{
"metadata": {
"name": "",
"signature": "sha256:6afa9a01ae55a53807ec01423714e6c6f113931ad8642d6e60f895f0e33b2225"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 3: Electric fields and Capacitors"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.1, Page 79"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"Q = 50e-03; # Electric charge, C\n",
"A = 600e-06; # Area of plate, m^2\n",
"\n",
"#Calculations\n",
"# Solving for electric field density, D\n",
"D = Q/A; # Electric field density, C/m^2\n",
"\n",
"#Result\n",
"print \"The density of the electric field existing between the plates = %4.1f C/m-square\"%D\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The density of the electric field existing between the plates = 83.3 C/m-square\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.2, Page 80"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"A = 400e-06; # Cross-sectional area of plate, m^2\n",
"I = 50e-06; # Source current, A\n",
"t = 3; # Flow time of current, s\n",
"\n",
"#Calculations\n",
"# Since electric current is the rate of flow of charge i.e I = Q/t, solving for Q \n",
"Q = I*t; # Amount of charge on plates, C\n",
"#Solving for density of the electric field between the plates\n",
"D = Q/A; # Electric field density, C/m^2\n",
"\n",
"#Results\n",
"print \"The charge on the plates = %3d micro-coloumb\"%(Q/1e-06);\n",
"print \"The density of the electric field between the plates = %5.3f C/m-square\"%D\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The charge on the plates = 150 micro-coloumb\n",
"The density of the electric field between the plates = 0.375 C/m-square\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.3, Page 83"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"d = 3e-03; # Thickness of dielectric, m\n",
"Q = 35e-03; # Electric charge on plates, C\n",
"V = 150; # Supply voltage, V\n",
"A = 144e-06; # Cross-sectional area of plates, m^2\n",
"\n",
"#Calculations\n",
"# Part (a)\n",
"# Since electric field strength(E) = potential gradient therefore we have\n",
"E = V/d; # Electric field strength, V/m\n",
"# Part (b)\n",
"# Solving for electric field density, D\n",
"D = Q/A; # Electric field density, C/m^2\n",
"\n",
"#Results\n",
"print \"The electric field strength = %2d kV/m\"%(E*1e-03);\n",
"print \"The flux density = %5.1f C/m^2\"%D\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The electric field strength = 50 kV/m\n",
"The flux density = 243.1 C/m^2\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.4, Page 83"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"d = 4e-03; # Thickness of air, m\n",
"Q = 2e-04; # Electric charge on plates, C\n",
"V = 125; # Supply voltage, V\n",
"D = 15; # Electric field density, coulomb-per-metre-square\n",
"\n",
"#Calculations\n",
"# Part (a)\n",
"# Since electric field strength(E) = potential gradient, therefore we have \n",
"E = V/d; # Electric field strength, V/m\n",
"# Part (b)\n",
"# Since D = Q/A, solving for A\n",
"A = Q/D; # Cross-sectional area of plates, m^2\n",
"# Part (c)\n",
"# Since Q = C*V, solving for C\n",
"C = Q/V; # Capacitance of the plates, F\n",
"\n",
"#Results\n",
"print \"The electric field strength between the plates = %5.2f kV/m\"%(E*1e-03);\n",
"print \"The csa of the field between the plates = %4.1f mm^2\"%(A/1e-06);\n",
"print \"The capacitance of the plates = %3.1f micro-coulomb\"%(C/1e-06);\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The electric field strength between the plates = 31.25 kV/m\n",
"The csa of the field between the plates = 13.3 mm^2\n",
"The capacitance of the plates = 1.6 micro-coulomb\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.5, Page 86"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"A = 6e-04; # Cross-sectional area of plates, m^2\n",
"d = 5e-04; # Thickness of mica sheet, m\n",
"Epsilon_r = 5.8; # Relative permittivity, unitless\n",
"Epsilon_0 = 8.854e-12; # Permittivity of Free Space\n",
"V = 200; # Potential difference, V\n",
"\n",
"#Calculations\n",
"# Part (a)\n",
"# Since absolute permittivity, Epsilon = C*(d/A), therefore solving for d & putting Epsilon = Epsilon_0*Epsilon_r \n",
