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| author | hardythe1 | 2014-07-25 13:33:31 +0530 |
|---|---|---|
| committer | hardythe1 | 2014-07-25 13:33:31 +0530 |
| commit | 1c1ea29e3e213559fef5f928df109b7d17c21f24 (patch) | |
| tree | a1d38eb039e19d9e84b6e769cec04d59e7179e66 /Electrical_Circuit_Theory_And_Technology/chapter_35.ipynb | |
| parent | efb9ead5d9758d5d0bed7a22069320b14f972e40 (diff) | |
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removing unwanted and adding book
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diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_35.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_35.ipynb new file mode 100644 index 00000000..d4ffa9bf --- /dev/null +++ b/Electrical_Circuit_Theory_And_Technology/chapter_35.ipynb @@ -0,0 +1,627 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:a02e331c8801853e98b04884a3697abba7050e1171bfe09d4dc7ba891e3a3f21"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1>Chapter 35: Maximum power transfer theorems and impedance matching</h1>"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 1, page no. 620</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "import cmath\n",
+ "#initializing the variables:\n",
+ "rv = 120;# in volts\n",
+ "thetav = 0;# in degrees\n",
+ "Z = 15 + 1j*20;# in ohm\n",
+ "\n",
+ " #calculation: \n",
+ " #voltage\n",
+ "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
+ " #maximum power transfer occurs when R = mod(Z)\n",
+ "R = abs(Z)\n",
+ " #the total circuit impedance\n",
+ "ZT = Z + R\n",
+ " #Current I flowing in the load is given by\n",
+ "I = V/ZT\n",
+ "Imag = abs(I)\n",
+ " #maximum power delivered\n",
+ "P = R*I**2\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n (a)maximum power transfer occurs when R is \",R,\" ohm\"\n",
+ "print \"\\n (b) maximum power delivered is \",abs(P),\" W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " (a)maximum power transfer occurs when R is 25.0 ohm\n",
+ "\n",
+ " (b) maximum power delivered is 180.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 2, page no. 620</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "import cmath\n",
+ "#initializing the variables:\n",
+ "rv = 120;# in volts\n",
+ "thetav = 0;# in degrees\n",
+ "Z = 15 + 1j*20;# in ohm\n",
+ "\n",
+ " #calculation: \n",
+ " #voltage\n",
+ "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
+ " #maximum power transfer occurs when X = -1*imag(Z) and R = real(Z)\n",
+ "z = Z.real - 1j*Z.imag\n",
+ " #Total circuit impedance at maximum power transfer condition,\n",
+ "ZT = Z + z\n",
+ " #Current I flowing in the load is given by\n",
+ "I = V/ZT\n",
+ "Imag = abs(I)\n",
+ " #maximum power delivered\n",
+ "P = Z.real*I**2\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n (a)maximum power transfer occurs when Z is \",z.real,\" + (\", z.imag,\")i ohm\"\n",
+ "print \"\\n (b) maximum power delivered is \",abs(P),\" W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " (a)maximum power transfer occurs when Z is 15.0 + ( -20.0 )i ohm\n",
+ "\n",
+ " (b) maximum power delivered is 240.0 W"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 3, page no. 620</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "import cmath\n",
+ "#initializing the variables:\n",
+ "rv = 200;# in volts\n",
+ "thetav = 0;# in degrees\n",
+ "R1 = 100;# in ohm\n",
+ "C = 1E-6;# in farad\n",
+ "f = 1000;# in Hz\n",
+ "\n",
+ " #calculation: \n",
+ " #voltage\n",
+ "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
+ " #Capacitive reactance, Xc\n",
+ "Xc = 1/(2*math.pi*f*C)\n",
+ " #Hence source impedance,\n",
+ "z = R1*(1j*Xc)/(R1 + 1j*Xc)\n",
+ " #maximum power transfer is achieved when R = mod(z)\n",
+ "R = abs(z)\n",
+ " #Total circuit impedance at maximum power transfer condition,\n",
+ "ZT = z + R\n",
+ " #Current I flowing in the load is given by\n",
