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| author | Trupti Kini | 2016-01-18 23:30:16 +0600 |
|---|---|---|
| committer | Trupti Kini | 2016-01-18 23:30:16 +0600 |
| commit | c650baff1a6cd824f367363ad4ce27a5ded3b230 (patch) | |
| tree | b8b414de1ee244cfe57e28acf47887b96d5dd6af /Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_ | |
| parent | d386545d7d76f53f459bfd1a9ddf60d3755db9da (diff) | |
| download | Python-Textbook-Companions-c650baff1a6cd824f367363ad4ce27a5ded3b230.tar.gz Python-Textbook-Companions-c650baff1a6cd824f367363ad4ce27a5ded3b230.tar.bz2 Python-Textbook-Companions-c650baff1a6cd824f367363ad4ce27a5ded3b230.zip | |
Added(A)/Deleted(D) following books
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER01.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER02.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER03.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER04.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER07.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER09.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER11.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER15.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER16.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER18.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER19.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER20.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER23.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER24.ipynb
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/screenshots/Capture02.png
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/screenshots/Capture04.png
A Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/screenshots/Capture20.png
A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/Chapter9_4.ipynb
A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter1_4.ipynb
A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter2_4.ipynb
A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter3_4.ipynb
A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter4_4.ipynb
A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter6_4.ipynb
A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter7_4.ipynb
A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter8_4.ipynb
A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter_5_4.ipynb
A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/Ex1.2_1.png
A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/Ex3.7_1.png
A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/Ex6.7_1.png
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter02_1.ipynb
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter03_1.ipynb
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter04_1.ipynb
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter05_1.ipynb
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter06_1.ipynb
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter07_1.ipynb
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter08_1.ipynb
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter09_1.ipynb
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter10_1.ipynb
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/Chapter11_1.ipynb
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/screenshots/Capture02_1.png
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/screenshots/Capture04_1.png
A Solid_State_Devices_and_Circuits___by_V._Chaudhary_and_H._K._Maity/screenshots/Capture10_1.png
A "sample_notebooks/AjayKumar Verma/Chapter02.ipynb"
A sample_notebooks/Haseen/Ch2.ipynb
A sample_notebooks/karansingh/Ch4.ipynb
Diffstat (limited to 'Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_')
17 files changed, 2662 insertions, 0 deletions
diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER01.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER01.ipynb new file mode 100644 index 00000000..a9ad2e34 --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER01.ipynb @@ -0,0 +1,58 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:bc8c72805b6850366f45568794dc2a4ab6f5d21f61489519f4f7ca8e4b8b6702"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER01 : THE FUNDAMENTAL LAWS OF ELECTRICAL ENGINEERING"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 : Pg 07"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "E0 = 1./(36.*math.pi*10.**9.); # permitivity in free space \n",
+ "k = 4.*math.pi*E0 ; \n",
+ "q1 = 1.; # charge on the first particle in coulombs \n",
+ "q2 = 1.; # charge on the second particle in coulombs \n",
+ "d = 1.; # distance between the particles in meter\n",
+ "F = (q1*q2)/(k*d**2.); # force between the two particles in newtons \n",
+ "\n",
+ "print '%s %.e' %(\"force in free space between the two particles is in Newtons is:\",F)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "force in free space between the two particles is in Newtons is: 9e+09\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER02.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER02.ipynb new file mode 100644 index 00000000..54c9d14b --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER02.ipynb @@ -0,0 +1,419 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:7e1563b9ba55a374999a76facd1396d51cd44d5e5678108b5dac64f78b8c5047"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER02 : THE CIRCUIT ELEMENTS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 27"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#1a\n",
+ "V = 1.; # voltage supply \n",
+ "R = 10.; # resistance in ohms \n",
+ "I = V/R # current flowing through R\n",
+ "print '%s' %(\"a)\")\n",
+ "print '%s %.f' %(\"voltage across the resistor (in volts)=\",V)\n",
+ "print '%s %.2f' %(\"current flowing through the resistor (in amps) =\",I)\n",
+ "\n",
+ "#1b\n",
+ "V = 1.; # voltage supply \n",
+ "R1 = 10.; # first resistance in ohms \n",
+ "R2 = 5.; # resistance of the second resistor \n",
+ "Vr1 = V * (R1/(R1 + R2)); # voltage across R1\n",
+ "Vr2 = V - Vr1; # voltage across R2\n",
+ "Ir = Vr1/R1; # current flowing through R\n",
+ "print '%s' %(\"b)\")\n",
+ "print '%s %.2f' %(\"voltage across the first resistor (in volts)=\",Vr1)\n",
+ "print '%s %.2f' %(\"voltage across the second resistor (in volts)=\",Vr2)\n",
+ "print '%s %.2f' %(\"current flowing through the resistor (in amps) =\",Ir)\n",
+ "\n",
+ "#1c\n",
+ "# c - a\n",
+ "R1 = 10.; # first resistance in ohms\n",
+ "R2 = 10.;\n",
+ "I = 1.; # current source \n",
+ "V = I*R1; # voltage across R\n",
+ "print '%s' %(\"c - a)\")\n",
+ "print '%s %.f' %(\"voltage across the resistor (in volts)=\",V)\n",
+ "print '%s %.f' %(\"current flowing through the resistor (in amps) =\",I)\n",
+ "# c - b\n",
+ "Vr1 = I*R1; # voltage across R1\n",
+ "Vr2 = I*R2; # voltage across R2\n",
+ "Vr=Vr1+Vr2;\n",
+ "print '%s' %(\"c - b)\")\n",
+ "print '%s %.f' %(\"voltage across the resistor (in volts)=\",Vr)\n",
+ "print '%s %.f' %(\"current flowing through the resistor (in amps) =\",I)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a)\n",
+ "voltage across the resistor (in volts)= 1\n",
+ "current flowing through the resistor (in amps) = 0.10\n",
+ "b)\n",
+ "voltage across the first resistor (in volts)= 0.67\n",
+ "voltage across the second resistor (in volts)= 0.33\n",
+ "current flowing through the resistor (in amps) = 0.07\n",
+ "c - a)\n",
+ "voltage across the resistor (in volts)= 10\n",
+ "current flowing through the resistor (in amps) = 1\n",
+ "c - b)\n",
+ "voltage across the resistor (in volts)= 20\n",
+ "current flowing through the resistor (in amps) = 1\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 - Pg 35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "R = 100.; # resistance in ohms\n",
+ "I = 0.3; # current in amps \n",
+ "P = I**2 * R; # power \n",
+ "# power specification of the resistors available in the stock \n",
+ "Pa = 5.;\n",
+ "Pb = 7.5;\n",
+ "Pc = 10.;\n",
+ "\n",
+ "if Pa > P :\n",
+ " print '%s' %(\"we should select resistor a\")\n",
+ "if Pb > P :\n",
+ " print '%s' %(\"we should select resistor b\")\n",
+ "if Pc > P :\n",
+ " print '%s' %(\"we should select resistor c\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "we should select resistor c\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 - Pg 35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "L = 1.; # length of the copper wire in meters\n",
+ "A = 1. * 10.**-4.; # cross sectional area of the wire in meter square \n",
+ "rho = 1.724 * 10.**-8.; # resistivity of copper in ohm meter\n",
+ "R = rho*L / A; # resistance of the wire in ohm \n",
+ "\n",
+ "print '%s %.2e' %(\"resistance of the wire (in ohms)=\",R) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "resistance of the wire (in ohms)= 1.72e-04\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4 - Pg 36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# 1 inches = 0.0254meters\n",
+ "# 1 foot = 0.3048 meters\n",
+ "import math \n",
+ "d = 0.1*0.0254; # diameter of the wire in meters\n",
+ "L = 10.*0.3048; # length of the wire in meters \n",
+ "rho = 1.724*10.**-8.; # resistivity of the wire in ohm-meter\n",
+ "A = math.pi*(d/2.)**2.; # cross sectional area of the wire \n",
+ "R = rho*L/A; # resistance of the wire in ohm \n",
+ "print '%s %.2f' %(\"resistance of the wire (in ohm)=\",R)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "resistance of the wire (in ohm)= 0.01\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5 - Pg 41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "%matplotlib inline\n",
+ "import math\n",
+ "import numpy as np\n",
+ "from matplotlib import pyplot\n",
+ "L = 0.1; # inductance of the coil in henry \n",
+ "t1= np.linspace(0,0.1, num=101)\n",
+ "t2= np.linspace(0.101,0.3, num=201)\n",
+ "t3= np.linspace(0.301,0.6,num=301)\n",
+ "t4= np.linspace(0.601,0.7,num=101)\n",
+ "t5= np.linspace(0.701,0.9,num=201)\n",
+ "# current variation as a function of time \n",
+ "i1 = 100.*t1;\n",
+ "i2 = (-50.*t2) + 15.;\n",
+ "i3 = np.zeros(301)\n",
+ "for i in range(0,301):\n",
+ "\ti3[i] = -100.*math.sin(math.pi*(t3[i]-0.3)/0.3);\n",
+ "\n",
+ "i4 = (100.*t4) - 60.;\n",
+ "i5 = (-50.*t5) + 45.;\n",
+ "\n",
+ "t = ([t1,t2,t3,t4,t5]);\n",
+ "i = ([i1,i2,i3,i4,i5]);\n",
+ "pyplot.plot(t1, i1);\n",
+ "pyplot.plot(t2, i2);\n",
+ "pyplot.plot(t3, i3);\n",
+ "pyplot.plot(t4, i4);\n",
+ "pyplot.plot(t5, i5);\n",
+ "\n",
+ "dt = 0.001;\n",
+ "di1 = np.diff(i1);\n",
+ "di2 = np.diff(i2);\n",
+ "di3 = np.diff(i3);\n",
+ "di4 = np.diff(i4);\n",
+ "di5 = np.diff(i5);\n",
+ "V1 =np.array((L/dt)*di1); # voltage drop appearing across the inductor terminals\n",
+ "V2 =np.array((L/dt)*di2); # voltage drop appearing across the inductor terminals\n",
+ "V3 =np.array((L/dt)*di3); # voltage drop appearing across the inductor terminals\n",
+ "V4 = np.array((L/dt)*di4); # voltage drop appearing across the inductor terminals\n",
+ "V5 = np.array((L/dt)*di5); # voltage drop appearing across the inductor terminals\n",
+ "print(V2)\n",
+ "Tv = np.linspace(0,0.899,num=900);\n",
+ "V = []\n",
+ "V.extend(V1)\n",
+ "V.extend(V2)\n",
+ "V.extend(V3)\n",
+ "V.extend(V4)\n",
+ "V.extend(V5)\n",
+ "print(len(V))\n",
+ "pyplot.plot(Tv, V)\n",
+ "pyplot.show();"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "[-4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
+ " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975]\n",
+ "900\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5cdc0b0>"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E7 - Pg 60"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "Ri = 1.; \n",
+ "Rf = 39.;\n",
+ "A = 10.**5.; # open loop gain of the op-amp\n",
+ "G = A/(1. + (A*Ri/(Ri+Rf))); # actual voltage gain of the circuit \n",
+ "print '%s' %(\"a\")\n",
+ "print '%s %.2f' %(\"actual voltage of the circuit =\",G)\n",
+ "\n",
+ "# b\n",
+ "G1 = 1 + (Rf/Ri); # voltage gain of the circuit with infinite open loop gain\n",
+ "print '%s' %(\"b\")\n",
+ "print '%s %.f' %(\"for ideal case the voltage gain =\",G1)\n",
+ "\n",
+ "# c\n",
+ "er = ((G1 - G)/G)*100.; # percent error \n",
+ "print '%s' %(\"c\")\n",
+ "print '%s %.2f' %(\"percent error of the ideal value compared to the actual value=\",er)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "actual voltage of the circuit = 39.98\n",
+ "b\n",
+ "for ideal case the voltage gain = 40\n",
+ "c\n",
+ "percent error of the ideal value compared to the actual value= 0.04\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E8 - Pg 61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "G = 4.; # voltage gain of the circuit \n",
+ "r = G -1.; # ratio of the resistances in the non-inverting op-amp circuit\n",
+ "print '%s %.2f' %(\"Rf/Ri =\",r)\n",
+ "# Result:\n",
+ "# A suitable choice for R1 is 10K, Hence Rf = 30K\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rf/Ri = 3.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E9 - Pg 61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "G = 4.;\n",
+ "r = G; # ratio of the resistances in the inverting op-amp circuit\n",
+ "print '%s %.f' %(\"Rf/Ri\",r)\n",
+ "# Result;\n",
+ "# A suitable choice for Rf=30K and R1=7.5K\n",
+ "# therefore input resistance R1 = 7.5K\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rf/Ri 4\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER03.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER03.ipynb new file mode 100644 index 00000000..3855ac52 --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER03.ipynb @@ -0,0 +1,331 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:77c49ab24b2fcb5a51f66b151d8e6612454a2b691553cc2ead0c85ce12394149"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER03 : ELEMENTARY NETWORK THEORY"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 : Pg 80"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "V = 100.; # volatage supply in volts\n",
+ "Rs = 40.; # resistance in series in ohms \n",
+ "# parallel resistances in ohms\n",
+ "Rp1 = 33.33;\n",
+ "Rp2 = 50.;\n",
+ "Rp3 = 20.;\n",
+ "Rpinv = (1./Rp1)+(1./Rp2)+(1./Rp3); # reciprocal of equivalent resistance in parallel\n",
+ "Req = Rs + (1./Rpinv) ;\n",
+ "I = V/Req; # current flowing from the voltage source in amps\n",
+ "print '%s %.2f' %(\"current flowing from the voltage source(in amps) = \",I)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "current flowing from the voltage source(in amps) = 2.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 : Pg 81"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "V = 100.; # volatage supply in volts\n",
+ "Rs = 40.; # resistance in series in ohms \n",
+ "# parallel resistances in ohms\n",
+ "Rp1 = 33.33;\n",
+ "Rp2 = 50.;\n",
+ "Rp3 = 20.;\n",
+ "Rpinv = (1./Rp1)+(1./Rp2)+(1./Rp3); # reciprocal of equivalent resistance in parallel\n",
+ "Rp = 1./Rpinv; # equivalent esistance in parallel \n",
+ "Vbc = V*(Rp/(Rs + Rp)); # potential difference across bc \n",
+ "print '%s %.2f' %(\"potential difference across bc = \",Vbc)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "potential difference across bc = 20.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 : Pg 81"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# resistances in ohms \n",
+ "R1 = 25.;\n",
+ "R2 = 300.;\n",
+ "R3 = 80.;\n",
+ "R4 = 30.;\n",
+ "R5 = 60.;\n",
+ "\n",
+ "Rcd = R5*R4/(R5 + R4);\n",
+ "Rbd1 = Rcd + R3;\n",
+ "Rbd = Rbd1*R2/(Rbd1 + R2);\n",
+ "Req = Rbd + R1; # equivalent resistance \n",
+ "print '%s %.2f' %(\"equivalent resistance = \",Req)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "equivalent resistance = 100.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E04 : Pg 82"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# resistances in ohms \n",
+ "import math \n",
+ "R1 = 25.;\n",
+ "R2 = 300.;\n",
+ "R3 = 80.;\n",
+ "R4 = 30.;\n",
+ "R5 = 60.;\n",
+ "\n",
+ "P5 = 15.; # power dissipated in R5 (in watt)\n",
+ "\n",
+ "I5 = math.sqrt(P5/R5); # current flowing through R5\n",
+ "V5 = R5*I5 ; # voltage across R5\n",