"C = ( Epsilon_r*Epsilon_0*A )/d; # Capacitance, F\n",
"# Part (b)\n",
"# Since electric field strength(E) = potential gradient, therefore we have \n",
"E = V/d; # Electric field strength, V/m\n",
"\n",
"#Results\n",
"print \"The capacitance of the capacitor = %5.2f pF\"%(C/1e-12);\n",
"print \"Electric field strength = %3d kV/m\"%(E*1e-03);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The capacitance of the capacitor = 61.62 pF\n",
"Electric field strength = 400 kV/m\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.6, Page 86"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"C = 0.224e-09; #Capacitance, F\n",
"A = 5625e-06; # Cross-sectional area of plates, m^2\n",
"Epsilon_r = 2.5; # Relative permittivity\n",
"Epsilon_0 = 8.854e-12; # Permittivity of Free Space\n",
"\n",
"#Calculations\n",
"# Since absolute permittivity, Epsilon = C*(d/A), therefore solving for d & putting Epsilon = Epsilon_0*Epsilon_r\n",
"d = ( Epsilon_r*Epsilon_0*A )/C; # Thickness of waxed paper dielectric, m\n",
"\n",
"#Result\n",
"print \"The thickness of paper required = %3.2f mm\"%(d/1e-03);\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The thickness of paper required = 0.56 mm\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.7, Page 86"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"C = 4.7e-08; #Capacitance, F\n",
"A = 4e-04; # Cross-sectional area of plates, m^2\n",
"d = 1e-04; # Thickness of dielectric, m\n",
"Epsilon_0 = 8.854e-12; # Permittivity of Free Space\n",
"\n",
"#Calculations\n",
"# Since absolute permittivity, Epsilon = C*(d/A), therefore solving for Epsilon_r & putting Epsilon = Epsilon_0*Epsilon_r\n",
"Epsilon_r = (C*d)/(Epsilon_0*A); # Relative permittivity\n",
"\n",
"#Result\n",
"print \"Relative permittivity = %4d\"%Epsilon_r\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Relative permittivity = 1327\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.8, Page 87"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"V = 180; # Potential difference, V\n",
"d = 3e-03; # Thickness of dielectric, m\n",
"A = 4.2e-04; # Cross-sectional area of plates, m^2\n",
"Epsilon_r = 3.5; # Relative permittivity\n",
"Epsilon_0 = 8.854e-12; # Permittivity of Free Space\n",
"\n",
"#Calculations\n",
"# Since absolute permittivity, Epsilon = C*(d/A), therefore solving for C & putting Epsilon = Epsilon_0*Epsilon_r \n",
"C = ( Epsilon_r*Epsilon_0*A )/d; # Capacitance, F\n",
"# Since C = Q/V, solving for Q\n",
"Q = C*V; # Electric charge, C\n",
"# Using D = Q/A,\n",
"D = Q/A; # Electric field density, C/m^2\n",
"\n",
"#Results\n",
"print \"The flux thus produced = %3.2f nC.\"%(Q/1e-09)\n",
"print \"The flux density thus produced. = %3.2f micro-coulomb-per-metre-square\"%(D/1e-06); \n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The flux thus produced = 0.78 nC.\n",
"The flux density thus produced. = 1.86 micro-coulomb-per-metre-square\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.9, Page 89"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"C_1 = 4.7e-06; #Capacitance, F\n",
"C_2 = 3.9e-06; #Capacitance, F\n",
"C_3 = 2.2e-06; #Capacitance, F\n",
"\n",
"#Calculations\n",
"# The resulting capacitance of parallerly connected capacitors is the sum of the individual capacitance present \n",
"#in the circuit\n",
"C = C_1 + C_2 + C_3; # Resulting capacitance of the circuit, F\n",
"\n",
"#Result\n",
"print \"The resulting capacitance of the combination = %4.1f micro-farad\"%(C/1e-06);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The resulting capacitance of the combination = 10.8 micro-farad\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.10, Page 90"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"C_1 = 6e-06; #Capacitance, F\n",
"C_2 = 4e-06; #Capacitance, F\n",
"V = 150; # Supply voltage, V\n",
"\n",
"#Calculations\n",
"# Part (a)\n",
"# The reciprocal of the resulting capacitance of capacitors connected in series is the sum of the reciprocal \n",
"#of the individual capacitances present in the circuit i.e 1/C = 1/C1 + 1/C2, solving for C\n",
