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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"<h1>Chapter 45: Transients and Laplace\n",
"transforms</h1>"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"<h3>Example 1, page no. 903</h3>"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate (a) the initial value of current flowing, \n",
"#(b) the value of current 150 ms after connection, \n",
"#(c) the value of capacitor voltage 80 ms after connection, and \n",
"#(d) the time after connection when the resistor voltage is 35 V.\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"C = 500E-9;# in Farad\n",
"R = 100000;# in Ohm\n",
"V = 50;# in VOlts\n",
"ti = 0.15;# in sec\n",
"tc = 0.08;# in sec\n",
"Vrt = 35;# in Volts\n",
"\n",
" #calculation:\n",
" #Initial current, \n",
"i0 = (V/R)\n",
" #when time t = 150ms current is\n",
"i150 = (V/R)*math.e**(-1*ti/(R*C))\n",
" #capacitor voltage, Vc\n",
"Vc = V*(1 - math.e**(-1*tc/(R*C)))\n",
" #time, t\n",
"tvr = -1*R*C*math.log(Vrt/V)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n initial value of current flowing is \",round(i0*1E3,2),\"mA\"\n",
"print \"\\n current flowing at t = 150ms is \",round(i150*1E6,2),\"uA\"\n",
"print \"\\n value of capacitor voltage at t = 80ms is \",round(Vc,2),\" V\"\n",
"print \"\\n the time after connection when the resistor voltage is 35 V is \",round(tvr*1E3,2),\"msec\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" initial value of current flowing is 0.5 mA\n",
"\n",
" current flowing at t = 150ms is 24.89 uA\n",
"\n",
" value of capacitor voltage at t = 80ms is 39.91 V\n",
"\n",
" the time after connection when the resistor voltage is 35 V is 17.83 msec\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"<h3>Example 2, page no. 905</h3>"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate the p.d. across the capacitor after 20 s\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"C = 5E-6;# in Farad\n",
"R = 2000000;# in Ohm\n",
"V = 200;# in VOlts\n",
"tc = 20;# in sec\n",
"\n",
" #calculation:\n",
" #capacitor voltage, Vc\n",
"Vc = V*(math.e**(-1*tc/(R*C)))\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n value of capacitor voltage at t = 20s is \",round(Vc,2),\" V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" value of capacitor voltage at t = 20s is 27.07 V"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"<h3>Example 3, page no. 907</h3>"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine (a) the final value of current, (b) the value of current after 4 ms, \n",
"#(c) the value of the voltage across the resistor after 6 ms,\n",
"#(d) the value of the voltage across the inductance after 6 ms, and \n",
"#(e) the time when the current reaches 15 A.\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"L = 0.05;# in Henry\n",
"R = 5;# in Ohm\n",
"V = 110;# in VOlts\n",
"ti = 0.004;# in sec\n",
"tvr = 0.006;# in sec\n",
"tvl = 0.006;# in sec\n",
"it = 15;# in amperes\n",
"\n",
" #calculation:\n",
" #steady state current i\n",
"i = V/R\n",
" #when time t = 4ms current is\n",
"i4 = (V/R)*(1 - math.e**(-1*ti*R/L))\n",
" #resistor voltage, VR\n",
"VR6 = V*(1 - math.e**(-1*tvr*R/L))\n",
" #inductor voltage, VL\n",
"VL6 = V*(math.e**(-1*tvl*R/L))\n",
" #time, t\n",
"ti = (-1*L/R)*math.log(1 - it*R/V)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n steady state current i is \",round(i,2),\" A\"\n",
"print \"\\n when time t = 4ms current is is \",round(i4,2),\" A\"\n",
"print \"\\n value of resistor voltage at t = 6ms is \",round(VR6,2),\" V\"\n",
"print \"\\n value of inductor voltage at t = 6ms is \",round(VL6,2),\" V\"\n",
"print \"\\n the time after connection when the current is 15 V is \",round(ti,5),\" sec\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" steady state current i is 22.0 A\n",
