculate Theta\n", + "\n", + "Theta = atan(-(Xl/R))*180/pi\n", + "print 'The Phase angle (Theta z) = %0.2f Degree'%Theta" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 26_3 Page No. 826" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Cutoff Frequency for RC High-Pass Filter = 10610.33 Hertz\n", + "i.e 10.61 kHz\n", + "The Cutoff Frequency for RL High-Pass Filter = 2387.32 Hertz\n", + "approx 2.39 kHz\n" + ] + } + ], + "source": [ + "from math import pi,sqrt\n", + "# Calculate the cutoff frequency for (a) the RC high-pass filter# (b) the RL high-pass filter\n", + "\n", + "# Given data\n", + "\n", + "R = 1.5*10**3# # Resistor=1.5 kOhms\n", + "L = 100.*10**-3 # Inductor=100 mHenry\n", + "C = 0.01*10**-6# # Capacitor=0.01 uFarad\n", + "\n", + "# To calculate fc for RC high-pass filter\n", + "\n", + "fc = 1./(2*pi*R*C)#\n", + "print 'The Cutoff Frequency for RC High-Pass Filter = %0.2f Hertz'%fc\n", + "print 'i.e 10.61 kHz'\n", + "\n", + "# To calculate fc for RL high-pass filter\n", + "\n", + "fc1 = R/(2*pi*L)#\n", + "print 'The Cutoff Frequency for RL High-Pass Filter = %0.2f Hertz'%fc1\n", + "print 'approx 2.39 kHz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 26_4 Page No. 827" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Cutoff Frequency for RC High-Pass filter = 159.15 Hertz\n", + "i.e 159 Hz\n", + "The Cutoff Frequency for RC High-Pass filter = 1591.55 Hertz\n", + "i.e 1.59 kHz\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# Calculate the cutoff frequencies fc1 and fc2.\n", + "\n", + "#Given data\n", + "\n", + "R1 = 1.*10**3# # Resistor 1=1 kOhms\n", + "C1 = 1.*10**-6# # Capacitor 1=1 uFarad\n", + "R2 = 100.*10**3# # Resistor 2=100 kOhms\n", + "C2 = 0.001*10**-6# # Capacitor 2=0.001 uFarad\n", + "\n", + "# To calculate fc1 for RC high-pass filter\n", + "\n", + "fc1 = 1/(2*pi*R1*C1)#\n", + "print 'The Cutoff Frequency for RC High-Pass filter = %0.2f Hertz'%fc1\n", + "print 'i.e 159 Hz'\n", + "\n", + "# To calculate fc2 for RC high-pass filter\n", + "\n", + "fc2 = 1/(2*pi*R2*C2)#\n", + "print 'The Cutoff Frequency for RC High-Pass filter = %0.2f Hertz'%fc2\n", + "print 'i.e 1.59 kHz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 26_5 Page No. 828" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Notch Frequency for RC Low-Pass filter = 7957.75 Hertz\n", + "i.e 7.96 kHz\n", + "The Required Value of 2R1 = 2000.00 Ohms\n", + "i.e 2 kohms\n", + "The Required Value of 2C1 = 2.00e-08 Ohms\n", + "0.02 uF\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# Calculate the notch frequency fn if R1 is 1 kOhms\u0004 and C1 is\u0005 0.01 \u0002uF. Also, calculate the required values for 2R1 and 2C1 in the low-pass filter.\n", + "\n", + "# Given data\n", + "\n", + "R1 = 1.*10**3# # Resistor 1=1 kOhms\n", + "C1 = 0.01*10**-6# # Capacitor 1=0.01 uFarad\n", + "\n", + "# To calculate Notch frequency fn for RC low-pass filter\n", + "\n", + "fn = 1/(4*pi*R1*C1)#\n", + "print 'The Notch Frequency for RC Low-Pass filter = %0.2f Hertz'%fn\n", + "print 'i.e 7.96 kHz'\n", + "\n", + "A = 2*R1#\n", + "print 'The Required Value of 2R1 = %0.2f Ohms'%A\n", + "print 'i.e 2 kohms'\n", + "\n", + "B = 2*C1#\n", + "print 'The Required Value of 2C1 = %0.2e Ohms'%B\n", + "print '0.02 uF'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 26_6 Page No. 829" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Power Gain of Amplifier = 20.00 dB\n" + ] + } + ], + "source": [ + "from math import log10\n", + "# A certain amplifier has an input power of 1 W and an output power of 100 W.Calculate the dB power gain of the amplifier.\n", + "\n", + "# Given data\n", + "\n", + "Pi = 1.# # Input power=1 Watts\n", + "Po = 100.# # Output power=100 Watts\n", + "\n", + "N = 10*log10(Po/Pi)#\n", + "print 'The Power Gain of Amplifier = %0.2f dB'%N" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 26_7 Page No. 830" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Attenuation offered by the Filter = -13.01 dB\n" + ] + } + ], + "source": [ + "from math import log10\n", + "# The input power to a filter is 100 mW, and the output power is 5 mW. Calculate the attenuation, in decibels, offered by the filter.\n", + "\n", + "# Given data\n", + "\n", + "Pi = 100.*10**-3# # Input power=1 Watts\n", + "Po = 5.*10**-3# # Output power=100 Watts\n", + "\n", + "N = 10*log10(Po/Pi)#\n", + "print 'The Attenuation offered by the Filter = %0.2f dB'%N" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 26_8 Page No. 832" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Attenuation at 0 Hz = 0 dB\n", + "The Attenuation at 1.592 kHz = -3.01 dB\n", + "The Attenuation at 15.92 kHz = -20.05 dB\n" + ] + } + ], + "source": [ + "from math import pi,log10,sqrt\n", + "# Calculate the attenuation, in decibels, at the following frequencies: (a) 0 Hz# (b) 1.592 kHz# (c) 15.92 kHz. (Assume that Vin is 10 V p-p at all frequencies.)\n", + "\n", + "# Given data\n", + "\n", + "f1 = 0# # Frequency 1=0 Hz\n", + "f2 = 1.592*10**3# # Frequency 2=1.592 kHz (cutoff frequency)\n", + "f3 = 15.92*10**3# # Frequency 3=15.92 kHz\n", + "Vi = 10.# # Voltage input=10 Volts(p-p)\n", + "R = 10.*10**3# # Resistor 1=10 kOhms\n", + "C = 0.01*10**-6# # Capacitor 1=0.01 uFarad \n", + "\n", + "Vo1 = Vi#\n", + "Vo2 = 0.707*Vi#\n", + "\n", + "# At 0 Hz\n", + "\n", + "N1 = 20*log10(Vo1/Vi)#\n", + "print 'The Attenuation at 0 Hz = %0.f dB'%N1\n", + "\n", + "#At 1.592 kHz (cutoff frequency)\n", + "\n", + "N2 = 20*log10(Vo2/Vi)#\n", + "print 'The Attenuation at 1.592 kHz = %0.2f dB'%N2\n", + "\n", + "# At 15.92 kHz\n", + "\n", + "Xc = 1./(2*pi*f3*C)#\n", + "\n", + "A = R*R#\n", + "B = Xc*Xc#\n", + "\n", + "Zt = sqrt (A+B)#\n", + "\n", + "N3 = 20*log10(Xc/Zt)#\n", + "print 'The Attenuation at 15.92 kHz = %0.2f dB'%N3" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter27.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter27.ipynb new file mode 100644 index 00000000..77d3bb82 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter27.ipynb @@ -0,0 +1,712 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 27 : Diodes and Diodes Applications" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_1 Page No. 864" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Forward Resistance at Point A = 59.09 Ohms\n", + "Approx 59.1 Ohms\n", + "The Forward Resistance at Point B = 31.11 Ohms\n" + ] + } + ], + "source": [ + "# For the diode curve, calculate the dc resistance, RF, at points A and B.\n", + "\n", + "# Given data\n", + "\n", + "Vf1 = 0.65# # Forward votage 1=0.65 Volts\n", + "If1 = 11*10**-3 # Forward current 1=11 mAmps\n", + "Vf2 = 0.7# # Forward votage 2=0.7 Volts\n", + "If2 = 22.5*10**-3 # Forward current 2=22.5 mAmps\n", + "\n", + "Rf1 = Vf1/If1#\n", + "print 'The Forward Resistance at Point A = %0.2f Ohms'%Rf1\n", + "print 'Approx 59.1 Ohms'\n", + "\n", + "Rf2 = Vf2/If2#\n", + "print 'The Forward Resistance at Point B = %0.2f Ohms'%Rf2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_2 Page No. 865" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Bulk Resistance = 0.40 Ohms\n" + ] + } + ], + "source": [ + "# A silicon diode has a forward voltage drop of 1.1 V for a forward diode current, If, of 1 A. Calculate the bulk resistance, Rb.\n", + "\n", + "# Given data\n", + "\n", + "Vf1 = 1.1# # Forward votage 1=1.1 Volts\n", + "If1 = 1. # Forward current 1=1 Amps\n", + "Vf2 = 0.7# # Fwd. vltg. 2=0.7 Volts (min working vltg of diode is 0.7 V)\n", + "If2 = 0 # Forward current=0 Amps\n", + "\n", + "delV = Vf1-Vf2# # diff. between max. min. Voltages\n", + "delI = If1-If2# # diff. between max. min. Currents\n", + "\n", + "Rb = delV/delI#\n", + "print 'The Bulk Resistance = %0.2f Ohms'%Rb" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_3 Page No. 868" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Load Voltage of First Approximation = 10.00 Volts(dc)\n", + "The Load Current of First Approximation = 0.10 Amps\n", + "i.e 100 mAmps\n", + "The Load Voltage of Second Approximation = 9.30 Volts\n", + "The Load Current of Second Approximation = 0.09 Amps\n", + "i.e 93 mAmps\n", + "The Load Current of Third Approximation = 0.09 Amps\n", + "i.e 90.73 mAmps\n", + "The Load Voltage of Third Approximation 9.07 Volts\n" + ] + } + ], + "source": [ + "# Solve for the load voltage and current using the first, second, and third diode approximations.\n", + "\n", + "# Given data\n", + "\n", + "Rl = 100.# # Load resistance=100 Ohms\n", + "Rb = 2.5# # Resistance=2.5 Ohms\n", + "Vin = 10.# # Input voltage=10 Volts\n", + "Vb = 0.7# # Voltage=0.7 Volts\n", + "\n", + "\n", + "# Using first approximation\n", + "\n", + "Vl1 = Vin\n", + "print 'The Load Voltage of First Approximation = %0.2f Volts(dc)'%Vl1\n", + "\n", + "Il1 = Vl1/Rl#\n", + "print 'The Load Current of First Approximation = %0.2f Amps'%Il1\n", + "print 'i.e 100 mAmps'\n", + "\n", + "# Using second approximation\n", + "\n", + "Vl2 = Vin-Vb\n", + "print 'The Load Voltage of Second Approximation = %0.2f Volts'%Vl2\n", + "\n", + "Il2 = Vl2/Rl#\n", + "print 'The Load Current of Second Approximation = %0.2f Amps'%Il2\n", + "print 'i.e 93 mAmps'\n", + "\n", + "# Using third approximation\n", + "\n", + "Il3 = (Vin-Vb)/(Rl+Rb)#\n", + "print 'The Load Current of Third Approximation = %0.2f Amps'%Il3\n", + "print 'i.e 90.73 mAmps'\n", + "\n", + "Vl3 = Il3*Rl#\n", + "print 'The Load Voltage of Third Approximation %0.2f Volts'%Vl3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_4 Page No. 875" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Secondary Voltage = 40.00 Volts(ac)\n", + "The DC Voltage = 17.76 Volts\n", + "The Load Current = 0.18 Amps\n", + "The DC Diode Current = 0.18 Amps\n", + "The PIV for Diode-1 = 56.56 Volts\n", + "The Output Frequency = 60.00 Hertz\n" + ] + } + ], + "source": [ + "# If the turns ratio Np:Ns is 3:1, calculate the following: Vs, Vdc, Il, Idiode, PIV for D1, and fout.\n", + "\n", + "# Given data\n", + "\n", + "Vp = 120.# # Primary voltage=120 Vac\n", + "A = 3/1.# # Turns ratio Np:Ns=3:1\n", + "B = 1/3.# # Turns ratio Ns:Np=1:3\n", + "Rl = 100.# # Load resistance=100 Ohms\n", + "fi = 60.# # Input frequency=60\n", + "\n", + "Vs = B*Vp#\n", + "print 'The Secondary Voltage = %0.2f Volts(ac)'%Vs\n", + "\n", + "Vspk = (Vs*1.414)#\n", + "\n", + "C = Vspk-0.7#\n", + "\n", + "Vdc = 0.318*C#\n", + "print 'The DC Voltage = %0.2f Volts'%Vdc\n", + "\n", + "Il = Vdc/Rl#\n", + "print 'The Load Current = %0.2f Amps'%Il\n", + "\n", + "Idiode = Il#\n", + "print 'The DC Diode Current = %0.2f Amps'%Idiode\n", + "\n", + "PIV = Vspk#\n", + "print 'The PIV for Diode-1 = %0.2f Volts'%PIV\n", + "\n", + "fo =fi#\n", + "print 'The Output Frequency = %0.2f Hertz'%fo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_5 age No. 878" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The DC Voltage = 17.54 Volts\n", + "The Load Current = 0.18 Amps\n", + "i.e 175.4 mAmps\n", + "The DC Diode Current = 0.09 Amps\n", + "i.e 87.7 mAmps\n", + "The PIV for Diode-1 = 55.86 Volts\n", + "The Output Frequency = 120.00 Hertz\n" + ] + } + ], + "source": [ + "# If the turns ratio Np:Ns is\u0006 3:1, calculate the following: Vdc, Il, Idiode, PIV for D1, and fout.\n", + "\n", + "# Given data\n", + "\n", + "Vp = 120. # Primary voltage=120 Vac\n", + "A = 3/1. # Turns ratio Np:Ns = 3:1\n", + "B = 1/3. # Turns ratio Ns:Np = 1:3\n", + "Rl = 100. # Load resistance=100 Ohms\n", + "\n", + "Vs = B*Vp#\n", + "Vspk = 1.414*(Vs/2)#\n", + "Vopk = Vspk-0.7#\n", + "\n", + "Vdc = 0.636*Vopk#\n", + "print 'The DC Voltage = %0.2f Volts'%Vdc\n", + "\n", + "Il = Vdc/Rl#\n", + "print 'The Load Current = %0.2f Amps'%Il\n", + "print 'i.e 175.4 mAmps'\n", + "\n", + "Idiode = Il/2#\n", + "print 'The DC Diode Current = %0.2f Amps'%Idiode\n", + "print 'i.e 87.7 mAmps'\n", + "\n", + "C = (Vspk*2)-0.7#\n", + "\n", + "PIV = C#\n", + "print 'The PIV for Diode-1 = %0.2f Volts'%PIV\n", + "\n", + "f =120#\n", + "print 'The Output Frequency = %0.2f Hertz'%f" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_6 Page No. 880" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The DC Voltage = 35.08 Volts\n", + "The Load Current = 0.35 Amps\n", + "i.e 350.8 mAmps\n", + "The DC Diode Current = 0.1754 Amps\n", + "i.e 175.4 mAmps\n", + "The PIV for each Diode = 55.86 Volts\n", + "The Output Frequency = 120.00 Hertz\n" + ] + } + ], + "source": [ + "# If the turns ratio Np:Ns is\u0006 3:1, calculate the following: Vdc, Il, Idiode, PIV for each diode, and fout.\n", + "\n", + "# Given data\n", + "\n", + "Vp = 120.# # Primary voltage=120 Vac\n", + "A = 3./1# # Turns ratio Np:Ns = 3:1\n", + "B = 1./3# # Turns ratio Ns:Np = 1:3\n", + "Rl = 100.# # Load resistance=100 Ohms\n", + "\n", + "Vs = B*Vp#\n", + "Vspk = 1.414*(Vs)#\n", + "Vopk = Vspk-1.4#\n", + "\n", + "Vdc = 0.636*Vopk#\n", + "print 'The DC Voltage = %0.2f Volts'%Vdc\n", + "\n", + "Il = Vdc/Rl#\n", + "print 'The Load Current = %0.2f Amps'%Il\n", + "print 'i.e 350.8 mAmps'\n", + "\n", + "Idiode = Il/2#\n", + "print 'The DC Diode Current = %0.4f Amps'%Idiode\n", + "print 'i.e 175.4 mAmps'\n", + "\n", + "C = Vspk-0.7#\n", + "\n", + "PIV = C#\n", + "print 'The PIV for each Diode = %0.2f Volts'%PIV\n", + "\n", + "f =120#\n", + "print 'The Output Frequency = %0.2f Hertz'%f" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_7 Page No. 883" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Ripple Voltage for Turns Ratio Np:Ns=4:1 = 5.21 Volts(p-p)\n", + "Approx 5.21 Volts(p-p)\n", + "The DC Voltage for Turns Ratio Np:Ns=4:1 = 39.12 Volts\n", + "Approx 39.12 Volts\n", + "The Ripple Voltage for Turns Ratio Np:Ns=2:1 = 2.69 Volts(p-p)\n", + "Approx 2.69 Volts(p-p)\n", + "The DC Voltage for Turns Ratio Np:Ns=2:1 = 40.38 Volts\n", + "Approx 40.38 Volts\n" + ] + } + ], + "source": [ + "from math import exp\n", + "# Assume the transformer turns ratio Np:Ns = 4:1 in Fig. 27–21 a and 2:1 in Fig. 27–22a. Compare Vripple and Vdc if C =\u0006 500 uF and Rl =\u0006 250\u0007.\n", + " \n", + "# Given data\n", + "\n", + "A1 = 4./1# # Turns ratio Np:Ns=4:1\n", + "B1 = 1./4# # Turns ratio Ns:Np=1:4\n", + "A2 = 2./1# # Turns ratio Np:Ns=2:1\n", + "B2 = 1./2# # Turns ratio Ns:Np=1:2\n", + "Vp = 120.# # Primary voltage=120 Vac\n", + "Vb = 0.7# # \n", + "t1 = 16.67*10**-3# # Charging Time of Capacitor of Turns ratio Np:Ns=4:1=16.67 mSec\n", + "t2 = 8.33*10**-3# # Charging Time of Capacitor of Turns ratio Np:Ns=4:1=8.33 mSec\n", + "Rl = 250.# # Load resistance=250 Ohms\n", + "C = 500.*10**-6# # Capacitor=500 uFarad\n", + "\n", + "# Calculations for Turns Ratio Np:Ns=4:1\n", + "\n", + "Vs1 = B1*Vp#\n", + "Vspk1 = Vs1*1.414#\n", + "Vopk1 = Vspk1 - Vb#\n", + "D = -t1/(Rl*C)#\n", + "\n", + "Vrp1 = Vopk1*(1-exp(D))#\n", + "print 'The Ripple Voltage for Turns Ratio Np:Ns=4:1 = %0.2f Volts(p-p)'%Vrp1\n", + "print 'Approx 5.21 Volts(p-p)'\n", + "\n", + "Vdc1 = Vopk1-(Vrp1/2)#\n", + "print 'The DC Voltage for Turns Ratio Np:Ns=4:1 = %0.2f Volts'%Vdc1\n", + "print 'Approx 39.12 Volts'\n", + "\n", + "# Calculations for Turns Ratio Np:Ns = 2:1\n", + "\n", + "Vs2 = B2*Vp#\n", + "V2 = Vs2/2#\n", + "V2pk2 = V2*1.414\n", + "Vopk2 = V2pk2 - Vb#\n", + "E = -t2/(Rl*C)#\n", + "\n", + "Vrp2 = Vopk2*(1-(exp(E)))\n", + "print 'The Ripple Voltage for Turns Ratio Np:Ns=2:1 = %0.2f Volts(p-p)'%Vrp2\n", + "print 'Approx 2.69 Volts(p-p)'\n", + "\n", + "Vdc2 = Vopk2-(Vrp2/2)#\n", + "print 'The DC Voltage for Turns Ratio Np:Ns=2:1 = %0.2f Volts'%Vdc2\n", + "print 'Approx 40.38 Volts'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_8 Page No. 885" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The LED Current = 0.01 Amps\n", + "i.e 10 mAmps\n" + ] + } + ], + "source": [ + "# Calculate the LED current.\n", + "\n", + "# Given data\n", + "\n", + "Vin = 24.# # Input voltage=24 Volts\n", + "Vled = 2.# # Voltage drop at LED=2 Volts\n", + "Rs = 2.2*10**3# # Source Resistance=2.2 kOhms\n", + "\n", + "Iled = (Vin-Vled)/Rs#\n", + "print 'The LED Current = %0.2f Amps'%Iled\n", + "print 'i.e 10 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_9 Page No. 888" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistance Rs, Required to Provide an LED Current of 25 mA = 880.00 Ohms\n" + ] + } + ], + "source": [ + "# Calculate the resistance Rs, required to provide an LED current of 25 mA.\n", + "\n", + "# Given data\n", + "\n", + "Vin = 24.# # Input voltage=24 Volts\n", + "Vled = 2.# # Voltage drop at LED=2 Volts\n", + "Iled = 25.*10**-3# # LED Current=25 mAmps\n", + "\n", + "Rs = (Vin-Vled)/Iled#\n", + "print 'The Resistance Rs, Required to Provide an LED Current of 25 mA = %0.2f Ohms'%Rs" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_10 Page No. 889" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Maximum Rated Current of Zener = 0.10 Amps\n", + "i.e 100 mAmps\n" + ] + } + ], + "source": [ + "# Calculate the maximum rated zener current for a 1 W, 10 V zener.\n", + "\n", + "# Given data\n", + "\n", + "Pzm = 1.# # Power rating of zener= 1 Watts\n", + "Vz = 10.# # Voltage rating of zener= 10 Volts\n", + "\n", + "Izm = Pzm/Vz#\n", + "print 'The Maximum Rated Current of Zener = %0.2f Amps'%Izm\n", + "print 'i.e 100 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_11 Page No. 890" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Zener Current = 0.01 Amps\n", + "i.e 15 mAmps\n" + ] + } + ], + "source": [ + "# If Vz=10 V, calculate Iz.\n", + "\n", + "# Given data\n", + "\n", + "Vin = 25.# # Input voltage=25 Volts\n", + "Vz = 10.# # Zener voltage=10 Volts\n", + "Rs = 1.*10**3# # Source Resistance=1 kOhms\n", + "\n", + "Iz = (Vin-Vz)/Rs#\n", + "print 'The Zener Current = %0.2f Amps'%Iz\n", + "print 'i.e 15 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_12 Page No. 891" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Source Current = 0.075 Amps\n", + "i.e 75 mAmps\n", + "The Load Current = 0.03 Amps\n", + "i.e 30 mAmps\n", + "The Zener Current = 0.045 Amps\n", + "i.e 45 mAmps\n", + "The Power Dissipation of Zener = 0.3375 Watts\n", + "i.e 337.5 mWatts\n" + ] + } + ], + "source": [ + "# If R L increases to 250 Ohms, calculate the following: Is, Il, Iz, and Pz.\n", + "\n", + "# Given data\n", + "\n", + "Vin = 25# # Input voltage=25 Volts\n", + "Vz = 7.5# # Zener voltage=7.5 Volts\n", + "Rl = 250# # Load Resistance=250 Ohms\n", + "Is = 75*10**-3# # Source current=75 mAmps\n", + "\n", + "print 'The Source Current = %0.3f Amps'%Is\n", + "print 'i.e 75 mAmps'\n", + "\n", + "Il = Vz/Rl#\n", + "print 'The Load Current = %0.2f Amps'%Il\n", + "print 'i.e 30 mAmps'\n", + "\n", + "Iz = Is-Il#\n", + "print 'The Zener Current = %0.3f Amps'%Iz\n", + "print 'i.e 45 mAmps'\n", + "\n", + "Pz = Vz*Iz#\n", + "print 'The Power Dissipation of Zener = %0.4f Watts'%Pz\n", + "print 'i.e 337.5 mWatts'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 27_13 Page No. 892" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Source Current = 0.06 Amps.\n", + "i.e 60 mAmps\n", + "The Load Current for 200 ohms Load = 0.05 Amps.\n", + "i.e 50 mAmps\n", + "The Zener Current for 200 ohms Load = 0.01 Amps.\n", + "i.e 10 mAmps\n", + "The Load Current for 500 ohms Load = 0.02 Amps.\n", + "i.e 20 mAmps\n", + "The Zener Current for 500 ohms load = 0.04 Amps.\n", + "i.e 40 mAmps\n" + ] + } + ], + "source": [ + "# Calculate Is, Il and Iz for (a)Rl=200 ohms# (b)Rl=500 ohms.\n", + "\n", + "# Given data\n", + "\n", + "Vin = 16.# # Vin=16 Volts given\n", + "Vz = 10.# # Vz=10 Volts given\n", + "Rs = 100.# # Source Resistance = 100 ohms given\n", + "Rla = 200.# # Load Resistance A = 200 ohms given\n", + "Rlb = 500.# #Load Resistance B = 500 ohms given\n", + "\n", + "# For Rl 200 ohms\n", + "\n", + "Is = (Vin-Vz)/Rs#\n", + "print 'The Source Current = %0.2f Amps.'%Is\n", + "print 'i.e 60 mAmps'\n", + "\n", + "Ila = Vz/Rla#\n", + "print 'The Load Current for 200 ohms Load = %0.2f Amps.'%Ila\n", + "print 'i.e 50 mAmps'\n", + "\n", + "Iza = Is-Ila\n", + "print 'The Zener Current for 200 ohms Load = %0.2f Amps.'%Iza\n", + "print 'i.e 10 mAmps'\n", + "\n", + "# For Rl 500 ohms\n", + "\n", + "Ilb = Vz/Rlb#\n", + "print 'The Load Current for 500 ohms Load = %0.2f Amps.'%Ilb\n", + "print 'i.e 20 mAmps'\n", + "\n", + "Izb = Is-Ilb\n", + "print 'The Zener Current for 500 ohms load = %0.2f Amps.'%Izb\n", + "print 'i.e 40 mAmps'" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter28.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter28.ipynb new file mode 100644 index 00000000..67d92401 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter28.ipynb @@ -0,0 +1,723 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 28 : Bipolar Junction Transistors" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_1 Page No. 910" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Emitter Current Ie = 5.00 Amps\n" + ] + } + ], + "source": [ + "# A transistor has the following currents: Ib is\u0002 20 mA and Ic is 4.98 A. Calculate Ie.\n", + "\n", + "# Given data\n", + "\n", + "Ib = 20*10**-3# # Base current=20 mAmps\n", + "Ic = 4.98# # Collector current=4.98 Amps\n", + "\n", + "Ie = Ic+Ib#\n", + "print 'The Emitter Current Ie = %0.2f Amps'%Ie" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_2 Page No. 912" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Collector Current Ic = 0.09804 Amps\n", + "i.e 98.04 mAmps\n" + ] + } + ], + "source": [ + "# A transistor has the following currents: Ie is\u0002 100 mA, Ib is\u0002 1.96 mA. Calculate Ic.\n", + "\n", + "# Given data\n", + "\n", + "Ie = 100.0*10**-3# # Emitter current=100 mAmps\n", + "Ib = 1.96*10**-3# # Base current=4.98 Amps\n", + "\n", + "Ic = Ie-Ib#\n", + "print 'The Collector Current Ic = %0.5f Amps'%Ic\n", + "print 'i.e 98.04 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_3 Page No. 913" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Base Current Ib = 0.0010 Amps\n", + "i.e 1 mAmps\n" + ] + } + ], + "source": [ + "# A transistor has the following currents: Ie is\u0002 50 mA, Ic is\u0002 49 mA. Calculate Ib.\n", + "\n", + "# Given data\n", + "\n", + "Ie = 50.0*10**-3# # Emitter current=50 mAmps\n", + "Ic = 49.0*10**-3# # Collector current=20 mAmps\n", + "\n", + "Ib = Ie-Ic#\n", + "print 'The Base Current Ib = %0.4f Amps'%Ib\n", + "print 'i.e 1 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_4 Page No. 914" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Alpha(dc) = 0.9960\n" + ] + } + ], + "source": [ + "# A transistor has the following currents: Ie is\u0002 15 mA, Ib is\u0002 60 u\u0004A. Calculate \u0002Alpha(dc).\n", + "\n", + "# Given data\n", + "\n", + "Ie = 15.*10**-3# # Emitter current=15 mAmps\n", + "Ib = 60.*10**-6# # Base current=60 uAmps\n", + "\n", + "Ic = Ie-Ib#\n", + "\n", + "Adc = Ic/Ie#\n", + "print 'The Value of Alpha(dc) = %0.4f'%Adc" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_5 Page No. 916" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Beta(dc) = 200\n" + ] + } + ], + "source": [ + "# A transistor has the following currents: Ic is\u0002 10 mA and Ib is 50 uA. Calculate Beta(dc).\n", + "\n", + "# Given data\n", + "\n", + "Ic = 10.*10**-3# # Collector current=10 mAmps\n", + "Ib = 50.*10**-6# # Base current=50 uAmps\n", + "\n", + "Bdc = Ic/Ib#\n", + "print 'The Value of Beta(dc) = %0.f'%Bdc" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_6 Page No. 918" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Collector Current Ic = 0.01125 Amps\n", + "i.e 11.25 mAmps\n" + ] + } + ], + "source": [ + "# A transistor has Beta(dc) of 150 and Ib of 75 uAmps. Calculate Ic.\n", + "\n", + "# Given data\n", + "\n", + "Bdc = 150.# # Beta(dc)=150\n", + "Ib = 75.*10**-6# # Base current=75 uAmps\n", + "\n", + "Ic = Bdc*Ib#\n", + "print 'The Collector Current Ic = %0.5f Amps'%Ic\n", + "print 'i.e 11.25 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_7 Page No. 920" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Alpha(dc) = 0.9901\n" + ] + } + ], + "source": [ + "# A transistor has Beta(dc) of 100. Calculate Alpha(dc).\n", + "\n", + "# Given data\n", + "\n", + "Bdc = 100.0# # Beta(dc)=100\n", + "\n", + "Adc = Bdc/(1+Bdc)#\n", + "print 'The Value of Alpha(dc) = %0.4f'%Adc" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_8 Page No. 922" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Beta(dc) =199.00\n" + ] + } + ], + "source": [ + "# A transistor has Alpha(dc) of 0.995. Calculate Beta(dc).\n", + "\n", + "# Given data\n", + "\n", + "Adc = 0.995# # Alpha(dc)=100\n", + "\n", + "Bdc = Adc/(1-Adc)#\n", + "print 'The Value of Beta(dc) =%0.2f'%Bdc" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_9 Page No. 922" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Power Dissipation = 0.05 Watts\n", + "i.e 50 mWatts\n" + ] + } + ], + "source": [ + "# Calculate Pd if Vcc is 10 V and Ib is 50 uAmps. Assume Beta(dc) is 100.\n", + "\n", + "# Given data\n", + "\n", + "Bdc = 100.# # Beta(dc)=100\n", + "Ib = 50.*10**-6# # Base current=50 uAmps\n", + "Vcc = 10.# # Supply voltage=10 Volts\n", + "\n", + "Vce = Vcc\n", + "\n", + "Ic = Bdc*Ib#\n", + "\n", + "Pd = Vce*Ic#\n", + "print 'The Power Dissipation = %0.2f Watts'%Pd\n", + "print 'i.e 50 mWatts'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_10 Page No. 922" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Maximum Allowable Collector Current Ic(max) = 0.025 Amps\n", + "i.e 25 mAmps\n" + ] + } + ], + "source": [ + "# The transistor has a power rating of 0.5 W. If Vce is\u0002 20 V, calculate the maximum allowable collector current, Ic, that can exist without exceeding the transistor’s power rating.\n", + "\n", + "# Given data\n", + "\n", + "Pdmax = 0.5# # Power dissipation(max)=0.5 Watts\n", + "Vce = 20.# # Voltage (collector to emitter)=20 Volts\n", + "\n", + "Ic = Pdmax/Vce#\n", + "print 'The Maximum Allowable Collector Current Ic(max) = %0.3f Amps'%Ic\n", + "print 'i.e 25 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_11 Page No. 923" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Power Rating at 50°C = 0.28 Watts\n", + "i.e 280 mWatts\n" + ] + } + ], + "source": [ + "# Assume that a transistor has a power rating Pd(max) of 350 mW at an ambient temperature Ta of 25°C. The derating factor is 2.8 mW/°C. Calculate the power rating at 50°C.\n", + "\n", + "# Given data\n", + "\n", + "f = 2.8*10**-3# # Derating factor=2.8 mW/°C\n", + "Pd = 350.*10**-3# # Power dissipation(max)=350 mWatts\n", + "Ta = 25.# # Ambient Temperature=25°C\n", + "Tp = 50.# # Power rating at 50°C\n", + "\n", + "delT = Tp-Ta# # Difference between max and min temp\n", + "\n", + "delPd = delT*f#\n", + "\n", + "Prat = Pd-delPd#\n", + "print 'The Power Rating at 50°C = %0.2f Watts'%Prat\n", + "print 'i.e 280 mWatts'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_12 Page No. 923" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Base Current = 0.0000 Amps.\n", + "Approx 28.97 mAmps\n", + "The Collector Current = 0.0043 Amps\n", + "Approx 4.35 mAmps\n", + "The Voltage Collector-Emitter = 5.48 Volts\n", + "Q(5.480769,0.004346)\n", + "\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fbd5f1197d0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "from matplotlib.pyplot import plot,xlabel,ylabel,show,title\n", + "# Solve for Ib, Ic, Vce. Also, Construct a dc load line showing the valuse of Ic(sat), Vce(off), Icq, Vceq.\n", + "\n", + "# Given data\n", + "Vcc = 12.# # Supply voltage=12 Volts\n", + "Vbe = 0.7# # Base-Emitter Voltage=0.7 Volts\n", + "Rb = 390.*10**3# # Base Resistor=390K Ohms\n", + "Rc = 1.5*10**3# # Collector Resistor=1.5K Ohms\n", + "B = 150.# # Beta(dc)=150\n", + "\n", + "Ib = (Vcc-Vbe)/Rb#\n", + "print 'The Base Current = %0.4f Amps.'%Ib\n", + "print 'Approx 28.97 mAmps'\n", + "\n", + "Icq = B*Ib#\n", + "print 'The Collector Current = %0.4f Amps'%Icq\n", + "print 'Approx 4.35 mAmps'\n", + "\n", + "Vceq = Vcc-(Icq*Rc)#\n", + "print 'The Voltage Collector-Emitter = %0.2f Volts'%Vceq\n", + "\n", + "# For DC load line\n", + "\n", + "Icsat = (Vcc/Rc)#\n", + "Vceoff = Vcc#\n", + "\n", + "Vce1=[Vceoff, Vceq ,0]\n", + "Ic1=[0 ,Icq ,Icsat]\n", + "\n", + "#To plot DC load line\n", + "\n", + "print \"Q(%f,%f)\\n\"%(Vceq,Icq)\n", + "plot(Vce1, Ic1)\n", + "plot(Vceq,Icq)\n", + "plot(0,Icq)\n", + "plot(Vceq,0)\n", + "plot(0,Icsat)\n", + "plot(Vceoff,0)\n", + "xlabel(\"Vce in volt\")\n", + "ylabel(\"Ic in Ampere\")\n", + "title(\"DC Load-line for Base-Biased Transistor Circuit\")\n", + "show() " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_13 Page No. 924" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Base Voltage = 2.61 Volts\n", + "The Emmiter Voltage = 1.91 Volts\n", + "The Collector Voltage = 10.65 Volts\n", + "Approx 10.65 Volts\n", + "The Collector-Emitter Voltage = 8.74 Volts\n", + "Approx 8.74 Volts\n", + "The Current Ic(sat) = 0.01 Amps\n", + "i.e 9.52 mAmps\n", + "The Voltage Vce(off) = 18.00 Volts\n", + "Q(8.737067,0.004901)\n", + "\n" + ] + }, + { + "data": { + "image/png": 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jH4NmhI5Xk9KWmcTWITH6Ap+lFFVVZBKh/T3x758qWtCrMf5LXbx+/etfJx6DYiqsuBST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\nMkOBRwDcfTawq5ntUcW6W9a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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fbd5f078490>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "from matplotlib.pyplot import plot,xlabel,ylabel,show,title\n", + "\n", + "# Solve for Vb, Ve, Ic, Vc, and Vce. Also, calculate Ic(sat) and Vce(off). Finally, construct a dc load line showing the values of Ic(sat), Vce(off), Icq, and Vceq.\n", + "\n", + "# Given data\n", + "\n", + "R1 = 33.*10**3# # Resistor 1=33 kOhms\n", + "R2 = 5.6*10**3# # Resistor 2=5.6 kOhms\n", + "Rc = 1.5*10**3# # Collector resistance=1.5 kOhms\n", + "Re = 390.# # Emitter resistance=390 Ohms\n", + "Bdc = 200.# # Beta(dc)= 200\n", + "Vcc = 18.# # Supply voltage = 18 Volts\n", + "Vbe = 0.7# # Base-Emmiter Voltage=0.7 Volts\n", + "\n", + "Vb = Vcc*(R2/(R1+R2))#\n", + "print 'The Base Voltage = %0.2f Volts'%Vb\n", + "\n", + "Ve = Vb-Vbe#\n", + "print 'The Emmiter Voltage = %0.2f Volts'%Ve\n", + "\n", + "Ie = Ve/Re# # Emitter current\n", + "\n", + "Ic = Ie#\n", + "\n", + "Vc = Vcc-(Ic*Rc)#\n", + "print 'The Collector Voltage = %0.2f Volts'%Vc\n", + "print 'Approx 10.65 Volts'\n", + "\n", + "Vce = Vcc-(Ic*(Rc+Re))#\n", + "print 'The Collector-Emitter Voltage = %0.2f Volts'%Vce\n", + "print 'Approx 8.74 Volts'\n", + "\n", + "Icsat = Vcc/(Rc+Re)#\n", + "print 'The Current Ic(sat) = %0.2f Amps'%Icsat\n", + "print 'i.e 9.52 mAmps'\n", + "\n", + "Vceoff = Vcc#\n", + "print 'The Voltage Vce(off) = %0.2f Volts'%Vceoff\n", + "\n", + "Icq = Ic\n", + "Vceq = Vce\n", + "\n", + "Vce1=[Vcc, Vceq, 0]\n", + "Ic1=[0, Icq, Icsat]\n", + "\n", + "#To plot DC load line\n", + "\n", + "print \"Q(%f,%f)\\n\"%(Vceq,Icq)\n", + "plot(Vce1, Ic1)\n", + "plot(Vceq,Icq)\n", + "plot(0,Icq)\n", + "plot(Vceq,0)\n", + "plot(0,Icsat)\n", + "plot(Vceoff,0)\n", + "xlabel(\"Vce in Volt\")\n", + "ylabel(\"Ic in mAmps\")\n", + "title(\"DC Load-line for Voltage Divider-Biased Transistor Circuit\")\n", + "show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_14 Page No. 925" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Base Voltage = -1.90 Volts\n", + "Approx -1.9 Volts\n", + "The Emitter Voltage = -1.20 Volts\n", + "Approx -1.2 Volts\n", + "The Collector Current = 0.00 Amps\n", + "Approx 2.4 mAmps\n", + "The Collector Voltage = -7.21 Volts\n", + "The Collector-Emitter Voltage = -6.01 Volts\n" + ] + } + ], + "source": [ + "# For the pnp transistor, solve for Vb, Ve, Ic, Vc, and Vce.\n", + "\n", + "# Given data\n", + "\n", + "R1 = 33.*10**3# # Resistor1=33 kOhms\n", + "R2 = 6.2*10**3# # Resistor2=6.2 kOhms\n", + "Rc = 2.*10**3# # Collector resistance=2 kOhms\n", + "Re = 500.# # Emitter resistance=500 Ohms\n", + "Vcc = 12.# # Supply voltage=12 Volts\n", + "Vbe = 0.7# # Base-Emmiter Voltage=0.7 Volts\n", + "\n", + "\n", + "Vb = -Vcc*(R2/(R1+R2))#\n", + "print 'The Base Voltage = %0.2f Volts'%Vb\n", + "print 'Approx -1.9 Volts'\n", + "\n", + "Ve = Vb-(-Vbe)#\n", + "print 'The Emitter Voltage = %0.2f Volts'%Ve\n", + "print 'Approx -1.2 Volts'\n", + "\n", + "Ic = -(Ve/Re)# # Ic =~ Ie\n", + "print 'The Collector Current = %0.2f Amps'%Ic\n", + "print 'Approx 2.4 mAmps'\n", + "\n", + "Vc = -Vcc+(Ic*Rc)\n", + "print 'The Collector Voltage = %0.2f Volts'%Vc\n", + "\n", + "Vce = -Vcc+(Ic*(Rc+Re))#\n", + "print 'The Collector-Emitter Voltage = %0.2f Volts'%Vce" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 28_15 Page No. 926" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Emitter current = 0.0053 Amps\n", + "i.e 5.3 mAmps\n", + "The Collector voltage = 7.05 Volts\n" + ] + } + ], + "source": [ + "# Calculate Ie and Vc\n", + "\n", + "# Given data\n", + "\n", + "Vee = 6.# # Supply voltage at emitter=6 Volts\n", + "Vcc = 15.# # Supply voltage at collector=15 Volts\n", + "Vbe = 0.7# # Base-Emmiter Voltage=0.7 Volts\n", + "Rc = 1.5*10**3# # Collector resistance=1.5 kOhms\n", + "Re = 1.*10**3# # Emitter resistance=1 kOhms\n", + "\n", + "Ie = (Vee-Vbe)/Re#\n", + "print 'The Emitter current = %0.4f Amps'%Ie\n", + "print 'i.e 5.3 mAmps'\n", + "\n", + "Ic = Ie# # Ic =~ Ie\n", + "\n", + "Vc = Vcc-Ic*Rc#\n", + "print 'The Collector voltage = %0.2f Volts'%Vc" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter29.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter29.ipynb new file mode 100644 index 00000000..040d35f9 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter29.ipynb @@ -0,0 +1,595 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 29 : Transistor Amplifiers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 29_1 Page No. 940" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Ac Resistance with R=10 kOhms = 26.88 Ohms\n", + "The Ac Resistance with R=5 kOhms = 13.44 Ohms\n", + "The Ac Resistance with R=1 kOhms = 2.69 Ohms\n", + "Approx 2.69 Ohms\n" + ] + } + ], + "source": [ + "# For the diode circuit, calculate the ac resistance, rac, for the following values of R: (a) 10 kOhms, (b) 5 kOhms, and (c) 1 kOhms. Use the second approximation of a diode.\n", + "\n", + "# Given data\n", + "\n", + "R1 = 10.*10**3# # Resistance 1=10 kOhms\n", + "R2 = 5.*10**3# # Resistance 2=5 kOhms\n", + "R3 = 1.*10**3# # Resistance 3=1 kOhms\n", + "Vdc = 10.# # DC supply=10 Volts\n", + "V = 0.7# # Starting voltage of diode=0.7 Volts\n", + "A = 25.*10**-3# # Constant\n", + "\n", + "# For R=10 kOhms\n", + "\n", + "Id1 = (Vdc-V)/R1#\n", + "\n", + "rac1 = A/Id1#\n", + "print 'The Ac Resistance with R=10 kOhms = %0.2f Ohms'%rac1\n", + "\n", + "# For R=5 kOhms\n", + "\n", + "Id2 = (Vdc-V)/R2#\n", + "\n", + "rac2 = A/Id2#\n", + "print 'The Ac Resistance with R=5 kOhms = %0.2f Ohms'%rac2\n", + "\n", + "# For R=1 kOhms\n", + "\n", + "Id3 = (Vdc-V)/R3#\n", + "\n", + "rac3 = A/Id3#\n", + "print 'The Ac Resistance with R=1 kOhms = %0.2f Ohms'%rac3\n", + "print 'Approx 2.69 Ohms'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 29_2 Page No. 941" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage Gain Av =200.00\n" + ] + } + ], + "source": [ + "#A common-emitter amplifier circuit has an input of 25 mVp-p and an output of 5 Vp-p. Calculate Av.\n", + "\n", + "# Given data\n", + "\n", + "Vin = 25*10**-3# # Input voltage=25 mVolts(p-p)\n", + "Vo = 5# # Output voltage=5 Volts(p-p).\n", + "\n", + "Av = Vo/Vin#\n", + "print 'The Voltage Gain Av =%0.2f'%Av" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 29_3 Page No. 942" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Output Voltage = 1.50 Volts(p-p)\n" + ] + } + ], + "source": [ + "# assume Av still equals 300. If vin is\u0003 5 mVp-p, calculate Vout.\n", + "\n", + "# Given data\n", + "\n", + "Vin = 5*10**-3# # Input voltage=5 mVolts(p-p)\n", + "Av = 300# # Voltage gain=300\n", + "\n", + "Vo = Av*Vin#\n", + "print 'The Output Voltage = %.2f Volts(p-p)'%Vo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 29_4 Page No. 948" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage Gain Av =89.96\n", + "Approx 90\n" + ] + } + ], + "source": [ + "# Assume that re varies from 3.33 Ohms to 6.67 Ohms as the temperature of the transistor changes. Calculate the variation in the voltage gain, Av.\n", + "\n", + "# Given data\n", + "\n", + "rl = 600# # Load resistance=600 Ohms\n", + "re = 6.67# # Internal emitter resistance=6.67 Ohms\n", + "\n", + "Av = rl/re#\n", + "print 'The Voltage Gain Av =%0.2f'%Av\n", + "print 'Approx 90'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 29_5 Page No. 949" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage Gain Av(max) when r`e=3.33 Ohms: 9.47\n", + "The Voltage Gain Av(min) when r`e=6.67 Ohms: 9.00\n" + ] + } + ], + "source": [ + "# Assume that r'e varies from 3.33 Ohms\u0003 to 6.67 Ohms. Calculate the minimum and maximum values for Av.\n", + "\n", + "# Given data\n", + "\n", + "rl = 600.# # Load resistance=600 Ohms\n", + "re1 = 3.33# # Internal emitter resistance=3.33 Ohms\n", + "re2 = 6.67# # Internal emitter resistance=6.67 Ohms\n", + "rE = 60.# # Emitter resistance=60 Ohms\n", + "\n", + "Av1 = rl/(re1+rE)#\n", + "print \"The Voltage Gain Av(max) when r`e=3.33 Ohms: %0.2f\"%Av1\n", + "\n", + "Av2 = rl/(re2+rE)#\n", + "print 'The Voltage Gain Av(min) when r`e=6.67 Ohms: %0.2f'%Av2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 29_6 Page No. 950" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage Gain Av =0.996\n", + "i.e 996 mVolts(p-p)\n", + "The Output Voltage = 0.700 Volts(p-p)\n" + ] + } + ], + "source": [ + "# Find the exact value of Av. Also, find Vout.\n", + "\n", + "# Given data\n", + "\n", + "rl = 909.# # Load resistance=909 Ohms\n", + "re = 3.35# # Internal emitter resistance=3.35 Ohms\n", + "Vin = 1.# # Input voltage=1 Volts(p-p)\n", + "\n", + "Av = rl/(re+rl)#\n", + "print 'The Voltage Gain Av =%0.3f'%Av\n", + "print 'i.e 996 mVolts(p-p)'\n", + "Vo = Av*Vin#\n", + "print 'The Output Voltage = %0.3f Volts(p-p)'%V\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 29_7 Page No. 951" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Input impedence = 2485.72 Ohms\n", + "i.e 2.48 kOhms\n" + ] + } + ], + "source": [ + "# Calculate Zin.\n", + "\n", + "# Given data\n", + "\n", + "rl = 909.# # Load resistance=909 Ohms\n", + "re = 3.35# # Internal emitter resistance=3.35 Ohms\n", + "B = 100.# # Beta=100\n", + "R1 = 4.7*10**3# # Resistance1=4.7 kOhms\n", + "R2 = 5.6*10**3# # Resistance2=5.6 kOhms\n", + "\n", + "Zibase = B*(re+rl)#\n", + "A = (R1*R2)/(R1+R2)#\n", + "\n", + "Zin = (Zibase*A)/(A+Zibase)#\n", + "print 'The Input impedence = %0.2f Ohms'%Zin\n", + "print 'i.e 2.48 kOhms'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 29_8 Page No. 952" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Base Voltage = 9.00 Volts\n", + "The Emitter Voltage = 8.30 Volts\n", + "The Collector current = 0.01 Amps\n", + "i.e 5.53 mAmps\n", + "The Collector Voltage = 20.00 Volts\n", + "The Collector-Emmiter Voltage = 11.70 Volts\n", + "The AC emmiter resistance = 4.52 Ohms\n", + "Approx 4.52 Ohms\n", + "The Voltage gain =0.99\n", + "The Input Base Impedence = 120903.61 Ohms\n", + "i.e 120.9 kOhms\n", + "The Input Impedence = 9150.71 Ohms\n", + "i.e 9.15 kOhms\n", + "The AC base voltage = 4.69 Volts(p-p)\n", + "The AC output voltage = 4.66 Volts(p-p)\n", + "Q(11.700000,0.005533)\n", + "\n" + ] + }, + { + "data": { + "image/png": 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+/TV06hSu9S6ST9JJIs+Y2fUZLMDYGliZsL0qeiydffasomw7oLeZvWBm5WbW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\nQUREqlawy56IiEjuFdyM9XQ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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fb8844408d0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "from matplotlib.pyplot import plot,xlabel,title,ylabel,show\n", + "# Calculate the following quantities: Vb, Ve, Ic, Vc, Vce, r'e, Zin(base), Zin, Av, vb, and vout. Also, plot the dc load line.\n", + "\n", + "# Given data\n", + "\n", + "R1 = 22.*10**3# # Resistance1=22 kOhms\n", + "R2 = 18.*10**3# # Resistance2=18 kOhms\n", + "Rg = 600.# # Generator resistance=600 Ohms\n", + "Re = 1.5*10**3# # Emitter resistance=1.5 kOhms\n", + "Rl = 1.*10**3# # Load resistance=1 kOhms\n", + "Vcc = 20.# # Supply Voltage=20 Volts\n", + "Vbe = 0.7# # Voltage Base-Emitter=0.7 Volts\n", + "B = 200.# # Beta=200\n", + "vin = 5.# # Input Voltage=5 Volts(p-p)\n", + "\n", + "# Calculate the DC quantities first:\n", + "\n", + "Vb = Vcc*(R2/(R1+R2))#\n", + "print 'The Base Voltage = %0.2f Volts'%Vb\n", + "\n", + "Ve = Vb-Vbe#\n", + "print 'The Emitter Voltage = %0.2f Volts'%Ve\n", + "\n", + "Ie = Ve/Re#\n", + "Ic = Ie# # Ic =~ Ie\n", + "print 'The Collector current = %0.2f Amps'%Ic\n", + "print 'i.e 5.53 mAmps'\n", + "\n", + "Vc = Vcc# # Since the collector is tied directly to Vcc\n", + "print 'The Collector Voltage = %0.2f Volts'%Vc\n", + "\n", + "Vce = Vcc-Ve#\n", + "print 'The Collector-Emmiter Voltage = %0.2f Volts'%Vce\n", + "\n", + "Icsat = Vcc/Re#\n", + "\n", + "Vceoff = Vcc#\n", + "\n", + "# Now, calculate AC quantities:\n", + "\n", + "a = 25.*10**-3#\n", + "\n", + "re = a/Ie#\n", + "print 'The AC emmiter resistance = %0.2f Ohms'%re\n", + "print 'Approx 4.52 Ohms'\n", + "\n", + "b = Re*Rl#\n", + "c = Re+Rl#\n", + "rl = b/c#\n", + "\n", + "Av = rl/(rl+re)#\n", + "print 'The Voltage gain =%0.2f'%Av\n", + "\n", + "Zinbase = B*(re+rl)#\n", + "print 'The Input Base Impedence = %0.2f Ohms'%Zinbase\n", + "print 'i.e 120.9 kOhms'\n", + "\n", + "d = 1./Zinbase#\n", + "e = 1./R1#\n", + "f = 1./R2#\n", + "\n", + "Zin = (d+e+f)**-1\n", + "print 'The Input Impedence = %0.2f Ohms'%Zin\n", + "print 'i.e 9.15 kOhms'\n", + "\n", + "vb = vin*(Zin/(Zin+Rg))#\n", + "print 'The AC base voltage = %0.2f Volts(p-p)'%vb\n", + "\n", + "vout = Av*vb#\n", + "print 'The AC output voltage = %0.2f Volts(p-p)'%vout\n", + "\n", + "Icq = Ic\n", + "Vceq = Vce\n", + "\n", + "Vce1=[Vcc, Vceq, 0]\n", + "Ic1=[0 ,Icq ,Icsat]\n", + "\n", + "#To plot DC load line\n", + "\n", + "print \"Q(%f,%f)\\n\"%(Vceq,Icq)\n", + "plot(Vce1, Ic1)\n", + "plot(Vceq,Icq)\n", + "plot(0,Icq)\n", + "plot(Vceq,0)\n", + "plot(0,Icsat)\n", + "plot(Vceoff,0)\n", + "xlabel(\"Vce in Volt\")\n", + "ylabel(\"Ic in mAmps\")\n", + "title(\"DC Load-line for Emitter Follower Circuit\")\n", + "show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 29_9 Page No. 963" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Emmiter current = 0.0046 Amps\n", + "i.e 4.61 mApms\n", + "The Collector-Base Voltage = 8.08 Volts\n", + "Approx 8.09 Volts\n", + "The AC emmiter resistance = 5.42 Ohms\n", + "The Voltage gain =138.33\n", + "The AC output voltage = 3.46 Volts(p-p)\n", + "Approx 3.46 Volts(p-p)\n", + "The Input Impedence = 5.41 Ohms\n" + ] + } + ], + "source": [ + "# Calculate the following: Ie, Vcb, r'\u0005e, Av, vout and zin.\n", + "\n", + "# Given data\n", + "\n", + "Rc = 1.5*10**3# # Collector resistance=1.5 kOhms\n", + "Re = 1.8*10**3# # Emitter resistance=1.8 kOhms\n", + "Rl = 1.5*10**3# # Load resistance=1.5 kOhms\n", + "Vcc = 15.# # +ve Supply Voltage=15 Volts\n", + "Vee = 9.# # -ve Supply Voltage=9 Volts\n", + "Vbe = 0.7# # Voltage Base-Emitter=0.7 Volts\n", + "vin = 25.*10**-3# # Input Voltage=25 mVolts(p-p)\n", + "\n", + "\n", + "Ie = (Vee-Vbe)/Re#\n", + "print 'The Emmiter current = %0.4f Amps'%Ie\n", + "print 'i.e 4.61 mApms'\n", + "\n", + "Ic = Ie# # Ic =~ Ie\n", + "\n", + "Vcb = Vcc-(Ic*Rc)#\n", + "print 'The Collector-Base Voltage = %0.2f Volts'%Vcb\n", + "print 'Approx 8.09 Volts'\n", + "\n", + "a = 25.*10**-3#\n", + "\n", + "re = a/Ie#\n", + "print 'The AC emmiter resistance = %0.2f Ohms'%re\n", + "\n", + "b = Rc*Rl#\n", + "c = Rc+Rl#\n", + "\n", + "rl = b/c#\n", + "\n", + "Av = rl/re#\n", + "print 'The Voltage gain =%0.2f'%Av\n", + "\n", + "vout = Av*vin#\n", + "print 'The AC output voltage = %0.2f Volts(p-p)'%vout\n", + "print 'Approx 3.46 Volts(p-p)'\n", + "\n", + "d = Re*re\n", + "e = Re+re\n", + "\n", + "Zin = d/e#\n", + "print 'The Input Impedence = %0.2f Ohms'%Zin" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 29_10 Page No. 968" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The AC output voltage = 1.45 Volts(p-p)\n" + ] + } + ], + "source": [ + "# Calculate the ac output voltage, vout.\n", + "\n", + "# Given data\n", + "\n", + "Rc = 1.2*10**3# # Collector resistance=1.2 kOhms\n", + "Re = 2.2*10**3# # Emitter resistance=2.2 kOhms\n", + "Rl = 3.3*10**3# # Load resistance=3.3 kOhms\n", + "Rg = 600.# # Generator Resistance=600 Ohms\n", + "Vcc = 12.# # +ve Supply Voltage=15 Volts\n", + "Vee = 12.# # -ve Supply Voltage=9 Volts\n", + "Vbe = 0.7# # Voltage Base-Emitter=0.7 Volts\n", + "vin = 1.# # Input Voltage=1 Volts(p-p)\n", + "\n", + "Ie = (Vee-Vbe)/Re#\n", + "\n", + "a = 25*10**-3#\n", + "re = a/Ie#\n", + "\n", + "b = Rc*Rl#\n", + "c = Rc+Rl#\n", + "rl = b/c#\n", + "\n", + "Av = rl/re#\n", + "\n", + "d = Re*re\n", + "e = Re+re\n", + "Zin = d/e#\n", + "\n", + "ve = vin*(Zin/(Zin+Rg))#\n", + "\n", + "vout = Av*ve#\n", + "print 'The AC output voltage = %0.2f Volts(p-p)'%vout" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter3.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter3.ipynb new file mode 100644 index 00000000..099cbe37 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter3.ipynb @@ -0,0 +1,636 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3 : Ohm's Law" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_1 Page No. 88" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current I = 15.00 Amps\n" + ] + } + ], + "source": [ + "# A heater with the resistance of 8 Ohms\u0003 is connected across the 120-V power line. How much is current I?\n", + "\n", + "# Given data\n", + "\n", + "V = 120# # Voltage of Power line=120 Volts\n", + "R = 8# # Heater Resistance=8 Ohms\n", + "\n", + "I = V/R#\n", + "print 'The Current I = %0.2f Amps'%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_2 Page No. 88" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current I = 0.05 Amps\n" + ] + } + ], + "source": [ + "# A small lightbulb with a resistance of 2400 Ohms\u0003 is connected across the 120-V power line. How much is current I?\n", + "\n", + "# Given data\n", + "\n", + "V = 120.# # Voltage of Power line=120 Volts\n", + "R = 2400# # Lightbulb Resistance=2400 Ohms\n", + "\n", + "I = V/R#\n", + "print 'The Current I = %0.2f Amps'%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_3 Page No. 88" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage = 30.00 Volts\n" + ] + } + ], + "source": [ + "# If a 12-Ohms\u0003 resistor is carrying a current of 2.5 A, how much is its voltage?\n", + "\n", + "# Given data\n", + "\n", + "I = 2.5# # Current=2.5 Amps\n", + "R = 12# # Resistance=12 Ohms\n", + "\n", + "V = I*R#\n", + "print 'The Voltage = %0.2f Volts'%V" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_4 Page No. 89" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistance = 75.00 ohms\n" + ] + } + ], + "source": [ + "# How much is the resistance of a lightbulb if it draws 0.16 A from a 12-V battery?\n", + "\n", + "# Given data\n", + "\n", + "V = 12# # Voltage of Battery=12 Volts\n", + "I = 0.16# # Current drawn form Battery=0.16 Amps\n", + "\n", + "R = V/I\n", + "print 'The Resistance = %0.2f ohms'%R" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_5 Page No. 89" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage = 40.00 Volts\n" + ] + } + ], + "source": [ + "# The I of 8 mA flows through a 5-kOhms Resistor. How much is the IR voltage?\n", + "\n", + "# Given data\n", + "\n", + "I = 8*10**-3# # Current flowing through Resistor=8m Amps\n", + "R = 5*10**3# # Resistance=5k Ohms\n", + "\n", + "V = I*R#\n", + "print 'The Voltage = %0.2f Volts'%V" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_6 Page No. 90" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current I = 0.005 Amps\n", + "i.e 5 mAmps\n" + ] + } + ], + "source": [ + "# How much current is produced by 60 V across 12 kOhms?\n", + "\n", + "# Given data\n", + "\n", + "V = 60.0# # Voltage=60 Volts\n", + "R = 12*10**3# # Resistance=12k Ohms\n", + "\n", + "I = V/R#\n", + "print 'The Current I = %0.3f Amps'%I\n", + "print 'i.e 5 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_7 Page No. 91" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Power used = 1200.00 Watts\n", + "OR 1.2 kW\n" + ] + } + ], + "source": [ + "# A toaster takes 10 A from the 120-V power line. How much power is used?\n", + "\n", + "# Given data\n", + "\n", + "V = 120# # Voltage of Power line=120 Volts\n", + "I = 10# # Current drawn from Powerline=10 Amps\n", + "\n", + "P = V*I#\n", + "print 'The Power used = %0.2f Watts'%P\n", + "print 'OR 1.2 kW'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_8 Page No. 91" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current I = 2.50 Amps\n" + ] + } + ], + "source": [ + "# How much current flows in the filament of a 300-W bulb connected to the 120-V power line?\n", + "\n", + "# Given Data\n", + "\n", + "V = 120.0# # Voltage of Power line=120 Volts\n", + "P = 300# # Power of Bulb=300 Watts\n", + "\n", + "I = P/V#\n", + "print 'The Current I = %0.2f Amps'%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_9 Page No. 91" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current I = 0.50 Amps\n", + "OR 500 mA\n" + ] + } + ], + "source": [ + "# How much current flows in the filament of a 60-W bulb connected to the 120-V power line?\n", + "\n", + "# Given Data\n", + "\n", + "V = 120# # Voltage of Power line=120 Volts\n", + "P = 60.0# # Power of Bulb=60 Watts\n", + "\n", + "I = P/V#\n", + "print 'The Current I = %0.2f Amps'%I\n", + "print 'OR 500 mA'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_10 Page No. 93" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Cost = 4.32 $\n" + ] + } + ], + "source": [ + "# Asuming that the cost of electricity is 6 cent per kWh, how much will it cost to light a 100-W lightbulb for 30 days?\n", + "\n", + "h = 24*30# # Total hours = 24 hrs * 30 days\n", + "\n", + "kWh = 0.1*h# # 100W=0.1kW\n", + "\n", + "Cost = kWh*0.06# # 6 cent = $0.06\n", + "\n", + "print 'Cost = %0.2f $'%Cost" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_11 Page No. 95" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Power = 200.00 Watts\n" + ] + } + ], + "source": [ + "# Calculate the power in a circuit where the source of 100 V produces 2 A in a 50 Ohms Resistor.\n", + "\n", + "# Given data\n", + "\n", + "I = 2# # Current=2 Amps\n", + "R = 50# # Resistance=50 Ohms\n", + "V = 100# # Voltage Source=100 Volts\n", + "\n", + "P = I*I*R#\n", + "print 'The Power = %0.2f Watts'%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_12 Page No. 95" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Power = 400.00 Watts\n" + ] + } + ], + "source": [ + " \n", + "# Calculate the power in a circuit where the source of 100 V produces 4 A in a 25 Ohms Resistor.\n", + "\n", + "# Given data\n", + "\n", + "I = 4# # Current=4 Amps\n", + "R = 25# # Resistance=25 Ohms\n", + "V = 100# # Voltage Source=100 Volts\n", + "\n", + "P = I*I*R#\n", + "print 'The Power = %0.2f Watts'%P" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_13 Page No. 96" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current I = 5.00 Amps\n" + ] + } + ], + "source": [ + "# How much current is needed for a 600-W, 120-V toaster?\n", + "\n", + "# Given data\n", + "\n", + "V = 120.0# # Applied Voltage=120 Volts\n", + "P = 600# # Power of toaster=600 Watts\n", + "\n", + "I = P/V#\n", + "print 'The Current I = %0.2f Amps'%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_14 Page No. 97" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistance = 24.00 Ohms\n" + ] + } + ], + "source": [ + "# How much is the resistance of a 600-W, 120-V toaster?\n", + "\n", + "# Given data\n", + "\n", + "V = 120.0# # Applied Voltage=120 Volts\n", + "P = 600# # Power of toaster=600 Watts\n", + "\n", + "R = (V*V)/P#\n", + "print 'The Resistance = %0.2f Ohms'%R" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_15 Page No. 97" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current I = 5.00 Amps\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "# How much current is needed for a 24 Ohms\u0003 Resistor that dissipates 600 W?\n", + "\n", + "# Given data\n", + "\n", + "R = 24.0# # Resistance=24 Ohms\n", + "P = 600.0# # Power=600 Watts\n", + "\n", + "I = sqrt(P/R)#\n", + "print 'The Current I = %0.2f Amps'%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_16 Page No. 98" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistor value = 1500.00 Ohms\n", + "i.e 1.5 kohms\n", + "The Power = 0.60 Watts\n", + "OR 600 mW\n" + ] + } + ], + "source": [ + "# Determine the required resistance and appropriate wattage rating of a resistor to meet the following requirements: The resistor must have a 30-V IR drop when its current is 20 mA. The resistors available have the following wattage ratings: 1⁄8, 1⁄4, 1⁄2, 1, and 2 W.\n", + "\n", + "# Given data\n", + "\n", + "I = 20.0*10**-3# # Current=20m Amps\n", + "V = 30.0# # Voltage Drop=30 Volts\n", + "\n", + "R = V/I#\n", + "print 'The Resistor value = %0.2f Ohms'%R\n", + "print 'i.e 1.5 kohms'\n", + "\n", + "P = I*I*R#\n", + "print 'The Power = %0.2f Watts'%P\n", + "print 'OR 600 mW'\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 3_17 Page No. 99" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistor value = 1500000 Ohms\n", + "i.e 1.5 Mohms\n", + "The Power = 0.03375 Watts\n", + "i.e 33.75 mW\n" + ] + } + ], + "source": [ + "#Determine the required resistance and appropriate wattage rating of a carbonfilm resistor to meet the following requirements: The resistor must have a 225-V IR drop when its current is 150 uA. The resistors available have the following wattage ratings: 1⁄8, 1⁄4, 1⁄2, 1, and 2 W.\n", + "\n", + "# Given data\n", + "\n", + "I = 150.0*10**-6# # Current=150 uAmps\n", + "V = 225# # Voltage Drop=225 Volts\n", + "\n", + "R = V/I#\n", + "print 'The Resistor value = %0.f Ohms'%R\n", + "print 'i.e 1.5 Mohms'\n", + "\n", + "P = I*I*R#\n", + "print 'The Power = %0.5f Watts'%P\n", + "print 'i.e 33.75 mW'" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter30.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter30.ipynb new file mode 100644 index 00000000..5b2dbf97 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter30.ipynb @@ -0,0 +1,578 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 30 : Field Effect Transistors" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 30_1 Page No. 984" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Id for Vgs is 0 Volts = 0.01 Amps\n", + "i.e 10 mAmps\n", + "The Value of Id for Vgs is -0.5 Volts = 0.0077 Amps\n", + "i.e 7.65 mAmps\n", + "The Value of Id for Vgs is -1 Volts = 0.0056 Amps\n", + "i.e 5.62 mAmps\n", + "The Value of Id for Vgs is -2 Volts = 0.0025 Amps\n", + "i.e 2.5 mAmps\n", + "The Value of Id for Vgs is -3 Volts = 0.0006 Amps\n", + "i.e 0.625 mAmps\n" + ] + } + ], + "source": [ + "# Determine Id for each value of Vgs (a) 0V# (b) -0.5V# (c) -1V (d) -2V (e) -3V\n", + "\n", + "# Given Data\n", + "\n", + "Vgs1 = 0# # Voltage Gate-Source 1=0 Volts\n", + "Vgs2 = -0.5# # Voltage Gate-Source 2=-0.5 Volts\n", + "Vgs3 = -1.# # Voltage Gate-Source 3=-1 Volts\n", + "Vgs4 = -2.# # Voltage Gate-Source 4=-2 Volts\n", + "Vgs5 = -3.# # Voltage Gate-Source 5=-3 Volts\n", + "Vgsoff = -4.# # Voltage Gate-Source(off)=-4 Volts\n", + "Idss = 10.