+ "I = V/ZT\n",
+ "Imag = abs(I)\n",
+ " #maximum power transferred,\n",
+ "P = R*Imag**2\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n (a)maximum power transfer occurs when R is \",round(R,2),\" ohm\"\n",
+ "print \"\\n (b) maximum power delivered is \",round(P,2),\" W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " (a)maximum power transfer occurs when R is 84.67 ohm\n",
+ "\n",
+ " (b) maximum power delivered is 127.9 W"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 4, page no. 621</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "import cmath\n",
+ "#initializing the variables:\n",
+ "rv = 60;# in volts\n",
+ "thetav = 0;# in degrees\n",
+ "R1 = 4;# in ohm\n",
+ "XL = 10;# in ohm\n",
+ "Xc = 7;# in ohm\n",
+ "R2 = XL*1j;# in ohm\n",
+ "R3 = -1j*Xc;# in ohm\n",
+ "\n",
+ " #calculation: \n",
+ " #voltage\n",
+ "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
+ " #maximum power transfer is achieved when\n",
+ "R = (R1**2 + (XL - Xc)**2)**0.5\n",
+ " #Hence source impedance,\n",
+ "ZT = R1 + R2 + R3 + R\n",
+ " #Current I flowing in the load is given by\n",
+ "I = V/ZT\n",
+ "Imag = abs(I)\n",
+ " #maximum power transferred,\n",
+ "P = R*Imag**2\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n (a)maximum power transfer occurs when R is \",R,\" ohm\"\n",
+ "print \"\\n (b) maximum power delivered is \",P,\" W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " (a)maximum power transfer occurs when R is 5.0 ohm\n",
+ "\n",
+ " (b) maximum power delivered is 200.0 W"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 5, page no. 622</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "import cmath\n",
+ "#initializing the variables:\n",
+ "V = 20;# in volts\n",
+ "R1 = 5;# in ohm\n",
+ "R2 = 15;# in ohm\n",
+ "\n",
+ "#calculation: \n",
+ " #R is removed from the network as shown in Figure 35.6\n",
+ " #P.d. across AB, E\n",
+ "E = (R2/(R1 + R2))*V\n",
+ " #Impedance \u2018looking-in\u2019 at terminals AB with the source removed is given by\n",
+ "r = R1*R2/(R1 + R2)\n",
+ " #The equivalent Th\u00b4evenin circuit supplying terminals AB is shown in Figure 35.7. \n",
+ " #From condition (2), for maximum power transfer\n",
+ "R = r\n",
+ " #Current I flowing in the load is given by\n",
+ "I = E/(R + r)\n",
+ " #maximum power transferred,\n",
+ "P = R*I**2\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n (a)maximum power transfer occurs when R is \",R,\" ohm\"\n",
+ "print \"\\n (b) maximum power delivered is \",P,\" W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " (a)maximum power transfer occurs when R is 3.75 ohm\n",
+ "\n",
+ " (b) maximum power delivered is 15.0 W"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 6, page no. 622</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "import cmath\n",
+ "#initializing the variables:\n",
+ "rv = 100;# in volts\n",
+ "thetav = 30;# in degrees\n",
+ "R1 = 5;# in ohm\n",
+ "R2 = 5;# in ohm\n",
+ "R3 = 10j;# in ohm\n",
+ "\n",
+ "#calculation: \n",
+ " #voltage\n",
+ "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180)\n",
+ " #Resistance R and reactance X are removed from the network as shown in Figure 35.9\n",
+ " #P.d. across AB,\n",
+ "E = ((R2 + R3)/(R1 + R2 + R3))*V\n",
+ " #With the source removed the impedance, z, \u2018looking in\u2019 at terminals AB is given by:\n",
+ "z = (R2 + R3)*R1/(R1 + R2 + R3)\n",
+ " #The equivalent Th\u00b4evenin circuit is shown in Figure 35.10. From condition 3, \n",
+ " #maximum power transfer is achieved when X = -1*imag(z) and R = real(z)\n",
+ "X = -1*z.imag\n",
+ "R = z.real\n",
+ "Z = R + 1j*X\n",
+ " #Current I flowing in the load is given by\n",
+ "I = E/(z + Z)\n",
+ "Imag = abs(I)\n",
+ " #maximum power transferred,\n",
+ "P = R*Imag**2\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n (a)maximum power transfer occurs when R is \",R,\" ohm and X is \", X,\" ohm\"\n",
