{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Chapter 40: Field theory
"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 1, page no. 725
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the capacitance per metre length of the system\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"e0 = 8.85E-12; \n",
"er = 2.8;\n",
"l = 1;# in m\n",
"\n",
"#calculation: \n",
" #From Figure 40.9\n",
"m = 16;# number of parallel squares measured along each equipotential\n",
"n = 6;# the number of series squares measured along each line of force\n",
"C = e0*er*l*m/n\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n capacitance is \",round(C*1E12,2),\"pFarad.\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" capacitance is 66.08 pFarad."
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 2, page no. 725
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#determine the capacitance of a 100 m length of the cable.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"e0 = 8.85E-12; \n",
"er = 3.4;\n",
"l = 100;# in m\n",
"\n",
" #calculation: \n",
" #From Figure 40.10\n",
"m = 13;# number of parallel squares measured along each equipotential\n",
"n = 4;# the number of series squares measured along each line of force\n",
"C = e0*er*l*m/n\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n capacitance is \",round(C*1E9,2),\"nFarad.\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" capacitance is 9.78 nFarad."
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 3, page no. 726
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the capacitance per metre length of the cable\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"e0 = 8.85E-12; \n",
"er = 2.7;\n",
"ri = 0.0005;# in m\n",
"ro = 0.006;# in m\n",
"\n",
" #calculation: \n",
" #capacitance C\n",
"C = 2*math.pi*e0*er/(math.log(ro/ri))\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n capacitance is \",round(C*1E12,2),\"pFarad.\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" capacitance is 60.42 pFarad."
]
}
],
"prompt_number": 3
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 4, page no. 727
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the internal diameter of the sheath.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"C = 80E-12;# in Farads\n",
"e0 = 8.85E-12; \n",
"er = 3.5;\n",
"d0 = 0.008;# in m\n",
"\n",
" #calculation: \n",
" #internal diameter\n",
"di = d0*(math.e**(2*math.pi*e0*er/C))\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n internal diameter is \",round(di,2),\" m.\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" internal diameter is 0.09 m."
]
}
],
"prompt_number": 4
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 5, page no. 728
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine (a) the capacitance per kilometre length of the cable, \n",
"#(b) the dielectric stress at a radius of 30 mm, and\n",
"#(c) the maximum and minimum values of dielectric stress.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"e0 = 8.85E-12; \n",
"er = 3.5;\n",
"di = 0.08;# in m\n",
"d0 = 0.032;# in m\n",
"r = 0.03;# in m\n",
"V = 40000;# in Volts\n",
"\n",
"#calculation: \n",
" #capacitance C\n",
"C = 2*math.pi*e0*er/(math.log(di/d0))\n",
" #dielectric stress at radius r,\n",
"E = V/(r*math.log(di/d0))\n",
" #maximum dielectric stress,\n",
"Emax = V/((d0/2)*(math.log((di/d0))))\n",
" #minimum dielectric stress,\n",
"Emin = V/((di/2)*(math.log((di/d0))))\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n capacitance is \",round(C*1E12,2),\"pF/km\"\n",
"print \"\\n dielectric stress at radius r is \",round(E,2),\"V/m\"\n",
"print \"\\n maximum dielectric stress, is \",round(Emax,2),\"V/m minimum dielectric stress \",round( Emin,2),\"V/m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" capacitance is 212.4 pF/km\n",
"\n",
" dielectric stress at radius r is 1455142.22 V/m\n",
"\n",
" maximum dielectric stress, is 2728391.67 V/m minimum dielectric stress 1091356.67 V/m"
]
}
],
"prompt_number": 5
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 6, page no. 729