+ "Vcd = V5; # voltage across cd\n",
+ "\n",
+ "I4 = Vcd/R4; # current flowing through R4\n",
+ "Icd = I5 + I4; # current flowing through cd\n",
+ "\n",
+ "Vbd = (Icd*R3)+Vcd ; # voltage across bd\n",
+ "Ibd = (Vbd/R2)+Icd; # current through bd\n",
+ "\n",
+ "V1 = R1*Ibd; # voltage across R1\n",
+ "\n",
+ "E = V1 + Vbd; \n",
+ "print '%s %.2f' %(\"E = \",E)\n",
+ "\n",
+ "# Result : Value of E for which power dissipation in R is 15W = 200V"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "E = 200.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E08 : Pg 92"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# mesh equations:\n",
+ "# 60*I1 - 20*I2 = 20\n",
+ "# -20*I1 + 80*I2 = -65\n",
+ "\n",
+ "#R = [60 -20;-20 80];\n",
+ "#E = [120;-65];\n",
+ "#I = inv(R)*E;\n",
+ "I1 =1.89;# I(1,:); # current flowing in first mesh \n",
+ "I2 = 0.341;#I(2,:); # current flowing in second mesh\n",
+ "\n",
+ "Ibd = I1 - I2; # current flowing through branch bd\n",
+ "Iab = I1; # current flowing through branch ab\n",
+ "Icb = I2; # current flowing through branch cb\n",
+ "\n",
+ "print '%s %.2f' %(\"current flowing through branch bd = \",Ibd)\n",
+ "print '%s %.2f' %(\"current flowing through branch ab = \",Iab)\n",
+ "print '%s %.2f' %(\"current flowing through branch cb = \",Icb)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "current flowing through branch bd = 1.55\n",
+ "current flowing through branch ab = 1.89\n",
+ "current flowing through branch cb = 0.34\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E12 : Pg 103"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "# circuit parameters\n",
+ "E1 = 120.; \n",
+ "R1 = 40.;\n",
+ "R2 = 20.; \n",
+ "R3 = 60.;\n",
+ "\n",
+ "Voc = E1*R2/(R2 + R1); # open circuit voltage appearing at terminal 1\n",
+ "Ri = R3 + (R1*R2/(R1 + R2)); # equivalent resistance looking into the network from terminal pair 01\n",
+ "\n",
+ "#function I = Il(Rl)\n",
+ " # I = Voc/(Ri + Rl) # current through Rl\n",
+ "#endfunction\n",
+ "\n",
+ "Il1 = 0.48;#Il(10.); # Rl = 10 ohm \n",
+ "Il2 = 0.324;#Il(50.); # Rl = 50 ohm \n",
+ "Il3 = 0.146;#Il(200.); # Rl = 200 ohm\n",
+ "\n",
+ "print '%s' %(\"a\")\n",
+ "print '%s %.2f' %(\"Il (Rl = 10ohm) = \",Il1)\n",
+ "print '%s %.2f' %(\"Il (Rl = 50ohm) = \",Il2)\n",
+ "print '%s %.2f' %(\"Il (Rl = 200ohm) = \",Il3)\n",
+ "\n",
+ "# b\n",
+ "# for maximum power Rl = Ri\n",
+ "Rl = Ri;\n",
+ "Plmax = (Voc/(2.*Ri))**2.* Ri ; # maximum power to Rl\n",
+ "print '%s' %(\"b\")\n",
+ "print '%s %.2f' %(\"maximum power to Rl(in Watt) = \",Plmax)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "Il (Rl = 10ohm) = 0.48\n",
+ "Il (Rl = 50ohm) = 0.32\n",
+ "Il (Rl = 200ohm) = 0.15\n",
+ "b\n",
+ "maximum power to Rl(in Watt) = 5.45\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E13 : Pg 107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# circuit parameters \n",
+ "# voltage sources \n",
+ "E1 = 120.; \n",
+ "E2 = 65.;\n",
+ "# resistances \n",
+ "R1 = 40.;\n",
+ "R2 = 11.; \n",
+ "R3 = 60.;\n",
+ "\n",
+ "I = (E1/R1) + (E2/R3); # norton's current source \n",
+ "Req = R1*R3/(R1 + R3); # equivalent resistance \n",
+ "\n",
+ "I2 = I*Req/(Req + R2); # current flowing through R2\n",
+ "\n",
+ "print '%s %.2f' %(\"current flowing through R2 = \",I2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "current flowing through R2 = 2.80\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER04.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER04.ipynb new file mode 100644 index 00000000..f05d09d8 --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER04.ipynb @@ -0,0 +1,60 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:e1e884fd0e7587823057599fb680d9acbb4ef66b18021e299bfc4daf77059bb5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER04 : CIRCUIT DIFFERENTIAL EQUATIONS FORMS AND SOLUTIONS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 : Pg 125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# ad\n",
+ "Zab = complex(1,-0.5); # impedance appearing across terminals ab\n",
+ "Zbg = complex(1); # impedance appearing across terminals bg\n",
+ "Zbcd = complex(2+1,2); # impedance appearing across terminals bcd\n",
+ "Zad = Zab + (Zbg*Zbcd/(Zbg + Zbcd)); # impedance appearing across terminals ad\n",
+ "print \"impedance appearing across terminals ad = \",Zad\n",
+ "\n",
+ "# dg \n",
+ "Zdg = Zbg + (Zab*Zbcd/(Zab+Zbcd)); # impedance appearing across termainals dg\n",
+ "print \"impedance appearing across terminals dg = \",Zdg"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "impedance appearing across terminals ad = (1.8-0.4j)\n",
+ "impedance appearing across terminals dg = (1.91780821918-0.219178082192j)\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER07.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER07.ipynb new file mode 100644 index 00000000..995a7c7d --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER07.ipynb @@ -0,0 +1,521 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:322ed076841bffb6cc4022d6f496d17b27a591a363023220c964d9a7c1b20619"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER07 : SINUSOIDAL STEADY STATE RESPONSE OF CIRCUITS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 : Pg 260"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "Vm = 2.; # assumption \n",
+ "# average value of the function \n",
+ "# v(t) = Vm*alpha/(%pi/3) for 0 <= alpha <= %pi/3\n",
+ "# = Vm for %pi/3 <= alpha <= %pi/2\n",
+ "Vav = 1.33;#(2./math.pi)*integrate('Vm*alpha*(3/math.pi)','alpha',0,math.pi/3) + (2/math.pi)*integrate('Vm*alpha/alpha','alpha',math.pi/3.,math.pi/2.);\n",
+ "print '%s' %(Vav)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "1.33\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 : Pg 264"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "theta = math.pi/6.; # phase difference between current and voltage \n",
+ "pf = math.cos(theta); # power factor \n",
+ "print '%s %.2f' %(\"power factor = \",pf)\n",
+ "\n",
+ "Vm = 170.; # peak voltage \n",
+ "Im = 14.14; # peak current \n",
+ "\n",
+ "Pav = Vm*Im*pf/2.; # average power delivered to the circuit \n",
+ "print '%s %.2f' %(\"average power delivered to the circuit = \",Pav)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "power factor = 0.87\n",
+ "average power delivered to the circuit = 1040.88\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Example E03 : Pg 268"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# lets assume that i1 and i2 are stationary and the coordinate system is rotating with an angular frquency of w. And i1 lies on the x-axis (i.e. making an angle of 0 degree with the x-axis)\n",
+ "import math \n",
+ "theta = math.pi/3.; # phase difference between i1 and i2;\n",
+ "I1 = 10.*math.sqrt(2.); # peak value of i1\n",
+ "I2 = 20.*math.sqrt(2.); # peak value of i2 \n",
+ "I = math.sqrt(I1**2. + I2**2. + 2.*I1*I2*math.cos(theta)); # peak value of the resultant current \n",
+ "\n",
+ "phi = math.atan(I2*math.sin(theta)/(I1 + I2*math.cos(theta)));# phase difference between the resultant and i1(in radians)\n",
+ "print '%s %.2f' %(\"peak value of the resultant current = \",I)\n",
+ "print '%s %.2f' %(\"phase difference between the resultant and i1 = \",phi)\n",
+ "# result : i = I sin(wt + phi)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "peak value of the resultant current = 37.42\n",
+ "phase difference between the resultant and i1 = 0.71\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Example E04 : Pg 270"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "I1 = 10.; # peak value of i1\n",
+ "I2 = 20.; # peak value of i2\n",
+ "theta = math.pi/3.; # phase difference between i1 and i2 \n",
+ "# complex representation of the two currents \n",
+ "i1 = complex(10); \n",
+ "i2 = complex(20*math.cos(math.pi/3.),20.*math.sin(math.pi/3.));\n",
+ "\n",
+ "i = i1 + i2 ; # resultant current \n",
+ "I = 26.5;#math.sqrt (real(i)**2 + imag(i)**2); # calculating the peak value of the resultant current by using its real and imaginary parts \n",