*10**-3 # Idss = 10m Amps\n", + "\n", + "a = (1-(Vgs1/Vgsoff))\n", + "b = (1-(Vgs2/Vgsoff))\n", + "c = (1-(Vgs3/Vgsoff))\n", + "d = (1-(Vgs4/Vgsoff))\n", + "e = (1-(Vgs5/Vgsoff))\n", + "\n", + "# Vgs = 0 Volts\n", + "\n", + "Id1 = Idss*a*a\n", + "print 'The Value of Id for Vgs is 0 Volts = %0.2f Amps'%Id1\n", + "print 'i.e 10 mAmps'\n", + "\n", + "# Vgs = -0.5 Volts\n", + "\n", + "Id2 = Idss*b*b\n", + "print 'The Value of Id for Vgs is -0.5 Volts = %0.4f Amps'%Id2\n", + "print 'i.e 7.65 mAmps'\n", + "\n", + "# Vgs = -1 Volts\n", + "\n", + "Id3 = Idss*c*c\n", + "print 'The Value of Id for Vgs is -1 Volts = %0.4f Amps'%Id3\n", + "print 'i.e 5.62 mAmps'\n", + "\n", + "# Vgs = -2 Volts\n", + "\n", + "Id4 = Idss*d*d\n", + "print 'The Value of Id for Vgs is -2 Volts = %0.4f Amps'%Id4\n", + "print 'i.e 2.5 mAmps'\n", + "\n", + "# Vgs = -3 Volts\n", + "\n", + "Id5 = Idss*e*e\n", + "print 'The Value of Id for Vgs is -3 Volts = %0.4f Amps'%Id5\n", + "print 'i.e 0.625 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 30_2 Page No. 985" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Id = 1.2500e-04 Amps using Minimum Values\n", + "i.e 125 uAmps\n", + "The Value of Vds = 1.88 Volts using Minimum Values\n", + "The Value of Id = 0.0132 Amps using Maximum Values\n", + "i.e 13.2 mAmps\n", + "The Value of Vds = -11.20 Volts using Maximun Values\n", + "The Value of Vds(p) = 6.50 Volts using Maximun Values\n" + ] + } + ], + "source": [ + "# Find the minimim and maximum value of Id and Vds if Vgs=-1.5 Volts\n", + "\n", + "# Given Data\n", + "\n", + "Idssmin = 2.*10**-3# # Idss(min)=2m Amp\n", + "Idssmax = 20.*10**-3# # Idss(max)=20m Amp\n", + "Vgs = -1.5# # Voltage Gate-Source=-1.5V\n", + "Vgsoffmin = -2.# # Voltage Gate-Source(off)(min)=-2 Volts\n", + "Vgsoffmax = -8.# # Voltage Gate-Source(off)(max)=-8 Volts\n", + "Vdd = 2.0# # Supply Voltage(Drain)=20 Volts\n", + "Rd = 1.*10**3# # Drain Resistance=1k Ohms\n", + "\n", + "a = 1-(Vgs/Vgsoffmin)#\n", + "b = 1-(Vgs/Vgsoffmax)#\n", + "\n", + "# Calculation using Minimum Values\n", + "\n", + "Id1 = Idssmin*a*a#\n", + "print 'The Value of Id = %0.4e Amps using Minimum Values'%Id1\n", + "print 'i.e 125 uAmps'\n", + "\n", + "Vds1 = Vdd-Id1*Rd#\n", + "print 'The Value of Vds = %0.2f Volts using Minimum Values'%Vds1\n", + "\n", + "# Calculation using Maximum Values\n", + "\n", + "Id2 = Idssmax*b*b#\n", + "print 'The Value of Id = %0.4f Amps using Maximum Values'%Id2\n", + "print 'i.e 13.2 mAmps'\n", + "\n", + "Vds2 = Vdd-Id2*Rd#\n", + "print 'The Value of Vds = %0.2f Volts using Maximun Values'%Vds2\n", + "\n", + "Vp = -Vgsoffmax#\n", + "\n", + "Vdsp = Vp+Vgs#\n", + "print 'The Value of Vds(p) = %0.2f Volts using Maximun Values'%Vdsp" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 30_3 Page No. 989" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Drain Voltage Vd = 5.00 Volts\n" + ] + } + ], + "source": [ + "# Calculate the value of Vd\n", + "\n", + "# Given Data\n", + "\n", + "Vs = 1.# # Voltage at Resistor Rs=1 Volts\n", + "Rs = 200.# # Source Resistor=200 Ohms\n", + "Vdd = 10.# # Supply Voltage(Drain)=10 Volts\n", + "Rd = 1.*10**3# # Drain Resistor=1k Ohms\n", + "\n", + "Is=Vs/Rs#\n", + "\n", + "Id = Is#\n", + "\n", + "Vd = Vdd-Id*Rd#\n", + "print 'The Drain Voltage Vd = %0.2f Volts'%Vd," + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 30_4 Page No. 991" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Vg = 3.06 Volts\n", + "i.e 3 Volts\n", + "The Value of Vs = 4.06 Volts\n", + "i.e 4 Volts\n", + "The Value of Id = 5.08e-03 Amps.\n", + "i.e 5 mAmps\n", + "The Value of Vd = 9.92 Volts\n", + "Approx 10 Volts\n" + ] + } + ], + "source": [ + "# Calculate Vg, Vs, Id, Vd.\n", + "\n", + "# Given Data\n", + "\n", + "R1 = 390.*10**3# # Resistor 1=390k Ohms\n", + "R2 = 100.*10**3# # Resistor 2=100k Ohms\n", + "Rd = 1.*10**3# # Drain Resistor=1k Ohms\n", + "Vdd = 15.# # Supply Voltage(Drain)=15 Volts\n", + "Vgs = -1.# # Voltage Gate-Source=-1 Volts\n", + "Rs = 800.# # Source Resistor=800 Ohms\n", + "\n", + "Vg = (R2/(R1+R2))*Vdd#\n", + "print 'The Value of Vg = %0.2f Volts'%Vg\n", + "print 'i.e 3 Volts'\n", + "\n", + "Vs = Vg-Vgs#\n", + "print 'The Value of Vs = %0.2f Volts'%Vs\n", + "print 'i.e 4 Volts'\n", + "\n", + "Id = Vs/Rs#\n", + "print 'The Value of Id = %0.2e Amps.'%Id\n", + "print 'i.e 5 mAmps'\n", + "\n", + "Vd = Vdd-Id*Rd\n", + "print 'The Value of Vd = %0.2f Volts'%Vd\n", + "print 'Approx 10 Volts'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 30_5 Page No. 992" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Drain Current Id = 6.50e-03 Amps\n", + "i.e 6.5 mAmps\n", + "The Drain Voltage Vd = 8.50 Voltage\n" + ] + } + ], + "source": [ + "# Calculate the value Drain Current Id and Drain Voltage Vd.\n", + "\n", + "# Given Data\n", + "\n", + "Vdd = 15# # Supply Voltage(Drain)=15 Volts\n", + "Vbe = 0.7# # Voltage Base-Emitter=0.7 Volts\n", + "Re = 2.2*10**3# # Emitter Resistor=2.2 kOhms\n", + "Rd = 1*10**3# # Drain Resistor=1 kOhms\n", + "Vee = 15# # Supply Voltage(Emitter)=15 Volts\n", + "\n", + "\n", + "Ic = (Vee-Vbe)/Re#\n", + "\n", + "Id = Ic#\n", + "print 'The Drain Current Id = %0.2e Amps'%Id\n", + "print 'i.e 6.5 mAmps'\n", + "\n", + "Vd = Vdd-Id*Rd#\n", + "print 'The Drain Voltage Vd = %0.2f Voltage'%Vd" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 30_6 Page No. 997" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage Gain Av =4.89\n", + "Approx 4.875\n", + "The Output Voltage Vo = 0.978 Volts(p-p)\n" + ] + } + ], + "source": [ + "# Calculate the Voltage Gain Av and Output Voltage Vo\n", + "\n", + "# Given Data\n", + "\n", + "Rd = 1.5*10**3# # Drain Resistor=1.5 kOhms\n", + "Rl = 10*10**3# # Load Resistor=10 kOhms\n", + "Idss = 10*10**-3# # Idss=10 mAmps\n", + "Vgs = -1# # Voltage Gate-Source=-1 Volts\n", + "Vgsoff = -4.# # Voltage Gate-Source(off)=-4 Volts\n", + "Vin = 0.2# # Input Voltage=0.2 Volts(p-p)\n", + "\n", + "gmo = 2*Idss/(-Vgsoff)#\n", + "\n", + "gm = gmo*(1-(Vgs/Vgsoff))#\n", + "\n", + "rl = (Rd*Rl)/(Rd+Rl)#\n", + "\n", + "Av = gm*rl#\n", + "print 'The Voltage Gain Av =%0.2f'%Av\n", + "print 'Approx 4.875'\n", + "\n", + "Vo = Av*Vin\n", + "print 'The Output Voltage Vo = %0.3f Volts(p-p)'%Vo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 30_7 Page No. 998" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage Gain Av =0.37\n", + "The Output Voltage Vo = 0.37 Volts(p-p)\n", + "The Output Impedence Zo = 143.28 Ohms\n", + "Approx 143.5 Ohms\n" + ] + } + ], + "source": [ + "# Calculate Av, Vo & Zo.\n", + "\n", + "# Given Data\n", + "\n", + "Rs = 240.# # Source Resistor=240 Ohms\n", + "Rl = 1.8*10**3# # Load Resistor=1.8 kOhms\n", + "Vgsoff = -8.# # Voltage Gate-Source(off)=-8 Volts\n", + "Vgs = -2.# # Voltage Gate-Source=-2 Volts\n", + "Idss = 15.*10**-3 # Idss=15 mAmps.\n", + "Vin = 1.# # Input Voltage=1 Volts(p-p)\n", + "\n", + "rl = ((Rs*Rl)/(Rs+Rl))#\n", + "gmo = 2*Idss/-Vgsoff#\n", + "gm = gmo*(1-(Vgs/Vgsoff))#\n", + "\n", + "Av = gm*rl/(1+gm*rl)#\n", + "print 'The Voltage Gain Av =%0.2f'%Av\n", + "\n", + "Vo = Av*Vin#\n", + "print 'The Output Voltage Vo = %0.2f Volts(p-p)'%Vo\n", + "\n", + "A = (1/gm)#\n", + "Zo = ((Rs*A)/(Rs+A))#\n", + "print 'The Output Impedence Zo = %0.2f Ohms'%Zo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 30_8 Page No. 1000" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage Gain Av =4.17\n", + "The Output Voltage = 4.17e-02 Volts(p-p)\n", + "Approx 41.6 mVolts(p-p)\n", + "The Output Impedence Zi = 114.29 Ohms\n", + "Approx 114 Ohms\n" + ] + } + ], + "source": [ + "#Calculate Av, Vo, Zin.\n", + "\n", + "# Given Data\n", + "\n", + "Rd = 1.2*10**3# # Drain Resistor=1.2 kOhms\n", + "Rl = 15.*10**3# # Load Resistor=15 kOhms\n", + "gm = 3.75*10**-3# # Transconductance=3.75 mSiemens\n", + "Vin = 10.*10**-3# # Input Voltage=10 mVpp\n", + "Rs = 200.# # Source Resistor=200 Ohms\n", + "\n", + "rl = ((Rd*Rl)/(Rd+Rl))#\n", + "\n", + "Av = gm*rl#\n", + "print 'The Voltage Gain Av =%0.2f'%Av\n", + "\n", + "Vo = Av*Vin#\n", + "print 'The Output Voltage = %0.2e Volts(p-p)'%Vo\n", + "print 'Approx 41.6 mVolts(p-p)'\n", + "\n", + "A = (1/gm)#\n", + "\n", + "Zi = ((Rs*A)/(Rs+A))#\n", + "print 'The Output Impedence Zi = %0.2f Ohms'%Zi\n", + "print 'Approx 114 Ohms'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 30_9 Page No. 1001" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Id for Vgs is 2 Volts = 2.25e-02 Amps\n", + "i.e 22.5 mAmps\n", + "The Value of Id for Vgs is -2 Volts = 2.50e-03 Amps\n", + "i.e 2.5 mAmps\n", + "The Value of Id for Vgs is 0 Volts = 1.00e-02 Amps\n", + "i.e 10 mAmps\n" + ] + } + ], + "source": [ + "#Determine Id for each value of Vgs (a) 2V# (b) -2V# (c) 0V\n", + "\n", + "# Given Data\n", + "Vgs1 = 2.# # Voltage Gate-Source 1=2 Volts\n", + "Vgs2 = -2.# # Voltage Gate-Source 2=-2 Volts\n", + "Vgs3 = 0# # Voltage Gate-Source 3=0 Volts\n", + "Vgsoff = -4.# # Voltage Gate-Source(off)=-4 Volts\n", + "Idss = 10.*10**-3# # Idss = 10m Amps\n", + "\n", + "a = (1-(Vgs1/Vgsoff))#\n", + "b = (1-(Vgs2/Vgsoff))#\n", + "c = (1-(Vgs3/Vgsoff))#\n", + "\n", + "# Vgs = 2 Volts\n", + "\n", + "Id1 = Idss*a*a#\n", + "print 'The Value of Id for Vgs is 2 Volts = %0.2e Amps'%Id1\n", + "print 'i.e 22.5 mAmps'\n", + "\n", + "# Vgs = -2 Volts\n", + "\n", + "Id2 = Idss*b*b#\n", + "print 'The Value of Id for Vgs is -2 Volts = %0.2e Amps'%Id2\n", + "print 'i.e 2.5 mAmps'\n", + "\n", + "# Vgs = 0 Volts\n", + "\n", + "Id3 = Idss*c*c#\n", + "print 'The Value of Id for Vgs is 0 Volts = %0.2e Amps'%Id3\n", + "print 'i.e 10 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 30_10 Page No. 1002" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Drain Resistance = 500.00 Ohms\n", + "A 470 Ohms resistor would provide the proper biasing voltage at the gate\n" + ] + } + ], + "source": [ + "# Calculate the value of Rd to provide an Id(on) of 10m Amps.\n", + "\n", + "# Given Data\n", + "\n", + "Vdd = 15.# # Suppy Voltage(Drain)=15 Volts\n", + "Vgson = 10.# # Voltage Gate-Source(on)=10 Volts\n", + "Idon = 10.*10**-3# # Drain Current(on)=10m Amps\n", + "\n", + "Rd = (Vdd-Vgson)/Idon#\n", + "print 'The Drain Resistance = %0.2f Ohms'%Rd\n", + "\n", + "print 'A 470 Ohms resistor would provide the proper biasing voltage at the gate'" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter31.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter31.ipynb new file mode 100644 index 00000000..6cbfbe90 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter31.ipynb @@ -0,0 +1,472 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 31 : Power Amplifiers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 31_1 Page No. 1019" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of Icq = 7.95e-03 Amps\n", + "i.e Approx 7.91 mAmps\n", + "The value of Vceq = 10.14 Volts\n", + "The Power Dissipation = 8.06e-02 Watts\n", + "i.e 80.6 mWatts\n", + "The value of Ic(sat) = 1.61e-02 Amps\n", + "i.e 16.1 mAmps\n", + "The value of Vce(off) = 20.00 Volts\n", + "Q(10.138406,0.007953)\n", + "\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f383ca8da10>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "from matplotlib.pyplot import plot,show,title,xlabel,ylabel\n", + "# Calculate the following dc quantities Icq, Vceq, Pd, Ic(sat) and Vce(off). Also draw the dc load line\n", + "\n", + "# Given Data\n", + "\n", + "R1 = 18.*10**3# # Resistor 1=18k Ohms\n", + "R2 = 2.7*10**3# # Resistor 2=2.7k Ohms\n", + "Vcc = 20.# # Supply Voltage(Collector)=20 Volts\n", + "Vbe = 0.7# # Voltage Base-Emitter=0.7 Volts\n", + "Re = 240.# # Emitter Resistor=240 Ohms\n", + "Rc = 1.*10**3# # Collector Resistor=1k Ohms\n", + "\n", + "Vb = Vcc*(R2/(R1+R2))#\n", + "\n", + "Ve = Vb-Vbe#\n", + "\n", + "#Ie = Ic#\n", + "\n", + "Icq = Ve/Re#\n", + "print 'The value of Icq = %0.2e Amps'%Icq\n", + "print 'i.e Approx 7.91 mAmps'\n", + "\n", + "Vceq = Vcc-Icq*(Rc+Re)#\n", + "print 'The value of Vceq = %0.2f Volts'%Vceq\n", + "\n", + "Pd = Vceq*Icq#\n", + "print 'The Power Dissipation = %0.2e Watts'%Pd\n", + "print 'i.e 80.6 mWatts'\n", + "\n", + "Icsat = Vcc/(Rc+Re)#\n", + "print 'The value of Ic(sat) = %0.2e Amps'%Icsat\n", + "print 'i.e 16.1 mAmps'\n", + "\n", + "Vceoff = Vcc#\n", + "print 'The value of Vce(off) = %0.2f Volts'%Vceoff\n", + "\n", + "# For DC load line\n", + "\n", + "Vce1=[Vceoff ,Vceq, 0]\n", + "Ic1=[0, Icq, Icsat]\n", + "\n", + "#To plot DC load line\n", + "\n", + "print \"Q(%f,%f)\\n\"%(Vceq,Icq)\n", + "plot(Vce1, Ic1)\n", + "plot(Vceq,Icq)\n", + "plot(0,Icq)\n", + "plot(Vceq,0)\n", + "plot(0,Icsat)\n", + "plot(Vceoff,0)\n", + "xlabel(\"Vce in volt\")\n", + "ylabel(\"Ic in Ampere\")\n", + "title(\"DC Load-line for Common-Emitter Class A Amplifier Circuit\")\n", + "show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 31_2 Page No. 1024" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage Gain Av =189.84\n", + "Approx 190\n", + "The Output Voltage = 4.75 Volts\n", + "The Load Power = 1.88e-03 Watts\n", + "i.e Approx 1.88 mWatts\n", + "The Dc Input Power = 1.78e-01 Watts\n", + "i.e Approx 177.4 mWatts\n", + "The Efficiency in % =1.06\n", + "Approx 1%\n", + "The Y-axis Value of AC Load-line is ic(sat) = 2.49e-02 Amps\n", + "i.e 24.89 mAmps\n", + "The X-axis value of AC Load-line is vce(off) = 14.94 Volts\n", + "Q(10.190000,0.007910)\n", + "\n" + ] + }, + { + "data": { + "image/png": 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8hAwPu7vPrc8Bm4oKE2kqGzfC+PFhed68MLy9SDHKazWXu89JvQFPABpoQiTq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\nvjUubwVKGitgEZFCc+SRsHp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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f383c6dc490>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "from matplotlib.pyplot import plot,show,title,xlabel,ylabel\n", + "\n", + "# Claculate the following AC quantities Av, Vout, Pl, Pcc and percent efficiency. Also calculate the endpoints of ac loadline\n", + "\n", + "# Given data\n", + "\n", + "Icq = 7.91*10**-3# # Collector Currect(Q-point)=7.91 mAmps\n", + "Rl = 1.5*10**3# # Load Resistor=1.5 kOhms\n", + "Rc = 1.*10**3# # Collector Resistor=1 kOhms\n", + "Vin = 25.*10**-3# # Input Voltage=25 mVolts(p-p)\n", + "R1 = 18.*10**3# # Resistor 1=18 kOhms\n", + "R2 = 2.7*10**3# # Resistor 2=2.7 kOhms\n", + "Vcc = 20.# # Supply Voltage(Collector)=20 Volts\n", + "Vceq = 10.19# # Voltage Colector-Emitter(Q-point)=10.19 Volts\n", + "\n", + "rc = (25.*10**-3)/Icq#\n", + "rl = (Rc*Rl)/(Rc+Rl)\n", + "\n", + "Av = rl/rc#\n", + "print 'The Voltage Gain Av =%0.2f'%Av\n", + "print 'Approx 190'\n", + "\n", + "Vout = Av*Vin#\n", + "print 'The Output Voltage = %0.2f Volts'%Vout\n", + "\n", + "Pl = (Vout*Vout)/(8*Rl)#\n", + "print 'The Load Power = %0.2e Watts'%Pl\n", + "print 'i.e Approx 1.88 mWatts'\n", + "\n", + "Ivd = Vcc/(R1+R2)#\n", + "# Ic = Icq\n", + "Icc = Ivd+Icq#\n", + "\n", + "Pcc = Vcc*Icc#\n", + "print 'The Dc Input Power = %0.2e Watts'%Pcc\n", + "print 'i.e Approx 177.4 mWatts'\n", + "\n", + "efficiency = ((Pl/Pcc)*100)#\n", + "print 'The Efficiency in %% =%0.2f'%efficiency\n", + "print 'Approx 1%'\n", + "\n", + "# Endpoints of AC load line\n", + "\n", + "icsat = Icq+(Vceq/rl)#\n", + "print 'The Y-axis Value of AC Load-line is ic(sat) = %0.2e Amps'%icsat\n", + "print 'i.e 24.89 mAmps'\n", + "\n", + "vceoff = Vceq+Icq*rl#\n", + "print 'The X-axis value of AC Load-line is vce(off) = %0.2f Volts'%vceoff\n", + "\n", + "# For AC load line\n", + "\n", + "Vce1=[vceoff, Vceq, 0]\n", + "Ic1=[0 ,Icq ,icsat]\n", + "\n", + "#To plot AC load line\n", + "\n", + "print \"Q(%f,%f)\\n\"%(Vceq,Icq)\n", + "plot(Vce1, Ic1)\n", + "plot(Vceq,Icq)\n", + "plot(0,Icq)\n", + "plot(Vceq,0)\n", + "plot(0,icsat)\n", + "plot(vceoff,0)\n", + "xlabel(\"Vce in volt\")\n", + "ylabel(\"Ic in Ampere\")\n", + "title(\"AC Load-line for Common-Emitter Class A Amplifier Circuit\")\n", + "show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 31_3 Page No. 1025" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Load Power = 6.25 Watts\n", + "The DC Input Power = 9.54 Watts\n", + "The Efficiency in % =65.51\n", + "The Maximum Power Dissipation = 1.80 Watts\n" + ] + } + ], + "source": [ + "#Calculate the following quantities: Pl, Pcc, Pdmax & percent efficiency\n", + "\n", + "# Given data\n", + "\n", + "Vin = 20.# # Input Voltage=20 Volts(p-p)\n", + "Vopp = 20.# # Output Voltage(p-p)=20 Volts(p-p)\n", + "Vcc = 24.# # Supply Voltage(Collector)=24 Volts\n", + "Vop = 10.# # Output Voltage(peak)=10 Volts\n", + "Rl = 8.# # Load Resistor=8 Ohms\n", + "\n", + "Vopp1 = Vopp*Vopp#\n", + "Pl = (Vopp1/(8*Rl))#\n", + "print 'The Load Power = %0.2f Watts'%Pl\n", + "\n", + "Icc = ((Vop/Rl)*0.318)#\n", + "\n", + "Pcc = Vcc*Icc\n", + "print 'The DC Input Power = %0.2f Watts'%Pcc\n", + "\n", + "eff = ((Pl/Pcc)*100)#\n", + "print 'The Efficiency in %% =%0.2f'%eff\n", + "\n", + "Pd = (Vcc*Vcc)/(40*Rl)#\n", + "print 'The Maximum Power Dissipation = %0.2f Watts'%Pd" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 31_4 Page No. 1037" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Load Power = 39.00 Watts\n", + "The DC Input Power = 57.24 Watts\n", + "The Efficiency in % = 68.13\n" + ] + } + ], + "source": [ + "# Calculate the following quantities Pl, Pcc & percent efficiency\n", + "\n", + "# Given data\n", + "\n", + "Rl = 8# # Load Resistor=8 Ohms\n", + "Vopp = 50# # Output Voltage(p-p)=50 Volts(p-p)\n", + "Vcc = 30# # Supply Voltage(Collector)=30 Volts\n", + "Vopk = Vopp/2# # Output Voltage(peak)\n", + "\n", + "Pl = (Vopp*Vopp)/(8*Rl)#\n", + "print 'The Load Power = %0.2f Watts'%Pl\n", + "\n", + "Pcc = Vcc*0.636*(Vopk/Rl)#\n", + "print 'The DC Input Power = %0.2f Watts'%Pcc\n", + "\n", + "efficiency = ((Pl/Pcc)*100)#\n", + "print 'The Efficiency in %% = %0.2f'%efficiency" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 31_5 Page No. 1038" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resonant Frequency = 2.00e+06 Hertz\n", + "i.e 2 MHz\n", + "The DC Bias Voltage at Base = 0.80 Volts\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "# Calculate the fr of LC tank circuit and dc bias voltage at base\n", + "\n", + "# Given data\n", + "\n", + "L = 100*10**-6# # Inductor=100 uHenry\n", + "C = 63.325*10**-12# # Capacitor=63.325 pFarad\n", + "Vin = 1.5# # Input Voltage(peak)=1.5 Volts\n", + "Vbe = 0.7# # Voltage Base-Emitter=0.7 Volts\n", + "\n", + "A = sqrt(L*C)#\n", + "fr = 1./(2*3.14*A)#\n", + "print 'The Resonant Frequency = %0.2e Hertz'%fr\n", + "print 'i.e 2 MHz'\n", + "\n", + "Vdc = (Vin-Vbe)#\n", + "print 'The DC Bias Voltage at Base = %0.2f Volts'%Vdc" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 31_6 Page No. 1039" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Minimum Base Reisitance Rb to Provide Clamping Action = 500 Ohms\n" + ] + } + ], + "source": [ + "# Calculate the minimum base reisitance Rb, necessary to provide clamping action\n", + "\n", + "# Given data\n", + "\n", + "C = 0.01*10**-6# # Capacitor=0.01 uFarad\n", + "fr = 2.*10**6# # Resonant Frequency=2 MHertz\n", + "\n", + "fin = fr\n", + "T = 1/fin\n", + "\n", + "Rb = 10*T/C\n", + "print 'The Minimum Base Reisitance Rb to Provide Clamping Action = %0.f Ohms'%Rb" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 31_7 Page No. 1040" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Bandwidth = 45120 Hertz\n", + "i.e Approx 45 kHz\n" + ] + } + ], + "source": [ + "# Calculate the Bandwidth\n", + "\n", + "# Given data\n", + "\n", + "L = 100*10**-6# # Inductor=100 uHenry\n", + "fr = 2*10**6# # Resonant Frequency=2 MHertz\n", + "ri = 12.56# # Resistance of Coil=12.56 Ohms\n", + "Rp = 100*10**3# # Rp=100 kOhms\n", + "\n", + "Xl = 2*3.14*fr*L#\n", + "Qcoil = Xl/ri#\n", + "Ztank = Qcoil*Xl#\n", + "\n", + "A = Ztank#\n", + "B = Rp#\n", + "C = (A*B)/(A+B)#\n", + "Qckt = C/Xl#\n", + "\n", + "BW = fr/Qckt#\n", + "print 'The Bandwidth = %0.f Hertz'%BW\n", + "print 'i.e Approx 45 kHz'" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter32.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter32.ipynb new file mode 100644 index 00000000..bc93187e --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter32.ipynb @@ -0,0 +1,71 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 32 : Thyristors" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 32_1 Page No. 1058" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Frequency of the Emmiter Voltage Waveform = 49.61 Hertz\n" + ] + } + ], + "source": [ + "from math import log\n", + "# Calculate the frequency of the emmiter voltage waveform. Assume n=0.6\n", + "\n", + "# Given data\n", + "\n", + "Rt = 220*10**3# # Resistor Rt=220k Ohms\n", + "Ct = 0.1*10**-6# # Capacitor Ct=0.1u Farad\n", + "n = 0.6# # Constant\n", + "\n", + "A = 1./(1-n)#\n", + "T = Rt*Ct*log(A)#\n", + "\n", + "f = 1./T#\n", + "print 'The Frequency of the Emmiter Voltage Waveform = %0.2f Hertz'%f" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter33.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter33.ipynb new file mode 100644 index 00000000..c663815f --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter33.ipynb @@ -0,0 +1,1000 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 33 : Operational Amplifiers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_1 Page No. 1072" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Differential Voltage Gain =143.00\n", + "The Ac Output Voltage = 1.43 Volts(p-p)\n" + ] + } + ], + "source": [ + "# Calculate the differential voltage gain, Ad, and the ac output voltage, Vout.\n", + "\n", + "# Given data\n", + "\n", + "Vin = 10*10**-3# # Input voltage=10 mVolts(p-p)\n", + "Rc = 10*10**3# # Collector resistance=10 kOhms\n", + "Ie = 715.*10**-6# # Emitter current=715 uAmps\n", + "\n", + "re = (25*10**-3)/Ie#\n", + "\n", + "Ad = Rc/(2*re)#\n", + "print 'The Differential Voltage Gain =%0.2f'%Ad\n", + "\n", + "Av = Ad\n", + "\n", + "Vo = Av*Vin#\n", + "print 'The Ac Output Voltage = %0.2f Volts(p-p)'%Vo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_2 Page No. 1073" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Common-Mode Voltage Gain Acm = 0.50\n", + "The Commom-Mode Rejection Ratio = 49.12 dB\n" + ] + } + ], + "source": [ + "from math import log10\n", + "# calculate the common-mode voltage gain, ACM, and the CMRR (dB).\n", + "\n", + "# Given data\n", + "\n", + "Rc = 10*10**3# # Collector resistance=10 kOhms\n", + "Re = 10.*10**3# # Emitter resistance=10 kOhms\n", + "Ad = 142.86# # Differential gain=142.86\n", + "\n", + "Acm = Rc/(2*Re)#\n", + "print 'The Common-Mode Voltage Gain Acm = %0.2f'%Acm\n", + "\n", + "CMRR = 20*log10(Ad/Acm)#\n", + "print 'The Commom-Mode Rejection Ratio = %0.2f dB'%CMRR" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_3 Page No. 1074" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Output Frequency = 7.96e+04 Hertz\n", + "i.e 79.6 kHz\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# Calculate fmax for an op amp that has an Sr of 5 V/u\u0002s and a peak output voltage of 10 V.\n", + "\n", + "# Given data\n", + "\n", + "Vpk = 10.# # Peak output voltage=10 Volts\n", + "Sr = 5./10**-6# # Slew rate=5 V/us\n", + "\n", + "\n", + "fo = Sr/(2*pi*Vpk)#\n", + "print 'The Output Frequency = %0.2e Hertz'%fo\n", + "print 'i.e 79.6 kHz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_4 Page No. 1075" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Closed-Loop Voltage Gain Acl =-10.00\n", + "The Output Voltage = 10.00 Volts(p-p)\n", + "The -ve sign indicates that input and output voltages are 180° out-of-phase\n" + ] + } + ], + "source": [ + "# calculate the closed-loop voltage gain, Acl, and the output voltage, Vout.\n", + "\n", + "# Given data\n", + "\n", + "Vin = 1.# # Input voltage=1 Volts(p-p)\n", + "Rf = 10.*10**3# # Feedback resistance=10 kOhms\n", + "Ri = 1.*10**3# # Input resistance=1 kOhms\n", + "\n", + "Acl = -(Rf/Ri)#\n", + "print 'The Closed-Loop Voltage Gain Acl =%0.2f'%Acl\n", + "\n", + "Vo = -Vin*Acl#\n", + "print 'The Output Voltage = %0.2f Volts(p-p)'%Vo\n", + "print 'The -ve sign indicates that input and output voltages are 180° out-of-phase'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_5 Page No. 1076" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Differential Input Voltage = 1.00e-04 Volts(p-p)\n", + "i.e 100 uVolts(p-p)\n" + ] + } + ], + "source": [ + "#If Avol equals 100,000, calculate the value of Vid.\n", + "\n", + "# Given data\n", + "\n", + "Avol = 100000.# # Open loop voltage gain=100,000\n", + "Vo = 10.# # Output voltage=10 Volts(p-p)\n", + "\n", + "Vid = Vo/Avol#\n", + "print 'The Differential Input Voltage = %0.2e Volts(p-p)'%Vid\n", + "print 'i.e 100 uVolts(p-p)'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_6 Page No. 1078" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Input Impedence = 1.00e+03 Ohms\n", + "i.e 1 kOhms\n", + "The Closed Loop Output Impedence = 0.01 Ohms\n" + ] + } + ], + "source": [ + "# calculate Zin and Zout(CL). Assume AVOL is\u0004 100,000 and Zout(OL) is\u0004 75 Ohms.\n", + "\n", + "# Given data\n", + "\n", + "Avol = 100000.# # Open loop voltage gain=100,000\n", + "Rf = 10.*10**3# # Feedback resistance=10 kOhms\n", + "Ri = 1.*10**3# # Input resistance=1 kOhms\n", + "Zool = 75.# # Output impedence (open-loop)=75 Ohms\n", + "\n", + "Zi = Ri#\n", + "print 'The Input Impedence = %0.2e Ohms'%Zi\n", + "print 'i.e 1 kOhms'\n", + "\n", + "Beta = Ri/(Ri+Rf)#\n", + "\n", + "A = Avol*Beta#\n", + "\n", + "Zocl = Zool/(1+A)#\n", + "print 'The Closed Loop Output Impedence = %0.2f Ohms'%Zocl" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_7 Page No. 1083" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Output Frequency = 1.59e+04 Hertz\n", + "i.e 15.915 kHz\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# Calculate the 5-V power bandwidth.\n", + "\n", + "# Given data\n", + "\n", + "Vo = 10.# # Output voltage=10 Volts(p-p)\n", + "Sr = 0.5/10**-6# # Slew rate=0.5 V/us\n", + "\n", + "Vpk = Vo/2#\n", + "\n", + "fo = Sr/(2*pi*Vpk)#\n", + "print 'The Output Frequency = %0.2e Hertz'%fo\n", + "print 'i.e 15.915 kHz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_8 Page No. 1085" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Closed-Loop Voltage Gain Acl =11.00\n", + "The Output Voltage = 11.00 Volts(p-p)\n" + ] + } + ], + "source": [ + "# Calculate the closed-loop voltage gain, Acl, and the output voltage, Vout.\n", + "\n", + "# Given data\n", + "\n", + "Vin = 1# # Input voltage=1 Volts(p-p)\n", + "Rf = 10.*10**3# # Feedback resistance=10 kOhms\n", + "Ri = 1.*10**3# # Input resistance=1 kOhms\n", + "\n", + "Acl = 1+(Rf/Ri)#\n", + "print 'The Closed-Loop Voltage Gain Acl =%0.2f'%Acl\n", + "\n", + "Vo = Vin*Acl#\n", + "print 'The Output Voltage = %0.2f Volts(p-p)'%Vo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_9 Page No. 1089" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Input Impedence Closed-Loop = 1.82e+10 Ohms\n", + "i.e 18 GOhms\n", + "The Closed-Loop Output Impedence = 0.01 Ohms\n" + ] + } + ], + "source": [ + "# Calculate Zin(CL) and Zout(CL). Assume Rin is\u0004 2 MOhms\u0006, Avol is 100,000, and Zout(OL) is 75 Ohms.\n", + "\n", + "# Given data\n", + "\n", + "Avol = 100000.# # Open loop voltage gain=100,000\n", + "Ri = 2.*10**6# # Input resistance=2 MOhms\n", + "B = 0.0909# # Beta=0.0909\n", + "Zool = 75.# # Output impedence (open-loop)=75 Ohms\n", + "\n", + "Zicl = Ri*(1+Avol*B)#\n", + "print 'The Input Impedence Closed-Loop = %0.2e Ohms'%Zicl\n", + "print 'i.e 18 GOhms'\n", + "\n", + "A = Avol*B#\n", + "\n", + "Zocl = Zool/(1+A)#\n", + "print 'The Closed-Loop Output Impedence = %0.2f Ohms'%Zocl" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_10 Page No. 1090" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Input impedence closed-loop = 2.00e+11 Ohms\n", + "i.e 200 GOhms\n", + "The Closed loop Output Impedence = 0.001 Ohms\n" + ] + } + ], + "source": [ + "# Assume Rin is 2 MOhms, Avol is 100,000, and Zout(OL) is 75 Ohms. Calculate Zin(CL) and Zout(CL)\n", + "\n", + "# Given data\n", + "\n", + "Avol = 100000.# # Open loop voltage gain=100,000\n", + "Ri = 2.0*10**6# # Input resistance=2 MOhms\n", + "B = 1.0# # Beta=1\n", + "Zool = 75.# # Output impedence (open-loop)=75 Ohms\n", + "\n", + "Zicl = Ri*(1+Avol*B)#\n", + "print 'The Input impedence closed-loop = %0.2e Ohms'% Zicl\n", + "print 'i.e 200 GOhms'\n", + "\n", + "A = Avol*B#\n", + "\n", + "Zocl = Zool/(1+A)#\n", + "print 'The Closed loop Output Impedence = %0.3f Ohms'%Zocl" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_11 Page No. 1091" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Closed-Loop Voltage Gain Acl =-10.00\n", + "The Output Voltage = 7.50 Volts\n" + ] + } + ], + "source": [ + "# Calculate the closed-loop voltage gain, Acl, and the dc voltage at the op-amp output terminal.\n", + "\n", + "# Given data\n", + "\n", + "V = 15.# # Voltage at +ve terminal of op-amp=15 Volts\n", + "Rf = 10.*10**3# # Feedback resistance=10 kOhms\n", + "Ri = 1.*10**3# # Input resistance=1 kOhms\n", + "R1 = 10.*10**3# # Resistance1=10 kOhms\n", + "R2 = 10.*10**3# # Rsistance2=10 kOhms\n", + "\n", + "Acl = -(Rf/Ri)#\n", + "print 'The Closed-Loop Voltage Gain Acl =%0.2f'%Acl\n", + "\n", + "Vo = V*(R2/(R1+R2))#\n", + "print 'The Output Voltage = %0.2f Volts'%Vo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_12 Page No. 1095" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Output Voltage 1 Volts\n" + ] + } + ], + "source": [ + "# Calculate the output voltage, Vout.\n", + "\n", + "# Given data\n", + "\n", + "V1 = 1# # Input voltage1=1 Volts\n", + "V2 = -5# # Input voltage2=-5 Volts\n", + "V3 = 3# # Input voltage3=3 Volts\n", + "\n", + "Vo = -(V1+V2+V3)#\n", + "print 'The Output Voltage %0.f Volts'%Vo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_13 Page No. 1097" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Output Voltage = 3.00 Volts\n" + ] + } + ], + "source": [ + "# Calculate the output voltage, Vout.\n", + "\n", + "# Given data\n", + "\n", + "V1 = 0.5# # Input voltage1=0.5 Volts\n", + "V2 = -2.0# # Input voltage2=-2 Volts\n", + "Rf = 10.*10**3# # Feedback resistance=10 kOhms\n", + "R1 = 1.*10**3# # Resistance1=1 kOhms\n", + "R2 = 2.5*10**3# # Rsistance2=2.5 kOhms\n", + "\n", + "A = Rf/R1#\n", + "B = Rf/R2#\n", + "\n", + "Vo = -(A*V1+B*V2)#\n", + "print 'The Output Voltage = %0.2f Volts'%Vo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_14 Page No. 1101" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Output Voltage of Case A = -12.50 Volts\n", + "The Output Voltage of Case B = 10.00 Volts\n", + "The Output Voltage of Case C = -0.00 Volts\n" + ] + } + ], + "source": [ + "# Calculate the output voltage, Vout, if (a) Vx is 1 Vdc and Vy is -0.25 Vdc, (b) -Vx is 0.5 Vdc and Vy is 0.5 Vdc, (c) Vx is 0.3 V and Vy is 0.3 V.\n", + "\n", + "# Given data\n", + "\n", + "Rf = 10.*10**3# # Feedback resistance=10 kOhms\n", + "R1 = 1.*10**3# # Resistance1=1 kOhms\n", + "Vx1 = 1.# # Input voltage Vx1 at -ve terminal of op-amp=1 Volts\n", + "Vy1 = -0.25# # Input voltage Vy1 at +ve terminal of op-amp=-0.25 Volts\n", + "Vx2 = -0.5# # Input voltage Vx2 at -ve terminal of op-amp=-0.5 Volts\n", + "Vy2 = 0.5# # Input voltage Vy2 at +ve terminal of op-amp=0.5 Volts\n", + "Vx3 = 0.3# # Input voltage Vx3 at -ve terminal of op-amp=0.3 Volts\n", + "Vy3 = 0.3# # Input voltage Vy3 at +ve terminal of op-amp=0.3 Volts\n", + "\n", + "A = -Rf/R1#\n", + "\n", + "# Case A\n", + "\n", + "Voa = A*(Vx1-Vy1)#\n", + "print 'The Output Voltage of Case A = %0.2f Volts'%Voa\n", + "\n", + "# Case B\n", + "\n", + "Voa = A*(Vx2-Vy2)#\n", + "print 'The Output Voltage of Case B = %0.2f Volts'%Voa\n", + "\n", + "# Case C\n", + "\n", + "Voa = A*(Vx3-Vy3)#\n", + "print 'The Output Voltage of Case C = %0.2f Volts'%Voa" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_15 Page No. 1102" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Output of Differential Amplifier = 5.00 Volts\n" + ] + } + ], + "source": [ + "# Assume that Rd increases to 7.5 k\u0006 due to an increase in the ambient temperature. Calculate the output of the differential amplifier. Note: Rb is 5 kOhms\u0006.\n", + "\n", + "# Given data\n", + "\n", + "Vi = 5.# # Voltage input=5 Volts(dc)\n", + "Rf = 10.*10**3# # Feedback resistance=10 kOhms\n", + "R1 = 1.*10**3# # Resistance1=1 kOhms\n", + "Ra = 5.*10**3# # Resistance A at wein bridge=5 kOhms\n", + "Rb = 10.*10**3# # Resistance B at wein bridge=10 kOhms\n", + "Rc = 5.*10**3# # Resistance C at wein bridge=5 kOhms\n", + "Rd = 7.5*10**3# # Resistance D at wein bridge=7.5 kOhms\n", + "\n", + "Vx = Vi*(Ra/Rb)#\n", + "Vy = Vi*(Rd/(Rd+Rc))#\n", + "A = -Rf/R1\n", + "\n", + "Vo = A*(Vx-Vy)#\n", + "print 'The Output of Differential Amplifier = %0.2f Volts'%Vo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_16 Page No. 1103" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Cutoff Frequency = 1.59e+03 Hertz\n", + "i.e 1.591 kHz\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# Calculate the cutoff frequency, fc.\n", + "\n", + "# Given data\n", + "\n", + "Rf = 10.*10**3# # Feedback resistance=10 kOhms\n", + "Cf = 0.01*10**-6# # Feedback capacitance=0.01 uFarad\n", + "\n", + "fc = 1./(2.*pi*Rf*Cf)#\n", + "print 'The Cutoff Frequency = %0.2e Hertz'%fc\n", + "print 'i.e 1.591 kHz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_17 Page No. 1104" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Closed-Loop Voltage Gain at 0 Hz =-10.00\n", + "The Closed-Loop Voltage Gain at 1 MHz =-0.02\n" + ] + } + ], + "source": [ + "from math import pi,sqrt\n", + "# Calculate the Voltage gain, Acl at (a)0 Hz and (b) 1 MHz\n", + "\n", + "# Given data\n", + "\n", + "f1 = 1.*10**6# # Frequency=1 MHertz\n", + "Rf = 10.*10**3# # Feedback resistance=10 kOhms\n", + "R1 = 1.*10**3# # Resistance1=1 kOhms\n", + "Cf = 0.01*10**-6# # Feedback capacitance=0.01 uFarad\n", + "\n", + "# At 0 Hz, Xcf = infinity ohms, So, Zf=Rf \n", + "\n", + "Acl = -Rf/R1#\n", + "print 'The Closed-Loop Voltage Gain at 0 Hz =%0.2f'%Acl\n", + "\n", + "# At 1 MHz\n", + "\n", + "Xcf = 1/(2*pi*f1*Cf)#\n", + "\n", + "A = (Rf*Rf)#\n", + "B = (Xcf*Xcf)#\n", + "\n", + "Zf = ((Xcf*Rf)/sqrt(A+B))#\n", + "\n", + "Acl1 = -Zf/R1#\n", + "print 'The Closed-Loop Voltage Gain at 1 MHz =%0.2f'%Acl1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_18 Page No. 1105" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage Gain at 0 Hz = 20.00 dB\n", + "The Voltage Gain at 1.591 kHz = 16.99 dB\n", + "approx 17dB\n" + ] + } + ], + "source": [ + "from math import log10,pi,sqrt\n", + "# Calculate the dB voltage gain, at (a)0 Hz and (b) 1.591 kHz\n", + "\n", + "# Given data\n", + "\n", + "f1 = 1.591*10**3# # Frequency=1.591 kHertz\n", + "Rf = 10.*10**3# # Feedback resistance=10 kOhms\n", + "Ri = 1.*10**3# # Input resistance=1 kOhms\n", + "Cf = 0.01*10**-6# # Feedback capacitance=0.01 uFarad\n", + "\n", + "# At 0 Hz, Xcf = infinity ohms, So, Zf=Rf \n", + "\n", + "A = Rf/Ri\n", + "\n", + "Acl = 20*log10(A)#\n", + "print 'The Voltage Gain at 0 Hz = %0.2f dB'%Acl\n", + "\n", + "# At 1.591 kHz\n", + "\n", + "Xcf = 1/(2*pi*f1*Cf)#\n", + "B = (Rf*Rf)#\n", + "C = (Xcf*Xcf)#\n", + "Zf = (Xcf*Rf/sqrt(B+C))#\n", + "D = Zf/Ri#\n", + "\n", + "Acl1 = 20*log10(D)#\n", + "print 'The Voltage Gain at 1.591 kHz = %0.2f dB'%Acl1\n", + "print 'approx 17dB'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_19 Page No. 1106" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Cutoff Frequency = 1591.55 Hertz\n", + "i.e 1.591 kHz\n" + ] + } + ], + "source": [ + "from math import pi\n", + "# Calculate the cutoff frequency, fc.\n", + "\n", + "# Given data\n", + "\n", + "Ri = 1.*10**3# # Input resistance=10 kOhms\n", + "Ci = 0.1*10**-6# # Input capacitance=0.01 uFarad\n", + "\n", + "fc = 1/(2*pi*Ri*Ci)#\n", + "print 'The Cutoff Frequency = %0.2f Hertz'%fc\n", + "print 'i.e 1.591 kHz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_20 Page No. 1118" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Output Current = 5.00e-03 Amps\n", + "i.e 5 mAmps\n" + ] + } + ], + "source": [ + "# Vin is 5 V, R is 1 kOhms , and Rl is 100 Ohms . Calculate the output current, Iout.\n", + "\n", + "# Given data\n", + "\n", + "Vin = 5.# # Input votage=5 Volts\n", + "Ri = 1.*10**3# # Input resistance=1 kOhms\n", + "Rl = 100.# # Load resistance=100 Ohms\n", + "\n", + "Io = Vin/Ri#\n", + "print 'The Output Current = %0.2e Amps'%Io\n", + "print 'i.e 5 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_21 Page No. 1120" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Output Voltage = 1.50 Volts\n" + ] + } + ], + "source": [ + "# Iin is 1.5 mA, R is 1 kOhms, and Rl is 10 kOhms. Calculate Vout.\n", + "\n", + "# Given data\n", + "\n", + "Iin = 1.5*10**-3# # Input votage=5 Volts\n", + "Ri = 1.*10**3# # Input resistance=1 kOhms\n", + "Rl = 100.# # Load resistance=100 Ohms\n", + "\n", + "Vo = Iin*Ri#\n", + "print 'The Output Voltage = %0.2f Volts'%Vo" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_22 Page No. 1121" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Upper Trigger Point = 0.129 Volts\n", + "i.e 128.7 mVolts\n", + "The Lower Trigger Point = -0.129 Volts\n", + "i.e -128.7 mVolts\n", + "The Hysterisis Voltage = 0.257 Volts\n", + "i.e 257.4 mVolts\n" + ] + } + ], + "source": [ + "# R1 is 1 kOhms and R2 is 100 kOhms . Calculate UTP, LTP, and VH.\n", + "\n", + "# Given data\n", + "\n", + "R1 = 1.*10**3# # Resistance1=1 kOhms\n", + "R2 = 100.*10**3# # Resistance2=100 kOhms\n", + "Vcc = 15.# # Applied votage=15 Volts\n", + "Vsat = 13.# # Assume Saturation voltage=13 Volts\n", + "\n", + "Beta = R1/(R1+R2)#\n", + "\n", + "Utp = Beta*Vsat#\n", + "print 'The Upper Trigger Point = %0.3f Volts'%Utp\n", + "print 'i.e 128.7 mVolts'\n", + "\n", + "Ltp = -Beta*Vsat#\n", + "print 'The Lower Trigger Point = %0.3f Volts'%Ltp\n", + "print 'i.e -128.7 mVolts'\n", + "\n", + "Vh = Utp-Ltp#\n", + "print 'The Hysterisis Voltage = %0.3f Volts'%Vh\n", + "print 'i.e 257.4 mVolts'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 33_23 Page No. 1124" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Minimum value of required Capacitor = 1.00e-04 Farads\n", + "i.e 100 uFarad\n" + ] + } + ], + "source": [ + "# Rl is 1 kOhms and the frequency of the input voltage equals 100 Hz. Calculate the minimum value of C required.\n", + "\n", + "# Given data\n", + "\n", + "f = 100.# # Applied frequency=100 Hertz\n", + "Rl = 1.*10**3# # Load resistance=1 kOhms\n", + "\n", + "T = 1./f#\n", + "\n", + "C = (10*T)/Rl#\n", + "print 'The Minimum value of required Capacitor = %0.2e Farads'%C\n", + "print 'i.e 100 uFarad'" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter4.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter4.ipynb new file mode 100644 index 00000000..b71eae51 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter4.ipynb @@ -0,0 +1,336 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4 : Series Circuits" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 4_1 Page No. 117" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Combined Series Resistance = 20 Ohms\n" + ] + } + ], + "source": [ + "# Two resistances R1 and R2 of 5 Ohms\u0004 each and R3 of 10 Ohms\u0004 are in series. How much is Rt?\n", + "\n", + "# Given data\n", + "\n", + "R1 = 5# # Resistor 1=5 Ohms\n", + "R2 = 5# # Resistor 2=5 Ohms\n", + "R3 = 10# # Resistor 3=10 Ohms\n", + "\n", + "Rt = R1+R2+R3#\n", + "print 'The Combined Series Resistance = %0.f Ohms'%Rt" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 4_2 Page No. 117" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current in Resistor R3 connected in Series = 4 Amps\n" + ] + } + ], + "source": [ + "#With 80 V applied across the series string, how much is the current in R3?\n", + "\n", + "# Given data\n", + "\n", + "Rt = 20# # Total Resistance=20 Ohms\n", + "Vt = 80# # Applied Voltage=80 Volts\n", + "\n", + "I = Vt/Rt#\n", + "print 'The Current in Resistor R3 connected in Series = %0.f Amps'%I" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 4_3 Page No. 119" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The combined series resistance = 60 ohms\n", + "The current = 0.20 Amps\n", + "i.e 200 mA\n", + "The Voltage Drop of Resistor R1 = 2.00 Volts\n", + "The Voltage Drop of Resistor R2 = 4.00 Volts\n", + "The Voltage Drop of Resistor R3 = 6.00 Volts\n" + ] + } + ], + "source": [ + "# Solve for Rt, I and the individual resistor voltage drops at R1, R2, R3.\n", + "\n", + "# Given data\n", + "\n", + "R1 = 10.# # Resistor 1=10 Ohms\n", + "R2 = 20.# # Resistor 2=20 Ohms\n", + "R3 = 30.# # Resistor 3=30 Ohms\n", + "Vt = 12.0# # Applied Voltage=12 Volts\n", + "\n", + "Rt = R1+R2+R3#\n", + "print 'The combined series resistance = %0.f ohms'%Rt\n", + "\n", + "I = Vt/Rt#\n", + "print 'The current = %0.2f Amps'%I\n", + "print 'i.e 200 mA'\n", + "\n", + "V1 = I*R1\n", + "print 'The Voltage Drop of Resistor R1 = %0.2f Volts'%V1\n", + "\n", + "V2 = I*R2\n", + "print 'The Voltage Drop of Resistor R2 = %0.2f Volts'%V2\n", + "\n", + "V3 = I*R3\n", + "print 'The Voltage Drop of Resistor R3 = %0.2f Volts'%V3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 4_4 Page No. 123" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Applied Voltage Vt = 280 Volts\n" + ] + } + ], + "source": [ + "# A voltage source produces an IR drop of 40 V across a 20 Ohms R1, 60 V across a 30 Ohms\u0004 R2, and 180 V across a 90 Ohms\u0004 R3, all in series. According to Kirchhoff’s voltage law, how much is the applied voltage Vt ?\n", + "\n", + "# Given data\n", + "\n", + "V1 = 40# # Voltage drop at R1=40 Volts\n", + "V2 = 60# # Voltage drop at R2=60 Volts\n", + "V3 = 180# # Voltage drop at R3=180 Volts\n", + "\n", + "Vt = V1+V2+V3#\n", + "print 'The Applied Voltage Vt = %0.f Volts'%Vt" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 4_5 Page No. 123" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage Drop across Resistor R2 = 80 Volts\n" + ] + } + ], + "source": [ + "# An applied Vt of 120 V produces IR drops across two series resistors R 1 and R 2 If the voltage drop across R1 is 40 V, how much is the voltage drop across R2?\n", + "\n", + "# Given data\n", + "\n", + "V1 = 40# # Voltage drop at R1=40 Volts\n", + "Vt = 120# # Applied Voltage=120 Volts\n", + "\n", + "V2 = Vt-V1#\n", + "print 'The Voltage Drop across Resistor R2 = %0.f Volts'%V2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 4_6 Page No. 131" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage Drop of Resistor R1 = 6.00 Volts\n", + "The Voltage Drop of Resistor R2 = 4.80 Volts\n", + "The Voltage Drop of Resistor R3 = 7.20 Volts\n", + "The Voltage Drop of Resistor R4 = 6.00 Volts\n", + "The Resistor R3 is defective since it is open circuit and drops all the voltage arround it\n" + ] + } + ], + "source": [ + "# Assume that the series circuit in Fig. 4–20 has failed. A technician troubleshooting the circuit used a voltmeter to record the following resistor voltage drops. V1=0 V# V2=0 V# V3=24 V# V4=0 V. Based on these voltmeter readings, which component is defective and what type of defect is it? (Assume that only one component is defective.)\n", + "\n", + "# Given data\n", + "\n", + "R1 = 150.# # Resistor 1=150 Ohms\n", + "R2 = 120.# # Resistor 2=120 Ohms\n", + "R3 = 180.# # Resistor 3=180 Ohms\n", + "R4 = 150.# # Resistor 4=150 Ohms\n", + "Vt = 24.# # Applied Voltage=24 Volts\n", + "\n", + "Rt = R1+R2+R3+R4#\n", + "\n", + "I = Vt/Rt#\n", + "\n", + "V1 = I*R1\n", + "print 'The Voltage Drop of Resistor R1 = %0.2f Volts'%V1\n", + "\n", + "V2 = I*R2\n", + "print 'The Voltage Drop of Resistor R2 = %0.2f Volts'%V2\n", + "\n", + "V3 = I*R3\n", + "print 'The Voltage Drop of Resistor R3 = %0.2f Volts'%V3\n", + "\n", + "V4 = I*R4\n", + "print 'The Voltage Drop of Resistor R4 = %0.2f Volts'%V4\n", + "\n", + "print 'The Resistor R3 is defective since it is open circuit and drops all the voltage arround it'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 4_7 Page No. 133" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Calculated from the Circuit\n", + "The Voltage Drop of Resistor R1 = 6.00 Volts\n", + "The Voltage Drop of Resistor R2 = 4.80 Volts\n", + "The Voltage Drop of Resistor R3 = 7.20 Volts\n", + "The Voltage Drop of Resistor R4 = 6.00 Volts\n" + ] + } + ], + "source": [ + "# Assume that the series circuit has failed. A technician troubleshooting the circuit used a voltmeter to record the following resistor voltage drops: V1 \u0005 8 V#V2 \u0005 6.4 V#V3 \u0005 9.6 V#V4 \u0005 0 V. Based on the voltmeter readings, which component is defective and what type of defect is it? (Assume that only one component is defective.)\n", + "\n", + "# Given data\n", + "\n", + "R1 = 150.# # Resistor 1=150 Ohms\n", + "R2 = 120.# # Resistor 2=120 Ohms\n", + "R3 = 180.# # Resistor 3=180 Ohms\n", + "R4 = 150.# # Resistor 4=150 Ohms\n", + "Vt = 24.# # Applied Voltage=24 Volts\n", + "\n", + "print 'Calculated from the Circuit'\n", + "\n", + "Rt = R1+R2+R3+R4#\n", + "\n", + "I = Vt/Rt#\n", + "\n", + "V1 = I*R1\n", + "print 'The Voltage Drop of Resistor R1 = %0.2f Volts'%V1\n", + "\n", + "V2 = I*R2\n", + "print 'The Voltage Drop of Resistor R2 = %0.2f Volts'%V2\n", + "\n", + "V3 = I*R3\n", + "print 'The Voltage Drop of Resistor R3 = %0.2f Volts'%V3\n", + "\n", + "V4 = I*R4\n", + "print 'The Voltage Drop of Resistor R4 = %0.2f Volts'%V4\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter5.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter5.ipynb new file mode 100644 index 00000000..64665511 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter5.ipynb @@ -0,0 +1,291 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5 : Parallel Circuits" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 5_1 Page No. 149" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current Resistor R1 = 0.015 Amps\n", + "i.e 15 mAmps\n", + "The Current Resistor R2 = 0.03 Amps\n", + "i.e 25 mAmps\n" + ] + } + ], + "source": [ + "# Solve for branch currents I1 and I2.\n", + "\n", + "R1 = 1.*10**3# # Resistor 1=1*10**3 Ohms\n", + "R2 = 600.# # Resistor 2=600 Ohms\n", + "Va = 15.# # Applied Voltage=15 Volts\n", + "\n", + "I1 = Va/R1#\n", + "print 'The Current Resistor R1 = %0.3f Amps'%I1\n", + "print 'i.e 15 mAmps'\n", + "\n", + "I2 = Va/R2#\n", + "print 'The Current Resistor R2 = %0.2f Amps'%I2\n", + "print 'i.e 25 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 5_2 Page No. 150" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Total Current in the Mainline = 11 Amps\n" + ] + } + ], + "source": [ + "# An R1 of 20 Ohms\u0002, an R2 of 40 Ohms\u0002, and an R3 of 60 Ohms\u0002 are connected in parallel across the 120-V power line. Using Kirchhoff’s current law, determine the total current It.\n", + "\n", + "# Given data\n", + "\n", + "R1 = 20.# # Resistor 1=20 Ohms\n", + "R2 = 40.# # Resistor 2=40 Ohms\n", + "R3 = 60.# # Resistor 3=60 Ohms\n", + "Va = 120.# # Applied Voltage=120 Volts\n", + "\n", + "I1 = Va/R1#\n", + "I2 = Va/R2#\n", + "I3 = Va/R3#\n", + "\n", + "It = I1+I2+I3\n", + "print 'The Total Current in the Mainline = %0.f Amps'%It" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 5_3 Page No. 151" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current in R2 branch = 5 Amps\n" + ] + } + ], + "source": [ + "# Two branches R1 and R2 across the 120-V power line draw a total line current It of 15 A. The R1 branch takes 10 A. How much is the current I2 in the R2 branch?\n", + "\n", + "# Given data\n", + "\n", + "I1 = 10# # Current in R1 branch=10 Amps\n", + "It = 15# # Total Current=15 Amps\n", + "\n", + "I2 = It-I1#\n", + "print 'The Current in R2 branch = %0.f Amps'%I2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 5_4 Page No. 152" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Total Current = 0.6008 Amps\n", + "i.e 600.8 mAmps\n" + ] + } + ], + "source": [ + "# Three parallel branch currents are 0.1 A, 500 mA, and 800 \u0002A. Using Kirchhoff’s current law, calculate It.\n", + "\n", + "\n", + "# Given data\n", + "\n", + "I1 = 0.1# # Branch Current 1=0.1 Amps\n", + "I2 = 0.5# # Branch Current 2=500m Amps\n", + "I3 = 800*10**-6# # Branch Current 3=800u Amps\n", + "\n", + "It = I1+I2+I3#\n", + "print 'The Total Current = %0.4f Amps'%It\n", + "print 'i.e 600.8 mAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 5_5 Page No. 153" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Equivalent Resistance Req = 9 Ohms\n" + ] + } + ], + "source": [ + "# Two branches, each with a 5-A current, are connected across a 90-V source. How much is the equivalent resistance Req?\n", + "\n", + "# Given data\n", + "\n", + "I1 = 5# # Branch Current 1=5 Amps\n", + "I2 = 5# # Branch Current 2=5 Amps\n", + "Va = 90# # Applied Voltage=90 Volts\n", + "\n", + "It = I1+I2#\n", + "Req = Va/It#\n", + "print 'The Equivalent Resistance Req = %0.f Ohms'%Req" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 5_6 Page No. 158" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Value of Rx = 60.00 Ohms\n" + ] + } + ], + "source": [ + "# What Rx in parallel with 40 Ohms\u0002 will provide an Req of 24 Ohms?\n", + "\n", + "# Given data\n", + "\n", + "R = 40.0# # Resistance=40 Ohms\n", + "Req = 24.0# # Equivqlent Resistance=24 Ohms\n", + "\n", + "Rx = (R*Req)/(R-Req)#\n", + "print 'The Value of Rx = %0.2f Ohms'%Rx" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 5_7 Page No. 158" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of R = 50000 Ohms\n", + "i.e 50 kohms\n" + ] + } + ], + "source": [ + "# What R in parallel with 50 kOhms will provide an Req of 25 kOhms\u0002\n", + "\n", + "# Given data\n", + "\n", + "R1 = 50*10**3# # R1=50k Ohms\n", + "Req = 25*10**3# # Req=25k Ohms\n", + "\n", + "R = (R1*Req)/(R1-Req)#\n", + "print 'The value of R = %0.f Ohms'%R\n", + "print 'i.e 50 kohms'" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter6.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter6.ipynb new file mode 100644 index 00000000..5ad41975 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter6.ipynb @@ -0,0 +1,141 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6 : Series-Parallel Crcuits" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 6_1 Page No. 194" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Unknown Resistance Rx = 564.20 ohms\n" + ] + } + ], + "source": [ + "# The Current in M1 reads 0 \u0002A with the standard resistor RS adjusted to 5642 Ohms\u0002. What is the value of the unknown resistor Rx?\n", + "\n", + "# Given data\n", + "\n", + "Rs = 5642.# # Standard Resistor=5642 Ohms\n", + "R1 = 1.*10**3# # Resistor 1=1k Ohms\n", + "R2 = 10.*10**3# # Resistor 2=10k Ohms\n", + "\n", + "Rx = Rs*(R1/R2)#\n", + "print 'The Unknown Resistance Rx = %0.2f ohms'%Rx" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 6_2 Page No. 195" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Unknown Resistance Rx(max) = 999.90 Ohms\n" + ] + } + ], + "source": [ + "# what is the maximum unknown resistance Rx that can be measured for the ratio arm values shown?\n", + "\n", + "# Given data\n", + "\n", + "Rsmax = 9999.# # Standard Resistor(max)=9999 Ohms\n", + "R1 = 1.*10**3# # Resistor 1=1k Ohms\n", + "R2 = 10.*10**3# # Resistor 2=10k Ohms\n", + "\n", + "Rxmax = Rsmax*(R1/R2)#\n", + "print 'The Unknown Resistance Rx(max) = %0.2f Ohms'%Rxmax" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 6_3 Page No. 197" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistance R(AB) = 100.00 ohms\n" + ] + } + ], + "source": [ + "# Assume that the series-parallel circuit in Fig. 6–15a has failed. A technician troubleshooting the circuit has measured the following voltages: V1 = 10.8 V# VAB = 9 V# V4 = 16.2 V. These voltage readings are shown in Fig. 6–15b. Based on the voltmeter readings shown, which component is defective and what type of defect does it have?\n", + "\n", + "# Given data\n", + "\n", + "V1 = 10.8# # Voltage at R1=10.8 Volts\n", + "Vab = 9.0# # Voltage at point (AB)=9 Volts\n", + "V4 = 16.2# # Voltage at R4=16.2 Volts\n", + "R1 = 120.0# # Resistor 1=120 Ohms\n", + "\n", + "\n", + "It = V1/R1#\n", + "Rab = Vab/It#\n", + "print 'The Resistance R(AB) = %0.2f ohms'%Rab\n", + "\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter7.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter7.ipynb new file mode 100644 index 00000000..b7757e4c --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter7.ipynb @@ -0,0 +1,70 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 7 : Voltage Dividers and Current Dividers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 7_1 Page No. 199" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage Drop across each Resistor = 60.00 Volts\n" + ] + } + ], + "source": [ + "# Three 50 Ohms resistors R1, R2 and R3 are in series across an applied voltage of 180 V. How much is the IR voltage drop across each resistor?\n", + "\n", + "# Given data\n", + "\n", + "R1 = 50*10**3# # Resistor 1=50k Ohms\n", + "R2 = 50*10**3# # Resistor 2=50k Ohms\n", + "R3 = 50.*10**3# # Resistor 3=50k Ohms\n", + "Vt = 180.# # Applied Voltage=180 Volts\n", + "\n", + "R = R1 # R = R1 = R2 = R3\n", + "Rt = R1+R2+R3#\n", + "V = Vt*(R/Rt)#\n", + "print 'The Voltage Drop across each Resistor = %0.2f Volts'%V" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter8.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter8.ipynb new file mode 100644 index 00000000..95d4e266 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter8.ipynb @@ -0,0 +1,106 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8 : Analog & Digital Multimeters" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 8_1 Page No. 251" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Current through Shunt at Full Scale Deflection = 0.000950 Amps\n", + "i.e 950 uAmps\n" + ] + } + ], + "source": [ + "# A shunt extends the range of a 50-u\u0002A meter movement to 1 mA. How much is the current through the shunt at full-scale deflection?\n", + "\n", + "# Given data\n", + "\n", + "It = 1*10**-3# # Total Current=1 mAmps\n", + "Im = 50*10**-6# # Current (cause of meter movement)=50 uAmps\n", + "\n", + "Is = It-Im#\n", + "print 'The Current through Shunt at Full Scale Deflection = %0.6f Amps'%Is\n", + "print 'i.e 950 uAmps'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 8_2 Page No. 252" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Shunt Resistance Rs needed to extend the range to 500 uA = 111.11 Ohms\n" + ] + } + ], + "source": [ + "# A 50 u\u0002A meter movement has an Rm of 1000 Ohms\u0002. What Rs is needed to extend the range to 500 uA?\n", + "\n", + "# Given data\n", + "\n", + "It = 500*10**-6# # Total Current=500u Amps\n", + "Im = 50*10**-6# # Current (cause of meter movement)=50 uAmps\n", + "rm = 1000# # Resistance of moving coil=1000 Ohms\n", + "\n", + "Is = It-Im#\n", + "Vm = Im*rm#\n", + "\n", + "Rs = Vm/Is#\n", + "print 'The Shunt Resistance Rs needed to extend the range to 500 uA = %0.2f Ohms'%Rs\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter9.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter9.ipynb new file mode 100644 index 00000000..c836f96b --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/Chapter9.ipynb @@ -0,0 +1,126 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9 : Kirchoff's Laws" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 9_1 Page No. 268" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Branch 3 Current I3 = 4.5 Amps\n" + ] + } + ], + "source": [ + "# Apply Kirchhoff’s current law to solve for the unknown current, I3.\n", + "\n", + "# Given data\n", + "\n", + "I1 = 2.5# # Branch 1 Current=2.5 Amps\n", + "I2 = 8# # Branch 2 Current=8 Amps\n", + "I4 = 6# # Branch 3 Current=6 Amps\n", + "I5 = 9# # Branch 4 Current=9 Amps\n", + "\n", + "# I1+I2+I3-I4-I5 = 0 Sum of all currents at node is ZERO\n", + "# I1+I2+I3 = I4+I5 Total Incomming Current = Total Outgoing Current\n", + "\n", + "I3 = I4+I5-I1-I2#\n", + "print 'The Branch 3 Current I3 = %0.1f Amps'%I3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. 9_2 Page No. 275" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage V(AG) = 7.20 Volts\n", + "The Voltage V(BG) = -1.80 Volts\n" + ] + } + ], + "source": [ + "# Apply Kirchhoff’s voltage law to solve for the voltages V(AG) & V(BG).\n", + "\n", + "# Given data\n", + "\n", + "V1 = 18.# # Source Voltage 1=18 Volts\n", + "V2 = 18.# # Source Voltage 2=18 Volts\n", + "R1 = 120.# # Resistor 10=120 Ohms\n", + "R2 = 100.# # Resistor 2=100 Ohms\n", + "R3 = 180.# # Resistor 3=180 Ohms\n", + "\n", + "Vt = V1+V2#\n", + "Rt = R1+R2+R3#\n", + "\n", + "I = Vt/Rt#\n", + "\n", + "VR1 = I*R1#\n", + "VR2 = I*R2#\n", + "VR3 = I*R3#\n", + "\n", + "# V1+V2-VR1-VR2-VR3=0 Sum of all Voltages in loop is ZERO\n", + "# V1+V2 = VR1+VR2+VR3 Total Applied Voltage = Total Dropped Voltage in Resistors\n", + "\n", + "Vt = VR1+VR2+VR3#\n", + "\n", + "VAG = VR2+VR3-V2#\n", + "print 'The Voltage V(AG) = %0.2f Volts'%VAG\n", + "\n", + "VBG = V1-VR1-VR2#\n", + "print 'The Voltage V(BG) = %0.2f Volts'%VBG" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/ChapterI.ipynb b/Grob's_Basic_Electronics_by_M._E._Schultz/ChapterI.ipynb new file mode 100644 index 00000000..3f57bc74 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/ChapterI.ipynb @@ -0,0 +1,347 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter - I : Introduction to powers of 10" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. I_9 Page No. 9" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The addition of 170*10**3 and 23*10**4 =4.00E+05\n" + ] + } + ], + "source": [ + "# Add 170*10**3 and 23*10**4. Express the final answer in scientific notation.\n", + "\n", + "# Given data\n", + "\n", + "A = 170*10**3# # Variable 1\n", + "B = 23*10**4# # Variable 2\n", + "\n", + "C = A+B#\n", + "print 'The addition of 170*10**3 and 23*10**4 =%0.2E'%C" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. I_10 Page No. 9" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The substraction of 250*10**3 and 1.5*10**6 =1.25E+06\n" + ] + } + ], + "source": [ + "# Substract 250*10**3 and 1.5*10**6. Express the final answer in scientific notation.\n", + "\n", + "# Given data\n", + "\n", + "A = 1.5*10**6# # Variable 1\n", + "B = 250*10**3# # Variable 2\n", + "\n", + "C = A-B#\n", + "print 'The substraction of 250*10**3 and 1.5*10**6 =%0.2E'%C" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. I_11 Page No. 10" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The multiplication of 3*10**6 by 150*10**2 =4.50E+10\n" + ] + } + ], + "source": [ + "# Multiply 3*10**6 by 150*10**2. Express the final answer in scientific notation.\n", + "\n", + "# Given data\n", + "\n", + "A = 3*10**6# # Variable 1\n", + "B = 150*10**2# # Variable 2\n", + "\n", + "C = A*B#\n", + "print 'The multiplication of 3*10**6 by 150*10**2 =%0.2E'%C" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. I_12 Page No. 10" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The division of 5.0*10**7 by 2.0*10**4 =2.50E+03\n" + ] + } + ], + "source": [ + "# Divide 5.0*10**7 by 2.0*10**4. Express the final answer in scientific notation.\n", + "\n", + "# Given data\n", + "\n", + "A = 5.0*10**7# # Variable 1\n", + "B = 2.0*10**4# # Variable 2\n", + "\n", + "C = A/B#\n", + "print 'The division of 5.0*10**7 by 2.0*10**4 =%0.2E'%C" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. I_13 Page No. 10" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The reciprocal of 10**5 =1.00e-05\n", + "i.e 10**-5\n", + "The reciprocal of 10-3 = 1.00e+03\n", + "i.e 10**3\n" + ] + } + ], + "source": [ + "# Find the reciprocals for the following powers of 10: (a) 10**5 (b) 10**-3.\n", + "\n", + "# Given data\n", + "\n", + "A = 10**5# # Variable 1\n", + "B = 10**-3# # Variable 2\n", + "\n", + "C = 1./A#\n", + "print 'The reciprocal of 10**5 =%0.2e'%C\n", + "print 'i.e 10**-5'\n", + "\n", + "D = 1./B#\n", + "print 'The reciprocal of 10-3 = %0.2e'%D\n", + "print 'i.e 10**3'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. I_14 Page No. 11" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The square of 3.0*10**4 =9.00E+08\n" + ] + } + ], + "source": [ + "# Square 3.0*10**4. Express the answer in scientific notation.\n", + "\n", + "# Given data\n", + "\n", + "A = 3.0*10**4# # Variable 1\n", + "\n", + "B = A*A#\n", + "print 'The square of 3.0*10**4 =%0.2E'%B" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. I_15 Page No. 11" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The squareroot of 4*10**6 = 2.00E+03\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "# Find the squareroot of 4*10**6. Express the answer in scientific notation.\n", + "\n", + "# Given data\n", + "\n", + "A = 4*10**6# # Variable 1\n", + "\n", + "B = sqrt(A)#\n", + "print 'The squareroot of 4*10**6 = %0.2E'%B" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. I_16 Page No. 12" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The squareroot of 90*10**5 =3.00E+03\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "# Find the squareroot of 90*10**5. Express the answer in scientific notation.\n", + "\n", + "# Given data\n", + "\n", + "A = 90*10**5# # Variable 1\n", + "\n", + "B = sqrt(A)#\n", + "print 'The squareroot of 90*10**5 =%0.2E'% B" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example No. I_17 Page No. 12" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The multiplication of 40*10**-3 by 5*10**6 =200000.00\n", + "i.e 200.000*10**03 OR 200E03\n" + ] + } + ], + "source": [ + "# Show the keystrokes for multiplying 40*10**-3 by 5*10**6.\n", + "\n", + "# Given data\n", + "\n", + "A = 40*10**-3# # Variable 1\n", + "B = 5*10**6# # Variable 2\n", + "\n", + "C = A*B#\n", + "print 'The multiplication of 40*10**-3 by 5*10**6 =%0.2f'%C\n", + "print 'i.e 200.000*10**03 OR 200E03'" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/screenshots/DCLoadLine.png b/Grob's_Basic_Electronics_by_M._E._Schultz/screenshots/DCLoadLine.png Binary files differnew file mode 100644 index 00000000..214d7c48 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/screenshots/DCLoadLine.png diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/screenshots/EmitterCurrent.png b/Grob's_Basic_Electronics_by_M._E._Schultz/screenshots/EmitterCurrent.png Binary files differnew file mode 100644 index 00000000..0f981bd1 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/screenshots/EmitterCurrent.png diff --git a/Grob's_Basic_Electronics_by_M._E._Schultz/screenshots/freqNperiod.png b/Grob's_Basic_Electronics_by_M._E._Schultz/screenshots/freqNperiod.png Binary files differnew file mode 100644 index 00000000..e4989287 --- /dev/null +++ b/Grob's_Basic_Electronics_by_M._E._Schultz/screenshots/freqNperiod.png diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter10_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter10_3.ipynb new file mode 100644 index 00000000..c3f165d1 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter10_3.ipynb @@ -0,0 +1,938 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:08446d35c254f7719d0bb2f900fc25303d15f487208d5c6fccc1dc48e1acf8aa"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 10 Magnets and Earth's Magnetism"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.1 Page no 558"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "F=0.8*10**-3*9.8 #N\n",
