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| author | hardythe1 | 2014-07-25 12:35:04 +0530 |
|---|---|---|
| committer | hardythe1 | 2014-07-25 12:35:04 +0530 |
| commit | 6846da7f30aadc7b3812538d1b1c9ef2e465a922 (patch) | |
| tree | 0ac5e382d6515013c6a1f23d3529dc46e03d0015 /Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint_3.ipynb | |
| parent | dccd504f6bd2f5e97c54d1f0b0d2a99f83497ce5 (diff) | |
| download | Python-Textbook-Companions-6846da7f30aadc7b3812538d1b1c9ef2e465a922.tar.gz Python-Textbook-Companions-6846da7f30aadc7b3812538d1b1c9ef2e465a922.tar.bz2 Python-Textbook-Companions-6846da7f30aadc7b3812538d1b1c9ef2e465a922.zip | |
removing unwanted:
Diffstat (limited to 'Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint_3.ipynb')
| -rwxr-xr-x | Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint_3.ipynb | 303 |
1 files changed, 0 insertions, 303 deletions
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint_3.ipynb deleted file mode 100755 index 6732d70c..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint_3.ipynb +++ /dev/null @@ -1,303 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:cc9bf89f9ef283e2e254a84fd5c837c0ed5af6c8784b676db04bdba61dcf814e"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 30: Introduction to network analysis</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 536</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 100;# in volts\n",
- "rv2 = 50;# in volts\n",
- "thetav1 = 0;# in degrees\n",
- "thetav2 = 90;# in degrees\n",
- "R1 = 25;# in ohm\n",
- "R2 = 20;# in ohm\n",
- "R3 = 10;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #voltage\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- " #The branch currents and their directions are labelled as shown in Figure 30.4\n",
- " #Two loops are chosen. loop ABEF, and loop BCDE\n",
- " #using kirchoff rule in 3 loops\n",
- " #two eqns obtained\n",
- " #(R1 + R2)*I1 + R2*I2 = V1\n",
- " #R2*I1 + (R2 + R3)*I2 = V2\n",
- "I1 = (3*V1 - 2*V2)/(3*(R1 + R2) - 2*(R2))\n",
- "I2 = (V2 - R2*I1)/(R2 + R3)\n",
- "I = I1 + I2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current, I1 is \",round(I1.real,2),\" + (\",round( I1.imag,2),\")i A, \\n current, I2 is \",round(I2.real,2),\" + (\",round( I2.imag,2),\")i A and \"\n",
- "print \" total current, I is \",round(I.real,2),\" + (\",round( I.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current, I1 is 3.16 + ( -1.05 )i A, \n",
- " current, I2 is -2.11 + ( 2.37 )i A and \n",
- " total current, I is 1.05 + ( 1.32 )i A\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 537</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "#initializing the variables:\n",
- "V = 8;# in volts\n",
- "R1 = 1;# in ohm\n",
- "R2 = 2;# in ohm\n",
- "R3 = 3;# in ohm\n",
- "R4 = 4;# in ohm\n",
- "R5 = 5;# in ohm\n",
- "R6 = 6;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #Currents and their directions are assigned as shown in Figure 30.6.\n",
- " #Three loops are chosen since three unknown currents are required. The choice of loop directions is arbitrary.\n",
- " #loop ABCDE, and loop EDGF and loop DCHG\n",
- " #using kirchoff rule in 3 loops\n",
- " #three eqns obtained\n",
- " #R5*I1 + (R6 + R4)*I2 - R4*I3 = V\n",
- " #-1*R1*I1 + (R6 + R1)*I2 + R2*I3 = 0\n",
- " # R3*I1 - (R3 + R4)*I2 + (R2 + R3 + R4)*I3 = 0\n",
- "#using determinants\n",
- "d1 = [[V, (R6 + R4), -1*R4],[0, (R6 + R1), R2], [0, (-1*(R3 + R4)), (R2 + R3 + R4)]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[R5, V, -1*R4],[-1*R1, 0, R2],[ R3, 0, (R2 + R3 + R4)]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d3 = [[R5, (R6 + R4), V],[-1*R1, (R6 + R1), 0],[ R3, (-1*(R3 + R4)), 0]]\n",
- "D3 = numpy.linalg.det(d3)\n",
- "d = [[R5, (R6 + R4), -1*R4],[-1*R1, (R6 + R1), R2],[ R3, (-1*(R3 + R4)), (R2 + R3 + R4)]]\n",
- "D = numpy.linalg.det(d)\n",
- "I1 = D1/D\n",