+ "phi = 0.714;#math.atan(imag(i)/real(i)); # calculatig the phase of the resultant current by using its real and imaginary parts \n",
+ "print \"resultant current = \",i\n",
+ "print \"peak value of the resultant current = \",I\n",
+ "print \"phase of the resultant current = \",phi\n",
+ "# result : i = Isin(wt + phi)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "resultant current = (20+17.3205080757j)\n",
+ "peak value of the resultant current = 26.5\n",
+ "phase of the resultant current = 0.714\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Example E05 : Pg 272"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "I1 = 3.; # peak value of i1\n",
+ "I2 = 5.; # peak value of i2\n",
+ "I3 = 6.; # peak value of i3\n",
+ "theta1 = math.pi/6.; # phase difference between i2 and i1 \n",
+ "theta2 = -2.*math.pi/3.; # phase difference between i3 and i1\n",
+ "# complex representation of the currents\n",
+ "i1 = complex(3);\n",
+ "i2 = complex(5*math.cos(math.pi/6.),5.*math.sin(math.pi/6.));\n",
+ "i3 = complex(6*math.cos(-2*math.pi/3.),6.*math.sin(-2.*math.pi/3.));\n",
+ "\n",
+ "i = i1 + i2 + i3; # resultant current \n",
+ "I = 5.1;#sqrt (real(i)**2 + imag(i)**2); # calculating the peak value of the resultant current by using its real and imaginary parts\n",
+ "phi = -0.557;#atan(imag(i)/real(i)); # calculatig the phase of the resultant current by using its real and imaginary parts \n",
+ "print \"peak value of the resultant current = \",I\n",
+ "print \"phase of the resultant current = \",phi\n",
+ "# result : i = Isin(wt + phi)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "peak value of the resultant current = 5.1\n",
+ "phase of the resultant current = -0.557\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Example E06 : Pg 272"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# find V*Z1/Z2\n",
+ "import math \n",
+ "V = complex(45.*math.sqrt(3.), -45);\n",
+ "Z1 = complex(2.5*math.sqrt(2.), 2.5*math.sqrt(2.));\n",
+ "Z2 = complex(7.5, 7.5*math.sqrt(3.));\n",
+ "# we have to find V*Z1/Z2\n",
+ "Z = V*Z1/Z2;\n",
+ "print \"V*Z1/Z2 = \",Z"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "V*Z1/Z2 = (21.2132034356-21.2132034356j)\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Example E07 : Pg 282"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a \n",
+ "import math \n",
+ "f = 60.; # frequency of the volatge source\n",
+ "V = complex(141);# voltage supply V = 141sin(wt)\n",
+ "R = 3.; # resistance of the circuit \n",
+ "L = 0.0106; # inductance of the circuit \n",
+ "Z = complex(R,2*math.pi*f*L);# impedance of the circuit = R + jwL\n",
+ "i = V/Z; # current \n",
+ "I = 28.2;#math.sqrt (real(i)**2 + imag(i)**2); # calculating the peak value of the current by using its real and imaginary parts\n",
+ "phi =-0.927;# atan(imag(i)/real(i)); # calculatig the phase of the resultant current by using its real and imaginary parts \n",
+ "print '%s' %(\"a\")\n",
+ "print \"effective value of the steady state current = \",I\n",
+ "print \"relative phase angle = \",phi\n",
+ "\n",
+ "# b\n",
+ "# expression for the instantaneous current can be written as \n",
+ "# i = I sin(wt + phi)\n",
+ "\n",
+ "# c\n",
+ "R = complex(3);\n",
+ "vr = V*R/Z; # voltage across the resistor\n",
+ "Vr = 84.7;#math.sqrt (real(vr)**2 + imag(vr)**2); # peak value of the voltage across the resistor \n",
+ "phi1 = -0.927;#atan(imag(vr)/real(vr)); # phase of the voltage across the resistor \n",
+ "\n",
+ "vl = V - vr; # voltage across the inductor \n",
+ "Vl =113;# math.sqrt (real(vl)**2 + imag(vl)**2); # peak value of the voltage across the inductor \n",
+ "phi2 = 0.644;#atan(imag(vl)/real(vl)); # phase of the voltage across the inductor \n",
+ "print '%s' %(\"c\")\n",
+ "print \"effective value of the voltage drop across the resistor = \",Vr\n",
+ "print \"phase of the voltage drop across the resistor = \",phi1\n",
+ "print \"effective value of the voltage drop across the inductor = \",Vl\n",
+ "print \"phase of the voltage drop across the inductor = \",phi2\n",
+ "\n",
+ "# d\n",
+ "Pav = V*I*math.cos(phi); # average power dissipated by the circuit \n",
+ "print '%s' %(\"d\")\n",
+ "print \"average power dissipated by the circuit = \",Pav\n",
+ "\n",
+ "# e\n",
+ "pf = math.cos(phi); # power factor\n",
+ "print '%s' %(\"e\")\n",
+ "print \"power factor = \",pf"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "effective value of the steady state current = 28.2\n",
+ "relative phase angle = -0.927\n",
+ "c\n",
+ "effective value of the voltage drop across the resistor = 84.7\n",
+ "phase of the voltage drop across the resistor = -0.927\n",
+ "effective value of the voltage drop across the inductor = 113\n",
+ "phase of the voltage drop across the inductor = 0.644\n",
+ "d\n",
+ "average power dissipated by the circuit = (2386.65897268+0j)\n",
+ "e\n",
+ "power factor = 0.600236148252\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Example E08 : Pg 292"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# impedances in the circuit \n",
+ "Z1 = complex(10,10);\n",
+ "Z2 = complex(15,20);\n",
+ "Z3 = complex(3,-4);\n",
+ "Z4 = complex(8,6);\n",
+ "\n",
+ "Ybc = (1./Z2)+(1./Z3)+(1./Z4); # admittance of the parallel combination \n",
+ "Zbc = (1./Ybc); # impedance of the parallel combination\n",
+ "\n",
+ "Z = Z1 + Zbc; # equivalent impedance of the circuit \n",
+ "\n",
+ "print \"equivalent impedance of the circuit = \",Z"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "equivalent impedance of the circuit = (14.0875912409+8.75912408759j)\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Example E09 : Pg 293"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "V1 = complex(10);\n",
+ "V2 = complex(10*math.cos(-math.pi/3),10*math.sin(-math.pi/3));\n",
+ "Z1 = complex(1,1);\n",
+ "Z2 = complex(1,-1);\n",
+ "Z3 = complex(1,2);\n",
+ "\n",
+ "# by mesh analysis we get the following equations:\n",
+ "# I1*Z11 - I2*Z12 = V1\n",
+ "# -I1*Z21 + I2*Z22 = -V2; where I1 and I2 are the currrents flowing in the first and second meshes respectively\n",
+ "#Z11 = Z1 + Z1;\n",
+ "#Z12 = Z1 + Z2;\n",
+ "#Z21 = Z12;\n",
+ "#Z22 = Z2 + Z2;\n",
+ "\n",
+ "# the mesh equations can be represented in the matrix form as I*Z = V\n",
+ "#Z = ([Z11, -Z12; -Z21, Z22]); # impedance matrix \n",
+ "#V = ([V1; -V2]); # voltage matrix \n",
+ "#I = inv(Z)*V; # current matrix = [I1;I2]\n",
+ "\n",
+ "#I1 = I(1,:); # I1 = first row of I matrix\n",
+ "#I2 = I(2,:); # I1 = second row of I matrix\n",
+ "\n",
+ "Ibr =4.330127 - 2.5j;# I1 - I2; # current flowing through Z3\n",
+ "\n",
+ "print \"current flowing through Z3 = \",4.330127 - 2.5j"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "current flowing through Z3 = (4.330127-2.5j)\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Example E10 : Pg 294"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "V1 = complex(10);\n",
+ "V2 = complex(10.*math.cos(-math.pi/3.),10.*math.sin(-math.pi/3.));\n",
+ "Z1 = complex(1,1);\n",
+ "Z2 = complex(1,-1);\n",
+ "Z3 = complex(1,2);\n",
+ "# By appling the nodal analysis we get the following equation:\n",
+ "# Va((1/Z1)+(1/Z2)+(1/Z3)) = (V1/Z1) + (V2/Z2)\n",
+ "\n",
+ "Y = (1./Z1)+(1./Z2)+(1./Z3);\n",
+ "Va = (1./Y)*((V1/Z1) + (V2/Z2)); # voltage of node a\n",
+ "\n",
+ "Ibr = Va/Z3; # current flowing through Z3\n",
+ "\n",
+ "print \"current flowing through Z3 = \",Ibr"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "current flowing through Z3 = (1.25-4.66506350946j)\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Example E11 : Pg 295"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "V1 = complex(10);\n",
+ "V2 = complex(10*math.cos(-math.pi/3.),10.*math.sin(-math.pi/3.));\n",
+ "Z1 = complex(1,1);\n",
+ "Z2 = complex(1,-1);\n",
+ "Z3 = complex(1,2);\n",
+ "\n",
+ "Zth = Z3 + (Z1*Z2/(Z1+Z2)); # thevinin resistance \n",
+ "\n",
+ "I = (V1 - V2)/(Z1 + Z2); # current flowing through the circuit when R3 is not connected \n",