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate (a) the core and inner sheath radii for the most economical cable,\n",
"#(b) the capacitance per metre length, and (c) the charging current per kilometre run.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"e0 = 8.85E-12; \n",
"er = 3.5;\n",
"V = 60000;# in Volts\n",
"f = 50;# in Hz\n",
"Em = 10E6;# in V/m\n",
"\n",
"\n",
"#calculation: \n",
" #core radius, a\n",
"a = V/Em\n",
" #internal sheath radius,\n",
"b = a*math.e**1\n",
" #capacitance\n",
"C = 2*math.pi*e0*er/(math.log(b/a))\n",
" #Charging current\n",
"I = V*2*math.pi*f*C\n",
" #charging current per kilometre\n",
"Ipkm = I*1000\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n core radius is \",round(a*1000,2),\"mm and internal sheath radius \",round(b*1000,1),\"mm\"\n",
"print \"\\n capacitance is \",round(C*1E12,0),\"pF/m\"\n",
"print \"\\n the charging current per kilometre \",round(Ipkm,2),\" A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" core radius is 6.0 mm and internal sheath radius 16.3 mm\n",
"\n",
" capacitance is 195.0 pF/m\n",
"\n",
" the charging current per kilometre 3.67 A"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 7, page no. 730
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine for a 1 km length of the cable (a) the capacitance, (b) the charging current and (c) the power loss.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"e0 = 8.85E-12; \n",
"er = 2.5;\n",
"di = 0.08;# in m\n",
"d0 = 0.025;# in m\n",
"r = 1000;# in m\n",
"V = 132000;# in Volts\n",
"f = 50;# in Hz\n",
"de = 3.5E-3;# rad.\n",
"\n",
" #calculation:\n",
" #core radius, a\n",
"a = d0/2\n",
" #internal sheath radius,\n",
"b = di/2\n",
" #capacitance\n",
"C = 2*math.pi*e0*er*1E3/(math.log(b/a))\n",
" #Charging current\n",
"I = V*2*math.pi*f*C\n",
" #power loss\n",
"P = (2*math.pi*f*C*math.tan(de))*V**2\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)capacitance for a 1 km length is \",round(C*1E6,2),\"uF\"\n",
"print \"\\n (b)the charging current \",round(I,2),\"A/km\"\n",
"print \"\\n (c)power loss \",round(P,2),\" W\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)capacitance for a 1 km length is 0.12 uF\n",
"\n",
" (b)the charging current 4.96 A/km\n",
"\n",
" (c)power loss 2289.78 W"
]
}
],
"prompt_number": 7
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 8, page no. 732
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#determine the capacitance of the cable per metre length by the method of curvilinear squares\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"e0 = 8.85E-12; \n",
"er = 3.2;\n",
"di = 0.06;# in m\n",
"d0 = 0.020;# in m\n",
"\n",
" #calculation:\n",
" #core radius, a\n",
"a = d0/2\n",
" #internal sheath radius,\n",
"b = di/2\n",
" #capacitance\n",
"C = 2*math.pi*e0*er/(math.log(b/a))\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n capacitance per m of length is \",round(C*1E9,2),\"nF\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" capacitance per m of length is 0.16 nF"
]
}
],
"prompt_number": 8
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 9, page no. 736
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the capacitance of the line if the total length is 200 m.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"e0 = 8.85E-12; \n",
"er = 1;\n",
"D = 0.05;# in m\n",
"d = 0.005;# in m\n",
"l = 200;# in m\n",
"\n",
" #calculation:\n",
" #capacitance\n",
"C = math.pi*e0*er/(math.log(D/(d/2)))\n",
" #capacitance of a 200 m length\n",
"C200 = C*l\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n capacitance of a 200 m length is \",round(C200*1E6,5),\"uF\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" capacitance of a 200 m length is 0.00186 uF"
]
}
],
"prompt_number": 9
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 10, page no. 736
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine, for a 1 km length of line, (a) the capacitance of the conductors, \n",