+ "print \"\\n (b) maximum power delivered is \",round(P,2),\" W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " (a)maximum power transfer occurs when R is 3.75 ohm and X is -1.25 ohm\n",
+ "\n",
+ " (b) maximum power delivered is 416.67 W"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 7, page no. 624</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "import cmath\n",
+ "#initializing the variables:\n",
+ "Ro = 448;# in ohm\n",
+ "tr = 8;# turn ratio N1/N2\n",
+ "\n",
+ " #calculation: \n",
+ " #The equivalent input resistance r of the transformer must be Ro for maximum power transfer.\n",
+ "r = Ro\n",
+ "RL = r*(1/tr)**2\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n the optimum value of load resistance is \",RL,\" ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " the optimum value of load resistance is 7.0 ohm"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 8, page no. 624</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "import cmath\n",
+ "#initializing the variables:\n",
+ "Zo = 450 + 1j*60;# in ohm\n",
+ "ZL = 40 + 1j*19;# in ohm\n",
+ "\n",
+ " #calculation: \n",
+ " #transformer turns ratio tr = (N1/N2)\n",
+ "Zomag = abs(Zo)\n",
+ "ZLmag = abs(ZL)\n",
+ "tr = (Zomag/ZLmag)**0.5\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n the transformer turns ratio is \",round(tr,2),\"\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " the transformer turns ratio is 3.2 "
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 9, page no. 625</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "import cmath\n",
+ "#initializing the variables:\n",
+ "V1 = 240;# in volts\n",
+ "V2 = 1920;# in volts\n",
+ "R1 = 5;# in ohms\n",
+ "R2 = 1600;# in ohms\n",
+ "\n",
+ "#calculation: \n",
+ " #The network is shown in Figure 35.12.\n",
+ " #turn ratio N1/N2 = V1/V2\n",
+ "tr = V1/V2\n",
+ " #Equivalent input resistance of the transformer,\n",
+ "RL = R2\n",
+ "r = RL*tr**2\n",
+ " #Total input resistance,\n",
+ "Rin = R1 + r\n",
+ " #primary current, I1\n",
+ "I1 = V1/Rin\n",
+ " #For an ideal transformer V1/V2 = I2/I1\n",
+ "I2 = I1*(V1/V2)\n",
+ " #Power dissipated in the load resistance\n",
+ "P = RL*I2**2\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n (a) primary current flowing is \",I1,\" A\"\n",
+ "print \"\\n (b) Power dissipated in the load resistance is \",P,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " (a) primary current flowing is 8.0 A\n",
+ "\n",
+ " (b) Power dissipated in the load resistance is 1600.0 W"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 10, page no. 625</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "import cmath\n",
+ "#initializing the variables:\n",
+ "rv = 30;# in volts\n",
+ "thetav = 0;# in degrees\n",
+ "r = 20000;# in ohms\n",
+ "tr = 20;# turn ratio\n",
+ "\n",
+ "#calculation:\n",
+ " #voltage\n",
+ "V = rv*math.cos(thetav*math.pi/180) + 1j*rv*math.sin(thetav*math.pi/180) \n",
+ " #The network diagram is shown in Figure 35.13.\n",
+ " #For maximum power transfer, r1 must be equal to\n",
+ "r1 = r\n",
+ " #load resistance RL\n",
+ "RL = r1/tr**2\n",
+ " #The total input resistance when the source is connected to the matching transformer is\n",
+ "RT = r + r1\n",
+ " #Primary current\n",
+ "I1 = V/RT\n",
+ " #N1/N2 = I2/I1\n",
+ "I2 = I1*tr\n",
+ " #Power dissipated in load resistance RL is given by\n",
+ "P = RL*I2**2\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n (a)the value of the load resistance is \",RL,\" ohm\"\n",
+ "print \"\\n (b) Power dissipated in the load resistance is \",abs(P*1000),\"mW\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " (a)the value of the load resistance is 50.0 ohm\n",
+ "\n",
+ " (b) Power dissipated in the load resistance is 11.25 mW"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |