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"worksheets": [
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Chapter 29: parallel resonance and Q-factor
"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 1, page no. 521
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine (a) the resonant frequency, (b) the dynamic resistance, \n",
"#(c) the current at resonance, and (d) the circuit Q-factor at resonance.\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"R = 10;# in ohms\n",
"L = 0.005;# IN Henry\n",
"C = 0.25e-6;# IN fARADS\n",
"V = 50;#in volts\n",
"\n",
"#calculation:\n",
" #Resonant frequency, for parallel\n",
"fr = ((1/(L*C) - ((R**2)/(L**2)))**0.5)/(2*math.pi)\n",
" #dynamic resistance\n",
"Rd = L/(C*R)\n",
" #Current at resonance\n",
"Ir = V/Rd\n",
"wr = 2*math.pi*fr\n",
" #Q-factor at resonance, Q = wr*L/R\n",
"Qr = wr*L/R\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)Resonance frequency is \",round(fr,2),\" Hz\\n\"\n",
"print \"\\n (b)dynamic resistance \",round(Rd,2),\" ohm\\n\"\n",
"print \"\\n (c)Current at resonance, Ir is \",round(Ir,2),\" A\\n\"\n",
"print \"\\n (d)Q-factor at resonance is \",round(Qr,2),\"\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)Resonance frequency is 4490.31 Hz\n",
"\n",
"\n",
" (b)dynamic resistance 2000.0 ohm\n",
"\n",
"\n",
" (c)Current at resonance, Ir is 0.02 A\n",
"\n",
"\n",
" (d)Q-factor at resonance is 14.11 \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 2, page no. 521
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the resonant frequency for the network\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"RL1 = 0;# in ohms\n",
"RL2 = 30;# in ohms\n",
"L = 0.100;# IN Henry\n",
"C = 40e-6;# IN fARADS\n",
"V = 50;#in volts\n",
"\n",
"#calculation:\n",
" #for RL1\n",
" #Resonant frequency,\n",
"wr1 = (1/(L*C))**0.5\n",
"fr1 = wr1/(2*math.pi)\n",
" #for RL2\n",
" #Resonant frequency,\n",
"wr2 = (1/(L*C) - ((RL2**2)/(L**2)))**0.5\n",
"fr2 = wr2/(2*math.pi)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)Resonance frequency at RL = 0 is \",round(fr1,2),\" Hz\"\n",
"print \"\\n (b)Resonance frequency at RL = 30 ohm is \",round(fr2,2),\" Hz\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)Resonance frequency at RL = 0 is 79.58 Hz\n",
"\n",
" (b)Resonance frequency at RL = 30 ohm is 63.66 Hz\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 3, page no. 523
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine for the condition when the supply current is a minimum, \n",
"#(a) the capacitance of the capacitor, (b) the dynamic resistance, \n",
"#(c) the supply current, (d) the Q-factor, (e) the bandwidth,\n",
"#(f) the upper and lower \u00033 dB frequencies, and (g) the value of the circuit impedance at the \u00033 dB frequencies\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"R = 150;# in ohms\n",
"L = 0.120;# IN Henry\n",
"V = 20;#in volts\n",
"fr = 4000;# in Hz\n",
"\n",
"#calculation:\n",
" #capacitance, C\n",
"C = 1/(L*((2*math.pi*fr)**2 + ((R**2)/(L**2))))\n",
"Rd = L/(C*R)\n",
" #Current at resonance\n",
"Ir = V/Rd\n",
"wr = 2*math.pi*fr\n",
" #Q-factor at resonance, Q = wr*L/R\n",
"Qr = wr*L/R\n",
" #bandwidth,.(f2 \u2212 f1)\n",
"bw = fr/Qr\n",
" #upper half-power frequency, f2\n",
"f2 = (bw + ((bw**2) + 4*(fr**2))**0.5)/2\n",
" #lower half-power frequency, f1\n",
"f1 = f2 - bw\n",
" #impedance at the \u22123 dB frequencies\n",
"Z = Rd/(2**0.5)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)the capacitance of the capacitor,C is \",round(C*1E6,2),\"uF\"\n",
"print \"\\n (b)dynamic resistance \",round(Rd,2),\"ohm\\n\"\n",
"print \"\\n (c)Current at resonance, Ir is \",round(Ir*1000,2),\"mA\\n\"\n",
"print \"\\n (d)Q-factor at resonance is \",round(Qr,2),\"\\n\"\n",
"print \"\\n (e)bandwidth is \",round(bw,2),\" Hz\\n\"\n",