+ "d=0.1 #m\n",
+ "u=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "m=math.sqrt(F*d**2/(u*5))\n",
+ "m1=5*m\n",
+ "\n",
+ "#Result\n",
+ "print\"Strength of pole M1 is\", round(m,2),\"Am\"\n",
+ "print\"Strength of pole M2 is\",round(m1,1),\"Am\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Strength of pole M1 is 12.52 Am\n",
+ "Strength of pole M2 is 62.6 Am\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.2 Page no 559"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "F=14.4*10**-4 #N\n",
+ "d=0.05 #m\n",
+ "F1=1.6*10**-4\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "u=4*math.pi*10**-7\n",
+ "m=math.sqrt((F*4*math.pi*d**2)/u)\n",
+ "d1=1/(math.sqrt((F1*4*math.pi)/(u*m**2)))\n",
+ "\n",
+ "#Result\n",
+ "print \"Distance is\",d1,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance is 0.15 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.5 Page no 560"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "M=8\n",
+ "d=0.2\n",
+ "\n",
+ "#Calculation\n",
+ "B=u*2*M/(4*math.pi*d**3)\n",
+ "Beqa=B/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Magnetic induction at axial point\", B*10**4,\"*10**-4 T\"\n",
+ "print\"(ii) Magnetic induction at equatorial point is\",Beqa*10**4,\"*10**-4 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnetic induction at axial point 2.0 *10**-4 T\n",
+ "(ii) Magnetic induction at equatorial point is 1.0 *10**-4 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.6 Page no 560"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=6.4*10**6 #m\n",
+ "B=0.4*10**-4 #T\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "M=(B*4*math.pi*d**3)/u\n",
+ "\n",
+ "#Result\n",
+ "print\"Earth's dipole moment is\", round(M*10**-23,2)*10**23,\"Am**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Earth's dipole moment is 1.05e+23 Am**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.7 Page no 560"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=0.40\n",
+ "d=0.5\n",
+ "\n",
+ "#Calculation\n",
+ "Beqa=u*M/(4*math.pi*d**3)\n",
+ "Baxial=2*Beqa\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of axial field is\", Baxial,\"T\"\n",
+ "print\"Magnitude of equatorial field is\",Beqa,\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of axial field is 6.4e-07 T\n",
+ "Magnitude of equatorial field is 3.2e-07 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.8 Page no 560"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19\n",
+ "f=6.8*10**15\n",
+ "n=1\n",
+ "r=0.53*10**-10\n",
+ "\n",
+ "#Calculation \n",
+ "import math\n",
+ "I=e*f\n",
+ "M=n*I*math.pi*r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Equivalent magnetic moment is\", round(M*10**24,1)*10**-24,\"Am**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equivalent magnetic moment is 9.6e-24 Am**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.9 Page no 561"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=50\n",
+ "r=0.2 #m\n",
+ "I=12 #A\n",
+ "\n",
+ "#Calculation\n",
+ "B=(u*n*I)/(2.0*r)\n",
+ "M=n*I*math.pi*r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Magnetic field at the centre of the coil is\", round(B*10**3,3),\"*10**-3 T\"\n",
+ "print\"(ii) Magnetic moment is\",round(M,1),\"Am**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnetic field at the centre of the coil is 1.885 *10**-3 T\n",
+ "(ii) Magnetic moment is 75.4 Am**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 58
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.10 Page no 561"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=7.5*10**-4 #m**2\n",
+ "I=12 #A\n",
+ "\n",
+ "#Calculation\n",
+ "M=A*I\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the magnetic moment is\", M*10**3,\"*10**-3 Am**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the magnetic moment is 9.0 *10**-3 Am**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 62
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.11 Page no 561"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=100\n",
+ "I=0.1 #A\n",
+ "r=0.05\n",
+ "B=1.5 #T\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "M=n*I*math.pi*r**2\n",
+ "W=2*M*B\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the coil is\", round(M,4),\"Am**2\"\n",
+ "print\"Workdone is\",round(W,4),\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the coil is 0.0785 Am**2\n",
+ "Workdone is 0.2356 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 70
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.12 Page no 561"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=10\n",
+ "I=3\n",
+ "A=7.85*10**-3\n",
+ "B=10**-2 #T\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "M=n*I*A\n",
+ "U1=-M*B*math.cos(0)\n",
+ "Uf=-M*B*math.cos(90)\n",
+ "w=-U1\n",
+ "t=M*B*math.sin(90*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Work done is\", round(t*10**3,1),\"*10**-3 Nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Work done is 2.4 *10**-3 Nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 81
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.13 Page no 562"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=4.8*10**-2 #J/T\n",
+ "a=30 #degree\n",
+ "B=3*10**-2 #t\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "t=M*B*math.sin(a*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Torque acting on the needle is\", round(t*10**4,1),\"*10**-4 Nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Torque acting on the needle is 7.2 *10**-4 Nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 86
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.14 Page no 562"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=0.2 #T\n",
+ "a=30 #degree\n",
+ "t=0.06 #Nm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "M=t/(B*math.sin(a*3.14/180.0))\n",
+ "U=M*B*math.cos(1*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Magnetic moment of the magnet is\", round(M,1),\"Am**2\"\n",
+ "print\"(ii) Orientation of the magnet is\", round(U,0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnetic moment of the magnet is 0.6 Am**2\n",
+ "(ii) Orientation of the magnet is 0.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 97
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.15 Page no 562"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "a=30 #degree\n",
+ "B=800*10**-4 #T\n",
+ "t=0.016 #Nm\n",
+ "A=2*10**-4 #m**2\n",
+ "n=1000 #turns\n",
+ "\n",
+ "#Calculation\n",
+ "M=t/(B*math.sin(a*3.14/180.0))\n",
+ "W=2*M*B\n",
+ "I=M/(n*A)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Magnetic moment of the magnet is\", round(M,2),\"Am**2\"\n",
+ "print\"(b) Work done is\", round(W,3),\"J\"\n",
+ "print\"(c) Current flowing through the solenoid is\", round(I,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Magnetic moment of the magnet is 0.4 Am**2\n",
+ "(b) Work done is 0.064 J\n",
+ "(c) Current flowing through the solenoid is 2.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 108
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.16 Page no 563"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "t=6.70\n",
+ "n=10.0\n",
+ "I=7.5*10**-6 #Kgm**2\n",
+ "M=6.7*10**-2 #Am**2\n",
+ "\n",
+ "#Calculation\n",
+ "T=t/n\n",
+ "B=(4*math.pi**2*I)/(M*T**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the magnetic field is\", round(B,2),\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the magnetic field is 0.01 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 113
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.18 Page no 569"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "t=1.2*10**-3 #nm\n",
+ "M=60\n",
+ "H=40*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=t/(M*H)\n",
+ "a=math.asin(A)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of the declination is\", round(a,0),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of the declination is 30.0 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 126
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.19 Page no 569"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=math.sqrt(3)\n",
+ "\n",
+ "#calculation\n",
+ "import math\n",
+ "a=math.atan(V)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of dip is\", round(a,0),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of dip is 60.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 131
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.20 Page no 569"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "H=0.28 #G\n",
+ "V=0.40 #G\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=V/H\n",
+ "a=math.atan(A)*180/3.14\n",
+ "R=math.sqrt(H**2+V**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Angle of dip is\", round(a,0),\"Degree\"\n",
+ "print\"(ii) Earth's total magnetic field is\", round(R,2),\"G\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Angle of dip is 55.0 Degree\n",
+ "(ii) Earth's total magnetic field is 0.49 G\n"
+ ]
+ }
+ ],
+ "prompt_number": 140
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.22 Page no 570"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "H=0.40\n",
+ "a=18 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "R=H/(math.cos(a*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of earth's magnetic field is\", round(R,2),\"G\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of earth's magnetic field is 0.42 G\n"
+ ]
+ }
+ ],
+ "prompt_number": 145
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.24 Page no 571"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=45 #Degree\n",
+ "b=60 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=math.tan(a*3.14/180.0)/(math.cos(b*3.14/180.0))\n",
+ "a=math.atan(A)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Apparant dip is\", round(a,1),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Apparant dip is 63.4 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 152
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.25 Page no 574"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=1.6 #Am**2\n",
+ "d=0.20 #m\n",
+ "u=10**-7 #N/A**2\n",
+ "\n",
+ "#Calculation\n",
+ "H=u*2*M/(d**3)\n",
+ "\n",
+ "#Result\n",
+ "print\"Horizontal component of the earth's magnetic field is\", H,\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Horizontal component of the earth's magnetic field is 4e-05 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 155
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.26 Page no 574"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=0.05 #m\n",
+ "d=0.12 #m\n",
+ "H=0.34*10**-4 #T\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "u=4*math.pi*10**-7\n",
+ "M=(4*math.pi*H*(d**2+l**2)**1.5)/u\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic moment of the magnet is\", round(M,3),\"J/T\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic moment of the magnet is 0.747 J/T\n"
+ ]
+ }
+ ],
+ "prompt_number": 162
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.27 Page no 577"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=7*10**-2 #m\n",
+ "H=2*10**-5 #T\n",
+ "n=50\n",
+ "\n",
+ "#calculation\n",
+ "import math\n",
+ "l=(2*r*H*math.tan(45*180/3.14))/u*n\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of current is\", round(l*10**-3,3),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of current is 0.043 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 172
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.28 Page no 577"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "K=0.095 #A\n",
+ "n=50\n",
+ "r=10*10**-2 #m\n",
+ "\n",
+ "#Calculation\n",
+ "H=K*u*n/(2.0*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"Horizontal component of earth's magnetic field is\", round(H*10**4,3),\"*10**-4 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Horizontal component of earth's magnetic field is 0.298 *10**-4 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 178
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.30 Page no 577"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=30 #degree\n",
+ "b=45 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "m=(2*math.tan(a*3.14/180.0))/(math.tan(b*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"Ratio of number of turns of the tangent galvanometers\", round(m,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Ratio of number of turns of the tangent galvanometers 1.155\n"
+ ]
+ }
+ ],
+ "prompt_number": 188
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter28_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter28_3.ipynb new file mode 100644 index 00000000..afb57de5 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter28_3.ipynb @@ -0,0 +1,59 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:511d2d405e1ede92783e0ff6e1c085ebc325e49ab2eff49fe438f3081d70cb4d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter28 Digital Electronics"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Example 28.3 page no 1497"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=1\n",
+ "\n",
+ "#Calculation\n",
+ "A=a*2**5+a*2**4+a*2**0\n",
+ "\n",
+ "#Result\n",
+ "print\"equivilant decimal is\",A"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "equivilant decimal is 49\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter29_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter29_3.ipynb new file mode 100644 index 00000000..9f1a9134 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter29_3.ipynb @@ -0,0 +1,526 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:8b87017ad47964520c2ead85c01092623a6e1bffcb1688b462701d086beba4f8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter29 Communication System"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.1 page no 1550"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "c=3*10**8\n",
+ "f=30.0*10**6\n",
+ "f1=300*10**6\n",
+ "f2=3000*10**6\n",
+ "\n",
+ "#Calculation\n",
+ "l=c/f\n",
+ "l1=l/2.0\n",
+ "l2=c/f1\n",
+ "l3=l2/2.0\n",
+ "l4=c/f2\n",
+ "l5=l4/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) length of half wave dipole antenna at 30 MHz is\",l1,\"m\"\n",
+ "print\"(ii) length of half wave dipole antenna at 300 MHz is\",l3,\"m\"\n",
+ "print\"(iii) length of half wave dipole antenna at 3000 MHz is\",15,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) length of half wave dipole antenna at 30 MHz is 5.0 m\n",
+ "(ii) length of half wave dipole antenna at 300 MHz is 0.5 m\n",
+ "(iii) length of half wave dipole antenna at 3000 MHz is 15 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.2 page no 1550"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "c=3*10**8 #m/s\n",
+ "f=3*10**4\n",
+ "f1=6*10**6\n",
+ "f2=5*10**7\n",
+ "\n",
+ "#Calculation\n",
+ "l=c/(4.0*f)\n",
+ "l1=c/(4.0*f1)\n",
+ "l2=c/(4.0*f2)\n",
+ "\n",
+ "#Result \n",
+ "print\"(i) length of quarter wave dipole antenna at 3*10**4 is\",l,\"m\"\n",
+ "print\"(ii) length of quarter wave dipole antenna at 6*10**6 is\",l1,\"m\"\n",
+ "print\"(iii) length of quarter wave dipole antena at 5*10**7 is\",l2,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) length of quarter wave dipole antenna at 3*10**4 is 2500.0 m\n",
+ "(ii) length of quarter wave dipole antenna at 6*10**6 is 12.5 m\n",
+ "(iii) length of quarter wave dipole antena at 5*10**7 is 1.5 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.3 page no 1551"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "AM=16 #mV\n",
+ "AM1=4 #mV\n",
+ "\n",
+ "#Calculation\n",
+ "Vmax=AM/2.0\n",
+ "Vmin=AM1/2.0\n",
+ "Ma=(Vmax-Vmin)/(Vmax+Vmin)\n",
+ "\n",
+ "#Result\n",
+ "print\" The modulation factor is\",Ma"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The modulation factor is 0.6\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.4 page no 1551"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Es=50 #V\n",
+ "Ec=100.0 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Ma=Es/Ec\n",
+ "\n",
+ "#Result\n",
+ "print\"The modulation factor\",Ma"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The modulation factor 0.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.6 page no 1552"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Pc=500 #watts\n",
+ "\n",
+ "#Calculation\n",
+ "Ps=(1/2.0)*(Pc)\n",
+ "Pt=Pc+Ps\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) sideband power is\",Ps,\"W\"\n",
+ "print\"(ii) power of AM wave is\",Pt,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) sideband power is 250.0 W\n",
+ "(ii) power of AM wave is 750.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.7 page no 1552"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Pc=50\n",
+ "Ma=0.8\n",
+ "Ma1=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "Ps=(1/2.0)*Ma**2*Pc\n",
+ "Ps1=(1/2.0)*Ma1**2*Pc\n",
+ "\n",
+ "#Result\n",
+ "print\"total sideband at 80% is\",Ps,\"KW\"\n",
+ "print\"total sideband at 10% is\",Ps1,\"KW\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "total sideband at 80% is 16.0 KW\n",
+ "total sideband at 10% is 0.25 KW\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.8 page no 1552"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Fc=500 #KHz\n",
+ "Fs=1 #KHz\n",
+ "\n",
+ "#Calculation\n",
+ "A1=Fc+Fs\n",
+ "A2=Fc-Fs\n",
+ "B=A1-A2\n",
+ "\n",
+ "#Result\n",
+ "print\"sideband frequancies are\",A1,\"KHz and\",A2,\"KHz\"\n",
+ "print\"bandwidth required is\",B,\"KHz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "sideband frequancies are 501 KHz and 499 KHz\n",
+ "bandwidth required is 2 KHz\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.9 page no 1552"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "F=10**10 #Hz\n",
+ "D=8*10**3 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "B=2/100.0*10**10\n",
+ "C=B/D\n",
+ "\n",
+ "#Result\n",
+ "print\"No. of telephones channels are\",C*10**-4,\"10**4\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "No. of telephones channels are 2.5 10**4\n"
+ ]
+ }
+ ],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.10 page no 1553"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=800.0*10**-7\n",
+ "C=3.0*10**8\n",
+ "f1=4.5*10**6 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "f=C/L\n",
+ "d=(1/100.0)*f\n",
+ "E=d/L\n",
+ "G=d/f1\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) number of channels for audio signal is\",round(E*10**-14,1),\"*10**8\"\n",
+ "print\"(ii) number of channels for video tv signal is\",round(G*10**-3,1),\"*10**5\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) number of channels for audio signal is 4.7 *10**8\n",
+ "(ii) number of channels for video tv signal is 8.3 *10**5\n"
+ ]
+ }
+ ],
+ "prompt_number": 80
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.11 Page no 1565"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=6400*10**3 #m\n",
+ "h=160\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "d=math.sqrt(2*R*h)\n",
+ "h2=4*h\n",
+ "\n",
+ "#Result\n",
+ "print\"Height is\", h2,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Height is 640 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 85
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.12 Page no 1565"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=6.4*10**6 #m\n",
+ "h=110\n",
+ "\n",
+ "#Calculation \n",
+ "import math\n",
+ "d=(math.sqrt(2*R*h))*10**-3\n",
+ "P=math.pi*d**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Population covered is\", round(P*10**-3,1),\"*10**6\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Population covered is 4.4 *10**6\n"
+ ]
+ }
+ ],
+ "prompt_number": 93
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.13 Page no 1565"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=6.4*10**6 #m\n",
+ "hr=50 #m\n",
+ "ht=32 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "d=math.sqrt(2*R*ht)+math.sqrt(2*R*hr)\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum distance is\", round(d*10**-3,1),\"Km\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum distance is 45.5 Km\n"
+ ]
+ }
+ ],
+ "prompt_number": 99
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 29.14 Page no 1566"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=300\n",
+ "R=6.4*10**6 #m\n",
+ "N=10**12\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "d=math.sqrt(2*R*h)\n",
+ "fc=9*N**0.5\n",
+ "\n",
+ "#Result\n",
+ "print\"fc=\", fc*10**-6,\"MHz\"\n",
+ "print\"5 MHz comes via ionospheric propogation.and 100 MHz signal comes via satellite transmission.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "fc= 9.0 MHz\n",
+ "5 MHz comes via ionospheric propogation.and 100 MHz signal comes via satellite transmission.\n"
+ ]
+ }
+ ],
+ "prompt_number": 107
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter11_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter11_3.ipynb new file mode 100644 index 00000000..11322719 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter11_3.ipynb @@ -0,0 +1,467 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:f9cf5cb53006209575af5d70cc318cffdaba99568e9075844fb3e2f1810bf06f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 11 Classification of magnetic materials"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.1 Page no 614"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19\n",
+ "f=6.8*10**15\n",
+ "n=1\n",
+ "r=0.53*10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=e*f\n",
+ "M=n*I*math.pi*r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Equivalent magnetic moment is\", round(M*10**24,1)*10**-24,\"Am**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equivalent magnetic moment is 9.6e-24 Am**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.2 Page no 615"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=240\n",
+ "R=474.0\n",
+ "r=12.5*10**-2\n",
+ "N=500\n",
+ "ur=5000\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=E/R\n",
+ "I1=2*math.pi*r\n",
+ "H=(N*I)/I1\n",
+ "u=4*math.pi*10**-7\n",
+ "B=u*ur*H\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The magnetising force is\", round(H,0),\"AT/m\"\n",
+ "print\"(ii) The magnetic flux density is\",round(B,2),\"Wb/m**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The magnetising force is 322.0 AT/m\n",
+ "(ii) The magnetic flux density is 2.03 Wb/m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.3 Page no 615"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r1=11\n",
+ "r2=12\n",
+ "B=2.5 #T\n",
+ "a=3000\n",
+ "I=0.70 #A\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=((r1+r2)/2.0)*10**-2\n",
+ "n=a/(2*math.pi*r)\n",
+ "ur=B*2*math.pi*r/(4*math.pi*10**-7*a*I)\n",
+ "\n",
+ "#Result\n",
+ "print\"Relative permeability is\", round(ur,1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Relative permeability is 684.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.4 Page no 616"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=0.5 #m\n",
+ "N=500\n",
+ "I1=0.15 #A\n",
+ "a=5000\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "H=(N*I1)/I\n",
+ "B=4*math.pi*10**-7*H\n",
+ "B1=B*a\n",
+ "I3=(B1-(H*4*math.pi*10**-7))/(4.0*math.pi*10**-7)\n",
+ "\n",
+ "#Result\n",
+ "print round(I3*10**-5,1),\"*10**5 A/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "7.5 *10**5 A/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.5 Page no 616"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=0.6\n",
+ "H=360.0\n",
+ "\n",
+ "#Calculation\n",
+ "u=B/H\n",
+ "x=(u-1*4*math.pi*10**-7)/(4.0*math.pi*10**-7)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Permeability is\",round(u*10**3,2),\"*10**-3 T/A m\"\n",
+ "print\"(ii) Susceptibility of the material is\",round(x,0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Permeability is 1.67 *10**-3 T/A m\n",
+ "(ii) Susceptibility of the material is 1325.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.6 Page no 616"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=8.0*10**22 #Am**2\n",
+ "R=64*10**5 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=(3*M)/(4.0*math.pi*R**3)\n",
+ "\n",
+ "#Result\n",
+ "print\"Earth's magnetisation is\", round(I,1),\"A/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Earth's magnetisation is 72.9 A/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 52
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.7 Page no 616"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "N=1800\n",
+ "l=0.6\n",
+ "I=0.9 #A\n",
+ "ur=500\n",
+ "n1=6.02*10**26\n",
+ "a=55.85\n",
+ "y=7850\n",
+ "\n",
+ "#Calculation\n",
+ "n=N/l\n",
+ "H=n*I\n",
+ "I1=(ur-1)*H\n",
+ "B=4*math.pi*10**-7*ur*H\n",
+ "x=(y*n1)/a\n",
+ "X=I1/x\n",
+ "\n",
+ "#Result\n",
+ "print\"Average magnetic moment per iron atom is\", round(X*10**23,2)*10**-23,\" A m**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average magnetic moment per iron atom is 1.59e-23 A m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 63
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.8 Page no 617"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=8.4 #g\n",
+ "d=7200.0\n",
+ "f=50 #Hz\n",
+ "E=3.2*10**4\n",
+ "t=30*60.0\n",
+ "\n",
+ "#Calculation\n",
+ "V=M/d\n",
+ "P=E/t\n",
+ "E1=P/(V*f)\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy dissipated per unit volume is\", round(E1,0),\"J/m**3/cycle\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy dissipated per unit volume is 305.0 J/m**3/cycle\n"
+ ]
+ }
+ ],
+ "prompt_number": 70
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.9 Page no 617"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "H=4*10**3 #A/m\n",
+ "a=60\n",
+ "b=0.12\n",
+ "\n",
+ "#Calculation\n",
+ "n=a/b\n",
+ "I=H/n\n",
+ "\n",
+ "#Result\n",
+ "print\"Current should be sent through the solenoid is\", I,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current should be sent through the solenoid is 8.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 73
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.10 Page no 617"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "x=1.68*10**-4\n",
+ "T1=293\n",
+ "T2=77.4\n",
+ "\n",
+ "#Calculation\n",
+ "x1=(x*T1)/T2\n",
+ "\n",
+ "#Result\n",
+ "print\"Susceptibility is\", round(x1*10**4,2),\"*10**-4\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Susceptibility is 6.36 *10**-4\n"
+ ]
+ }
+ ],
+ "prompt_number": 79
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.11 Page no 617"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=10**-6 #m\n",
+ "d=7.9 #g\n",
+ "a=6.023*10**23\n",
+ "n=55.0\n",
+ "M1=9.27*10**-24\n",
+ "\n",
+ "#Calculation\n",
+ "V=l**2\n",
+ "M=V*d\n",
+ "N=(a*M)/n\n",
+ "Mmax=N*M1\n",
+ "Imax=Mmax/V*10**-4\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of iron atom is\",round(N*10**-10,2),\"*10**10 atoms\"\n",
+ "print\"Magnetisation of the dipole is\",round(Imax*10**5,0),\"*10**5 A/m\"\n",
+ "print\"Maximum possible dipole moment is\",round(Mmax*10**13,0)*10**-13,\"A m**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of iron atom is 8.65 *10**10 atoms\n",
+ "Magnetisation of the dipole is 8.0 *10**5 A/m\n",
+ "Maximum possible dipole moment is 8e-13 A m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 113
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter12_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter12_3.ipynb new file mode 100644 index 00000000..c7cfe98a --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter12_3.ipynb @@ -0,0 +1,1049 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:58b6e0817b83a6d5c266f15dbe5c66e891c1c2ed8d1ef60dafaf43f98ff2d620"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 12 Electromagnetic induction"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.1 Page no 665"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=20 #mWb\n",
+ "a1=-20 #mWb\n",
+ "t=2*10**-3 #s\n",
+ "N=100\n",
+ "\n",
+ "#Calculation\n",
+ "a2=(a1-a)*10**-3\n",
+ "e=(-N*a2)/t\n",
+ "\n",
+ "#Result\n",
+ "print\"Average e.m.f induced in the coil is\", e,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average e.m.f induced in the coil is 2000.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.2 Page no 665"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=5*10**-2 #m\n",
+ "N=1\n",
+ "B=0.35\n",
+ "t=0.12 #S\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=math.pi*r**2\n",
+ "a1=B*A\n",
+ "a2=-B*a1\n",
+ "e=(N*a1)/t\n",
+ "\n",
+ "#Result\n",
+ "print round(e,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "0.02 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.3 Page no 665"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=10**-2 #m**2\n",
+ "a=45 #degree\n",
+ "B1=0.1 #T\n",
+ "R=0.5 #ohm\n",
+ "t=0.7 #S\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a1=B1*A*math.cos(a*3.14/180.0)\n",
+ "a2=0\n",
+ "a3=a1-a2\n",
+ "e=a3/t\n",
+ "I=e/R\n",
+ "\n",
+ "#Result\n",
+ "print\"Current during this time interval is\", round(I*10**3,1),\"*10**-3 A\"\n",
+ "print\"Magnitude of induced emf is\",round(e*10**3,0),\"*10**-3 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current during this time interval is 2.0 *10**-3 A\n",
+ "Magnitude of induced emf is 1.0 *10**-3 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.4 Page no 666"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=1.2*10**-3 #A\n",
+ "N=1.0\n",
+ "R=10 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "e=I*R\n",
+ "a=e/N\n",
+ "\n",
+ "#Result\n",
+ "print\"Necessary rate is\", a*10**2,\"*10**-2 Wb/second\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Necessary rate is 1.2 *10**-2 Wb/second\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.5 Page no 666"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=10**-1 #m\n",
+ "B=3.0*10**-5 #T\n",
+ "t=0.25 #S\n",
+ "N=500\n",
+ "R=2 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a1=B*math.pi*r**2*math.cos(0*3.14/180.0)\n",
+ "a2=B*math.pi*r**2*math.cos(180*3.14/180.0)\n",
+ "a3=a1-a2\n",
+ "e=(N*a3)/t\n",
+ "I=e/R\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the emf is\", round(e*10**3,1),\"*10**-3 V\"\n",
+ "print\"Current induced in the coil is\",round(I*10**3,1),\"*1)**-3 A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the emf is 3.8 *10**-3 V\n",
+ "Current induced in the coil is 1.9 *1)**-3 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.6 Page no 666"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=10**-2 #V\n",
+ "B=5*10**-5 #T\n",
+ "r=0.5 #m\n",
+ "N=1\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=math.pi*r**2\n",
+ "n=(e*N)/(math.pi*r**2*B)\n",
+ "\n",
+ "#Result\n",
+ "print\"Rate of rotation of the blade is\", round(n,1),\"revolutions/second\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate of rotation of the blade is 254.6 revolutions/second\n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.7 Page no 667"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=12\n",
+ "b=7\n",
+ "t=2\n",
+ "\n",
+ "#Calculation\n",
+ "e=((a*t)+b)*10**-3\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Magnitude of induced emf is\", e*10**3,\"mV\"\n",
+ "print\"(ii) The current induced in the coil will be anticlockwise\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnitude of induced emf is 31.0 mV\n",
+ "(ii) The current induced in the coil will be anticlockwise\n"
+ ]
+ }
+ ],
+ "prompt_number": 59
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.8 Page no 673"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=1 #T\n",
+ "l=0.5 #m\n",
+ "v=40 #m/s\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "e=B*l*v*math.sin(60*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"emf induced in the conductor is\", round(e,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "emf induced in the conductor is 17.32\n"
+ ]
+ }
+ ],
+ "prompt_number": 63
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.9 Page no 673"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "g=9.8\n",
+ "h=10\n",
+ "B=1.7*10**-5\n",
+ "l=1 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "v=math.sqrt(2*g*h)\n",
+ "e=B*l*v\n",
+ "\n",
+ "#Result\n",
+ "print\"Potential difference between its end is\", e*10**4,\"*10**4 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Potential difference between its end is 2.38 *10**4 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.10 Page no 673"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=72 *(5/18.0) #Km/h\n",
+ "B=40*10**-6 #T\n",
+ "A=40\n",
+ "l=2 #m\n",
+ "t=1.0\n",
+ "N=1\n",
+ "\n",
+ "#Calculation\n",
+ "A=l*v\n",
+ "a=B*A\n",
+ "e=N*a/t\n",
+ "\n",
+ "#Result\n",
+ "print\"e.m.f generated in the axle of the car\", e*10**3,\"mV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "e.m.f generated in the axle of the car 1.6 mV\n"
+ ]
+ }
+ ],
+ "prompt_number": 77
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.11 Page no 673"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "w=1000/60.0\n",
+ "r=0.3\n",
+ "B=0.5 #T\n",
+ "\n",
+ "#Calculation\n",
+ "v=w*r\n",
+ "vav=v/2.0\n",
+ "e=B*r*vav\n",
+ "\n",
+ "#Result\n",
+ "print\"e.m.f induced is\",e,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "e.m.f induced is 0.375 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 84
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.12 Page no "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.5 #m\n",
+ "n=2 #r.p.s\n",
+ "B=0.4*10**-4 #T\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "w=2*math.pi*n\n",
+ "e=0.5*B*r**2*w\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of induced e.m.f between the axle and rim is\", round(e*10**5,2)*10**-5,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of induced e.m.f between the axle and rim is 6.28e-05 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 91
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.13 Page no 674"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=1 #m\n",
+ "B=1\n",
+ "f=50\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "e=math.pi*R**2*B*f\n",
+ "\n",
+ "#Result\n",
+ "print\"e.m.f between the centre and the matallic ring is\", round(e,1),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "e.m.f between the centre and the matallic ring is 157.1 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 96
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.14 Page no 679"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=500\n",
+ "a=1.4*10**-4 #Wb\n",
+ "l=2.5 #A\n",
+ "\n",
+ "#Calculation\n",
+ "L=(N*a)/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Inductance of the coil is\", L*10**3,\"mH\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Inductance of the coil is 28.0 mH\n"
+ ]
+ }
+ ],
+ "prompt_number": 100
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.15 Page no 679"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=130*10**-3 #H\n",
+ "I1=20 #mA\n",
+ "I2=28 #mA\n",
+ "t=140.0*10**-3 #S\n",
+ "\n",
+ "#Calculation\n",
+ "l=I2-I1\n",
+ "e=(-L*l)/t\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of induced e.m.f is\", round(e,2),\"*10**-3 V\"\n",
+ "print\"Direction oppose the increase in current\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of induced e.m.f is -7.43 *10**-3 V\n",
+ "Direction oppose the increase in current\n"
+ ]
+ }
+ ],
+ "prompt_number": 106
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.16 Page no 679"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=4000\n",
+ "l=0.6 #m\n",
+ "r=16*10**-4 #m\n",
+ "\n",
+ "#Calculation\n",
+ "u=4*math.pi*10**-7\n",
+ "L=(u*N**2*((math.pi*r)/4.0))/l\n",
+ "Liron=N*L\n",
+ "\n",
+ "#Result\n",
+ "print\"Inductance of the solenoid is\", round(Liron,0),\"H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Inductance of the solenoid is 168.0 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 115
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.17 Page no 679"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=10.0 #H\n",
+ "e=300 #V\n",
+ "t=10**-2 #S\n",
+ "\n",
+ "#Calculation\n",
+ "dl=(e*t)/L\n",
+ "a=e*t\n",
+ "\n",
+ "#Result\n",
+ "print\"Charge in magnetic flux is\", a,\"Wb\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge in magnetic flux is 3.0 Wb\n"
+ ]
+ }
+ ],
+ "prompt_number": 119
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.18 Page no 680"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=10*10**-3\n",
+ "I=4*10**-3\n",
+ "N=200.0\n",
+ "\n",
+ "#Calculation\n",
+ "N1=L*I\n",
+ "a=N1/N\n",
+ "\n",
+ "#Result\n",
+ "print\"Total flux linked with the coil is\", N1,\"Wb\"\n",
+ "print\"Magnetic flux through the cross section of the coil is\",a,\"Wb\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total flux linked with the coil is 4e-05 Wb\n",
+ "Magnetic flux through the cross section of the coil is 2e-07 Wb\n"
+ ]
+ }
+ ],
+ "prompt_number": 125
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.19 Page no 680"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=500*10**-3\n",
+ "I1=20*10**-3 #A\n",
+ "I2=10*10**-3 #A\n",
+ "\n",
+ "#Calculation\n",
+ "U1=0.5*L*I1**2\n",
+ "U2=0.5*L*I2**2\n",
+ "\n",
+ "#Result\n",
+ "print \"Magnetic energy stored in the coil is\",U1*10**6,\"*10**-4 J\"\n",
+ "print\"New value of energy is\",U2,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic energy stored in the coil is 100.0 *10**-4 J\n",
+ "New value of energy is 2.5e-05 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 134
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.20 Page no 680"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=12\n",
+ "R=30.0 #ohm\n",
+ "L=0.22 \n",
+ "\n",
+ "#Calculation\n",
+ "I0=E/R\n",
+ "I=I0/2.0\n",
+ "P=E*I\n",
+ "dl=(E-(I*R))/L\n",
+ "du=L*I*dl\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Energy being delivered by the battery is\", P,\"W\"\n",
+ "print\"(ii) ENergy being stored in the magnetic field of inductor is\",du,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Energy being delivered by the battery is 2.4 W\n",
+ "(ii) ENergy being stored in the magnetic field of inductor is 1.2 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 144
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.21 Page no 680"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=2.0 #H\n",
+ "i=2 #A\n",
+ "\n",
+ "#Calculation\n",
+ "U=0.5*L*i**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Amount of energy spent during the period is\", U,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Amount of energy spent during the period is 4.0 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 147
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.22 Page no 686"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1500 #V\n",
+ "dl=3 #A\n",
+ "dt=0.001 #s\n",
+ "\n",
+ "#Calculation\n",
+ "M=(e*dt)/dl\n",
+ "\n",
+ "#Result\n",
+ "print\"Mumtual induction between the two coils is\", M,\"H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mumtual induction between the two coils is 0.5 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 150
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.23 Page no 686"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N2=1000\n",
+ "I1=5.0 #A\n",
+ "a2=0.4*10**-4 #Wb\n",
+ "dl=-24 #A\n",
+ "dt=0.02 #S\n",
+ "\n",
+ "#Calculation\n",
+ "M=(N2*a2)/I1\n",
+ "eb=(-M*dl)/dt\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Mutual induction between A and B is\", M,\"H\"\n",
+ "print\"(ii) e.m.f induced by the coil is\", eb"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Mutual induction between A and B is 0.008 H\n",
+ "(ii) e.m.f induced by the coil is 9.6\n"
+ ]
+ }
+ ],
+ "prompt_number": 155
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.24 Page no 687"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=1200\n",
+ "A=12*10**-4 #m**2\n",
+ "r=0.15 #m\n",
+ "N2=300\n",
+ "a=0.05\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "u=4*math.pi*10**-7\n",
+ "L=(u*N**2*A)/(2*math.pi*r)\n",
+ "M=(u*N*N2*A)/(2*math.pi*r)\n",
+ "dl=2/a\n",
+ "e=M*dl\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Self inductance of the toroid is\", round(L*10**3,1),\"*10**-3 H\"\n",
+ "print\"(ii) Induced e.m.f. in the second coil is\",round(e,3),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Self inductance of the toroid is 2.3 *10**-3 H\n",
+ "(ii) Induced e.m.f. in the second coil is 0.023 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 168
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.25 Page no 688"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=2.0\n",
+ "a1=20*10**-2\n",
+ "x=0.15\n",
+ "A2=0.3*10**-2\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "u=4*math.pi*10**-7\n",
+ "B1=(u*I*a1**2)/(2.0*(a1**2+x**2)**1.5)\n",
+ "a=B1*math.pi*A2**2\n",
+ "M=a/I\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Flux linking the bigger loop is\", round(a*10**11,1)\n",
+ "print\"(ii) Mutual induction between the two loops is\",round(M*10**11,2),\"!0**-11 H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Flux linking the bigger loop is 9.1\n",
+ "(ii) Mutual induction between the two loops is 4.55 !0**-11 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 181
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.26 Page no 688"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=0.5 #m\n",
+ "n=20 #turns\n",
+ "r=50 #cm\n",
+ "A1=40*10**-4 #m**2\n",
+ "n1=25\n",
+ "A2=25*10**-4 #m**2\n",
+ "\n",
+ "#Calculation\n",
+ "u=4*math.pi*10**-7\n",
+ "N=n*r\n",
+ "N2=n1*r\n",
+ "M=(u*N*N2*A2)/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Mutual induction of the system is\",round(M*10**3,2),\"*10**-3 H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mutual induction of the system is 7.85 *10**-3 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 188
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter13_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter13_3.ipynb new file mode 100644 index 00000000..8800aa9c --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter13_3.ipynb @@ -0,0 +1,1642 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:2cfb52e11f318cb94f2cd1051460c5732d92997d5b28187a631460f42051172b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 13 Alternating currents"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.1 Page no 721"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I0=141.4 #A\n",
+ "w=314\n",
+ "t=3*10**-3 #s\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "f=w/(2*math.pi)\n",
+ "T=1/f\n",
+ "I=-I0*t*math.sin(314*180/3.14)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The maximum value is\",I0,\"A\"\n",
+ "print\"(ii) Frequency is\",round(f,0),\"Hz\"\n",
+ "print\"(iii) Time period is\",round(T,2),\"S\"\n",
+ "print\"(iv) The instantaneous value is\", round(I*10**3,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The maximum value is 141.4 A\n",
+ "(ii) Frequency is 50.0 Hz\n",
+ "(iii) Time period is 0.02 S\n",
+ "(iv) The instantaneous value is 411.54 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.3 Page no 721"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=220 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "E0=math.sqrt(2)*E\n",
+ "Emean=2*E0/math.pi\n",
+ "\n",
+ "#Result\n",
+ "print\"Average e.m.f during a positive half cycle is\", round(Emean,0),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average e.m.f during a positive half cycle is 198.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.4 Page no 721"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=2\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=math.sqrt(A**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"r.m.s Value of current is\", I,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "r.m.s Value of current is 2.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.5 Page no 722"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I0=120 #A\n",
+ "a=360.0\n",
+ "b=96\n",
+ "c=120.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "t=1/a\n",
+ "I=I0*math.sin(math.pi/3.0)\n",
+ "a1=b/c\n",
+ "a2=math.asin(a1)\n",
+ "t=a2/(c*math.pi)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Instantaneous value after 1/360 second is\",round(I,2),\"A\"\n",
+ "print\"(ii) Time taken to reach 96 A for the first time is\", round(t,5),\"S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Instantaneous value after 1/360 second is 103.92 A\n",
+ "(ii) Time taken to reach 96 A for the first time is 0.00246 S\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.7 Page no 726"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E0=60\n",
+ "R=20.0 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Ev=E0/(math.sqrt(2))\n",
+ "Iv=Ev/R\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) A.C ammeter will\",round(Iv,2),\"A\"\n",
+ "print\"(ii) Average value of a.c over one cycle is zero\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) A.C ammeter will 2.12 A\n",
+ "(ii) Average value of a.c over one cycle is zero\n"
+ ]
+ }
+ ],
+ "prompt_number": 73
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.8 Page no 726"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E0=250 #V\n",
+ "I0=10 #A\n",
+ "\n",
+ "#Calculation\n",
+ "P=E0*I0\n",
+ "P1=P/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Peak power is\", P,\"W\"\n",
+ "print\"(ii) Average power is\",P1,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Peak power is 2500 W\n",
+ "(ii) Average power is 1250.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.9 Page no 726"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ev=120.0\n",
+ "P=1000 #W\n",
+ "Ev1=240\n",
+ "\n",
+ "#Calculation\n",
+ "Iv=P/Ev\n",
+ "I0=math.sqrt(2)*Iv\n",
+ "R=Ev/Iv\n",
+ "P=Ev1**2/R\n",
+ "\n",
+ "#Result\n",
+ "print\"Resistance is\", R,\"ohm \\nPeak current is\",P,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance is 14.4 ohm \n",
+ "Peak current is 4000.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 88
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.11 Page no 729"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Xl=220 #ohm\n",
+ "L=0.7 #H\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "f=Xl/(2*math.pi*L)\n",
+ "\n",
+ "#Result\n",
+ "print\"Frequency is\", round(f,0),\"HZ\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Frequency is 50.0 HZ\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.12 Page no 729"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=50 #Hz\n",
+ "I=1.4\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "E=2*math.pi*f*I*2*math.cos(2*math.pi*f)\n",
+ "Ev=E/math.sqrt(2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Potential difference across the coil is\", round(E,0),\"cos 100*math.pi*t\"\n",
+ "print\"(ii) r.m.s value of p.d across the coil is\", round(Ev,1),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Potential difference across the coil is 880.0 cos 100*math.pi*t\n",
+ "(ii) r.m.s value of p.d across the coil is 622.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.13 Page no 729"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=50 #Hz\n",
+ "L=2 \n",
+ "Ev=12 #V\n",
+ "L1=6\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xl=2*math.pi*f*L\n",
+ "Iv=Ev/Xl\n",
+ "Xl1=2*math.pi*f*L1\n",
+ "Iv1=Ev/Xl1\n",
+ "\n",
+ "#Result\n",
+ "print\"Current flows when the inductance is changed to 6 H\", round(Iv1,4),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current flows when the inductance is changed to 6 H 0.0064 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.14 Page no 729"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ev=200 #V\n",
+ "I0=0.9 #A\n",
+ "f=50 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "E0=math.sqrt(2)*Ev\n",
+ "Xl=E0/I0\n",
+ "L=Xl/(2*math.pi*f)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of inductance is\", round(L,0),\"H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of inductance is 1.0 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.15 Page no 732"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=318*10**-6 #F\n",
+ "Ev=230 #V\n",
+ "f=50 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xc=1/(2*math.pi*f*C)\n",
+ "Iv=Ev/Xc\n",
+ "E0=math.sqrt(2)*Ev\n",
+ "I0=math.sqrt(2)*Iv\n",
+ "w=2*math.pi*f\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The capacitive reactance is\",round(Xc,0),\"ohm\"\n",
+ "print\"(ii) r.m.s value of circuit current is\",round(Iv,0),\"A\"\n",
+ "print\"(iii) Equation for voltage is\",round(E0,2),\"sin 314 t\"\n",
+ "print\"Equation for current is\",round(I0,2),\"sin (314 t+pi/2)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The capacitive reactance is 10.0 ohm\n",
+ "(ii) r.m.s value of circuit current is 23.0 A\n",
+ "(iii) Equation for voltage is 325.27 sin 314 t\n",
+ "Equation for current is 32.5 sin (314 t+pi/2)\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.16 Page no 732"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=1 #H\n",
+ "Xl=3142.0 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "f=Xl/(2*math.pi*L)\n",
