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| author | Trupti Kini | 2016-01-27 23:30:14 +0600 |
|---|---|---|
| committer | Trupti Kini | 2016-01-27 23:30:14 +0600 |
| commit | 39d8a08165d74513440d0bd6ca8bf32ee9d32729 (patch) | |
| tree | 2af2aad7fc86dcae852869f1210fc3f6b18d57af | |
| parent | 0c755162b4449d3eac5cbfcc12d320b4be545b6b (diff) | |
| download | Python-Textbook-Companions-39d8a08165d74513440d0bd6ca8bf32ee9d32729.tar.gz Python-Textbook-Companions-39d8a08165d74513440d0bd6ca8bf32ee9d32729.tar.bz2 Python-Textbook-Companions-39d8a08165d74513440d0bd6ca8bf32ee9d32729.zip | |
Added(A)/Deleted(D) following books
A Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter1.ipynb
A Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter10.ipynb
A Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter11.ipynb
A Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter2.ipynb
A Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter3.ipynb
A Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter4.ipynb
A Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter5.ipynb
A Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter6.ipynb
A Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter8.ipynb
A Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter9.ipynb
A Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/screenshots/chap1.png
A Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/screenshots/chap2.png
A Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/screenshots/chap3.png
13 files changed, 9645 insertions, 0 deletions
diff --git a/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter1.ipynb b/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter1.ipynb new file mode 100644 index 00000000..fa197e42 --- /dev/null +++ b/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter1.ipynb @@ -0,0 +1,1313 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1: D C Circuit analysis" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1: pg 9" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current through 4 ohm resistor in A is 3.82\n", + "The current through 2 ohm resistor in A is 4.36\n", + "The current through 6 ohm resistor in A is -0.55\n", + "That is 0.55 A current flows in 6 ohm resistor from C to B\n" + ] + } + ], + "source": [ + "#pg 9\n", + "#calculate the current through all resistors\n", + "import numpy\n", + "from numpy import matrix\n", + "from numpy.linalg import inv\n", + "# Given data\n", + "R1=4.;# in ohm\n", + "R2= 6.;# in ohm\n", + "R3= 2.;# in ohm\n", + "V1= 24.;# in V\n", + "V2= 12.;# in V\n", + "# Applying KVL in Mesh ABEFA, V1 = (R1+R3)*I1 - R3*I2 (i)\n", + "# Applying KVL in Mesh BCDEB, V2 = R3*I1 - (R2+R3)*I2 (ii)\n", + "#calculations\n", + "A= numpy.matrix([[(R1+R3), R3],[-R3, -(R2+R3)]]);# assumed\n", + "B= numpy.matrix([V1,V2]);# assumed\n", + "I= B*inv(A);# Solving equations by matrix multiplication\n", + "I1= I[0,0];# in A\n", + "I2= I[0,1];# in A\n", + "print \"The current through 4 ohm resistor in A is\",round(I1,2)\n", + "# current through 2 ohm resistor \n", + "Ix= I1-I2;# in A\n", + "print \"The current through 2 ohm resistor in A is\",round(Ix,2)\n", + "print \"The current through 6 ohm resistor in A is\",round(I2,2)\n", + "print \"That is \",round(abs(I2),2),\" A current flows in 6 ohm resistor from C to B\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2: pg 11" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current through 10 ohm resistor in A is 3.333\n", + "The current through 5 ohm resistor in A is 13.33\n", + "The current through 20 ohm resistor in A is 10.0\n" + ] + } + ], + "source": [ + "#pg 11\n", + "#calculate the current through all resistors\n", + "# Given data\n", + "V = 100.;# in V\n", + "I3= 10.;# in A\n", + "R1 = 10.;# in ohm \n", + "R2 = 5.;# in ohm\n", + "# I1 = (V - V_A)/R1\n", + "# I2 = (V_A-0)/R2\n", + "# Using KCL at note A, I1-I2+I3=0 or\n", + "#calculations\n", + "V_A= (R1*R2)/(R1+R2)*(I3+V/R1);# in V\n", + "I1 = (V - V_A)/R1;# in A\n", + "I2 = (V_A-0)/R2;# in A\n", + "#results\n", + "print \"The current through 10 ohm resistor in A is\",round(I1,3)\n", + "print \"The current through 5 ohm resistor in A is\",round(I2,2)\n", + "print \"The current through 20 ohm resistor in A is\",I3\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3: pg 16" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The equivalent current in A is 5.0\n", + "The equivalent voltage in V is 50\n" + ] + } + ], + "source": [ + "#pg 16\n", + "#calculate the equivalent current and voltage\n", + "# Given data\n", + "# Part (a)\n", + "V = 30.;# in V\n", + "R = 6;# in ohm\n", + "#calculations\n", + "I = V/R;# the equivalent current in A\n", + "print \"The equivalent current in A is\",I\n", + "# Part (b)\n", + "I = 10;# in A\n", + "R = 5;# in ohm\n", + "V = I*R;# the equivalent voltage in V\n", + "print \"The equivalent voltage in V is\",V\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4: pg 17" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current in A is 0.0\n" + ] + } + ], + "source": [ + "#pg 17\n", + "#calculate the current\n", + "# Given data\n", + "R1= 6.;# in ohm\n", + "R2= 2.;# in ohm\n", + "R3= 5.;# in ohm\n", + "I2= 4.;# in A\n", + "V=24.;#in V\n", + "# Applying KVL to the loop ABCDA, -R1*I1-R3*I+V=0 (i)\n", + "# but I1= I+I2 , so from eq(i)\n", + "#calculations\n", + "I= (V-R1*I2)/(R1+R3);# in A\n", + "#results\n", + "print \"The current in A is\",I\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5: pg 17" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of I1 in A is : 0.7926\n", + "The value of I2 in A is : 0.2949\n" + ] + } + ], + "source": [ + "#pg 17\n", + "#calculate the value of current in different branches\n", + "# Given data\n", + "import numpy\n", + "R1= 40.;# in ohm\n", + "R2= 20.;# in ohm\n", + "R3= 25.;# in ohm\n", + "R4= 60.;# in ohm\n", + "R5= 50.;# in ohm\n", + "V1= 120.;# in V\n", + "V2= 60.;# in V\n", + "V3= 40.;# in V\n", + "#calculations\n", + "# Applying KVL in Mesh ABEFA, we get -I1*(R1+R2+R3)+I2*R3=V2-V1 (i)\n", + "# Applying KVL in Mesh BCEDB, we get R3*I1-I2*(R3+R4+R5)= V3-V2 (ii)\n", + "A= numpy.matrix([[-(R1+R2+R3), R3],[R3, -(R3+R4+R5)]]);\n", + "B= numpy.matrix([V2-V1, V3-V2]);\n", + "I= B*numpy.linalg.inv(A);#Solving eq(i) and (ii) by Matrix method\n", + "I1= I[0,0];# in A\n", + "I2= I[0,1];# in A\n", + "#results\n", + "print \"The value of I1 in A is : \",round(I1,4)\n", + "print \"The value of I2 in A is : \",round(I2,4)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6: pg 18" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of I1 in A is : 3.091\n", + "The value of I2 in A is : 2.545\n", + "The value of I_L in A is : 5.636\n" + ] + } + ], + "source": [ + "#pg 18\n", + "#calculate the value of current in all branches\n", + "import numpy\n", + "# Given data\n", + "R1= 2.;# in ohm\n", + "R2= 4.;# in ohm\n", + "R3= 6.;# in ohm\n", + "V1= 4.;# in V\n", + "V2= 44.;# in V\n", + "#calculations\n", + "#Applying KVL in ABEFA : -R1*I1 + R2*I2 = V1 (i)\n", + "#Applying KVL in BCDEB: R3*I1 + I2*(R2+R3)=V2 (ii)\n", + "A= ([[-R1, R2],[R3, (R2+R3)]]); # assumed\n", + "B= ([[V1],[V2]]);# assumed\n", + "I=numpy.dot(numpy.linalg.inv(A),B);# Solving eq(i) and (ii) by Matrix method\n", + "I1= I[0];# in A\n", + "I2= I[1];# in A\n", + "I_L= I1+I2;# in A\n", + "#results\n", + "print \"The value of I1 in A is : \",round(I1,3)\n", + "print \"The value of I2 in A is : \",round(I2,3)\n", + "print \"The value of I_L in A is : \",round(I_L,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7: pg 19" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of I1 in A is : 0.317\n", + "The value of I2 in A is : 0.117\n" + ] + } + ], + "source": [ + "#pg 19\n", + "#calculate the value of current\n", + "# Given data\n", + "import numpy\n", + "R1= 1;# in ohm\n", + "R2= 1;# in ohm\n", + "R3= 2;# in ohm\n", + "R4= 1;# in ohm\n", + "R5= 1;# in ohm\n", + "V1= 1.5;# in V\n", + "V2= 1.1;# in V\n", + "#calculations\n", + "#Applying KVL in ABCFA : I1*(R1+R2+R3) + R3*I2 = V1 (i)\n", + "#Applying KVL in BCDEB: R3*I1 + I2*(R3+R4+R5)=V2 (ii)\n", + "A= ([[(R1+R2+R3), R3],[R3, (R3+R4+R5)]]);\n", + "B= ([[V1], [V2]]);\n", + "I= numpy.dot(numpy.linalg.inv(A),B);# Solving eq(i) and (ii) by Matrix method\n", + "I1= I[0];# in A\n", + "I2= I[1];# in A\n", + "#results\n", + "print \"The value of I1 in A is : \",round(I1,3)\n", + "print \"The value of I2 in A is : \",round(I2,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8: pg 20" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current through 6 ohm resistance in A is : 1.5493\n" + ] + } + ], + "source": [ + "#pg 20\n", + "#calculate the current\n", + "# Given data\n", + "import numpy\n", + "R1= 2.;# in ohm\n", + "R2= 4.;# in ohm\n", + "R3= 1.;# in ohm\n", + "R4= 6.;# in ohm\n", + "R5= 4.;# in ohm\n", + "V1= 10.;# in V\n", + "V2= 20.;# in V\n", + "#calculations\n", + "#Applying KVL in ABGHA : I1*(R1+R2) - R2*I2 = V1 (i)\n", + "#Applying KVL in BCFGB : I1*R5-I2*(R3+R4+R5)+I3*R4 = 0 (ii)\n", + "#Applying KVL in CDEFC: R4*I2-I3*(R2+R4)=V2 (iii)\n", + "A= ([[(R1+R2), -R5, 0],[R2, -(R3+R4+R5), R4],[0, R4, -(R2+R4)]]);\n", + "B= ([[V1], [0], [V2]]);\n", + "I= numpy.dot(numpy.linalg.inv(A),B);# Solving eq(i), (ii) and (iii) by Matrix method\n", + "I1= I[0];# in A\n", + "I2= I[1];# in A\n", + "I3= I[2];# in A\n", + "I6_ohm_resistor= I2-I3;#The current through 6 ohm resistance in A\n", + "#results\n", + "print \"The current through 6 ohm resistance in A is : \",round(I6_ohm_resistor,4)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9: pg 21" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current through 30 ohm resistance in A is : 2.56\n", + "The current through 60 ohm resistance in A is : 2.72\n", + "The current through 50 ohm resistance in A is : -3.38\n", + "The current through 20 ohm resistance in A is : -3.54\n", + "The current through 40 ohm resistance in A is : -0.15\n", + "Note: In the book there is a mistake in eq(iii), the R.H.S of eq(iii) should be -24 not -240. Since they divide the L.H.S of eq(iii) by 10 and R.H.S not divided, So the answer in the book is wrong\n" + ] + } + ], + "source": [ + "#pg 21\n", + "#calculate the current through resistances\n", + "# Given data\n", + "import numpy\n", + "R1= 30.;# in ohm\n", + "R2= 40.;# in ohm\n", + "R3= 20.;# in ohm\n", + "R4= 60.;# in ohm\n", + "R5= 50.;# in ohm\n", + "V= 240.;# in V\n", + "#calculations\n", + "#Applying KVL in ABDA : I1*-(R1+R2+R3) + R2*I2+R3*I3 =0 (i)\n", + "#Applying KVL in BCDB : I1*R2+I2*-(R2+R4+R5)+I3*R5 = 0 (ii)\n", + "#Applying KVL in CFEADC: I1*R3+ R5*I2+I3*-(R3+R5)=-V (iii)\n", + "A= ([[-(R1+R2+R3), R2, R3], [R2, -(R2+R4+R5), R5], [R3, R5, -(R3+R5)]]);\n", + "B= ([[0], [0], [-V]]);\n", + "I= numpy.dot(numpy.linalg.inv(A),B);# Solving eq(i), (ii) and (iii) by Matrix method\n", + "I1= I[0];# in A\n", + "I2= I[1];# in A\n", + "I3= I[2];# in A\n", + "I30_ohm_resistor= I1;# in A\n", + "I60_ohm_resistor= I2;# in A\n", + "I50_ohm_resistor= I2-I3;# in A\n", + "I20_ohm_resistor= I1-I3;# in A\n", + "I40_ohm_resistor= I1-I2;# in A\n", + "#results\n", + "print \"The current through 30 ohm resistance in A is : \",round(I30_ohm_resistor,2)\n", + "print \"The current through 60 ohm resistance in A is : \",round(I60_ohm_resistor,2)\n", + "print \"The current through 50 ohm resistance in A is : \",round(I50_ohm_resistor,2)\n", + "print \"The current through 20 ohm resistance in A is : \",round(I20_ohm_resistor,2)\n", + "print \"The current through 40 ohm resistance in A is : \",round(I40_ohm_resistor,2)\n", + "\n", + "print'Note: In the book there is a mistake in eq(iii), the R.H.S of eq(iii) should be -24 not -240. Since they divide the L.H.S of eq(iii) by 10 and R.H.S not divided, So the answer in the book is wrong' \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10: pg 22" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current in R3 in A is : 2.7619\n", + "The current in R4 in A is : 1.9048\n" + ] + } + ], + "source": [ + "#pg 22\n", + "#calculate the current\n", + "# Given data\n", + "import numpy\n", + "R1= 5.;# in ohm\n", + "R2= 5.;# in ohm\n", + "R3= 10.;# in ohm\n", + "R4= 10.;# in ohm\n", + "R5= 5.;# in ohm\n", + "V1= 50.;# in V\n", + "V2= 20.;# in V\n", + "#calculations\n", + "#Applying KCL at node A: VA*(R1*R3+R3*R2+R2*R1)+VB*-R1*R3 = V1*R2*R3 (i)\n", + "#Applying KCL at node B: VA*R4*R5+VB*-(R2*R4+R4*R5+R5*R2) = -V2*R2*R4 (ii)\n", + "\n", + "A=([[(R1*R3+R2*R3+R2*R1), -R4*R5], [-R1*R3, (R2*R4+R4*R5+R5*R2)]])\n", + "B= ([[V1*R2*R3], [V2*R2*R4]]);\n", + "V= numpy.dot(numpy.linalg.inv(A),B);# Solving eq(i) and (ii) by Matrix method\n", + "VA= V[0];# in V\n", + "VB= V[1];# in V\n", + "I_through_R3= VA/R3;# in A\n", + "I_through_R4= VB/R4;# in A\n", + "#results\n", + "print \"The current in R3 in A is : \",round(I_through_R3,4)\n", + "print \"The current in R4 in A is : \",round(I_through_R4,4)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11: pg 23" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of I1 in A is : [-5.75]\n", + "The value of I2 in A is : [ 9.25]\n", + "The value of I3 in A is : [-3.5]\n", + "The value of I4 in A is : [ 5.5]\n", + "The value of I5 in A is : [-9.]\n" + ] + } + ], + "source": [ + "#pg 23\n", + "#calculate the current\n", + "# Given data\n", + "import numpy\n", + "R1= 1.;# in ohm\n", + "R2= 1.;# in ohm\n", + "R3= 0.5;# in ohm\n", + "R4= 2.;# in ohm\n", + "R5= 1.;# in ohm\n", + "V1= 15.;# in V\n", + "V2= 20.;# in V\n", + "#calculations\n", + "#Applying KCL at node A: VA*(R1*R2+R2*R3+R3*R1)+VB*-R1*R2 = V1*R2*R3 (i)\n", + "#Applying KCL at node B: VA*R4*R5+VB*-(R3*R4+R4*R5+R5*R3) = V2*R3*R4 (ii)\n", + "A=([[2*(R1*R2+R2*R3+R3*R1), -R4*R5], [2*R1*R2, -(R3*R4+R4*R5+R5*R3)]])\n", + "B= ([[2*V1*R2*R3], [-V2*R3*R4]]);\n", + "V= numpy.dot(numpy.linalg.inv(A),B);# Solving eq(i) and (ii) by Matrix method\n", + "VA= V[0];# in V\n", + "VB= V[1];# in V\n", + "I1= (VA-V1)/R1;# in A\n", + "I2= VA/R2;# in A\n", + "I3= (VA-VB)/R3;# in A\n", + "I4= VB/R4;# in A\n", + "I5= (VB-V2)/R5;# in A\n", + "#results\n", + "print \"The value of I1 in A is : \",I1\n", + "print \"The value of I2 in A is : \",I2\n", + "print \"The value of I3 in A is : \",I3\n", + "print \"The value of I4 in A is : \",I4\n", + "print \"The value of I5 in A is : \",I5\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12: pg 24" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of I1 in A is : 1.0\n", + "The value of I2 in A is : 0.0\n", + "The value of I3 in A is : 1.0\n" + ] + } + ], + "source": [ + "#pg 24\n", + "#calculate the value of current\n", + "# Given data\n", + "V1 = 12.;# in V\n", + "V2 = 10.;# in V\n", + "VB = 0.;# in V\n", + "R1 = 2.;# in ohm\n", + "R2 = 1.;# in ohm\n", + "R3 = 10.;# in ohm\n", + "#calculations\n", + "# Using KCL at node A :\n", + "VA= (V1*R2*R3+V2*R3*R1)/(R1*R2+R2*R3+R3*R1);# in V\n", + "I1 = (V1-VA)/R1;# in A\n", + "I2 = (V2-VA)/R2;# in A\n", + "I3 = (VA-VB)/R3;# in A\n", + "#results\n", + "print \"The value of I1 in A is : \",I1\n", + "print \"The value of I2 in A is : \",I2\n", + "print \"The value of I3 in A is : \",I3\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13: pg 25" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage at node 1 in volts is : [ 2.]\n", + "The voltage at node 2 in volts is : [ 4.]\n" + ] + } + ], + "source": [ + "#pg 25\n", + "#calculate the voltage at all nodes\n", + "# Given data\n", + "import numpy\n", + "R1= 1.;# in ohm\n", + "R2= 2.;# in ohm\n", + "R3= 2.;# in ohm\n", + "R4= 1.;# in ohm\n", + "I1= 1.;# in A\n", + "I5= 2.;# in A\n", + "#calculations\n", + "# Using KCL at node 1: V1*(R2+R3)-V2*R2= I1*R2*R3 (i)\n", + "# Using KCL at node 2: V1*R4-V2*(R3+R4)= -I5*(R3*R4) (ii)\n", + "A= ([[(R2+R3), -R4], [R2, -(R3+R4)]]);\n", + "B= ([[I1*R2*R3], [-2*I5*R3*R4]]);\n", + "V= numpy.dot(numpy.linalg.inv(A),B);# Solving eq(i) and (ii) by Matrix method\n", + "V1= V[0];# in V\n", + "V2= V[1];# in V\n", + "#results\n", + "print\"The voltage at node 1 in volts is : \",V1\n", + "print \"The voltage at node 2 in volts is : \",V2\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14: pg 26" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of I1 in A is : 1.667\n", + "The value of I2 in A is : 1.56\n" + ] + } + ], + "source": [ + "#pg 26\n", + "# Given data\n", + "#calculate the value of current\n", + "import numpy\n", + "R1= 2.;# in ohm\n", + "R2= 6.;# in ohm\n", + "R3= 3.;# in ohm\n", + "V1= 10.;# in V\n", + "V2= 6.;# in V\n", + "V3= 2.;# in V\n", + "#calculations\n", + "#Applying KVL in ABEFA : I1*(R1+R2) - R2*I2=V1-V2 (i)\n", + "#Applying KVL in BCDEB : -I1*R2+I2*(R2+R3)=V2-V3 (ii)\n", + "A= ([[(R1+R2), -R2], [-R2, (R2+R3)]]);\n", + "B= ([[(V1-V2)], [(V2-V3)]]);\n", + "I= numpy.dot(numpy.linalg.inv(A),B);# Solving eq(i), and (ii) by Matrix method\n", + "I1= I[0];# in A\n", + "I2= I[1];# in A\n", + "#results\n", + "print \"The value of I1 in A is : \",round(I1,3)\n", + "print \"The value of I2 in A is : \",round(I2,2)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15: pg 26" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of I1 in A is : 0.606\n", + "The value of I2 in A is : 0.545\n" + ] + } + ], + "source": [ + "#pg 26\n", + "#calculate the value of current\n", + "# Given data\n", + "import numpy\n", + "R1= 2.;# in ohm\n", + "R2= 6.;# in ohm\n", + "R3= 4.;# in ohm\n", + "R4= 3.;# in ohm\n", + "R5= 5.;# in ohm\n", + "V1= 10.;# in V\n", + "V2= 6.;# in V\n", + "V3= 2.;# in V\n", + "#calculations\n", + "#Applying KVL in ABEFA : I1*(R1+R2+R3) - R2*I2 = V1-V2 (i)\n", + "#Applying KVL in BCDEB : I1*-R2+I2*(R2+R4+R5) =V2-V3 (ii)\n", + "A= ([[(R1+R2+R3), -R2], [-R2, (R2+R4+R5)]]);\n", + "B= ([[(V1-V2)], [(V2-V3)]]);\n", + "I= numpy.dot(numpy.linalg.inv(A),B);# Solving eq(i) and (ii) by Matrix method\n", + "I1= I[0];# in A\n", + "I2= I[1];# in A\n", + "#results\n", + "print \"The value of I1 in A is : \",round(I1,3)\n", + "print \"The value of I2 in A is : \",round(I2,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16: pg 27" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current through 10 ohm resistor in A