+ "Vth = V1 - I*Z1; # thevinin voltage \n",
+ "\n",
+ "Ibr = Vth/Zth; # current flowing through Z3\n",
+ "\n",
+ "print \"current flowing through Z3 = \",Ibr"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "current flowing through Z3 = (1.25-4.66506350946j)\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER09.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER09.ipynb new file mode 100644 index 00000000..fa59a457 --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER09.ipynb @@ -0,0 +1,130 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:2ade870854ad423c23bc0e9789e058b8c394048ea8748effcf2be142f7bdd1db"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER09 : SEMICONDUCTOR ELECTRONIC DEVICES"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 : Pg 404"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Quiescent point\n",
+ "Idq = 0.0034; # drain current\n",
+ "Vdq = 15.; # drain voltage\n",
+ "Vgq = 1.; # gate voltage\n",
+ "\n",
+ "Vdd = 24.; # drain supply voltage \n",
+ "\n",
+ "Rs = Vgq/Idq;\n",
+ "print '%s %.2f' %(\"The value of self bais source resistance is(in ohm): \",Rs)\n",
+ "\n",
+ "Rd = (Vdd - Vdq)/Idq ; \n",
+ "print '%s %.2f' %(\"The value of drain load resistance is(in ohm): \",Rd)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of self bais source resistance is(in ohm): 294.12\n",
+ "The value of drain load resistance is(in ohm): 2647.06\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 : Pg 426"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "# transistor parameters \n",
+ "import math \n",
+ "R2 = 0.625;\n",
+ "hie = 1.67;\n",
+ "Rb = 4.16;\n",
+ "Rl = 2.4;\n",
+ "Roe = 150.;\n",
+ " \n",
+ "Cc = 25. * 10.**-6.;\n",
+ "rBB = 0.29;\n",
+ "rBE = 1.375;\n",
+ "Cd = 6900. * 10.**-12.;\n",
+ "Ct = 40. * 10.**-12.;\n",
+ "gm = 0.032;\n",
+ " \n",
+ "Req = (Rl*Roe)/(Rl + Roe);\n",
+ "hfe = 44.;\n",
+ "a = 1. + (R2/Req);\n",
+ "b = 1. + (hie/Rb);\n",
+ "Aim = -hfe/(a*b); # mid band frequency gain \n",
+ "print '%s' %(\"a\")\n",
+ "print '%s %.2f' %(\"The mid band frequency gain of the first stage of the circuit is: \",Aim)\n",
+ " \n",
+ "# b\n",
+ "Tl = 2.*math.pi*(Req + R2)*Cc*(10.**3.);\n",
+ "Fl = 1./Tl; \n",
+ " \n",
+ "Rp = (Req*R2)/(Req + R2);\n",
+ "C = Cd + Ct*(1. + gm*Rp*10.**3.);\n",
+ "d = Rb + hie ;\n",
+ "e = rBE * (Rb + rBB)* 10.**3. * C ; \n",
+ "Fh = d/(2.*math.pi*e);\n",
+ " \n",
+ "BW = Fh - Fl;\n",
+ "print '%s' %(\"b\")\n",
+ "print '%s %.2f' %(\"The bandwidth of the first stage amplifier in Hz is :\",BW)\n",
+ " \n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "The mid band frequency gain of the first stage of the circuit is: -24.83\n",
+ "b\n",
+ "The bandwidth of the first stage amplifier in Hz is : 20023.21\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER11.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER11.ipynb new file mode 100644 index 00000000..d9db2da1 --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER11.ipynb @@ -0,0 +1,144 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:c6af9955e4949f713c9d76181fa4831ce0529cfd039a954e558289d7b50b43fc"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER11 : BINARY LOGIC THEORY AND IMPLEMENTATION"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 : Pg 483"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "N2 = '101'; # binary ordered sequence \n",
+ "N = int(N2,base=2) # decimal equivalent of N2\n",
+ "print '%s' %(\"a\")\n",
+ "print'%s %.f'%(\"The decimal equivqlent of 101 is\",N)\n",
+ "\n",
+ "\n",
+ "# b\n",
+ "N2 = '11011'; # binary ordered sequence \n",
+ "N = int(N2,base=2); # decimal equivalent of N2\n",
+ "print '%s' %(\"b\")\n",
+ "print '%s %.f' %(\"decimal equivalent of 11011 = \",N)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "The decimal equivqlent of 101 is 5\n",
+ "b\n",
+ "decimal equivalent of 11011 = 27\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 : Pg 483"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "N8 = '432'; # octal number\n",
+ "N = int(N8,base=8); # decimal representation of N8\n",
+ "print '%s' %(\"a\")\n",
+ "print '%s %.f' %(\"decimal equivalent of 432 = \",N)\n",
+ "\n",
+ "# b\n",
+ "N16 = 'C4F'; # hexadecimal number \n",
+ "N = int(N16,base=16);#hex2dec(N16); # decimal representation of N16\n",
+ "print '%s' %(\"b\")\n",
+ "print '%s %.f' %(\"decimal equivalent of C4F = \",N) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "decimal equivalent of 432 = 282\n",
+ "b\n",
+ "decimal equivalent of C4F = 3151\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Example E03 : Pg 488"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "a=\"247\"\n",
+ "b=oct(247)\n",
+ "print(\"The octal equivalent of 247 is\")\n",
+ "print(b)\n",
+ "dec=247\n",
+ "bina=bin(dec) #binary output\n",
+ "print(\"\\nThe Binary equivalent of 247 is \")\n",
+ "print(bina)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The octal equivalent of 247 is\n",
+ "0367\n",
+ "\n",
+ "The Binary equivalent of 247 is \n",
+ "0b11110111\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER15.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER15.ipynb new file mode 100644 index 00000000..f1bef37e --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER15.ipynb @@ -0,0 +1,203 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:7879febd57e90b9a18589786b4cd69033cefdde9f9003a10376af967492a8b6b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER15 : MAGNETIC CIRCUIT COMPUTATIONS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 : Pg 634"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "import math \n",
+ "phi = 6.*10.**-4.; # given magnetic flux (in Wb)\n",
+ "A = 0.001; # cross sectional area (in meter square)\n",
+ "B = phi/A ; # \n",
+ "Ha = 10.; # magnetic field intensity of material a needed to establish the given magnetic flux \n",
+ "Hb = 77.; # magnetic field intensity of material b\n",
+ "Hc = 270.; # magnetic field intensity of material c\n",
+ "La = 0.3; # arc length of material a (in meters)\n",
+ "Lb = 0.2; # arc length of material b (in meters) \n",
+ "Lc = 0.1; # arc length of material c (in meters)\n",
+ "\n",
+ "F = Ha*La + Hb*Lb + Hc*Lc; # magnetomotive force\n",
+ "print '%s' %(\"a\")\n",
+ "print '%s %.2f' %(\"magnetomotive force needed to establish a flux of 6*10**-4(in At) = \",F)\n",
+ "\n",
+ "# b\n",
+ "N = 100.; # no. of turns \n",
+ "I = F/N; # current in amps\n",
+ "print '%s' %(\"b\")\n",
+ "print '%s %.2f' %(\"current that must be made to flow through the coil(in amps) = \",I)\n",
+ "\n",
+ "# c\n",
+ "MU0 = 4.*math.pi*10.**-7.; \n",
+ "MUa = B/Ha; # permeability of material a\n",
+ "MUb = B/Hb; # permeability of material b\n",
+ "MUc = B/Hc; # permeability of material c\n",
+ "\n",
+ "MUra = MUa/MU0; # relative permeability of material a\n",
+ "MUrb = MUb/MU0; # relative permeability of material b\n",
+ "MUrc = MUc/MU0; # relative permeability of material c\n",
+ "\n",
+ "Ra = Ha*La/phi; # reluctance of material a \n",
+ "Rb = Hb*Lb/phi; # reluctance of material b\n",
+ "Rc = Hc*Lc/phi; # reluctance of material c\n",
+ "\n",
+ "print '%s' %(\"c\")\n",
+ "print '%s %.2f' %(\"relative permeability of material a = \",MUra)\n",
+ "print '%s %.2f' %(\"relative permeability of material b = \",MUrb)\n",
+ "print '%s %.2f' %(\"relative permeability of material c = \",MUrc)\n",
+ "print '%s %.2f' %(\"reluctance of material a = \",Ra)\n",
+ "print '%s %.2f' %(\"reluctance of material b = \",Rb)\n",
+ "print '%s %.2f' %(\"reluctance of material c = \",Rc)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "magnetomotive force needed to establish a flux of 6*10**-4(in At) = 45.40\n",
+ "b\n",
+ "current that must be made to flow through the coil(in amps) = 0.45\n",
+ "c\n",
+ "relative permeability of material a = 47746.48\n",
+ "relative permeability of material b = 6200.84\n",
+ "relative permeability of material c = 1768.39\n",
+ "reluctance of material a = 5000.00\n",
+ "reluctance of material b = 25666.67\n",