"#(b) the value of charge carried by each conductor, and (c) the charging current\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"e0 = 8.85E-12; \n",
"er = 1;\n",
"D = 1.2;# in m\n",
"r = 0.004;# in m\n",
"f = 50;# in Hz\n",
"V = 15000;# in Volts\n",
"l = 1000;# in m\n",
"\n",
" #calculation:\n",
" #capacitance\n",
"C = math.pi*e0*er/(math.log(D/r))\n",
" #capacitance of a 1 km length\n",
"Cpkm = C*l\n",
" #Charge Q\n",
"Q = Cpkm*V\n",
" #Charging current\n",
"I = V*2*math.pi*f*Cpkm\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n capacitance per 1km length is \",round(Cpkm*1E9,2),\"nF\"\n",
"print \"\\n Charge Q is \",round(Q*1E6,2),\"uC\"\n",
"print \"\\n Charging current is \",round(I,2),\" A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" capacitance per 1km length is 4.87 nF\n",
"\n",
" Charge Q is 73.12 uC\n",
"\n",
" Charging current is 0.02 A"
]
}
],
"prompt_number": 10
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 11, page no. 737
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#determine (a) the maximum value required for the capacitance per metre length,\n",
"#and (b) the maximum diameter of each conductor if their distance between centres is 1.25 m.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"e0 = 8.85E-12; \n",
"er = 1;\n",
"I = 0.015;# in Amperes\n",
"d = 1.25;# in m\n",
"r = 800;# in m\n",
"f = 50;# in Hz\n",
"V = 10000;# in Volts\n",
"\n",
" #calculation:\n",
" #capacitance\n",
"C = I/(2*math.pi*f*V)\n",
" #required maximum value of capacitance\n",
"Cmax = C/r\n",
" #maximum diameter of each conductor\n",
"D = 2*d/(math.e**(math.pi*e0*er/Cmax))\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n required maximum value of capacitance is \",round(Cmax*1E12,2),\"pF/m\"\n",
"print \"\\nthe maximum diameter of each conductor is \",round(D,2),\" m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" required maximum value of capacitance is 5.97 pF/m\n",
"\n",
"the maximum diameter of each conductor is 0.02 m"
]
}
],
"prompt_number": 11
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 12, page no. 739
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the energy stored in a 10 nF capacitor when charged to 1 kV, and \n",
"#the average power developed if this energy is dissipated in 10 \u03bcs\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"e0 = 8.85E-12; \n",
"er = 1;\n",
"C = 10E-9;# in Farad\n",
"V = 1000;# in Volts\n",
"t = 10E-6;# in sec\n",
"\n",
" #calculation:\n",
" #energy stored,Wf\n",
"Wf = C*V*V/2\n",
" #average power developed\n",
"Pav = Wf/t\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n the energy stored is \",Wf,\"J\"\n",
"print \"\\nthe average power developed is \",Pav,\" W\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" the energy stored is 0.005 J\n",
"\n",
"the average power developed is 500.0 W"
]
}
],
"prompt_number": 12
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 13, page no. 739
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#determine (a) the voltage across the plates and (b) the capacitance of the capacitor.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"e0 = 8.85E-12; \n",
"er = 1;\n",
"Q = 5E-3;# in Coulomb\n",
"W = 0.625;# in Joules\n",
"\n",
" #calculation:\n",
" #voltage across the plates\n",
"V = 2*W/Q\n",
" #Capacitance C\n",
"C = Q/V\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n voltage across the plates is \",V,\" V\"\n",
"print \"\\n Capacitance C is \",C*1E6,\"uF\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" voltage across the plates is 250.0 V\n",
"\n",
" Capacitance C is 20.0 uF"
]
}
],
"prompt_number": 13
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 14, page no. 740
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#determine (a) the required thickness of the ceramic dielectric, \n",
"#(b) the area of plate required if the relative permittivity of the ceramic is 10, and \n",
"#(c) the maximum energy stored by the capacitor.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"e0 = 8.85E-12; \n",
"er = 10;\n",
"C = 0.01E-6;# in Farad\n",
"E = 10E6;# in V/m\n",
"V = 2500;# in Volts\n",