is : [ 0.625]\n" + ] + } + ], + "source": [ + "#pg 27\n", + "#calculate the current\n", + "# Given data\n", + "import numpy\n", + "R1= 10.;# in ohm\n", + "R2= 5.;# in ohm\n", + "R3= 5.;# in ohm\n", + "R4= 5.;# in ohm\n", + "V2= 10.;# in V\n", + "I= 1.;# in A\n", + "#calculations\n", + "V1= R4*I;# in V\n", + "#Applying KVL in ABEFA : I1*(R1+R2+R3) + R1*I2 = V1 (i)\n", + "#Applying KVL in BCDEB : I1*R1+I2*(R1+R4) =V2 (ii)\n", + "A= ([[(R1+R2+R3), R1], [R1, (R1+R4)]]);\n", + "B= ([[V1], [V2]]);\n", + "I= numpy.dot(numpy.linalg.inv(A),B);# Solving eq(i) and (ii) by Matrix method\n", + "I1= I[0];# in A\n", + "I2= I[1];# in A\n", + "I10_ohm= I1+I2;# in A\n", + "#results\n", + "print \"The current through 10 ohm resistor in A is : \",I10_ohm\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17: pg 28" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current through 10 ohm resistor from right to left in A is : 0.702\n" + ] + } + ], + "source": [ + "#pg 28\n", + "#calculate the current\n", + "# Given data\n", + "import numpy\n", + "R1= 4.;# in ohm\n", + "R2= 5.;# in ohm\n", + "R3= 10.;# in ohm\n", + "R4= 6.;# in ohm\n", + "R5= 4.;# in ohm\n", + "V1= 15.;# in V\n", + "V2= 30.;# in V\n", + "#calculations\n", + "#Applying KCL at node A: VA*(R1*R2+R2*R3+R3*R1)+VB*-R1*R2 = V1*R1*R3 (i)\n", + "#Applying KCL at node B: VA*R4*R5+VB*-(R3*R4+R4*R5+R5*R3) = -V2*R3*R4 (ii)\n", + "A=([[(R1*R2+R2*R3+R3*R1), -R4*R5], [R1*R2, -(R3*R4+R4*R5+R5*R3)]])\n", + "B= ([[V1*R1*R3], [-V2*R3*R4]]);\n", + "V= numpy.dot(numpy.linalg.inv(A),B);# Solving eq(i) and (ii) by Matrix method\n", + "VA= V[0];# in V\n", + "VB= V[1];# in V\n", + "I10_ohm= abs((VA-VB)/R3);# in A\n", + "#results\n", + "print \"The current through 10 ohm resistor from right to left in A is : \",round(I10_ohm,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19: pg 29" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of I1 in A is : 1.0\n", + "The value of I2 in A is : [ 0.6]\n", + "The value of I3 in A is : [ 0.4]\n", + "The value of I4 in A is : [ 0.6]\n", + "The value of I5 in A is : [ 0.2]\n", + "The value of I6 in A is : [ 0.3]\n", + "The value of I7 in A is : 0.5\n" + ] + } + ], + "source": [ + "#pg 29\n", + "#calculate the current\n", + "# Given data\n", + "import numpy\n", + "R1= 10.;# in ohm\n", + "R2= 10.;# in ohm\n", + "R3= 20.;# in ohm\n", + "R4= 20.;# in ohm\n", + "R5= 20.;# in ohm\n", + "V= 10.;# in V\n", + "I1= 1.;# in A\n", + "I7=0.5;# in A\n", + "#calculations\n", + "#Applying KCL at node A: VA*(R1+R2)+VB*-R1 = I1*R1*R2 (i)\n", + "#Applying KCL at node B: VA*R3*R4+VB*-(R2*R3+R3*R4+R4*R2)+VC*R2*R3 = V*R2*R4 (ii)\n", + "#Applying KCL at node C: -VB*R5+VC*(R4+R5)=I7*R4*R5 (iii)\n", + "A=([[(R1+R2), -R1, 0], [R3*R4, -(R2*R3+R3*R4+R4*R2), R2*R3],[0, -R5, (R4+R5)]])\n", + "B= ([[I1*R1*R2], [V*R2*R4], [I7*R4*R5]]);\n", + "Value= numpy.dot(numpy.linalg.inv(A),B);# Solving eq(i), (ii) and (iii) by Matrix method\n", + "VA= Value[0];# in V\n", + "VB= Value[1];# in V\n", + "VC= Value[2]\n", + "I2= VA/R1;# in A\n", + "I3= (VA-VB)/R2;# in A\n", + "I4= (VB+V)/R3;# in A\n", + "I5= (VC-VB)/R4;# in A\n", + "I6= VC/R5;# in A\n", + "#results\n", + "print \"The value of I1 in A is : \",I1\n", + "print \"The value of I2 in A is : \",I2\n", + "print \"The value of I3 in A is : \",I3\n", + "print \"The value of I4 in A is : \",I4\n", + "print \"The value of I5 in A is : \",I5\n", + "print \"The value of I6 in A is : \",I6\n", + "print \"The value of I7 in A is : \",I7\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20: pg 31" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current through 8 ohm resistance in A is : 0.8767\n" + ] + } + ], + "source": [ + "#pg 31\n", + "#calculate the current\n", + "# Given data\n", + "import numpy\n", + "R1 = 3.;# in ohm\n", + "R2 = 8.;# in ohm\n", + "R3 = 4.;# in ohm\n", + "R4 = 12.;# in ohm\n", + "R5 = 14.;# in ohm\n", + "V1 = 10.;# in V\n", + "V2 = 3.;# in V\n", + "V3 = 6.;# in V\n", + "#Applying KCL at node A: VA*(R1*R2+R2*R3+R3*R1)+VB*-R1*R2 = V1*R2*R3+V2*R1*R2 (i)\n", + "#Applying KCL at node B: VA*R4*R5+VB*-(R3*R4+R4*R5+R5*R3) = V2*R4*R5-V3*R3*R4 (ii)\n", + "A=([[(R1*R2+R2*R3+R3*R1), R4*R5], [-R1*R2, -(R3*R4+R4*R5+R5*R3)]])\n", + "B= ([(V1*R2*R3+V2*R1*R2), (V2*R4*R5-V3*R3*R4)])\n", + "V= numpy.dot(B,numpy.linalg.inv(A));# Solving eq(i) and (ii) by Matrix method\n", + "VA= V[0];# in V\n", + "VB= V[1];# in V\n", + "I8_ohm= VA/R2;#The current through 8 ohm resistance in A\n", + "#results\n", + "print \"The current through 8 ohm resistance in A is : \",round(I8_ohm,4)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21: pg 32" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current drawn from the source in A is : 12.64\n" + ] + } + ], + "source": [ + "#pg 32\n", + "#calculate the current\n", + "# Given data\n", + "V= 100.;# in V\n", + "R12 = 3.;# in ohm\n", + "R31 = 2.;# in ohm\n", + "R23 = 4.;# in ohm\n", + "R4= 6.;# in ohm\n", + "R5=2.;# in ohm\n", + "R6= 5.;# in ohm\n", + "#calculations\n", + "R1 = (R12*R31)/(R12+R23+R31);# in ohm\n", + "R2 = (R31*R23)/(R12+R23+R31);# in ohm\n", + "R3 = (R23*R12)/(R12+R23+R31);# in ohm\n", + "R_S= R6+R1;# in ohm\n", + "R_P1= R2+R4;# in ohm\n", + "R_P2= R3+R5;# in ohm\n", + "R_P= R_P1*R_P2/(R_P1+R_P2);# in ohm\n", + "R= R_P+R_S;# in ohm\n", + "I= V/R;# in A\n", + "#results\n", + "print \"The current drawn from the source in A is : \",round(I,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 22: pg 33" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Part (i) : Using by KVL\n", + "The current through 10 ohm resistance in A is : 3.33\n", + "The current through 5 ohm resistance in A is : 13.33\n", + "The current through 20 ohm resistance in A is : 10.0\n", + "Part (ii) : Using by KVL\n", + "The current through 10 ohm resistance in A is : -3.33\n", + "The current through 5 ohm resistance in A is : 13.33\n", + "The current through 20 ohm resistance in A is : 10.0\n" + ] + } + ], + "source": [ + "#pg 33\n", + "#calculate the current using both methods\n", + "# Given data\n", + "R1= 10.;# in ohm\n", + "R2= 5.;# in ohm\n", + "R3= 20.;# in ohm\n", + "V= 100.;# in V\n", + "I2= 10.;# in A\n", + "#calculations\n", + "# Applying KVL in ABEFA : -R1*I1-R2*(I1+I2)+V= 0\n", + "I1= (V-R2*I2)/(R1+R2);# in A\n", + "I10_ohm= I1;#current through 10 ohm resistance in A\n", + "I5_ohm= I1+I2;#current through 5 ohm resistance in A\n", + "I20_ohm= I2;#current through 20 ohm resistance in A\n", + "print \"Part (i) : Using by KVL\"\n", + "print \"The current through 10 ohm resistance in A is : \",round(I10_ohm,2)\n", + "print \"The current through 5 ohm resistance in A is : \",round(I5_ohm,2)\n", + "print \"The current through 20 ohm resistance in A is : \",I20_ohm\n", + "# Applying KCL at node A :\n", + "VA= (V*R2+I2*R1*R2)/(R1+R2);# in V\n", + "I10_ohm= (VA-V)/R1;# in A\n", + "I5_ohm= VA/R2;# in A\n", + "I20_ohm= I2;# in A\n", + "print \"Part (ii) : Using by KVL\"\n", + "print \"The current through 10 ohm resistance in A is : \",round(I10_ohm,2)\n", + "print \"The current through 5 ohm resistance in A is : \",round(I5_ohm,2)\n", + "print \"The current through 20 ohm resistance in A is : \",I20_ohm\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23: pg 34" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current through 2 ohm resistor in A is : 5.0\n", + "The voltage across 2 ohm resistor in V is : 10.0\n" + ] + } + ], + "source": [ + "#pg 34\n", + "#calculate the current and voltage\n", + "# Given data\n", + "import numpy\n", + "R1= 5.;# in ohm\n", + "R2= 10.;# in ohm\n", + "R3= 3.;# in ohm\n", + "R4= 2.;# in ohm\n", + "V1= 10.;# in V\n", + "V2= 20.;# in V\n", + "I= 5.;# in A\n", + "#Applying KCL at node A: VA*(R1+R2)+VB*-R1 =I*R1*R2+V1*R1 (i)\n", + "#Applying KCL at node B: VA*R3*R4+VB*-(R2*R3+R4*R3+R4*R2) =V1*R3*R4+V2*R2*R3 (ii)\n", + "A=([[(R1+R2), R3*R4], [-R1, -(R3*R2+R4*R3+R4*R2)]])\n", + "B= ([(I*R1*R2+V1*R1), (V1*R3*R4+V2*R2*R3)]);\n", + "V= numpy.dot(B,numpy.linalg.inv(A));# Solving eq(i) and (ii) by Matrix method\n", + "VA= V[0];# in V\n", + "VB= V[1];# in V\n", + "I4= (VB+V2)/R4;# in A\n", + "V4= R4*I4;# in V\n", + "#results\n", + "print \"The current through 2 ohm resistor in A is : \",I4\n", + "print \"The voltage across 2 ohm resistor in V is : \",V4\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 24: pg 36" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage across 6 ohm resistor in V is : 12.0\n" + ] + } + ], + "source": [ + "#pg 36\n", + "#calculate the voltage\n", + "# Given data\n", + "import numpy\n", + "R1= 6.;# in ohm\n", + "R2= 12.;# in ohm\n", + "R3= 2.;# in ohm\n", + "R4= 6.;# in ohm\n", + "V2= 12.;# in V\n", + "V3= 30.;# in V\n", + "#calculations\n", + "#Applying KVL in ABEFA : I1*(R1+R2) - R2*I2=V3-V2 (i)\n", + "#Applying KVL in BCDEB : -I1*R2+I2*(R1+R2+R3)=V2 (ii)\n", + "A= ([[(R1+R2), -R2], [-R2, (R1+R2+R3)]]);\n", + "B= ([(V3-V2), (V2)]);\n", + "I= numpy.dot(B,numpy.linalg.inv(A));# Solving eq(i), and (ii) by Matrix method\n", + "I1= I[0];# in A\n", + "I2= I[1];# in A\n", + "V1= I2*R1;#voltage across 6 ohm resistor in V\n", + "#results\n", + "print \"The voltage across 6 ohm resistor in V is : \",V1\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25: pg 36" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance between the point B and C in ohm is : 1.3\n" + ] + } + ], + "source": [ + "#pg 36\n", + "#calculate the resistance\n", + "# Given data\n", + "R1 = 6.;# in ohm\n", + "R2 = 2.;# in ohm\n", + "R3 = 2.;# in ohm\n", + "R4 = 4.;# in ohm\n", + "R5 = 4.;# in ohm\n", + "R6 = 6.;# in ohm\n", + "#calculations\n", + "R12= R1*R2/(R1+R2);# in ohm\n", + "R34= R3*R4/(R3+R4);# in ohm\n", + "R56= R5*R6/(R5+R6);# in ohm\n", + "# Resistance between the point B and C\n", + "R_BC= (R12+R34)*R56/((R12+R34)+R56);# in ohm\n", + "#results\n", + "print \"The resistance between the point B and C in ohm is : \",round(R_BC,1)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 26: pg 37" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage across R1 resistor in V is : 30.0\n", + "The voltage across R2 resistor in V is : 30.0\n" + ] + } + ], + "source": [ + "#pg 37\n", + "#calculate the voltage across resistors\n", + "# Given data\n", + "R1 = 10.;# in ohm\n", + "R2 = 10.;# in ohm\n", + "R4 = 80.;# in ohm\n", + "V1= 100.;# in V\n", + "I2= 0.5;# in A\n", + "#calculations\n", + "V2= I2*R4;# in V\n", + "# Applying KVL : -R1*I1-V2+V1-R1*I2=0\n", + "I1= (V1-V2)/(R1+R2);# in A\n", + "V_R1= I1*R1;#voltage across R1 resistor in V\n", + "V_R2= I1*R2;#voltage across R2 resistor in V\n", + "#results\n", + "print \"The voltage across R1 resistor in V is : \",V_R1\n", + "print \"The voltage across R2 resistor in V is : \",V_R2\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 27: pg 38" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of I1 in A is : 2.06\n", + "The value of I2 in A is : -3.24\n" + ] + } + ], + "source": [ + "#pg 38\n", + "#calculate the currents\n", + "# Given data\n", + "import numpy\n", + "R1 = 8.;# in ohm\n", + "R2 = 4.;# in ohm\n", + "R3 = 4.;# in ohm\n", + "R4 = 4.;# in ohm\n", + "R5 = 8.;# in ohm\n", + "R6 = 8.;# in ohm\n", + "I=10.;# in A\n", + "V= 20.;# in V\n", + "#calculations\n", + "# Applying KVL in ABEFA : I1*(R1+R2+R3)+I2*(R3)= I*R2-V (i)\n", + "# Applying KVL in BCDEB : I1*R3-I2*(R3+R4+R5)= R4*I+V (ii)\n", + "A= ([[(R1+R2+R3), R3], [R3, -(R3+R4+R5)]]);\n", + "B= ([I*R2-V, R4*I+V]);\n", + "I= numpy.dot(B,numpy.linalg.inv(A));## Solving equations by matrix multiplication\n", + "I1= I[0];# in A\n", + "I2= I[1];# in A\n", + "#results\n", + "print \"The value of I1 in A is : \",round(I1,2)\n", + "print \"The value of I2 in A is : \",round(I2,2)\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter10.ipynb b/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter10.ipynb new file mode 100644 index 00000000..b4ea59b0 --- /dev/null +++ b/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter10.ipynb @@ -0,0 +1,935 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 10: DC Machines" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1: pg 329" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The emf generated in V is 624.0\n" + ] + } + ], + "source": [ + "#pg 329\n", + "#calculate the emf generated\n", + "# Given data\n", + "A = 2.;# in wavewound\n", + "N = 1200.;# in rpm\n", + "phi = 0.02;# in Wb\n", + "n = 65.;# no of slots\n", + "P = 4.;\n", + "#calculations\n", + "Z = n*12;# total number of conductor\n", + "# Emf equation\n", + "Eg = (N*P*phi*Z)/(60*A);# in V\n", + "#results\n", + "print \"The emf generated in V is\",Eg\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2: pg 329" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The numbers of conductors when armature is lap wound 880.0\n", + "The numbers of conductors when armature is wave wound 220.0\n" + ] + } + ], + "source": [ + "#pg 329\n", + "#calculate the number of conductors\n", + "# Given data\n", + "P = 8.;\n", + "N = 1200.;# in rpm\n", + "phi = 25.;# in mWb\n", + "phi = phi * 10**-3;# in Wb\n", + "Eg = 440.;# in V\n", + "A = P;\n", + "#calculations\n", + "# Eg = (N*P*phi*Z)/(60*A);\n", + "Z = (Eg*60*A)/(phi*N*P);# in conductors\n", + "print \"The numbers of conductors when armature is lap wound\",Z\n", + "A = 2;\n", + "# Eg = (N*P*phi*Z)/(60*A);\n", + "Z = (Eg*60*A)/(phi*N*P);# in conductors\n", + "print \"The numbers of conductors when armature is wave wound \",Z\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3: pg 330" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The induced voltage in V is 138.24\n" + ] + } + ], + "source": [ + "#pg 330\n", + "#calculate the induced voltage\n", + "# Given data\n", + "P = 4;\n", + "phi = 20;# in mWb\n", + "phi = phi * 10**-3;# in Wb\n", + "A = 4;\n", + "P = A;\n", + "N =720.;# in rpm\n", + "n = 144.;# no of slots in slots\n", + "n1 = 2.;# no of coils \n", + "n2 = 2;# no of turns in turns\n", + "#calculations\n", + "Z = n*n1*n2;# total number of conductor\n", + "# Generated emf\n", + "E = (N*P*phi*Z)/(60*A);# in V\n", + "#results\n", + "print \"The induced voltage in V is\",E\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4: pg 330" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Part (i) : The generated emf in V is 125.0\n", + "Part (ii) : The generated emf in V is 135.0\n" + ] + } + ], + "source": [ + "#pg 330\n", + "#calculate the emf\n", + "# Given data\n", + "Eg1 = 100.;# in V\n", + "phi1 = 20.;# in mWb\n", + "phi1 = phi1 * 10**-3;# in Wb\n", + "N1 = 800.;# in rpm\n", + "N2 = 1000.;# in rpm\n", + "#calculations\n", + "# Eg1/Eg2 = (phi1/phi2) * (N1/N2) but phi1 = phi2\n", + "Eg2 = (Eg1*N2)/N1;# in V\n", + "print \"Part (i) : The generated emf in V is\",Eg2\n", + "phi2 = 24;# in mWb\n", + "phi2 = phi2 * 10**-3;# in Wb\n", + "N2 = 900;# in rpm\n", + "# Eg1/Eg2 = (phi1/phi2) * (N1/N2) ;\n", + "Eg2 = (Eg1*N2*phi2)/(N1*phi1);# in V\n", + "print \"Part (ii) : The generated emf in V is\",Eg2\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5: pg 331" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total power developed by the armature in kW is 31.43\n" + ] + } + ], + "source": [ + "#pg 331\n", + "#calculate the power\n", + "# Given data\n", + "P = 30;# in kW\n", + "P = P * 10**3;# in W\n", + "V = 300.;# in V\n", + "Ra = 0.05;# in ohm\n", + "Rsh = 100;# in ohm\n", + "#calculations\n", + "# p = V*I_L;\n", + "I_L = P/V;# in A\n", + "Ish = V/Rsh;# in A\n", + "Ia = I_L+Ish;# in A\n", + "Eg = V + (Ia*Ra);# in V\n", + "# power developed by armature \n", + "power = (Eg*Ia);# in W\n", + "power = power * 10**-3;# in kW\n", + "#results\n", + "print \"The total power developed by the armature in kW is\",round(power,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6: pg 331" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The power developed in the armature in kW is 25.3\n" + ] + } + ], + "source": [ + "#pg 331\n", + "#calculate the power\n", + "# Given data\n", + "V = 200;# in V\n", + "Ra = 0.5;# in ohm\n", + "Rsh = 200;# in ohm\n", + "P = 20;# in kW\n", + "P = P * 10**3;# in W\n", + "#calculations\n", + "# P = V*I_L;\n", + "I_L =P/V;# in A\n", + "Ish = V/Rsh;# in A\n", + "Ia = I_L+Ish;# in A\n", + "Eg = V + (Ia*Ra);# in V\n", + "# power developed in the armature \n", + "power = Eg*Ia;# in W\n", + "power = power * 10**-3;# in kW\n", + "#results\n", + "print \"The power developed in the armature in kW is\",round(power,1)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7: pg 332" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total armature current in A is 31.25\n", + "The generated emf in V is 105.125\n" + ] + } + ], + "source": [ + "#pg 332\n", + "#calculate the armature current and emf\n", + "# Given data\n", + "P = 60.;\n", + "A =P;\n", + "Vbrush = 2;# in V/brush\n", + "Vt = 100.;# in V\n", + "Ra = 0.1;# in ohm\n", + "Rsh = 80;# in ohm\n", + "#calculations\n", + "Ish = Vt/Rsh;# in A\n", + "Ilamp = P/Vt;# in A\n", + "I_L = 50*Ilamp;# in A\n", + "# Armature current\n", + "Ia = I_L+Ish;# in A\n", + "# Evaluation of generated emf\n", + "Eg = Vt + (Ia*Ra) + Vbrush;# in V\n", + "#results\n", + "print \"The total armature current in A is\",Ia\n", + "print \"The generated emf in V is\",Eg\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8: pg 332" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The generated voltage for long shunt in V is 503.3\n", + "The generated voltage for short shunt in V is 501.2\n" + ] + } + ], + "source": [ + "#pg 332\n", + "#calculate the generated voltage\n", + "# Given data\n", + "V = 440.;# in V\n", + "I_L =40.;# in A\n", + "Rse = 1.;# in ohm\n", + "Rsh = 200.;# in ohm\n", + "Ra = 0.5;# in ohm\n", + "#calculations\n", + "Ish = V/Rsh;# in A\n", + "Ia = I_L+Ish;# in A\n", + "Eg = V + (Ia*(Ra+Rse));# in V\n", + "print \"The generated voltage for long shunt in V is\",Eg\n", + "#Voltage across shunt field, Vsh = V + Ise*Rse = V + (I_L*Rse);\n", + "Vsh = V+(I_L*Rse);# in V\n", + "Ish = Vsh/Rsh;# in A\n", + "Ia =I_L+Ish;# in A\n", + "Eg = V + (I_L*Rse) + (Ia*Ra);# in V\n", + "print \"The generated voltage for short shunt in V is\",Eg\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9: pg 341" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The back emf in V is 428.0\n" + ] + } + ], + "source": [ + "#pg 341\n", + "#calculate the back emf\n", + "# Given data\n", + "V = 440.;# in V\n", + "I = 80.;# in A\n", + "Rse = 0.025;# in ohm\n", + "Ra = 0.1;# in ohm\n", + "Bd = 2.;# brush drop in V\n", + "#calculations\n", + "Ia = I;# in A\n", + "Ise = I;# in A\n", + "Eb = V - (Ia*(Ra+Rse)) - Bd;# in V\n", + "#results\n", + "print \"The back emf in V is\",Eb\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10: pg 341" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The armature current in A is 18.75\n", + "The back emf in V is 244.375\n" + ] + } + ], + "source": [ + "#pg 341\n", + "#calculate the armature current and back emf\n", + "# Given data\n", + "V = 250.;# in V\n", + "I_L = 20;# in A\n", + "Ra = 0.3;# in ohm\n", + "Rsh = 200;# in ohm\n", + "#calculations\n", + "Ish = V/Rsh;# in A\n", + "# I_L = Ia+Ish;\n", + "Ia = I_L-Ish;# inA\n", + "Eb = V-(Ia*Ra);# in V\n", + "#results\n", + "print \"The armature current in A is\",Ia\n", + "print \"The back emf in V is\",Eb\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11: pg 342" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The speed in rpm is 1374.0\n" + ] + } + ], + "source": [ + "#pg 342\n", + "#calculate the speed in rpm\n", + "# Given data\n", + "P = 4.;\n", + "A = 2.;#(wave connected)\n", + "Z = 200.;\n", + "V=250.;# in V\n", + "phi = 25.;# in mWb\n", + "phi = phi * 10**-3;# in Wb\n", + "Ia = 60;# in A\n", + "I_L = 60;# in A\n", + "Ra = 0.15;# in ohm\n", + "Rse = 0.2;# in ohm\n", + "#calculations\n", + "#V = Eb + (Ia*Ra) + (Ia*Rse);\n", + "Eb = V - (Ia*Ra) - (Ia*Rse);# in V\n", + "# Eb = (phi*P*N*Z)/(60*A);\n", + "N = (Eb*60*A)/(phi*P*Z);# in rpm\n", + "#results\n", + "print \"The speed in rpm is\",N\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12: pg 343" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The armature resistance in ohm is 0.342\n", + "The armature current in A is 701.5\n" + ] + } + ], + "source": [ + "#pg 343\n", + "#calculate the resistance and current\n", + "# Given data\n", + "Eb = 227.;# in V\n", + "Rsh = 160.;# in ohm\n", + "Ish = 1.5;# in A\n", + "I_L = 39.5;# in A\n", + "#calculations\n", + "V = Ish*Rsh;# in V\n", + "Ia = I_L-Ish;# in A\n", + "#V = Eb + (Ia*Ra);\n", + "Ra = (V-Eb)/Ia;# in ohm\n", + "Ia = V/Ra;# in A\n", + "#results\n", + "print \"The armature resistance in ohm is\",round(Ra,3)\n", + "print \"The armature current in A is\",round(Ia,1)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13: pg 343" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The ratio of speed as a generator to speed as a motor is 1.105\n" + ] + } + ], + "source": [ + "#pg 343\n", + "#calculate the ratio of speed\n", + "# Given data\n", + "V = 230;# in V\n", + "Ra = 0.115;# in ohm\n", + "Rsh = 115.;# in ohm\n", + "I_L = 100.;# inA\n", + "#calculations\n", + "Ish =V/Rsh;# in A\n", + "Ia = I_L + Ish;# in A\n", + "Eg = V + (Ia*Ra);# in V\n", + "Ia = I_L-Ish;# in A\n", + "Eb = V - (Ia*Ra);# in V\n", + "# The ratio of speed as a generator to speed as a motor \n", + "NgBYNm = Eg/Eb;\n", + "#results\n", + "print \"The ratio of speed as a generator to speed as a motor is\",round(NgBYNm,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14: pg 344" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The induced voltage in V is 138.24\n" + ] + } + ], + "source": [ + "#pg 344\n", + "#calculate the induced voltage\n", + "# Given data\n", + "P = 4;\n", + "slots = 144.;\n", + "phi = 20.;# in mWb\n", + "phi = phi * 10**-3;# in Wb\n", + "N = 720.;# in rpm\n", + "A = 4.;\n", + "P =4.;\n", + "n1 = 2;# in coil/slot\n", + "n2 = 2;# in turns/coil\n", + "#calculations\n", + "Z = slots*n1*n2;# total number of conductor\n", + "Eg = (N*P*phi*Z)/(60*A);# in V\n", + "#results\n", + "print \"The induced voltage in V is\",Eg\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15: pg 344" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The emf when lap is connected in V is 200.0\n", + "The emf when wave is connected in V is 800.0\n" + ] + } + ], + "source": [ + "#pg 344\n", + "#calculate the emf\n", + "# Given data\n", + "P = 8;\n", + "phi = 0.1;# in Wb\n", + "Z = 400.;\n", + "N =300.;# in rpm\n", + "#calculations\n", + "Eg = (N*phi*Z)/(60);# in V (A = p)\n", + "print \"The emf when lap is connected in V is\",Eg\n", + "# For A=2, connected armature\n", + "A = 2;\n", + "Eg = (N*phi*P*Z)/(60*A);# in V\n", + "print \"The emf when wave is connected in V is\",Eg\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16: pg 345" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The power developed in armature in kW is 20.7\n" + ] + } + ], + "source": [ + "#pg 345\n", + "#calculate the power \n", + "# Given data\n", + "P_L = 20;# in kW\n", + "P_L = P_L * 10**3;# in W\n", + "V = 200;# in V\n", + "Ra = 0.05;# in ohm\n", + "Rsh = 200.;# in ohm\n", + "#calculations\n", + "# P_L = V*I_L;\n", + "I_L = P_L/V;# in A\n", + "Ish = V/Rsh;# in A\n", + "Ia = I_L+Ish;# in A\n", + "Eg = V + (Ia*Ra);# in V\n", + "Pa = Eg*Ia;# in W\n", + "Pa = Pa * 10**-3;# in kW\n", + "#results\n", + "print \"The power developed in armature in kW is\",round(Pa,1)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17: pg 345" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The speed when 30 A current through the armature in rpm is 615.38\n" + ] + } + ], + "source": [ + "#pg 345\n", + "#calculate the speed \n", + "# Given data\n", + "N1 = 600.;# inrpm\n", + "I_L1 = 60.;# in A\n", + "V = 230.;# in V\n", + "Rsh = 115.;# in ohm\n", + "Ra= 0.2;# in ohm\n", + "Ia2 = 30.;# in A\n", + "#calculations\n", + "Ish = V/Rsh;# in A\n", + "Ia1 = I_L1 - Ish;# in A\n", + "Eb1 = V-(Ia1*Ra);# in V\n", + "Eb2 = V - (Ia2*Ra);# in V\n", + "# N1/N2 = Eb1/Eb2;\n", + "N2 = (N1*Eb2)/Eb1;# in rpm\n", + "#results\n", + "print \"The speed when 30 A current through the armature in rpm is\",round(N2,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18: pg 346" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The speed of the motor in rpm is 565.2\n" + ] + } + ], + "source": [ + "#pg 346\n", + "#calculate the speed of motor\n", + "# Given data\n", + "P = 6;\n", + "A = 6.;\n", + "Z = 500.;\n", + "Ra = 0.05;# in ohm\n", + "Rsh =25.;# in ohm\n", + "V = 100.;# in V\n", + "I_L = 120.;# in A\n", + "phi = 2*10**-2;# in Wb\n", + "#calculations\n", + "Ish = V/Rsh;# in A\n", + "Ia = I_L-Ish;# in A\n", + "Eb = V - (Ia*Ra);# in V\n", + "# Eb = (N*P*phi*Z)/(60*A);\n", + "N = (Eb*60*A)/(P*phi*Z);# in rpm\n", + "#results\n", + "print \"The speed of the motor in rpm is\",N\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19: pg 346" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The change in emf in percent is 96.0\n" + ] + } + ], + "source": [ + "#pg 346\n", + "#calculate the change in emf\n", + "# Given ata\n", + "N1 = 1;\n", + "N2 = 1.2*N1;\n", + "phi1 = 1;\n", + "phi2 = 0.8*phi1;\n", + "Eg1BYEg2 = (N1/N2) * (phi1/phi2);\n", + "Eg1 = 1;# assumed\n", + "# The change in emf \n", + "#calculations\n", + "Eg2 = (Eg1*phi2*N2)/(phi1*N1);\n", + "Eg2 = Eg2 * 100;# in %\n", + "#results\n", + "print \"The change in emf in percent is\",Eg2\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20: pg 347" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total armature power developed in kW is 26.255\n", + "Total armature power developed when working as a motor in kW is 23.8046\n" + ] + } + ], + "source": [ + "#pg 347\n", + "#calculate the armature power\n", + "# Given data\n", + "Pout = 25.;# in kW\n", + "Pout = Pout*10**3;# in W\n", + "Vt = 250.;# in V\n", + "Ra = 0.06;# in ohm\n", + "Rsh = 100.;# in ohm\n", + "#calculations\n", + "# Pout = Vt*I_L;\n", + "I_L = Pout/Vt;# in A\n", + "Ish = Vt/Rsh;# in A\n", + "Ia = I_L+Ish;# in A\n", + "Eg = Vt + (Ia*Ra);# in V\n", + "# Total armature power developed when working as a generator \n", + "Pdeveloped = Eg*Ia;# in W\n", + "Pdeveloped = Pdeveloped * 10**-3;# in kW\n", + "print \"Total armature power developed in kW is\",round(Pdeveloped,3)\n", + "Ia = I_L-Ish;# in A\n", + "Eb = Vt - (Ia*Ra);# in V\n", + "# Total armature power developed when working as a motor \n", + "Pdeveloped = Eb*Ia;# in W\n", + "Pdeveloped = Pdeveloped * 10**-3;# in kW\n", + "print \"Total armature power developed when working as a motor in kW is\",round(Pdeveloped,4)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21: pg 347" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The useful flux per mole when armature is LAP connected in Wb is 0.11\n", + "The useful flux per mole when armature is WAVE connected in Wb is 0.055\n" + ] + } + ], + "source": [ + "#pg 347\n", + "#calculate the useful flux\n", + "# Given data\n", + "P = 4.;\n", + "A = 4.;\n", + "Turns = 100.;\n", + "N = 600.;# in rpm\n", + "Eg = 220.;# in V\n", + "n = 2.;# no of total conductors\n", + "Z = n*Turns;\n", + "#calculations\n", + "# Eg = (N*P*phi*Z)/(60*A);\n", + "phi = (Eg*60*A)/(N*P*Z);# in Wb\n", + "print \"The useful flux per mole when armature is LAP connected in Wb is\",phi\n", + "A = 2;\n", + "# Eg = (N*P*phi*Z)/(60*A);\n", + "phi = (Eg*60*A)/(N*P*Z);# in Wb\n", + "print \"The useful flux per mole when armature is WAVE connected in Wb is\",phi\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter11.ipynb b/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter11.ipynb new file mode 100644 index 00000000..ab730f25 --- /dev/null +++ b/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter11.ipynb @@ -0,0 +1,610 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 11: Induction Motors" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1: pg 363" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The speed of the motor in rpm is 1440.0\n", + "The synchronous speed in rpm is 1500.0\n", + "The rotor current frequency in Hz is 16.7\n" + ] + } + ], + "source": [ + "#pg 363\n", + "#calculate the synchronous speed, speed and rotor current\n", + "# Given data\n", + "P = 4.;\n", + "f = 50.;# in Hz\n", + "s = 4.;\n", + "#calculations\n", + "Ns = (120*f)/P;# in rpm\n", + "# s = ((Ns-N)/Ns)*100;\n", + "N = Ns - ( (s*Ns)/100. );# in rpm\n", + "print \"The speed of the motor in rpm is\",N\n", + "N = 1000.;# in rpm\n", + "s = ((Ns-N)/Ns);\n", + "f_desh= s*f;# in Hz\n", + "#results\n", + "print \"The synchronous speed in rpm is\",Ns\n", + "print \"The rotor current frequency in Hz is\",round(f_desh,1)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2: pg 363" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The slip in percentage is 4.0\n", + "The speed of the motor in rpm is 1440.0\n", + "The answer is a bit different due to rounding off error in textbook\n" + ] + } + ], + "source": [ + "#pg 363\n", + "#calculate the slip and speed\n", + "# Given data\n", + "f = 50.;# in Hz\n", + "P = 4.;\n", + "f_DASH = 2;# in Hz\n", + "#calculations\n", + "# f_DASH = s*f;\n", + "s = (f_DASH/f)*100;# in %\n", + "N_S = (120*f)/P;# in rpm\n", + "# s = (N_S-N)/N_S;\n", + "N = N_S - (s/100*N_S);# in rpm\n", + "#results\n", + "print \"The slip in percentage is\",s\n", + "print \"The speed of the motor in rpm is\",N\n", + "print 'The answer is a bit different due to rounding off error in textbook'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3: pg 364" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The synchronous speed in rpm is 1000.0\n", + "No load speed in rpm is 990.0\n", + "The full load speed in rpm is 970.0\n", + "The frequency of rotor current in Hz is 50.0\n", + "The frequency of rotor current at full load in Hz is 1.5\n", + "Note : The calculated value of Nnl is wrong and value of Nfl is correct but at last they printed wrong.\n" + ] + } + ], + "source": [ + "#pg 364\n", + "#calculate the speed, frequency\n", + "# Given data\n", + "P = 6.;\n", + "f = 50.;# in Hz\n", + "Snl = 1./100;\n", + "Sfl = 3./100;\n", + "#calculations\n", + "N_S = (120.*f)/P;# in rpm\n", + "print \"The synchronous speed in rpm is\",N_S\n", + "Nnl = N_S*(1-Snl);# in rpm\n", + "print \"No load speed in rpm is\",Nnl\n", + "Nfl = N_S*(1-Sfl);# in rpm.. correction \n", + "print \"The full load speed in rpm is\",Nfl\n", + "# frequency of rotor current \n", + "s = 1;\n", + "Sf = s*f;# in Hz\n", + "print \"The frequency of rotor current in Hz is\",Sf\n", + "# frequency of rotor current at full load \n", + "f_r = Sfl * f;# in Hz\n", + "print \"The frequency of rotor current at full load in Hz is\",f_r\n", + "\n", + "print 'Note : The calculated value of Nnl is wrong and value of Nfl is correct but at last they printed wrong.'\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4: pg 364" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The numbers of pole is : 4.0\n", + "The percentage slip is : 4.0\n" + ] + } + ], + "source": [ + "#pg 364\n", + "#calculate the numbers and percentage slip\n", + "# Given data\n", + "Pa= 12.;\n", + "N= 1440.;# in rpm\n", + "Na= 500.;# in rpm\n", + "Nm= 1450.;# in rpm\n", + "#calculations\n", + "fa= Pa*Na/120;# in Hz\n", + "Pm= round(120*fa/Nm);\n", + "# Synchronous speed of motor\n", + "Ns= 120*fa/Pm;# in rpm\n", + "s= (Ns-N)/Ns*100;# in percentage\n", + "#results\n", + "print \"The numbers of pole is : \",Pm\n", + "print \"The percentage slip is : \",s\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5: pg 365" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The frequency of rotor in Hz is 1.5\n", + "The magnitude of induced emf in V is 119.8\n", + "The magnitude of induced emf in the running condition in V is 3.594\n" + ] + } + ], + "source": [ + "#pg 365\n", + "#calculate the frequency, induced emf\n", + "# Given data\n", + "from math import sqrt\n", + "K = 1./2;\n", + "P = 4.;\n", + "f = 50.;# in Hz\n", + "N = 1445.;# in rpm\n", + "E1line = 415.;# in V\n", + "N = 1455.;# in rpm\n", + "#calculations\n", + "Ns = (120*f)/P;# in rpm\n", + "s = (Ns-N)/Ns*100;# in %\n", + "f_r = s/100*f;# in Hz\n", + "print \"The frequency of rotor in Hz is\",f_r\n", + "E1ph = E1line/sqrt(3);# in V\n", + "# E2ph/E1ph = K;\n", + "E2ph = E1ph*K;# in V\n", + "print \"The magnitude of induced emf in V is\",round(E2ph,1)\n", + "E2r = s/100*E2ph;# in V\n", + "print \"The magnitude of induced emf in the running condition in V is\",round(E2r,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6: pg 366" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of Ns in rpm is 1500.0\n", + "The rotor speed when slip is 4 percent in rpm is 1440.0\n", + "The rotor frequency when rotor runs at 600 rpm in Hz is 30.0\n" + ] + } + ], + "source": [ + "#pg 366\n", + "#calculate the rotor speed, frequency and Ns\n", + "# Given data\n", + "P = 4.;\n", + "S =4./100;\n", + "f = 50.;# in Hz\n", + "#calculations\n", + "Ns = (120*f/P);# in rpm\n", + "print \"The value of Ns in rpm is\",Ns\n", + "# The rotor speed when slip is 4 %\n", + "N = Ns*(1-S);# in rpm\n", + "print \"The rotor speed when slip is 4 percent in rpm is\",N\n", + "# The rotor speed when rotor runs at 600 rpm\n", + "N1 = 600;# in rpm\n", + "s1 = ((Ns-N1)/Ns)*100;# in %\n", + "f_r = (s1/100)*f;# in Hz\n", + "print \"The rotor frequency when rotor runs at 600 rpm in Hz is\",f_r\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7: pg 366" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Number of poles is 6.0\n", + "The percentage of full load slip in percent is 5.0\n", + "The rotor induced voltage in V is 5.0\n", + "The frequency at full load in Hz is 2.5\n" + ] + } + ], + "source": [ + "#pg 366\n", + "#calculate the poles, full load slip, induced voltage and frequency\n", + "# Given data\n", + "V_L = 230.;# in V\n", + "f = 50.;# in Hz\n", + "N = 950.;# in rpm\n", + "E2 = 100.;# in V\n", + "Ns =1000.;# in rpm\n", + "#calculations\n", + "# Ns = 120*f/P;\n", + "P = (120*f)/Ns;\n", + "print \"The Number of poles is\",P\n", + "s = ((Ns-N)/Ns)*100;# %s in %\n", + "print \"The percentage of full load slip in percent is\",s\n", + "# The rotor induced voltage at full load\n", + "E2r = (s/100)*E2;# in V\n", + "print \"The rotor induced voltage in V is\",E2r\n", + "# The rotor frequency at full load\n", + "f_r = (s/100)*f;# in Hz\n", + "print \"The frequency at full load in Hz is\",f_r\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8: pg 367" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The number of poles in the machine is 4.0\n", + "Speed of rotation air gap field in rpm is 1500.0\n", + "Produced emf in rotor in V is 352.0\n", + "The frequency of rotor current in Hz is 1.67\n" + ] + } + ], + "source": [ + "#pg 367\n", + "#calculate the poles, speed, emf and frequency\n", + "# Given data\n", + "V = 440.;# in V\n", + "f = 50.;# in Hz\n", + "N = 1450.;# in rpm\n", + "Ns = 1450.;# in rpm\n", + "Nr = 1450.;# in rpm\n", + "#calculations\n", + "P = round((120*f)/Ns);\n", + "print \"The number of poles in the machine is\",P\n", + "P = 4;\n", + "Ns = (120*f)/P;# in rpm\n", + "print \"Speed of rotation air gap field in rpm is\",Ns\n", + "k = 0.8/1;\n", + "# Pemf = k*E1 = k*V;\n", + "Pemf = k*V;# produced emf in rotor in V\n", + "print \"Produced emf in rotor in V is\",Pemf\n", + "s = ((Ns-Nr)/Ns)*100;# in %\n", + "Ivoltage = k*(s/100)*V;# rotor induces voltage in V\n", + "f_r = (s/100)*f;# in Hz\n", + "print \"The frequency of rotor current in Hz is \",round(f_r,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9: pg 367" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The full load slip in percent is 4.0\n", + "The corresponding speed in rpm is 720.0\n" + ] + } + ], + "source": [ + "#pg 367\n", + "#calculate the full load slip, speed\n", + "# Given data\n", + "P = 8.;\n", + "f = 50.;# in Hz\n", + "f_r = 2.;# in Hz\n", + "#calculations\n", + "# f_r = s*f;\n", + "s = (f_r/f)*100;# in %\n", + "# s = Ns-N/Ns;\n", + "Ns = (120*f)/P;# in rpm\n", + "N = Ns*(1-(s/100));# in rpm\n", + "#results\n", + "print \"The full load slip in percent is\",s\n", + "print \"The corresponding speed in rpm is\",N\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10: pg 368" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The speed at which maximum torque is developed in rpm is 1440.0\n" + ] + } + ], + "source": [ + "#pg 368\n", + "#calculate the speed\n", + "# Given data\n", + "R2 = 0.024;# in per phase\n", + "X2 = 0.6;# in ohm per phase\n", + "#calculations\n", + "s = R2/X2;\n", + "f = 50;# in Hz\n", + "P = 4;\n", + "Ns = (120*f)/P;# in rpm\n", + "# Speed corresponding to maximum torque\n", + "N = Ns*(1-s);# in rpm\n", + "#results\n", + "print \"The speed at which maximum torque is developed in rpm is\",N\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11: pg 368" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The synchronous speed in rpm is : 1800.0\n", + "The rotor speed in rpm is 1746.0\n", + "The rotor current frequency in Hz is 1.8\n", + "The rotor magnetic field rotates at speed in rpm is 54.0\n" + ] + } + ], + "source": [ + "#pg 368\n", + "#calculate the speed, frequency\n", + "# Given data\n", + "P = 4.;\n", + "f =60.;# in Hz\n", + "s = 0.03;\n", + "#calculations\n", + "Ns = (120*f)/P;# in rpm\n", + "N = Ns*(1-s);# in rpm\n", + "print \"The synchronous speed in rpm is : \",Ns\n", + "print \"The rotor speed in rpm is\",N\n", + "f_r = s*f;# in Hz\n", + "print \"The rotor current frequency in Hz is\",f_r\n", + "# Rotor magnetic field rorats at speed \n", + "Rm = (120*f_r)/P;# in rpm\n", + "print \"The rotor magnetic field rotates at speed in rpm is\",Rm\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12: pg 369" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The slip in percent is 4.0\n", + "The frequency of rotor induced emf in Hz is 2.0\n", + "The number of poles is 6.0\n", + "Speed of rotor field with respect to rotor structure in rpm is 40.0\n" + ] + } + ], + "source": [ + "#pg 369\n", + "#calculate the frequency, poles and speed\n", + "# Given data\n", + "N = 960.;# in rpm\n", + "f = 50.;# in Hz\n", + "Ns = 1000.;# in rpm\n", + "#calculations\n", + "s = ((Ns-N)/Ns)*100;# %s in %\n", + "print \"The slip in percent is\",s\n", + "f_r = (s/100)*f;# in Hz\n", + "print \"The frequency of rotor induced emf in Hz is\",f_r\n", + "# Ns = (120*f)/P;\n", + "P = (120*f)/Ns;\n", + "print \"The number of poles is\",P\n", + "# Speed of rotor field with respect to rotor structure \n", + "s1 = (120*f_r)/P;# in rpm\n", + "print \"Speed of rotor field with respect to rotor structure in rpm is\",s1\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13: pg 369" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The full load speed in rpm is 1440.0\n" + ] + } + ], + "source": [ + "#pg 369\n", + "#calculate the full load speed \n", + "# Given data\n", + "P = 4.;\n", + "f = 50.;# in Hz\n", + "Sfl = 4./100;\n", + "#calculations\n", + "Ns = (120*f)/P;# in rpm\n", + "# The full load speed, Sfl = (Ns-Nfl)/Ns;\n", + "Nfl = Ns - (Sfl*Ns);# in rpm\n", + "#results\n", + "print \"The full load speed in rpm is\",Nfl\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter2.ipynb b/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter2.ipynb new file mode 100644 index 00000000..2f855750 --- /dev/null +++ b/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter2.ipynb @@ -0,0 +1,1414 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2: Network Theorems" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1: pg 48" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current in A is 2.0\n" + ] + } + ], + "source": [ + "#pg 48\n", + "#calculate the current\n", + "# Given data\n", + "R1 = 6.;# in ohm\n", + "R2 = 6.;# in ohm\n", + "R3 = 6.;# in ohm\n", + "V = 24.;# in V\n", + "#calculations\n", + "R_T =R1+R1*R2/(R1+R2);# in ohm\n", + "I_T = V/R_T;# in A\n", + "I1 = (R1/(R1+R2))*I_T;# in A\n", + "V = 12;# in V\n", + "I_T = V/R_T;# in A\n", + "I2 = (R1/(R1+R2))*I_T;# in A\n", + "I = I1+I2;# in A\n", + "#results\n", + "print \"The current in A is\",I\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2: pg 51" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current through 12 ohm resistor in A is 0.59\n" + ] + } + ], + "source": [ + "#pg 51\n", + "#calculate the current\n", + "# Given data\n", + "R1 = 5.;# in ohm\n", + "Vth= 10.;# in ohm\n", + "R2 = 7.;# in ohm\n", + "R3=10.;# in ohm\n", + "R_L = 12.;# in ohm\n", + "V = 20.;# in ohm\n", + "#calculations\n", + "Vth = (Vth*V)/(R1+R3);# in V\n", + "Rth = R2 + ((Vth*R1)/(Vth+R1));# in ohm\n", + "# The current through 12 ohm resistor \n", + "I = Vth/(Rth+R_L);# in A\n", + "#results\n", + "print \"The current through 12 ohm resistor in A is\",round(I,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3: pg 54" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current through 12 ohm resistor in A is 0.561\n" + ] + } + ], + "source": [ + "#pg 54\n", + "#calculate the current\n", + "# Given data\n", + "R1 = 6.;# in ohm\n", + "R2 = 7.;# in ohm\n", + "R3 = 4.;# in ohm\n", + "R_L = 12.;# in ohm\n", + "V = 30.;# in V\n", + "#calculations\n", + "Vth = (R3*V)/(R3+R1);# in V\n", + "Rth = R2 + ((R3*R1)/(R3+R1)) ;# in ohm\n", + "I_N = Vth/Rth;# in A\n", + "#The current through 12 ohm resistor \n", + "I = (I_N*Rth)/(Rth+R_L);# in ohm\n", + "#results\n", + "print \"The current through 12 ohm resistor in A is\",round(I,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4: pg 54" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of load resistance in ohm is 10.33\n", + "The magnitude of maximum power in W is 4.3\n" + ] + } + ], + "source": [ + "#pg 57\n", + "#calculate the resistance and max power\n", + "# Given data\n", + "R1 = 5.;# in ohm\n", + "R2 = 10.;# in ohm\n", + "R3 = 7.;# in ohm\n", + "V = 20.;# in V\n", + "#calculations\n", + "Vth = R2*V/(R1+R2);# in V\n", + "Rth = R3 + ((R2*R1)/(R2+R1));# in ohm\n", + "R_L = Rth;# in ohm\n", + "Pmax = (Vth**2)/(4*R_L);# in W\n", + "#results\n", + "print \"The value of load resistance in ohm is\",round(R_L,2)\n", + "print \"The magnitude of maximum power in W is\",round(Pmax,1)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5: pg 59" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current across 4 ohm resistor in A is 1.53\n" + ] + } + ], + "source": [ + "#pg 59\n", + "#calculate the current\n", + "# Given data\n", + "V1 = 12.;# in V\n", + "V2 = 10.;# in V\n", + "R1 = 6.;# in ohm\n", + "R2 = 7.;# in ohm\n", + "R3 = 4.;# in ohm\n", + "#calculations\n", + "R_T = R1 + ( (R2*R3)/(R2+R3) );# in ohm\n", + "I_T = V1/R_T;# in A\n", + "I1 = (R2/(R2+R3))*I_T;# in A\n", + "R_T = R2 + ( (R1*R3)/(R1+R3) );# in ohm\n", + "I_T = V2/R_T;# in A\n", + "I2 = (R1*I_T)/(R1+R3);# in A\n", + "# current across 4 ohm resistor \n", + "I = I1+I2;# in A\n", + "#results\n", + "print \"The current across 4 ohm resistor in A is\",round(I,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6: pg 60" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current in the branch AB of the circuit in A is 0.1\n" + ] + } + ], + "source": [ + "#pg 60\n", + "#calculate the current\n", + "# Given data\n", + "R1 = 2.;# in ohm\n", + "R2 = 3.;# in ohm\n", + "R3 = 1.;# in ohm\n", + "R4= 2.;# in ohm\n", + "V1 = 4.2;# in V\n", + "V2 = 3.5;# in V\n", + "#calculations\n", + "R_T =R1+R3+R2*R4/(R2+R4);# in ohm\n", + "I_T = V1/R_T;# in A\n", + "I1 = (R1/(R1+R2))*I_T;# in A\n", + "R = R1+R3;# in ohm\n", + "R_desh = (R*R2)/(R+R2);# in ohm\n", + "R_T = R_desh+R1;# in ohm\n", + "I_T = V2/R_T;# in A\n", + "I2 = (R2/(R2+R))*I_T;# in A\n", + "# current in the branch AB \n", + "I = I2-I1;# in A\n", + "#results\n", + "print \"The current in the branch AB of the circuit in A is\",I\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7: pg 62" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current through in 8 ohm resistor in A is 1.6\n" + ] + } + ], + "source": [ + "#pg 62\n", + "#calculate the current\n", + "# Given data\n", + "R1 = 2.;# in ohm\n", + "R2 = 4.;# in ohm\n", + "R3 = 8.;# in ohm\n", + "Ig = 2.;# in A\n", + "V = 20.;# in V\n", + "#calculations\n", + "R_T = R1+R3;# in ohm\n", + "I1 = V/R_T;# in A\n", + "I2 = (R1/(R1+R3))*Ig;# in A\n", + "# current through in 8 ohm resistor \n", + "I = I1-I2;# in A\n", + "#results\n", + "print \"The current through in 8 ohm resistor in A is\",I\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8: pg 63" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current through 16 ohm resistor in A is 1.1029\n" + ] + } + ], + "source": [ + "#pg 63\n", + "#calculate the current\n", + "# Given data\n", + "R1 = 4.;# in ohm\n", + "R2 = 24.;# in ohm\n", + "R_L = 16.;# in ohm\n", + "V1 = 20.;# in V\n", + "V2 = 30.;# in V\n", + "#calculations\n", + "# V1-R1*I-R2*I-V2 = 0;\n", + "I= (V1-V2)/(R1+R2)\n", + "# V1-R1*I-Vth = 0;\n", + "Vth = V1-R1*I;# in V\n", + "Rth = (R1*R2)/(R1+R2);# in ohm\n", + "# current through 16 ohm resistor \n", + "I_L = Vth/(Rth+R_L);# in A\n", + "#results\n", + "print \"The current through 16 ohm resistor in A is\",round(I_L,4)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9: pg 64" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current through 6 ohm resistance in A is 1.0\n" + ] + } + ], + "source": [ + "#pg 64\n", + "#calculate the current\n", + "# Given data\n", + "R1 = 6.;# in ohm\n", + "R2 = 4.;# in ohm\n", + "R3 = 3.;# in ohm\n", + "R_L = 6.;# in ohm\n", + "V1 = 6.;# in V\n", + "V2 = 15.;# in V\n", + "#calculations\n", + "# V1 - R1*I - R3*I -V2 = 0\n", + "I= (V1-V2)/(R1+R3);\n", + "# Vth - R3*I -V2 = 0;\n", + "Vth =V2+R3*I;# in V\n", + "Rth = ((R1*R3)/(R1+R3)) + R2;# in ohm\n", + "# current through 6 ohm resistance \n", + "I_L = Vth/(Rth+R_L);# in A\n", + "#results\n", + "print \"The current through 6 ohm resistance in A is\",I_L\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10: pg 65" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current through 10 ohm resistance in A is 0.3235\n" + ] + } + ], + "source": [ + "#pg 65\n", + "#calculate the current\n", + "# Given data\n", + "R1 = 8.;# in ohm\n", + "R2 = 5.;# in ohm\n", + "R3 = 2.;# in ohm\n", + "R_L = 10.;# in ohm\n", + "V1= 20.;# in V\n", + "V2= 12.;# in V\n", + "#calculations\n", + "# V1-R3*I - R2*I = 0;\n", + "I = V1/(R2+R3);# in A\n", + "# Vth + V2 - R3*I = 0;\n", + "Vth = R3*I - V2;# in V\n", + "Rth = ((R2*R3)/(R2+R3)) + R1;# in ohm\n", + "# current through 10 ohm resistance \n", + "I_L = abs(Vth)/(Rth+R_L);# in A\n", + "#results\n", + "print \"The current through 10 ohm resistance in A is\",round(I_L,4)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11: pg 66" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current in 5 ohm resistance in A is 1.4\n" + ] + } + ], + "source": [ + "#pg 66\n", + "#calculate the current\n", + "# Given data\n", + "R1 = 4.;# in ohm\n", + "R2 = 3.;# in ohm\n", + "R3 = 2.;# in ohm\n", + "R_L = 5.;# in ohm\n", + "I = 6.;# in A\n", + "V = 15.;# in V\n", + "#calculations\n", + "# V-R1*I1-R3*(I1+I) = 0;\n", + "I1 = (V-R3*I)/(R1+R3);# in A\n", + "I = I1 + I;# in A\n", + "Vth = R3*I;# in V\n", + "Rth = ((R1*R3)/(R1+R3)) + R2;# in ohm\n", + "# current in 5 ohm resistance \n", + "I_L = Vth/(Rth+R_L);# in A\n", + "#results\n", + "print \"The current in 5 ohm resistance in A is\",round(I_L,1)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12: pg 68" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of Vth in volts is : 356.0\n", + "The value of Rth in ohm is : 40.0\n" + ] + } + ], + "source": [ + "#pg 68\n", + "#calculate the value of voltage and resistance\n", + "# Given data\n", + "R1 = 8.;# in ohm\n", + "R2 = 32.;# in ohm\n", + "V = 60.;# in V\n", + "I1= 5.;# in A\n", + "I2= 3.;# in A\n", + "#calculations\n", + "# Vth-R1*I1-(I1+I2)*R2-V=0\n", + "Vth= R1*I1+(I1+I2)*R2+V\n", + "Rth = R1+R2;# in ohm\n", + "#results\n", + "print \"The value of Vth in volts is : \",Vth\n", + "print \"The value of Rth in ohm is : \",Rth\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13: pg 69" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current through 6 ohm resistor in A is 1.0\n" + ] + } + ], + "source": [ + "#pg 69\n", + "#calculate the current\n", + "# Given data\n", + "R1 = 6.;# in ohm\n", + "R2 = 4.;# in ohm\n", + "R3 = 3.;# in ohm\n", + "R_L = 6.;# in ohm\n", + "V1 = 6.;# in V\n", + "V2= 15.;# in V\n", + "#calculations\n", + "# V1 - R1*I - R3*I -V2 = 0;\n", + "I= (V1-V2)/(R1+R3)\n", + "Vth = V2 + (R3*I);# in V\n", + "Rth = ((R1*R3)/(R1+R3)) + R2;# in ohm\n", + "I_N = Vth/Rth;# in A\n", + "# current through 6 ohm resistor \n", + "I = (I_N*Rth)/(Rth+R_L);# in A\n", + "#results\n", + "print \"The current through 6 ohm resistor in A is\",I\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14: pg 70" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current through 10 ohm resistace in A is 0.3235\n" + ] + } + ], + "source": [ + "#pg 70\n", + "#calculate the current\n", + "# Given data\n", + "R1 = 5.;# in ohm\n", + "R2 = 2.;# in ohm\n", + "R3 = 8.;# in ohm\n", + "V1 = 20.;# in V\n", + "V2 = 12.;# in V\n", + "#calculations\n", + "# V1-R2*I-R1*I = 0;\n", + "I = V1/(R1+R2);# in A\n", + "# Vth + V2 - R2*I = 0;\n", + "Vth = (R2*I) - V2;# in V\n", + "Rth = ((R1*R2)/(R1+R2)) + R3;# in ohm\n", + "I_N = Vth/Rth;# in A\n", + "R_L = 10;# in ohm\n", + "# current through 10 ohm resistace \n", + "I = (abs(I_N)*Rth)/(Rth+R_L);# in A\n", + "#results\n", + "print \"The current through 10 ohm resistace in A is\",round(I,4)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15: pg 71" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current through 5 ohm resistor in A is 1.393\n" + ] + } + ], + "source": [ + "#pg 71\n", + "#calculate the current\n", + "# Given data\n", + "V = 15.;# in V\n", + "R1 = 4.;# in ohm\n", + "R2 = 3.;# in ohm\n", + "R3 = 2.;# in ohm\n", + "R_L = 5.;# in ohm\n", + "Ig = 6.;# in A\n", + "#calculations\n", + "# V - R1*I1 - R3*(I1+Ig) = 0;\n", + "I1 = (V-R3*Ig)/(R1+R3);# in A\n", + "I = I1 + Ig;# in A\n", + "Vth = R3*I;# in V\n", + "Rth = ((R1*R3)/(R1+R3)) + R2;# in ohm\n", + "I_N = Vth/Rth;# in A\n", + "# current through 5 ohm resistor \n", + "I = (I_N*Rth)/(Rth+R_L);# in A\n", + "#results\n", + "print \"The current through 5 ohm resistor in A is\",round(I,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16: pg 72" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of R in ohm is 1.29\n" + ] + } + ], + "source": [ + "#pg 72\n", + "#calculate the resistance\n", + "# Given data\n", + "V = 6.;# in V\n", + "R1 = 2.;# in ohm\n", + "R2 = 1.;# in ohm\n", + "R3 = 3.;#in ohm\n", + "R4 = 2.;# in ohm\n", + "#calculations\n", + "Rth=(R1*R2/(R1+R2)+R3)*R4/((R1*R2/(R1+R2)+R3)+R4)\n", + "R_L = Rth;# in ohm\n", + "#results\n", + "print \"The value of R in ohm is\",round(R_L,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17: pg 73" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of load resistance in ohm is 9.0\n", + "The power delivered to the load in W is 2.778\n" + ] + } + ], + "source": [ + "#pg 73\n", + "#calculate the power and load resistance\n", + "# Given data\n", + "R1 = 10.;# in ohm\n", + "R2 = 10.;# in ohm\n", + "R3 = 4.;# in ohm\n", + "V = 20.;# in V\n", + "#calculations\n", + "# V - R1*I1 - R2*I1 = 0;\n", + "I1 = V/(R1+R2);# in A\n", + "Vth = R1*I1;# in V\n", + "Rth =R1*R2/(R1+R2)+R3\n", + "R_L = Rth;# in ohm\n", + "Pmax = (Vth**2)/(4*Rth);# in W\n", + "#results\n", + "print \"The value of load resistance in ohm is\",R_L\n", + "print \"The power delivered to the load in W is\",round(Pmax,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18: pg 74" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current through 6 ohm resistor in A is 9.09\n" + ] + } + ], + "source": [ + "#pg 74\n", + "#calculate the current\n", + "# Given data\n", + "R1 = 3.;# in ohm\n", + "R2 = 9.;# in ohm\n", + "R3 = 6.;# in ohm\n", + "V1 = 120.;# in V\n", + "V2 = 60.;# in V\n", + "#calculations\n", + "R = (R3*R2)/(R3+R2);# in ohm\n", + "R_T = R+R1;# in ohm\n", + "I_T = V1/R_T;# in A\n", + "I1 = (R2/(R2+R3)) * I_T;# in A\n", + "R_T = 2 + R2;# in ohm\n", + "I_T = V2/R_T;# in A\n", + "I2 = (R1/(R1+R3)) * I_T;# in A\n", + "# current through 6 ohm resistor \n", + "I = I1-I2;# in A\n", + "#results\n", + "print \"The current through 6 ohm resistor in A is\",round(I,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19: pg 75" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current in 8 ohm resistor in A is 5.03\n" + ] + } + ], + "source": [ + "#pg 75\n", + "#calculate the current\n", + "# Given data\n", + "R1 = 36.;# in ohm\n", + "R2 = 12.;# in ohm\n", + "R3 = 8.;# in ohm\n", + "V1 = 90.;# in V\n", + "V2 = 60.;# in V\n", + "#calculations\n", + "R_T = (R2*R3)/(R2+R3)+R1;# in ohm\n", + "I_T = V1/R_T;# in A\n", + "I1 = (R2/(R2+R3)) * I_T;# in A\n", + "R = (R1*R3)/(R1+R3);# in ohm\n", + "R_T = R2+R;# in ohm\n", + "I_T = V2/R_T;# in A \n", + "I2 = (R1/(R1+R3))*I_T;# in A\n", + "Ra = (R1*R2)/(R1+R2);# in ohm asumed\n", + "I_T = 2;# in A\n", + "I3 = (Ra/(Ra+R3))*I_T;# in A\n", + "# current in 8 ohm resistor \n", + "I = I1+I2+I3;# in A\n", + "#results\n", + "print \"The current in 8 ohm resistor in A is\",round(I,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20: pg 77" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Thevenins voltage in V is 45.0\n", + "The Thevenins resistance in ohm is 12.5\n" + ] + } + ], + "source": [ + "#pg 77\n", + "#calculate the thevenins voltage\n", + "# Given data\n", + "R1 = 5.;# in ohm\n", + "R2 = 10.;# in ohm\n", + "R3 = 5.;# in ohm\n", + "V1 = 60.;# in v\n", + "V2 = 30.;# in V\n", + "#calculations\n", + "#-R1*i1 - R3*i1 - V2+V1 = 0;\n", + "i1 = (V2-V1)/(R1+R3);# in A\n", + "V_acrossR3 = R3*i1;# in V\n", + "Vth = V_acrossR3+V1;# in V\n", + "V_AB =Vth;# in V\n", + "R = (R1*R3)/(R1+R3);# in ohm\n", + "Rth = R2+R;# in ohm\n", + "#results\n", + "print \"The Thevenins voltage in V is\",V_AB\n", + "print \"The Thevenins resistance in ohm is\",Rth\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21: pg 78" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current flowing through 5 ohm resistor in A is 0.321\n" + ] + } + ], + "source": [ + "#pg 78\n", + "#calculate the current\n", + "# Given data\n", + "R1 = 4.;# in ohm\n", + "R2 = 3.;# in ohm\n", + "R3 = 2.;# in ohm\n", + "R_L = 5.;# in ohm\n", + "V = 15.;# in V\n", + "I2 = 6.;# in A\n", + "#calculations\n", + "# -R1*I1 - R3*I1 + R3*I2 + V = 0;\n", + "I1 = (V+R3*I2)/(R1+R3);# in A\n", + "Vth = I2/R3;# in V\n", + "V_CD = Vth;# in V\n", + "Rth = (R1*R3)/(R1+R3)+R2;# in ohm\n", + "I = Vth/(Rth+R_L);# in A\n", + "#results\n", + "print \"The current flowing through 5 ohm resistor in A is\",round(I,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 22: pg 80" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of Rth in ohm is 9.0\n", + "The value of I_N in A is 0.667\n" + ] + } + ], + "source": [ + "#pg 80\n", + "#calculate the resistance and current\n", + "# Given data\n", + "R1 = 20.;# in ohm\n", + "R2 = 5.;# in ohm\n", + "R3 = 3.;# in ohm\n", + "R4 = 2.;# in ohm\n", + "V = 30.;# in V\n", + "I1=4.;# in A\n", + "#calculations\n", + "V1= I1*R3;# in V\n", + "# R1*I -R2*I+V = 0;\n", + "I = V/(R1+R2);# in A\n", + "V_acrossR2= R2*I;# in V\n", + "V_AB = V_acrossR2-V1;# in V\n", + "Vth = abs(V_AB);# in V\n", + "Rth = (R1*R2)/(R1+R2)+R3+R4;# in ohm\n", + "I_N = Vth/Rth;# in A\n", + "#results\n", + "print \"The value of Rth in ohm is\",Rth\n", + "print \"The value of I_N in A is\",round(I_N,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23: pg 81" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of Vth in volts is : 72.0\n", + "The value of Rth in ohm is : 7.0\n" + ] + } + ], + "source": [ + "#pg 81\n", + "#calculate the resistance and voltage\n", + "# Given data\n", + "R1 = 2.;# in ohm\n", + "R2 = 4.;# in ohm\n", + "R3 = 6.;# in ohm\n", + "R4 = 4.;# in ohm\n", + "V = 16.;# in v\n", + "I1= 8.;# in A\n", + "I2= 16;# in A\n", + "#calculations\n", + "V1= I1*R2;# in V\n", + "V2= I2*R3;# in V\n", + "# Applying KVL : R2*I+V1+R3*I-V2+V+R1*I\n", + "I= (V2-V1-V)/(R1+R2+R3);# in A\n", + "Vth= V2-R3*I;# in V\n", + "Rth= (R1+R2)*R3/((R1+R2)+R3)+R4;# in ohm\n", + "#results\n", + "print \"The value of Vth in volts is : \",Vth\n", + "print \"The value of Rth in ohm is : \",Rth\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 24: pg 82" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of load resistance in ohm is 2.43\n" + ] + } + ], + "source": [ + "#pg 82\n", + "#calculate the load resistance\n", + "# Given data\n", + "R1 = 3.;# in ohm\n", + "R2 = 2.;# in ohm\n", + "R3 = 1.;# in ohm\n", + "R4 = 8.;# in ohm\n", + "R5 = 2.;# in ohm\n", + "V = 10.;# in V\n", + "#calculations\n", + "R = ((R1+R2)*R5)/((R1+R2)+R5);# in ohm\n", + "Rth = R + R3;# in ohm\n", + "R_L = Rth;# in ohm\n", + "#results\n", + "print \"The value of load resistance in ohm is\",round(R_L,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25: pg 84" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of R_L in ohm is 25.0\n", + "The value of maximum power in W is 1806.25\n" + ] + } + ], + "source": [ + "#pg 84\n", + "#calculate the resistance and max power\n", + "# Given data\n", + "V = 250.;# in V\n", + "R1 = 10.;# in ohm\n", + "R2 = 10.;# in ohm\n", + "R3 = 10.;# in ohm\n", + "R4 = 10.;# in ohm\n", + "I2 = 20.;# in A.\n", + "#calculations\n", + "#Applying KVL in GEFHG : -R1*I1-R2*I1-R2*I2 + V = 0;\n", + "I1= (V-R2*I2)/(R1+R2);# in A\n", + "V_AB= R3*I2+V-R1*I1;# in V\n", + "Vth = V_AB;# in V\n", + "Rth = (R1*R2)/(R1+R2)+R3+R4;# in ohm\n", + "R_L = Rth;# in ohm\n", + "Pmax = (Vth**2)/(4*R_L);#maximum power in W\n", + "#results\n", + "print \"The value of R_L in ohm is\",R_L\n", + "print \"The value of maximum power in W is\",Pmax\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 26: pg 85" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current in A is 1.5\n", + "Note: At last, there is calculation error to find the value of I, so the answer in the book is wrong.\n" + ] + } + ], + "source": [ + "#pg 85\n", + "#calculate the current\n", + "# Given data\n", + "R1 = 2.;# in ohm\n", + "R2 = 4.;# in ohm\n", + "R_L = 4.;# in ohm\n", + "V1 = 6.;# in v\n", + "V2 = 12.;# in V\n", + "#calculations\n", + "# -R2*Ix -R1*Ix-V1+V2= 0;\n", + "Ix = (V2-V1)/(R1+R2);# in A\n", + "Vth = V1+R1*Ix;# in V\n", + "Rth = (R1*R2)/(R1+R2);# in ohm\n", + "I_N = Vth/Rth;# in A\n", + "I = (I_N*Rth)/(Rth+R_L);# in A\n", + "#results\n", + "print \"The current in A is\",I\n", + "\n", + "print 'Note: At last, there is calculation error to find the value of I, so the answer in the book is wrong.'\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 27: pg 86" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current in 4 ohm resistor in A is : 4.0\n" + ] + } + ], + "source": [ + "#pg 86\n", + "#calculate the current\n", + "# Given data\n", + "import numpy\n", + "R1 = 3.;# in ohm\n", + "R2 = 6.;# in ohm\n", + "R_L = 4.;# in ohm\n", + "V = 27.;# in V\n", + "I=3.;# in A\n", + "#calculations\n", + "# -I1+I2= I (i)\n", + "# Applying KVL: I1*R1+I2*R2=V (ii)\n", + "A= ([[-1, R1], [1, R2]]);\n", + "B= ([I, V])\n", + "I= numpy.dot(B,numpy.linalg.inv(A));# Solving eq(i) and (2) by Matrix method\n", + "I1= I[0];# in A\n", + "I2= I[1];# in A\n", + "Vth= R2*I2;# in V\n", + "Rth= R1*R2/(R1+R2);# in ohm\n", + "# current in 4 ohm resistor \n", + "I= Vth/(Rth+R_L);# in A\n", + "#results\n", + "print \"The current in 4 ohm resistor in A is : \",I\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 28: pg 88" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current in 20 ohm resistor in A is 2.274\n" + ] + } + ], + "source": [ + "#pg 88\n", + "#calculate the current\n", + "# Given data\n", + "R1 = 20.;# in ohm\n", + "R2 = 12.;# in ohm\n", + "R3 = 8.;# in ohm\n", + "V1 = 90.;# in V\n", + "V2 = 60.;# in V\n", + "#calculations\n", + "R_T = R1 + ((R2*R3)/(R2+R3));# in ohm\n", + "I_T = V1/R_T;# in A\n", + "I1 = I_T;# in A\n", + "R_T = R2 + ((R1*R3)/(R1+R3));# in ohm\n", + "I_T = V2/R_T;# in A\n", + "I2 = (R3/(R3+R1))*I_T;# in A\n", + "R_T = R1 + ((R2*R3)/(R2+R3));# in ohm\n", + "I_T = 2;# in A (given)\n", + "R = (R2*R3)/(R2+R3);# in ohm\n", + "I3 = (R/(R1+R))*I_T;# in A\n", + "# current in 20 ohm resistor \n", + "I20 = I1-I2-I3;# in A\n", + "#results\n", + "print \"The current in 20 ohm resistor in A is\",round(I20,3)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 29 :pg 89" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current through R2 resistor in A is 2.1818\n" + ] + } + ], + "source": [ + "#pg 89\n", + "#calculate the current\n", + "# Given data\n", + "R1 = 10.;# in ohm\n", + "R2 = 20.;# in ohm\n", + "R3 = 60.;# in ohm\n", + "R4 = 30.;# in ohm\n", + "E1 = 120.;# in V\n", + "E2 = 60.;# in V\n", + "#calculations\n", + "R_T = ((R2*R3)/(R2+R3)) + R4+R1;# in ohm\n", + "I_T = E1/R_T;# in A\n", + "I1 = (R3/(R2+R3))*I_T;# in A\n", + "R_T = ( ((R1+R4)*R2)/((R1+R4)+R2) ) + R3;# in ohm\n", + "I_T = E2/R_T;# in A\n", + "I2 = ((R1+R4)/(R1+R4+R2))*I_T;# in A\n", + "# current through R2 resistor \n", + "I= I1+I2;# in A\n", + "#results\n", + "print \"The current through R2 resistor in A is\",round(I,4)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 30: pg 91" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total current through upper 4 ohm resistor in A is : 0.0\n", + "Total current through lower 4 ohm resistor in A is : 3.75\n", + "Total current through 8 ohm resistor in A is : 3.75\n" + ] + } + ], + "source": [ + "#pg 91\n", + "#calculate the current in all cases\n", + "# Given data\n", + "R1 = 4.;# in ohm\n", + "R2 = 4.;# in ohm\n", + "R3 = 8.;# in ohm\n", + "Ig = 3.;# in A \n", + "V = 15.;# in V\n", + "I3 = 0.;# in A\n", + "#calculations\n", + "I1 = R1/(R1+R2)*Ig;# in A\n", + "I2 = -I1;# in A\n", + "R_T = ((R1+R2)*R3)/((R1+R2)+R3);# in ohm\n", + "I_T = V/R_T;# in A\n", + "I_2= R3/(R1+R2+R3)*I_T;# in A\n", + "I_1 = I_2;# in A\n", + "# Total current through upper 4 ohm resistor \n", + "tot_cur_up_4ohm= I1+I2;# in A\n", + "# Total current through lower 4 ohm resistor \n", + "tot_cur_low_4ohm= I_1+I_2;# in A\n", + "# Total current through 8 ohm resistor \n", + "tot_cur_8ohm= I3+I_T;# in A\n", + "#results\n", + "print \"Total current through upper 4 ohm resistor in A is : \",tot_cur_up_4ohm\n", + "print \"Total current through lower 4 ohm resistor in A is : \",tot_cur_low_4ohm\n", + "print \"Total current through 8 ohm resistor in A is : \",tot_cur_8ohm\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 31: pg 92" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total current through upper in 5 ohm resistor in A is 0.0\n", + "The total current through lower in 5 ohm resistor in A is 2.0\n", + "The total current through in 10 ohm resistor in A is 1.0\n" + ] + } + ], + "source": [ + "#pg 92\n", + "#calculate the total current in all cases\n", + "# Given data\n", + "R1 = 5.;# in ohm\n", + "R2 = 5.;# in ohm\n", + "R3 = 10.;# in ohm\n", + "V = 10.;# in V\n", + "Ig = 2.;# in A \n", + "#calculations\n", + "I2 = (R1/R3)*Ig;# in A\n", + "I1 = I2;# in A\n", + "I3 = 0;# in A\n", + "R_T = ((R1+R2)*R3)/((R1+R2)+R3);# in ohm\n", + "I_T = V/R_T;# in A\n", + "I_2 = (R3/((R1+R2)+R3))*I_T;# in A\n", + "I_1 = I_2;# in A\n", + "I_3 = I_1;# in A\n", + "# Total current through upper in 5 ohm resistor \n", + "tot_cur_up_5ohm = I1-I2;# in A\n", + "# Total current through lower in 5 ohm resistor \n", + "tot_cur_low_5ohm = I_1+I_2;# in A\n", + "# Total current through 10 ohm resistor \n", + "tot_cur_10ohm = I3+I_3;# in A\n", + "#results\n", + "print \"The total current through upper in 5 ohm resistor in A is\",tot_cur_up_5ohm\n", + "print \"The total current through lower in 5 ohm resistor in A is\",tot_cur_low_5ohm\n", + "print \"The total current through in 10 ohm resistor in A is\",tot_cur_10ohm\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter3.ipynb b/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter3.ipynb new file mode 100644 index 00000000..952cf6cb --- /dev/null +++ b/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter3.ipynb @@ -0,0 +1,647 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3: AC Fundamentals" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2: pg 119" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum value of current in A is 141.4\n", + "The frequency in Hz is 50.0\n", + "The time period in sec is 0.02\n", + "The instantaneous value in A is 114.36\n" + ] + } + ], + "source": [ + "#pg 119\n", + "#calculate the max. current, frequency and instantaneous current\n", + "import math\n", + "# Given data\n", + "Im = 141.4;# in A\n", + "t = 3;# in ms\n", + "t = t * 10**-3;# in sec\n", + "#calculations\n", + "omega = 314;# in rad/sec\n", + "# omega = 2*%pi*f;\n", + "f = round(omega/(2*math.pi));# in Hz\n", + "T = 1/f;# in sec\n", + "i = 141.4 * math.sin(omega*t);# in A\n", + "#results\n", + "print \"The maximum value of current in A is\",Im\n", + "print \"The frequency in Hz is\",f\n", + "print \"The time period in sec is\",T\n", + "print \"The instantaneous value in A is\",round(i,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3: pg 119" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of current after 1/360 sec in A is 103.92\n", + "The time taken to reach 96 A for the first time in sec is 0.00246\n" + ] + } + ], + "source": [ + "#pg 119\n", + "#calculate the current and time taken\n", + "# Given data\n", + "from math import sin,asin,pi\n", + "f = 60.;# in Hz\n", + "Im = 120.;# in A\n", + "t = 1/360.;# in sec\n", + "#calculations\n", + "omega = 2*pi*f;# in rad/sec\n", + "i = Im*sin(omega*t);# in A\n", + "print \"The value of current after 1/360 sec in A is\",round(i,2)\n", + "i = 96;# in A\n", + "# i = Im*sind(omega*t);\n", + "t = (asin(i/Im))/omega;# in sec\n", + "print \"The time taken to reach 96 A for the first time in sec is\",round(t,5)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4: pg 120" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The average value in A is 15.0\n", + "The RMS value in A is 17.8\n", + "The form factor is 1.19\n", + "The peak factor is 1.69\n" + ] + } + ], + "source": [ + "#pg 120\n", + "#calculate the average, RMS values of current. also find form and peak factors\n", + "from math import sqrt\n", + "# Given data\n", + "i1 = 0;# in A\n", + "i2 = 10.;# in A\n", + "i3 = 20.;# in A\n", + "i4 = 30.;# in A\n", + "i5 = 20.;#in A\n", + "i6 = 10.;# in A\n", + "n = 6.;# unit less\n", + "#calculations\n", + "Iav = (i1+i2+i3+i4+i5+i6)/n;# in A\n", + "print \"The average value in A is\",Iav\n", + "Irms = sqrt(( (i1**2) + (i2**2) + (i3**2) + (i4**2) + (i5**2) + (i6**2) )/n);# in A\n", + "print \"The RMS value in A is\",round(Irms,1)\n", + "k_f = Irms/Iav;# unit less\n", + "print \"The form factor is\",round(k_f,2)\n", + "Im = 30;# in A\n", + "k_p = Im/Irms;# unit less\n", + "print \"The peak factor is\",round(k_p,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5: pg 121" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The phase difference in degree is 105.0\n" + ] + } + ], + "source": [ + "#pg 121\n", + "#calculate the phase difference\n", + "# Given data\n", + "theta1 = 60.;# in degree\n", + "theta2 = -45.;# in degree\n", + "#calculations\n", + "# phase difference \n", + "phi = theta1-theta2;# in degree\n", + "#results\n", + "print \"The phase difference in degree is\",phi\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6: pg 121" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The sum of V1 and V2 is : \n", + "87.178 sin (theta -23.41 ) V\n", + "The difference of V1 and V2 is : \n", + "52.915 sin (theta+ 40.893 ) V\n" + ] + } + ], + "source": [ + "#pg 121\n", + "#calculate the sum and difference\n", + "# Given data\n", + "import cmath,numpy\n", + "from math import cos,sin,atan,pi\n", + "V1= 60*(cos(0*pi/180) + 1j*sin(0*pi/180));# in V\n", + "V2= 40*(cos(-pi/3.) +1j*sin(-pi/3));# in V\n", + "#calculations\n", + "add_V= V1+V2;# in V\n", + "diff_V= V1-V2;# in V\n", + "#results\n", + "print \"The sum of V1 and V2 is : \"\n", + "print round(abs(add_V),3),\" sin (theta\",round(numpy.angle(add_V)*180/pi,2),\") V\"\n", + "print \"The difference of V1 and V2 is : \"\n", + "print round(abs(diff_V),3),\" sin (theta+\",round(numpy.angle(diff_V)*180/pi,3),\") V\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7: pg 121" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The average value of output voltage in volts is : 0.318 *Vo or Vo/pi\n", + "The R.M.S value of output voltage in volts is : 0.5 *Vo or Vo/2\n", + "The form factor is : 1.57\n" + ] + } + ], + "source": [ + "#pg 121\n", + "#calculate the average, rms value of current and form factor\n", + "# Given data\n", + "import math,scipy\n", + "from scipy import integrate\n", + "from math import cos,sqrt,pi,sin\n", + "Vo= 1;# in V (assumed)\n", + "#calculations\n", + "fun1 = lambda theta:Vo*sin(theta)\n", + "fun2= lambda theta: Vo**2*(1-cos(2*theta))/2\n", + "Vav= scipy.integrate.quad(fun1,0,pi)[0]/(2*pi);\n", + "Vrms= sqrt(scipy.integrate.quad(fun2,0,pi)[0])*sqrt(1./(2*pi));\n", + "kf= Vrms/Vav;\n", + "#results\n", + "print \"The average value of output voltage in volts is : \",round(Vav,3),\"*Vo or Vo/pi\"\n", + "print \"The R.M.S value of output voltage in volts is : \",Vrms,\"*Vo or Vo/2\"\n", + "print \"The form factor is : \",round(kf,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8: pg 123" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The average value of voltage in V is 6.67\n", + "The R.M.S value of voltage in V is 11.5\n" + ] + } + ], + "source": [ + "#pg 123\n", + "#calculate the average and rms value of voltage\n", + "# Given data\n", + "from scipy import integrate\n", + "from math import sqrt\n", + "T = 0.3;# in sec\n", + "V = 20;# in V\n", + "#calculations\n", + "fun1 = lambda t:1\n", + "Vav = 1/T*V*integrate.quad(fun1,0,0.1)[0]\n", + "Vrms =sqrt(1/T*V**2*integrate.quad(fun1,0,0.1)[0]) \n", + "#results\n", + "print \"The average value of voltage in V is\",round(Vav,2)\n", + "print \"The R.M.S value of voltage in V is\",round(Vrms,1)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9: pg 123" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rectangular form of the voltage in V is : 61.24 +1j* 35.36\n", + "Polar form of the voltage :\n", + "Magnitude of voltage in V is : 70.7107 V\n", + "Angle is : 30.0 degree\n" + ] + } + ], + "source": [ + "#pg 123\n", + "#calculate the polar and rectangular form of voltage\n", + "# Given data\n", + "import cmath,numpy\n", + "from math import pi,sqrt,cos,sin\n", + "Vm = 100;# in V\n", + "phi = pi/6;# in degree\n", + "#calculations\n", + "Vrms = Vm/(sqrt(2.));# in V\n", + "# Rectangular form of the voltage \n", + "RectForm= Vrms*(cos(phi) + 1j*sin(phi))\n", + "#results\n", + "print \"Rectangular form of the voltage in V is : \",round(RectForm.real,2),\"+1j*\",round(RectForm.imag,2)\n", + "print \"Polar form of the voltage :\"\n", + "print \"Magnitude of voltage in V is : \",round(abs(RectForm),4),\" V\"\n", + "print \"Angle is : \",round(numpy.angle(RectForm)*180/pi,3),\" degree\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10: pg 123" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The R.M.S. value of the resultant in volts is : 216.42\n" + ] + } + ], + "source": [ + "#pg 123\n", + "#calculate the rms value \n", + "# Given data\n", + "import cmath,math,scipy\n", + "from scipy.linalg import expm\n", + "from math import sqrt,pi\n", + "from cmath import exp\n", + "V1= 100/sqrt(2)*exp(1j*0.*pi/180);# in V\n", + "V2= 200/sqrt(2)*exp(1j*60.*pi/180);# in V\n", + "V3= 50/sqrt(2)*exp(1j*-90.*pi/180);# in V\n", + "V4= 150/sqrt(2)*exp(1j*-45.*pi/180);# in V\n", + "#calculations\n", + "# The R.M.S. value of the resultant \n", + "V_R= V1.real+V2.real+V3.real+V4.real;# in V\n", + "#results\n", + "print \"The R.M.S. value of the resultant in volts is : \",round(V_R,2)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11: pg 124" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of current after 1/200 sec in A is 10\n", + "The time to reach 10 A in ms is 1.9357\n", + "The average value in A is 9.555\n" + ] + } + ], + "source": [ + "#pg 124\n", + "#calculate the current and time\n", + "from math import pi,sin,asin\n", + "# Given data\n", + "Im = 15.;# in A\n", + "f = 60.;# in Hz\n", + "#calculations\n", + "omega = 2*pi * f;# in rad/sec\n", + "t = 1/200.;# in sec\n", + "i = Im*sin(omega*t);# in A\n", + "i = 10;# in A\n", + "# i = Im*sind(omega*t);\n", + "print \"The value of current after 1/200 sec in A is\",i\n", + "t = (asin(i/Im))/omega;# in sec\n", + "t = t * 10**3;# in ms\n", + "print \"The time to reach 10 A in ms is\",round(t,4)\n", + "Iav = Im*0.637;# in A\n", + "print \"The average value in A is\",Iav\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12: pg 125" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum value of current in A is 42.42\n", + "The frequency in Hz is 100.0\n", + "The rms value in A is 30.0\n", + "The average value in A is 27.0\n", + "The form factor is 1.11\n" + ] + } + ], + "source": [ + "#pg 125\n", + "#calculate the current, frequency, form factor\n", + "# Given data\n", + "from math import sqrt,pi,sin\n", + "Im = 42.42;# in A\n", + "omega = 628;# in rad/sec\n", + "t = 1/6.977;# in sec assumed \n", + "#calculations\n", + "i = Im*sin(omega*t/180*pi);# in A\n", + "print \"The maximum value of current in A is\",round(i,2)\n", + "# omega = 2*%pi*f;\n", + "f = omega/(2*pi);# in Hz\n", + "print \"The frequency in Hz is\",round(f)\n", + "Irms = Im/(sqrt(2));# in A\n", + "print \"The rms value in A is\",round(Irms)\n", + "Iav = (2*Im)/pi;# in A\n", + "print \"The average value in A is\",round(Iav)\n", + "k_f = Irms/Iav;\n", + "print \"The form factor is\",round(k_f,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13: pg 125 " + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The power factor is 0.866\n", + "The R.M.S value of current in A is 15.5563\n", + "The frequency in Hz is 50.0\n" + ] + } + ], + "source": [ + "#pg 125\n", + "#calculate the power factor, rms value and frequency\n", + "# Given data\n", + "from math import cos,pi,sqrt\n", + "phi = pi/6;\n", + "#calculations\n", + "# Power factor\n", + "powerfactor = cos(phi);# in lag\n", + "print \"The power factor is\",round(powerfactor,3)\n", + "Im = 22.;# in A\n", + "# The R.M.S value of current \n", + "Irms = Im/sqrt(2);# in A\n", + "print \"The R.M.S value of current in A is\",round(Irms,4)\n", + "omega = 314;# in rad/sec\n", + "# omega = 2*%pi*f;\n", + "f = omega/(2*pi);# in Hz\n", + "print \"The frequency in Hz is\",round(f)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14: pg 126" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The R.M.S value of current in A is : 70.7107\n", + "The average value of current in A is : 63.662\n", + "The form factor is : 1.11\n" + ] + } + ], + "source": [ + "#pg 126\n", + "#calculate tghe form factor, average and rms value of current\n", + "# Given data\n", + "from math import sqrt,sin,cos,pi\n", + "from scipy import integrate\n", + "Im= 100.;# in A\n", + "#calculations\n", + "fun1 = lambda theta:1-cos(2*theta)\n", + "Irms= sqrt(Im**2/2*integrate.quad(fun1,0,pi)[0]/pi);# in A\n", + "print \"The R.M.S value of current in A is : \",round(Irms,4)\n", + "fun2 = lambda theta:sin(theta)\n", + "Iav= Im*integrate.quad(fun2,0,pi)[0]/pi;# in A\n", + "print \"The average value of current in A is : \",round(Iav,4)\n", + "# The form factor \n", + "kf= Irms/Iav;\n", + "print \"The form factor is : \",round(kf,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15: pg 127" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The form factor is : 1.155\n" + ] + } + ], + "source": [ + "#pg 127\n", + "# Given data\n", + "from math import sqrt\n", + "from scipy import integrate\n", + "A= 2*10;# area under curve for a cycle\n", + "B= 2;# base of half cycle\n", + "#calculations\n", + "Vav= 1./2*A/B;# in V\n", + "# For line AB\n", + "y1= 0;\n", + "y2= 10.;\n", + "x1= 0;\n", + "x2= 1.;\n", + "m_for_AB= (y2-y1)/(x2-x1);\n", + "# For line BC\n", + "y1= 10.;\n", + "y2= 0;\n", + "x1= 1;\n", + "x2= 2;\n", + "m_for_BC= (y2-y1)/(x2-x1);\n", + "fun1=lambda t:(m_for_AB*t)**2\n", + "fun2= lambda t:(m_for_BC*t+20)**2\n", + "Vrms= sqrt((integrate.quad(fun1,0,1)[0]+integrate.quad(fun2,1,2)[0])/2.);# in V\n", + "kf= Vrms/Vav;\n", + "print \"The form factor is : \",round(kf,3)\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter4.ipynb b/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter4.ipynb new file mode 100644 index 00000000..aafaef41 --- /dev/null +++ b/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter4.ipynb @@ -0,0 +1,1556 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4: Analysis of Single Phase AC Circuit" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1: pg 167" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The currrent in A is 23.0\n", + "The power consumed in W is 5290.0\n", + "Voltage equation : v = 325.27 sin ( 314.0 t)\n", + "Current equation : i = 32.53 sin ( 314.0 t)\n" + ] + } + ], + "source": [ + "# Exa 4.1\n", + "#pg 167\n", + "#calculate the current and power consumed\n", + "# Given data\n", + "from math import sqrt,pi\n", + "R = 10.;# inohm\n", + "V = 230.;# in V\n", + "f = 50.;# in Hz\n", + "#calculations\n", + "I = V/R;# in A\n", + "print \"The currrent in A is\",I\n", + "P =V*I;# in W\n", + "print \"The power consumed in W is\",P\n", + "Vm = sqrt(2)*V;# in V\n", + "Im =sqrt(2)*I;# in A\n", + "omega = 2*pi*f;# in rad/sec\n", + "#Equation for voltage: V = Vm*sind(omega*t) \n", + "#Equation for current: i = Im*sind(omega*t)\n", + "print \"Voltage equation : v = \",round(Vm,2),\" sin (\",round(omega),\" t)\"\n", + "print \"Current equation : i = \",round(Im,2),\" sin (\",round(omega),\" t)\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2: pg 167" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Instantaneous power taken by resistor in watts is : \n", + "450.0 (1+cos(2*omega*t))\n", + "The average power in watts is : 450.0\n" + ] + } + ], + "source": [ + "# Exa 4.2\n", + "#pg 167\n", + "#calculate the instantaneous and average power\n", + "# Given data\n", + "from math import pi,cos\n", + "R = 100.;# in ohm\n", + "i= '3*cos(omega*t)';# in A\n", + "#calculations\n", + "A= R*3**2;# assumed\n", + "P= R*3**2/2.*(1+cos(pi/2));# in watts\n", + "#results\n", + "print \"Instantaneous power taken by resistor in watts is : \"\n", + "print round(A/2.),\" (1+cos(2*omega*t))\"\n", + "print \"The average power in watts is : \",P\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3: pg 168" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Inductive reactance in ohm is 23.0\n", + "Inductance of the coil in H is 0.0732\n", + "Voltage equation : v = 325.27 sin ( 314.0 t)\n", + "Current equation : i = 14.14 sin ( 314.0 t - pi/2)\n" + ] + } + ], + "source": [ + "# Exa 4.38\n", + "#pg 168\n", + "from math import pi,sqrt,sin\n", + "# Given data\n", + "I = 10.;# in A\n", + "V = 230.;# in V\n", + "f = 50.;# in Hz\n", + "#calculations\n", + "X_L = V/I;# in ohm\n", + "# X_L = 2*%pi*f*L;\n", + "Vrms = V;# in V\n", + "Irms = I;# in A\n", + "Vm = Vrms*sqrt(2);# in V\n", + "Im = Irms*sqrt(2);# in A\n", + "omega = 2*pi*f;# in rad/sec\n", + "L = X_L/(2*pi*f);# in H\n", + "#results\n", + "#Equation for voltage: V = Vm*sind(omega*t) \n", + "#Equation for current: i = Im*sind(omega*t)\n", + "print \"Inductive reactance in ohm is\",X_L\n", + "print \"Inductance of the coil in H is\",round(L,4)\n", + "print \"Voltage equation : v = \",round(Vm,2),\" sin (\",round(omega),\" t)\"\n", + "print \"Current equation : i = \",round(Im,2),\" sin (\",round(omega),\" t - pi/2)\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4: pg 168" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The capacitive reactance in ohm is 10.0\n", + "The R.M.S value of current in A is 23.0\n", + "Voltage equation : v = 325.27 sin ( 314.0 t)\n", + "Current equation : i = 32.5 sin ( 314.0 t + pi/2)\n" + ] + } + ], + "source": [ + "# Exa 4.4\n", + "#pg 168\n", + "#calculate the voltage and current equations\n", + "# GIven data\n", + "from math import pi,sqrt\n", + "C = 318.;# in muF\n", + "C = C * 10**-6;# in F\n", + "V = 230.;# in V\n", + "f = 50.;# in Hz\n", + "#calculations\n", + "X_C = 1/(2*pi*f*C);# in ohm\n", + "print \"The capacitive reactance in ohm is\",round(X_C)\n", + "I = V/X_C;# in A\n", + "print \"The R.M.S value of current in A is\",round(I)\n", + "Vrms = V;# in V\n", + "Irms = I;# in A\n", + "Vm = Vrms*sqrt(2);# in V\n", + "Im = Irms*sqrt(2);# in A\n", + "omega = 2*pi*f;# in rad/sec\n", + "# V = Vm*sind(omega*t);\n", + "# i = Im*sind((omega*t)+(%pi/2));\n", + "#Equation for voltage: V = Vm*sind(omega*t) \n", + "#Equation for current: i = Im*sind(omega*t)\n", + "print \"Voltage equation : v = \",round(Vm,2),\" sin (\",round(omega),\" t)\"\n", + "print \"Current equation : i = \",round(Im,2),\" sin (\",round(omega),\" t + pi/2)\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5: pg 169" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The circuit current in A is 18.85\n", + "The phase angle in degree is 55.0\n", + "The power factor is 0.574\n", + "The power consumed in W is 2488.49\n" + ] + } + ], + "source": [ + "# Exa 4.5\n", + "#pg 169\n", + "#calculate the power consumed, phase angle and circuit current\n", + "# Given data\n", + "from math import pi,sqrt,cos,atan\n", + "R = 7;# in ohm\n", + "L = 31.8;# in mH\n", + "L = L * 10**-3;# in H\n", + "V = 230.;# in V\n", + "f = 50.;# in Hz\n", + "#calculations\n", + "X_L = 2*pi*f*L;# in ohm\n", + "Z = sqrt( (R**2)+(X_L**2) );# in ohm\n", + "I = V/Z;# in A\n", + "print \"The circuit current in A is\",round(I,2)\n", + "# tand(phi) = X_L/R;\n", + "phi = atan(X_L/R);# in radians lag\n", + "print \"The phase angle in degree is\",round(phi*180/pi,0)\n", + "# Power factor\n", + "powerfactor = cos(phi);# in lag\n", + "print \"The power factor is\",round(powerfactor,3)\n", + "P = V*I*cos(phi);# in W\n", + "print \"The power consumed in W is\",round(P,2)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6: pg 169" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of R in ohm is : 23.04\n", + "The value of L in H is : 0.055\n" + ] + } + ], + "source": [ + "# Exa 4.6\n", + "#pg 169\n", + "#calculate the value of resistance and inductance\n", + "# Given data\n", + "from math import acos,cos,pi\n", + "from cmath import exp\n", + "P = 400.;# in W\n", + "f = 50.;# in Hz\n", + "V = 120.;# in V\n", + "phi= acos(0.8);# in radians\n", + "# P =V*I*cos(phi);\n", + "#calculations\n", + "I = P/(V*cos(phi));# in A\n", + "Z= V/I;# in ohm\n", + "Z= Z*exp(1j*phi);# ohm\n", + "R= Z.real;# in ohm\n", + "XL= Z.imag;# in ohm\n", + "# Formula XL= 2*%pi*f*L\n", + "L= XL/(2*pi*f);# in H\n", + "#results\n", + "print \"The value of R in ohm is : \",round(R,2)\n", + "print \"The value of L in H is : \",round(L,3)\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7: pg 170" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The active component of current in A is : 8.66\n", + "The reactive component of current in A is : 5.0\n", + "The active power in W is : 1732.9\n", + "The reactive power in VAR is : 1000.0\n", + "Note: There is calculation error to evaluate the value of P, so the answer in the book is wrong.\n" + ] + } + ], + "source": [ + "# Exa 4.7\n", + "#pg 170\n", + "#calculate the active and reactive components of power and current\n", + "from math import sin,cos,acos,sqrt,pi\n", + "# Given data\n", + "R = 17.32;# in ohm\n", + "L = 31.8;# in mH\n", + "L = L * 10**-3;# in H\n", + "V = 200.;# in V\n", + "f = 50.;# in Hz\n", + "#calculations\n", + "X_L = 2*pi*f*L;# in ohm\n", + "Z = sqrt( (R**2) + (X_L**2) );# in ohm\n", + "I = V/Z;# in A\n", + "phi =acos( R/Z);# in radians\n", + "ActiveCom= I*cos(phi);# in A\n", + "ReactiveCom= I*sin(phi);# in A\n", + "print \"The active component of current in A is : \",round(ActiveCom,2)\n", + "print \"The reactive component of current in A is : \",round(ReactiveCom)\n", + "P= V*I*cos(phi);# in W\n", + "print \"The active power in W is : \",round(P,1)\n", + "Q= V*I*sin(phi);# in VAR\n", + "print \"The reactive power in VAR is : \",round(Q)\n", + "\n", + "print 'Note: There is calculation error to evaluate the value of P, so the answer in the book is wrong.'\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8: pg 170" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage across 20 ohm resistor in V is : 152.7\n", + "The voltage across 200 muF capacitor in V is 121.54\n", + "The voltage across the circuit in V is 195.2\n" + ] + } + ], + "source": [ + "#pg 170\n", + "#calculate the voltage in all cases\n", + "from math import sin,sqrt,pi\n", + "# Given data\n", + "R = 20.;# in ohm\n", + "C = 200.;# in muF\n", + "C=C*10**-6\n", + "f =50.;# in Hz\n", + "#I = 10.8 sin(314*t)\n", + "Im = 10.8;# in A\n", + "#calculations\n", + "I = Im/sqrt(2);# in A\n", + "V_R = I*R;# in V\n", + "print \"The voltage across 20 ohm resistor in V is : \",round(V_R,1)\n", + "#Vc = I*X_C and X_C = 1/omega*C;\n", + "omega = 2*pi*f;# in rad/sec\n", + "Vc = I * 1/(omega*C);# in V\n", + "print \"The voltage across 200 muF capacitor in V is\",round(Vc,2)\n", + "V = sqrt( (V_R**2) + (Vc**2) );# in V\n", + "print \"The voltage across the circuit in V is\",round(V,1)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9: pg 171" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Part (a)\n", + "The value of resistance in ohm is : 12.0\n", + "The value of inductance in mH is : 79.58\n", + "Part (b)\n", + "The value of resistance in ohm is : 0.0\n", + "The value of inductance in muF is : 44.21\n", + "Part (c)\n", + "The value of resistance in ohm is : 10.0\n", + "The value of inductance in mH is : 45.94\n" + ] + } + ], + "source": [ + "#pg 171\n", + "#calculate the resistance and inductance in all cases\n", + "from math import pi,sqrt\n", + "import cmath\n", + "# Given data\n", + "f= 60.;# in Hz\n", + "print \"Part (a)\"\n", + "Z= 12.+30*1j;\n", + "R= Z.real # in ohm\n", + "XL= Z.imag;# in ohm\n", + "# Formula XL= 2*%pi*f*L\n", + "#calculations\n", + "L= XL/(2*pi*f);# in H\n", + "L= L*10**3;# in mH\n", + "print \"The value of resistance in ohm is : \",R\n", + "print \"The value of inductance in mH is : \",round(L,2)\n", + "L= L*10**-3;# in H\n", + "print \"Part (b)\"\n", + "Z= 0-60*1j;\n", + "R= Z.real;# in ohm\n", + "XC= (abs(Z.imag));# in ohm\n", + "# Formula XC= 1/(2*%pi*f*C)\n", + "C= 1/(2*pi*XC*f);# in H\n", + "C= C*10**6;# in muF\n", + "print \"The value of resistance in ohm is : \",R\n", + "print \"The value of inductance in muF is : \",round(C,3)\n", + "C= C*10**-6;# in F\n", + "print \"Part (c)\"\n", + "Z= 20*cmath.exp(60*1j*pi/180.)\n", + "R= Z.real;# in ohm\n", + "XL= Z.imag;# in ohm\n", + "# Formula XL= 2*%pi*f*L\n", + "L= XL/(2*pi*f);# in H\n", + "L= L*10**3;# in mH\n", + "print \"The value of resistance in ohm is : \",R\n", + "print \"The value of inductance in mH is : \",round(L,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10: pg 171" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The power factor is : 0.4327\n", + "Supply voltage : \n", + "Magnitude is : 249.5776 V and angle is : -64.359 degree\n", + "The voltage across resistance in V is : 108.0\n", + "The voltage across capacitance in V is : 225.0\n", + "The active power in W is : 97.2\n", + "The reactive power in VAR is : -202.5\n" + ] + } + ], + "source": [ + "#pg 171\n", + "#calculate the power factor and voltage in all cases\n", + "import math,cmath\n", + "# Given data\n", + "R = 120.;# in ohm\n", + "XC = 250.;# in ohm\n", + "I = 0.9;# in A\n", + "#calculations\n", + "Z= R-1j*XC;# in ohm\n", + "phi= cmath.phase(Z)\n", + "V=I*Z;# in V\n", + "VR = I*R;# in V\n", + "VC= I*XC;# in V\n", + "P= abs(V)*I*math.cos(phi);# in W\n", + "Q= abs(V)*I*math.sin(phi);# in VAR\n", + "print \"The power factor is : \",round(math.cos(phi),4)\n", + "print \"Supply voltage : \"\n", + "print \"Magnitude is : \",round(abs(V),4),\" V and angle is : \",round(cmath.phase(V)*180/math.pi,3),\" degree\"\n", + "print \"The voltage across resistance in V is : \",VR\n", + "print \"The voltage across capacitance in V is : \",VC\n", + "print \"The active power in W is : \",P\n", + "print \"The reactive power in VAR is : \",Q\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11: pg 172" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The impedance in ohm is 449.3\n", + "The current in A is 0.512\n", + "The phase angle between current and voltage is : 89.7 lead\n", + "The power factor is : 0.00556 lead\n", + "The power consumed in W is 0.655\n" + ] + } + ], + "source": [ + "#pg 172\n", + "#calculate the power, impedance, current and phase angle\n", + "import math,cmath\n", + "from math import pi,sqrt,cos,acos,atan\n", + "# Given data\n", + "V = 230.;# in V\n", + "f = 50.;# in Hz\n", + "L = 0.06;# in H\n", + "R = 2.5;# in ohm\n", + "C = 6.8;# in microF\n", + "#calculations\n", + "C = C * 10**-6;# in F\n", + "X_L = 2*pi*f*L;# in ohm\n", + "X_C = 1/(2*pi*f*C);# in ohm\n", + "Z = sqrt( (R**2) + ((X_L-X_C)**2) );# in ohm\n", + "print \"The impedance in ohm is\",round(Z,1)\n", + "I = V/Z;# in A\n", + "print \"The current in A is\",round(I,3)\n", + "# tan(phi) = (X_L-X_C)/R;\n", + "phi = atan( (X_L-X_C)/R );# in lead\n", + "print \"The phase angle between current and voltage is : \",round(abs(phi)*180/pi,1),\" lead\"\n", + "phi = acos(R/Z);\n", + "print \"The power factor is : \",round(math.cos(phi),5),\" lead\"\n", + "P = V*I*cos(phi);# in W\n", + "print \"The power consumed in W is\",round(P,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12: pg 173" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resonant frequency in Hz is 1591549.43\n", + "The current at resonance in A is 0.1\n", + "The voltage across L at resonance in V is 100.0\n", + "The voltage across C at resonance in V is 100.0\n", + "The Q-factor is : 10.0\n" + ] + } + ], + "source": [ + "#pg 173\n", + "#calculate the voltage, current, frequency\n", + "from math import pi,sqrt\n", + "# GIven data\n", + "R = 100.;# in ohm\n", + "L = 100.;# in muH\n", + "L = L * 10**-6;# in H\n", + "C = 100.;# in pF\n", + "C = C * 10**-12;# in F\n", + "V = 10;# in V\n", + "#calculations\n", + "# The resonant frequency \n", + "f_r = 1/(2*pi*sqrt(L*C));# in Hz \n", + "print \"The resonant frequency in Hz is\",round(f_r,2)\n", + "# current at resonance \n", + "Ir = V/R;# in A\n", + "print \"The current at resonance in A is\",Ir\n", + "X_L = 2*pi*f_r*L;# in ohm \n", + "# voltage across L at resonance \n", + "V_L = Ir*X_L;# in V\n", + "print \"The voltage across L at resonance in V is\",V_L\n", + "X_C = X_L;# in ohm\n", + "# voltage across C at resonance \n", + "V_C = Ir*X_C;# in V\n", + "print \"The voltage across C at resonance in V is\",V_C\n", + "Q= 1/R*sqrt(L/C);\n", + "print \"The Q-factor is : \",Q\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13: pg 174" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The frequency at resonace in Hz is 56.2698\n", + "The current in A is 10.0\n", + "The power in W is 1000.0\n", + "The voltage across R in V is 100.0\n", + "The voltage across L in V is 707.107\n", + "The voltage across C in V is 707.107\n", + "The quality factor is 7.071\n", + "The half power frequencies are : 52.2909 Hz and 60.2486 Hz\n", + "The bandwidth in Hz is : 7.9577\n" + ] + } + ], + "source": [ + "#pg 174\n", + "#calculate the quality,frequency,current,power \n", + "from math import pi,sqrt\n", + "# Given data\n", + "R = 10.;# in ohm\n", + "L = 0.2;# in H\n", + "C = 40.;# in muF\n", + "C = C * 10**-6;# in F\n", + "V = 100;# in V\n", + "#calculations\n", + "f_r = 1/(2*pi*sqrt(L*C));# in Hz\n", + "print \"The frequency at resonace in Hz is\",round(f_r,4)\n", + "Im = V/R;# in A\n", + "print \"The current in A is\",Im\n", + "Pm = (Im**2)*R;# in W\n", + "print \"The power in W is\",Pm\n", + "# voltage across R \n", + "V_R = Im*R;# in V\n", + "print \"The voltage across R in V is\",V_R\n", + "X_L = 2*pi*f_r*L;# in ohm\n", + "# voltage across L \n", + "V_L = Im*X_L;# in V\n", + "print \"The voltage across L in V is\",round(V_L,3)\n", + "X_C = 1/(2*pi*f_r*C);# in ohm\n", + "# voltage across C \n", + "V_C = Im*X_C;# in V\n", + "print \"The voltage across C in V is\",round(V_C,3)\n", + "omega = 2*pi*f_r;# in rad/sec\n", + "Q = (omega*L)/R;\n", + "print \"The quality factor is\",round(Q,3)\n", + "del_F = R/(4*pi*L);\n", + "f1 = f_r-del_F;# in Hz\n", + "f2 = f_r+del_F;# in Hz\n", + "print \"The half power frequencies are : \",round(f1,4),\" Hz and \",round(f2,4),\" Hz\"\n", + "BW = f2-f1;# in Hz\n", + "print \"The bandwidth in Hz is : \",round(BW,4)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14: pg 175" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The bandwidth in kHz is 106.1\n" + ] + } + ], + "source": [ + "#pg 175\n", + "#calculate the bandwidth\n", + "from math import sqrt,pi\n", + "# Given data\n", + "R = 10.;# in ohm\n", + "L = 15.;# in muH\n", + "L = L * 10**-6;# in H\n", + "C = 100;# in pF\n", + "C = C * 10**-12;# in F\n", + "#calculations\n", + "f_r = 1/(2*pi*sqrt(L*C));# in Hz\n", + "X_L = 2*pi*f_r*L;# in ohm\n", + "Q = X_L/R;# in ohm\n", + "BW = f_r/Q;# in Hz\n", + "BW = BW * 10**-3;# in kHz\n", + "#results\n", + "print \"The bandwidth in kHz is\",round(BW,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15: pg 175" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resonant frequency in kHz is 159.0\n", + "The quality factor is 100.0\n", + "The half point frequencies are : 158.4 Hz and 160.0 Hz\n" + ] + } + ], + "source": [ + "#pg 175\n", + "#calculate the quality, frequencies\n", + "# Given data\n", + "from math import pi,sqrt\n", + "R = 1000.;# in ohm\n", + "L = 100.;# in mH\n", + "L = L * 10**-3;# in H\n", + "C = 10;# in muF\n", + "C = C * 10**-12;# in F\n", + "#calculations\n", + "f_r = 1/(2*pi*sqrt(L*C));# in Hz\n", + "print \"The resonant frequency in kHz is\",round(f_r*10**-3)\n", + "Q = (1/R)*(sqrt(L/C));\n", + "print \"The quality factor is\",Q\n", + "f1 = f_r - R/(4*pi*L);# in Hz\n", + "f1 = f1 * 10**-3;# in kHz\n", + "f2 = f_r + R/(4*pi*L);# in Hz\n", + "f2 = f2 * 10**-3;# in kHz\n", + "print \"The half point frequencies are : \",round(f1,1),\" Hz and \",round(f2,1),\" Hz\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16: pg 176" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The line current in A is 25.73\n", + "The power factor is : 0.447 lag\n", + "The power consumed in W is 2645.0\n" + ] + } + ], + "source": [ + "#pg 176\n", + "#calculate the line current, power factor and power consumed\n", + "# Given data\n", + "from math import sqrt,cos,acos,pi\n", + "R = 20.;# in ohm\n", + "L = 31.8;# in mH\n", + "L = L * 10**-3;# in H\n", + "V = 230.;# in V\n", + "f = 50.;# in Hz\n", + "#calculations\n", + "I_R = V/R;# in A\n", + "X_L = 2*pi*f*L;# in ohm\n", + "I_L = V/X_L;# in A\n", + "I = sqrt( (I_R**2) + (I_L**2) );# in A\n", + "phi= acos( I_R/I);\n", + "P = V*I*cos(phi);# in W\n", + "#results\n", + "print \"The line current in A is\",round(I,2)\n", + "print \"The power factor is : \",round(cos(phi),3),\" lag\"\n", + "print \"The power consumed in W is\",P\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17: pg 177" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The line current in A is 6.42\n", + "The power factor is : 0.964 lag\n", + "The power consumed in W is 1236.97\n" + ] + } + ], + "source": [ + "#pg 177\n", + "#calculate the line current, power\n", + "# Given data\n", + "from math import pi,sqrt,atan,sin,cos\n", + "import math\n", + "C = 50.;# in muF\n", + "C = C * 10**-6;# in F\n", + "R = 20.;# in ohm\n", + "L = 0.05;# in H\n", + "V = 200.;# in V\n", + "f = 50.;# in Hz\n", + "#calculations\n", + "X_C = 1/(2*pi*f*C);# in ohm\n", + "Z1 = X_C;# in ohm\n", + "I1 = V/X_C;# in A\n", + "X_L = 2*pi*f*L;# in ohm\n", + "Z2 = sqrt( (R**2) + (X_L**2) );# in ohm\n", + "I2 = V/Z2;# in A\n", + "# tan(phi2) = X_L/R;\n", + "phi2 = atan(X_L/R);# in degree\n", + "phi1 = 90*math.pi/180.;# in degree\n", + "I_cos_phi = I1*cos(phi1) + I2*cos(phi2);# in A \n", + "I_sin_phi = I1*sin(phi1) - I2*sin(phi2);# in A \n", + "phi= atan(I_sin_phi/I_cos_phi);# in radians\n", + "I= sqrt(I_cos_phi**2+I_sin_phi**2);# in A\n", + "P= V*I*cos(phi);# in W\n", + "#results\n", + "print \"The line current in A is\",round(I,2)\n", + "print \"The power factor is : \",round(cos(phi),3),\" lag\"\n", + "print \"The power consumed in W is\",round(P,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18: pg 178" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The phase angle in degrees is : 4.786\n", + "The power factor is : 0.997 lag\n" + ] + } + ], + "source": [ + "#pg 178\n", + "#calculate the phase angle and power factor\n", + "# Given data\n", + "import cmath,math\n", + "V= 68+154*1j;# in V\n", + "I1= 10+14*1j;# in A\n", + "I2= 2+8*1j;# in A\n", + "#calculations\n", + "I= I1+I2;# in A\n", + "phi= cmath.phase(V) - cmath.phase(I);# in radians\n", + "#results\n", + "print \"The phase angle in degrees is : \",round(phi*180./math.pi,3)\n", + "print \"The power factor is : \",round(math.cos(phi),3),\" lag\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19: pg 178" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Supply current : \n", + "Magnitude is : 3.93 A\n", + "Angle : -15.9 degrees\n", + "Power factor is : 0.962 lag\n" + ] + } + ], + "source": [ + "#pg 178\n", + "#calculate the supply current, power factor\n", + "# Given data\n", + "from math import pi,cos\n", + "import cmath\n", + "R1 = 50.;# in ohm\n", + "L = 318.;# in mH\n", + "L = L * 10**-3;# in H\n", + "R2 = 75.;# in ohm\n", + "C = 159.;# in muF\n", + "C =C * 10**-6;# in F\n", + "V = 230.;# in V\n", + "f = 50.;# in Hz\n", + "#calculations\n", + "XL= 2*pi*f*L;# in ohm\n", + "Z1= R1+XL*1j;# in ohm\n", + "I1= V/Z1;# in A\n", + "XC= 1/(2*pi*f*C);# in ohm\n", + "Z2= R2-1j*XC;# in ohm\n", + "I2= V/Z2;# in A\n", + "I= I1+I2;# in A\n", + "phi= cmath.phase(I)# in radians\n", + "#results\n", + "print \"Supply current : \"\n", + "print \"Magnitude is : \",round(abs(I),2),\" A\"\n", + "print \"Angle : \",round(phi*180/pi,1),\" degrees\"\n", + "print \"Power factor is : \",round(cos(phi),3),\" lag\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20: pg 179" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total admittance of the circuit : \n", + "Magnitude is : 0.02155 mho\n", + "Angle is : -44.4 degrees\n", + "The supply current : \n", + "Magnitude is : 5.39 A\n", + "Angle is : -44.4 degrees\n", + "Power factor is : 0.71 degrees lag\n" + ] + } + ], + "source": [ + "#pg 179\n", + "#calculate the admittance and supply current\n", + "# Given data\n", + "import cmath\n", + "from math import pi,cos\n", + "V=250;# in V\n", + "Z1= 70.7+70.7*1j;# in ohm\n", + "Z2= 120+160*1j;# in ohm\n", + "Z3= 120+90*1j;# in ohm\n", + "Y1= 1/Z1;# in S\n", + "Y2= 1/Z2;# in S\n", + "Y3= 1/Z3;# in S\n", + "Y_T= Y1+Y2+Y3;# in S\n", + "#calculations\n", + "phi= cmath.phase(Y_T)*180./pi;# in degrees\n", + "I= V*Y_T;# in A\n", + "#results\n", + "print \"Total admittance of the circuit : \"\n", + "print \"Magnitude is : \",round(abs(Y_T),5),\" mho\"\n", + "print \"Angle is : \",round(phi,1),\" degrees\"\n", + "print \"The supply current : \"\n", + "print \"Magnitude is : \",round(abs(I),2),\" A\"\n", + "print \"Angle is : \",round(phi,1),\" degrees\"\n", + "print \"Power factor is : \",round(cos(phi*pi/180.),2),\" degrees lag\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21: pg 180" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The impedance is : (5.7735026919-3.33333333333j) ohm\n", + "The resistance is : 5.7735 ohm\n", + "The reactance is : 3.3333 ohm\n", + "The power is : 649.52 W\n", + "The power factor is : 0.866 leading\n" + ] + } + ], + "source": [ + "#pg 180\n", + "#calculate the impedance, resistance, reactance, power and power factor\n", + "# Given data\n", + "from math import sqrt,atan,cos,pi\n", + "from cmath import exp\n", + "import cmath\n", + "Vm = 100.;# in V\n", + "phi1= 30.;# in degrees\n", + "Im = 15.;# in A\n", + "phi2= 60.;# in degrees\n", + "V= Vm/sqrt(2)*exp(phi1*1j*pi/180);# in V\n", + "I= Im/sqrt(2)*exp(phi2*1j*pi/180);# in A\n", + "Z= V/I;# in ohm\n", + "R= Z.real;# in ohm\n", + "XC= abs(Z.imag);# in ohm\n", + "phi= cmath.phase(Z)*180/pi;# in degrees\n", + "P= abs(V)*abs(I)*cos(phi*pi/180);# in W\n", + "#results\n", + "print \"The impedance is : \",Z,\" ohm\"\n", + "print \"The resistance is : \",round(R,4),\" ohm\"\n", + "print \"The reactance is : \",round(XC,4),\" ohm\"\n", + "print \"The power is : \",round(P,2),\" W\"\n", + "print \"The power factor is : \",round(cos(phi*pi/180),3),\" leading\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 22: pg 180" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of pure inductance in H is 1.0186\n", + "Note: There is calculation error to find the value of V_L, So the answer in the book is wrong and coding is correct.\n" + ] + } + ], + "source": [ + "#pg 180\n", + "#calculate the pure inductance\n", + "# Given data\n", + "from math import sqrt,pi\n", + "P = 100.;# in W\n", + "V = 120.;# in V\n", + "f= 50.;# in Hz\n", + "V = 200;# in V\n", + "V_R = 120;# in V\n", + "#calculations\n", + "I = P/V;# in A\n", + "V_L = sqrt( (V**2) - (V_R**2) );# in V\n", + "# V_L = I*X_L;\n", + "X_L = V_L/I;# in ohm\n", + "# X_L = 2*%pi*f*L;\n", + "L = X_L/(2*pi*f);# in H\n", + "#results\n", + "print \"The value of pure inductance in H is\",round(L,4)\n", + "\n", + "print 'Note: There is calculation error to find the value of V_L, So the answer in the book is wrong and coding is correct.'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23: pg 181" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The circuit admittance is : 0.137 mho\n", + "The circuit impedance is : 7.3 ohm\n", + "The power consumed in W is : 7235.0\n", + "The power factor is : 0.999 lead\n", + "The answer is a bit different due to rounding off error from textbook\n" + ] + } + ], + "source": [ + "#pg 181\n", + "#calculate the circuit admittance, impedance, power \n", + "# Given data\n", + "import cmath,math\n", + "from cmath import exp,phase\n", + "from math import pi,cos\n", + "V=230.;# in V\n", + "f= 50.;# in Hz\n", + "Z1= 10.*exp(-30*1j*pi/180);# in ohm\n", + "Z2= 20.*exp(60*1j*pi/180);# in ohm\n", + "Z3= 40.*exp(0*1j*pi/180);# in ohm\n", + "#calculations\n", + "Y1= 1/Z1;# in S\n", + "Y2= 1/Z2;# in S\n", + "Y3= 1/Z3;# in S\n", + "Y= Y1+Y2+Y3;# in S\n", + "phi= phase(Y)*180./pi;# in degrees\n", + "Z=1/Y;# in ohm\n", + "P= V**2*abs(Y);# in W\n", + "#results\n", + "print \"The circuit admittance is : \",round(abs(Y),3),\" mho\"\n", + "print \"The circuit impedance is : \",round(abs(Z),1),\" ohm\"\n", + "print \"The power consumed in W is : \",round(P)\n", + "print \"The power factor is : \",round(cos(phi*pi/180.),3),\" lead\"\n", + "print 'The answer is a bit different due to rounding off error from textbook'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 24: pg 181" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of P1 in W is : 737.705\n", + "The value of P2 in W is : 1438.525\n" + ] + } + ], + "source": [ + "#pg 181\n", + "#calculate the value of P1,P2\n", + "import cmath\n", + "# Given data\n", + "Z1= 10+15*1j;# in ohm\n", + "Z2= 6-8*1j;# in ohm\n", + "R1= 10.;# in ohm\n", + "R2= 6.;# in ohm\n", + "I_T= 15.;# in A\n", + "#calculations\n", + "I1= I_T*Z2/(Z1+Z2);# in A\n", + "I2= I_T*Z1/(Z1+Z2);# in A\n", + "P1= (abs(I1))**2*R1;# in W\n", + "P2= (abs(I2))**2*R2;# in W\n", + "#results\n", + "print \"The value of P1 in W is : \",round(P1,3)\n", + "print \"The value of P2 in W is : \",round(P2,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25: pg 182" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Impedence of the entire circuit : \n", + "Magnitude is : 16.9671 ohm\n", + "Angle is : 61.868 degrees\n", + "Current flowing through the condensor : \n", + "Magnitude is : 13.556 ohm\n", + "Angle is : -61.868 degree\n", + "Power factor of the circuit is : 0.4715 lag\n", + "The voltage across the condensor in V is : 308.2071\n" + ] + } + ], + "source": [ + "#pg 182\n", + "#calculate the impedance, current and voltage\n", + "# Given data\n", + "from math import pi,cos\n", + "from cmath import phase\n", + "R = 8.;# in ohm\n", + "L = 0.12;# in H\n", + "C = 140.;# in muF\n", + "C = C * 10**-6;# in F\n", + "V = 230.;# in V\n", + "f = 50.;# in Hz\n", + "#calculations\n", + "XL = 2*pi*f*L;# in ohm\n", + "XC= 1/(2*pi*f*C);# in ohm\n", + "Z= R+1j*XL-1j*XC;# in ohm\n", + "I= V/Z;# in A\n", + "phi= phase(I)# in radians\n", + "PowerFactor= cos(phi);\n", + "VC= abs(I)*XC;# in V\n", + "#results\n", + "print \"Impedence of the entire circuit : \"\n", + "print \"Magnitude is : \",round(abs(Z),4),\" ohm\"\n", + "print \"Angle is : \",round(phase(Z)*180/pi,3),\" degrees\"\n", + "print \"Current flowing through the condensor : \"\n", + "print \"Magnitude is : \",round(abs(I),3),\" ohm\"\n", + "print \"Angle is : \",round(phase(I)*180/pi,3),\" degree\"\n", + "print \"Power factor of the circuit is : \",round(cos(phi),4),\" lag\"\n", + "print \"The voltage across the condensor in V is : \",round(VC,4)\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 26: pg 183" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The half power frequencies are : 169.9829 Hz and 185.8984 Hz\n" + ] + } + ], + "source": [ + "#pg 183\n", + "#calculate the half power frequencies\n", + "# Given data\n", + "from math import sqrt,pi\n", + "R = 10;# in ohm\n", + "L = 0.1;# in H\n", + "C = 8;# in muF\n", + "C = C * 10**-6;# in F\n", + "#calculations\n", + "f_r = 1/(2*pi*sqrt(L*C));# in Hz\n", + "Q = (1/R) * (sqrt(L/C));\n", + "del_F = R/(4*pi*L);\n", + "# The half power frequencies \n", + "f1 = f_r - del_F;# in Hz\n", + "f2 = f_r+del_F;# in Hz\n", + "#results\n", + "print \"The half power frequencies are : \",round(f1,4),\" Hz and \",round(f2,4),\" Hz\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 27: pg 183" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of capacitance in muF is 97.94\n" + ] + } + ], + "source": [ + "#pg 183\n", + "#calculate the capacitance\n", + "# Given data\n", + "from math import pi\n", + "R = 15.;# in ohm\n", + "X_L = 10.;# in ohm\n", + "f_r = 50.;# in Hz\n", + "#calculations\n", + "# X_L = 2*%pi*f_r*L;\n", + "L = X_L/(2*pi*f_r);# in H\n", + "# value of capacitance \n", + "C = 1/( L*( ((f_r*2*pi)**2)+((R**2)/(L**2)) ));# in F\n", + "C = C*10**6;# in muF\n", + "#results\n", + "print \"The value of capacitance in muF is\",round(C,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 28: pg 183" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of current : \n", + "The magnitude in A is : 69.0\n", + "The phase angle in degree is : -53.13\n", + "The power drawn from the source in W is : 9522.0\n" + ] + } + ], + "source": [ + "#pg 183\n", + "#calculate the current, phase angle and power\n", + "# Given data\n", + "import cmath,math\n", + "Z1= 3+4*1j;# in ohm\n", + "Z2= 6+8*1j;# in ohm\n", + "V= 230.;# in V\n", + "#calculations\n", + "I1= V/Z1;# in A\n", + "I2= V/Z2;# in A\n", + "I_T= I1+I2;# in A\n", + "phi= cmath.phase(I_T);# in degrees\n", + "P= V*abs(I_T)*math.cos(phi);# in V\n", + "#results\n", + "print \"The value of current : \"\n", + "print \"The magnitude in A is : \",abs(I_T)\n", + "print \"The phase angle in degree is : \",round(phi*180/math.pi,2)\n", + "print \"The power drawn from the source in W is : \",P\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 29: pg 184" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The admittance in mho is : (0.16-0.12j)\n", + "The admittance in mho is : (0.06+0.08j)\n", + "Total circuit impedance : \n", + "Magnitude : 10.0 ohm\n", + "Angle : 53.13 degrees\n", + "The total power supplied in W is : 600.0\n" + ] + } + ], + "source": [ + "#pg 184\n", + "#calculate the admittance and impedance\n", + "# Given data\n", + "import cmath\n", + "from math import cos,pi\n", + "Z1= 1.6+1j*7.2;# in ohm\n", + "Z2= 4+1j*3;# in ohm\n", + "Z3= 6-1j*8;# in ohm\n", + "V= 100.;# in V\n", + "Y2= 1/Z2;# in mho\n", + "#calculations\n", + "Y3= 1/Z3;# in mho\n", + "print \"The admittance in mho is : \",Y2\n", + "print \"The admittance in mho is : \",Y3\n", + "ZT= Z1+1/(Y2+Y3);\n", + "phi = cmath.phase(ZT)\n", + "print \"Total circuit impedance : \"\n", + "print \"Magnitude : \",abs(ZT),\" ohm\"\n", + "print \"Angle : \",round(phi*180/pi,2),\" degrees\"\n", + "IT= V/ZT;# in A\n", + "PT= V*abs(IT)*cos(phi);# in W\n", + "print \"The total power supplied in W is : \",PT\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 30: pg 185" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of capacitance in muF is 20.2642\n", + "The voltage across the capacitance in V 3927.0\n", + "The Q factor of the circuit is 39.27\n" + ] + } + ], + "source": [ + "#pg 185\n", + "#calculate the capacitance, voltage and Q factor\n", + "# Given data\n", + "from math import pi\n", + "R = 4;# in ohm\n", + "L = 0.5;# in H\n", + "V = 100.;# in V\n", + "f = 50.;# in Hz\n", + "#calculations\n", + "X_L = 2*pi*f*L;# in ohm\n", + "X_C = X_L;# in ohm\n", + "# X_C = 1/(2*%pi*f*C);\n", + "C = 1/(X_C*2*pi*f);# in F\n", + "C = C * 10**6;# in F\n", + "I = V/R;# in A]\n", + "V_C = I*X_C;# in V\n", + "omega = 2*pi*f;# in rad/sec\n", + "Q = (omega*L)/R;\n", + "#results\n", + "print \"The value of capacitance in muF is\",round(C,4)\n", + "print \"The voltage across the capacitance in V\",round(V_C)\n", + "print \"The Q factor of the circuit is\",round(Q,2)\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter5.ipynb b/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter5.ipynb new file mode 100644 index 00000000..5b49dff1 --- /dev/null +++ b/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter5.ipynb @@ -0,0 +1,838 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5: Three phase AC Circuits" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1: pg 200" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The line current in A is 9.24\n", + "The power factor is : 0.8 degrees lag.