+ "C=1/(2.0*math.pi*f*Xl)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Value of frequency is\", round(f,0),\"ohm\"\n",
+ "print\"(ii) Capacity of a condenser is\",round(C*10**6,2),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Value of frequency is 500.0 ohm\n",
+ "(ii) Capacity of a condenser is 0.1 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 59
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.17 Page no 732"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=50*10**-6 #F\n",
+ "V=230 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "q=C*V*math.sqrt(2)\n",
+ "E=0.5*C*(V*math.sqrt(2))**2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Maximum charge on the capacitor is\", round(q*10**3,2),\"*10**-3 C\"\n",
+ "print\"(ii) The maximum energy stored in the capacitor is\", round(E,2),\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Maximum charge on the capacitor is 16.26 *10**-3 C\n",
+ "(ii) The maximum energy stored in the capacitor is 2.65 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 69
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.18 Page no 732"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I0=10 #A\n",
+ "w=314\n",
+ "L=5\n",
+ "\n",
+ "#Calculation\n",
+ "E=0.5*L*I0**2\n",
+ "E0=w*L*I0\n",
+ "C=(E*2)/(E0**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Capacitance of the capacitor is\",round(C*10**6,2),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance of the capacitor is 2.03 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 76
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.19 Page no 737"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=50\n",
+ "L=31.8*10**-3 #H\n",
+ "R=7.0 #ohm\n",
+ "Ev=230 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xl=2*math.pi*f*L\n",
+ "Z=math.sqrt(R**2+Xl**2)\n",
+ "Iv=Ev/Z\n",
+ "T=Xl/R\n",
+ "a=math.atan(T)*180/3.14\n",
+ "a1=math.cos(a)*3.14/180.0\n",
+ "P=Iv**2*R\n",
+ "t=55*math.pi/(180.0*3.14)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Circuit current is\",round(Iv,2),\"A\"\n",
+ "print\"(ii) Phase angle is\", round(a,0),\"lag\"\n",
+ "print\"(iii) Power factor is\", round(a1*10**3,3),\"lag\"\n",
+ "print\"(iv) Power consumed is\",round(P,0),\"W\"\n",
+ "print\"Time lag between voltage maximum and current maximum is\",round(t*10**1,2),\"*10**-3 S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Circuit current is 18.85 A\n",
+ "(ii) Phase angle is 55.0 lag\n",
+ "(iii) Power factor is 0.554 lag\n",
+ "(iv) Power consumed is 2488.0 W\n",
+ "Time lag between voltage maximum and current maximum is 3.06 *10**-3 S\n"
+ ]
+ }
+ ],
+ "prompt_number": 121
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.20 Page no 737"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=400 #W\n",
+ "Ev=250 #V\n",
+ "Iv=2.5 #A\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=P/(Ev*Iv)\n",
+ "Z=Ev/Iv\n",
+ "R=Z*a\n",
+ "Xl=math.sqrt(Z**2-R**2)\n",
+ "L=Xl/(2*math.pi*f)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The power factor is\",a,\"lag\"\n",
+ "print\"(ii) Resistance of the coil is\", R,\"ohm\"\n",
+ "print\"(iii) Inductance of the coil is\", round(L,3),\"H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The power factor is 0.64 lag\n",
+ "(ii) Resistance of the coil is 64.0 ohm\n",
+ "(iii) Inductance of the coil is 0.245 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 134
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.21 Page no 738"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vr=150 #V\n",
+ "R=75.0 #ohm\n",
+ "f=50 #Hz\n",
+ "L=318*10**-3 #H\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Iv=Vr/R\n",
+ "Xl=2*math.pi*f*L\n",
+ "Vl=Iv*Xl\n",
+ "Z=math.sqrt(R**2+Xl**2)\n",
+ "Ev=Iv*Z\n",
+ "a=Xl/R\n",
+ "a1=math.atan(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print \"(i) The supply voltage is\",round(Ev,0),\"V\"\n",
+ "print\"(ii) The phase angle is\",round(a1,2),\"degree lag\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The supply voltage is 250.0 V\n",
+ "(ii) The phase angle is 53.13 degree lag\n"
+ ]
+ }
+ ],
+ "prompt_number": 149
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.22 Page no 738"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=60 #W\n",
+ "Ev=100.0 #V\n",
+ "Ev1=220 #v\n",
+ "f=50 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "Iv=P/Ev\n",
+ "Vr=Ev1-Ev\n",
+ "R=Vr/Iv\n",
+ "Xl=Vl/Iv\n",
+ "Vl=math.sqrt(Ev1**2-Ev**2)\n",
+ "L=Xl/(2*math.pi*f)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The value of non inductive resistance is\", R,\"ohm\"\n",
+ "print\"(ii) Pure inductance is\",round(L,2),\"H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The value of non inductive resistance is 200.0 ohm\n",
+ "(ii) Pure inductance is 1.04 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 163
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.23 Page no 739"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f1=50.0\n",
+ "L=1\n",
+ "E=100 #V\n",
+ "I=1.0 #A\n",
+ "Iv=0.5 #A\n",
+ "f=0\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xl=2*math.pi*f*L\n",
+ "R=E/I\n",
+ "Z=Ev/Iv\n",
+ "Xl1=math.sqrt(Z**2-R**2)\n",
+ "L=Xl1/(2.0*math.pi*f1)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of resistance is\",R ,\"ohm\"\n",
+ "print\"The value of impedence is\",Z,\"ohm\"\n",
+ "print\"Inductance of the coil is\",round(L,2),\"H\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of resistance is 100.0 ohm\n",
+ "The value of impedence is 200.0 ohm\n",
+ "Inductance of the coil is 0.55 H\n"
+ ]
+ }
+ ],
+ "prompt_number": 183
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.24 Page no 742"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=10**6 #A\n",
+ "f=50 #Hz\n",
+ "C=79.5\n",
+ "R=30 #ohm\n",
+ "Ev=100 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xc=I/(2*math.pi*f*C)\n",
+ "Z=math.sqrt(R**2+Xc**2)\n",
+ "Iv=Ev/Z\n",
+ "a=Xc/R\n",
+ "a1=math.atan(a)*180/3.14\n",
+ "w=2*math.pi*f\n",
+ "I0=math.sqrt(2)*Iv\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Circuit impedence is\", round(Z,0),\"ohm\"\n",
+ "print\"(ii) Circuit current is\",round(Iv,0),\"A\"\n",
+ "print\"(iii) Phase angle is\",round(a1,0),\"degree lead\"\n",
+ "print\"(iv) Equation for the instantaneous value of current is\",round(I0,3),\"sin(\",round(w,0),\"t+\",round(a1,0),\"degree)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Circuit impedence is 50.0 ohm\n",
+ "(ii) Circuit current is 2.0 A\n",
+ "(iii) Phase angle is 53.0 degree lead\n",
+ "(iv) Equation for the instantaneous value of current is 2.827 sin( 314.0 t+ 53.0 degree)\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.25 Page no 742"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=80 #W\n",
+ "V=100.0 #v\n",
+ "V1=200 #V\n",
+ "f=50 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Iv=P/V\n",
+ "Vc=math.sqrt(V1**2-V**2)\n",
+ "Xc=Vc/Iv\n",
+ "C=1/(2.0*math.pi*f*Xc)\n",
+ "\n",
+ "#Result\n",
+ "print\"Capcitance of a capacitor is\", round(C*10**6,1),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capcitance of a capacitor is 14.7 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.26 Page no 742"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ev=200 #V\n",
+ "Iv=10.0\n",
+ "f=50 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "z=Ev/Iv\n",
+ "R=z*math.cos(30*3.14/180.0)\n",
+ "Xc=z*math.sin(30*3.14/180.0)\n",
+ "C=1/(2.0*math.pi*f*Xc)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Value of resistance is\", round(R,2),\"ohm\"\n",
+ "print\"(ii) Capacitive reactance is\", round(Xc,0),\"ohm\"\n",
+ "print\"(iii) Capacitance of the circuit is\", round(C*10**6,0),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Value of resistance is 17.32 ohm\n",
+ "(ii) Capacitive reactance is 10.0 ohm\n",
+ "(iii) Capacitance of the circuit is 318.0 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.27 Page no 743"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Iv=5 #A\n",
+ "R=10 #ohm\n",
+ "Ev=60 #V\n",
+ "C=400 #micro F\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Vr=Iv*R\n",
+ "Vc=math.sqrt(Ev**2-Vr**2)\n",
+ "Xc=Vc/Iv\n",
+ "f=1/(2.0*math.pi*C*Xc)\n",
+ "a=Vc/Vr\n",
+ "a1=math.atan(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of supplied frequency is\", round(f*10**6,0),\"Hz\"\n",
+ "print\"Phase angle between circuit current and supply voltage is\",round(a1,1),\"degree lead\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of supplied frequency is 60.0 Hz\n",
+ "Phase angle between circuit current and supply voltage is 33.6 degree lead\n"
+ ]
+ }
+ ],
+ "prompt_number": 65
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.28 Page no 743"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=200 #ohm\n",
+ "C=15*10**-6 #F\n",
+ "Ev=220 #V\n",
+ "f=50 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xc=1/(2*math.pi*f*C)\n",
+ "Z=math.sqrt(R**2+Xc**2)\n",
+ "Iv=Ev/Z\n",
+ "Vr=Iv*R\n",
+ "Vc=Iv*Xc\n",
+ "V=Vr+Vc\n",
+ "Vrc=math.sqrt(Vr**2+Vc**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The current in the circuit is\", round(Iv,3),\"A\"\n",
+ "print\"(b) Voltage across the resistor and capacitor is\",Vrc,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " (a) The current in the circuit is 0.754 A\n",
+ "(b) Voltage across the resistor and capacitor is 220.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 83
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.29 Page no 746"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=10.0 #ohm\n",
+ "R2=5.0 #ohm\n",
+ "R3=15 #ohm\n",
+ "Ev=200\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "R=R1+R2+R3\n",
+ "X=R3-(R1+R3)\n",
+ "Z=math.sqrt(R**2+R1**2)\n",
+ "Iv=Ev/Z\n",
+ "T=X/R\n",
+ "a=-math.atan(T)*180/3.14\n",
+ "b=math.cos(a*3.14/180.0)\n",
+ "P=Iv**2*R\n",
+ "print\"(i) Circuit current is\",round(Iv,2) ,\"A\"\n",
+ "print\"(ii) Circuit phase angle is\",round(a,2),\"degree lead\"\n",
+ "print\"(iii)Phase angle between applied voltage and circuit current \",round(b,3),\"lead\"\n",
+ "print\"(iv)Power consumed is\",P,\"W\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Circuit current is 6.32 A\n",
+ "(ii) Circuit phase angle is 18.44 degree lead\n",
+ "(iii)Phase angle between applied voltage and circuit current 0.949 lead\n",
+ "(iv)Power consumed is 1200.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 116
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.30 Page no. 746"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "F=50 #HZ\n",
+ "L=0.06 \n",
+ "C=6.8\n",
+ "l=10**6\n",
+ "R=2.5\n",
+ "Ev=230 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math \n",
+ "Xl=2*math.pi*F*L\n",
+ "Xc=l/(2*math.pi*F*C)\n",
+ "Z=math.sqrt(R**2+(Xl-Xc)**2)\n",
+ "Iv=Ev/Z\n",
+ "a=(Xl-Xc)/R\n",
+ "a2=-math.atan(a)*180.0/3.14\n",
+ "P=R/Z\n",
+ "P1=Ev*Iv*P\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Circuit impedence is\",round(Z,1),\"ohm\"\n",
+ "print\"(ii) Circuit current is\",round(Iv,3),\"A\"\n",
+ "print\"(iii) Phase angle is \",round(a2,1),\"degree lead\" \n",
+ "print\"(iv) Power factor is\",round(P,4),\"lead\"\n",
+ "print\"(v) Power consumed is\",round(P1,2),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Circuit impedence is 449.3 ohm\n",
+ "(ii) Circuit current is 0.512 A\n",
+ "(iii) Phase angle is 89.7 degree lead\n",
+ "(iv) Power factor is 0.0056 lead\n",
+ "(v) Power consumed is 0.66 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 146
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.31 Page no. 746"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=65 #degree\n",
+ "b=20 #degree\n",
+ "w=3000\n",
+ "L=0.01\n",
+ "E0=400 #V\n",
+ "I=10\n",
+ "f=50\n",
+ "\n",
+ "#calculation\n",
+ "import math\n",
+ "a=a-b\n",
+ "Xl=w*L\n",
+ "Z=E0/(I*math.sqrt(2))\n",
+ "R=Z/math.sqrt(2)\n",
+ "Xc=Xl-R\n",
+ "C=1/(w*Xc*10**-6)\n",
+ "\n",
+ "#Result\n",
+ "print\" The value of C is\" ,round(C,1),\"microF\"\n",
+ "print\" The value of R is\",R,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The value of C is 33.3 microF\n",
+ " The value of R is 20.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 161
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.32 Page no 747"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=50 #Hz\n",
+ "L=0.03\n",
+ "R=8 #ohm\n",
+ "Ev=240 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xl=2*math.pi*f*L\n",
+ "Z=math.sqrt(R**2+Xl**2)\n",
+ "Iv=Ev/Z\n",
+ "P=Iv**2*R\n",
+ "a=R/Z\n",
+ "Xc=2*Xl\n",
+ "C=1/(2*math.pi*f*Xc)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The value of current is\",round(Iv,2) ,\"A\"\n",
+ "print\" The value of power is\",round(P,0),\"W\"\n",
+ "print \" Power factor is\",round(a,2),\"lag\"\n",
+ "print\"(ii) The vaue of capacitance is\", round(C*10**6,0),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The value of current is 19.41 A\n",
+ " The value of power is 3015.0 W\n",
+ " Power factor is 0.65 lag\n",
+ "(ii) The vaue of capacitance is 169.0 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 186
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.33 Page no 747"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vr=65 #V\n",
+ "R=100.0 #ohm\n",
+ "Vl=204\n",
+ "Vc=415\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Iv=Vr/R\n",
+ "Xl=Vl/Iv\n",
+ "L=Xl/(2*math.pi*f)\n",
+ "Xc=Vc/Iv\n",
+ "C=1/(2*math.pi*f*Xc)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Circuit current is\", Iv,\"A\"\n",
+ "print\"(ii) Inductance is\",round(L,0),\"H\"\n",
+ "print\"(iii) The value of capacitance is\",round(C*10**6,0),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Circuit current is 0.65 A\n",
+ "(ii) Inductance is 1.0 H\n",
+ "(iii) The value of capacitance is 5.0 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 201
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.34 Page no 752"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=100*10**-12 #F\n",
+ "L=100*10**-6 #H\n",
+ "Ev=10\n",
+ "R=100.0 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "fr=1/(2*math.pi*math.sqrt(L*C))\n",
+ "Iv=Ev/R\n",
+ "Vl=Iv*2*math.pi*fr*L\n",
+ "Vc=Iv/(2.0*math.pi*fr*C)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Resonant frequency is\", round(fr*10**-6,2),\"*10**6 HZ\"\n",
+ "print\"(ii) Current at resonance is\", Iv,\"A\"\n",
+ "print\"(iii) Voltage across L and C is\", Vc,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Resonant frequency is 1.59 *10**6 HZ\n",
+ "(ii) Current at resonance is 0.1 A\n",
+ "(iii) Voltage across L and C is 100.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 218
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.35 Page no 752"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=50 #Hz\n",
+ "L=0.5\n",
+ "Ev=100 #v\n",
+ "R=4 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "C=1/(4*math.pi**2*f**2*L)\n",
+ "Ir=Ev/R\n",
+ "Vr=Ir*2*math.pi*f*L\n",
+ "Vc=Ir/(2*math.pi*f*C)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The capacitance is\", round(C*10**6,2),\"micro F\"\n",
+ "print\"(ii) The voltage across inductance and capacitance is\", round(Vc,0),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The capacitance is 20.26 micro F\n",
+ "(ii) The voltage across inductance and capacitance is 3927.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 229
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.36 Page no 752"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=50 #Hz\n",
+ "L=0.318 #H\n",
+ "Iv=2.3\n",
+ "R=100 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "C=1/((2*math.pi*f)**2*L)\n",
+ "Vl=Iv*2*math.pi*f*C*10**4\n",
+ "P=Iv**2*R\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The value of capacitor is\", round(C*10**6,1),\"micro F\"\n",
+ "print\"(ii) Voltage across the inductor is\", round(Vl,0),\"V\"\n",
+ "print\"(iii)Total power consumed is\",P,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The value of capacitor is 31.9 micro F\n",
+ "(ii) Voltage across the inductor is 230.0 V\n",
+ "(iii)Total power consumed is 529.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 245
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.37 Page no 753"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E0=283 #V\n",
+ "f=50 #Hz\n",
+ "R=3.0 #ohm\n",
+ "L=25.48*10**-3 #h\n",
+ "C=796*10**-6 #F\n",
+ "Xl=8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Xc=1/(2*math.pi*f*C)\n",
+ "Z=math.sqrt(R**2+(Xl-Xc)**2)\n",
+ "a=math.atan(Xc/R)*180/3.14\n",
+ "Iv=(E0/math.sqrt(2))/Z\n",
+ "P=Iv**2*R\n",
+ "a1=math.cos(a*180/3.14)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The inpedence of the circuit is\", round(Z,0),\"ohm\"\n",
+ "print\"(b) The phase difference is\", round(a,1),\"degree\"\n",
+ "print\"(c) The power dissipated is\", round(P,0),\"W\"\n",
+ "print\"(d) Power factor is\", round(a1,1),\"lag\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The inpedence of the circuit is 5.0 ohm\n",
+ "(b) The phase difference is 53.1 degree\n",
+ "(c) The power dissipated is 4804.0 W\n",
+ "(d) Power factor is 0.8 lag\n"
+ ]
+ }
+ ],
+ "prompt_number": 270
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.38 Page no 753"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=25.48*10**-3 #H\n",
+ "C=796*10**-6\n",
+ "R=3.0 #ohm\n",
+ "E0=283\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "fr=1/(2.0*math.pi*math.sqrt(L*C))\n",
+ "Iv=(E0/math.sqrt(2))/R\n",
+ "P=Iv**2*R\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Frequency of the source is\", round(fr,1),\"Hz\"\n",
+ "print\"(b) The value of impedence is\",R,\"ohm\"\n",
+ "print\"The value of current is\",round(Iv,1),\"A\"\n",
+ "print\"The power dissipated is\",round(P,0),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Frequency of the source is 35.3 Hz\n",
+ "(b) The value of impedence is 3.0 ohm\n",
+ "The value of current is 66.7 A\n",
+ "The power dissipated is 13348.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 287
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.39 Page no 757"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=1200*10**-12 #F\n",
+ "E=500\n",
+ "L=0.075 #H\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "q0=C*E\n",
+ "I0=q0/(math.sqrt(L*C))\n",
+ "f=1/(2*math.pi*math.sqrt(L*C))\n",
+ "T=1/f\n",
+ "U=q0**2/(2.0*C)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The initial charge onthe capcitor is\",q0,\"c\"\n",
+ "print\"(ii) The maximum current is\",round(I0*10**3,0),\"mA\"\n",
+ "print\"(iii) The value of frequency is\", round(f*10**-3,0),\"*10**3 Hz\"\n",
+ "print\"Time period is\", round(T*10**5,0),\"*10**-5 S\"\n",
+ "print\"(iv) Total energy is\",U*10**4,\"*10**-4 J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The initial charge onthe capcitor is 6e-07 c\n",
+ "(ii) The maximum current is 63.0 mA\n",
+ "(iii) The value of frequency is 17.0 *10**3 Hz\n",
+ "Time period is 6.0 *10**-5 S\n",
+ "(iv) Total energy is 1.5 *10**-4 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 315
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.40 Page no 758"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L=8*10**-6 #H\n",
+ "C=0.02*10**-6 #F\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/(2*math.pi*math.sqrt(L*C))\n",
+ "w=c/f\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength is\", round(w*10**-2,2),\"*10**2 m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength is 7.54 *10**2 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 321
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter14_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter14_3.ipynb new file mode 100644 index 00000000..7bb51839 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter14_3.ipynb @@ -0,0 +1,629 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:1bd22700738f4a2a80468f70e18e63c26ad56b1a125cfb69784af1d3eb280a8a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 14 Electrical devices"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.1 Page no 787"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=100\n",
+ "A=10**-2 #m**2\n",
+ "B=0.5 #T\n",
+ "f=500/60.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "w=2*math.pi*f\n",
+ "E0=N*A*B*w\n",
+ "E=E0*math.sin(60*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum emf produced in the coil is\", round(E0,2),\"V\"\n",
+ "print\"Instantaneous value of e.m.f. is\",round(E,1),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum emf produced in the coil is 26.18 V\n",
+ "Instantaneous value of e.m.f. is 22.7 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.2 Page no 787"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=50\n",
+ "A=2.5\n",
+ "B=0.3 #T\n",
+ "w=60\n",
+ "R=500 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "E0=N*B*A*w\n",
+ "I0=E0/R\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Maximum current drawn from the gnerator is\",I0,\"A\"\n",
+ "print\"(ii) The maximum flux through the coil is zero and when induced current is maximum, flux through the coil is zero\"\n",
+ "print\"(iii) Yes the generator will work because the basic condition for induced an .m.f. is that there should be relative motion between coil and magnetic field\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Maximum current drawn from the gnerator is 4.5 A\n",
+ "(ii) The maximum flux through the coil is zero and when induced current is maximum, flux through the coil is zero\n",
+ "(iii) Yes the generator will work because the basic condition for induced an .m.f. is that there should be relative motion between coil and magnetic field\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.3 Page no 787"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=150\n",
+ "A=2*10**-2 #m**2\n",
+ "B=0.15 #T\n",
+ "f=60\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "w=2*math.pi*f\n",
+ "E0=N*A*B*w\n",
+ "\n",
+ "#Result\n",
+ "print\"Peak value of e.m.f is\", round(E0,0),\"V\"\n",
+ "print\"Average value of induced e.m.f is zero\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Peak value of e.m.f is 170.0 V\n",
+ "Average value of induced e.m.f is zero\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.4 Page no 787"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=100\n",
+ "A=3\n",
+ "B=0.04 #T\n",
+ "w=60\n",
+ "R=500 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "E0=N*A*B*w\n",
+ "I0=E0/R\n",
+ "P=E0*I0\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum power dissipated in the coil is\", P,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum power dissipated in the coil is 1036.8 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.5 Page no 788"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=100\n",
+ "A=0.10 #m**2\n",
+ "f=0.5 #Hz\n",
+ "B=0.01 #T\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "w=2*math.pi*f\n",
+ "E0=N*A*B*w\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum voltage generated in the coil is\", round(E0,3),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum voltage generated in the coil is 0.314 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.6 Page no 792"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=240 #V\n",
+ "I=5 #A\n",
+ "R=4 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Eb=V-(I*R)\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of back e.m.f is\", Eb,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of back e.m.f is 220 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.7 Page no 792"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=20 #A\n",
+ "R=2 #ohm\n",
+ "n=0.5 \n",
+ "P=2000 #W\n",
+ "\n",
+ "#Calculation\n",
+ "P1=P/n\n",
+ "V=P1/I\n",
+ "Eb=V-(I*R)\n",
+ "\n",
+ "#Result\n",
+ "print\"The back e.m.f is\", Eb,\"V \\nSupply voltage is\",V,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The back e.m.f is 160.0 V \n",
+ "Supply voltage is 200.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.8 Page no 793"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=100 #V\n",
+ "I=6 #A\n",
+ "V1=0.7\n",
+ "\n",
+ "#Calculation\n",
+ "Pin=V*I\n",
+ "R=(V1*Pin)/I**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Armature resistance is\", round(R,2),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Armature resistance is 11.67 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.10 Page no 793"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=200 #V\n",
+ "I=5 #A\n",
+ "R=8.5 #ohm \n",
+ "\n",
+ "#Calculation\n",
+ "Eb=V-(I*R)\n",
+ "Pi=V*I\n",
+ "P0=Eb*I\n",
+ "n=(P0*100)/Pi\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Back e.m.f of motor is\", Eb,\"V\"\n",
+ "print\"(ii) Power input is\",Pi,\"W\"\n",
+ "print\"(iii) Output power is\",P0,\"W\"\n",
+ "print\"(iv) Efficiency of motor is\",n,\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Back e.m.f of motor is 157.5 V\n",
+ "(ii) Power input is 1000 W\n",
+ "(iii) Output power is 787.5 W\n",
+ "(iv) Efficiency of motor is 78.75 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 61
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.11 Page no 796"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vp=200 #V\n",
+ "n=200.0\n",
+ "Ip=2 #A\n",
+ "\n",
+ "#Calculation\n",
+ "Vs=Vp*n\n",
+ "Is=(Ip*V)/Vs\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Voltage developed in the secondary is\", Vs,\"V\"\n",
+ "print\"(ii) The current in the secondary is\",Is ,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Voltage developed in the secondary is 40000.0 V\n",
+ "(ii) The current in the secondary is 0.01 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 69
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.12 Page no 796"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vp=220.0 #V\n",
+ "Is=5 #A\n",
+ "n=20\n",
+ "\n",
+ "#Calculation\n",
+ "Vs=Vp*n\n",
+ "Ip=(Vs*Is)/Vp\n",
+ "P=Vs*Is\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Voltage across secondary is\",Vs,\"V\"\n",
+ "print\"(ii) The current in primary is\",Ip,\"A\"\n",
+ "print\"(iii) The power output is\",P*10**-3,\"K W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Voltage across secondary is 4400.0 V\n",
+ "(ii) The current in primary is 100.0 A\n",
+ "(iii) The power output is 22.0 K W\n"
+ ]
+ }
+ ],
+ "prompt_number": 79
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.13 Page no 797"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=120*10**3 #W\n",
+ "R=0.4 #ohm\n",
+ "Ev=240.0 #V\n",
+ "Ev1=24000.0 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Iv=P/Ev\n",
+ "P=Iv**2*R\n",
+ "Iv1=P/Ev1\n",
+ "P1=Iv1**2*R\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Power loss at 240 V is\", P*10**-3,\"K W\"\n",
+ "print\"(ii) Power loss at 24000 V is\", round(P1,0),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Power loss at 240 V is 100.0 K W\n",
+ "(ii) Power loss at 24000 V is 7.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 94
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14 Page no 797"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Np=5000\n",
+ "Vp=2200 #V\n",
+ "Vs=220 #V\n",
+ "Pout=8 #K W\n",
+ "n=0.9\n",
+ "\n",
+ "#Calculation\n",
+ "Ns=(Vs*Np)/Vp\n",
+ "Pin=Pout/n\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The number of turns in the secondary is\", Ns\n",
+ "print\"(ii) Input power is\",round(Pin,1),\"K W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The number of turns in the secondary is 500\n",
+ "(ii) Input power is 8.9 K W\n"
+ ]
+ }
+ ],
+ "prompt_number": 99
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.15 Page no 797"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vp=220.0 #V\n",
+ "Vs=22 #V\n",
+ "Z=220 #ohm\n",
+ "Is=0.1\n",
+ "\n",
+ "#Calclation\n",
+ "Ip=(Vs*Is)/Vp\n",
+ "\n",
+ "#Result\n",
+ "print\"Current drawn is\", Ip,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current drawn is 0.01 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 104
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.16 Page no 798"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vs=24 #v\n",
+ "R=9.6 #ohm\n",
+ "Vp=120.0\n",
+ "\n",
+ "#Calculation\n",
+ "Is=Vs/R\n",
+ "Ip=(Vs*Is)/Vp\n",
+ "P1=Vs*Is\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Current in the secondary coil is\", Is,\"A\"\n",
+ "print\"(ii) Current in primary coil is\",Ip ,\"A\"\n",
+ "print\"Power used is\",P1,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Current in the secondary coil is 2.5 A\n",
+ "(ii) Current in primary coil is 0.5 A\n",
+ "Power used is 60.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 116
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter15_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter15_3.ipynb new file mode 100644 index 00000000..111cd390 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter15_3.ipynb @@ -0,0 +1,381 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:977816390da876a89acf9dab4a43ac1149d2a8f7e4f57b74a89090971103e376"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 15 Electromagnetic waves"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.1 Page no 836"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=8.854*10**-12 #C**2/N/m**2\n",
+ "A=10**-4 #m**2\n",
+ "E=3*10**6 #V/ms\n",
+ "\n",
+ "#Calculation\n",
+ "Id=e*A*E\n",
+ "\n",
+ "#Result\n",
+ "print\"Displacement current is\", round(Id*10**9,1)*10**-9,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Displacement current is 2.7e-09 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.2 Page no 836"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Id=1 #A\n",
+ "C=10.0**-6 #F\n",
+ "\n",
+ "#Calculation\n",
+ "V=Id/C\n",
+ "\n",
+ "#Result\n",
+ "print\"Instantaneous current is\", V,\"V/S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Instantaneous current is 1000000.0 V/S\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.3 Page no 836"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=0.15 #A\n",
+ "R=0.12 #m\n",
+ "r=0.065 #m\n",
+ "r1=0.15 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=math.pi*R**2\n",
+ "u=4*math.pi*10**-7\n",
+ "B=(u*I*r)/(2*math.pi*R**2)\n",
+ "B1=(u*I)/(2*math.pi*r1)\n",
+ "Bmax=(u*I)/(2*math.pi*R)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) (a) Magnetic field on the axis is zero\"\n",
+ "print\"(b) Magnetic field at r=6.5 cm is\",round(B*10**7,2)*10**-7 ,\"T\"\n",
+ "print\"(c) Magnetic field at r=15 cm is\", B1,\"T\"\n",
+ "print\"(ii) Distance is\", Bmax,\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) (a) Magnetic field on the axis is zero\n",
+ "(b) Magnetic field at r=6.5 cm is 1.35e-07 T\n",
+ "(c) Magnetic field at r=15 cm is 2e-07 T\n",
+ "(ii) Distance is 2.5e-07 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.5 Page no 837"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.05 #m\n",
+ "E=10**12 #V/m/s\n",
+ "e=8.854*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Id=e*math.pi*r**2*E\n",
+ "\n",
+ "#Result\n",
+ "print\"Displacement current is\", round(Id,4),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Displacement current is 0.0695 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.7 Page no 846 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=100 #v\n",
+ "c=3.0*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=E/c\n",
+ "u=4.0*math.pi*10**-7\n",
+ "H=B/u\n",
+ "U=e*E**2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Value of B is\", round(B*10**7,2)*10**-7,\" T\"\n",
+ "print\"(ii) Value of H is\",round(H,3),\"A/m\"\n",
+ "print\"(iii) Energy density is\",U,\"J/m**3"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Value of B is 3.33e-07 T\n",
+ "(ii) Value of H is 0.265 A/m\n",
+ "(iii) Energy density is 8.854e-08\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.8 Page no 847"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E0=8*10**-4 #v\n",
+ "c=3.0*10**8\n",
+ "w=6*10**6\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B0=E0/c\n",
+ "f=w/(2.0*math.pi)\n",
+ "l=c/f\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of the wave is\", round(l*10**-4,4),\"m\"\n",
+ "print\"Frequency is\",round(f*10**-6,3),\"*10**10 Hz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of the wave is 0.0314 m\n",
+ "Frequency is 0.955 *10**10 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.9 Page no 847"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=6.3 #V/m\n",
+ "c=3.0*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "B=E/c\n",
+ "\n",
+ "#Result\n",
+ "print\"B=\", B,\"K^ Tesla\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "B= 2.1e-08 K^ Tesla\n"
+ ]
+ }
+ ],
+ "prompt_number": 59
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.10 Page no 847"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=18 #W/cm**2\n",
+ "A=20 #cm**2\n",
+ "t=30*60\n",
+ "\n",
+ "#Calculation\n",
+ "U=f*A*t\n",
+ "P=U/c\n",
+ "F=P/t\n",
+ "P1=2*P\n",
+ "F1=P1/t\n",
+ "\n",
+ "#Result\n",
+ "print\"Average force exerted on the surface is\", F1,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average force exerted on the surface is 2.4e-06 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 65
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.11 Page no 847"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=3.0 #m\n",
+ "n=0.025\n",
+ "P=100 #w\n",
+ "C=3*10**8\n",
+ "\n",
+ "#Calculation \n",
+ "import math\n",
+ "A=4*math.pi*r**2\n",
+ "I=(n*P)/A\n",
+ "E0=math.sqrt((2*I)/(e*C))\n",
+ "B0=E0/C\n",
+ "\n",
+ "#Result\n",
+ "print\"Peak value of electric field is\", round(E0,2),\"V/m\"\n",
+ "print\"Peak value of magnetic field is\",round(B0*10**8,2)*10**-8,\"T\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Peak value of electric field is 4.08 V/m\n",
+ "Peak value of magnetic field is 1.36e-08 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 76
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter16_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter16_3.ipynb new file mode 100644 index 00000000..f74b4c1b --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter16_3.ipynb @@ -0,0 +1,369 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:badd30ad9356c0b0c69f3c0e035f97e954d74c789af08b85cb880fbe7e95ff6b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 16 Reflection of light"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Exaple 16.1 Page no 890"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "HF=2\n",
+ "EF=1.9\n",
+ "\n",
+ "#Calculation\n",
+ "L=0.5*HF\n",
+ "DB=0.5*EF\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum height of mirror is\", L,\"m\"\n",
+ "print\"Bottom edge of the mirror is\",DB,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum height of mirror is 1.0 m\n",
+ "Bottom edge of the mirror is 0.95 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.2 Page no 890"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-15.0 #cm\n",
+ "f=-10 #cm\n",
+ "o=2.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/((1.0/f)-(1.0/u))\n",
+ "m=v/u\n",
+ "I=-m*o\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image is\", v,\"cm\"\n",
+ "print\"Size of the image is\",I,\"cm\"\n",
+ "print\"Nature of the image isreal and inverted \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the image is -30.0 cm\n",
+ "Size of the image is -4.0 cm\n",
+ "Nature of the image isreal and inverted \n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.3 Page no 891"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-10.0 #cm\n",
+ "f=-15.0\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/((1/f)-(1/u))\n",
+ "m=-v/u\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Image position is\", v,\"cm\"\n",
+ "print\"(ii) Magnification is\", m"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Image position is 30.0 cm\n",
+ "(ii) Magnification is 3.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.4 Page no 891"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=12.0\n",
+ "v=4.0\n",
+ "\n",
+ "#Calculation\n",
+ "u=1/((1/f)-(1/v))\n",
+ "m=-v/u\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Object position is\", u,\"cm\"\n",
+ "print\"(ii) Magnification is\", round(m,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Object position is -6.0 cm\n",
+ "(ii) Magnification is 0.67\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.5 Page no 891"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=36 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "f=-R/2.0\n",
+ "u=(2*f)/3.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the object is\", u,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the object is -12.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.6 Page no 892"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=20 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=R/2.0\n",
+ "u=-f\n",
+ "v=-u/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the object is\",u,\"cm\"\n",
+ "print\"Position of the image is\",v,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the object is -10.0 cm\n",
+ "Position of the image is 5.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.7 Page no 892"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=-15.0\n",
+ "\n",
+ "#Calculation\n",
+ "u=(1/(1/f)/3.0)*4\n",
+ "v=u/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of object is\" ,u,\"cm\"\n",
+ "print\"When the image is virtual\",v,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of object is -20.0 cm\n",
+ "When the image is virtual -10.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.8 Page no 892"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=30 #ohm\n",
+ "u=-10.0 \n",
+ "h1=5\n",
+ "\n",
+ "#Calculation\n",
+ "f=R/2.0\n",
+ "v=1/((1/f)-(1/u))\n",
+ "h2=(-v*h1)/u\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image is\", v,\"cm\"\n",
+ "print\"Size of the image is\",h2,\"cm\"\n",
+ "print\"The image is virtual\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the image is 6.0 cm\n",
+ "Size of the image is 3.0 cm\n",
+ "The image is virtual\n"
+ ]
+ }
+ ],
+ "prompt_number": 58
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.9 Page no 893"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=-10.0 #cm\n",
+ "u=-25.0 #cm\n",
+ "h1=3\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/((1/f)-(1/u))\n",
+ "h2=(-v*h1)/u\n",
+ "A=h2**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Area enclosed by the image of the wire is\", A,\"cm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Area enclosed by the image of the wire is 4.0 cm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter17_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter17_3.ipynb new file mode 100644 index 00000000..6ddbfa96 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter17_3.ipynb @@ -0,0 +1,1299 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:7117da667e9242c9a19d8eb5f355fd755219357cacc372784f1e3ef0c539e46b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 17 Refraction of the light"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.1 Page no 918"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u1=1.50\n",
+ "u2=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "sinr=u1*math.sin(50*3.14/180.0)/u2\n",
+ "a=math.asin(sinr)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of refraction is\", round(a,1),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of refraction is 59.8 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.2 Page no 918"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u1=1.0\n",
+ "u2=1.526\n",
+ "i=45 #degree\n",
+ "#Calculation\n",
+ "sinr=(u1*math.sin(i*3.14/180.0))/u2\n",
+ "r=math.asin(sinr)*180/3.14\n",
+ "d=i-r\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of deviation is\", round(d,2),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of deviation is 17.39 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.3 Page no 918"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "c=3.0*10**8\n",
+ "u=1.5\n",
+ "f=6*10**14 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "v=c/u\n",
+ "l=c/f\n",
+ "lm=v/f\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Wavelength of light in air is\", l,\"m\"\n",
+ "print\"(ii) Wavelength of light in glass is\",round(lm*10**7,1)*10**-7,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Wavelength of light in air is 5e-07 m\n",
+ "(ii) Wavelength of light in glass is 3.3e-07 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.4 Page no 919"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ug=1.5\n",
+ "uw=1.3\n",
+ "vw=2.25*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "vg=(uw*vw)/ug\n",
+ "\n",
+ "#Result\n",
+ "print\"Speed of the light in glass is\", vg*10**-8,\"*10**8 m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Speed of the light in glass is 1.95 *10**8 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.5 Page no 919"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.6\n",
+ "t=8\n",
+ "t1=4.5\n",
+ "u1=1.5\n",
+ "t2=6\n",
+ "u2=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "d=t*(1-(1/u))\n",
+ "d1=t1*(1-(1/u1))\n",
+ "d2=t2*(1-(1/u2))\n",
+ "D=d+d1+d2\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of mark from the bottom is\", round(D,0),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of mark from the bottom is 6.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.6 Page no 919"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "uw=1.33\n",
+ "uo=1.20\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "uow=uw/uo\n",
+ "sinr=(math.sin(30*3.14/180.0))/uow\n",
+ "r=math.asin(sinr)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of refraction in water is\", round(r,1),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of refraction in water is 26.8 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.7 Page no 920"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=2.0*10**8 #m/s\n",
+ "c=3*10**8 #m/s\n",
+ "d=6.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "ug=c/v\n",
+ "a=d/ug\n",
+ "D=d-a\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance through which ink dot appears to be raised is\", D,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance through which ink dot appears to be raised is 2.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 54
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.8 Page no 924"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ug=1.5\n",
+ "uw=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "u1=ug/uw\n",
+ "sinC=1/u1\n",
+ "C=math.asin(sinC)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Critical angle is\", round(C,2),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Critical angle is 62.49 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 60
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.9 Page no 924"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=1.5*10**8\n",
+ "c=3.0*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=v/c\n",
+ "C=math.asin(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of critical angle is\", round(C,0),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of critical angle is 30.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 65
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.10 Page no 924"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "uw=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "a=1/uw\n",
+ "b=math.sin(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of refraction is\", round(b,0),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of refraction is 39.0 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.11 Page no 924"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=4\n",
+ "b=6.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=a/b\n",
+ "B=math.atan(A)*180/3.14\n",
+ "ur=1/(math.sin(B*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"Refrective index of the liquid is\", round(ur,1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Refrective index of the liquid is 1.8\n"
+ ]
+ }
+ ],
+ "prompt_number": 79
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.12 Page no 925"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=52 #Degree\n",
+ "b=33 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "u2=(math.sin(a*3.14/180.0))/(math.sin(b*3.14/180.0))\n",
+ "C=1/u2\n",
+ "A=math.asin(C)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of refrection is\", round(A,1),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of refrection is 43.7 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 88
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.13 Page no 932"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-240.0\n",
+ "R=15.0 #cm\n",
+ "u1=1.33\n",
+ "u2=1.5\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/((((u2-u1)/R)+(u1/u))/u2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image is\", round(v,0),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the image is 259.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 94
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.14 Page no 932"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-9.0 #cm\n",
+ "y=1\n",
+ "y1=1.5\n",
+ "R=-15.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/(((y-y1)/R)-(y1/-u))\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of distance is\",v,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of distance is -7.5 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.15 Page no 933 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-15 #cm\n",
+ "y1=1\n",
+ "y2=1.5\n",
+ "R=-7.5 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/(((y1-y2)/R)-(y2/-u))\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image is\",v,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the image is -30.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.16 Page no 933"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-60.0 #cm\n",
+ "R=25.0 #cm\n",
+ "y1=1\n",
+ "y2=1.5\n",
+ "\n",
+ "#Calcution\n",
+ "v=1/((((y2-y1)/R)+(y1/u))/y2)\n",
+ "P=(y2-y1)/(R*10**-2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image is\", v,\"cm\"\n",
+ "print\"Power of the refracting surface is\", P,\"dioptre\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the image is 450.0 cm\n",
+ "Power of the refracting surface is 2.0 dioptre\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.17 Page no 934"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u1=1\n",
+ "u2=1.5\n",
+ "R=1\n",
+ "\n",
+ "#Calculation\n",
+ "x=(u1+u2)/(u2-u1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance of the object is\", x,\"R\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance of the object is 5.0 R\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.18 Page no 934"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=7.5 #cm\n",
+ "u1=1\n",
+ "u2=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/(((u1-u2)/R))\n",
+ "\n",
+ "#Result\n",
+ "print\"It gets focused at\", round(v,1),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "It gets focused at -22.7 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.19 Page no 935"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u1=1\n",
+ "u2=1.5\n",
+ "u=-10\n",
+ "v=-40 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "R=-v*(u2-u1)/(u1+u2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Curvature given to the bounding surface is\", R,\"cm (Convex)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Curvature given to the bounding surface is 8.0 cm (Convex)\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.20 Page no 935"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u1=1\n",
+ "u2=1.5\n",
+ "v=100 #cm\n",
+ "R=20.0 #cm\n",
+ "a=3\n",
+ "b=200.0\n",
+ "\n",
+ "#Calculation\n",
+ "u1=(u2-u1)/R\n",
+ "u2=-1/(u1-(a/b))\n",
+ "d=-u2+R\n",
+ "\n",
+ "#Result\n",
+ "print\"The object distance from the centre of curvature is\", d,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The object distance from the centre of curvature is 120.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.21 Page no 952"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ug=1.5\n",
+ "R1=50.0 #cm\n",
+ "R2=-50.0 #cm\n",
+ "uw=9/8.0\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/((ug-1)*((1/R1)+(1/R1)))\n",
+ "f1=1/((uw-1)*((1/R1)+(1/R1)))\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Focal length in air is\", f,\"cm\"\n",
+ "print\"(ii) Focal lenth in water is\", f1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Focal length in air is 50.0 cm\n",
+ "(ii) Focal lenth in water is 200.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 50
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.22 Page no 953"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "fa=20 #cm\n",
+ "ug=9/8.0\n",
+ "uw=3/2.0\n",
+ "\n",
+ "#Calculation\n",
+ "a=(uw-1)/(ug-1)\n",
+ "fw=a*fa\n",
+ "f=fw-fa\n",
+ "\n",
+ "#Result\n",
+ "print\"Change in focal length is\", f,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Change in focal length is 60.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 54
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.23 Page no 953"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.56\n",
+ "R1=20.0 #cm\n",
+ "u1=-10.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/((u-1)*(2/R1))\n",
+ "v=1/((1/u1)+(1/f))\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image formed is\", round(v,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the image formed is -22.73\n"
+ ]
+ }
+ ],
+ "prompt_number": 63
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.24 Page no 953"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.47\n",
+ "\n",
+ "#Calculation\n",
+ "u1=u\n",
+ "\n",
+ "#Result\n",
+ "print\"The liquid is not water because refractive index of water is 1.33\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The liquid is not water because refractive index of water is 1.33\n"
+ ]
+ }
+ ],
+ "prompt_number": 65
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.25 Page no 954"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=18 #cm\n",
+ "u=1.5\n",
+ "\n",
+ "#Calculation\n",
+ "R=(u-1)*f\n",
+ "\n",
+ "#Result\n",
+ "print\"Radius of the curvature is\", R,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of the curvature is 9.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.26 Page no 954"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-25.0 #cm\n",
+ "f=10.0 #cm\n",
+ "h1=5\n",
+ "\n",
+ "#Calculation\n",
+ "v=1/((1/f)+(1/u))\n",
+ "h2=(v*h1)/u\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image is\", round(v,2),\"cm\"\n",
+ "print\"Size of the image is\",round(h2,2),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the image is 16.67 cm\n",
+ "Size of the image is -3.33 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.27 Page no 954"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=-15.0 #cm\n",
+ "v=-10.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "u=1/((1/v)-1/f)\n",
+ "\n",
+ "#Result\n",
+ "print\"The object is placed at a distance of\", u,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The object is placed at a distance of -30.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 81
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.28 Page no 954"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=-20.0 #cm\n",
+ "u=-60.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/((1/v)-(1/u))\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length of the lens is\", f,\"cm\"\n",
+ "print\"The lens is diverging\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length of the lens is -30.0 cm\n",
+ "The lens is diverging\n"
+ ]
+ }
+ ],
+ "prompt_number": 85
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.29 Page no 955"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-10.0 #cm\n",
+ "m=-3.0\n",
+ "\n",
+ "#Calculation\n",
+ "v=m*u\n",
+ "f=1/((1/v)-(1/u))\n",
+ "\n",
+ "#Result\n",
+ "print\"Image formed at\",v,\"cm\"\n",
+ "print\"Focal length is\",f,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Image formed at 30.0 cm\n",
+ "Focal length is 7.5 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 94
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.30 Page no 955"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P1=6\n",
+ "P2=-2.0\n",
+ "\n",
+ "#Calculation\n",
+ "P=P1+P2\n",
+ "f=1/P\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length of the combination is\", f*10**2,\"cm\"\n",
+ "print\"Power of the combinationis\",P,\"D\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length of the combination is 25.0 cm\n",
+ "Power of the combinationis 4.0 D\n"
+ ]
+ }
+ ],
+ "prompt_number": 102
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.31 Page no 955"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f1=20.0 #cm\n",
+ "f2=-40.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/((1/f1)+(1/f2))\n",
+ "P=1/f\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length is\", f,\"cm\"\n",
+ "print\"Power is\",P*10**2,\"D\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length is 40.0 cm\n",
+ "Power is 2.5 D\n"
+ ]
+ }
+ ],
+ "prompt_number": 107
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.32 Page no 955"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=2.0\n",
+ "b=1\n",
+ "\n",
+ "#Calculation\n",
+ "u=(b/a)+b\n",
+ "\n",
+ "#Result\n",
+ "print\"Refractive index of the material is\", u"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Refractive index of the material is 1.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 109
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.33 Page no 955"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=-0.2 #m\n",
+ "v=0.3 #m\n",
+ "\n",
+ "#Calculation\n",
+ "u=1/((1/v)-(1/f))\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the point is\", u,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the point is 0.12 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 113
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17.35 Page no 957"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u1=-30.0 #cm\n",
+ "f1=10.0\n",
+ "u2=10\n",
+ "f2=-10.0\n",
+ "\n",
+ "#calculation\n",
+ "v1=1/((1/u1)+(1/f1))\n",
+ "v2=1/((1/u2)+(1/f2))\n",
+ "v3=-u1\n",
+ "\n",
+ "#Result\n",
+ "print\"Position of the image for first lens is\", v1,\"cm\"\n",
+ "print\"Position of the image for second lens is\", round(v2*10**-2,0),\"cm\"\n",
+ "print\"Position of the image for third lens is\", v3,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Position of the image for first lens is 15.0 cm\n",
+ "Position of the image for second lens is -0.0 cm\n",
+ "Position of the image for third lens is 30.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 127
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter18_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter18_3.ipynb new file mode 100644 index 00000000..46d58979 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter18_3.ipynb @@ -0,0 +1,549 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:b23935cae4f05cd3030e505584dd90917a65a9efe1b245c771989ab0310cdeb5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 18 Dispersion of light"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.1 Page no 986"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=60 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=math.sqrt(2)*math.sin(30*3.14/180.0)\n",
+ "b=math.asin(a)*180/3.14\n",
+ "c=(b*2)-A\n",
+ "i=(A+c)/2.0\n",
+ "r=A/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Angle of minimum deviation is\", round(c,0),\"Degree\"\n",
+ "print\"(ii) Angle of incidence is\", round(i,0),\"Degree\"\n",
+ "print\"(iii) The angle of refraction is\", r,\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Angle of minimum deviation is 30.0 Degree\n",
+ "(ii) Angle of incidence is 45.0 Degree\n",
+ "(iii) The angle of refraction is 30.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.2 Page no 986"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=51 #Degree\n",