"C = ( C_1*C_2 )/(C_1 + C_2); # Resulting capacitance, F\n",
"# Part (b)\n",
"Q = V*C; # Electric charge on the capacitors, C\n",
"# Part (c)\n",
"V_1 = Q/C_1; # P.d across capacitor C_1, V\n",
"V_2 = Q/C_2; # P.d across capacitor C_2, V\n",
"\n",
"#Results\n",
"print \"The total capacitance of the combination = %3.1f micro-farad\"%(C/1e-06);\n",
"print \"The charge on each capacitor = %3d micro-coulomb\"%(Q/1e-06);\n",
"print \"The p.d. developed across %1d micro-farad capacitor = %2d V\"%(C_1/1e-06, V_1);\n",
"print \"The p.d. developed across %1d micro-farad capacitor = %2d V\"%(C_2/1e-06, V_2);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The total capacitance of the combination = 2.4 micro-farad\n",
"The charge on each capacitor = 360 micro-coulomb\n",
"The p.d. developed across 6 micro-farad capacitor = 60 V\n",
"The p.d. developed across 4 micro-farad capacitor = 90 V\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.11, Page 91"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"C_1 = 3e-06; #Capacitance, F\n",
"C_3 = 12e-06; #Capacitance, F\n",
"C_2 = 6e-06; #Capacitance, F\n",
"V = 400; # Supply voltage, V\n",
"\n",
"#Calculations\n",
"# The reciprocal of the resulting capacitance of capacitors connected in series is the sum of the reciprocal of the \n",
"#individual capacitances present in the circuit i.e 1/C = 1/C1 + 1/C2 + 1/C_3, solving for C\n",
"C = (C_1 * C_2 * C_3)/( C_1*C_2 + C_2*C_3 + C_3*C_1); # Resulting capacitance, F\n",
"Q = V*C; # Electric charge on the capacitors, C\n",
"# Part (c)\n",
"V_1 = Q/C_1; # P.d across capacitor C_1, V\n",
"V_2 = Q/C_2; # P.d across capacitor C_2, V\n",
"V_3 =Q/C_3; # P.d across capacitor C_2, V\n",
"\n",
"#Results\n",
"print \"P.d across capacitor %1d micro-farad = %5.1f V\"%(C_1/1e-06, V_1);\n",
"print \"P.d across capacitor %1d micro-farad = %5.1f V\"%(C_2/1e-06, V_2);\n",
"print \"P.d across capacitor %2d micro-farad = %4.1f V\"%(C_3/1e-06, V_3);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"P.d across capacitor 3 micro-farad = 228.6 V\n",
"P.d across capacitor 6 micro-farad = 114.3 V\n",
"P.d across capacitor 12 micro-farad = 57.1 V\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.12, Page 92"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"V = 200; # Supply voltage, voltage\n",
"C_AB = 2.; # Capacitance across branch AB, micro-farad\n",
"C_BC = 3.; # Capacitance across branch BC, micro-farad\n",
"C_CD = 6.; # Capacitance across branch CD, micro-farad\n",
"C_EF = 8.; # Capacitance across branch EF, micro-farad\n",
"C_BD = 4.; # Capacitance across branch EF, micro-farad\n",
"\n",
"#Calculations\n",
"# Part (a)\n",
"# Since 3-micro-farad & 6-micro-farad capacitors are in series & the reciprocal of the resulting capacitance of \n",
"#capacitors connected in series is the sum of the reciprocal of the individual capacitances present in the circuit, \n",
"#therefore i.e 1/C = 1/C1 + 1/C2 \n",
"C_BCD = ( C_BC*C_CD )/(C_BC+C_CD); # Resulting capacitance across branch BCD, micro-farad\n",
"#Since C_BCD & 4-micro-farad capacitors are in parallel & the resulting capacitance of parallerly connected capacitors \n",
"#is the sum of the individual capacitance present in the circuit\n",
"C_BD = C_BCD + C_BD; # Resulting capacitance across branch BD, micro-farad\n",
"# Since 2-micro-farad & C_BD capacitors are in series & the reciprocal of the resulting capacitance of capacitors \n",
"#connected in series is the sum of the reciprocal of the individual capacitances present in the circuit, therefore, \n",
"#we have\n",
"C_AD = (C_BD*C_AB)/(C_BD+C_AB); # Resulting capacitance across branch AD, micro-farad\n",
"#Since C_AD & C_EF capacitors are in parallel & the resulting capacitance of parallerly connected capacitors is the \n",
"#sum of the individual capacitance present in the circuit\n",
"C = C_AD + C_EF; # Resulting capacitance of the circuit, micro-farad\n",
"Q = V*C; # Electric charge drawn from the supply, C\n",
"\n",
"# Part (b)\n",
"Q_EF = V*C_EF; # The charge on the 8 micro-farad capacitor, micro-coulomb\n",
"\n",
"# Part (c)\n",