"\n",
" #calculation:\n",
" #thickness of ceramic dielectric,\n",
"d = V/E\n",
" #cross-sectional area of plate\n",
"A = C*d/(e0*er)\n",
" #Maximum energy stored,\n",
"W = C*V*V/2\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n thickness of ceramic dielectric is \",d*1000,\"mm\"\n",
"print \"\\n cross-sectional area of plate, is \",round(A,2),\"m2\"\n",
"print \"\\n Maximum energy stored is \",round(W,3),\" J\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" thickness of ceramic dielectric is 0.25 mm\n",
"\n",
" cross-sectional area of plate, is 0.03 m2\n",
"\n",
" Maximum energy stored is 0.031 J"
]
}
],
"prompt_number": 14
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 15, page no. 740
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate the energy stored per cubic metre of the dielectric.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"e0 = 8.85E-12; \n",
"er = 2.3;\n",
"A = 0.02;# in m2\n",
"C = 400E-12;# in Farad\n",
"V = 100;# in Volts\n",
"\n",
" #calculation:\n",
" #energy stored per unit volume of dielectric,\n",
"W = ((C*V)**2)/(2*e0*er*A**2)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n energy stored per unit volume of dielectric is \",round(W,2),\" J/m3\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" energy stored per unit volume of dielectric is 0.1 J/m3"
]
}
],
"prompt_number": 15
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 16, page no. 744
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the inductance of the cable per metre length.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"u0 = 4*math.pi*1E-7; \n",
"ur = 1;\n",
"a = 0.001;# in m\n",
"b = 0.004;# in m\n",
"\n",
" #calculation:\n",
" #inductance L\n",
"L = (u0*ur/(2*math.pi))*(0.25 + math.log(b/a))\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n inductance L is \",round(L*1E6,2),\"uH/m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" inductance L is 0.33 uH/m"
]
}
],
"prompt_number": 16
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 17, page no. 744
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#determine the diameter of the sheath.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"u0 = 4*math.pi*1E-7; \n",
"ur = 1;\n",
"da = 0.010;# in m\n",
"L = 4E-7;# in H/m\n",
"\n",
" #calculation:\n",
" #diameter of the sheath\n",
"db = da*(math.e**(L/(u0*ur/(2*math.pi))))\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n diameter of the sheath is \",round(db,2),\" m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" diameter of the sheath is 0.07 m"
]
}
],
"prompt_number": 17
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 18, page no. 745
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine for the cable (a) the inductance, assuming nonmagnetic materials, and \n",
"#(b) the capacitance, assuming a dielectric of relative permittivity 3.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"u0 = 4*math.pi*1E-7; \n",
"ur = 1;\n",
"e0 = 8.85E-12;\n",
"er = 3;\n",
"da = 0.010;# in m\n",
"db = 0.025;# in m\n",
"l = 7500;# in m\n",
"\n",
"#calculation:\n",
" #inductance per metre length\n",
"L = (u0*ur/(2*math.pi))*(0.25 + math.log(db/da))\n",
" #Since the cable is 7500 m long,\n",
"L7500 = L*7500\n",
" #capacitance C\n",
"C = 2*math.pi*e0*er/(math.log(db/da))\n",
" #//Since the cable is 7500 m long,\n",
"C7500 = C*7500\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\ninductance is \",round(L7500*1000,2),\" mH\"\n",
"print \"\\ncapCItance is \",round(C7500*1E6,2),\"uF\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"inductance is 1.75 mH\n",
"\n",
"capCItance is 1.37 uF"
]
}
],
"prompt_number": 18
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 19, page no. 748
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the inductance of the line per metre length ignoring internal linkages\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"u0 = 4*math.pi*1E-7; \n",
"ur = 1;\n",
"e0 = 8.85E-12;\n",
"er = 3;\n",
"D = 1.2;# in m\n",
"a = 0.008;# in m\n",
"\n",
" #calculation:\n",
" #inductance per metre length\n",