- "I2 = D2/D\n",
- "I3 = D3/D \n",
- "#Current in the 2 ohm resistance\n",
- "I = I1 - I2 + I3\n",
- "#power dissipated in the 3 ohm resistance\n",
- "P3 = R3*I**2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)current through 2 ohm resistor is \",round(I2,3),\" A\"\n",
- "print \"\\n (b)power dissipated in the 3 ohm resistor is \",round(P3,3),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)current through 2 ohm resistor is 0.203 A\n",
- "\n",
- " (b)power dissipated in the 3 ohm resistor is 1.267 W"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 539</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "E1 = 5 + 0j;# in volts\n",
- "E2 = 2 + 4j;# in volts\n",
- "Z1 = 3 + 4j;# in ohm\n",
- "Z2 = 2 - 5j;# in ohm\n",
- "Z3 = 6 + 8j;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #Currents I1 and I2 with their directions are shown in Figure 30.8.\n",
- " #Two loops are chosen with their directions both clockwise.loop ABEF and loop BCDE,\n",
- " #using kirchoff rule in 3 loops\n",
- " #two eqns obtained\n",
- " #(Z1 + Z3)*I1 - Z3*I2 = E1\n",
- " #-1*Z3*I1 + (Z2 + Z3)*I2 = E2\n",
- "I1 = ((Z2 + Z3)*E1 + Z3*E2)/((Z2 + Z3)*(Z1 + Z3) - Z3*Z3)\n",
- "I2 = -1*(E1 - (Z1 + Z3)*I1)/Z3\n",
- "I3 = I1 - I2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"current, I1 is \",round(I1.real,2),\" + (\",round( I1.imag,2),\")i A,\\n current, I2 is \",round(I2.real,2),\" + (\",round( I2.imag,2),\")i A and \"\n",
- "print \" current, I3 is \",round(I3.real,2),\" + (\",round( I3.imag,2),\")i A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "current, I1 is 0.57 + ( 0.62 )i A,\n",
- " current, I2 is 0.56 + ( 1.33 )i A and \n",
- " current, I3 is 0.01 + ( -0.71 )i A\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 541</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "from __future__ import division\n",
- "import math\n",
- "import numpy\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "rv1 = 10;# in volts\n",
- "rv2 = 12;# in volts\n",
- "rv3 = 15;# in volts\n",
- "thetav1 = 0;# in degrees\n",
- "thetav2 = 0;# in degrees\n",
- "thetav3 = 0;# in degrees\n",
- "R1 = 4;# in ohm\n",
- "R2 = -5j;# in ohm\n",
- "R3 = 8;# in ohm\n",
- "R4 = 4;# in ohm\n",
- "R5 = 3j;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #voltages\n",
- "V1 = rv1*math.cos(thetav1*math.pi/180) + 1j*rv1*math.sin(thetav1*math.pi/180)\n",
- "V2 = rv2*math.cos(thetav2*math.pi/180) + 1j*rv2*math.sin(thetav2*math.pi/180)\n",
- "V3 = rv3*math.cos(thetav3*math.pi/180) + 1j*rv3*math.sin(thetav3*math.pi/180)\n",
- " #Currents I1, I2 and I3 with their directions are shown in Figure 30.10.\n",
- " #Three loops are chosen. The choice of loop directions is arbitrary. loop ABGH, and loopBCFG and loop CDEF\n",
- "Z4 = R4 + R5\n",
- " #using kirchoff rule in 3 loops\n",
- " #three eqns obtained\n",
- " #R1*I1 + R2*I2 = V1 + V2\n",
- " #-1*R3*I1 + (R3 + R2)*I2 + R3*I3 = V2 + V3\n",
- " # -1*R3*I1 + R3*I2 + (R3 + Z4)*I3 = V3\n",
- " #using determinants\n",
- "d1 = [[(V1 + V2), R2, 0],[(V2 + V3), (R3 + R2), R3],[V3, R3, (R3 + Z4)]]\n",
- "D1 = numpy.linalg.det(d1)\n",
- "d2 = [[R1, (V1 + V2), 0],[-1*R3, (V2 + V3), R3],[-1*R3, V3, (R3 + Z4)]]\n",
- "D2 = numpy.linalg.det(d2)\n",
- "d3 = [[R1, R2, (V1 + V2)],[-1*R3, (R3 + R2), (V2 + V3)],[-1*R3, R3, V3]]\n",
- "D3 = numpy.linalg.det(d3)\n",
- "d = [[R1, R2, 0],[-1*R3, (R3 + R2), R3],[-1*R3, R3, (R3 + Z4)]]\n",
- "D = numpy.linalg.det(d)\n",
- "I1 = D1/D\n",
- "I2 = D2/D\n",
- "I3 = D3/D \n",
- "I3mag = abs(I3)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n magnitude of the current through (4 + i3)ohm impedance is \",round(I3mag,2),\" A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " magnitude of the current through (4 + i3)ohm impedance is 1.84 A"
- ]
- }
- ],
- "prompt_number": 8
- }
- ],
- "metadata": {}
- }
- ]
-}
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