+ "A=60 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "b=(A+a)/2.0\n",
+ "c=A/2.0\n",
+ "u=(math.sin(b*3.14/180.0))/(math.sin(c*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The refracting angle of the prism is\", A,\"Degree\"\n",
+ "print\"(ii) The refractive index of the material is\",round(u,4)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The refracting angle of the prism is 60 Degree\n",
+ "(ii) The refractive index of the material is 1.6485\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.3 Page no 987"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "i1=48 #Degree\n",
+ "A=60 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=A/2.0\n",
+ "u=math.sin(i1*3.14/180.0)/math.sin(r*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Refractive index of the material is\", round(u,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Refractive index of the material is 1.49\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.4 Page no 987"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=2.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=math.sqrt(a)/a\n",
+ "i=math.asin(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of incidence is\", round(i,0),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of incidence is 45.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.5 Page no 987"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.5\n",
+ "a=6 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "A=a/(u-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of the prism is\", A,\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of the prism is 12.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.6 Page no 987"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ug=1.5\n",
+ "r1=30 #Degree\n",
+ "ua=1.0\n",
+ "A=60 #Degree\n",
+ "A1=90 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "sin=(ug*math.sin(r1*3.14/180.0))/ua\n",
+ "i1=math.asin(sin)*180/3.14\n",
+ "a=(2*i1)-A\n",
+ "sin1=1/ug\n",
+ "r1=math.asin(sin1)*180/3.14\n",
+ "r2=A-r1\n",
+ "sin2=(ug*math.sin(r2*3.14/180.0))\n",
+ "i2=math.asin(sin2)*180/3.14\n",
+ "A3=A1+i2-A\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The angle of incidence for minimum deviation is\", round(i1,0),\"Degree\"\n",
+ "print\"(ii) The angle of minimum deviation is\", round(a,0)\n",
+ "print\"(iii) The angle of emergence of light at maximum deviation is\", round(i2,0),\"Degree\"\n",
+ "print\"(iv) Angle of maximum deviation is\", round(A3,0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The angle of incidence for minimum deviation is 49.0 Degree\n",
+ "(ii) The angle of minimum deviation is 37.0\n",
+ "(iii) The angle of emergence of light at maximum deviation is 28.0 Degree\n",
+ "(iv) Angle of maximum deviation is 58.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.7 Page no 991"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "uv=1.68\n",
+ "ur=1.56\n",
+ "A=18 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "A1=A*(uv-ur)\n",
+ "\n",
+ "#Result\n",
+ "print\"Angular dispersion is\", A1,\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angular dispersion is 2.16 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 71
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.8 Page no 991"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "av=3.32 #Degree\n",
+ "ar=3.22 #Degree\n",
+ "a=3.27 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "w=(av-ar)/a\n",
+ "\n",
+ "#Result\n",
+ "print\"Dispersive power of the flint glass is\", round(w,4)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Dispersive power of the flint glass is 0.0306\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.9 Page no 991"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ur=1.5155\n",
+ "uv=1.5245\n",
+ "\n",
+ "#Calculation\n",
+ "u=(ur+uv)/2.0\n",
+ "w=(uv-ur)/(u-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Dispersive power of the crown glass is\", round(w,4)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Dispersive power of the crown glass is 0.0173\n"
+ ]
+ }
+ ],
+ "prompt_number": 79
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.10 Page no 991"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "w=0.031\n",
+ "ur=1.645\n",
+ "ub=1.665\n",
+ "\n",
+ "#Calculation\n",
+ "u=1+((ub-ur))/w\n",
+ "\n",
+ "#Result\n",
+ "print\"Refrective index for yellow colour is\", round(u,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Refrective index for yellow colour is 1.645\n"
+ ]
+ }
+ ],
+ "prompt_number": 83
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.11 Page no 992"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=5 #Degree\n",
+ "uv=1.523\n",
+ "ur=1.515\n",
+ "uv1=1.688\n",
+ "ur1=1.650\n",
+ "\n",
+ "#Calculation\n",
+ "u=(uv+ur)/2.0\n",
+ "u1=(uv1+ur1)/2.0\n",
+ "A1=-((u-1)*A)/(u1-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of flint line is\",round(A1,2),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of flint line is -3.88 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 91
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.12 Page no 992"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "w=0.021\n",
+ "u=1.53\n",
+ "w1=0.045\n",
+ "u1=1.65\n",
+ "A1=4.20 #Degree\n",
+ "\n",
+ "#Calculation\n",
+ "A=-(w1*A1*(u1-1))/(w*(u-1))\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of the prism is\", round(A,2),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of the prism is -11.04 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18.13 Page no 992"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=72 #Degree\n",
+ "ab=56.4 #Degree\n",
+ "ar=53 #Degree\n",
+ "ay=54.6 #Degree\n",
+ "az=54\n",
+ "A11=60 #Degree\n",
+ "ab1=52.8 \n",
+ "A12=50.6\n",
+ "A13=51.9\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A1=(A+ay)/2.0\n",
+ "A2=A/2.0\n",
+ "ub=(math.sin(A1*3.14/180.0))/(math.sin(A2*3.14/180.0))\n",
+ "A3=(A+ar)/2.0\n",
+ "ur=(math.sin(A3*3.14/180.0))/(math.sin(A2*3.14/180.0))\n",
+ "A4=(A+az)/2.0\n",
+ "uy=(math.sin(A4*3.14/180.0))/(math.sin(A2*3.14/180.0))\n",
+ "w=(ub-ur)/(uy-1)\n",
+ "\n",
+ "#For flint glass prism\n",
+ "A5=(A11+ab1)/2.0\n",
+ "A51=A11/2.0\n",
+ "ub1=(math.sin(A5*3.14/180.0))/(math.sin(A51*3.14/180.0))\n",
+ "A6=(A11+A12)/2.0\n",
+ "ur1=(math.sin(A6*3.14/180.0))/(math.sin(A51*3.14/180.0))\n",
+ "A7=(A11+A13)/2.0\n",
+ "uy1=(math.sin(A7*3.14/180.0))/(math.sin(A51*3.14/180.0))\n",
+ "w1=(ub1-ur1)/(uy1-1)\n",
+ "w2=w/w1\n",
+ "\n",
+ "#Result\n",
+ "print\"The ratio of dispersive power of crown glass and flint glass prism is\", round(w2,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of dispersive power of crown glass and flint glass prism is 0.64\n"
+ ]
+ }
+ ],
+ "prompt_number": 122
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter19_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter19_3.ipynb new file mode 100644 index 00000000..3a7eac21 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter19_3.ipynb @@ -0,0 +1,841 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:91c393a3f1616e8ab4ec5b337712a2b4f8e8dca0635755aebf9e9a0db573e23b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 19 Optical instruments"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.1 Page no 1013"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=-75.0\n",
+ "u=0\n",
+ "\n",
+ "#Calculation\n",
+ "f=v\n",
+ "P=100/f\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length is\", f,\"cm\"\n",
+ "print\"Power of the lens is\",round(P,2),\"D\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length is -75.0 cm\n",
+ "Power of the lens is -1.33 D\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.2 Page no 1014"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-25.0 #cm\n",
+ "v=-150.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/((1/v)-1/u)\n",
+ "P=100/f\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length of the lens is\", f,\"cm\"\n",
+ "print\"Power of the lens is\",round(P,2),\"D\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length of the lens is 30.0 cm\n",
+ "Power of the lens is 3.33 D\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.3 Page no 1014"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-25.0 #cm\n",
+ "v=-50.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/((1/v)-1/u)\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length is\", f,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length is 50.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.4 Page no 1014"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=-80.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=v\n",
+ "P=100/f\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Power of the lens is\", P,\"D\"\n",
+ "print\"(b) No the corrective lens is concave and it reduces the size of the image. Because it bring the object at the far point of the eye\"\n",
+ "print\"(c) The myopic person may have a normal near point. He must keep the book at a distance greater than 25 cm.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Power of the lens is -1.25 D\n",
+ "(b) No the corrective lens is concave and it reduces the size of the image. Because it bring the object at the far point of the eye\n",
+ "(c) The myopic person may have a normal near point. He must keep the book at a distance greater than 25 cm.\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.5 Page no 1015"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=-75.0 #cm\n",
+ "u=-25.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=1/((1/v)-1/u)\n",
+ "P=100/f\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Power of the lens is\", round(P,2),\"D\"\n",
+ "print\"(b) The corrective lens produce a virtual imageof an object at 25 cm. The angular size of this image is the same as the object\"\n",
+ "print\"(c) A hypermetropic eye may have normal far point.Hence the person prefers not to use the spectacles for distant object\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Power of the lens is 2.67 D\n",
+ "(b) The corrective lens produce a virtual imageof an object at 25 cm. The angular size of this image is the same as the object\n",
+ "(c) A hypermetropic eye may have normal far point.Hence the person prefers not to use the spectacles for distant object\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.6 Pageno 1015"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=-0.8 #d\n",
+ "v1=-15.0 #cm \n",
+ "v2=-100.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f=100/P\n",
+ "u1=1/((1/v1)-1/f)\n",
+ "u2=1/((1/v2)-(1/f))\n",
+ "\n",
+ "#Result\n",
+ "print\"The person can see objects lying between\",round(-u1,0),\"cm and\",-u2,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The person can see objects lying between 17.0 cm and 500.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.7 Page no 1016"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-25 #cm\n",
+ "p=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "f=100/p\n",
+ "v=1/((1/f)+1/u)\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance is\",round(v,0),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance is -1.0 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.8 Page no 1016"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=-25.0 #cm\n",
+ "v=-90.0 #cm\n",
+ "\n",
+ "#calculation\n",
+ "f=1/((1/v)-1/u)\n",
+ "f1=(1/2.0)*10**2\n",
+ "u=1/((1/v)-1/f1)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) focal length is\",round(f,1),\"cm\"\n",
+ "print\"(ii) Distance is\",round(u,1),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) focal length is 34.6 cm\n",
+ "(ii) Distance is -32.1 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.9 Page no 1022"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "D=25\n",
+ "f=5.0 #cm\n",
+ "\n",
+ "#calculation\n",
+ "M=1+(D/f)\n",
+ "M1=D/f\n",
+ "\n",
+ "#Result\n",
+ "print\"The magnifying power if the final image is formed at the least distance is\",M\n",
+ "print\"The magnifying power if image is formed at infinity is\",M1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The magnifying power if the final image is formed at the least distance is 6.0\n",
+ "The magnifying power if image is formed at infinity is 5.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 61
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.10 Page no 1023"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f=4.80 #cm\n",
+ "a=1.20\n",
+ "v=-24.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "D=f/(a-1)\n",
+ "u=1/((1/v)-1/f)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The least distance of distinct vision is\",D,\"cm\"\n",
+ "print\"(ii) Distance from the lens is\",-u,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The least distance of distinct vision is 24.0 cm\n",
+ "(ii) Distance from the lens is 4.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 69
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.11 Page no 1023"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v0=15.0 #cm\n",
+ "f0=3.0 #cm\n",
+ "D=25\n",
+ "fe=9\n",
+ "\n",
+ "#Calculation\n",
+ "u0=1/((1/v0)-1/f0)\n",
+ "M=-(v0*D)/(u0*fe)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnifying power is\", round(M,1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnifying power is 11.1\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.12 Page no 1024"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P1=1.5 #D\n",
+ "P2=20.0 #D\n",
+ "u=-25.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "f2=100/P2\n",
+ "M=1+(D/f2)\n",
+ "f1=100/P1\n",
+ "v=1/((1/f1)+1/u)\n",
+ "M1=1-(v/f2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The maximum magnifying power together with his glasses\", M\n",
+ "print\"(ii) The maximum magnifying power without glasses\",M1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The maximum magnifying power together with his glasses 6.0\n",
+ "(ii) The maximum magnifying power without glasses 9.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 83
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.13 Page no 1024"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=16\n",
+ "d=-2.5 #cm\n",
+ "f0=0.4 #cm\n",
+ "D=25\n",
+ "\n",
+ "#Calculation\n",
+ "v0=l+d\n",
+ "u0=1/((1/v0)-1/f0)\n",
+ "M=-v0*D/(u0*d)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnifying power of the microscope is\", M"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnifying power of the microscope is -327.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 88
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.14 Page no 1025"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f0=1.0\n",
+ "u0=-1.1 #cm\n",
+ "D=25\n",
+ "fe=5.0\n",
+ "ve=25.0\n",
+ "\n",
+ "#Calculation\n",
+ "v0=1/((1/f0)+1/u0)\n",
+ "d=v0+fe\n",
+ "M=-(v0*D)/(u0*fe)\n",
+ "ue=-1/((1/ve)+1/fe)\n",
+ "D1=v0-ue\n",
+ "M1=-(v0/u0)*(1+(D/fe))\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Distance between the lenses when image is at infinity\", d,\"cm\"\n",
+ "print\"Magnifying power is\",M\n",
+ "print\"(ii) Distance between the lenses when image is at distinct vision\",round(D1,2),\"cm\"\n",
+ "print\"Magnifying Power is\",M1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Distance between the lenses when image is at infinity 16.0 cm\n",
+ "Magnifying power is 50.0\n",
+ "(ii) Distance between the lenses when image is at distinct vision 15.17 cm\n",
+ "Magnifying Power is 60.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 106
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.15 Page no 1032"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "f0=200 #cm\n",
+ "fe=5.0 #cm\n",
+ "D=25.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "M=(f0/fe)*(1+(fe/D))\n",
+ "M1=f0/fe\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Magnifying power when image is formed at near point is\", M\n",
+ "print\"(ii) Magnifying power when image is formed at infinity\",M1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnifying power when image is formed at near point is 48.0\n",
+ "(ii) Magnifying power when image is formed at infinity 40.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 110
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.16 Page no 1033"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "fe=3\n",
+ "M=4\n",
+ "\n",
+ "#Calculation\n",
+ "f0=fe*M\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length of the lenses is\" ,f0,\"cm and\",fe,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length of the lenses is 12 cm and 3 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 112
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.17 Page no 1033"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u0=-200.0 #cm\n",
+ "f0=30.0 #cm\n",
+ "fe=3\n",
+ "\n",
+ "#Calculation\n",
+ "v0=1/((1/f0)+1/u0)\n",
+ "a=v0+fe\n",
+ "\n",
+ "#Result\n",
+ "print\"Separation between the objective and eyepiece is\", round(a,1),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Separation between the objective and eyepiece is 38.3 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 118
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.18 Page no 1033"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ve=24.0\n",
+ "fe=8.0\n",
+ "f0=250.0\n",
+ "a=10\n",
+ "\n",
+ "#Calculation\n",
+ "ue=1/((1/ve)-(1/fe))\n",
+ "D=f0-ue\n",
+ "d=a/2.0\n",
+ "A=d/f0\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Distance between objective and eyepiece is\", D,\"cm\"\n",
+ "print\"(ii) Angle subtended by the sun at the objective is\",A,\"rad\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Distance between objective and eyepiece is 262.0 cm\n",
+ "(ii) Angle subtended by the sun at the objective is 0.02 rad\n"
+ ]
+ }
+ ],
+ "prompt_number": 126
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.19 Page no 1034"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=-20\n",
+ "R=-120\n",
+ "\n",
+ "#Calculation\n",
+ "f0=R/2.0\n",
+ "fe=f0/M\n",
+ "\n",
+ "#Result\n",
+ "print\"Focal length of eyepiece is\", fe,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Focal length of eyepiece is 3.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 131
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.20 Page no 1034"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "fa=180\n",
+ "f=3.5\n",
+ "fe=5.0\n",
+ "\n",
+ "#Calculation\n",
+ "d=fa+(2*f)+(2*f)+fe\n",
+ "M=-fa/fe\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnifying power of thetelescope is\", M"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnifying power of thetelescope is -36.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 135
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19.21 Page no 1034"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u0=-200.0 #cm\n",
+ "fa=50.0 #cm\n",
+ "ve=-25.0 #cm\n",
+ "fe=5.0 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "v0=1/((1/fa)+1/u0)\n",
+ "M0=v0/u0\n",
+ "ue=1/((1/ve)-1/fe)\n",
+ "Me=ve/ue\n",
+ "D=v0-ue\n",
+ "M=M0*Me\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Saparation between the objective and eyepiece is\", round(D,2),\"cm\"\n",
+ "print\"(ii) Magnification is\",M"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Saparation between the objective and eyepiece is 70.83 cm\n",
+ "(ii) Magnification is -2.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 147
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter1_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter1_3.ipynb new file mode 100644 index 00000000..f6180b5a --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter1_3.ipynb @@ -0,0 +1,543 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:9880f2d8505e271317a099910ead6c2116ce86fa0e83f56feb35ac33a1b96b23"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 1 Electric charge"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.1 Page no 9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=4.5*10**-19 #C\n",
+ "e=1.6*10**-19 #C\n",
+ "\n",
+ "#Calculation\n",
+ "n=q/e\n",
+ "\n",
+ "#Result\n",
+ "print\"n= \",round(n,1),\"This value of charge is not possible\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "n= 2.8 This value of charge is not possible\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.2 Page no 9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=3.2*10**-7 #C\n",
+ "e=1.6*10**-19 #C\n",
+ "\n",
+ "#Calculation\n",
+ "n=q/e\n",
+ "\n",
+ "#Result\n",
+ "print\"The required number of electrons is \",n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The required number of electrons is 2e+12\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.3 Page no 9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=19.2*10**-19\n",
+ "e=1.6*10**-19\n",
+ "me=9*10**-31 #Kg\n",
+ "\n",
+ "#Calculation\n",
+ "n=q/e\n",
+ "M=n*me\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The value of n=\",n,\"\\n(ii) Charge on silk=\",-q*10**19,\"*10**-19\"\n",
+ "print\"(iii) Mass=\",M,\"Therefore mass transferred is negligibly small\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The value of n= 12.0 \n",
+ "(ii) Charge on silk= -19.2 *10**-19\n",
+ "(iii) Mass= 1.08e-29 Therefore mass transferred is negligibly small\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.4 Page no 9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=16\n",
+ "n=6.023*10**23 #C\n",
+ "\n",
+ "#Calculation\n",
+ "W=2+a\n",
+ "A=((n*100)/W)*10\n",
+ "\n",
+ "#Result\n",
+ "print\"Total number of electrons in 100 g of water \", round(A,-23)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total number of electrons in 100 g of water 3.35e+25\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.5 Page no 9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=10**9\n",
+ "e=1.6*10**-19 #C\n",
+ "Q=1\n",
+ "\n",
+ "#Calculation\n",
+ "q=n*e\n",
+ "t=Q/q\n",
+ "\n",
+ "#Result\n",
+ "print (t*10**-9),\"10**9 S\"\n",
+ "print\"Time required is about 198 years\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "6.25 10**9 S\n",
+ "Time required is about 198 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.6 Page no 13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q1=20 #micro C\n",
+ "q2=-5 #micro C\n",
+ "a=9*10**9\n",
+ "r=0.1 \n",
+ "\n",
+ "#Calculation\n",
+ "q=q1+q2\n",
+ "q3=q/2.0\n",
+ "F=(a*q3*q3)/r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Force is \",round(F*10**-13,3),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force is 5.062 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 1.10 Page no 14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q=5*10**-6\n",
+ "r=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "F=(m*q*q)/r**2\n",
+ "C=2*F*math.cos(30)*(180/3.14)\n",
+ "\n",
+ "#Result\n",
+ "print\"Force on each charge is \", round(C,1)*10**-1,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force on each charge is 39.79 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.11 Page no 14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q=1\n",
+ "r=0.24\n",
+ "A=20\n",
+ "B=12.0\n",
+ "m1=10**-4\n",
+ "g=9.8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "F=(m*q**2)/r**2\n",
+ "AD=math.sqrt(A**2-B**2)\n",
+ "C=AD/B\n",
+ "F1=(1/C)*m1*g\n",
+ "Q=math.sqrt(F1/F)\n",
+ "\n",
+ "#Result\n",
+ "print\"Charge on each sphere\", round(Q*10**8,1),\"10**-8\",\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge on each sphere 6.9 10**-8 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 79
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.12 Page no 15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "F=3.7*10**-9 #C\n",
+ "q=1.6*10**-19 #c\n",
+ "m=9*10**9\n",
+ "r=5*10**-10\n",
+ "\n",
+ "#Calculation \n",
+ "import math\n",
+ "n=math.sqrt(F*r**2/(m*q**2))\n",
+ "\n",
+ "#Result\n",
+ "print round(n,0),\"electrons are missing from each icon\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "2.0 electrons are missing from each icon\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.14 Page no 16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19\n",
+ "m=9*10**9\n",
+ "G=6.67*10**-11\n",
+ "me=9.11*10**-31\n",
+ "mp=1.67*10**-27\n",
+ "r=10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "F0=(m*e**2)/(G*me*mp)\n",
+ "F1=(m*e**2)/(G*mp*mp)\n",
+ "F2=m*e**2/r**2\n",
+ "A1=F2/me\n",
+ "A2=F2/mp\n",
+ "\n",
+ "#Result\n",
+ "print\"(a)(i)strength of an electrons and protons\", round(F0*10**-39,1)*10**39\n",
+ "print\" (ii)Strength of two protons \",round(F1*10**-36,1)*10**36\n",
+ "print\"(b) Acceleration of electron is \",round(A1*10**-22,1)*10**22,\"m/s**2\"\n",
+ "print\" Acceleration of proton is \",round(A2*10**-19,1)*10**19,\"m/s*2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)(i)strength of an electrons and protons 2.3e+39\n",
+ " (ii)Strength of two protons 1.2e+36\n",
+ "(b) Acceleration of electron is 2.5e+22 m/s**2\n",
+ " Acceleration of proton is 1.4e+19 m/s*2\n"
+ ]
+ }
+ ],
+ "prompt_number": 112
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.16 Page no 19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9 #C\n",
+ "q1=10*10**-6\n",
+ "q2=5*10**-6\n",
+ "r=0.05\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "F1=m*q1*q2/r**2\n",
+ "F2=m*q1*q2/r**2\n",
+ "F3=math.sqrt(F1**2+F2**2+(2*F1*F2*math.cos(120)*180/3.14))\n",
+ "\n",
+ "#Result\n",
+ "print\"Resultant charge is \", round(F3*10**-1,0),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resultant charge is 176.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 132
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.17 Page no 20 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q1=1.2*10**-8\n",
+ "q2=1\n",
+ "r=0.03\n",
+ "r1=0.04\n",
+ "q3=1.6*10**-8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "F1=m*q1*q2/r**2\n",
+ "F2=m*q3*q2/r1**2\n",
+ "F3=math.sqrt(F1**2+F2**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Total force is \", F3*10**-5,\"10**5\",\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total force is 1.5 10**5 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 149
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.18 Page no 20"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q1=1\n",
+ "q2=100\n",
+ "r=10\n",
+ "q3=75 #C\n",
+ "r1=5\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "F=m*q1*q2/r**2 #along BA\n",
+ "F1=m*q1*q2/r**2 #along AC\n",
+ "F2=m*q3/(math.sqrt(r**2-r1**2)**2)\n",
+ "F3=math.sqrt(F1**2+F2**2)\n",
+ "X=F1/F2\n",
+ "\n",
+ "#Result\n",
+ "print\"Force experienced by 1 C Charge is \",round(F3*10**-9,2),\"N\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force experienced by 1 C Charge is 12.73 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 168
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter20_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter20_3.ipynb new file mode 100644 index 00000000..e96a7bf3 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter20_3.ipynb @@ -0,0 +1,330 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:49e10d509d6c3c83253662b249f2d9cebaf084cb6d339d2868de883d5e7038f4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 20 Photometry"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.1 Page no 1055"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=2.5*10**5 #lm/m**2\n",
+ "r=1.5*10**11 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l=E*r**2\n",
+ "a=4*math.pi*l\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Luminous intensity is\", l,\"cd\"\n",
+ "print\"(ii) Luminous flux of the sun is\",round(a*10**-28,3)*10**28,\"lm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Luminous intensity is 5.625e+27 cd\n",
+ "(ii) Luminous flux of the sun is 7.069e+28 lm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.2 Page no 1055"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I2=150\n",
+ "I1=75.0\n",
+ "E1=20\n",
+ "\n",
+ "#Calculation\n",
+ "E2=(I2*E1)/I1\n",
+ "\n",
+ "#Result\n",
+ "print\"Illumination is\", E2,\"lux\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Illumination is 40.0 lux\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.3 Page no 1056"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=35\n",
+ "e=5.0 #lumen/watt\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=4*math.pi*I\n",
+ "P=a/e\n",
+ "\n",
+ "#Result\n",
+ "print\"Power of the lamp is\", round(P,0),\"Watt\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power of the lamp is 88.0 Watt\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.4 Page no 1056"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=1260\n",
+ "r=8 #m\n",
+ "a1=6\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=a/(4.0*math.pi)\n",
+ "Ea=I/r**2\n",
+ "LB=math.sqrt(r**2+a1**2)\n",
+ "cos=r/LB\n",
+ "Eb=(I*cos)/LB**2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The illumination at a point immediately below the lamp is\", round(Ea,2),\"lux\"\n",
+ "print\"(ii) The illumination on the working plane is\",round(Eb,1),\"lux\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The illumination at a point immediately below the lamp is 1.57 lux\n",
+ "(ii) The illumination on the working plane is 0.8 lux\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.5 Page no 1056"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=6.0 #m\n",
+ "I=250 #cd\n",
+ "PQ=8\n",
+ "\n",
+ "#Calculation\n",
+ "Ep=I/r**2\n",
+ "LQ=math.sqrt(r**2+PQ**2)\n",
+ "cos=r/LQ\n",
+ "EQ=(I*cos)/LQ**2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Illumination at a point P is\", round(Ep,2),\"lux\"\n",
+ "print\"(ii) illumination at a point Q is\",EQ,\"lux\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Illumination at a point P is 6.94 lux\n",
+ "(ii) illumination at a point Q is 1.5 lux\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.6 Page no 1057"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "t1=2.5 #second\n",
+ "r1=0.5\n",
+ "r2=1\n",
+ "\n",
+ "#Calculation\n",
+ "t2=(t1*r2**2)/r1**2\n",
+ "\n",
+ "#Result\n",
+ "print\"exposure time is\",t2,\"s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "exposure time is 10.0 s\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.7 Page no 1058"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "i2=60\n",
+ "r2=105.0\n",
+ "r1=70\n",
+ "\n",
+ "#Calculation\n",
+ "i1=(i2*r1**2)/r2**2\n",
+ "\n",
+ "#Result\n",
+ "print\"The luminous intensity of the first lamp is\",round(i1,2),\"cd\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The luminous intensity of the first lamp is 26.67 cd\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20.8 Page no 1059"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ra=60\n",
+ "rb=45.0\n",
+ "a=40.0\n",
+ "\n",
+ "#Calculation\n",
+ "ia1=(ra**2)/(rb**2)\n",
+ "ia=(ra**2)/(a**2)\n",
+ "i=ia-ia1\n",
+ "A=(i*100)/ia\n",
+ "\n",
+ "#Result\n",
+ "print\"percentage of light is absorbed by the glass is\",round(A,0),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "percentage of light is absorbed by the glass is 21.0 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 52
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter21_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter21_3.ipynb new file mode 100644 index 00000000..5e2ded02 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter21_3.ipynb @@ -0,0 +1,383 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:5eb9e6d48b48ecf2d0c9ee8abbe7a462b6b60df5a09da8ebed0ac004de2a0383"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 21 Huygen Principle and interference "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.1 Page no 1090"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Goven\n",
+ "d=5*10**-3 #m\n",
+ "D=1.0 #m\n",
+ "b=0.1092*10**-3\n",
+ "\n",
+ "#Calculation\n",
+ "l=(d*b)/D\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of light used is\", l*10**10,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of light used is 5460.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.2 Page no 1090"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=6200*10**-10 #m\n",
+ "D=0.8\n",
+ "b=2.8*10**-3/4.0\n",
+ "\n",
+ "#Calculation\n",
+ "d=(l*D)/b\n",
+ "\n",
+ "#Result\n",
+ "print\"Separation of the two slit is\", round(d*10**3,1),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Separation of the two slit is 0.7 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.3 Page no 1090"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=62\n",
+ "l=5893\n",
+ "l1=4358.0\n",
+ "\n",
+ "#Calculation\n",
+ "n=(a*l)/l1\n",
+ "\n",
+ "#Result\n",
+ "print\"Fringes required is\", round(n,0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Fringes required is 84.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.4 Page no 1091"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=6000*10**-10 #m\n",
+ "D=0.800 #m\n",
+ "d=0.200*10**-3 #m\n",
+ "\n",
+ "#Calculation\n",
+ "x2=(3*l*D)/(2.0*d)\n",
+ "x21=(2*D*l)/d\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Distance of the second dark fringe is\", x2*10**2,\"cm\"\n",
+ "print\"(ii) Distance of the second dark fringe is\", x21*10**2,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Distance of the second dark fringe is 0.36 cm\n",
+ "(ii) Distance of the second dark fringe is 0.48 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.6 Page no 1091"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Imax=16\n",
+ "Imin=4\n",
+ "\n",
+ "#Calculation\n",
+ "r=Imax/Imin\n",
+ "\n",
+ "#Result\n",
+ "print\"Deduce the ratio of intensity is\", r,\":1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deduce the ratio of intensity is 4 :1\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.7 Page no 1092"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "b=2\n",
+ "u=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "b1=b/u\n",
+ "\n",
+ "#Result\n",
+ "print\"Fringe width is\", round(b1,1),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Fringe width is 1.5 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.8 Page no 1092"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "b2=0.4\n",
+ "b1=0.6\n",
+ "l1=5000\n",
+ "\n",
+ "#Calculation\n",
+ "l2=(b2*2*l1)/b1\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of the light is\", round(l2,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of the light is 6667.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.9 Page no 1092"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=0.125*10**-3 #m\n",
+ "l=4500*10**-10 #m\n",
+ "D=1 #m\n",
+ "\n",
+ "#Calculation\n",
+ "x2=(2*D*l)/d\n",
+ "d1=2*x2\n",
+ "\n",
+ "#Result\n",
+ "print\"Separation between the fringes is\", d1*10**3,\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Separation between the fringes is 14.4 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.10 Page no 1092"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Imax=121\n",
+ "Imin=81.0\n",
+ "\n",
+ "#Calculation\n",
+ "a=Imax/Imin\n",
+ "\n",
+ "#Result\n",
+ "print\"The ratio of intensity at the maxima and minima is\",round(a,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of intensity at the maxima and minima is 1.49\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21.13 Page no 1093"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=5.0 #m\n",
+ "d=1 #mm\n",
+ "\n",
+ "#Calculation\n",
+ "a=d/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Width of each slit is\", a,\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Width of each slit is 0.2 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter22_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter22_3.ipynb new file mode 100644 index 00000000..e91e3a27 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter22_3.ipynb @@ -0,0 +1,900 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:c280b16f38bc8dbf3b2a0607bdd0cfd4670bde51a104a8d547b07a434c49c7f5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 22 Diffraction and polarisation"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 22.1 Page no 1124"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "D=1 #m\n",
+ "l=5*10**-7 #m\n",
+ "d=0.1*10**-3 #m\n",
+ "\n",
+ "#Calculation\n",
+ "W=(2*D*l)/d\n",
+ "\n",
+ "#Result\n",
+ "print\"Width of the central maximum is\", W*10**2,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Width of the central maximum is 1.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.2 Page no 1124"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "D=1.60 #m\n",
+ "l=6328*10**-10 #m\n",
+ "w=4.0*10**-3\n",
+ "\n",
+ "#Calculation\n",
+ "d=(2*D*l)/w\n",
+ "\n",
+ "#Result\n",
+ "print\"Width of the slit is\", round(d*10**3,2),\"mm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Width of the slit is 0.51 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.3 Page no 1124"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=7500*10**-10\n",
+ "d=1.0*10**-6\n",
+ "c=20\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=l/d\n",
+ "b=math.asin(a)*180/3.14\n",
+ "A=2*b\n",
+ "x=c*math.tan(a*3.14/180.0)\n",
+ "w=2*x\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Width of central maximum is\", round(A,0),\"Degree\"\n",
+ "print\"(ii) Width of central maximum is\",round(w*10**2,0),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Width of central maximum is 97.0 Degree\n",
+ "(ii) Width of central maximum is 52.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.4 Page no 1125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=6.3*10**-7 #m\n",
+ "a=3.6 #Degree\n",
+ "n=10\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "d=(n*l)/math.sin(a*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Slit width is\", round(d*10**3,1),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Slit width is 0.1 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.5 Page no 1125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=5500*10**-10\n",
+ "d=0.01\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=l/d\n",
+ "b=math.asin(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angular deflection is\", round(b,4),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angular deflection is 0.0032 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.6 Page no 1125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "lr=660\n",
+ "d=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "l1=(2*lr)/d\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of lambda is\",l1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of lambda is 440.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.7 Page no 1126"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l1=5890*10**-10 #m\n",
+ "l2=5896*10**-10\n",
+ "d=2.0*10**-6 #m\n",
+ "D=2 #m\n",
+ "\n",
+ "#Calculation\n",
+ "x=(3*D*(l2-l1))/(2*d)\n",
+ "\n",
+ "#Result\n",
+ "print\"Spacing between the first maxima of two sodium lines is\",x*10**4,\"*10**-4 m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Spacing between the first maxima of two sodium lines is 9.0 *10**-4 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.8 Page no 1126"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=3*10**-3 #m\n",
+ "l=500*10**-9 #m\n",
+ "\n",
+ "#Calculation\n",
+ "Z=d**2/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance is\",Z,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance is 18.0 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.9 Page no 1126"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=2*10**-3 #m\n",
+ "l=600*10**-9 #m\n",
+ "\n",
+ "#Calculation\n",
+ "Z=d**2/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance is\",round(Z,2),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance is 6.67 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.10 Page no 1126"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=2*10**-3 #m\n",
+ "l=5000*10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "Z=d**2/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Fresnel Distance is\",Z,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Fresnel Distance is 8.0 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.11 Page no 1129"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=5.50*10**-7 #m\n",
+ "D=5.1\n",
+ "\n",
+ "#Calculation\n",
+ "a=(1.22*l)/D\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum angular separation is\", round(a*10**7,1)*10**-7,\"rad\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum angular separation is 1.3e-07 rad\n"
+ ]
+ }
+ ],
+ "prompt_number": 60
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.12 Page no 1130"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=6*10**-7 #m\n",
+ "D=0.6\n",
+ "l1=10**10 #m\n",
+ "r=10.0**4*9.46*10**15 #m\n",
+ "\n",
+ "#Calculation\n",
+ "a=(1.22*l)/D\n",
+ "a1=l1/r\n",
+ "\n",
+ "#Result\n",
+ "print round(a1*10**10,2)*10**-10,\"rad\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "1.06e-10 rad\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.13 Page no 1130"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=6000*10**-8\n",
+ "D=254.0\n",
+ "\n",
+ "#Calculation\n",
+ "a=(1.22*l)/D\n",
+ "\n",
+ "#Result\n",
+ "print\"Limt of resolution of a telescope is\",round(a*10**7,1)*10**-7,\"Radian\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Limt of resolution of a telescope is 2.9e-07 Radian\n"
+ ]
+ }
+ ],
+ "prompt_number": 70
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.14 Page no 1130"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "D=600.0 #cm\n",
+ "l=5.5*10**-5 #cm\n",
+ "d=3.8*10**10 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "a=(1.22*l)/D\n",
+ "x=d*a\n",
+ "\n",
+ "#Result\n",
+ "print\"Separation of two points is\", round(x,0),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Separation of two points is 4250.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.15 Page no 1130"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=10**-4 #cm\n",
+ "l=5.8*10**-5 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "Na=l/(2.0*d)\n",
+ "\n",
+ "#Result\n",
+ "print\"Numerical aperature of a microscope is\", Na"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Numerical aperature of a microscope is 0.29\n"
+ ]
+ }
+ ],
+ "prompt_number": 80
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.16 Page no 1131"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1\n",
+ "l=600*10**-9 #,\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "rp=(2*u*math.sin(30*3.14/180.0))/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Resolving power of a microscope is\", round(rp*10**-6,2),\"*10**6\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resolving power of a microscope is 1.67 *10**6\n"
+ ]
+ }
+ ],
+ "prompt_number": 86
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.17 Page no 1133"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l1=15*10**-10 #m\n",
+ "l=6563*10**-10\n",
+ "c=3*10**8 #m/s\n",
+ "\n",
+ "#Calculation\n",
+ "v=(c*l1)/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Speed of star is\", round(v*10**-5,2),\"*10**5 m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Speed of star is 6.86 *10**5 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 91
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.18 Page no 1133"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l1=0.032\n",
+ "l=100.0\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "v=-(l1*c)/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Velocity of star is\",v*10**-4,\"*10**4 m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Velocity of star is -9.6 *10**4 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.21 Page no 1142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=60 #Degree\n",
+ "a1=90\n",
+ "\n",
+ "import math\n",
+ "A=math.tan(a*3.14/180.0)\n",
+ "ap=a1-a\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Refractive index of the medium is\", round(A,3)\n",
+ "print\"(ii) The refracting angle is\",ap,\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Refractive index of the medium is 1.73\n",
+ "(ii) The refracting angle is 30 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 118
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.22 Page no 1142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=1.33\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "ap=math.atan(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of incidence is\", round(ap,0),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of incidence is 53.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 122
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.23 Page no 1142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.33\n",
+ "a=90\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "ap=math.atan(u)*180/3.14\n",
+ "A=a-ap\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle between the sun and the horizon is\", round(A,0),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle between the sun and the horizon is 37.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 127
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.24 Page no 1142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=1.5\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "ap=math.atan(u)*180/3.14\n",
+ "r=90-ap\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of refraction is\", round(r,1),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of refraction is 33.7 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 132
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.25 Page no 1143"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=30 #Degree\n",
+ "I=3 #A\n",
+ "I0=4.0\n",
+ "I1=1\n",
+ "\n",
+ "#Calculation\n",
+ "a=I/I0\n",
+ "a1=I1/I0\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Fraction of maximum light transferred for 30 degree is\", a\n",
+ "print\"(ii) Fraction of maximum light transferred for 60 degree is\", a1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Fraction of maximum light transferred for 30 degree is 0.75\n",
+ "(ii) Fraction of maximum light transferred for 60 degree is 0.25\n"
+ ]
+ }
+ ],
+ "prompt_number": 135
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 22.26 Page no 1143"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ap=60 #Degree\n",
+ "u=3\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=1/math.sqrt(u)\n",
+ "C=math.asin(a)*180/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Critical angle for this medium is\", round(C,2),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Critical angle for this medium is 35.28 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 140
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter23_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter23_3.ipynb new file mode 100644 index 00000000..7d0fa841 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter23_3.ipynb @@ -0,0 +1,893 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:38e55bd383948f67d919af3879ad291116d41c75f201f86aa7c1c2e80cc59941"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Cahpter 23 Dual nature of radiation and matter"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.1 Page no 1200"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.62*10**-34 #J\n",
+ "c=3*10**8 #m/s\n",
+ "l=4.0*10**-7 #m\n",
+ "\n",
+ "#Calculation\n",
+ "E=((h*c)/l)/1.6*10**-19\n",
+ "p=h/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of energy is\", round(E*10**38,1),\"ev\"\n",
+ "print\"Momentum of photon is\",p,\"kg m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of energy is 3.1 ev\n",
+ "Momentum of photon is 1.655e-27 kg m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.2 Page no 1200"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=75*1.6*10**-19 #J\n",
+ "h=6.62*10**-34 #J s\n",
+ "\n",
+ "#Calculation\n",
+ "f=E/h\n",
+ "l=(12400/E)*1.6*10**-19\n",
+ "f=c/(l*10**10)\n",
+ "\n",
+ "#Result\n",
+ "print\"Frequency of the photon is\", round(f*10**5,0)*10**15,\"Hz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Frequency of the photon is 1.8e+16 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.3 Page no 1200"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.62*10**-34 #Js\n",
+ "f=880*10**3 #Hz\n",
+ "E1=10*10**3\n",
+ "\n",
+ "#Calculation\n",
+ "E=h*f\n",
+ "n=E1/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of photons emitted per second is\", round(n*10**-31,3)*10**31"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of photons emitted per second is 1.717e+31\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.4 Page no 1200"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "w=1.8\n",
+ "h=6.63*10**-34\n",
+ "l=5000*10**-10\n",
+ "m=9.0*10**-31\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "W=12400/w\n",
+ "h1=(((h*c)/l)-(w*1.6*10**-19))\n",
+ "h2=h1/1.6*10**-19\n",
+ "vmax=math.sqrt((2*h1)/m)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Threshold wavelength is\",round(W,0),\"A\"\n",
+ "print\"(ii) Maximum K.E of emitted photoelectrons is\", round(h2*10**38,3),\"ev\"\n",
+ "print\"(iii) Maximum velocity is\",round(vmax*10**-5,0),\"*10**5 m/s\"\n",
+ "print\"(iv) If the intensity of light is doubled, K.E of emitted electrons will remain unchanged\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Threshold wavelength is 6889.0 A\n",
+ "(ii) Maximum K.E of emitted photoelectrons is 0.686 ev\n",
+ "(iii) Maximum velocity is 5.0 *10**5 m/s\n",
+ "(iv) If the intensity of light is doubled, K.E of emitted electrons will remain unchanged\n"
+ ]
+ }
+ ],
+ "prompt_number": 65
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.5 Page no 1201"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=2*10**-4\n",
+ "I=30*10**-2\n",
+ "t=1\n",
+ "E=6.62*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "n=(I*A)/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Rate at which photons strike the surface is\",round(n*10**-13,2)*10**13,\"photons/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate at which photons strike the surface is 9.06e+13 photons/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 71
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.6 Page no 1201"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.62*10**-34 #Js\n",
+ "c=3*10**8\n",
+ "l=4500*10**-10 #m\n",
+ "w=2.3\n",
+ "\n",
+ "#Calculation\n",
+ "E=(h*c)/l\n",
+ "E1=(E/1.6*10**-19)*10**38\n",
+ "K=E1-w\n",
+ "f0=(w*1.6*10**-19)/h\n",
+ "p=h/l\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The energy of photon is\", round(E1,1),\"ev\"\n",
+ "print\"(ii) The maximum kinetic energy of emitted electrons is\",round(K,1),\"ev\"\n",
+ "print\"(iii) Threshold frequency for sodium is\",round(f0*10**-14,1)*10**14,\"Hz\"\n",
+ "print\"(iv) Momentum of a photon is\",round(p*10**27,1)*10**-27,\"Kg m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The energy of photon is 2.8 ev\n",
+ "(ii) The maximum kinetic energy of emitted electrons is 0.5 ev\n",
+ "(iii) Threshold frequency for sodium is 5.6e+14 Hz\n",
+ "(iv) Momentum of a photon is 1.5e-27 Kg m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 100
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.7 Page no 1202"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=36.0*10**-8 #m\n",
+ "w0=2*1.6*10**-19 #J\n",
+ "h=6.62*10**-34 #Js\n",
+ "c=3*10**8\n",
+ "e=1.6*10**-19\n",
+ "m=9.0*10**-31\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l0=(h*c)/w0\n",
+ "E=(h*c)/l\n",
+ "E1=(E/1.6*10**-19)*10**38\n",
+ "K=E1-2\n",
+ "v0=K\n",
+ "vmax=math.sqrt(e*v0*2/m)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Threshold wavelength is\",round(l0*10**10,0),\"A\"\n",
+ "print\"(ii) Maximum kinetic energy of emitted photoelectrons is\", round(K,3),\"ev\"\n",
+ "print\"(iii) Stopping potential is\",round(v0,3),\"Volts\"\n",
+ "print\"(iv) Velocity is \",round(vmax*10**-5,2),\"*10**5 m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Threshold wavelength is 6206.0 A\n",
+ "(ii) Maximum kinetic energy of emitted photoelectrons is 1.448 ev\n",
+ "(iii) Stopping potential is 1.448 Volts\n",
+ "(iv) Velocity is 7.18 *10**5 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.8 Page no 1202"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.62*10**-34\n",
+ "c=3*10**8\n",
+ "l0=24.8*10**-8\n",
+ "a=1.2\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "w0=(h*c)/l0\n",
+ "w01=(w0/1.6*10**-19)*10**38\n",
+ "h1=w01+a\n",
+ "C=h1*e\n",
+ "l=(h*c)/C\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of incident light is\", round(l*10**10,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of incident light is 2000.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.9 Page no 1203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v1=16.5\n",
+ "V0=6.6 #V\n",
+ "f0=4.6*10**15 #Hz\n",
+ "f=2.2*10**15 #Hz\n",
+ "\n",
+ "#Calculation\n",
+ "h=(e*(v1-V0))/((f0-f))\n",
+ "\n",
+ "#Result\n",
+ "print\"Planck's constant is\", h"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Planck's constant is 6.6e-34\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.10 Page no 1203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.62*10**-34\n",
+ "f0=44*10**13 #Hz\n",
+ "a=11.5*10**14\n",
+ "b=4.4*10**14\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "w0=((h*f0)/1.6*10**-19)*10**38\n",
+ "h=3/(a-b)\n",
+ "h1=h*e\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Work function of the material is\", round(w0,2),\"ev\"\n",
+ "print\"(ii) Plank's constant is\", round(h1*10**34,2)*10**-34"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Work function of the material is 1.82 ev\n",
+ "(ii) Plank's constant is 6.76e-34\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.11 Page no 1204"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.6*10**-34\n",
+ "c=3*10**8\n",
+ "l=2000*10**-10\n",
+ "w0=4.2*1.6*10**-19\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "K=((h*c)/l)-w0\n",
+ "v0=K/e\n",
+ "l1=(h*c)/w0\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Potential difference is\", v0,\"V\"\n",
+ "print\"(ii) Wavelength of incident light is\", round(l1*10**10,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Potential difference is 1.9875 V\n",
+ "(ii) Wavelength of incident light is 2946.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.12 Page no 1204"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.6*10**-34\n",
+ "c=3*10**8\n",
+ "w0=2.39*1.6*10**-19\n",
+ "f1=4000.0 #A\n",
+ "f2=6000 #A\n",
+ "m=9.1*10**-31\n",
+ "e=1.9*10**-19\n",
+ "d=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l=(h*c)/w0\n",
+ "K=(12400/f1)-2.39\n",
+ "vmax=math.sqrt((2*K*1.6*10**-19)/m)\n",
+ "B=(m*vmax)/(e*d)\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum value of B is\", round(B*10**5,2)*10**-5,\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum value of B is 2.39e-05 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.13 Page no 1204"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "w0=4.4\n",
+ "\n",
+ "#Calculation\n",
+ "l=12400/w0\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of visible light is\", round(l,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of visible light is 2818.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.14 Page no 1205"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.625*10**-34\n",
+ "c=3*10**8\n",
+ "l=5600*10**-10\n",
+ "a=5\n",
+ "\n",
+ "#Calculation\n",
+ "E=(h*c)/l\n",
+ "n=a/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of visible photons emitted per second is\", round(n*10**-19,2)*10**19"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of visible photons emitted per second is 1.41e+19\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.15 Page no 1211"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=100\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l=12.27/math.sqrt(v)\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of an electron is\", l,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of an electron is 1.227 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.16 Page no 1212"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "h=6.62*10**-34\n",
+ "m=9*10**-31\n",
+ "v=10**5\n",
+ "mp=1.67*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "l=h/(m*v)\n",
+ "lp=h/(mp*v)\n",
+ "\n",
+ "#Result\n",
+ "print\"De-Broglie wavelength of electrons is\", round(l*10**10,1)*10**-10,\"m\"\n",
+ "print\"De-Broglie wavelength of protons is\",round(lp*10**10,4)*10**-10 ,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "De-Broglie wavelength of electrons is 7.36e-09 m\n",
+ "De-Broglie wavelength of protons is 3.96e-12 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 62
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.17 Page no 1212"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=500*1.6*10**-19\n",
+ "mp=1.67*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l=h/(math.sqrt(2*mp*E))\n",
+ "\n",
+ "#Result\n",
+ "print\"De-Broglie wavelength is\", round(l*10**12,2)*10**-12,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "De-Broglie wavelength is 1.28e-12 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.18 Page no 1212"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=150.0\n",
+ "mn=1.675*10**-27 #Kg\n",
+ "En=150*1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "le=12.27/math.sqrt(v)\n",
+ "ln=h/math.sqrt(2*mn*En)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) De-Broglie wavelength of electron is\",round(le,0),\"A\"\n",
+ "print\"(ii) De-Broglie wavelength of neutron is\", round(ln*10**10,4),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) De-Broglie wavelength of electron is 1.0 A\n",
+ "(ii) De-Broglie wavelength of neutron is 0.0233 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.19 Page no 1213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=2.0*10**-10 #m\n",
+ "h=6.62*10**-34\n",
+ "\n",
+ "#Calculation\n",
+ "p=h/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Momentum of electrons is\", p,\"Kg m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Momentum of electrons is 3.31e-24 Kg m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 81
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.20 Page no 1213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=1.4*10**-10 #m\n",
+ "h=6.63*10**-34\n",
+ "l1=2.0*10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "E=h*c*(1/l-1/l1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy of the scattered electron is\", round(E*10**16,2)*10**-16,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy of the scattered electron is 4.26e-16 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 87
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.22 Page no 1213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "me=9.11*10**-31 #Kg\n",
+ "lp=1.813*10**-4\n",
+ "vp=3\n",
+ "\n",
+ "#Calculation\n",
+ "mp=me/(lp*vp)\n",
+ "\n",
+ "#Result\n",
+ "print\"The particle's mass is\", round(mp*10**27,3)*10**-27,\"Kg. The particle is proton\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The particle's mass is 1.675e-27 Kg. The particle is proton\n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 23.23 Page no 1214"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=0.82*10**-10 #m\n",
+ "h=6.6*10**-34\n",
+ "m=9.1*10**-31\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "le=math.sqrt((h*l)/(2*c*m))\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength associated with the photoelectrons is\", round(le*10**10,4),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength associated with the photoelectrons is 0.0996 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 103
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter24_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter24_3.ipynb new file mode 100644 index 00000000..b7102dac --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter24_3.ipynb @@ -0,0 +1,874 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:6d1662d2dadbe072b20c80081401408d705c47c14e10e838032934acc7c20ff4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 24 Atoms"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.1 Page no 1264"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "k=7.68*10**6*1.6*10**-19 #J\n",
+ "e=1.6*10**-19\n",
+ "Z=29\n",
+ "m=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "r=(m*2*Z*e**2)/k\n",
+ "\n",
+ "#Result\n",
+ "print\"The distance of the closest approach is\",round(r*10**14,1)*10**-14,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The distance of the closest approach is 1.1e-14 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.2 Page no 1265"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=10 #degree\n",
+ "e=1.6*10**-19\n",
+ "Z=79\n",
+ "m=9*10**9\n",
+ "a=5.0*1.6*10**-13\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "b=(Z*e**2*(1/(math.tan(5*3.14/180.0)))*m)/a\n",
+ "\n",
+ "#Result\n",
+ "print\"Impact parameter is\", round(b*10**13,1)*10**-13,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Impact parameter is 2.6e-13 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.3 Page no 1265"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Z=79\n",
+ "m=9*10**9\n",
+ "e=1.6*10**-19\n",
+ "r=4.0*10**-14\n",
+ "\n",
+ "#Calculation\n",
+ "K=(m*2*Z*e**2)/(r*1.6*10**-13)\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy is\", round(K,2),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy is 5.69 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.4 Page no 1265"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=2.1*10**7 #m/s\n",
+ "a=4.8*10**7 #C/Kg\n",
+ "Z=79\n",
+ "e=1.6*10**-19\n",
+ "m=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "r0=(2*m*Z*e*a)/v**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance of the closest approach is\", round(r0*10**14,1)*10**-14,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance of the closest approach is 2.5e-14 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.6 Page no 1266"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Z=79\n",
+ "e=1.6*10**-19 #C\n",
+ "v=1.6*10**-12\n",
+ "m=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "b=(m*Z*e**2*(1/(math.tan(45*3.14/180.0))))/v\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Scattering angle is 180 degree\"\n",
+ "print\"(b) The value of scattering angle decreases\"\n",
+ "print\"(c) Impact parameter is\", round(b*10**14,1)*10**-14,\"m\"\n",
+ "print\"(d) The scattering of particle takes place due to charge on the nucleus\",\n",
+ "print\"(e) Scattering angle is increase with decrease in impact parameter\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Scattering angle is 180 degree\n",
+ "(b) The value of scattering angle decreases\n",
+ "(c) Impact parameter is 1.1e-14 m\n",
+ "(d) The scattering of particle takes place due to charge on the nucleus (e) Scattering angle is increase with decrease in impact parameter\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.7 Page no 1280"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=8.854*10**-12\n",
+ "h=6.62*10**-34\n",
+ "m=9*10**-31\n",
+ "e1=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r1=((e*h**2)/(math.pi*m*e1**2))*10**10\n",
+ "v1=e1**2/(2*e*h)\n",
+ "n=2*r1\n",
+ "\n",
+ "#Result\n",
+ "print\"Radius of first orbit is\", round(r1,2),\"A\"\n",
+ "print\"Velocity of electron is\",round(v1*10**-6,1),\"*10**6 m/s\"\n",
+ "print\"Size of hydrogen atom is\",round(n,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of first orbit is 0.54 A\n",
+ "Velocity of electron is 2.2 *10**6 m/s\n",
+ "Size of hydrogen atom is 1.07 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 62
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.8 Page no 1281"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=1.0\n",
+ "n1=2.0\n",
+ "n2=3.0\n",
+ "a=0.53*10**-10\n",
+ "Z=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "r1=(a*n)/Z\n",
+ "r2=(a*n1**2)/Z\n",
+ "r3=(a*n2**2)/Z\n",
+ "E1=(-13.6*Z**2)/n**2\n",
+ "E2=(-13.6*Z**2)/n1**2\n",
+ "E3=(-13.6*Z**2)/n2**2\n",
+ "E=E3-E1\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Radii of three lowest allowed orbits is\", round(r1*10**10,2),\"A,\",round(r2*10**10,2),\"A and\",r3*10**10,\"A\"\n",
+ "print\"(ii) Energy of three lowest allowed orbits is\",E1,\"ev,\",E2,\"ev and\",E3,\"ev\"\n",
+ "print\"Energy of the photon is\",E,\"ev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Radii of three lowest allowed orbits is 0.18 A, 0.71 A and 1.59 A\n",
+ "(ii) Energy of three lowest allowed orbits is -122.4 ev, -30.6 ev and -13.6 ev\n",
+ "Energy of the photon is 108.8 ev\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.9 Page no 1281"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=2.0\n",
+ "n1=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "E2=-13.6/n**2\n",
+ "E3=-13.6/n1**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Energies of two energy level is\",E2,\"ev and\",round(E3,2),\"ev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energies of two energy level is -3.4 ev and -1.51 ev\n"
+ ]
+ }
+ ],
+ "prompt_number": 86
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.10 Page no 1282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rh=1.097*10**7\n",
+ "\n",
+ "#Calculation\n",
+ "l=9/(8.0*Rh)\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of second line is\",round(l*10**10,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of second line is 1026.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 90
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.11 Page no 1282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rh=1.097*10**7\n",
+ "\n",
+ "#Calculation\n",
+ "l=4/Rh\n",
+ "\n",
+ "#Result\n",
+ "print\"Shortest wavelength is\",round(l*10**10,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shortest wavelength is 3646.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.12 Page no 1282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rh=1.097*10**7\n",
+ "\n",
+ "#Calculation\n",
+ "l=4/(3.0*Rh)\n",
+ "\n",
+ "#Result\n",
+ "print\"Longest wavelength is\",round(l*10**10,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Longest wavelength is 1215.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 99
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.13 Page no 1282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=1.0\n",
+ "h=6.62*10**-34\n",
+ "c=3*10**8\n",
+ "f=1.6*10**-19\n",
+ "Z=2\n",
+ "\n",
+ "#Calculation\n",
+ "E1=(-13.6*Z**2)/n**2\n",
+ "l=-(h*c)/(E1*f)\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum wavelength is\", round(l*10**10,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum wavelength is 228.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 104
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.14 Page no 1283"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=1\n",
+ "Z=1.0\n",
+ "a=0.53*10**-10\n",
+ "Z1=4.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "rh=(a*n)/Z**2\n",
+ "n1=math.sqrt((a*Z1/rh))\n",
+ "Eh=(-13.6*Z**2)/n**2\n",
+ "Ebe=(-13.6*Z1**2)/n1**2\n",
+ "E=Ebe/Eh\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy of two states is\",E"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy of two states is 4.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 112
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.15 Page no 1283"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Z=2\n",
+ "e=1.6*10**-19\n",
+ "e1=8.854*10**-12\n",
+ "n=3\n",
+ "h=6.62*10**-34\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "v=(Z*e**2)/(2*e1*n*h)\n",
+ "a=v/c\n",
+ "\n",
+ "#Result\n",
+ "print\"Speed of the electron is\",round(a,3 )"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Speed of the electron is 0.005\n"
+ ]
+ }
+ ],
+ "prompt_number": 116
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.16 Page no 1284"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=10**-10\n",
+ "R=10**-15\n",
+ "Rs=7*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "R1=r/R\n",
+ "Re=R1*Rs\n",
+ "\n",
+ "#Result\n",
+ "print\"Radius of the earth's orbit is\",Re,\"m. Thus the earth would be much farther away from the sun\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of the earth's orbit is 7e+13 m. Thus the earth would be much farther away from the sun\n"
+ ]
+ }
+ ],
+ "prompt_number": 118
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.17 Page no 1284"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=-13.6*1.9*10**-19 #J\n",
+ "m=9*10**9\n",
+ "e=1.6*10**-19\n",
+ "n=1\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "r=(-e**2*m)/(2.0*E)\n",
+ "v=c/(137*n)\n",
+ "\n",
+ "#Result\n",
+ "print\"Orbital radius is\", round(r*10**11,1)*10**-11,\"m\"\n",
+ "print\"Velocity of the electron is\",round(v*10**-6,1),\"*10**6 m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Orbital radius is 4.5e-11 m\n",
+ "Velocity of the electron is 2.2 *10**6 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 131
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.18 Page no 1284"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=2.2*10**6\n",
+ "r=5.3*10**-11\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "f=v/(2*math.pi*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"Initial frequency of light is\",round(f*10**-15,1)*10**15,\"Hz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Initial frequency of light is 6.6e+15 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 135
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.19 Page no 1285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=10 #Kg\n",
+ "T=2*60*60 #S\n",
+ "rn=8*10**6 #m\n",
+ "h=6.62*10**-34\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "vn=(2*math.pi*rn)/T\n",
+ "n=(2*math.pi*rn*vn)/h\n",
+ "\n",
+ "#Result\n",
+ "print\"Quantum number is\",round(n*10**-44,1)*10**45"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Quantum number is 5.3e+45\n"
+ ]
+ }
+ ],
+ "prompt_number": 139
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.20 Page no 1285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E2=18.70\n",
+ "E1=16.70\n",
+ "h=6.62*10**-34\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "E=E2-E1\n",
+ "l=(h*c)/(E*1.6*10**-19)\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength is\", round(l*10**9,0),\"nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength is 621.0 nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 148
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.21 Page no 1285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n1=2\n",
+ "n2=3\n",
+ "lb=6563\n",
+ "a=20\n",
+ "b=108.0\n",
+ "\n",
+ "#Calculation\n",
+ "l1=(lb*a)/b\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of first member is\",round(l1,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of first member is 1215.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 151
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.22 Page no 1285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rh=1.097*10**7 #/m\n",
+ "h=6.63*10**-34\n",
+ "c=3*10**8\n",
+ "n=2.0\n",
+ "n1=4.0\n",
+ "\n",
+ "#Calculation\n",
+ "E=(h*c*Rh*(1/n**2-1/n1**2))/1.6*10**-19\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum energy is\", round(E*10**38,2),\"ev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum energy is 2.56 ev\n"
+ ]
+ }
+ ],
+ "prompt_number": 158
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 24.23 Page no 1286"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rh=1.097*10**7\n",
+ "n2=4.0\n",
+ "n1=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "lm=1/(Rh*(1/n1**2-1/n2**2))\n",
+ "lm1=9/Rh\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength is\", round(lm1*10**9,1),\"nm. This wavelength is in infrared part\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength is 820.4 nm. This wavelength is in infrared part\n"
+ ]
+ }
+ ],
+ "prompt_number": 167
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter25_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter25_3.ipynb new file mode 100644 index 00000000..95708b68 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter25_3.ipynb @@ -0,0 +1,1188 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:efdc68d7aa35d22a94f64e5e8f01516d21c5f3fcea362a5520b7b1d532197f7c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 25 Nuclei"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.1 Page no 1312"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R0=1.2*10**-15 #m\n",
+ "A=208\n",
+ "A1=16\n",
+ "\n",
+ "#calculation\n",
+ "R=R0*A**0.33\n",
+ "R1=R0*A1**0.33\n",
+ "\n",
+ "#Result\n",
+ "print\"Nuclear radius of lead is\", round(R*10**15,1),\"fm\"\n",
+ "print\"Nuclear radius of oxygen is\", round(R1*10**15,0),\"fm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Nuclear radius of lead is 7.0 fm\n",
+ "Nuclear radius of oxygen is 3.0 fm\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.2 page no1312"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "me=9.1*10**-31\n",
+ "c=3*10**8\n",
+ "e=1.6*10**-19\n",
+ "mp=1.673*10**-27\n",
+ "mn=1.675*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "E=(me*c**2)/e\n",
+ "E1=(mp*c**2)/e\n",
+ "E2=(mn*c**2)/e\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Equivalent energy of electron is\",round(E*10**-6,2),\"Mev\"\n",
+ "print\"(ii) Equivalent energy of proton is\",round(E1*10**-6,1),\"Mev\"\n",
+ "print\"(iii) Equivalent energy of neutron is\",round(E2*10**-6,1),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Equivalent energy of electron is 0.51 Mev\n",
+ "(ii) Equivalent energy of proton is 941.1 Mev\n",
+ "(iii) Equivalent energy of neutron is 942.2 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.3 Page no 1312"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=10**-3 #m\n",
+ "c=3*10**8 #m/s\n",
+ "a=3.6*10**6 #J\n",
+ "\n",
+ "#Calculation\n",
+ "E=(m*c**2)/a\n",
+ "\n",
+ "#Result\n",
+ "print E*10**-7,\"*10**7 KWh\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "2.5 *10**7 KWh\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.4 Page no 1313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Z=17\n",
+ "A=35\n",
+ "Z1=92\n",
+ "A1=235\n",
+ "Z2=4\n",
+ "A2=9\n",
+ "\n",
+ "#Calculation\n",
+ "n=A-Z\n",
+ "n1=A1-Z1\n",
+ "n2=A2-Z2\n",
+ "\n",
+ "#Calculation\n",
+ "print\"Number of neutron in 17Cl35 is\",n\n",
+ "print\"Number of neutron in 92U235 is\",n1\n",
+ "print\"Number of neutron in 4Be9 is\",n2"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of neutron in 17Cl35 is 18\n",
+ "Number of neutron in 92U235 is 143\n",
+ "Number of neutron in 4Be9 is 5\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.5 Page no 1313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A2=235\n",
+ "A1=16.0\n",
+ "R1=3*10**-15 #m\n",
+ "\n",
+ "#Calculation\n",
+ "R=(A2/A1)**0.33\n",
+ "R2=R*R1\n",
+ "\n",
+ "#Result\n",
+ "print\"Nuclear radius is\", round(R2*10**15,3),\"fermi\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Nuclear radius is 7.281 fermi\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.6 Page no 1313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "me=55.85\n",
+ "u=1.66*10**-27 #Kg\n",
+ "R=1.2*10**-15 \n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "m=me*u\n",
+ "a=(3*u)/(4.0*math.pi*R**3)\n",
+ "\n",
+ "#Result\n",
+ "print\"Nuclear density is\", round(a*10**-17,2)*10**17,\"Kg/m**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Nuclear density is 2.29e+17 Kg/m**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.7 Page no 1317"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=4.001509 #a.m.u\n",
+ "N=1.008666\n",
+ "N1=1.007277\n",
+ "a=1.66*10**-27\n",
+ "c=3*10**8\n",
+ "e=1.6*10**-19\n",
+ "n=4.0\n",
+ "\n",
+ "#Calculation\n",
+ "A=2*N1+2*N\n",
+ "M1=A-M\n",
+ "Eb=M1*a*c**2/e\n",
+ "B=Eb/n\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Mass defect is\",M1,\"a.m.u\"\n",
+ "print\"(ii) Binding energy is\",round(Eb*10**-6,1),\"Mev\"\n",
+ "print\" Binding energy per nucleon is\",round(B*10**-6,2),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Mass defect is 0.030377 a.m.u\n",
+ "(ii) Binding energy is 28.4 Mev\n",
+ " Binding energy per nucleon is 7.09 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 61
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.8 Page no 1317"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ma=1.00893\n",
+ "m1=1.00813\n",
+ "m2=2.01473\n",
+ "a=931.5\n",
+ "a1=4.00389\n",
+ "\n",
+ "#Calculation\n",
+ "m=ma+m1-m2\n",
+ "Eb=m*a\n",
+ "m3=2*ma+2*m1-a1\n",
+ "Eb1=m3*a\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Binding energy when one neutron and one proton combined together is\", round(Eb,2),\"Mev\"\n",
+ "print\"(ii) Binding energy when two neutrons and two protons are combined is\",round(Eb1,1) ,\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Binding energy when one neutron and one proton combined together is 2.17 Mev\n",
+ "(ii) Binding eergy when two neutrons and two protons are combined is 28.2 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.10 Page no 1318"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=1.66*10**-27 #Kg\n",
+ "c=3*10**8\n",
+ "mp=1.00727\n",
+ "mn=1.00866\n",
+ "mo=15.99053\n",
+ "\n",
+ "#Calculation\n",
+ "E=(a*c**2)/1.6*10**-19\n",
+ "m1=8*mp+8*mn-mo\n",
+ "a1=m1*E\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy equivalent of one atomic mass unit is\", round(a1*10**32,1),\"Mev/c**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy equivalent of one atomic mass unit is 127.8 Mev/c**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.11 Page no 1318"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "mp=1.007825\n",
+ "mn=1.008665\n",
+ "m=39.962589\n",
+ "a2=931.5\n",
+ "Z=40.0\n",
+ "\n",
+ "#Calculation\n",
+ "E=20*mp+20*mn\n",
+ "m1=E-m\n",
+ "Eb=m1*a2\n",
+ "B=Eb/Z\n",
+ "\n",
+ "#Result\n",
+ "print\"Binding energy per nucleon is\", round(B,3),\"Mev/nucleon\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Binding energy per nucleon is 8.551 Mev/nucleon\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.12 Page no 1330"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "t=5000 #Days\n",
+ "t1=2000.0\n",
+ "a=0.693 \n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "dt=(a*t)/t1\n",
+ "N=math.log10(dt)\n",
+ "l=a*N/(t1)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The fraction remaining after 5000 days is\", round(N,3)\n",
+ "print\"(ii) The activity of sample after 5000 days is\",round(l*10**5,1),\"*10**8 Bq\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The fraction remaining after 5000 days is 0.239\n",
+ "(ii) The activity of sample after 5000 days is 8.3 *10**8 Bq\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.13 Page no 1330"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=3.67*10**10 #dis/second\n",
+ "r=226.0\n",
+ "A=6.023*10**23\n",
+ "\n",
+ "#Calculation\n",
+ "n=A/r\n",
+ "l=N/n\n",
+ "D=0.693/l\n",
+ "a=D/(3600.0*24.0*365.0)\n",
+ "\n",
+ "#Result\n",
+ "print\" Half life of radium is\",round(a,0),\"years\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Half life of radium is 1596.0 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.14 page no 1330"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N0=475\n",
+ "N=270.0\n",
+ "t=5.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=N0/N\n",
+ "l=math.log(a)/t\n",
+ "T=1/l\n",
+ "T1=0.693/l\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The decay constant is\",round(l,3),\"/minute\"\n",
+ "print\"(ii) Mean life is\",round(T,2),\"minute\"\n",
+ "print\"(iii) Half life is\",round(T1,2),\"minute\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The decay constant is 0.113 /minute\n",
+ "(ii) Mean life is 8.85 minute\n",
+ "(iii) Half life is 6.13 minute\n"
+ ]
+ }
+ ],
+ "prompt_number": 55
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.15 page no 1331"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "t=1500\n",
+ "N=0.01\n",
+ "N0=0.999\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "T=t*math.log(N)/math.log(0.5)\n",
+ "T1=t*math.log(N0)/math.log(0.5)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Years will reduce to 1 centigram is\",round(T,1),\"years\"\n",
+ "print\"(ii) Years will lose 1 mg is\",round(T1,2),\"years\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Years will reduce to 1 centigram is 9965.8 years\n",
+ "(ii) Years will lose 1 mg is 2.17 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 61
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.16 page no 1331"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=2*10**12\n",
+ "b=9.0*10**12\n",
+ "T=80\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "c=math.log(a/b)\n",
+ "t=-(c*T)/0.693\n",
+ "\n",
+ "#Result\n",
+ "print\"Time required is\",round(t,0),\"second\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time required is 174.0 second\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.17 page no 1332"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "T=6.0\n",
+ "A=6.023*10**23\n",
+ "W=99.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l=0.693/T\n",
+ "N0=A*10**-12/W\n",
+ "A0=l*N0\n",
+ "N=N0*(1/math.log10(l))\n",
+ "A1=-(l*N)\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\" Activity in the beginning and after one hour\",round(A1*10**-8,3),\"/h\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Activity in the beginning and after one hour 7.496 /h\n"
+ ]
+ }
+ ],
+ "prompt_number": 80
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.18 page no 1332"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "T=30.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l=0.693/T\n",
+ "T1=1/l\n",
+ "t=math.log(4)/l\n",
+ "t1=math.log(8)/l\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) average life is\",round(l,4),\"/day\"\n",
+ "print\"(ii) The time taken for 3/4 of the original no. to disintegrate is\",round(T1,2),\"days\"\n",
+ "print\"(iii) Time taken is\",round(t,0),\"days\"\n",
+ "print\"(iv) Time taken is\",round(t1,0),\"days\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) average life is 0.0231 /day\n",
+ "(ii) The time taken for 3/4 of the original no. to disintegrate is 43.29 days\n",
+ "(iii) Time taken is 60.0 days\n",
+ "(iv) Time taken is 90.0 days\n"
+ ]
+ }
+ ],
+ "prompt_number": 93
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.19 page no 1332"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=1620.0\n",
+ "l1=405.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "T=(1/l)+(1/l1)\n",
+ "t=math.log(4)/T\n",
+ "\n",
+ "#Result\n",
+ "print\"The time during which three-fourths of a sample will decay is\",round(t,0),\"years\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The time during which three-fourths of a sample will decay is 449.0 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 99
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.20 page no 1333"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=3.7*10**10 #disintegrations/s\n",
+ "A=6.02*10**23\n",
+ "B=234\n",
+ "\n",
+ "#Calculation\n",
+ "D=(C*B)/A\n",
+ "\n",
+ "#Result \n",
+ "print\"Mass ofuranium atoms disintegrated per second is\",round(D*10**11,3)*10**-11,\"g\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mass ofuranium atoms disintegrated per second is 1.438e-11 g\n"
+ ]
+ }
+ ],
+ "prompt_number": 105
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.21 page no 1333"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=0.075 #kg /mol\n",
+ "m=1.2*10**-6 #kg\n",
+ "A=6.0*10**23 #/mol\n",
+ "t=9.6*10**18\n",
+ "N=170\n",
+ "\n",
+ "#Calculation\n",
+ "n=(A*m)/M\n",
+ "l=N/t\n",
+ "T=0.693/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of K-40 atoms in the sample is\", n\n",
+ "print\"Half life of K-40 is\", round(T/(24.0*3600.0*365)*10**-9,3),\"*10**9 years\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of K-40 atoms in the sample is 9.6e+18\n",
+ "Half life of K-40 is 1.241 *10**9 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 122
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.22 Page no 1337"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "mp=232.03714\n",
+ "mn=228.02873\n",
+ "m0=4.002603\n",
+ "a=931.5\n",
+ "A=232.0\n",
+ "e=1.6*10**-19\n",
+ "m=1.66*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "M=mp-mn-m0\n",
+ "Q=M*a\n",
+ "K=(A-4)*Q/A\n",
+ "S=math.sqrt((2*K*e)/(4.0*m))\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Kinetic energy is\", round(K,1),\"Mev\"\n",
+ "print\"(ii) Speed of particle is\", round(S*10**-4,1),\"*10**7 m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Kinetic energy is 5.3 Mev\n",
+ "(ii) Speed of particle is 1.6 *10**7 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 139
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.23 Page no 1337"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "b=238\n",
+ "c=206\n",
+ "d=92\n",
+ "e=82\n",
+ "\n",
+ "#Calculation\n",
+ "a=(b-c)/4.0\n",
+ "A=-d+(2*a)+e\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The emission of alpha particle will reduce the mass number by 4a and charge number by 2a\"\n",
+ "print\"(ii) Number of alpha particle is\", a\n",
+ "print\"Number of beta particle is\",A"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The emission of alpha particle will reduce the mass number by 4a and charge number by 2a\n",
+ "(ii) Number of alpha particle is 8.0\n",
+ "Number of beta particle is 6.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 145
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.24 Page no 1338"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=218\n",
+ "b=84\n",
+ "\n",
+ "#Calculation\n",
+ "A=a-4\n",
+ "Z=b-2\n",
+ "\n",
+ "#Result\n",
+ "print\"Atomic number of new element formed is\", A\n",
+ "print\"Mass number of new element formed is\",Z"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Atomic number of new element formed is 214\n",
+ "Mass number of new element formed is 82\n"
+ ]
+ }
+ ],
+ "prompt_number": 148
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.27 Page no 1340"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "mp=10.016125\n",
+ "mn=4.003874\n",
+ "mp1=13.007490\n",
+ "mn1=1.008146\n",
+ "a=931.5\n",
+ "\n",
+ "#Calculation\n",
+ "Mr=mp+mn\n",
+ "Mp=mp1+mn1\n",
+ "Md=Mr-Mp\n",
+ "A=a*Md\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy released in the reaction is\",round(A,3),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy released in the reaction is 4.064 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 154
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.28 Page no 1345"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=10**6 #J/s\n",
+ "E=200*10**6*1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "N=a/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of fission per second is\", round(N*10**-16,2)*10**16"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of fission per second is 3.13e+16\n"
+ ]
+ }
+ ],
+ "prompt_number": 159
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.29 Page no 1345"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=3*10**8 #W\n",
+ "E=200*10**6*1.6*10**-19\n",
+ "a=235\n",
+ "m=6.023*10**23\n",
+ "\n",
+ "#Calculation\n",
+ "E1=P*3600\n",
+ "N=E1/E\n",
+ "M1=(a*N)/m\n",
+ "\n",
+ "#Result\n",
+ "print\"Mass of uranium fissioned per hour is\", round(M1,2),\"g\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mass of uranium fissioned per hour is 13.17 g\n"
+ ]
+ }
+ ],
+ "prompt_number": 166
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.30 Page no 1345"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=6.023*10**26\n",
+ "a=235.0\n",
+ "t=30 #Days\n",
+ "E=200*10**6*1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "N=(2/a)*m\n",
+ "A=N/(t*24*60.0*60.0)\n",
+ "P=E*A\n",
+ "\n",
+ "#Result\n",
+ "print\"Power output is\", round(P*10**-6,1),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power output is 63.3 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 173
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.31 Page no 1348"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=1.0076\n",
+ "mp=4.0039\n",
+ "a=931.5*10**6 #ev\n",
+ "\n",
+ "#Calculation\n",
+ "Mr=4*m\n",
+ "Md=Mr-mp\n",
+ "E=Md*a*1.6*10**-19\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy released is\", round(E*10**13,2)*10**-13,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy released is 3.95e-12 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 181
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 25.32 Page no 1349"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=6*10**-3 #Kg\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "E=a*c**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy liberated is\", E,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy liberated is 5.4e+14 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 184
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter26_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter26_3.ipynb new file mode 100644 index 00000000..98e47e10 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter26_3.ipynb @@ -0,0 +1,425 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:e41de143240c9df8b907c856d0ba61f830495897881ab9f990dfa099753c5c2e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 26 Semiconductors"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.1 Page no 1414"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=0.47\n",
+ "ue=0.39 #m**2/volt sec\n",
+ "uh=0.19 #m**2/volt sec\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "a1=1/a\n",
+ "ni=a1/(e*(ue+uh))\n",
+ "\n",
+ "#Result\n",
+ "print\"Intrinsic carrier conceentration is\", round(ni*10**-19,1)*10**19,\"/m**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Intrinsic carrier conceentration is 2.3e+19 /m**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.2 Page no 1414"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=0.01\n",
+ "e=1.6*10**-19\n",
+ "ue=0.39\n",
+ "\n",
+ "#Calculation\n",
+ "a1=1/a\n",
+ "Nd=a1/(e*ue)\n",
+ "\n",
+ "#Result\n",
+ "print\"Donor concentration is\", round(Nd*10**-21,1)*10**21,\"/m**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Donor concentration is 1.6e+21 /m**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.3 Page no 1414"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ni=2.5*10**19 #/m**3\n",
+ "e=1.6*10**19\n",
+ "ue=0.36 #m**2/volt sec\n",
+ "uh=0.17 \n",
+ "\n",
+ "#Calculation\n",
+ "a=ni*e*(ue+uh)\n",
+ "\n",
+ "#Result\n",
+ "print\"Conductivity is\", a*10**-38,\"S/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Conductivity is 2.12 S/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.4 Page no 1414"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ne=8*10**13 #/cm**3\n",
+ "nh=5*10**12 #/cm**3\n",
+ "ue=23000 #cm**2/vs\n",
+ "e=1.6*10**-19\n",
+ "uh=100 #cm**2/vs\n",
+ "\n",
+ "#Calculation\n",
+ "a=e*((ne*ue)+(nh*uh))\n",
+ "A1=1/a\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Since electron density is greater than the hole density, the semiconductor is n-type\"\n",
+ "print\"(ii) Resistivity of the sample is\", round(A1,3),\"ohm cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Since electron density is greater than the hole density, the semiconductor is n-type\n",
+ "(ii) Resistivity of the sample is 3.396 ohm cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.5 Page no 1415"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ni=1.5*10**16 #/m**3\n",
+ "nh=4.5*10**22 #/m**3\n",
+ "\n",
+ "#Calculation\n",
+ "ne=ni**2/nh\n",
+ "\n",
+ "#Result\n",
+ "print\"ne in the doped semiconductor is\",ne*10**-9,\"*10**9 /m**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "ne in the doped semiconductor is 5.0 *10**9 /m**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.6 Page no 1415 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=5890.0 #A\n",
+ "\n",
+ "#Calculation\n",
+ "E=12400/l\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum energy is\",round(E,1),\"ev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum energy is 2.1 ev\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.7 Page no 1415"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=6*10**19\n",
+ "b=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "A=a*b\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of holes is\",A"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of holes is 6e+12\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.8 Page no 1415"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=0.65\n",
+ "a=10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "l=(12400*a)/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum wavelength of electromagnetic radiation is\",round(l*10**6,1)*10**-6,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum wavelength of electromagnetic radiation is 1.9e-06 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.9 Page no 1416"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=5 #/ohm/cm\n",
+ "ue=3900 #cm**2/vs\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "Nd=a/(ue*e)\n",
+ "\n",
+ "#Result\n",
+ "print\"Number density of donor atom is\",round(Nd*10**-15,2)*10**15,\"/cm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number density of donor atom is 8.01e+15 /cm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.10 Page no 1416"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ni=1.5*10**16 #/m**3\n",
+ "a=5*10**28\n",
+ "b=10.0**6\n",
+ "\n",
+ "#Calculation\n",
+ "Ne=a/b\n",
+ "nh=ni**2/Ne\n",
+ "\n",
+ "#Result\n",
+ "print\"Number of Electrons is\",Ne,\"/m**3\"\n",
+ "print\"Number of holes is\",nh*10**-9,\"*10**9 /m**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of Electrons is 5e+22 /m**3\n",
+ "Number of holes is 4.5 *10**9 /m**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 50
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 26.11 Page no 1416"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=4.0*10*-8 #m\n",
+ "\n",
+ "#Calculation\n",
+ "a=2/1.6*10**-19\n",
+ "E=-a/d\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric field is\", round(E*10**22,0),\"*10**7 V/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric field is 4.0 *10**7 V/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 61
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter27_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter27_3.ipynb new file mode 100644 index 00000000..01e54494 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter27_3.ipynb @@ -0,0 +1,921 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:eeaae422be8c264750ed4950e51451be86906b82350f05dcb46e0f610199fb25"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 27 Semiconductor devices"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.1 Page no 1446"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=1.5 #V\n",
+ "Vd=0.5 #V\n",
+ "P=0.1 #W\n",
+ "\n",
+ "#Calculation\n",
+ "Imax=P/Vd\n",
+ "V=E-Vd\n",
+ "R1=V/Imax\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of resistance is\",R1,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of resistance is 5.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.2 Page no 1446"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=2 #V\n",
+ "R=10.0 #ohm\n",
+ "R1=20.0\n",
+ "\n",
+ "#Calculation\n",
+ "I=V/R\n",
+ "I1=V/R1\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Current drawn from battery is\", I,\"A\"\n",
+ "print\"(ii) Current drawn from point B is\",I1,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Current drawn from battery is 0.2 A\n",
+ "(ii) Current drawn from point B is 0.1 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.3 Page no 1446"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "Vl=15 #V\n",
+ "Rl=2.0*10**3\n",
+ "Iz=10 #mA\n",
+ "\n",
+ "#Calculation\n",
+ "Il=(Vl/Rl)*10**3\n",
+ "Ir=Iz+Il\n",
+ "Vr=Ir*10**-2*R1\n",
+ "V=Vr+Vl\n",
+ "\n",
+ "#Result\n",
+ "print\"Voltage is\", V,\"V\"\n",
+ "print\"Zener rating required is\",Ir,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage is 18.5 V\n",
+ "Zener rating required is 17.5 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.4 Page no 1447"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=10.0\n",
+ "V=230 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Vrpm=math.sqrt(2)*V\n",
+ "Vsm=Vrpm/N\n",
+ "Vdc=Vsm/math.pi\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The output dc voltage is\", round(Vdc,2),\"V\"\n",
+ "print\"(ii) Peak inverse voltage is\",round(Vsm,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The output dc voltage is 10.35 V\n",
+ "(ii) Peak inverse voltage is 32.53 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.5 Page no 1447"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vm=50 #V\n",
+ "rf=20.0\n",
+ "Rl=800 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Im=(Vm/(rf+Rl))*10**3\n",
+ "Idc=Im/math.pi\n",
+ "Irms=Im/2.0\n",
+ "P=(Irms/1000.0)**2*(rf+Rl)\n",
+ "P1=(Idc/1000.0)**2*Rl\n",
+ "V=Idc*Rl*10**-3\n",
+ "A=P1*100/P\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Im=\",round(Im,0),\"mA \\nIdc=\",round(Idc,1),\"mA \\nIrms=\",round(Irms,1),\"mA\"\n",
+ "print\"(ii) a.c power input is\",round(P,3),\"watt \\nd.c. power is\",round(P1,3),\"watt\"\n",
+ "print \"(iii) d.c. output voltage is\",round(V,2),\"Volts\"\n",
+ "print\"(iv) Efficiency of rectification is\", round(A,1),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Im= 61.0 mA \n",
+ "Idc= 19.4 mA \n",
+ "Irms= 30.5 mA\n",
+ "(ii) a.c power input is 0.762 watt \n",
+ "d.c. power is 0.301 watt\n",
+ "(iii) d.c. output voltage is 15.53 Volts\n",
+ "(iv) Efficiency of rectification is 39.5 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 67
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.6 Page no 1448"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "rf=20 #ohm\n",
+ "Rl=980\n",
+ "V=50 #v\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Vm=V*math.sqrt(2)\n",
+ "Im=(Vm/(rf+Rl))*10**3\n",
+ "Idc=(2*Im)/(math.pi)\n",
+ "Irms=Im/math.sqrt(2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) load current is\",round(Im,1),\"mA\"\n",
+ "print\"(ii) Mean load currant is\",round(Idc,0),\"mA\"\n",
+ "print\"(iii) R.M.S value of load current is\",Irms,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) load current is 70.7 mA\n",
+ "(ii) Mean load currant is 45.0 mA\n",
+ "(iii) R.M.S value of load current is 50.0 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 85
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.7 page no 1448"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=5.0\n",
+ "A=230 #V\n",
+ "B=2\n",
+ "Rl=100\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "V1=A/N\n",
+ "V2=V1*math.sqrt(2)\n",
+ "Vm=V2/B\n",
+ "Idc=2*Vm/(math.pi*Rl)\n",
+ "Vdc=Idc*Rl\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) d.c voltage output is\",round(Vdc,1),\"V\"\n",
+ "print\"(ii) peak inverse voltage is\",round(V2,0),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) d.c voltage output is 20.7 V\n",
+ "(ii) peak inverse voltage is 65.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 122
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.8 page no 1448"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Il=4.0 #mA\n",
+ "Vz=6 #V\n",
+ "E=10.0 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Lz=5*Il\n",
+ "L=Il+Lz\n",
+ "Rs=E-Vz\n",
+ "Rs1=Rs/(L*10**-3)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of series resister Rs\",round(Rs1,0),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of series resister Rs 167.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 123
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.9 page no 1449"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vf=0.3 #V\n",
+ "If=4.3*10**-3 #A\n",
+ "Vc=0.35\n",
+ "Va=0.25\n",
+ "Ic=6*10**-3\n",
+ "Ia=3*10**-3\n",
+ "\n",
+ "#Calculation\n",
+ "Rdc=Vf/If\n",
+ "Vf1=Vc-Va\n",
+ "If1=Ic-Ia\n",
+ "Rac=Vf1/If1\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) D.C. resistance is\",round(Rdc,2),\"ohm\"\n",
+ "print\"(ii) A.C. resistance is\",round(Rac,2),\"ohm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) D.C. resistance is 69.77 ohm\n",
+ "(ii) A.C. resistance is 33.33 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 139
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.10 page no 1462"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=0.9\n",
+ "Ie=1 #mA\n",
+ "\n",
+ "#Calculation\n",
+ "Ic=A*Ie\n",
+ "Ib=Ie-Ic\n",
+ "\n",
+ "#Result\n",
+ "print\"Base current is\",Ib,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Base current is 0.1 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 144
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.11 page no 1462"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=50\n",
+ "Ib=0.02 #mA\n",
+ "\n",
+ "#Calculation\n",
+ "Ic=B*Ib\n",
+ "Ie=Ib+Ic\n",
+ "\n",
+ "#Result\n",
+ "print\"Ie =\",Ie,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Ie = 1.02 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 147
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.12 page no 1462"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=49\n",
+ "Ie=12 #mA\n",
+ "Ib=240 #microA\n",
+ "\n",
+ "#Calculation\n",
+ "A=(B/1+B)*10**-2\n",
+ "Ic=A*Ie\n",
+ "Ic1=B*Ib\n",
+ "\n",
+ "#Result\n",
+ "print\" The value of Ic using A is\",Ic,\"mA\"\n",
+ "print\" The value of Ic using B is\",Ic1*10**-3,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The value of Ic using A is 11.76 mA\n",
+ " The value of Ic using B is 11.76 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 165
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.13 page no 1463"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=45.0\n",
+ "Ic=1 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Ib=Ic/B\n",
+ "\n",
+ "#Result\n",
+ "print\" The base current for common emitter connection is\",round(Ib,3),\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The base current for common emitter connection is 0.022 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 169
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.14 page no 1463"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vcc=8 #V\n",
+ "V=0.5 #V\n",
+ "Rc=800.0 #ohm\n",
+ "a=0.96\n",
+ "\n",
+ "#Calculation\n",
+ "Vce=Vcc-V\n",
+ "Ic=V/Rc*10**3\n",
+ "B=a/(1-a)\n",
+ "Ib=Ic/B\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Collector-emitter voltage is\",Vce,\"V\"\n",
+ "print\"(ii) Base current is\",round(Ib,3),\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Collector-emitter voltage is 7.5 V\n",
+ "(ii) Base current is 0.026 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 184
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.15 page no 1463"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=10\n",
+ "b=2\n",
+ "c=3\n",
+ "\n",
+ "#Calculation\n",
+ "Vce=a-b\n",
+ "Ic=c-b\n",
+ "Ro=Vce/Ic\n",
+ "\n",
+ "#Result\n",
+ "print\"The output resistance is\",Ro,\"k ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The output resistance is 8 k ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 188
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.16 page no 1464"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ic=4.0 #mA\n",
+ "Ib=30 #micro A\n",
+ "Ib1=20 #micro A\n",
+ "Vce=10 #V\n",
+ "c=4.5 #mA\n",
+ "d=3.0 #mA\n",
+ "\n",
+ "#Calculation\n",
+ "Ib2=Ib-Ib1\n",
+ "Ic1=c-d\n",
+ "Bac=Ic1/Ib2*10**3\n",
+ "Bdc=c/Ib*10**3\n",
+ "\n",
+ "#Result \n",
+ "print\"The value of Bac of the transister is\",Bdc\n",
+ "print\"The value of Bdc of the transister is\",Bdc"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Bac of the transister is 150.0\n",
+ "The value of Bdc of the transister is 150.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 249
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.17 page no 1464"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ri=665.0 #ohm\n",
+ "Ib=15.0 #micro A\n",
+ "Ic=2 #mA\n",
+ "Ro=5*10**3 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Bac=Ic/Ib*10**3\n",
+ "Av=Bac*(Ro/Ri)\n",
+ "\n",
+ "#Result\n",
+ "print\" The voltage gain is\", round(Av,0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The voltage gain is 1003.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 240
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.18 page no 1464"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given \n",
+ "Vbb=2.0 #v\n",
+ "Rc=2000 #ohm\n",
+ "B=100\n",
+ "Vbe=0.6 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Ic=Vbb/Rc*10**3\n",
+ "Ib=Ic/B\n",
+ "Ib1=10*Ib\n",
+ "Rb=(Vbb-Vbe)/Ib\n",
+ "Ic=B*Ib1\n",
+ "\n",
+ "#Result \n",
+ "print\"d.c. collector current is\",Ic,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "d.c. collector current is 10.0 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 236
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.19 page no 1465"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=10**10\n",
+ "e=1.6*10**-19\n",
+ "t=10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "Ie=(N*e)/t*10**3\n",
+ "Ib=(2/100.0)*Ie\n",
+ "Ic=Ie-Ib\n",
+ "c=Ic/Ie\n",
+ "B=Ic/Ib\n",
+ "#Result\n",
+ "print\"The current transfer ratio\",c\n",
+ "print\"current amplification factor is\",B"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current transfer ratio 0.98\n",
+ "current amplification factor is 49.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 257
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.20 page no 1465"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=200\n",
+ "b=50\n",
+ "c=17\n",
+ "d=5\n",
+ "e=4000\n",
+ "\n",
+ "#Calculation\n",
+ "Ib=(a-b)*10**-3\n",
+ "Ic=c-d\n",
+ "B=Ic/Ib\n",
+ "D=e/B\n",
+ "Ap=B**2*D\n",
+ "\n",
+ "#Result\n",
+ "print\" The value of current gain is\",B\n",
+ "print\" The value of resistance gain is\",D \n",
+ "print\" The value of power gain is\",Ap*10**-5,\"*10**5\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The value of current gain is 80.0\n",
+ " The value of resistance gain is 50.0\n",
+ " The value of power gain is 3.2 *10**5\n"
+ ]
+ }
+ ],
+ "prompt_number": 279
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.21 page no 1469"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "L1=58.6*10**-6 #H\n",
+ "C1=300.0*10**-12 #F\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "f=1/((2.0*math.pi)*math.sqrt(L1*C1))\n",
+ "\n",
+ "#Result\n",
+ "print\"Frequency of oscillation is\", round(f*10**-3,0),\"KHz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Frequency of oscillation is 1200.0 KHz\n"
+ ]
+ }
+ ],
+ "prompt_number": 294
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 27.22 Page no 1469"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Vbe=0.8 #V\n",
+ "Vcc=5 #V\n",
+ "Rc=1 #K ohm\n",
+ "b=250.0\n",
+ "Rb=100 #K ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Ic=Vcc/Rc\n",
+ "Ib=(Ic/b)*10**3\n",
+ "Vi=(Ib*Rb)+Vbe\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The minimum base current is\",Ib,\"micro A\"\n",
+ "print\"(ii) The input voltage is\",round(Vi*10**-3,0),\"V\"\n",
+ "print\"(iii) Between 0 V and 0.6 V,the transistor will switched off. Between 2.8 V and 5.0 V the transistor will switched on\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The minimum base current is 20.0 micro A\n",
+ "(ii) The input voltage is 2.0 V\n",
+ "(iii) Between 0 V and 0.6 V,the transistor will switched off. Between 2.8 V and 5.0 V the transistor will switched on\n"
+ ]
+ }
+ ],
+ "prompt_number": 309
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter2_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter2_3.ipynb new file mode 100644 index 00000000..3e6bc047 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter2_3.ipynb @@ -0,0 +1,684 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:44a7ee8bf26f9edf15a5d80a8ba17ba5be989b77884cdc1289de3bb40b0e7b99"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 2 Electric field"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.1 Page no 47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=1000\n",
+ "d=10.0*10**-3\n",
+ "m=4.8*10**-15\n",
+ "g=10\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "E=V/d\n",
+ "q=m*g/E\n",
+ "n=q/e\n",
+ "\n",
+ "#Result\n",
+ "print\"The number of electrons on the drop \", n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The number of electrons on the drop 3.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2 Page no 47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19\n",
+ "E=3*10**4\n",
+ "m=9.0*10**-31\n",
+ "y1=4*10**-2\n",
+ "m2=1.67*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "a=e*E/m\n",
+ "t=math.sqrt((2*y1)/a)\n",
+ "a2=e*E/m2\n",
+ "t2=math.sqrt((2*y1)/a2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Time t1=\", round(t*10**9,1)*10**-9,\"S\",\"\\nTime t2=\",round(t2*10**7,2)*10**-7,\"S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time t1= 3.9e-09 S \n",
+ "Time t2= 1.67e-07 S\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.3 Page no 48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=1 #m\n",
+ "m=9*10**9\n",
+ "q=500*10**-6\n",
+ "r1=0.3 #m\n",
+ "\n",
+ "#Calculation\n",
+ "E=m*q/r**2\n",
+ "E2=m*q/r1**2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Electric field intensity from the centre of the sphere \",E*10**-6,\"10**6\",\"N/C\"\n",
+ "print\"(ii) Electric field intensity at the surface of the sphere is \",E2*10**-7,\"10**7 N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Electric field intensity from the centre of the sphere 4.5 10**6 N/C\n",
+ "(ii) Electric field intensity at the surface of the sphere is 5.0 10**7 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.4 Page no 48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=2*10**-8\n",
+ "E=2*10**4\n",
+ "m=80*10**-6\n",
+ "g=9.8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=q*E/(m*g)\n",
+ "b=math.atan(a)*180/3.14\n",
+ "T=(q*E/(math.sin(b*3.14/180.0)))*10**-4\n",
+ "\n",
+ "#Result\n",
+ "print\"The angle is \", round(b,0),\"degree\"\n",
+ "print\"Tension in the thread of the pendulum is \", round(T*10**8,2),\"*10**-4 N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The angle is 27.0 degree\n",
+ "Tension in the thread of the pendulum is 8.8 *10**-4 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.5 Page no 48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "r=0.707\n",
+ "q=5*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "E=m*q/r**2 #along AO\n",
+ "E2=m*q/r**2 #along BO\n",
+ "E3=m*q/r**2 #along OD\n",
+ "E11=E+E2\n",
+ "E12=E2+E3\n",
+ "I=(2*E11*r)*10**-4\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric field at the centre of the sphere is \",round(I,2),\"*10**4 N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric field at the centre of the sphere is 25.46 *10**4 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.6 Page no 49"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=5*10**-9\n",
+ "x=0.15 #m\n",
+ "r=0.1 #m\n",
+ "a=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "E=(a*q*x)/((r**2+x**2))**1.5\n",
+ "\n",
+ "#Result\n",
+ "print\"Intensity of the electric field is \", round(E,0),\"N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Intensity of the electric field is 1152.0 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.7 Page no 49"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=10**-3\n",
+ "F=1\n",
+ "v0=20\n",
+ "v=0\n",
+ "\n",
+ "#Calculation\n",
+ "a=-F/m\n",
+ "s=v**2-v0**2/(2.0*a)\n",
+ "\n",
+ "#Result\n",
+ "print\"The distance is \", s,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The distance is 0.2 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.9 Page no 50"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q1=1/3.0*10**-7\n",