"print \"\\n (f)the upper half-power frequency, f2 is \",round(f2,2),\" Hz and \"\n",
"print \" the lower half-power frequency, f1 is \",round(f1,2),\" Hz\\n\"\n",
"print \"\\n (g)impedance at the -3 dB frequencies is \",round(Z,2),\" ohm\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)the capacitance of the capacitor,C is 0.01 uF\n",
"\n",
" (b)dynamic resistance 60788.85 ohm\n",
"\n",
"\n",
" (c)Current at resonance, Ir is 0.33 mA\n",
"\n",
"\n",
" (d)Q-factor at resonance is 20.11 \n",
"\n",
"\n",
" (e)bandwidth is 198.94 Hz\n",
"\n",
"\n",
" (f)the upper half-power frequency, f2 is 4100.71 Hz and \n",
" the lower half-power frequency, f1 is 3901.76 Hz\n",
"\n",
"\n",
" (g)impedance at the -3 dB frequencies is 42984.21 ohm\n",
"\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 4, page no. 525
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the resonant frequency of the network.\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"RL = 5;# in ohms\n",
"L = 0.002;# IN Henry\n",
"C = 25e-6;# IN fARADS\n",
"Rc = 3;# in ohms\n",
"\n",
"#calculation:\n",
" #Resonant frequency, for parallel\n",
"fr = (1/(2*math.pi*((L*C)**0.5)))*((RL**2 - (L/C))/(Rc**2 - (L/C)))**0.5\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n resonant frequency, fr is \",round(fr,2),\" Hz\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" resonant frequency, fr is 626.45 Hz"
]
}
],
"prompt_number": 4
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 5, page no. 525
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine for the parallel network the values of inductance L\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"RL = 3;# in ohms\n",
"fr = 1000;# in Hz\n",
"Xc = 10;# IN ohms\n",
"Rc = 4;# in ohms\n",
"\n",
"#calculation:\n",
"XL1 = (((Rc**2 + Xc**2)/Xc) + ((((Rc**2 + Xc**2)/Xc)**2) - 4*(RL**2))**0.5)/2\n",
"XL2 = (((Rc**2 + Xc**2)/Xc) - ((((Rc**2 + Xc**2)/Xc)**2) - 4*(RL**2))**0.5)/2\n",
"wr = 2*math.pi*fr\n",
" #inductance\n",
"L1 = XL1/wr\n",
"L2 = XL2/wr\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n inductance is either \",round(L1*1000,2),\"mH or \",round(L2*1000,2),\"mH\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" inductance is either 1.71 mH or 0.13 mH"
]
}
],
"prompt_number": 6
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 6, page no. 526
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the overall Q-factor of the parallel combination.\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"QL = 60;# Q-factor\n",
"Qc = 300;# Q-factor\n",
"\n",
"#calculation:\n",
"QT = QL*Qc/(QL + Qc)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n the overall Q-factor is \",round(QT,2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" the overall Q-factor is 50.0"
]
}
],
"prompt_number": 7
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 7, page no. 527
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine for the circuit (a) the Q-factor, (b) the dynamic resistance, and\n",
"#(c) the magnitude of the impedance when the supply frequency is 0.4% greater than the tuned frequency.\n",
"from __future__ import division\n",
"import math\n",
"import cmath\n",
"#initializing the variables:\n",
"C = 10.61E-9;# in Farad\n",
"bw = 500;# in Hz\n",
"fr = 150000;# in Hz\n",
"x = 0.004\n",
"\n",
"#calculation:\n",
" #Q-factor\n",
"Q = fr/bw\n",
"wr = 2*math.pi*fr\n",
" #dynamic resistance, RD\n",
"Rd = Q/(C*wr)\n",
"de = x\n",
"Z = Rd/(1 + (2*de*Q*1j))\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)Q-factor \",round(Q,2),\"\"\n",
"print \"\\n (b)dynamic resistance \",round(Rd,2),\"ohm\"\n",
"print \"\\n (c)magnitude of the impedance \",round(abs(Z),2),\"ohm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)Q-factor 300.0 \n",
"\n",
" (b)dynamic resistance 30000.93 ohm\n",
"\n",
" (c)magnitude of the impedance 11538.82 ohm"
]
}
],
"prompt_number": 8
}
],
"metadata": {}
}
]
}