"Q_AD = Q - Q_EF; # The charge on the 4 micro-farad capacitor, C\n",
"Q_BD = Q_AD; # Charge in series combination of capacitors, micro-farad\n",
"# Since Q = C*V, solving for V\n",
"V_BD = Q_BD/C_BD; # The p.d. across the 4F capacitor,V\n",
"\n",
"# Part(d)\n",
"Q_BCD = V_BD*C_BCD; # Electric charge across branch BCD, C\n",
"Q_BC = Q_BCD; # Electric charge, C\n",
"V_BC = Q_BC/C_BC; # The p.d. across the 3 micro-farad capacitor\n",
"\n",
"#Results\n",
"print \"The charge drawn from the supply = %3.1f mC\"%(Q/1e+03);\n",
"print \"The charge on the %1d micro-farad capacitor = %3.1f mC\"%(C_EF, Q_EF/1e+03);\n",
"print \"The p.d. across the %1d micro-farad capacitor= %2d V\"%(C_BD, V_BD);\n",
"print \"The p.d. across the %1d micro-farad capacitor = %5.2f V\"%(Q_BC, V_BC);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The charge drawn from the supply = 1.9 mC\n",
"The charge on the 8 micro-farad capacitor = 1.6 mC\n",
"The p.d. across the 6 micro-farad capacitor= 50 V\n",
"The p.d. across the 100 micro-farad capacitor = 33.33 V\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.13, Page 96"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"N = 20; # Number of plates in a capacitor\n",
"A = 6400e-06; # Cross - sectional area of plate, m^2\n",
"d = 1.5e-03; # Distance between plates, m\n",
"epsilon_r = 6.4; # Relative permittivity for mica\n",
"epsilon_o = 8.854e-12; # Relative permittivity for free space\n",
"\n",
"#Calculations\n",
"# Calculating the capacitance of the capacitor\n",
"C = ((epsilon_o)*(epsilon_r)*A*(N-1))/d; # Capacitance, F\n",
"\n",
"#Result\n",
"print \" The capacitance of the capacitor = %3.1f nF\"%(C/1e-09);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The capacitance of the capacitor = 4.6 nF\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.14, Page 96"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"N = 9; # Number of plates in a capacitor\n",
"A = 1200e-06; # Cross - sectional area of plate, m^2\n",
"C = 3e-10; # Capacitance, F\n",
"epsilon_r = 5; # Relative permittivity for mica\n",
"epsilon_o = 8.854e-12; # Relative permittivity for free space\n",
"\n",
"#Calculations\n",
"# Using the formula of capacitance, C = ((epsilon_o)*(epsilon_r)*A*(N-1))/d and solving for d, we have\n",
"d = ((epsilon_o)*(epsilon_r)*A*(N-1))/C; # Distance between plates, m\n",
"\n",
"#Result\n",
"print \"The thickness of mica between parallel plates of a capacitor = %4.2f mm\"%(d/1e-03);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The thickness of mica between parallel plates of a capacitor = 1.42 mm\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.15, Page 97"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"N = 11; # Number of plates in a capacitor\n",
"r = 25e-03; # Radius of circular plate, m\n",
"A = (math.pi*r**2); # Cross - sectional area of plate, m^2\n",
"d = 5e-04; # Distance between plates, m\n",
"epsilon_r = 1; # Relative permittivity for air\n",
"epsilon_o = 8.854e-12; # Relative permittivity for free space\n",
"\n",
"#Calculations\n",
"# Calculating the capacitance of the capacitor\n",
"C = ((epsilon_o)*(epsilon_r)*A*(N-1))/d; # Capacitance, F\n",
"\n",
"#Result\n",
"print \" The capacitance of the capacitor = %3.2f pF\"%(C/1e-10);\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The capacitance of the capacitor = 3.48 pF\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.16, Page 99"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"C_1 = 3e-06; # Capacitance, F\n",
"C_2 = 6e-06; # Capacitance, F\n",
"V_1 = 250; # Voltage across capacitor C_1, V\n",
"\n",
"#Calculations\n",
"# Since each capacitor will take charge according to its capacitance, so we have\n",
"Q = C_1*V_1; # Charge on first capacitor C_1, C\n",
"W_1 = 0.5*C_1*(V_1**2); # Energy stored, J\n",
"# When the two capacitors are connected in parallel the 3 micro-farad will share its charge with 6 micro-farad capacitor. Thus the total charge in the system will remain unchanged, but the total capacitance will now be different\n",
"C = C_1 + C_2; # Total capacitance, F\n",
"# Since Q = C*V, solving for V\n",
"V = Q/C; # Voltage across capacitor C_2, V\n",