+ "r=5*10**-2\n",
+ "F=58.8*10**-3\n",
+ "\n",
+ "#Calculation\n",
+ "q2=F*r**2/(q1*m)\n",
+ "\n",
+ "#Result\n",
+ "print\"Charge is \", q2,\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge is 4.9e-07 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.10 Page no 61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=4*10**5 #V/m\n",
+ "q=3.2*10**-19\n",
+ "a=2.4*10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "p=q*a\n",
+ "W=p*E*(1-(math.cos(180*180/3.14)))\n",
+ "\n",
+ "#Result\n",
+ "print\"Work done is \", W"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Work done is 2.79670959474e-23\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.11 Page no 61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=16*10**-19\n",
+ "a=3.9*10**-12\n",
+ "E=10**5\n",
+ "\n",
+ "#Calculation\n",
+ "p=q*a\n",
+ "U=-p*E\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The electric dipole moment \", p,\"Cm\"\n",
+ "print\"(ii) Potential energy of dipole in the stable equilibrium position \",U,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The electric dipole moment 6.24e-30 Cm\n",
+ "(ii) Potential energy of dipole in the stable equilibrium position -6.24e-25 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.12 Page no 61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=20*10**-6\n",
+ "a=10**-2\n",
+ "m=9*10**9\n",
+ "r=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "p=q*a\n",
+ "E=m*2*p/r**3\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric field intensity is \", E*10**-5,\"*10**5 N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric field intensity is 36.0 *10**5 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.13 Page no 62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=4*10**5\n",
+ "q=1*10**-6\n",
+ "a=3*10**-2\n",
+ "\n",
+ "#Calculation\n",
+ "t=q*a*E\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum torque on the dipole is \", t*10**2,\"*10**-2 Nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum torque on the dipole is 1.2 *10**-2 Nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 55
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.14 Page no 62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=1*10**-6\n",
+ "a=2*10**-2\n",
+ "E=10**5\n",
+ "\n",
+ "#Calculation\n",
+ "p=q*a\n",
+ "W=2*p*E\n",
+ "\n",
+ "#Result\n",
+ "print\"Work done in the rotation is \", W*10**3,\"*10**-3 J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Work done in the rotation is 4.0 *10**-3 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 60
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.15 Page no 62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=2*10**-6\n",
+ "a=0.1\n",
+ "m=9*10**9\n",
+ "r=0.5\n",
+ "\n",
+ "#Calculation\n",
+ "p=q*a\n",
+ "E=m*p/r**3\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric field intensity is \",E*10**-4,\"*10**4 N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric field intensity is 1.44 *10**4 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 65
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.16 Page no 62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "qa=2.5*10**-7\n",
+ "qb=-2.5*10**-7\n",
+ "a=15\n",
+ "b=15\n",
+ "\n",
+ "#Calculation\n",
+ "q=qa+qb\n",
+ "C=(a+b)*10**-2\n",
+ "E=qa*C\n",
+ "\n",
+ "#Result\n",
+ "print\"Total charge is \", q,\"\\nElectric dipole moment of the system is \",E,\"Cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total charge is 0.0 \n",
+ "Electric dipole moment of the system is 7.5e-08 Cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 71
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.17 Page no 63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "p=2*10**-8\n",
+ "m=9*10**9\n",
+ "r=1.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "b=3*math.cos(60*3.14/180.0)**2+1\n",
+ "a=p*math.sqrt(b)\n",
+ "E=(m*a)/r**3\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of electric intensity is\", round(E,1),\"N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of electric intensity is 238.2 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.18 Page no 63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "p=5*10**-8\n",
+ "m=9*10**9\n",
+ "r=0.15\n",
+ "\n",
+ "#Calculation\n",
+ "E=m*2*p/r**3\n",
+ "E1=m*p/r**3\n",
+ "\n",
+ "print\"(i) Electric field along AB is \", round(E*10**-5,2),\"*10**5 N/C\"\n",
+ "print\"(ii) Electric field along BA is \", round(E1*10**-5,2),\"*10**5 N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Electric field along AB is 2.67 *10**5 N/C\n",
+ "(ii) Electric field along BA is 1.33 *10**5 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 89
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter3_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter3_3.ipynb new file mode 100644 index 00000000..8aff7044 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter3_3.ipynb @@ -0,0 +1,1016 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:b7451f8a5d88007ceae262e7f31a9c2ef46acd4d68ca53e0c7b4ce898e623cd8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3 Electrostatic potential and flux"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.1 Page no 89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=300*10**-6 #c\n",
+ "V=6\n",
+ "\n",
+ "#Calculation\n",
+ "W=q*V\n",
+ "\n",
+ "#Result\n",
+ "print\"Work done is \", W*10**3,\"*10**-3 J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Work done is 1.8 *10**-3 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.2 Page no 89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given \n",
+ "Va=-10 #V\n",
+ "W=300 #J\n",
+ "q=3.0 #C\n",
+ "\n",
+ "#Calculation\n",
+ "V=(W/q)+Va\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of V is \", V,\"Volts\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of V is 90.0 Volts\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3 Page no 89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q=16*10**-10 #C\n",
+ "r=0.1\n",
+ "r1=0.06\n",
+ "q1=12*10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "Vb=m*q/r\n",
+ "Vb1=m*q/r1\n",
+ "V=Vb1-Vb\n",
+ "W=q1*V\n",
+ "\n",
+ "#Result\n",
+ "print\"Workdone is \", W*10**8,\"*10**-8 J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Workdone is 11.52 *10**-8 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.4 Page no 89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=3.4*10**-14 #m\n",
+ "n=47\n",
+ "q=1.6*10**-19 #C\n",
+ "m=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "V=m*n*q/r\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric potential at the surface of silver nucleus is \", round(V*10**-6,2),\"*10**6 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric potential at the surface of silver nucleus is 1.99 *10**6 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.5 Page no 90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q=4*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "V=2*q*m\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric potential is \", V*10**-3,\"*10**3 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric potential is 72.0 *10**3 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.9 Page no 91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q=250*10**-6\n",
+ "r=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "V=m*q/r\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric potential at the centre is \", V*10**-7,\"*10**7 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric potential at the centre is 2.25 *10**7 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.10 Page no 96"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=3*10**-16\n",
+ "g=9.8\n",
+ "d=5*10**-3\n",
+ "q=16.0*10**-18\n",
+ "\n",
+ "#Calculation\n",
+ "V=(m*g*d/q)*10\n",
+ "\n",
+ "#Result\n",
+ "print\"Voltage needed to balance an oil drop is \",round(V,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage needed to balance an oil drop is 9.19 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.12 Page no 96"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=1.6*10**-19 #C\n",
+ "V=3000 #V\n",
+ "r=5*10**-2 #m\n",
+ "g=9.8\n",
+ "\n",
+ "#Calculation\n",
+ "E=V/r\n",
+ "m=q*E/g\n",
+ "\n",
+ "#Result\n",
+ "print\"The mass of the particle is \", round(m*10**16,1),\"*10**-16 Kg\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mass of the particle is 9.8 *10**-16 Kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.13 Page no 100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**-9\n",
+ "q1=3*10**-9\n",
+ "q2=3*10**-9\n",
+ "q3=10**9\n",
+ "r=0.2\n",
+ "\n",
+ "#Calculation\n",
+ "W=m*((q1*q3/r)+(q2*q3/r))\n",
+ "\n",
+ "#Result\n",
+ "print\"Workdone is \", W,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Workdone is 2.7e-07 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.14 Page no 100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q=1.6*10**-19\n",
+ "r=10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "U=m*q**2/r\n",
+ "K=U/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Kinetic energy is \",K,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Kinetic energy is 1.152e-18 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.15 Page no 100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**-31\n",
+ "V=10**6\n",
+ "q=1.6*10**-19\n",
+ "a=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "K=m*V**2\n",
+ "r=a*q**2/K\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance of the closest approach is \", r,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance of the closest approach is 2.56e-10 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 57
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.17 Page no 101"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.53*10**-10 #m\n",
+ "q1=1.6*10**-19 #C\n",
+ "q2=-1.6*10**-19 #C\n",
+ "a=9*10**9\n",
+ "r1=1.06*10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "U=a*q1*q2/r\n",
+ "Ue=U/q1\n",
+ "K=-Ue/2.0\n",
+ "E=Ue+K\n",
+ "U1=(a*q1*q2/r1)/q1\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Potential energy of the system is \", round(Ue,1),\"eV\"\n",
+ "print\"(ii) Minimum amount of work required to free the elctrons ia \",round(E,1),\"ev\"\n",
+ "print\"(iii) Potential energyof the system is \",round(E,1) ,\"ev and work requiredto free the electrons is \",round(-E,1),\"eV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Potential energy of the system is -27.2 eV\n",
+ "(ii) Minimum amount of work required to free the elctrons ia -13.6 ev\n",
+ "(iii) Potential energyof the system is -13.6 ev and work requiredto free the electrons is 13.6 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 75
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.18 Page no 102"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=9*10**9\n",
+ "q1=7*10**-6 #C\n",
+ "q2=-2*10**-6\n",
+ "r=0.18\n",
+ "r1=0.09\n",
+ "A=9*10**5\n",
+ "\n",
+ "#Calculation\n",
+ "U=a*q1*q2/r\n",
+ "W=0-U\n",
+ "U1=(q1*A/r1)+(q2*A/r1)+U\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Electrostatic potential energy is \", round(U,1),\"J\"\n",
+ "print\"(b) Work required to seperate two charges is \",round(W,1),\"J\"\n",
+ "print\"(c) Electrostatic energy is \", U1,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Electrostatic potential energy is -0.7 J\n",
+ "(b) Work required to seperate two charges is 0.7 J\n",
+ "(c) Electrostatic energy is 49.3 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 93
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.20 Page no 103"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "p=6*10**-6\n",
+ "E=10**6\n",
+ "a=1\n",
+ "\n",
+ "#Calculation,\n",
+ "U1=-p*E*a\n",
+ "U2=(p*E*(math.cos(60)*180/3.14))*10**-2\n",
+ "U3=U2-U1\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat released by substance is \", round(U3,0),\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat released by substance is 3.0 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 103
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.21 Page no 109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=10**-7\n",
+ "e=8.854*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "a=q/e\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric flux through the surface of the cube is \", round(a*10**-4,2),\"Nm**2C-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric flux through the surface of the cube is 1.13 Nm**2C-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 108
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.22 Page no 109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=8.85*10**-6 \n",
+ "e=8.85*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "a=q/e\n",
+ "b=a/6.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric flux through each face is \", round(b*10**-5,2),\"Nm**2C-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric flux through each face is 1.67 Nm**2C-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 115
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.23 Page no 109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E0=2*10**3 #N/C\n",
+ "S=0.2\n",
+ "\n",
+ "#Calculation\n",
+ "a=(3/5.0)*E0*S\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric flux of the field is \", a,\"Nm**2C-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric flux of the field is 240.0 Nm**2C-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 118
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.24 Page no 109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.2\n",
+ "m=9*10**9\n",
+ "b=50\n",
+ "\n",
+ "import math\n",
+ "E=250*r\n",
+ "a=E*4*math.pi*r**2\n",
+ "q=b*r**2/m\n",
+ "\n",
+ "#Result\n",
+ "print\"Charge contained in a sphere is \", round(q*10**10,2)*10**-10,\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge contained in a sphere is 2.22e-10 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 128
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.25 Page no 110"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=0.1 #m\n",
+ "A=800\n",
+ "e=8.854*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "b=A*a**2.5*(math.sqrt(2)-1)\n",
+ "q=e*b\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The flux through the cube is \", round(b,2),\"Nm**2C-1\"\n",
+ "print\"The charge within the cube is \",round(q*10**12,2)*10**-12,\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The flux through the cube is 1.05 Nm**2C-1\n",
+ "The charge within the cube is 9.28e-12 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 139
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.26 Page no 111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=200\n",
+ "a=0.05\n",
+ "e=8.854*10**-12\n",
+ "d=3.14\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "b=E*math.pi*a**2\n",
+ "c=2*b\n",
+ "q=e*d\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Net outward flux through each flat face is \", round(b,2),\"Nm**2C-1\"\n",
+ "print\"(b) Flux through the side of cylinder is zero \"\n",
+ "print\"(c) Net outward flux through the cylinder is \", round(c,2),\"Nm**2C-1\"\n",
+ "print\"(d) The net charge in the cylinder is \",round(q*10**11,2)*10**-11,\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Net outward flux through each flat face is 1.57 Nm**2C-1\n",
+ "(b) Flux through the side of cylinder is zero \n",
+ "(c) Net outward flux through the cylinder is 3.14 Nm**2C-1\n",
+ "(d) The net charge in the cylinder is 2.78e-11 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 156
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.28 Page no 114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=5.8*10**-6 #C\n",
+ "r=8*10**-2 #m\n",
+ "e=8.854*10**-12\n",
+ "l=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "E=q/(2*math.pi*e*r*l)\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric field is \", round(E*10**-5,1),\"*10**5 N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric field is 4.3 *10**5 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.29 Page no 114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=9*10**4 #N/C\n",
+ "r=2*10**-2 #m\n",
+ "m=9*10**9\n",
+ "\n",
+ "#Calculation\n",
+ "a=r*E/(2.0*m)\n",
+ "print\"Linear charge density is \", a,\"Cm-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Linear charge density is 1e-07 Cm-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.30 Page no 115"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=10*10**-6 #C\n",
+ "r=0.1 #m\n",
+ "a=8.85*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "E=q/(4.0*math.pi*a*r**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Electric field intensity at a point 10 cm from the centre\", round(E*10**-6,0),\"*10**6 N/C\"\n",
+ "print\"(ii) Since the point is lying inside the shell, electric intensity at this point is zero\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Electric field intensity at a point 10 cm from the centre 9.0 *10**6 N/C\n",
+ "(ii) Since the point is lying inside the shell, electric intensity at this point is zero\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.31 Page no 118"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Z=79\n",
+ "e=1.6*10**-19\n",
+ "e0=8.854*10**-12\n",
+ "R=6.2*10**-15\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "q=Z*e\n",
+ "E=q/(4.0*math.pi*e0*R**2)\n",
+ "b=E/4.0\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The magnitude of the electric field at the surface of nucleus is \", round(E*10**-21,0)*10**21,\"N/C\"\n",
+ "print\"(ii) The magnitude of the electric field at a distance 2R from the centre of the nucleus is \",round(b*10**-21,2),\"*10**21 N/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The magnitude of the electric field at the surface of nucleus is 3e+21 N/C\n",
+ "(ii) The magnitude of the electric field at a distance 2R from the centre of the nucleus is 0.74 *10**21 N/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.32 Page no 119"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=8.854*10**-12\n",
+ "A=0.5\n",
+ "F=1.8*10**-12 #N\n",
+ "E=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "q=(2*e*A**2*F)/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Total charge on the sheet is \", round(q*10**6,0),\"micro C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total charge on the sheet is 50.0 micro C\n"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.33 Page no 119"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=5*10**-6\n",
+ "e=8.854*10**-12\n",
+ "r=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "b=-(a*math.pi*r**2*(math.cos(60)*180/3.14))/(2*e)\n",
+ "\n",
+ "#Result\n",
+ "print\"Electric flux through a circular area is \", round(b*10**-5,2),\"*10**3 Nm**2C-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric flux through a circular area is 4.84 *10**3 Nm**2C-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 54
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter4_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter4_3.ipynb new file mode 100644 index 00000000..5987ff7f --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter4_3.ipynb @@ -0,0 +1,1326 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:2ebe494197bc592ac147978ecceacaa802bf7a7b9283aeec109e01967ce4cfa8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 4 Capacitance "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.1 Page no 159"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "r=6.4*10**6 #m\n",
+ "\n",
+ "#Calculation\n",
+ "C=r/m\n",
+ "\n",
+ "#Result\n",
+ "print\"The capacitance of the earth is \", round(C*10**6,0),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The capacitance of the earth is 711.0 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.2 Page no 160"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "c=50*10**-12\n",
+ "V=10**4\n",
+ "\n",
+ "#Calculation\n",
+ "r=(m*c)*10**2\n",
+ "q=(c*V)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Radius of a isolated sphere is \",r,\"cm\"\n",
+ "print\"(ii) Charge of a isolated sphere is \", q*10**6,\"micro C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Radius of a isolated sphere is 45.0 cm\n",
+ "(ii) Charge of a isolated sphere is 0.5 micro C\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.3 Page no 160"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=3*10**-3 #m\n",
+ "m=9*10**9\n",
+ "q1=27*10**-12 #C\n",
+ "\n",
+ "#Calculation\n",
+ "R=3*r\n",
+ "C=R/m\n",
+ "V=q1/C\n",
+ "\n",
+ "#Result\n",
+ "print\"Capacitance of the bigger drop is \", C*10**12,\"pico F \\npotential of the bigger drop is \",V,\"Volts\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance of the bigger drop is 1.0 pico F \n",
+ "potential of the bigger drop is 27.0 Volts\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4 Page no 162"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "ra=0.09\n",
+ "rb=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "C=ra*rb/(m*(rb-ra))\n",
+ "\n",
+ "#Result\n",
+ "print\"Capacitance of the capacitor is \", C*10**12,\"pico F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance of the capacitor is 100.0 pico F\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.5 Page no 162"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=2 #cm\n",
+ "d=1.2\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "R=(d/r)*10**2\n",
+ "rab=(R*2)\n",
+ "x=r**2+4*rab\n",
+ "y=math.sqrt(x)\n",
+ "\n",
+ "#Result\n",
+ "print\"ra+rb=\", y,\"cm \\nra-rb=\",r ,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "ra+rb= 22.0 cm \n",
+ "ra-rb= 2 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 64
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.6 Page no 164"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=10**-3 #m\n",
+ "c=1 #F\n",
+ "e=8.854*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "A=c*d/e\n",
+ "\n",
+ "#Result\n",
+ "print\"Area is \", round(A*10**-8,1),\"*10**8 m**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Area is 1.1 *10**8 m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 69
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.7 Page no 164"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=0.02 #m**2\n",
+ "r=0.5 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "d=A/(4.0*math.pi*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance is \", round(d*10**3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance is 3.18 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.8 Page no 164"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=8.854*10**-12\n",
+ "K=6\n",
+ "A=30\n",
+ "d=2.0*10**-3\n",
+ "E=500\n",
+ "\n",
+ "#Calculation\n",
+ "C=e*K*A/d\n",
+ "V=E*d*10**3\n",
+ "q=C*V\n",
+ "\n",
+ "#Result\n",
+ "print\"Capacitance of a parallel plate \", round(q*10**3,3),\"micro C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance of a parallel plate 0.797 micro C\n"
+ ]
+ }
+ ],
+ "prompt_number": 87
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.9 Page no 165"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=300*10**-12\n",
+ "V=10*10**3\n",
+ "A=0.01\n",
+ "d=1*10**-3\n",
+ "\n",
+ "#Calculation\n",
+ "q=C*V\n",
+ "a=q/A\n",
+ "E=V/d\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Charge on each plate is \", q,\"C\"\n",
+ "print\"(ii) Electric flux density is \", a*10**4,\"10**-4 C/m**2\"\n",
+ "print\"(iii) Potential gradient is \", E,\"V/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Charge on each plate is 3e-06 C\n",
+ "(ii) Electric flux density is 3.0 10**-4 C/m**2\n",
+ "(iii) Potential gradient is 10000000.0 V/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 101
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.10 Page no 165"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A2=500 #cm**2\n",
+ "A1=100 #cm**2\n",
+ "d1=0.05 #cm\n",
+ "\n",
+ "#Calculation\n",
+ "d2=A2*d1/A1\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance between the plates of second capacitor is \", d2,\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance between the plates of second capacitor is 0.25 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 104
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.11 page no 167"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "c1=0.5 #micro F\n",
+ "c2=0.3 #micro F\n",
+ "c3=0.2 #micro F\n",
+ "\n",
+ "#Calculation\n",
+ "Cp=c1+c2+c3 \n",
+ "Cs=(1/c1)+(1/c2)+(1/c3)\n",
+ "\n",
+ "#Result\n",
+ "print\" The ratio ofmaximum capacitance to minimum capacitance is \",round (Cs,1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The ratio ofmaximum capacitance to minimum capacitance is 10.3\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.12 Page no 168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "c1=15.0 #micro F\n",
+ "c2=20.0 #micro F\n",
+ "V=10**-6\n",
+ "v1=600 #V \n",
+ "\n",
+ "#Calculation\n",
+ "Cs=c1*c2/(c1+c2)\n",
+ "Q=Cs*V*v1\n",
+ "Pd=(Q/c1)*10**6\n",
+ "Pd1=(Q/c2)*10**6\n",
+ "\n",
+ "#Result\n",
+ "print\"(i)charge on each capacitor is\",round(Q *10**3,2),\"10**-3 C\"\n",
+ "print\"(ii)P.D across15 micro Fcapacitor is\",round (Pd,1),\"V\"\n",
+ "print\" P.D across 20 micro F is\",round (Pd1,0),\"V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i)charge on each capacitor is 5.14 10**-3 C\n",
+ "(ii)p.D across15 micro Fcapacitor is 342.9 V\n",
+ " P.D across 20 micro F is 257.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.13 page no.168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ca=18 #micro F\n",
+ "Cb=4 #micro F\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "C=Ca*Cb\n",
+ "C12=math.sqrt(Ca**2-4*C)\n",
+ "C2=2*C12\n",
+ "\n",
+ "#Result\n",
+ "print\"The capacitance of capacitor C1 is\", C12,\"micro F\"\n",
+ "print\"The capacitance of capacitor C2 is\",C2,\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The capacitance of capacitor C1 is 6.0 micro F\n",
+ "The capacitance of capacitor C2 is 12.0 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.14 Page no 168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=750*10**-6\n",
+ "C1=15*10**-6\n",
+ "V2=20.0 #V\n",
+ "C3=8*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "V1=q/C1\n",
+ "V=V1+V2\n",
+ "q3=C3*V2\n",
+ "q2=q-q3\n",
+ "C2=q2/V2\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of V1 is \", V1,\"V\"\n",
+ "print\"The value of V is \",V,\"V\"\n",
+ "print\"The value of capacitance is\",C2*10**6,\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of V1 is 50.0 V\n",
+ "The value of V is 70.0 V\n",
+ "The value of capacitance is 29.5 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 52
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.15 Page no 169"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C2=9.0 #micro F\n",
+ "C3=9.0\n",
+ "C4=9.0\n",
+ "C1=3\n",
+ "V=10 #V\n",
+ "\n",
+ "#Calculation\n",
+ "C=1/((1/C2)+(1/C3)+(1/C4))\n",
+ "Cab=C1+C\n",
+ "q=Cab*V\n",
+ "\n",
+ "#Result\n",
+ "print\"Equivalent capacitance between point A and B is \", Cab,\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equivalent capacitance between point A and B is 6.0 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 60
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.17 Page no 169"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Cab=10 #micro F\n",
+ "C1=8.0 #micro F\n",
+ "C2=8.0\n",
+ "C3=8\n",
+ "C4=8\n",
+ "C5=12\n",
+ "V=400\n",
+ "\n",
+ "#Calculation\n",
+ "Cbc=((C1*C2)/(C1+C2))+C3+C4\n",
+ "Cac=Cab*Cbc/(Cab+Cbc)\n",
+ "Ccd=C1+C5\n",
+ "Cad=Cac*Ccd/(Cac+Ccd)\n",
+ "q=Cad*V\n",
+ "Vcd=q/Ccd\n",
+ "q1=C5*Vcd\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The equivalent capacitance between A and D is \", Cad,\"micro f\"\n",
+ "print\"(ii) The charge on 12 micro F capacitor is \",q1*10**-3,\"mC\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The equivalent capacitance between A and D is 5.0 micro f\n",
+ "(ii) The charge on 12 micro F capacitor is 1.2 mC\n"
+ ]
+ }
+ ],
+ "prompt_number": 84
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.20 Page no 171"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C1=5 #micro F\n",
+ "C2=6 #micro F\n",
+ "V=10 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Cp=C1+C2\n",
+ "q=Cp*V\n",
+ "\n",
+ "#Result\n",
+ "print\"Charge supplied by battery is \", q,\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge supplied by battery is 110 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 88
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.21 Page no 171"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C1=2 #micro F\n",
+ "C2=2 #micro F\n",
+ "C3=2\n",
+ "C4=2\n",
+ "\n",
+ "#Calculation\n",
+ "Cs=C1*C2/(C1+C2)\n",
+ "Cab=C3*C4/(C3+C4)\n",
+ "\n",
+ "#Result\n",
+ "print\"The capacitance of the Capacitors\", Cab,\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The capacitance of the Capacitors 1 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 93
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.22 Page no 171"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C1=10.0 #micro F\n",
+ "C2=10.0\n",
+ "C3=10.0\n",
+ "C4=10*10**-3\n",
+ "V=500 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Cs=1/((1/C1)+(1/C2)+(1/C3))\n",
+ "Cab=Cs+(C4*10**3)\n",
+ "Q=(C1*(500/3.0))*10**-3\n",
+ "Q1=C4*V\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) The equivalent capacitance of the network is\",round(Cab,1),\"micro F\"\n",
+ "print \"(b) The charge on 12 micro F Capacitor is\",Q1,\"*10**-3 C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The equivalent capacitance of the network is 13.3 micro F\n",
+ "(b) The charge on 12 micro F Capacitor is 5.0 *10**-3 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 115
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.23 Page no 172"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C4=6 #micro F\n",
+ "C5=12 \n",
+ "C1=8.0\n",
+ "C7=1\n",
+ "\n",
+ "#Calculation\n",
+ "Cs=C4*C5/(C4+C5)\n",
+ "C11=(C1*Cs)/(C1+Cs)\n",
+ "Cs1=C1*C7/(C1+C7)\n",
+ "Cp=C11+Cs1\n",
+ "C=1/(1-(1/Cp))\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of capacitance C is \", round(C,2),\"micro F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of capacitance C is 1.39 micro F\n"
+ ]
+ }
+ ],
+ "prompt_number": 129
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.24 Page no 175"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "K=5\n",
+ "l=0.2\n",
+ "c=10**-9 #F\n",
+ "b=15.4\n",
+ "a=15\n",
+ "pd=5000 #V\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "C=(K*l*c)/(41.1*math.log10(b/a))\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The capacitance of cylindrical capacitor is \", round(C*10**9,1)*10**-9,\"F\"\n",
+ "print\"(ii) The potential of the inner cylinder is equal to p.d. between two cylinders i.e potentila of inner cylinder is\",pd,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The capacitance of cylindrical capacitor is 2.1e-09 F\n",
+ "(ii) The potential of the inner cylinder is equal to p.d. between two cylinders i.e potentila of inner cylinder is 5000 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 138
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.25 Page no 179"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=5*10**-6\n",
+ "V=100\n",
+ "C1=3*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "q=C*V\n",
+ "Cp=C+C1\n",
+ "pd=q/Cp\n",
+ "\n",
+ "#Result\n",
+ "print\"P.D across the capacitor is \", pd,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "P.D across the capacitor is 62.5 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 143
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.26 Page no 179 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=250 #V\n",
+ "C1=6 #micro F\n",
+ "C2=4\n",
+ "Cp=10*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "pd=V*C1/(C1+C2)\n",
+ "q=pd*C2*10**-6\n",
+ "q1=2*q\n",
+ "pd1=q1/Cp\n",
+ "q2=C2*pd1\n",
+ "q3=C1*pd1\n",
+ "\n",
+ "#Result\n",
+ "print\"New potentila difference is \", pd1,\"V\"\n",
+ "print\"Charge on 4 micro F capacitor is \",q2,\"micro C\"\n",
+ "print\"Charge on 6 micro F capacitor is \",q3,\"micro C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "New potentila difference is 120.0 V\n",
+ "Charge on 4 micro F capacitor is 480.0 micro C\n",
+ "Charge on 6 micro F capacitor is 720.0 micro C\n"
+ ]
+ }
+ ],
+ "prompt_number": 156
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.28 Page no 180"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C1=16*10**-6 # F\n",
+ "C2=4 #micro F\n",
+ "V1=100 #V\n",
+ "Cp=20*10**-6 #f\n",
+ "\n",
+ "#Calculation\n",
+ "q=C1*V1\n",
+ "U1=0.5*C1*V1**2\n",
+ "V=q/Cp\n",
+ "U2=0.5*Cp*V**2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Potential difference across the capacitor is \", V,\"Volts\"\n",
+ "print\"(ii) The electrostatic energies before and after the capacitors are connected \",U2,\"J\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Potential difference across the capacitor is 80.0 Volts\n",
+ "(ii) The electrostatic energies before and after the capacitors are connected 0.064 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.29 Page no 180"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "V=3.0*10**6\n",
+ "r=2\n",
+ "\n",
+ "#Calculation\n",
+ "q=(V*r)/m\n",
+ "E=0.5*q*V\n",
+ "\n",
+ "#Result\n",
+ "print\"The heat generated is \", E,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The heat generated is 1000.0 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.30 Page no 180"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=12 #V\n",
+ "C=1.35*10**-10 #C\n",
+ "\n",
+ "#Calculation\n",
+ "q=C\n",
+ "\n",
+ "#Result\n",
+ "print\"Extra Charge supplied by battery is \", q,\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Extra Charge supplied by battery is 1.35e-10 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.31 Page no 181"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "C=100*10**-6 #F\n",
+ "V=500 #V\n",
+ "\n",
+ "#Calculation\n",
+ "q=V/2.0\n",
+ "E=0.5*(0.5*C*V**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Charge in the new stored energy is \", E,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge in the new stored energy is 6.25 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.32 Page no 187"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=2*10**-3 #m**2\n",
+ "d=0.01 #m\n",
+ "t=6*10**-3 #m\n",
+ "K=3\n",
+ "a=8.854*10**-12\n",
+ "\n",
+ "#Calculation\n",
+ "C=a*A/(d-t*(1-(1/3.0)))\n",
+ "\n",
+ "#Result\n",
+ "print\"The capacitance of the capacitor is \", round(C*10**12,2)*10**-12,\"F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The capacitance of the capacitor is 2.95e-12 F\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.33 Page no 188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=8.854*10**-12\n",
+ "A=2\n",
+ "t1=0.5*10**-3\n",
+ "t2=1.5*10**-3\n",
+ "t3=0.3*10**-3\n",
+ "K1=2.0\n",
+ "K2=4.0\n",
+ "K3=6.0\n",
+ "\n",
+ "#Calculation\n",
+ "C=(e*A)/((t1/K1)+(t2/K2)+(t3/K3))\n",
+ "\n",
+ "#Result\n",
+ "print\"The capacitance of the capacitor is \", round(C*10**6,3)*10**-6,\"F\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The capacitance of the capacitor is 2.6e-08 F\n"
+ ]
+ }
+ ],
+ "prompt_number": 52
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.34 Page no 188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=3 #mm\n",
+ "b=4.0 #mm\n",
+ "K1=5\n",
+ "\n",
+ "#Calaculation\n",
+ "K2=1/((a**2/b)-a/b)*K1\n",
+ "\n",
+ "#Result\n",
+ "print\"The relative permittivity of the additional dielectric is \", round(K2,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The relative permittivity of the additional dielectric is 3.33\n"
+ ]
+ }
+ ],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.35 Page no 188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=5\n",
+ "t=2\n",
+ "K=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "D=d+(t-t/K)\n",
+ "\n",
+ "#Result\n",
+ "print\"New seperaion between the plates are \", round(D,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "New seperaion between the plates are 6.33 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 62
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.36 Page no 189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "d=4\n",
+ "t=2\n",
+ "K=4.0\n",
+ "C1=50*10**-12 #f\n",
+ "V0=200 #V\n",
+ "\n",
+ "#Calculation\n",
+ "C=(d-t+(t/K))/d\n",
+ "q=C1*V0\n",
+ "V=V0*C\n",
+ "U=0.5*q*V\n",
+ "E=0.5*q*(V0-V)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Final charge on ach plate is \", q,\"C\"\n",
+ "print\"(ii) P.D batween the plates is \", V,\"volts\"\n",
+ "print\"(iii)Final energy in the capacitor is \", U,\"J\"\n",
+ "print\"(iv) Energy loss is \", E,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Final charge on ach plate is 1e-08 C\n",
+ "(ii) P.D batween the plates is 125.0 volts\n",
+ "(iii)Final energy in the capacitor is 6.25e-07 J\n",
+ "(iv) Energy loss is 3.75e-07 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.39 Page no 193"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=25*10**5\n",
+ "E=5.0*10**7\n",
+ "\n",
+ "#Calculation\n",
+ "r=V/E\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum radius of the spherical shell is \", r*100,\"cm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum radius of the spherical shell is 5.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter5_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter5_3.ipynb new file mode 100644 index 00000000..639e5486 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter5_3.ipynb @@ -0,0 +1,1741 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:76fc8177d7b8f6f9c96106003a3d3f70d960561edd9596ed73b03964026f4fb8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 5 Electric current and resistance "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.1 Page no 257"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=10**17\n",
+ "e=1.6*10**-19 #C\n",
+ "t=1.0 #S\n",
+ "\n",
+ "#Calculation\n",
+ "I=n*e/t\n",
+ "\n",
+ "#Result\n",
+ "print\"The magnitude of current in the wire is \",I*10**2,\"10**-2 A and direction is from left to right\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The magnitude of current in the wire is 1.6 10**-2 A and direction is from left to right\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.2 Page no 257"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "I=0.5\n",
+ "T=1\n",
+ "e=1.6*10**-19\n",
+ "t=60 #minute\n",
+ "\n",
+ "#Calculation\n",
+ "n=I*T/e\n",
+ "q=I*t**2\n",
+ "n1=q/e\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The number of electrons passing a cross section of the bulb is \",round(n*10**-18,1)*10**18,\"electrons/S\"\n",
+ "print\"(ii) The number of electrons is \",round(n1*10**-22,1)*10**22,\"electrons/hour\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The number of electrons passing a cross section of the bulb is 3.1e+18 electrons/S\n",
+ "(ii) The number of electrons is 1.1e+22 electrons/hour\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.3 Page no 257"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19 #C\n",
+ "f=6.8*10**15 #rev/sec\n",
+ "r=0.51*10**-10 #m\n",
+ "\n",
+ "#Calculation\n",
+ "I=e*f\n",
+ "\n",
+ "#Result\n",
+ "print\"The equivalent current is \", I*10**3,\"*10**-3 A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The equivalent current is 1.088 *10**-3 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5 Page no 267"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=10 #A\n",
+ "A=1 #m*m**2\n",
+ "e=1.6*10**-19 #C\n",
+ "n=10**28 #m**-3\n",
+ "\n",
+ "#Calculation\n",
+ "Vd=I/(n*A*e)\n",
+ "\n",
+ "#Result\n",
+ "print\"Drift velocity of the conduction electrons are \", Vd,\"m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Drift velocity of the conduction electrons are 6.25e-09 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.6 Page no 267"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=10 #A\n",
+ "A=4*10**-6 #m**2\n",
+ "e=1.6*10**-19 #C\n",
+ "n=8*10**28 #m**-3\n",
+ "l=4\n",
+ "\n",
+ "#Calculation\n",
+ "Vd=I/(n*A*e)\n",
+ "t=l/Vd\n",
+ "\n",
+ "#Result\n",
+ "print\"Time required by an electron is \", t*10**-4,\"*10**4 S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time required by an electron is 2.048 *10**4 S\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.7 page no 267"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=6.023*10**23\n",
+ "m=63.5*10**-3\n",
+ "d=9*10**3\n",
+ "A=10**-7 #m**2\n",
+ "e=1.6*10**-19 #C\n",
+ "I=1.5 #a\n",
+ "K=1.38*10**-23 #J/K\n",
+ "T=300 #K\n",
+ "Vd=1.1*10**-3\n",
+ "C=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "n=a*d/m\n",
+ "Vd=I/(n*A*e)\n",
+ "V=math.sqrt((3*K*T*a)/m)\n",
+ "V1=Vd/V\n",
+ "E=Vd/C\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Thermal speeds of copper atoms at ordinary temperatures are \", round(V1*10**6,2),\"*10**-6 m/s\"\n",
+ "print\"(ii) Speed of propagation of electric fild is \", round(E*10**12,1)*10**-12"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Thermal speeds of copper atoms at ordinary temperatures are 3.2 *10**-6 m/s\n",
+ "(ii) Speed of propagation of electric fild is 3.7e-12\n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.8 Page no 268"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=5\n",
+ "l=0.1\n",
+ "Vd=2.5*10**-4\n",
+ "\n",
+ "#Calculation\n",
+ "E=V/l\n",
+ "u=Vd/E\n",
+ "\n",
+ "#Result\n",
+ "print\"The electron mobility is \", u,\"m**2/V/C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electron mobility is 5e-06 m**2/V/C\n"
+ ]
+ }
+ ],
+ "prompt_number": 57
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.9 Page no 268"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=2.4\n",
+ "A=0.30*10**-6\n",
+ "m=9.1*10**-31\n",
+ "n=8.4*10**28\n",
+ "e=1.6*10**-19\n",
+ "E=7.5\n",
+ "\n",
+ "#Calculation\n",
+ "J=I/A\n",
+ "t=m*J/(n*e**2*E)\n",
+ "\n",
+ "#Result\n",
+ "print\"Average relaxation time is \", round(t*10**16,2)*10**-16,\"S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average relaxation time is 4.51e-16 S\n"
+ ]
+ }
+ ],
+ "prompt_number": 64
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.10 Page no 268"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.12*10**-2 #m\n",
+ "I=10\n",
+ "r1=0.08*10**-2 #m\n",
+ "I=10 #A\n",
+ "e=1.6*10**-19 #C\n",
+ "n=8.4*10**28\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=math.pi*(r**2)\n",
+ "J=I/A\n",
+ "A1=math.pi*r1**2\n",
+ "Vd=I/(e*n*A1)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Current density in the alluminium wire is \",round(J*10**-6,1),\"*10**6 A/m**2\"\n",
+ "print\"(ii) Drift velocity of electrons in the copper wire is \",round(Vd*10**4,1),\"*10**-4 m/S\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Current density in the alluminium wire is 2.2 *10**6 A/m**2\n",
+ "(ii) Drift velocity of electrons in the copper wire is 3.7 *10**-4 m/S\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.11 Page no 269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "D=0.13*10**-2\n",
+ "R=3.4 #ohms\n",
+ "l=10.0\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=(math.pi/4.0)*D**2\n",
+ "a=R*A/l\n",
+ "b=1/a\n",
+ "\n",
+ "#Result\n",
+ "print\"Conductivity of a material is \",round(b*10**-6,1),\"*10**6 S/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Conductivity of a material is 2.2 *10**6 S/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 93
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.12 Page no 269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A1=25.0 #mm**2\n",
+ "l2=1 #m\n",
+ "R2=1/58.0\n",
+ "A2=1\n",
+ "l1=1000\n",
+ "\n",
+ "#Calculation\n",
+ "R=(l1/l2)*(A2/A1)\n",
+ "R1=R*R2\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of resistance is \", round(R1,2),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of resistance is 0.69 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 98
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.13 Page no 269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=4.5\n",
+ "A1=1\n",
+ "A2=2.0\n",
+ "l2=3\n",
+ "l1=1.0\n",
+ "\n",
+ "#Calculation\n",
+ "R=(l2/l1)*(A1/A2)\n",
+ "R2=R*R1\n",
+ "\n",
+ "#Result\n",
+ "print\"The resistance of another wire is \", R2,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resistance of another wire is 6.75 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 103
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.14 Page no 269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=1\n",
+ "r1=0.5\n",
+ "R1=0.15 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A1=(math.pi/4.0)*r**2\n",
+ "A2=(math.pi/4.0)*r1**2\n",
+ "l=A1/A2\n",
+ "R=l*l\n",
+ "R2=R*R1\n",
+ "\n",
+ "#Result\n",
+ "print\"New resistance of the wire is \", R2,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "New resistance of the wire is 2.4 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 115
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.15 Page no 270"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=1.5 #ohm\n",
+ "e=1.6*10**-19 #C\n",
+ "t=1 #second\n",
+ "V=3 #V\n",
+ "\n",
+ "#Calculation\n",
+ "I=V/R\n",
+ "n=I*t/e\n",
+ "\n",
+ "#Result\n",
+ "print \"Number of electrons flowing through it in 1 S is \",n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of electrons flowing through it in 1 S is 1.25e+19\n"
+ ]
+ }
+ ],
+ "prompt_number": 118
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.16 Page no 270"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "ne=2.8*10**18\n",
+ "np=1.2*10**18\n",
+ "e=1.6*10**-19\n",
+ "t=1 #S\n",
+ "V=220\n",
+ "\n",
+ "#Calculation\n",
+ "q=(ne+np)*e\n",
+ "I=q/t\n",
+ "R=V/I\n",
+ "\n",
+ "#Result\n",
+ "print\"Effective resistance of the tube is \", round(R,0),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Effective resistance of the tube is 344.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 124
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.17 Page no 270"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=84 #g\n",
+ "d=10.5 #g/cm**3\n",
+ "a=1.6*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "V=m/d\n",
+ "s=V**(1/3.0)\n",
+ "R=a/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Resistance between the opposite faces is \", R,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance between the opposite faces is 8e-07 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 130
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.18 Page no 270"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l=1.001\n",
+ "A=1.001\n",
+ "\n",
+ "#Calculation\n",
+ "R=l*A\n",
+ "R1=R-1\n",
+ "A=R1*100\n",
+ "\n",
+ "#Result\n",
+ "print\"Percentage change in its resistance is \", round(A,1),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percentage change in its resistance is 0.2 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 137
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.19 Page no 271"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=0.45 #Kg\n",
+ "R=0.0014 #ohm\n",
+ "a=1.78*10**-8 #ohm\n",
+ "d=8.93*10**3 #Kg/m**3\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "l=math.sqrt(R*m/(a*d))\n",
+ "r=math.sqrt(m/(math.pi*d*1.99))\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of length is\",round(l,2),\"m\"\n",
+ "print\"The value of radius is \",round(r*10**3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of length is 1.99 m\n",
+ "The value of radius is 2.84 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 148
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.20 Page no 271"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R15=80 #ohm\n",
+ "a=0.004\n",
+ "\n",
+ "#Calculation\n",
+ "R0=R15/(1+15*a)\n",
+ "R50=R0*(1+a*50)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of resistance at 50 degree C is \", round(R50,2),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of resistance at 50 degree C is 90.57 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 154
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.21 Page no 271"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R20=20 #ohm\n",
+ "P=60 #W\n",
+ "V=120.0 #Volts\n",
+ "a=5*10**-3\n",
+ "\n",
+ "#Calculation\n",
+ "I=P/V\n",
+ "Rt=V/I\n",
+ "t=(((Rt/R20)-1)/a)+R20\n",
+ "\n",
+ "#Result\n",
+ "print\"Normal working temperature of the lamp is \", t,\"degree C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Normal working temperature of the lamp is 2220.0 degree C\n"
+ ]
+ }
+ ],
+ "prompt_number": 160
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.22 Page no 271"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R0=5 #ohm\n",
+ "R100=5.23 #ohm\n",
+ "Rt=5.795 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "t=((Rt-R0)/(R100-R0))*100\n",
+ "\n",
+ "#Result\n",
+ "print\"The temperature of the bath is \", round(t,2),\"degree C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The temperature of the bath is 345.65 degree C\n"
+ ]
+ }
+ ],
+ "prompt_number": 165
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.23 Page no 272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=15*10**-4 #m**2\n",
+ "a=7.6*10**-8 # ohm m\n",
+ "l=2000 #m\n",
+ "b=0.005 #degree/C\n",
+ "\n",
+ "#Calculation\n",
+ "R0=a*l/A\n",
+ "R50=R0*(1+(b*50))\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of resistance is \", round(R50,3),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of resistance is 0.127 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 172
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.24 Page no 272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=0.004\n",
+ "ac=0.0007\n",
+ "R0=100\n",
+ "\n",
+ "#Calculation\n",
+ "R=ac*R0/a\n",
+ "\n",
+ "#Result\n",
+ "print\"The resistance of a copper filament is \", R,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resistance of a copper filament is 17.5 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 175
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.28 Page no 281"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=4.0 #ohm\n",
+ "R2=4.0 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Rab=1/((1/R1)+(1/R2))\n",
+ "\n",
+ "#Result\n",
+ "print\"The equivalent resisatance is \", Rab,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The equivalent resisatance is 2.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 178
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.29 Page no 281"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=15 #ohm\n",
+ "R2=30 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "R=R1*R2/(R1+R2)\n",
+ "\n",
+ "#Result\n",
+ "print\"The equivqlent resistance between A and B is \", R,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The equivqlent resistance between A and B is 10 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 181
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.31 Page no 282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=5 #ohm\n",
+ "R2=9 #ohm\n",
+ "R3=14 #ohm\n",
+ "R4=11\n",
+ "R5=7\n",
+ "R6=18\n",
+ "R7=13\n",
+ "R8=22\n",
+ "V=22\n",
+ "\n",
+ "#Calculation\n",
+ "Rec=(R1+R2)*R3/(R1+R2+R3)\n",
+ "Rbe=(R4+R5)*R6/(R4+R5+R6)\n",
+ "Rae=(R7+R2)*R8/(R7+R2+R8)\n",
+ "I=V/Rae\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of current in the branch AF is \", I,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of current in the branch AF is 2 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 187
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.32 Page no 282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=12 #ohm\n",
+ "R2=6 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Rdg=R1*R2/(R1+R2)\n",
+ "Rch=R1*R2/(R1+R2)\n",
+ "Rab=Rdg+Rch\n",
+ "\n",
+ "#Result\n",
+ "print\"The equivalent resistance is \", Rab,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The equivalent resistance is 8 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 191
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.33 Page no 283"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rab=500.0 #ohm\n",
+ "Rl=500 #ohm\n",
+ "Rbc=1500 #ohm\n",
+ "E=50 #Volts\n",
+ "Rac=2000.0 #ohm\n",
+ "V=40\n",
+ "\n",
+ "#Calculation\n",
+ "R=Rbc*Rl/(Rbc+Rl)\n",
+ "I=E/(Rab+R)\n",
+ "Pd=I*Rab\n",
+ "Rl1=E-Pd\n",
+ "I1=E/Rac\n",
+ "R12=V/I1\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Potential difference across the road is \", round(Rl1,2),\"V\"\n",
+ "print\"(ii) Resistance at BC is \", R12,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Potential difference across the road is 21.43 V\n",
+ "(ii) Resistance at BC is 1600.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 206
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.35 Page no 284"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=5 #ohm\n",
+ "R2=5.0 #ohm\n",
+ "R=6\n",
+ "\n",
+ "#Calculation\n",
+ "n=(1/(R-R1)*R2)\n",
+ "\n",
+ "#Result\n",
+ "print\"There are\", n,\"resistance are in parallel\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "There are 5.0 resistance are in parallel\n"
+ ]
+ }
+ ],
+ "prompt_number": 210
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.36 Page no 284"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=20.0 #ohm\n",
+ "R2=10.0 #ohm\n",
+ "R4=10\n",
+ "\n",
+ "#Calculation\n",
+ "Rbd=(R1*R2)/(R1+R2)\n",
+ "Rae=R2+Rbd+R4\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of resistance is \", round(Rae,2),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of resistance is 26.67 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 218
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.37 Page no 284"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=2.0 #ohm\n",
+ "R2=3 #ohm\n",
+ "R3=2.8\n",
+ "E=6 #V\n",
+ "\n",
+ "#Calculation\n",
+ "Rab=R1*R2/(R1+R2)\n",
+ "Rt=Rab+R3\n",
+ "I=E/Rt\n",
+ "Vab=I*Rab\n",
+ "I1=Vab/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"The steady state current is \", I1,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The steady state current is 0.9 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 226
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.38 Page no 285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=3 #ohm\n",
+ "R2=3\n",
+ "R3=6\n",
+ "\n",
+ "#Calculation\n",
+ "Rad=(R1+R2)*R3/(R1+R2+R3)\n",
+ "Rae=(Rad+R1)*R3/(Rad+R1+R3)\n",
+ "Raf=(Rae+R1)*R3/(Rae+R1+R3)\n",
+ "Rab=(Raf+R1)*R2/(Rae+R1+R2)\n",
+ "\n",
+ "#Result\n",
+ "print\"the effective resistance between the point A and B is\", Rab,\"Ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the effective resistance between the point A and B is 2 Ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 234
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 39 Page no 285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R2=50.0 #ohm\n",
+ "R3=50.0 #ohm\n",
+ "R4=75.0 #ohm\n",
+ "E=4.75\n",
+ "R1=100\n",
+ "\n",
+ "#Calculation\n",
+ "Rbc=1/((1/R2)+(1/R3)+(1/R4))\n",
+ "R=R1+Rbc\n",
+ "I=E/R\n",
+ "R11=I*R1\n",
+ "Vbc=E-(I*R1)\n",
+ "I2=Vbc/R2\n",
+ "I3=Vbc/R3\n",
+ "I4=Vbc/R4\n",
+ "\n",
+ "#Result\n",
+ "print\"Equivalent resistance of the circuit is \", R,\"ohm\"\n",
+ "print\"Current in R2 is\",I2,\"A\"\n",
+ "print\"Current in R3 is\",I3,\"A\"\n",
+ "print\"Current in R4 is\",I4,\"A\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equivalent resistance of the circuit is 118.75 ohm\n",
+ "Current in R2 is 0.015 A\n",
+ "Current in R3 is 0.015 A\n",
+ "Current in R4 is 0.01 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 251
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.40 Page no 292"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=19\n",
+ "I1=0.5\n",
+ "I2=2 #A\n",
+ "r=2 \n",
+ "\n",
+ "#Calculation\n",
+ "E=V+I1*r\n",
+ "\n",
+ "#Result\n",
+ "print\"E.M.F is \", E,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "E.M.F is 20.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 254
