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"source": [
"# Chapter17 : Capacitive Reactance"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example No. 17_1 Page No. 530"
]
},
{
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"execution_count": 2,
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"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"
]
},
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"cell_type": "code",
"execution_count": 3,
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"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'"
]
}
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