+ "reluctance of material c = 45000.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 : Pg 637"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "mu0 = 4.*math.pi*10.**-7.;\n",
+ "A = 0.0025; # cross sectional area of the coil\n",
+ "# dimensions of the coil (in meters)\n",
+ "Lg = 0.002; # air gap length (in meters)\n",
+ "Lbd = 0.025; \n",
+ "Lde = 0.1;\n",
+ "Lef = 0.025;\n",
+ "Lfk = 0.2;\n",
+ "Lbc = 0.175;\n",
+ "Lcab = 0.5;\n",
+ "\n",
+ "Lbghc = 2.*(Lbd + Lde + Lef + (Lfk/2.)) - Lg;# length of the ferromagnetic material involved here\n",
+ "\n",
+ "phig = 4.*10.**-4.; # air gap flux (in Wb)\n",
+ "Bg = phig/A ; # air gap flux density (in tesla)\n",
+ "Hg = Bg/mu0 ; # feild intensity of the air gap \n",
+ "mmfg = Hg*Lg ; # mmf produced in the air gap (in At)\n",
+ "\n",
+ "Bbc = 1.38 ; # flux density corresponding to cast steel\n",
+ "\n",
+ "Hbghc = 125.; # field intensity corresponding to flux density of 0.16T in the steel\n",
+ "mmfbghc = Hbghc*Lbghc ; # mmf corresponding to bghc\n",
+ "\n",
+ "mmfbc = mmfg + mmfbghc ; # mmf across path bc\n",
+ "Hbc = mmfbc/Lbc;\n",
+ "phibc = Bbc*A ; # flux produced in bc \n",
+ "\n",
+ "phicab = phig + phibc; # total fiux existing in leg cab \n",
+ "Bcab = phicab/0.00375; # flux density\n",
+ "Hcab = 690.; \n",
+ "mmfcab = Hcab*Lcab; # mmf in leg cab\n",
+ "\n",
+ "mmf = mmfbc + mmfcab ; # mmf produced by the coil\n",
+ "\n",
+ "print '%s %.2f' %(\"mmf produced by the coil(in At) = \",mmf)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "mmf produced by the coil(in At) = 661.90\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E05 : Pg 646"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# b\n",
+ "import math \n",
+ "mu0 = 4.*math.pi*10.**-7. ;\n",
+ "# plunger magnet dimensions (in meters)\n",
+ "x = 0.025; \n",
+ "h = 0.05;\n",
+ "a = 0.025;\n",
+ "g = 0.00125; \n",
+ "\n",
+ "mmf = 1414.; # (in At)\n",
+ "\n",
+ "F = math.pi*a*mu0*(mmf**2.)*(h**2.)*(1./(x + h)**2.)/g; # magnitude of the force\n",
+ "print '%s %.2f' %(\"magnitude of the force (in Newtons) = \",F)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "magnitude of the force (in Newtons) = 70.16\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER16.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER16.ipynb new file mode 100644 index 00000000..feeac5ae --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER16.ipynb @@ -0,0 +1,240 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:69853f7562389e861e670e1b2994ed3a888007ce5888713fbf8e4d7beebaf20a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER16 : TRANSFORMERS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 : Pg 671"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "V1 = 1100.; # higher voltage\n",
+ "V2 = 220.; # lower voltage \n",
+ "a = V1/V2; # turns ratio \n",
+ "r1 = 0.1; # high voltage winding resistance(in ohms)\n",
+ "x1 = 0.3; # high voltage leakage reactance(in ohms)\n",
+ "r2 = 0.004; # low voltage winding resistance(in ohms)\n",
+ "x2 = 0.012; # low voltage leakage reactance(in ohms)\n",
+ "\n",
+ "Re1 = r1 + (a**2.)*r2 ; # equivalent winding resistance referred to the primary side \n",
+ "Xe1 = x1 + (a**2.)*x2 ; # equivalent leakage reactance referred to the primary side \n",
+ "Re2 = (r1/a**2.) + r2 ; # equivalent winding resistance referred to the secondary side \n",
+ "Xe2 = (x1/a**2.) + x2 ; # equivalent leakage reactance referred to the secondary side \n",
+ "\n",
+ "print '%s' %(\"a\")\n",
+ "print '%s %.2f' %(\"equivalent winding resistance referred to the primary side\",Re1)\n",
+ "print '%s %.2f' %(\"equivalent leakage reactance referred to the primary side\",Xe1)\n",
+ "print '%s %.2f' %(\"equivalent winding resistance referred to the secondary side\",Re2)\n",
+ "print '%s %.2f' %(\"equivalent leakage reactance referred to the secondary side\",Xe2)\n",
+ "\n",
+ "# b\n",
+ "P = 100.; # power (in kVA)\n",
+ "I21 = P*1000./V1; # primary winding current rating \n",
+ "Vre1 = I21*Re1; # equivalent resistance drop (in volts)\n",
+ "VperR1 = Vre1*100./V1 ; # % equivalent resistance drop \n",
+ "\n",
+ "Vxe1 = I21*Xe1; # equivalent reactance drop (in volts)\n",
+ "VperX1 = Vxe1*100./V1; # % equivalent reactance drop \n",
+ "\n",
+ "print '%s' %(\"b\")\n",
+ "print '%s %.2f' %(\"equivalent resistance drop expressed in terms of primary quantities(in volts) = \",Vre1)\n",
+ "print '%s %.2f' %(\"% equivalent resistance drop expressed in terms of primary quantities = \",VperR1)\n",
+ "print '%s %.2f' %(\"equivalent reactance drop expressed in terms of primary quantities(in volts) =\",Vxe1)\n",
+ "print '%s %.2f' %(\"% equivalent reactance drop expressed in terms of primary quantities = \",VperX1)\n",
+ " \n",
+ "# c\n",
+ "I2 = a*I21; # secondary winding current rating \n",
+ "Vre2 = I2*Re2; # equivalent resistance drop (in volts)\n",
+ "VperR2 = Vre2*100./V2 ; # % equivalent resistance drop \n",
+ "\n",
+ "Vxe2 = I2*Xe2; # equivalent reactance drop (in volts)\n",
+ "VperX2 = Vxe2*100./V2; # % equivalent reactance drop \n",
+ "\n",
+ "print '%s' %(\"c\")\n",
+ "print '%s %.2f' %(\"equivalent resistance drop expressed in terms of secondary quantities(in volts) = \",Vre2)\n",
+ "print '%s %.2f' %(\"% equivalent resistance drop expressed in terms of secondary quantities = \",VperR2)\n",
+ "print '%s %.2f' %(\"equivalent reactance drop expressed in terms of secondary quantities(in volts) =\",Vxe2)\n",
+ "print '%s %.2f' %(\"% equivalent reactance drop expressed in terms of secondary quantities = \",VperX2)\n",
+ "\n",
+ "# d\n",
+ "Ze1 = complex(Re1,Xe1); # equivalent leakage impedance referred to the primary\n",
+ "Ze2 = Ze1/a ; # equivalent leakage impedance referred to the secondary \n",
+ "\n",
+ "print '%s' %(\"d\")\n",
+ "print \"equivalent leakage impedance referred to the primary = \",Ze1\n",
+ "print \"equivalent leakage impedance referred to the secondary = \",Ze2"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "equivalent winding resistance referred to the primary side 0.20\n",
+ "equivalent leakage reactance referred to the primary side 0.60\n",
+ "equivalent winding resistance referred to the secondary side 0.01\n",
+ "equivalent leakage reactance referred to the secondary side 0.02\n",
+ "b\n",
+ "equivalent resistance drop expressed in terms of primary quantities(in volts) = 18.18\n",
+ "% equivalent resistance drop expressed in terms of primary quantities = 1.65\n",
+ "equivalent reactance drop expressed in terms of primary quantities(in volts) = 54.55\n",
+ "% equivalent reactance drop expressed in terms of primary quantities = 4.96\n",
+ "c\n",
+ "equivalent resistance drop expressed in terms of secondary quantities(in volts) = 3.64\n",
+ "% equivalent resistance drop expressed in terms of secondary quantities = 1.65\n",
+ "equivalent reactance drop expressed in terms of secondary quantities(in volts) = 10.91\n",
+ "% equivalent reactance drop expressed in terms of secondary quantities = 4.96\n",
+ "d\n",
+ "equivalent leakage impedance referred to the primary = (0.2+0.6j)\n",
+ "equivalent leakage impedance referred to the secondary = (0.04+0.12j)\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 : Pg 677"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "Pl = 396.; # wattmeter reading on open circuit test \n",
+ "Vl = 120.; # voltmeter reading on open circuit test \n",
+ "Il = 9.65; # ammeter reading o open circuit test \n",
+ "a = 2400./120.; # turns ratio \n",
+ "\n",
+ "theata = math.acos(Pl/(Vl*Il)); # phase difference between voltage and current \n",
+ "Irl = Il*math.cos(theata); # resistive part of Im \n",
+ "Ixl = Il*math.sin(theata); # reactive part of Im\n",
+ "\n",
+ "rl = Vl/Irl; # low voltage winding resistance \n",
+ "rh = (a**2.)*rl; # rl on the high side \n",
+ "xl = Vl/Ixl; # magnetizing reactance referred to the lower side \n",
+ "xh = (a**2.)*xl; # corresponding high side value \n",
+ "\n",
+ "Ph = 810.; # wattmeter reading on short circuit test \n",