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.41 Page no 292"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=1.5\n",
+ "a=1.5\n",
+ "r1=0.5 #ohm\n",
+ "r2=0.25\n",
+ "R=2.25 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "E=V+a\n",
+ "r=r1+r2\n",
+ "Rt=r+R\n",
+ "I=E/Rt\n",
+ "pd=V-(I*r1)\n",
+ "pd1=V-(I*r2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The circuit current is \",I,\"A\"\n",
+ "print\"(ii) P.D across the terminals of each cell is \",pd,\"V and \",pd1,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The circuit current is 1.0 A\n",
+ "(ii) P.D across the terminals of each cell is 1.0 V and 1.25 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 268
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.42 Page no 292"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=10\n",
+ "E=1.5\n",
+ "R=4 #ohm\n",
+ "r=0.1\n",
+ "a=8\n",
+ "\n",
+ "#Calculation\n",
+ "Emf=n*E\n",
+ "Rt=R+(n*r)\n",
+ "I=Emf/Rt\n",
+ "Emf1=(a*E)-(2*E)\n",
+ "I1=Emf1/Rt\n",
+ "I11=I-I1\n",
+ "\n",
+ "#Result\n",
+ "print\"Reduction in current is \", I11,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Reduction in current is 1.2 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 277
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.43 Page no 292"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Emf=2\n",
+ "Emf1=1.9\n",
+ "Emf2=1.8\n",
+ "R1=0.05\n",
+ "R2=0.06\n",
+ "R3=0.07\n",
+ "R0=5 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Emft=Emf+Emf1+Emf2\n",
+ "R=R1+R2+R3\n",
+ "Rt=R+R0\n",
+ "I=Emft/Rt\n",
+ "\n",
+ "#Result\n",
+ "print\"The reading of the ammeter is \", round(I,1),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The reading of the ammeter is 1.1 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 283
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.44 Page no 293"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=6.0 #ohm\n",
+ "R2=3\n",
+ "I=0.8 #A\n",
+ "a=24\n",
+ "\n",
+ "#Calculation\n",
+ "I1=I*(R1+R2)/R1\n",
+ "I11=I1-I\n",
+ "Rp=R1*R2/(R1+R2)\n",
+ "Rt=Rp+8\n",
+ "r=(a/I1)-10\n",
+ "V=I1*Rt\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Current in 6 ohm resistance is \", I11,\"A\"\n",
+ "print\"(ii) Internal resistance of the battery is \", r,\"ohm\"\n",
+ "print\"(iii) The terminal potential difference of the battery is \", V,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Current in 6 ohm resistance is 0.4 A\n",
+ "(ii) Internal resistance of the battery is 10.0 ohm\n",
+ "(iii) The terminal potential difference of the battery is 12.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 295
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.45 Page no 294"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=2 #ohm\n",
+ "R2=4\n",
+ "R3=6\n",
+ "E=8\n",
+ "r=1\n",
+ "\n",
+ "#Calculation\n",
+ "Rac=(R1+R2)*R3/(R1+R2+R3)\n",
+ "I=E/(Rac+r)\n",
+ "I1=I/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Internal resistance is \", I1,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Internal resistance is 1.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 299
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.46 Page no 294"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=1\n",
+ "R=2\n",
+ "\n",
+ "#Calculation\n",
+ "r=(E*R)-E\n",
+ "print\"The internal resisatnce of aech cell is \",r,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The internal resisatnce of aech cell is 1 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.47 Page no 294"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=15.0 # ohm\n",
+ "R2=15.0\n",
+ "E=2\n",
+ "V=1.6\n",
+ "\n",
+ "#Calculation\n",
+ "R=R1*R2/(R1+R2)\n",
+ "r=((E/V)-1)*R*4\n",
+ "\n",
+ "#Result\n",
+ "print\"Internal resisatnce of each cell is \", r,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Internal resisatnce of each cell is 7.5 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.48 Page no 295"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I1=1 #A\n",
+ "E=1.5\n",
+ "I2=0.6\n",
+ "R2=2.33 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "R=2*E/I1\n",
+ "R1=2*E/I2\n",
+ "r=R1-2*R2\n",
+ "\n",
+ "#Result\n",
+ "print\"Internal resisatnce of each battery is \", r,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Internal resisatnce of each battery is 0.34 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.49 Page no 295"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=4 #ohm\n",
+ "R2=4 #ohm\n",
+ "R3=12\n",
+ "R4=6.0\n",
+ "E=16\n",
+ "r=1 #ohm\n",
+ "\n",
+ "#calculation\n",
+ "Rab=R1*R2/(R1+R2)\n",
+ "Rcd=R3*R4/(R3+R4)\n",
+ "R=Rab+Rcd+1\n",
+ "I=E/(R+r)\n",
+ "I1=I/2.0\n",
+ "I3=I*R4/(R3+R4)\n",
+ "I4=I*R3/(R3+R4)\n",
+ "Vab=4*I1\n",
+ "Vbc=I*1\n",
+ "Vcd=12*I3\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) equivalent resistance of the network is \", R,\"ohm\"\n",
+ "print\"(ii) Circuit current is\", I,\"A , Current in R1 is\",I1,\"A , Current in R3 is\",round(I3,2),\"A , Current in R4 is \",round(I4,2)\n",
+ "print \"Voltage drop Vab is\",Vab,\"V \\nVbc is\",Vbc,\"V \\nVcd is\",Vcd,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) equivalent resistance of the network is 7.0 ohm\n",
+ "(ii) Circuit current is 2.0 A , Current in R1 is 1.0 A , Current in R3 is 0.67 A , Current in R4 is 1.33\n",
+ "Voltage drop Vab is 4.0 V \n",
+ "Vbc is 2.0 V \n",
+ "Vcd is 8.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter6_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter6_3.ipynb new file mode 100644 index 00000000..f1c6f4b7 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter6_3.ipynb @@ -0,0 +1,624 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:1bababaf98233cb133c1893aef0743afe5d48d8c1bcfe7a7487181ba9a8fde89"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 6 Electrical measurements"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.1 Page no 332"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=4\n",
+ "b=2.0\n",
+ "c=8\n",
+ "d=5\n",
+ "e=3.0\n",
+ "\n",
+ "#Calculation\n",
+ "I1=((a*c)+(b*e))/((b*c)+(d*e))\n",
+ "I2=(a-(2*I1))/e\n",
+ "V=(I1-I2)*5\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Current through each battery is\", round(I1,2),\"A and\",round(I2,2),\"A\"\n",
+ "print\"(ii) Terminal voltage is\",round(V,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Current through each battery is 1.23 A and 0.52 A\n",
+ "(ii) Terminal voltage is 3.55 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.4 Page no 334"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=10\n",
+ "b=5.0\n",
+ "c=9.0\n",
+ "d=19.0\n",
+ "\n",
+ "#Calculation\n",
+ "I2=(a-c)/((b*a)-(d*c))\n",
+ "I1=(1-(5*I2))/c\n",
+ "I=I1+I2\n",
+ "pd=I*10\n",
+ "\n",
+ "#Result\n",
+ "print\"Current through each cell is\", round(I,2),\"A\"\n",
+ "print\"Potential difference across 10 ohm wire is\",round(pd,3),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current through each cell is 0.11 A\n",
+ "Potential difference across 10 ohm wire is 1.074 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6 Page no 335"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=-3\n",
+ "b=4.0\n",
+ "c=3\n",
+ "\n",
+ "#Calculation\n",
+ "I1=a/(b+(c**2))\n",
+ "I2=-1-c*I1\n",
+ "I3=-(I1+I2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Current through each cell is\", round(I1,2),\"A ,\",round(I2,2),\"A and\",round(I3,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current through each cell is -0.23 A , -0.31 A and 0.54 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.7 Page no 336"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=15\n",
+ "b=4\n",
+ "c=12.0\n",
+ "d=10\n",
+ "\n",
+ "#Calculation\n",
+ "R=(a*b)/c\n",
+ "X=(d*R)/(d-R)\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of resistance is\", X,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of resistance is 10.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.8 Page no 336"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=4 #ohm\n",
+ "R2=3 #ohm\n",
+ "R3=2.0\n",
+ "R11=2.4 #ohm\n",
+ "E=6\n",
+ "\n",
+ "#Calculation\n",
+ "X=(R1*R2)/R3\n",
+ "R4=R2+X\n",
+ "R5=R1+R3\n",
+ "Rt=((R4*R5)/(R4+R5))+R11\n",
+ "I=E/Rt\n",
+ "\n",
+ "#Result\n",
+ "print\"the value of unknown resistance is\", X,\"ohm\"\n",
+ "print\"The current drawn by the circuit is\",I,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of unknown resistance is 6.0 ohm\n",
+ "The current drawn by the circuit is 1.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 59
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.9 Page no 337"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=10\n",
+ "b=7.0\n",
+ "c=5\n",
+ "d=4\n",
+ "e=8.0\n",
+ "\n",
+ "#Calculation\n",
+ "I1=(a+a)/(b+1)\n",
+ "I3=(c+(4*I1))/e\n",
+ "I2=(-a+(6*I3)+I1)/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Current I1=\",I1,\"A \\nI2=\",I2,\"A \\nI3=\",I3,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current I1= 2.5 A \n",
+ "I2= 1.875 A \n",
+ "I3= 1.875 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 70
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.11 Page no 338"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=28\n",
+ "b=5.0\n",
+ "c=2\n",
+ "\n",
+ "#Calculation\n",
+ "Rak=a/(b*c)\n",
+ "\n",
+ "#Result\n",
+ "print\"Total resistance from one end of vacant edge to other end is\", Rak,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total resistance from one end of vacant edge to other end is 2.8 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.12 Page no 345"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=10\n",
+ "l2=68.5\n",
+ "l1=58.3\n",
+ "\n",
+ "#Calculation\n",
+ "X=R*(l2/l1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of X is\", round(X,1),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of X is 11.7 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.13 Page no 346"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=2 #ohm\n",
+ "R1=2.4 #ohm\n",
+ "V=4 #V\n",
+ "E=1.5\n",
+ "\n",
+ "#Calculation\n",
+ "R11=R+R1\n",
+ "I=V/R11\n",
+ "Vab=I*R\n",
+ "K=Vab\n",
+ "l=E/K\n",
+ "\n",
+ "#Result\n",
+ "print\"Length for zero galvanometer deflection is\", l,\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Length for zero galvanometer deflection is 0.825 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 85
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.15 Page no 347"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "l1=33.7\n",
+ "l2=51.9\n",
+ "\n",
+ "#Calculation\n",
+ "S1=l1/(100-l1)\n",
+ "s11=l2/(100-l2)\n",
+ "s=((s11*12)/S1)-12\n",
+ "R=s*S1\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of R is\", round(R,2),\"ohm \\nValue of S is\",round(s,1),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of R is 6.85 ohm \n",
+ "Value of S is 13.5 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.16 Page no 347"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=0.4\n",
+ "b=0.6\n",
+ "lab=10\n",
+ "\n",
+ "#Calculation\n",
+ "K=a/b\n",
+ "Vab=K*lab\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Potentila gradient along AB is\",round(K,2),\"V/m\"\n",
+ "print \"(ii) P.D between point A and B is\",round(Vab,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Potentila gradient along AB is 0.67 V/m\n",
+ "(ii) P.D between point A and B is 6.67 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 103
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.17 Page no 348"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=990 #ohm\n",
+ "R=10.0 #ohm\n",
+ "E=2\n",
+ "l=1000 #mm\n",
+ "l1=400 #mm\n",
+ "\n",
+ "#Calculation\n",
+ "Rt=R1+R\n",
+ "I=E/Rt\n",
+ "pd=I*R\n",
+ "K=pd/l\n",
+ "pd1=K*l1\n",
+ "\n",
+ "#Result\n",
+ "print\"e.m.f. generated by the thermocouple is\", pd1,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "e.m.f. generated by the thermocouple is 0.008 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 111
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.18 Page no 348"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "AB=600 #cm\n",
+ "AC=500.0 #cm\n",
+ "l=40*10**-3 #A\n",
+ "E=2\n",
+ "r=10\n",
+ "\n",
+ "#Calculation\n",
+ "R=2*AB/(AC*l)\n",
+ "K=2/AC\n",
+ "AC1=AC-r\n",
+ "pd=K*AC1\n",
+ "Iv=(E-pd)/r\n",
+ "R1=pd/Iv\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The resistance of the whole wire is\", R,\"ohm\"\n",
+ "print\"(ii) Reading of voltmeter is\", pd,\"V\"\n",
+ "print\"(iii) Resistance of the voltmeter is\",R1,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The resistance of the whole wire is 60.0 ohm\n",
+ "(ii) Reading of voltmeter is 1.96 V\n",
+ "(iii) Resistance of the voltmeter is 490.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 124
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.20 Page no 350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=6\n",
+ "b=2\n",
+ "\n",
+ "#Calculation\n",
+ "R1=a/((b*b)-1)\n",
+ "R2=b*R1\n",
+ "\n",
+ "#Result\n",
+ "print\"Resistance R1 is\", R1,\"ohm \\nR2 is\",R2,\"Ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance R1 is 2 ohm \n",
+ "R2 is 4 Ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 130
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.21 Page no 350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=20 #ohm\n",
+ "L=10 #m\n",
+ "pd=10**-3 #V/m\n",
+ "V=10**-2 #Volts\n",
+ "\n",
+ "#Calculation\n",
+ "I=V/R\n",
+ "R11=(2/I)-R\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of resistance is\", R11,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of resistance is 3980.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 134
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter7_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter7_3.ipynb new file mode 100644 index 00000000..74f3f441 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter7_3.ipynb @@ -0,0 +1,735 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:47b2c0fcc74d4ba925e8938987dfe5c551c445c65c0145d8b28ae4df323cfc30"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 7 Heating effect of electric curent"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.1 Page no 373"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=240 #V\n",
+ "P=60\n",
+ "P1=100\n",
+ "\n",
+ "#Calculation\n",
+ "R=V**2/P\n",
+ "R1=V**2/P1\n",
+ "\n",
+ "#Result\n",
+ "print\"Resistance of a bulb for 60 W is\", R,\"ohm and for 100 W is\",R1,\"ohm\"\n",
+ "print\"Hence The resistance of 60 W,240 V bulb is more than that of 100 W,240 V bulb\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance of a bulb for 60 W is 960 ohm and for 100 W is 576 ohm\n",
+ "Hence The resistance of 60 W,240 V bulb is more than that of 100 W,240 V bulb\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.2 Page no 373"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=230 #v\n",
+ "P=100\n",
+ "t=20*60\n",
+ "V1=115 #V\n",
+ "\n",
+ "#Calculation\n",
+ "R=V**2/P\n",
+ "E=(V1**2*t)/R\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat and light energy is\", E,\"J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat and light energy is 30000 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.3 Page no 374"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=500 #W\n",
+ "V=200.0 #V\n",
+ "V1=240\n",
+ "\n",
+ "#Calculation\n",
+ "I=P/V\n",
+ "R=V1-V\n",
+ "R1=R/I\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of R=\",R1,\"ohm\"\n",
+ "print\"Current in a circuit is\",I,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R= 16.0 ohm\n",
+ "Current in a circuit is 2.5 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.4 Page no 374"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P1=100.0 #W\n",
+ "P=1100.0 #W\n",
+ "V=250\n",
+ "\n",
+ "#Calculation\n",
+ "P2=P-P1\n",
+ "R=V**2/P2\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of unknown resistance is\", R,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of unknown resistance is 62.5 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.6 Page no 374"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=220\n",
+ "P=200.0\n",
+ "P1=100\n",
+ "\n",
+ "#Calculation\n",
+ "R1=V**2/P\n",
+ "R2=V**2/P1\n",
+ "H=R1/R2\n",
+ "\n",
+ "#Result\n",
+ "print\"The ratio of heats genetated in them is\", H"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of heats genetated in them is 0.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.7 Page no 375"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=1\n",
+ "c=1\n",
+ "a=100 #W\n",
+ "b=15\n",
+ "t=7.5 #second\n",
+ "P=1 #KW\n",
+ "C=860 #Kcal\n",
+ "\n",
+ "#Calculation\n",
+ "A=m*c*(a-b)\n",
+ "B=P*t/60.0\n",
+ "D=B*C\n",
+ "n=A*a/D\n",
+ "\n",
+ "#Result\n",
+ "print\"Efficiency of the kettle is\", round(n,1),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Efficiency of the kettle is 79.1 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.8 Page no 375"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P1=9 #W\n",
+ "R1=8\n",
+ "R2=12.0\n",
+ "\n",
+ "#Calculation\n",
+ "P2=(P1*R1)/R2\n",
+ "\n",
+ "#Result\n",
+ "print\"Power dissipated in 12 ohm resistor is\", P2,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power dissipated in 12 ohm resistor is 6.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.9 Page no 375"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "H1=10\n",
+ "a=5.0\n",
+ "b=4.2\n",
+ "\n",
+ "#Calculation\n",
+ "I1=(H1*b)/(a*4)\n",
+ "A=I1*4/b\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat generated in 4 ohm resistor is\", A,\"cal/sec\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat generated in 4 ohm resistor is 2.0 cal/sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.10 Page no 375"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=12 #V\n",
+ "I=1 #A\n",
+ "r=0.5 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "P1=E*I\n",
+ "P2=I**2*r\n",
+ "P=P1-P2\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Rate of consumption of chemical energy is\", P1,\"W\"\n",
+ "print\"(ii) Rate Of energy dissipated inside the battery is\",P2,\"W\"\n",
+ "print\"(iv) Rate of energy dissipated in the resistor is\", P,\"W\"\n",
+ "print\"(v) Power output of the source is\",P,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Rate of consumption of chemical energy is 12 W\n",
+ "(ii) Rate Of energy dissipated inside the battery is 0.5 W\n",
+ "(iv) Rate of energy dissipated in the resistor is 11.5 W\n",
+ "(v) Power output of the source is 11.5 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 55
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.11 Page no 376"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P=110 #W\n",
+ "P1=100 #W\n",
+ "n=5\n",
+ "V=220 #V\n",
+ "t=2 #hours\n",
+ "n1=4\n",
+ "P2=1120 #W\n",
+ "m=1.5 #per KWh\n",
+ "\n",
+ "#Calculation\n",
+ "W=n*P1\n",
+ "W1=V*t\n",
+ "W2=n1*P\n",
+ "W3=W+W1+W2+P2\n",
+ "E=(W3*t)*10**-3\n",
+ "E2=E*30\n",
+ "B=m*E2\n",
+ "\n",
+ "#Result\n",
+ "print\"Electricity bill for the month of september is\", B,\"Rs\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electricity bill for the month of september is 225.0 Rs\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.12 Page no 376"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=220 #V\n",
+ "P=60.0 #W\n",
+ "P1=85 #w\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "R=V**2/P\n",
+ "V1=math.sqrt(P1*R)\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum voltage is\", round(V1,1),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum voltage is 261.9 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 75
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.13 Page no 376"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=200 #V\n",
+ "P=500.0 #W\n",
+ "V1=160 #v\n",
+ "\n",
+ "#Calculation\n",
+ "R=V**2/P\n",
+ "H=V1**2/R\n",
+ "P1=P-H\n",
+ "H1=P1*100/P\n",
+ "\n",
+ "#Result\n",
+ "print\"Heat percentage is\", H1,\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat percentage is 36.0 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.14 Page no 376"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "P1=500 #W\n",
+ "P2=100\n",
+ "\n",
+ "#Calculation\n",
+ "R=P1/P2\n",
+ "\n",
+ "#Result\n",
+ "print\"Since P1'=5P2', 100W bulb will glow brighter\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Since P1'=5P2', 100W bulb will glow brighter\n"
+ ]
+ }
+ ],
+ "prompt_number": 86
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.15 Page no 377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=900\n",
+ "w=100.0\n",
+ "c=1\n",
+ "a=80\n",
+ "b=4.2\n",
+ "V=210 #V\n",
+ "x=12\n",
+ "y=60\n",
+ "\n",
+ "#Calculation\n",
+ "Hout=(m+w)*c*a\n",
+ "Hin=(V*x*y)/b\n",
+ "Hin1=90/w*Hin\n",
+ "I=Hout/Hin1\n",
+ "\n",
+ "#Result\n",
+ "print\"Strength of the current is\", round(I,3),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Strength of the current is 2.469 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.16 Page no 377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=0.8\n",
+ "\n",
+ "#Calculation\n",
+ "H=a**2\n",
+ "H1=(1-H)*100\n",
+ "\n",
+ "#Result\n",
+ "print\"Decreased percentage is\", H1,\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Decreased percentage is 36.0 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 99
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.17 Page no 377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=14\n",
+ "b=60\n",
+ "c=24\n",
+ "d=7.0\n",
+ "\n",
+ "#Calculation\n",
+ "t=a*b/60.0\n",
+ "t1=(c/d)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Time in series is\", t,\"minute\"\n",
+ "print\"(ii) Time in parallel is\",round(t1,2),\"minute\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Time in series is 14.0 minute\n",
+ "(ii) Time in parallel is 3.43 minute\n"
+ ]
+ }
+ ],
+ "prompt_number": 108
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.19 Page no 378"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=0.5\n",
+ "R=100\n",
+ "t=30\n",
+ "a=4.2\n",
+ "m=200 #g\n",
+ "w=10 #g\n",
+ "\n",
+ "#Calculation\n",
+ "H=I**2*R*t*60/a\n",
+ "A=H/(m+w)\n",
+ "\n",
+ "#Result\n",
+ "print\"The rise of temperature is\", round(A,2),\"degree C\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The rise of temperature is 51.02 degree C\n"
+ ]
+ }
+ ],
+ "prompt_number": 113
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.20 Page no 379"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "c=4.2 #KJ/Kg/C\n",
+ "m=0.2 #Kg\n",
+ "a=90\n",
+ "b=20\n",
+ "t=30\n",
+ "V=230\n",
+ "\n",
+ "#calculation\n",
+ "d=a-b\n",
+ "H=c*m*d\n",
+ "P=H/t\n",
+ "I=P/V\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of current is\", round(I*10**3,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of current is 8.52 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 120
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter8_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter8_3.ipynb new file mode 100644 index 00000000..baadbaea --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter8_3.ipynb @@ -0,0 +1,643 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:db58e543ef86cd601814ac49a8404db7a1403e7140977a41ff4c6b1fc2ae61b9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 8 Magnetic field due to electric current"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.1 Page no 417"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=1.6*10**-19 #c\n",
+ "B=0.1 #T\n",
+ "v=5.0*10**6 #m/s\n",
+ "a=90 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Fm=q*v*B*math.sin(a)\n",
+ "\n",
+ "#Result\n",
+ "print\"Force on the proton is\", round(Fm*10**14,1)*10**-14,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force on the proton is 7.2e-14 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.2 Page no 417"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=1.0*10**29 #m**-3\n",
+ "e=1.6*10**-19 #C\n",
+ "A=2*10**-6 #m**2\n",
+ "I=5 #A\n",
+ "B=0.15 #T\n",
+ "a=90 #degree\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Vd=I/(n*e*A)\n",
+ "Fm=e*Vd*B*math.sin(a)\n",
+ "\n",
+ "#Result\n",
+ "print\"Force acting on each electron is\", round(Fm*10**24,2)*10**-24,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force acting on each electron is 3.35e-24 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.3 Page no 417"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=2*1.6*10**-19 #C\n",
+ "v=6*10**5 #m/s\n",
+ "B=0.2 #T\n",
+ "a=90 #degree\n",
+ "m=6.65*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "Fm=q*v*B*math.sin(a)\n",
+ "a=Fm/m\n",
+ "\n",
+ "#Result\n",
+ "print\"Force on alpha particle is\", round(Fm*10**14,2)*10**-14,\"N\"\n",
+ "print\"Acceleration of alpha particle is\",round(a*10**-12,2)*10**12,\"m/s**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force on alpha particle is 3.43e-14 N\n",
+ "Acceleration of alpha particle is 5.16e+12 m/s**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.4 Page no 426"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=60 #degree\n",
+ "u=4*3.14*10**-7 #T/A m\n",
+ "Bc=2\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "a=(Bc/2.0)/(math.tan(60)*180/3.14)\n",
+ "B1=(10**-7*math.tan(60)*(math.sin(60*180/3.14)+math.sin(60*180/3.14)))*10\n",
+ "B=3*B1\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic fieldat the centroid of the triangle is\", round(B*10**7,0),\"*10**-7 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic fieldat the centroid of the triangle is 10.0 *10**-7 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 64
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.5 Page no 426"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=20\n",
+ "I=1 #A\n",
+ "r=0.08 #m\n",
+ "u=4*3.14*10**-7 #T/A m\n",
+ "\n",
+ "#Calculation\n",
+ "B=u*n*I/(2*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the magnetic field is\", B*10**4,\"*10*4 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the magnetic field is 1.57 *10*4 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.6 Page no 426"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=10**-7\n",
+ "I=10*10**-2 #A\n",
+ "r=0.5\n",
+ "\n",
+ "#Calculation\n",
+ "B=u*I/r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic field on Y axis is\", B,\"K^ T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic field on Y axis is 4e-08 K^ T\n"
+ ]
+ }
+ ],
+ "prompt_number": 71
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.7 Page no 426"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=5 #A\n",
+ "l=0.01 #m\n",
+ "a=45 #degree\n",
+ "r=2 #m\n",
+ "u=10**-7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=(u*I*l*math.sin(a)*180/3.14)/r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic field is\", round(B*10**8,1)*10**-10,\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic field is 6.1e-10 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.8 Page no 427"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "u=4*3.14*10**-7 #T/A m\n",
+ "n=20\n",
+ "I=12 #A\n",
+ "r=0.1 #m\n",
+ "\n",
+ "#Calculation\n",
+ "B=u*n*I/(2*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic field at the centre of coil is\", round(B*10**3,1),\"*10**-3 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic field at the centre of coil is 1.5 *10**-3 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 87
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.12 Page no 429"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.02 #m\n",
+ "\n",
+ "#Calculation\n",
+ "B=u*I/(4*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"The magnitude of magnetic field is\", round(B*10**4,2),\"*10**-4 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The magnitude of magnetic field is 1.88 *10**-4 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 92
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.13 Page no 429"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v=4*10**6\n",
+ "r=0.5*10**-10\n",
+ "e=1.6*10**-19\n",
+ "t=1\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "f=v/(2.0*math.pi*r)\n",
+ "I=f*e/t\n",
+ "B=u*I/(2*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnetic field produced by the electrons is\", round(B,1),\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic field produced by the electrons is 25.6 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 98
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.15 Page no 430"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=100\n",
+ "I=5 #A\n",
+ "r=0.1 #m\n",
+ "x=0.05\n",
+ "\n",
+ "#Calculation\n",
+ "B=u*n*I/(2*r)\n",
+ "B1=(u*n*I*r**2)/(2.0*(r**2+x**2)**1.5)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Magnetic field at the centre of the coil is\",B*10**3,\"*10**-3 T\"\n",
+ "print\"(ii) The magnetic field at the point on the axis of the coil is\",round(B1*10**3,2),\"*10**-3 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnetic field at the centre of the coil is 3.14 *10**-3 T\n",
+ "(ii) The magnetic field at the point on the axis of the coil is 2.25 *10**-3 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 109
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.18 Page no 431"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=5*10**-2\n",
+ "I=50\n",
+ "e=1.6*10**-19\n",
+ "B1=10**7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=u*I/(2*math.pi*a)\n",
+ "F=e*B1*B\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Force on electron when velocity is towards the wire\", round(F*10**16,1)*10**-16,\"N\"\n",
+ "print\"(ii) Force on electron when velocity is parallel to the wire\", round(F*10**16,1)*10**-16,\"N\"\n",
+ "print\"(iii) Force on electron when velocity is perpendicular to the wire is zero\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Force on electron when velocity is towards the wire 3.2e-16 N\n",
+ "(ii) Force on electron when velocity is parallel to the wire 3.2e-16 N\n",
+ "(iii) Force on electron when velocity is perpendicular to the wire is zero\n"
+ ]
+ }
+ ],
+ "prompt_number": 118
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.20 Page no 432"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19\n",
+ "f=6.8*10**15\n",
+ "r=0.51*10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=e*f\n",
+ "B=(u*I)/(2*r)\n",
+ "M=1*I*math.pi*r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"The effective dipole moment is\",round(M*10**24,0)*10**-24,\"Am**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The effective dipole moment is 9e-24 Am**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 127
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.22 Page no 439"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=5*850/1.23\n",
+ "I=5.57 #A\n",
+ "\n",
+ "#calculation\n",
+ "u=4*math.pi*10**-7\n",
+ "B=u*n*I\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of magnetic field is\", round(B*10**3,1),\"*10**-3 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of magnetic field is 24.2 *10**-3 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.23 Page no 439"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r1=20\n",
+ "r2=25\n",
+ "I=2 #a\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=(r1+r2)/2.0\n",
+ "l=(2*math.pi*r)*10**-2\n",
+ "n=1500/l\n",
+ "B=u*n*I\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Magnetic field inside the toroid is\", round(B,3),\"T\"\n",
+ "print\"(ii) magnetic field outside the toroid is zero\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnetic field inside the toroid is 0.003 T\n",
+ "(ii) magnetic field outside the toroid is zero\n"
+ ]
+ }
+ ],
+ "prompt_number": 138
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.25 Page no 440"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=2 #A\n",
+ "R=5*10**-2 #m\n",
+ "r=3*10**-2 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=u*I*r/(2*math.pi*R**2)\n",
+ "\n",
+ "#Result\n",
+ "print round(B*10**6,1),\"*10**-6 T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "4.8 *10**-6 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 142
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter9_3.ipynb b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter9_3.ipynb new file mode 100644 index 00000000..7669d0e6 --- /dev/null +++ b/principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter9_3.ipynb @@ -0,0 +1,1740 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:02dc05916beb2a686e89acb729415599e9656d95fc169884e1d4a92b0e8ee888"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 9 Motion of charged particles in electric and magnetic motion"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.1 Page no 472"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=90 #V\n",
+ "d=2.0*10**-2\n",
+ "e=1.8*10**11\n",
+ "x=5*10**-2\n",
+ "v=10**7\n",
+ "\n",
+ "#Calculation\n",
+ "E=V/d\n",
+ "a=e*E\n",
+ "t=x/v\n",
+ "y=0.5*a*t**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Transverse deflection produced by electric field is\", round(y*10**2,1),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Transverse deflection produced by electric field is 1.0 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.2 Page no 473"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=500\n",
+ "d=2*10**-2 #m\n",
+ "v=3*10**7\n",
+ "x=6*10**-2\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "E=V/d\n",
+ "a=E*e\n",
+ "t=x/v\n",
+ "v1=a*t\n",
+ "T=v1/v\n",
+ "A=math.atan(T)*180.0/3.14\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle is\", round(A,1),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle is 16.7 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.3 Page no 474"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "x=10*10**-2\n",
+ "v=3*10**7\n",
+ "S=1.76*10**-3\n",
+ "a=1800\n",
+ "\n",
+ "#Calculation\n",
+ "t=x/v\n",
+ "e=S*2/(a*t**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Specific charge of the electron is\", e,\"C/Kg\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Specific charge of the electron is 1.76e+11\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.4 Page no 478"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**-31\n",
+ "v=3*10**7\n",
+ "q=1.6*10**-19 #C\n",
+ "B=6*10**-4\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=m*v/(q*B)\n",
+ "f=q*B/(2.0*math.pi*m)\n",
+ "E=(0.5*m*v**2)/1.6*10**-16\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy is\", round(E*10**32,2),\"Kev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy is 2.53 Kev\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.5 Page no 479"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**-31\n",
+ "e=1.6*10**-19\n",
+ "V=100\n",
+ "B=0.004\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=math.sqrt(2*m*e*V)/(e*B)\n",
+ "\n",
+ "#Result\n",
+ "print\"Radius of the path is\", round(r*10**3,1),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of the path is 8.4 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 50
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.6 Page no 479"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=1.67*10**-27\n",
+ "v=4*10**5\n",
+ "a=60\n",
+ "q=1.6*10**-19\n",
+ "B=0.3\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=(m*v*math.sin(a*3.14/180.0))/q*B\n",
+ "P=v*math.cos(a*3.14/180.0)*((2*math.pi*m)/(q*B))\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Radius of the helical path is\",round(r*10**3,1),\"cm\"\n",
+ "print\"(ii) Pitch of helix is\", round(P*10**2,2),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Radius of the helical path is 1.1 cm\n",
+ "(ii) Pitch of helix is 4.38 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 70
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.7 Page no 479"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "M=5*10**6 #ev\n",
+ "e=1.6*10**-19\n",
+ "m=1.6*10**-27\n",
+ "B=1.5\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "v=math.sqrt((2*M*e)/m)\n",
+ "F=q*v*B*math.sin(90*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the force is\", round(F*10**12,2)*10**-12,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the force is 7.59e-12 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.8 Page no 480"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=1.67*10**-27 #Kg\n",
+ "v=4*10**5\n",
+ "B=0.3 #T\n",
+ "q=1.6*10**-19 #C\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=m*v*math.sin(60*3.14/180.0)/(q*B)\n",
+ "P=2*math.pi*r*1/(math.tan(60*3.14/180.0))\n",
+ "\n",
+ "#Result\n",
+ "print\"Pitch of the helix is\", round(P*10**2,2),\"cm\"\n",
+ "print\"Radius of helical path is\",round(r*10**2,3),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Pitch of the helix is 4.38 cm\n",
+ "Radius of helical path is 1.205 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9 Page no 480"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=3.2*10**-19\n",
+ "B=1.2\n",
+ "r=0.45\n",
+ "m=6.8*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "v=(q*B*r)/m\n",
+ "f=v/(2.0*math.pi*r)\n",
+ "K=(0.5*m*v**2)/(1.6*10**-19)\n",
+ "V=K/2.0\n",
+ "\n",
+ "#Result\n",
+ "print\"Required potentila difference is\", round(V*10**-6,0),\"*10**6 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Required potentila difference is 7.0 *10**6 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.10 Page no 480"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=4\n",
+ "u=10**-7\n",
+ "a=0.2 #m\n",
+ "v=4*10**6\n",
+ "q=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "B=(u*2*I)/a\n",
+ "F=q*v*B\n",
+ "\n",
+ "#Result\n",
+ "print\"Force is\", F,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force is 2.56e-18 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.11 Page no 481"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19\n",
+ "a=10**6\n",
+ "\n",
+ "#Calculation\n",
+ "q=2*e\n",
+ "F=q*a\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude force acting on the particle is\", F"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude force acting on the particle is 3.2e-13\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.13 Page no 482"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=3.4*10**4 #V/m\n",
+ "B=2*10**-3 #Wb/m**2\n",
+ "m=9.1*10**-31\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "v=E/B\n",
+ "r=(m*v)/(e*B)\n",
+ "\n",
+ "#Result\n",
+ "print\"Radius of the circular path is\", round(r*10**2,1),\"*10**-2 m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of the circular path is 4.8 *10**-2 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.14 Page no 482"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=600 #V\n",
+ "d=3*10**-3 #m\n",
+ "v=2*10**6 #m/s\n",
+ "\n",
+ "#Calculation\n",
+ "B=V/(d*v)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of magnetic field is\", B,\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of magnetic field is 0.1 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.15 Page no 487"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=1.6*10**-19 #c\n",
+ "B=2 #T\n",
+ "m=1.66*10**-27 #Kg\n",
+ "K=5*10**6\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "f=(q*B)/(2.0*math.pi*m)\n",
+ "v=math.sqrt((2*K*q)/m)\n",
+ "r=(m*v)/(q*B)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The frequency needed for applied alternating voltage is\", round(f*10**-7,0),\"*10**7 HZ\"\n",
+ "print\"(ii) Radius of the cyclotron is\",round(r,2),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The frequency needed for applied alternating voltage is 3.0 *10**7 HZ\n",
+ "(ii) Radius of the cyclotron is 0.16 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.16 Page no 487"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=1.7 #T\n",
+ "q=1.6*10**-19 #c\n",
+ "r=0.5\n",
+ "m=1.66*10**-27\n",
+ "\n",
+ "#Calculation\n",
+ "K=((B**2*q**2*r**2)/(2.0*m))/q\n",
+ "\n",
+ "#Result\n",
+ "print\"Kinetic energy of proton is\", round(K*10**-6,0),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Kinetic energy of proton is 35.0 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.17 Page no 487"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=0.8\n",
+ "q=3.2*10**-19 #C\n",
+ "d=1.2\n",
+ "m=4*1.66*10**-27 #Kg\n",
+ "a=1.60*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r=d/2.0\n",
+ "K=(B**2*q**2*r**2)/(2.0*m*a)\n",
+ "v=(q*B*r)/m\n",
+ "f=(q*B)/(2.0*math.pi*m)\n",
+ "\n",
+ "#Result\n",
+ "print\"Frequency of alternating voltage is\", round(f*10**-7,2),\"*10**7 HZ\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Frequency of alternating voltage is 0.61 *10**7 HZ\n"
+ ]
+ }
+ ],
+ "prompt_number": 61
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.18 Page no 488"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=1.6*10**-19 #C\n",
+ "r=0.6 #m\n",
+ "m=1.67*10**-27 #Kg\n",
+ "f=10**7\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "B=(2*math.pi*m*f)/q\n",
+ "K=((B**2*q**2*r**2)/(2.0*m))/1.6*10**-13\n",
+ "\n",
+ "#Result\n",
+ "print\"Kinetic energy of the protons is\", round(K*10**26,1),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Kinetic energy of the protons is 7.4 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 67
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.19 Page no 493"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I=5 #A\n",
+ "l=0.06 #m\n",
+ "B=0.02 #T\n",
+ "a=90\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "F=I*B*l*math.sin(a*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Force is\", round(F,3),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force is 0.006 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 71
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.20 Page no 494"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=0.2 #Kg\n",
+ "I=2 #A\n",
+ "l=1.5 #m\n",
+ "g=9.8\n",
+ "\n",
+ "#Calculation\n",
+ "B=(m*g)/(I*l)\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of the magnetic field is\", round(B,2),\"T\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of the magnetic field is 0.65 T\n"
+ ]
+ }
+ ],
+ "prompt_number": 76
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.21 Page no 494"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r=0.002 #m\n",
+ "m=0.05\n",
+ "g=9.8\n",
+ "\n",
+ "#Calculation\n",
+ "u=4*math.pi*10**-7\n",
+ "f=u/(2*math.pi*r)\n",
+ "f1=m*g\n",
+ "I=math.sqrt(f1*f**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Current in each wire is\", I,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current in each wire is 70.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.22 Page no 494"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=0.04 #m\n",
+ "I1=20\n",
+ "I2=16\n",
+ "l=0.15\n",
+ "r1=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "u=4*math.pi*10**-7\n",
+ "F1=(u*I1*I2*l)/(2.0*math.pi*r)\n",
+ "F2=(u*I1*I2*l)/(2.0*math.pi*r1)\n",
+ "F=F1-F2\n",
+ "\n",
+ "#Result\n",
+ "print\"Net force on the loop is\", F*10**4,\"*10**-4 N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Net force on the loop is 1.44 *10**-4 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 89
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.23 Page no 495"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=0.3 #Kg\n",
+ "a=30 #degree\n",
+ "B=0.15 #T\n",
+ "g=9.8 #m/s**2\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "I=(m*g*math.tan(a*3.14/180.0))/B\n",
+ "\n",
+ "#Result\n",
+ "print\"value of current is\", round(I,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of current is 11.31 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 94
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.24 Page no 495"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "B=3*10**-5 #T\n",
+ "I=1 #A\n",
+ "\n",
+ "#Calculation\n",
+ "F=I*B*math.sin(90)\n",
+ "\n",
+ "#Result\n",
+ "print\"The direction of the force is downward i.e\", round(F*10**5,0),\"*10**-5 N/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The direction of the force is downward i.e 3.0 *10**-5 N/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 99
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Example 9.25 Page no 495"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=1.2*10**-3\n",
+ "B=0.6 #T\n",
+ "g=9.8 #m/s**2\n",
+ "r=0.05\n",
+ "b=3.8\n",
+ "\n",
+ "#Calculation\n",
+ "I=(m*g)/B\n",
+ "R=r*b\n",
+ "V=I*R\n",
+ "\n",
+ "#Result\n",
+ "print\"Potentila difference is\", round(V*10**3,1),\"*10**-3 V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Potentila difference is 3.7 *10**-3 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 105
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.26 Page no 496"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "I2=10 #A\n",
+ "r=0.1 #m\n",
+ "l=2 #m\n",
+ "I1=2\n",
+ "I2=10\n",
+ "r=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "u=4*math.pi*10**-7\n",
+ "F=u*I1*I2*I1/(2.0*math.pi*r)\n",
+ "\n",
+ "#Result\n",
+ "print\"Force on small conductor\", F,\"N\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force on small conductor 8e-05 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 109
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.27 Page no 500"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "A=10**-3 #m**\n",
+ "n=10\n",
+ "I=2 #A\n",
+ "B=0.1 #T\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "t=n*I*A*B*math.cos(0)\n",
+ "t1=n*I*A*B*math.cos(60*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Torque when magnetic field is parallel to the field\", round(t*10**3,0),\"*10**-3 Nm\"\n",
+ "print\"(ii) Torque when magnetic field is perpendicular to the field is zero\"\n",
+ "print\"(iii) Torque when magnetic field is 60 degree to the field is\",round(t1*10**3,1),\"*10**-3 Nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Torque when magnetic field is parallel to the field 2.0 *10**-3 Nm\n",
+ "(ii) Torque when magnetic field is perpendicular to the field is zero\n",
+ "(iii) Torque when magnetic field is 60 degree to the field is 1.0 *10**-3 Nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 121
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.28 Page no 500"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "r=7\n",
+ "I=10\n",
+ "B=100*10**-4\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=math.pi*r**2\n",
+ "t=I*A*B\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of maximum torque is\", round(t*10**-1,2),\"*10**-3 Nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of maximum torque is 1.54 *10**-3 Nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 127
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.29 Page no 501"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "N=10\n",
+ "I=0.06\n",
+ "r=0.05\n",
+ "n=1000\n",
+ "I2=25\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "A=math.pi*r**2\n",
+ "M=N*I*A\n",
+ "u=4*math.pi*10**-7\n",
+ "B=u*n*I2\n",
+ "t=M*B*math.sin(45*3.14/180.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Torgue is\", round(t*10**4,2),\"*10**-4 Nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Torgue is 1.05 *10**-4 Nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 134
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.30 Page no 501"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=100\n",
+ "l=3.2 \n",
+ "r=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "u=4*math.pi*10**-7\n",
+ "B=(u*n*l)/(2.0*r)\n",
+ "M=n*l*math.pi*r**2\n",
+ "t=M*B*math.sin(0)\n",
+ "t1=(M*B*math.sin(90*3.14/180.0))*10**3\n",
+ "w=math.sqrt((2*M*B*10**3)/r)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Field at the centre of the coil is\", round(B*10**3,0),\"*10**-3 T\"\n",
+ "print\"(b) Magnetic moment of the coil is\",round(M,0),\"Am**2\"\n",
+ "print\"(c) Magnitude of the torque on the coil in the initial position is\",t\n",
+ "print\" Magnitude of the torque on the coil in the final position is\",round(t1,0),\"Nm\"\n",
+ "print \"(d) Angular speed acquired by the coil is\",round(w,0),\"rad/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Field at the centre of the coil is 2.0 *10**-3 T\n",
+ "(b) Magnetic moment of the coil is 10.0 Am**2\n",
+ "(c) Magnitude of the torque on the coil in the initial position is 0.0\n",
+ " Magnitude of the torque on the coil in the final position is 20.0 Nm\n",
+ "(d) Angular speed acquired by the coil is 20.0 rad/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.31 Page no 505"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=125\n",
+ "I=20*10**-3 #A\n",
+ "B=0.5 #T\n",
+ "A=400*10**-6 #m**2\n",
+ "K=40*10**-6\n",
+ "\n",
+ "#Calculation\n",
+ "T=n*I*B*A\n",
+ "a=T/K\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Torque exerted is\", T*10**4,\"*10**-4 Nm\"\n",
+ "print\"(ii) Angular deflection of the coil is\", a,\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Torque exerted is 5.0 *10**-4 Nm\n",
+ "(ii) Angular deflection of the coil is 12.5 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.32 Page no 505"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "K=3*10**-9 #Nm/deg\n",
+ "a=36\n",
+ "n=60\n",
+ "B=9*10**-3 #T\n",
+ "A=5*10**-5 #m**2\n",
+ "\n",
+ "#Calculation\n",
+ "I=(K*a)/(n*B*A)\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum current is\", I*10**3,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum current is 4.0 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.33 Page no 506"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=30\n",
+ "B=0.25 #T\n",
+ "A=1.5*10**-3\n",
+ "K=10**-3\n",
+ "\n",
+ "#Calculation\n",
+ "S=(n*B*A)/K\n",
+ "\n",
+ "#Result\n",
+ "print\"Current sensitivity of the galvanometer is\", S,\"degree/A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current sensitivity of the galvanometer is 11.25 degree/A\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.35 Page no 509"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Ig=0.015 #A\n",
+ "G=5\n",
+ "I=1\n",
+ "V=15\n",
+ "\n",
+ "#Calculation\n",
+ "S=(Ig*G)/(I-Ig)\n",
+ "R=G*S/(G+S)\n",
+ "R1=(V/Ig)-G\n",
+ "R2=R1+G\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Resistance of ammeter of range 0-1 A is\", R,\"ohm\"\n",
+ "print\"(ii) Resistance of ammeter of range 0-15 A is\", R2,\"ohm\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Resistance of ammeter of range 0-1 A is 0.075 ohm\n",
+ "(ii) Resistance of ammeter of range 0-15 A is 1000.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.36 Page no 510"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=75 #mV\n",
+ "Ig=0.025 #A\n",
+ "I=25 #mA\n",
+ "I1=100\n",
+ "V1=750\n",
+ "\n",
+ "#Calculation\n",
+ "G=V/I\n",
+ "S=(Ig*G)/(I1-Ig)\n",
+ "R=(V1/Ig)-G\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Resistance for an ammeter of range 0-100 A is\", round(S,5),\"ohm\"\n",
+ "print\"(ii) Resistance for an ammeter of range 0-750 A is\", round(R,5),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Resistance for an ammeter of range 0-100 A is 0.00075 ohm\n",
+ "(ii) Resistance for an ammeter of range 0-750 A is 29997.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 58
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.37 Page no 510"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "Rg=60\n",
+ "R1=3.0\n",
+ "rs=0.02\n",
+ "\n",
+ "#Calculation\n",
+ "Rt=Rg+R1\n",
+ "I=R1/Rt\n",
+ "Rm=(Rg*rs)/(Rg+rs)\n",
+ "R2=Rm+R1\n",
+ "I1=R1/R2\n",
+ "I2=R1/R1\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The value of current is\", round(I,3),\"A\"\n",
+ "print\"(ii) The value of current is\", round(I1,2),\"A\"\n",
+ "print\"(iii) The value of current is\",I2,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The value of current is 0.048 A\n",
+ "(ii) The value of current is 0.99 A\n",
+ "(iii) The value of current is 1.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.38 Page no 511"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=100\n",
+ "v=1\n",
+ "a=1980\n",
+ "\n",
+ "#Calculation\n",
+ "Rm=a/(V-v)\n",
+ "\n",
+ "#Result\n",
+ "print\"Resistance of the voltmeter is\", Rm,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance of the voltmeter is 20 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 77
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.39 Page no 511"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=1200.0 #ohm\n",
+ "R2=600 #ohm\n",
+ "Vab=5 #V\n",
+ "V=35\n",
+ "\n",
+ "#Calculation\n",
+ "Rp=(R1*R2)/(R1+R2)\n",
+ "I=Vab/Rp\n",
+ "pd=V-Vab\n",
+ "R=pd/I\n",
+ "\n",
+ "#Result\n",
+ "print\"value of unknown resistance is\", R,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of unknown resistance is 2400.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 84
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.40 Page no 511"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R1=400 #ohm\n",
+ "R2=800.0\n",
+ "R3=10\n",
+ "V=6\n",
+ "R11=10000.0\n",
+ "R22=400\n",
+ "\n",
+ "#Calculation\n",
+ "Rt=R1+R2+R3\n",
+ "I=V/Rt\n",
+ "Rp=(R11*R22)/(R11+R22)\n",
+ "R=Rp+800\n",
+ "I1=V/R\n",
+ "Vab=I1*Rp\n",
+ "\n",
+ "#Result\n",
+ "print\"Hence the voltmeter will read\", round(Vab,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Hence the voltmeter will read 1.95 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 95
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.41 Page no 512"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V=2 #V\n",
+ "R=2000.0 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "I=V/R\n",
+ "pd=I*R\n",
+ "\n",
+ "#Result\n",
+ "print\"Reading of ammeter is\", I*10**3,\"mA \\nReading of voltmeter is\",pd,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Reading of ammeter is 1.0 mA \n",
+ "Reading of voltmeter is 2.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 101
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.42 Page no 512"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=3\n",
+ "G=100\n",
+ "R=200.0\n",
+ "n=30\n",
+ "\n",
+ "#Calculation\n",
+ "Ig=E/(G+R)\n",
+ "K=(Ig/n)*10**6\n",
+ "\n",
+ "#Result\n",
+ "print\"Figure of merit of the galvanometer is\", round(K,1),\"micro A/division\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Figure of merit of the galvanometer is 333.3 micro A/division\n"
+ ]
+ }
+ ],
+ "prompt_number": 108
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.43 Page no 513"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "V1=60 #ohm\n",
+ "V2=30\n",
+ "R=300.0\n",
+ "R1=1200\n",
+ "R2=400 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "V=V1-V2\n",
+ "I=V/R\n",
+ "R11=(R1*R)/(R1+R)\n",
+ "I=V1/(R11+R2)\n",
+ "V11=I*R11\n",
+ "\n",
+ "#Result\n",
+ "print\"Voltmeter will read\", V11,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltmeter will read 22.5 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 115
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.44 Page no 513"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "R=20.0 #K ohm\n",
+ "R2=1 #K ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Vr=(R*R2)/(R+R2)\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Voltmeter resistance is\", R,\"K ohm\"\n",
+ "print\"(ii) Voltmeter resistance is\",R2,\"K ohm\"\n",
+ "print\"(iii) Voltmeter resistance is\",round(Vr,2),\"K ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Voltmeter resistance is 20.0 K ohm\n",
+ "(ii) Voltmeter resistance is 1 K ohm\n",
+ "(iii) Voltmeter resistance is 0.95 K ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 123
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.45 Page no 514"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "s=20*10**-6\n",
+ "n=30\n",
+ "I=1 #A\n",
+ "G=25 #ohm\n",
+ "\n",
+ "#Calculation\n",
+ "Ig=s*n\n",
+ "S=Ig*G/(1-Ig)\n",
+ "Ra=G*S/(G+S)\n",
+ "\n",
+ "#Result\n",
+ "print\"Resistance of ammeter is\",Ra,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance of ammeter is 0.015 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 128
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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