"W = 0.5*C*(V**2); # Total energy stored by the combination, J\n",
"\n",
"#Results\n",
"print \"The charge and energy stored by %1d micro-F capacitor are %3.2f mC and %5.2f mJ respectively \"%(C_1/1e-06, Q/1e-03 , W_1/1e-03);\n",
"print \"The p.d. between the plates = %5.2f V\"%V\n",
"print \"The energy stored by the combination of %1d micro-F and %1d micro-F capacitors = %5.2f mJ\"%(C_1/1e-06, C_2/1e-06, W/1e-03);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The charge and energy stored by 3 micro-F capacitor are 0.75 mC and 93.75 mJ respectively \n",
"The p.d. between the plates = 83.33 V\n",
"The energy stored by the combination of 3 micro-F and 6 micro-F capacitors = 31.25 mJ\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.17, Page 100"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"V = 200; # Supply voltage, V\n",
"C_1 = 10e-06; # Capacitance, farad\n",
"C_2 = 6.8e-06; # Capacitance, farad\n",
"C_3 = 4.7e-06; # Capacitance, farad\n",
"\n",
"#Calculations\n",
"# Part (a)\n",
"# Since each capacitor will take charge according to its capacitance, so we have\n",
"Q_1 = V*C_1; # Charge stored on capacitor C_1, C\n",
"W_1 = 0.5*C_1*(V**2); # Energy stored on capacitor C_1, J\n",
"# Part (b)\n",
"# Since C_2 and C_3 are in series and hence, their equivalent capacitance is given by their series combination\n",
"C_4 = (C_2 * C_3)/(C_2 + C_3); # Equivalent capacitance of C_2 and C_3, F\n",
"# Since C_1 and C_4 are in parallel and hence, their equivalent capacitance is given by their parallel combination\n",
"C = C_1 + C_4; # Total capacitance of circuit, F\n",
"# Since Q = C*V, solving for V\n",
"V_1 = Q_1/C; # New p.d across C_1, V\n",
"W = 0.5*C*(V_1**2); # Total energy remaining in the circuit, J\n",
"energy_used = W_1 - W; # Energy, J\n",
"\n",
"#Results\n",
"print \"The charge and energy stored by %2d micro-F capacitor are %1d mC and %2.1f J respectively \"%(C_1/1e-06, Q_1/1e-03, W_1);\n",
"print \"The new p.d across %2d micro-F capacitor = %5.1f V\"%(C_1/1e-06,V_1);\n",
"print \"The amount of energy used in charging %3.1f micro-F and %3.2f micro-F capacitors from %2d micro-F capacitor = %4.3f J\"%(C_2/1e-06, C_3/1e-06, C_1/1e-06, energy_used/1e-03);\n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The charge and energy stored by 10 micro-F capacitor are 2 mC and 0.2 J respectively \n",
"The new p.d across 10 micro-F capacitor = 156.5 V\n",
"The amount of energy used in charging 6.8 micro-F and 4.70 micro-F capacitors from 10 micro-F capacitor = 43.495 J\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.18, Page 101"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"V = 400; # Supply voltage, V\n",
"E = 0.5e06; # Dielectric strength, V/m\n",
"\n",
"#Calculations\n",
"# Since E = V/d, solving for d\n",
"d = V/E; # Thickness of dielectric, m\n",
"\n",
"#Result\n",
"print \"The minimum thickness of dielectric required = %3.1fmm\"%(d/1e-03);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The minimum thickness of dielectric required = 0.8mm\n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.19, Page 101"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"C = 270e-12; # Capacitance, F\n",
"A = 60e-04; # Cross-sectional area of plate, m^2\n",
"E = 350e03; # Dielectric strength, V/m\n",
"epsilon_r = 2.1; # Relative permittivity\n",
"epsilon_o = 8.854e-12; # Permittivity of free space\n",
"\n",
"#Calculations\n",
"# Part (a)\n",
"# Since formula for capacitance, C = ((epsilon_o)*(eplison_r)*A)/d, solving for d\n",
"d = ((epsilon_o)*(epsilon_r)*A)/C; # Thickness of dielectric, m\n",
"# Part (b)\n",
"# Since E = V/d, solving for V\n",
"V = E*d; # Maximum possible working voltage, V\n",
"\n",
"#Results\n",
"print \"The thickness of Teflon sheet required = %5.4f mm\"%(d/1e-03);\n",
"print \"The maximum possible working voltage for the capacitor = %5.1f V\"%V;\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The thickness of Teflon sheet required = 0.4132 mm\n",
"The maximum possible working voltage for the capacitor = 144.6 V\n"
]
}
],
"prompt_number": 19
}
],
"metadata": {}
}
]
}