"\n",
" when time t = 4ms current is is 7.25 A\n",
"\n",
" value of resistor voltage at t = 6ms is 49.63 V\n",
"\n",
" value of inductor voltage at t = 6ms is 60.37 V\n",
"\n",
" the time after connection when the current is 15 V is 0.01145 sec"
]
}
],
"prompt_number": 3
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"<h3>Example 4, page no. 909</h3>"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine\n",
"#(a) the time for the current in the 2 H inductor to fall to 200 mA,\n",
"#and (b) the maximum voltage appearing across the resistor.\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"i = 5;# in Amperes\n",
"L = 2# in Henry\n",
"i1 = 0.2;# in Amperes\n",
"R = 10;# in Ohm\n",
"\n",
" #calculation:\n",
" #time t\n",
"ti = (-1*L/R)*math.log(i1/i)\n",
" #voltage across the resistor is a maximum \n",
"VRm = i*R\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n time t for the current in the 2 H inductor to fall to 200 mA is \",round(ti,3),\" sec\"\n",
"print \"\\n max voltage across the resistor is \",VRm,\" V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" time t for the current in the 2 H inductor to fall to 200 mA is 0.644 sec\n",
"\n",
" max voltage across the resistor is 50 V"
]
}
],
"prompt_number": 4
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"<h3>Example 5, page no. 911</h3>"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#(a) Determine whether the circuit is over, critical or underdamped. (b) If C D 5 nF, determine the state of damping.\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"L = 0.002# in Henry\n",
"R = 1000;# in Ohm\n",
"C1 = 5E-6;# in farad\n",
"C2 = 5E-9;# in farad\n",
"\n",
" #calculation:\n",
"a = (R/(2*L))**2\n",
"b = 1/(L*C1)\n",
"if (a>b):\n",
"\ts1 = \"overdamped\";\n",
"elif (a<b):\n",
"\ts1 = \"underdamped\";\n",
"else:\n",
"\ts1 = \"critically damped\";\n",
"c = 1/(L*C2)\n",
"if (a>c):\n",
"\ts2 = \"overdamped\";\n",
"elif (a<c):\n",
"\ts2 = \"underdamped\";\n",
"else:\n",
"\ts2 = \"critically damped\";\n",
"\t\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n circuit is \",s1\n",
"print \"\\n if C = 5 nF, circuit is \",s2"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" circuit is overdamped\n",
"\n",
" if C = 5 nF, circuit is underdamped"
]
}
],
"prompt_number": 5
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"<h3>Example 6, page no. 912</h3>"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#what value of capacitance will give critical damping ?\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"L = 0.002# in Henry\n",
"R = 1000;# in Ohm\n",
"\n",
" #calculation:\n",
"a = (R/(2*L))**2\n",
" #for critically damped\n",
"C = 4*L/R**2\n",
"\t\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n capacitance C is \",C*1E9,\"nF\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" capacitance C is 8.0 nF"
]
}
],
"prompt_number": 6
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"<h3>Example 7, page no. 913</h3>"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the nature of the response and obtain an expression for the current in the coil.\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"L = 1.5# in Henry\n",
"R = 90;# in Ohm\n",
"C = 5*1E-6; # in Farad\n",
"V = 10; # in Volts\n",
"\n",
"#calculation:\n",
"a = -1*R/(2*L)\n",
"b = (1/(L*C) - (R/(2*L))**2)**0.5\n",
"V0 = V\n",
"I0 = 0\n",
"A = V0\n",
"B = (I0 - C*a*V0)/(C*b)\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"Current, i = e^(\",a,\"t) (\",round((a*C*B - A*C*b),4),\"sin(\",round(b,1),\"t) + (\",round((-1*a*C*A + B*C*C*b),0),\"cos(\",round(b,1),\"t) Amps.\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"Current, i = e^( -30.0 t) ( -0.0183 sin( 363.9 t) + ( 0.0 cos( 363.9 t) Amps.\n"
]
}
],
"prompt_number": 2
}
],
"metadata": {}
}
]
}
|