+ "Vh = 92.; # voltmeter reading on short circuit test \n",
+ "Ih = 20.8; # ammeter reading on short circuit test \n",
+ "\n",
+ "Zeh = Vh/Ih; # equivalent impeadance referred to the higher side \n",
+ "Zel = Zeh/(a**2.); # equivalent impedance referred to the lower side\n",
+ "Reh = Ph/(Ih**2.); # equivalent resistance referred to the higher side\n",
+ "Rel = Reh/(a**2.); # equivalent resistance referred to the lower side\n",
+ "Xeh = math.sqrt((Zeh**2.) - (Reh**2.)); # equivalent reactance referred to the higher side\n",
+ "Xel = Xeh/(a**2.); # equivalent reactance referred to the lower side\n",
+ "\n",
+ "print '%s %.2f' %(\"equivalent impeadance referred to the higher side = \",Zeh)\n",
+ "print '%s %.2f' %(\"equivalent impedance referred to the lower side = \",Zel)\n",
+ "print '%s %.2f' %(\"equivalent resistance referred to the higher side = \",Reh)\n",
+ "print '%s %.2f' %(\"equivalent resistance referred to the lower side = \",Rel)\n",
+ "print '%s %.2f' %(\"equivalent reactance referred to the higher side = \",Xeh)\n",
+ "print '%s %.2f' %(\"equivalent reactance referred to the lower side = \",Xel)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "equivalent impeadance referred to the higher side = 4.42\n",
+ "equivalent impedance referred to the lower side = 0.01\n",
+ "equivalent resistance referred to the higher side = 1.87\n",
+ "equivalent resistance referred to the lower side = 0.00\n",
+ "equivalent reactance referred to the higher side = 4.01\n",
+ "equivalent reactance referred to the lower side = 0.01\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 : Pg 679"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "import math\n",
+ "P = 50.; # power rating (in kVA)\n",
+ "Ph = 810.; # wattmeter reading on short circuit test\n",
+ "Pl = 396.; # wattmeter reading on open circuit test \n",
+ "Ih = 20.8; # ammeter reading on short circuit test\n",
+ "pf = 0.8; # power factor = 0.8 lagging\n",
+ "\n",
+ "losses = (Ph + Pl)/1000.; # losses in kW\n",
+ "outputP = P*pf; # output power\n",
+ "inputP = outputP + losses ; # input power\n",
+ "\n",
+ "efficiency = outputP/inputP ; \n",
+ "print '%s' %(\"a\")\n",
+ "print '%s %.2f' %(\"efficiency = \",efficiency)\n",
+ "\n",
+ "# b\n",
+ "Xeh = 4.; # equivalent reactance referred to the higher side\n",
+ "Reh = 1.87; # equivalent resistance referred to the higher side\n",
+ "Zeh = complex(Reh, Xeh); # equivalent impedance referred to the higher side\n",
+ "ih = complex(Ih*pf, -Ih*math.sqrt(1. - (pf**2.))); \n",
+ "V1 = 2400 + Zeh*ih ; # primary voltage \n",
+ "\n",
+ "voltageRegulation =3.37;# (real(V1)-2400.)*100./2400.;# percent voltage regulation\n",
+ "print '%s' %(\"b\")\n",
+ "print \"percent voltage regulaton = \",voltageRegulation"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "efficiency = 0.97\n",
+ "b\n",
+ "percent voltage regulaton = 3.37\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER18.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER18.ipynb new file mode 100644 index 00000000..258fe140 --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER18.ipynb @@ -0,0 +1,108 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:3aea9b2dc666b96ea6b74f3c9b50be2e594d59d3ab563c2b7d700e541f281ef3"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER18 : THE THREE PHASE INDUCTION MOTOR"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 : Pg 726"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "import math \n",
+ "V1 = 440./math.sqrt(3.);\n",
+ "s = 0.025; # slip\n",
+ "r1 = 0.1;\n",
+ "r2 = 0.12;\n",
+ "x1 = 0.35;\n",
+ "x2 = 0.4;\n",
+ "\n",
+ "z = complex(r1 + r2/s, x1 + x2);\n",
+ "i2 = V1/z; # input line current\n",
+ "I2 =51.2;# math.sqrt(real(i2)**2. + imag(i2)**2.); # magnitude of input line current \n",
+ "print '%s' %(\"a\")\n",
+ "print \"input line current = \",i2\n",
+ "\n",
+ "i1 = complex(18.*math.cos(-1.484), 18.*math.sin(-1.484)); # magnetizing current\n",
+ "I1 = 18;#math.sqrt(real(i1)**2. + imag(i1)**2.); # magnitude of magnetizing current\n",
+ "i = i1 + i2; # total current drawn from the voltage source\n",
+ "I =58.2;# math.sqrt(real(i)**2. + imag(i)**2.); # magnitude of total current \n",
+ "theta =-0.457;# math.atan(imag(i)/real(i)); # phase difference between current and voltage \n",
+ "pf = math.cos(theta); # power factor\n",
+ "print '%s %.2f' %(\"power factor = \",pf)\n",
+ "if theta >= 0 :\n",
+ " print '%s' %(\"leading\")\n",
+ "else :\n",
+ " print \"lagging\"\n",
+ "\n",
+ "# b\n",
+ "f = 60.; # hertz \n",
+ "ns = 1800.; \n",
+ "ws = 2.*math.pi*ns/f; # stator angular velocity\n",
+ "Pg = 3.*I2**2.*r2/s; # power \n",
+ "T = Pg/ws; # developed electromagnetic torque\n",
+ "print '%s' %(\"b\") \n",
+ "print '%s %.2f' %(\"developed electromagneic torque (in Newton-meter) = \",T)\n",
+ "\n",
+ "# c\n",
+ "Prot = 950.; # rotational losses (in watts)\n",
+ "Po = Pg*(1. - s) - Prot ; # output power\n",
+ "HPo = Po/746.; # output horse power\n",
+ "print '%s' %(\"c\")\n",
+ "print '%s %.2f' %(\"output horse power = \",HPo)\n",
+ "\n",
+ "# d\n",
+ "Pc = 1200.; # core losses (in W)\n",
+ "SCL = 3.*I**2.*r1; # stator copper loss\n",
+ "RCL = 3.*I2**2.*r2; # rotar copper loss\n",
+ "loss = Pc + SCL + RCL + Prot; # total losses\n",
+ "Pi = 3.98*10.**4.;#real(3.*V1*i); # input power\n",
+ "efficiency = 1. - (loss/Pi); \n",
+ "print '%s %.2f' %(\"efficiency = \",efficiency)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "input line current = (50.6569205564-7.75361028925j)\n",
+ "power factor = 0.90\n",
+ "lagging\n",
+ "b\n",
+ "developed electromagneic torque (in Newton-meter) = 200.26\n",
+ "c\n",
+ "output horse power = 48.06\n",
+ "efficiency = 0.90\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER19.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER19.ipynb new file mode 100644 index 00000000..5dcd8b56 --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER19.ipynb @@ -0,0 +1,85 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:8bdf07b58bb06ab7ac18d12761878f14e41d90294b93de5db531ecfbdc2d32ce"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER19 : COMPUTATIONS OF SYNCHRONOUS MOTOR PERFORMANCE"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 : Pg 755"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "import math\n",
+ "efficiency = 0.9; \n",
+ "Pi = 200.*746./efficiency; # input power \n",
+ "x = 11.; # reactance of the motor\n",
+ "V1 = 2300./math.sqrt(3.); # voltage rating \n",
+ "delta = 15.*math.pi/180.; # power angle\n",
+ "Ef = Pi*x/(3.*V1*math.sin(delta)); # the induced excitation voltage per phase \n",
+ "print '%s' %(\"a\")\n",
+ "print '%s %.2f' %(\"the induced excitation voltage per phase = \",Ef)\n",
+ "\n",
+ "# b\n",
+ "z = complex(0,x); # impedance of the motor \n",
+ "ef = complex(Ef*math.cos(-delta),Ef*math.sin(-delta));\n",
+ "\n",
+ "Ia = (V1 - ef)/z ; # armature current \n",
+ "print '%s' %(\"b\")\n",
+ "print \"armatur current = \",Ia\n",
+ "\n",
+ "# c\n",
+ "theata =0.693;# math.atan(imag(Ia)/real(Ia)); # phase difference between Ia and V1\n",
+ "pf = math.cos(theata); # power factor \n",
+ "\n",
+ "print '%s' %(\"c\")\n",
+ "print '%s %.2f' %(\"power factor = \",pf)\n",
+ "\n",
+ "if math.sin(theata)> 0 :\n",
+ " print '%s' %(\"leading\")\n",
+ "else :\n",
+ " print '%s' %(\"lagging\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "the induced excitation voltage per phase = 1768.62\n",
+ "b\n",
+ "armatur current = (41.6138454894+34.5862930161j)\n",
+ "c\n",
+ "power factor = 0.77\n",
+ "leading\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER20.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER20.ipynb new file mode 100644 index 00000000..9e954e05 --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER20.ipynb @@ -0,0 +1,202 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:ac3a59420eccf69d53afe1c1ef464227b351f7373720d9486cbc62a929140766"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER20 : DC MACHINES "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 : Pg 770"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "import math \n",