"L = (u0*ur/(math.pi))*(math.log(D/a))\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\ninductance is \",round(L*1E6,2),\"uH/m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"inductance is 2.0 uH/m"
]
}
],
"prompt_number": 19
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 20, page no. 748
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine (a) the loop inductance, and \n",
"#(b) the capacitance of a 1 km length of single-phase twin line having conductors of diameter 10 mm and spaced 800 mm apart in air.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"u0 = 4*math.pi*1E-7; \n",
"ur = 1;\n",
"e0 = 8.85E-12;\n",
"er = 1;\n",
"l = 1000;# in m\n",
"D = 0.8;# in m\n",
"a = 0.01/2;# in m\n",
"\n",
" #calculation:\n",
" #inductance per metre length\n",
"L = (u0*ur/(math.pi))*(0.25 + math.log(D/a))\n",
" #Since the cable is 1000 m long,\n",
"L1k = L*l\n",
" #capacitance C\n",
"C = math.pi*e0*er/(math.log(D/a))\n",
" #//Since the cable is 1000 m long,\n",
"C1k = C*l\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\ninductance is \",round(L1k*1000,2),\" mH\"\n",
"print \"\\ncapcitance is \",round(C1k*1E9,2),\"nF\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"inductance is 2.13 mH\n",
"\n",
"capcitance is 5.48 nF\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 21, page no. 749
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the distance between their centres.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"L = 2.185E-6;# in H/m\n",
"u0 = 4*math.pi*1E-7; \n",
"ur = 1;\n",
"a = 0.012/2;# in m\n",
"\n",
" #calculation:\n",
" #distance D\n",
"D = a*math.e**((L*math.pi)/(u0*ur) - 0.25)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\ndistance D is \",round(D,2),\" m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"distance D is 1.1 m"
]
}
],
"prompt_number": 21
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 22, page no. 752
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#What value of current would double the energy stored?\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"L = 0.2;# in H\n",
"I = 0.05;# in Amperes\n",
"u0 = 4*math.pi*1E-7; \n",
"ur = 1;\n",
"\n",
"#calculation:\n",
" #energy stored in inductor\n",
"W = L*I*I/2\n",
" #current I\n",
"I = (2*2*W/L)**0.5\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\nenergy stored in inductor is \",round(W*1000,2),\"mJ\"\n",
"print \"\\ncurrent I is \",round(I,2),\"A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"energy stored in inductor is 0.25 mJ\n",
"\n",
"current I is 0.07 A"
]
}
],
"prompt_number": 22
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 23, page no. 752
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#determine the total energy stored in the magnetic field of the airgap.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"B = 0.05;# in Tesla\n",
"A = 500E-6;# in m2\n",
"l = 0.002;# in m\n",
"u0 = 4*math.pi*1E-7; \n",
"\n",
"#calculation:\n",
" #energy stored\n",
"W = (B**2)/(2*u0)\n",
" #Volume of airgap\n",
"v = A*l\n",
" #energy stored in airgap\n",
"W = W*v\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\nenergy stored in the airgap is \",round(W*1E6,2),\"uJ\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"energy stored in the airgap is 994.72 uJ"
]
}
],
"prompt_number": 23
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 24, page no. 752
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the strength of a uniform electric fi\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"B = 0.8;# in Tesla\n",
"A = 500E-6;# in m2\n",
"l = 0.002;# in m\n",
"u0 = 4*math.pi*1E-7; \n",
"ur = 1;\n",
"e0 = 8.85E-12;\n",
"er = 1;\n",
"\n",
"#calculation:\n",
" #energy stored in mag. field\n",
"W = (B**2)/(2*u0)\n",
" #electric field\n",
"E = (2*W/(e0*er))**0.5\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\nelectric field strength is \",round(E/1E6,2),\"MV/m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
"electric field strength is 239.89 MV/m"
]
}
],
"prompt_number": 24
}
],
"metadata": {}
}
]
}