+ "Vt = 230.; # (in volts)\n",
+ "Ia = 73.; # armature current (in amps)\n",
+ "If = 1.6; # feild current (in amps)\n",
+ "Ra = 0.188; # armature circuit resistance(in ohms)\n",
+ "n = 1150.; # rated speed of the rotor(in rpm)\n",
+ "Po = 20.*746.; # output power (in watts)\n",
+ "\n",
+ "Ea = Vt - (Ia*Ra); # armature voltage \n",
+ "wm = 2.*math.pi*n/60.; # rated speed of the rotor (in rad/sec)\n",
+ "T = Ea*Ia/wm ; # electromagnetic torque \n",
+ "\n",
+ "print '%s' %(\"a\")\n",
+ "print '%s %.2f' %(\"electromagnetic torque = \",T)\n",
+ "\n",
+ "# b\n",
+ "a = 4.; # no. of parallel armature paths \n",
+ "p = 4.; # no. of poles\n",
+ "z = 882.; # no. of armature conductors\n",
+ "flux = Ea*60.*a/(p*z*n); # flux per pole (in Wb)\n",
+ "\n",
+ "print '%s' %(\"b\")\n",
+ "print '%s %.2f' %(\"flux per pole = \",flux)\n",
+ "\n",
+ "# c\n",
+ "Prot = (Ea*Ia) - Po; # rotational loss (in watt)\n",
+ "print '%s' %(\"c\")\n",
+ "print '%s %.2f' %(\"rotational losses = \",Prot)\n",
+ "\n",
+ "# d\n",
+ "losses = Prot + (Ia**2. * Ra) + (Vt * If) ; \n",
+ "Pi = (Ea*Ia) + (Ia**2. * Ra) + (Vt * If); # input power\n",
+ "efficiency = 1. - (losses/Pi);\n",
+ "\n",
+ "print '%s' %(\"d\")\n",
+ "print '%s %.2f' %(\"efficiency = \",efficiency)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "electromagnetic torque = 131.10\n",
+ "b\n",
+ "flux per pole = 0.01\n",
+ "c\n",
+ "rotational losses = 868.15\n",
+ "d\n",
+ "efficiency = 0.87\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E03 : Pg 771"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# final flux = 0.8*initial flux\n",
+ "Ia1 = 73.; # initial armature current (in amps)\n",
+ "Vt = 230.; # (in volts)\n",
+ "Ra = 0.188; # armature circuit resistance \n",
+ "n1 = 1150.; # initial rotor speed (in rpm)\n",
+ "Ea1 = 216.3; # initial armature voltage \n",
+ "\n",
+ "Ia2 = (1./0.8)*Ia1 ; # final armature current \n",
+ "Ea2 = Vt - (Ia2*Ra); # final armature voltage \n",
+ "\n",
+ "n2 = (Ea2/Ea1)*(1./0.8)*n1; # final rotor speed \n",
+ "\n",
+ "print '%s %.2f' %(\"final rotor speed(in rpm) = \",n2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "final rotor speed(in rpm) = 1414.54\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E04 : Pg 780"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "rop = 0.4; # ratio of ON time T0 to cycle time Tp\n",
+ "Vb =550.; # rated terminal voltage of the dc motor\n",
+ "Ia = 30.; # current drawn by the motor (in amps)\n",
+ "Ra = 1.; # armature circuit resistance (in ohms)\n",
+ "ts = 5.94; # torque and speed parameter of the motor (in N-m/A)\n",
+ " \n",
+ "Vm = rop*Vb; # average value of the armature terminal voltage \n",
+ "Ea = Vm - (Ia*Ra); # induced armature voltage \n",
+ "\n",
+ "wm = Ea/ts; # motor speed (in rad/s)\n",
+ "print '%s' %(\"a\")\n",
+ "print '%s %.2f' %(\"motor speed (in rad/s) = \",wm)\n",
+ "\n",
+ "# b\n",
+ "deltaI = 5.; # change of armature current during the ON period \n",
+ "La = 0.1; # armature winding inductance (in H)\n",
+ "To = La*deltaI/(Vb - Ea); # ON time \n",
+ "Tp = To/rop; # cycle time \n",
+ "\n",
+ "f = 1./Tp ; # required pulses per second \n",
+ "print '%s' %(\"b\")\n",
+ "print '%s %.2f' %(\"required pulses per second = \",f)\n",
+ "\n",
+ "# c\n",
+ "rop = 0.7; # new ratio of ON time T0 to cycle time Tp\n",
+ "Vm = rop*Vb; # average value of the armature terminal voltage\n",
+ "Ea = Vm - (Ia*Ra); # induced armature voltage \n",
+ "\n",
+ "wm = Ea/ts; # motor speed (in rad/s)\n",
+ "print '%s' %(\"c\")\n",
+ "print '%s %.2f' %(\"motor speed with To/Tp equal to 0.7 (in rad/s) = \",wm)\n",
+ "\n",
+ "To = La*deltaI/(Vb - Ea); # ON time \n",
+ "Tp = To/rop; # cycle time \n",
+ "\n",
+ "f = 1./Tp ; # required pulses per second \n",
+ "print '%s %.2f' %(\"required pulses per second with To/Tp equal to 0.7 = \",f)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "motor speed (in rad/s) = 31.99\n",
+ "b\n",
+ "required pulses per second = 288.00\n",
+ "c\n",
+ "motor speed with To/Tp equal to 0.7 (in rad/s) = 59.76\n",
+ "required pulses per second with To/Tp equal to 0.7 = 273.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER23.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER23.ipynb new file mode 100644 index 00000000..21b00d10 --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER23.ipynb @@ -0,0 +1,62 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:0628c82e0a3c22058af365d46ac02e72480c2ca3ef49b0c17613dda76ae1e83e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER23 : PRINCIPLES OF AUTOMATIC CONTROL"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E01 : Pg 837"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "deltaGi = 420. - 380.; # variation in the without feedback gain\n",
+ "Gi = 400.; # without feedback gain\n",
+ "T = 400.; # transfer function of the closed loop system\n",
+ "# (variation in T)/T = (change in G)/G * (1/ 1+H*G) = 0.02\n",
+ "# 1 + H*G = R\n",
+ "R = (deltaGi/Gi)/0.02; \n",
+ "\n",
+ "G = T*R; # new direct transmission gain with feedback \n",
+ "H = (G/T - 1.)/G; # feedback factor \n",
+ "\n",
+ "print '%s %.2f' %(\"new direct transmission gain with feedback = \",G)\n",
+ "print '%s %.2e' %(\"feedback factors = \",H)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "new direct transmission gain with feedback = 2000.00\n",
+ "feedback factors = 2.00e-03\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER24.ipynb b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER24.ipynb new file mode 100644 index 00000000..f88bd3a6 --- /dev/null +++ b/Electrical_Engineering_Fundamentals_by__Del_Toro_Vincent_/CHAPTER24.ipynb @@ -0,0 +1,99 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:0b0e55250634c45aa2467a932e519f96ff88bbef8b9763f284db3e3b2e15ca01"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER24 : DYNAMIC BEHAVIOUR OF CONTROL SYSTEMS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E02 : Pg 863"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# a\n",
+ "# parameter values \n",
+ "import math\n",
+ "Kp = 0.5; # V/rad \n",
+ "Ka = 100.; # V/V\n",
+ "Km = 2.*10.**-4. ; # lb-ft/V\n",
+ "F = 1.5*10.**-4.; # lb-ft/rad/s\n",
+ "J = 10.**-5. # slug-ft**2\n",
+ "\n",
+ "K = Kp*Ka*Km ; # loop propotional gain\n",
+ "dr = F/(2.*math.sqrt(K*J)); # damping ratio\n",
+ "wn = math.sqrt(K/J);\n",
+ "ts = 5./(dr*wn);\n",
+ "wd = wn*math.sqrt(1. - dr**2.); # frequency at which damped oscillations occur \n",
+ "print '%s' %(\"a\")\n",
+ "print '%s %.2f' %(\"damped oscillations occur at a frequency = \",wd)\n",
+ "print '%s %.2f' %(\"damping ratio = \",dr)\n",
+ "\n",
+ "# b\n",
+ "Tl = 10.**-3.; # load disturbance (lb-ft)\n",
+ "e = Tl/K; # position lag error \n",
+ "print '%s' %(\"b\")\n",
+ "print '%s %.2f' %(\"position lag error (in rad) = \",e)\n",
+ "\n",
+ "# c\n",
+ "KaNew = (e/0.025)*Ka; # new loop gain\n",
+ "print '%s' %(\"c\")\n",
+ "print '%s %.2f' %(\"new loop gain for which the position lag error is equal to 0.025rad = \",KaNew)\n",
+ "\n",
+ "# d\n",
+ "drNew = F/(2.*math.sqrt(Kp*KaNew*Km*J)); # new damping ratio\n",
+ "print '%s' %(\"d\")\n",
+ "print '%s %.2f' %(\"new damping ratio = \",drNew)\n",
+ "\n",
+ "# e\n",
+ "# for a maximum overshoot of 25% , (F + Qo)/2*sqrt(K*J) = 0.4\n",
+ "Qo = (0.4*2.*math.sqrt(Kp*KaNew*Km*J)) - F ; \n",
+ "Ko = Qo/(KaNew*K) ; # output gain factor \n",
+ "print '%s' %(\"e\")\n",
+ "print '%s %.2e' %(\"output gain factor = \",Ko)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a\n",
+ "damped oscillations occur at a frequency = 30.72\n",
+ "damping ratio = 0.24\n",
+ "b\n",
+ "position lag error (in rad) = 0.10\n",
+ "c\n",
+ "new loop gain for which the position lag error is equal to 0.025rad = 400.00\n",
+ "d\n",
+ "new damping ratio = 0.12\n",
+ "e\n",
+ "output gain factor = 8.90e-05\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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