\n", + "The power supplied in W is 5120.0\n" + ] + } + ], + "source": [ + "#pg 200\n", + "#calculate the line current, power factor and power supplied\n", + "# Given data\n", + "from math import cos,acos,sqrt\n", + "R = 20.;# in ohm\n", + "X_L = 15.;# in ohm\n", + "V_L = 400.;# in V\n", + "f = 50.;# in Hz\n", + "#calculations\n", + "V_Ph = V_L/sqrt(3);# in V\n", + "Z_Ph = sqrt( (R**2) + (X_L**2) );# in ohm\n", + "I_Ph = V_Ph/Z_Ph;# in A\n", + "I_L = I_Ph;# in A\n", + "print \"The line current in A is\",round(I_L,2)\n", + "# pf = cos(phi) = R_Ph/Z_Ph;\n", + "R_Ph = R;# in ohm\n", + "phi= acos(R_Ph/Z_Ph);\n", + "# Power factor\n", + "pf= cos(phi);# in radians\n", + "print \"The power factor is : \",pf,\"degrees lag.\"\n", + "P = sqrt(3)*V_L*I_L*cos(phi);# in W\n", + "print \"The power supplied in W is\",P\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2: pg 201" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The line voltage in V is 400.0\n", + "The phase voltage in V is 230.94\n", + "The line current in A is : 11.55\n", + "The line current in A is : 11.55\n", + "Power factor is : 0.8 lagging\n", + "The power absorbed in W is : 6400.0\n" + ] + } + ], + "source": [ + "#pg 201\n", + "#calculate the line and phase voltage, current and power\n", + "# Given data\n", + "from math import sqrt,cos\n", + "import cmath\n", + "R_Ph = 16.;# in ohm\n", + "X_L = 12.;# in ohm\n", + "V_L = 400.;# in V\n", + "f = 50.;# in Hz\n", + "#calculations\n", + "V_Ph = V_L/sqrt(3);# in V\n", + "Z_Ph = R_Ph + 1j*X_L;# in ohm\n", + "I_Ph= V_Ph/Z_Ph;# in A\n", + "I_L= I_Ph;# in A\n", + "phi= cmath.phase(I_L);\n", + "cos_phi= R_Ph/abs(Z_Ph);\n", + "P= sqrt(3)*V_L*abs(I_L)*cos_phi;# in W\n", + "#results\n", + "print \"The line voltage in V is\",V_L\n", + "print \"The phase voltage in V is\",round(V_Ph,2)\n", + "print \"The line current in A is : \",round(abs(I_L),2)\n", + "print \"The line current in A is : \",round(abs(I_Ph),2)\n", + "print \"Power factor is : \",cos_phi,\" lagging\"\n", + "print \"The power absorbed in W is : \",P\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3: pg 202" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Resistance in ohm is 4.27\n", + "The inductance in H is 0.0665\n" + ] + } + ], + "source": [ + "#pg 202\n", + "#calculate the resistance and inductance\n", + "# Given data\n", + "from math import sqrt,cos,pi,acos\n", + "P = 1.5;# in kW\n", + "P = P * 10**3;# in W\n", + "pf = 0.2;# in lag\n", + "V_L = 400;# in V\n", + "f = 50;# in Hz\n", + "#calculations\n", + "phi= acos(pf);\n", + "V_Ph = V_L/sqrt(3);# in V\n", + "# P = sqrt(3)*V_L*I_L*cos(phi);\n", + "I_L = P/(sqrt(3)*V_L*cos(phi));# in A\n", + "I_Ph = I_L;# in A\n", + "Z_Ph = V_Ph/I_Ph;# in ohm\n", + "R_Ph = Z_Ph*cos(phi);# in ohm\n", + "X_Ph = sqrt( (Z_Ph**2) - (R_Ph**2) );# in ohm\n", + "L_Ph = X_Ph/(2*pi*f);# in H\n", + "#results\n", + "print \"The Resistance in ohm is\",round(R_Ph,2)\n", + "print \"The inductance in H is\",round(L_Ph,4)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4: pg 203" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The line current in A is 94.91\n", + "The total power absorbed in kW is 45.04\n", + "Note: To evaluate the value of P, the wrong value of I_L is put , so the calculated value of P in the book is not correct\n" + ] + } + ], + "source": [ + "#pg 203\n", + "#calculate the line current and total power\n", + "# Given data\n", + "from math import sqrt,pi,cos,acos\n", + "R = 5;# in ohm\n", + "L =0.02;# in H\n", + "V_L = 440.;# in V\n", + "f = 50.;# in Hz\n", + "#calculations\n", + "X_L = 2*pi*f*L;# in ohm\n", + "Z_Ph = sqrt( (R**2)+(X_L**2) );# in ohm\n", + "V_Ph = V_L;# in V\n", + "I_Ph = V_Ph/Z_Ph;# in A\n", + "I_L = sqrt(3)*I_Ph;# in A\n", + "phi = acos(R/Z_Ph);# in lag\n", + "P = sqrt(3)*V_L*I_L*cos(phi);# in W\n", + "P= P*10**-3;# in kW\n", + "#results\n", + "print \"The line current in A is\",round(I_L,2)\n", + "print \"The total power absorbed in kW is\",round(P,2)\n", + "\n", + "print 'Note: To evaluate the value of P, the wrong value of I_L is put , so the calculated value of P in the book is not correct'\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5: pg 203" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The phase current in A is 10.0\n", + "The resistance of coil in ohm is 32.0\n", + "The inductance of coil in mH is 76.4\n", + "The power drawn by each coil in W is 3200.0\n" + ] + } + ], + "source": [ + "#pg 203\n", + "#calculate the phase current, resistance, inductance and power\n", + "# Given data\n", + "from math import acos,sqrt,pi,cos\n", + "V_L = 400.;# in V\n", + "f = 50.;# in Hz\n", + "I_L = 17.32;# in A\n", + "pf = 0.8;# in lag\n", + "#calculations\n", + "I_Ph = I_L/sqrt(3);# in A\n", + "print \"The phase current in A is\",round(I_Ph)\n", + "V_Ph = V_L;# in V\n", + "Z_Ph = V_Ph/I_Ph;# in ohm\n", + "phi = acos(pf)# in lag\n", + "R_Ph = Z_Ph*cos(phi);# in ohm\n", + "print \"The resistance of coil in ohm is\",round(R_Ph)\n", + "X_Ph = sqrt( (Z_Ph**2) - (R_Ph**2) );# in ohm\n", + "# X_Ph = 2*%pi*f*L;\n", + "L = X_Ph/(2*pi*f);# in H\n", + "L = L * 10**3;# in mH\n", + "print \"The inductance of coil in mH is\",round(L,1)\n", + "P = V_Ph*I_Ph*cos(phi);# in W\n", + "print \"The power drawn by each coil in W is\",round(P)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6: pg 208" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The power factor of the load is : 0.89 lag.\n" + ] + } + ], + "source": [ + "#pg 208\n", + "#calculate the power factor\n", + "from math import atan,sqrt,cos\n", + "# Given data\n", + "W1 = 1000.;# in W\n", + "W2 = 550.;# in W\n", + "#calculations\n", + "phi = (atan( sqrt(3)*((W1-W2)/(W1+W2)) ));# in radians\n", + "# power factor \n", + "pf= cos(phi);# lag\n", + "print \"The power factor of the load is : \",round(pf,2),\" lag.\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7: pg 208" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Part (i) : Power factor is : 0.6934 lagging\n", + "Part (ii) : Power factor is : 0.327 lagging\n" + ] + } + ], + "source": [ + "#pg 208\n", + "#calculate the power factor in both cases\n", + "# Given data\n", + "from math import cos,atan,sqrt\n", + "W1 = 2000.;# in W\n", + "W2 = 500.;# in W\n", + "#calculations\n", + "phi = (atan( sqrt(3)*((W1-W2)/(W1+W2)) ));# in lag\n", + "# power factor \n", + "pf= cos(phi);# lagging\n", + "print \"Part (i) : Power factor is : \",round(pf,4),\" lagging\"\n", + "W2 = -W2;# in W\n", + "phi = (atan( sqrt(3)*((W1-W2)/(W1+W2)) ));# in lag\n", + "# power factor \n", + "pf= cos(phi);# lagging\n", + "print \"Part (ii) : Power factor is : \",round(pf,3),\" lagging\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8: pg 208" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The power factor is : 0.404 lag.\n" + ] + } + ], + "source": [ + "#pg 208\n", + "#calculate the power factor\n", + "# Given data\n", + "from math import atan,sqrt,cos\n", + "W1 = 375.;# in W\n", + "W2 = -50.;# in W\n", + "#calculations\n", + "# tan(phi) = sqrt(3)*((W1-W2)/(W1+W2));\n", + "phi = atan(sqrt(3)*((W1-W2)/(W1+W2)));# in degree\n", + "# power factor \n", + "pf= cos(phi);# lag\n", + "#results\n", + "print \"The power factor is : \",round(pf,3),\" lag.\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9: pg 209" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The power input in kW is 400.0\n", + "The power factor is 0.756\n", + "The line current in A is 152.75\n", + "The power output in kW is 360.0\n" + ] + } + ], + "source": [ + "#pg 209\n", + "#calculate the power input,factor,output and line current\n", + "# Given data\n", + "from math import atan,cos,sqrt\n", + "W1 = 300.;# in kW\n", + "W2 = 100.;# in kW\n", + "V_L= 2000.;# in V\n", + "Eta= 90/100.;\n", + "#calculations\n", + "P = W1+W2;# in kW\n", + "# tan(phi) = sqrt(3)*((W1-W2)/(W1+W2));\n", + "phi = atan(sqrt(3)*((W1-W2)/(W1+W2)));\n", + "pf = cos(phi);# power factor\n", + "# P = sqrt(3)*V_L*I_L*cosd(phi);\n", + "I_L = (P*10**3)/(sqrt(3)*V_L*pf);# in A\n", + "output = P*Eta;# in kW\n", + "#results\n", + "print \"The power input in kW is\",P\n", + "print \"The power factor is\",round(pf,3)\n", + "print \"The line current in A is\",round(I_L,2)\n", + "print \"The power output in kW is\",output\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10: pg 209" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The phase current in A is 20.0\n", + "The impedance of load in ohm is 11.55\n", + "The power factor is : 0.866 lag.\n" + ] + } + ], + "source": [ + "#pg 209\n", + "#calculate the phase current,impedance and power factor\n", + "# Given data\n", + "from math import sqrt,acos,cos\n", + "P = 12.;# in kW\n", + "P = P * 10**3;# in W\n", + "V_L = 400.;# in V\n", + "I_L = 20.;# in A\n", + "I_Ph = I_L;# in A\n", + "#calculations\n", + "V_Ph = V_L/sqrt(3);# in V\n", + "Z_Ph = V_Ph/I_Ph;# in ohm\n", + "# P = sqrt(3)*V_L*I_L*cos(phi);\n", + "phi= acos(P/(sqrt(3)*V_L*I_L));# in lag\n", + "# power factor\n", + "pf= cos(phi);# lag\n", + "#results\n", + "print \"The phase current in A is\",I_Ph\n", + "print \"The impedance of load in ohm is\",round(Z_Ph,2)\n", + "print \"The power factor is : \",round(pf,3),\" lag.\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11: pg 210" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The line current in A is : 23.09\n", + "Power factor is : 0.8 lagging\n", + "The three phase power in W is : 12800.0\n", + "The three phase volt-amperes in VA is : 16000.0\n" + ] + } + ], + "source": [ + "#pg 210\n", + "#calculate the line current, power factor, three phase power and volt amperes\n", + "# Given data\n", + "import cmath\n", + "from math import cos,sqrt\n", + "Z_Ph= 8+6*1j;# in ohm\n", + "V_L= 400;# in V\n", + "#calculations\n", + "V_Ph= V_L/sqrt(3);# in V\n", + "I_Ph= V_Ph/Z_Ph;# in A\n", + "I_L= I_Ph;# in A\n", + "phi= cmath.phase(I_L);# in radians\n", + "print \"The line current in A is : \",round(abs(I_L),2)\n", + "# power factor\n", + "pf= cos(phi);# lagging\n", + "print \"Power factor is : \",pf,\" lagging\"\n", + "P= sqrt(3)*V_L*abs(I_L)*cos(phi);# in W\n", + "print \"The three phase power in W is : \",P\n", + "S= sqrt(3)*V_L*abs(I_L);# in VA.\n", + "print \"The three phase volt-amperes in VA is : \",S\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12: pg 211" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The power in kW is : 15.0\n", + "The power factor of the load is : 0.3273\n" + ] + } + ], + "source": [ + "#pg 211\n", + "#calculate the power and power factor\n", + "# Given data\n", + "from math import atan,cos,sqrt\n", + "W1 = 20.;# in kW\n", + "W2 = -5.;# in kW\n", + "#calculations\n", + "P = W1+W2;# in kW\n", + "phi = (atan( sqrt(3)*((W1-W2)/(W1+W2)) ));# in lag\n", + "# Power factor of the load\n", + "pf= cos(phi)\n", + "#results\n", + "print \"The power in kW is : \",P\n", + "print \"The power factor of the load is : \",round(pf,4)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13: pg 211" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The reading of first wattmeter in W is : 4000.0\n", + "The reading of second wattmeter in W is : -3810.0\n", + "The answer in textbook is wrong. please check using a calculator\n" + ] + } + ], + "source": [ + "#pg 211\n", + "#calculate the readings on both meters\n", + "# Given data\n", + "from math import atan,sqrt,cos,pi\n", + "V_L = 400.;# in V\n", + "I_L = 10.;# in A\n", + "W2= 1.;# assumed\n", + "#calculations\n", + "W1= 2*W2;\n", + "phi= atan(sqrt(3)*(W1-W2)/(W1+W2))*180/pi;\n", + "W1= V_L*I_L*cos(30-phi);# in W\n", + "W2= V_L*I_L*cos(30+phi);# in W\n", + "#results\n", + "print \"The reading of first wattmeter in W is : \",W1\n", + "print \"The reading of second wattmeter in W is : \",round(W2)\n", + "print 'The answer in textbook is wrong. please check using a calculator'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14: pg 212" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The phase current in A is 10.0\n", + "The resistance of the coil in ohm is : 32.0\n", + "The inductance of the coil in mH is : 76.3966\n", + "The power drawn by each coil in W is : 3200.0\n" + ] + } + ], + "source": [ + "#pg 212\n", + "#calculate the phase current, resistance, inductance and power\n", + "# Given data\n", + "from math import acos,pi,cos,sqrt\n", + "from cmath import phase,exp\n", + "V_L = 400;# in V\n", + "f = 50;# in Hz\n", + "I_L = 17.32;# in A\n", + "#calculations\n", + "phi = acos(0.8);\n", + "I_Ph =I_L/sqrt(3);# in A\n", + "print \"The phase current in A is\",round(I_Ph)\n", + "V_Ph=V_L;# in V\n", + "Z_Ph = V_Ph/I_Ph;# in ohm\n", + "Z_Ph= Z_Ph*exp(phi*1j);# in ohm\n", + "R= Z_Ph.real;# in ohm\n", + "XL= Z_Ph.imag;# in ohm\n", + "L= XL/(2*pi*f);# in H\n", + "L= L*10**3;# in mH\n", + "print \"The resistance of the coil in ohm is : \",round(R)\n", + "print \"The inductance of the coil in mH is : \",round(L,4)\n", + "# The power drawn by each coil\n", + "P_Ph= V_Ph*I_Ph*cos(phi);# in W\n", + "print \"The power drawn by each coil in W is : \",round(P_Ph)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15: pg 212" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The reading of first wattmeter in kW is 23.8352\n", + "The reading of second wattmeter in kW is 6.1648\n" + ] + } + ], + "source": [ + "#pg 212\n", + "#calculate the readings of both wattmeters\n", + "# Given data\n", + "from math import sqrt,acos,cos,pi\n", + "P = 30;# in kW\n", + "pf = 0.7;\n", + "#calculations\n", + "# cosd(phi) = pf;\n", + "phi = acos(pf)*180/pi;# in degree\n", + "# P = sqrt(3)*V_L*I_L*cosd(phi);\n", + "theta = 30.;# in degree\n", + "V_LI_L = P/(sqrt(3)*cos(phi*pi/180));\n", + "W1 = V_LI_L*cos((theta-phi)*pi/180);# in kW\n", + "W2 = V_LI_L*cos((theta+phi)*pi/180);# in kW\n", + "#results\n", + "print \"The reading of first wattmeter in kW is\",round(W1,4)\n", + "print \"The reading of second wattmeter in kW is\",round(W2,4)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16: pg 213" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The power factor is : 0.3936 leading\n", + "The resistance in ohm is : 1.6667\n", + "The capacitance in muF is : 817.8438\n", + "The load is capacitive in nature.\n" + ] + } + ], + "source": [ + "#pg 213\n", + "#calculate the power factor,resistance, capacitance \n", + "# Given data\n", + "from math import cos,pi,acos,sqrt\n", + "from cmath import exp\n", + "P = 18.;# in kW\n", + "P= P*10**3;# in W\n", + "I_L = 60.;# in A\n", + "V_L = 440.;# in V\n", + "f= 50.;# in Hz\n", + "#calculations\n", + "# P = sqrt(3)*V_L*I_L*cosd(phi);\n", + "phi= acos(P/(sqrt(3)*V_L*I_L));# in radians\n", + "I_L= I_L*exp(phi*1j);# in A\n", + "I_Ph= I_L;# in A\n", + "V_Ph= V_L/sqrt(3);# in V\n", + "Z_Ph= V_Ph/I_Ph;# in ohm\n", + "R= Z_Ph.real;# in ohm\n", + "XC=abs(Z_Ph.imag);# in ohm\n", + "C = 1/(2*pi*f*XC);# in F\n", + "C=C*10**6;# in muF\n", + "# Power factor\n", + "pf= cos(phi);# lead\n", + "#results\n", + "print \"The power factor is : \",round(pf,4),\" leading\"\n", + "print \"The resistance in ohm is : \",round(R,4)\n", + "print \"The capacitance in muF is : \",round(C,4)\n", + "print \"The load is capacitive in nature.\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17: pg 213" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The power factor is : 0.86603\n", + "The line current in A is 20.0\n", + "The impedance of each phase in ohm is : (10+5.7735026919j)\n", + "The resistance of each phase in ohm is : 10.0\n", + "The inductance of each phase in H is : 0.01838\n" + ] + } + ], + "source": [ + "#pg 213\n", + "#calculate the power factor, line current, impedance, resistance and inductance\n", + "# Given data\n", + "from math import cos,pi,acos,sqrt,atan\n", + "from cmath import exp\n", + "V_L = 400.;# in V\n", + "f = 50.;# in Hz\n", + "W1 = 8000.;# in W\n", + "W2 = 4000.;# in W\n", + "#calculations\n", + "W = W1+W2;# in W\n", + "phi =(atan( sqrt(3)*((W1-W2)/(W1+W2)) ));# in lag\n", + "P = W;# in W\n", + "# P = sqrt(3)*V_L*I_L*cosd(phi);\n", + "I_L = P/(sqrt(3)*V_L*cos(phi));# in A\n", + "V_Ph = V_L/sqrt(3);# in V\n", + "I_Ph = I_L;# in A\n", + "Z_Ph = V_Ph/I_Ph;# in ohm\n", + "Z_Ph= Z_Ph*exp(phi*1j);# ohm\n", + "R_Ph= Z_Ph.real;# in ohm\n", + "XL_Ph= Z_Ph.imag;# in ohm\n", + "L_Ph= XL_Ph/(2*pi*f);# in H\n", + "# power factor\n", + "pf= cos(phi);\n", + "#results\n", + "print \"The power factor is : \",round(pf,5)\n", + "print \"The line current in A is\",I_L\n", + "print \"The impedance of each phase in ohm is : \",Z_Ph\n", + "print \"The resistance of each phase in ohm is : \",R_Ph\n", + "print \"The inductance of each phase in H is : \",round(L_Ph,5)\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter6.ipynb b/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter6.ipynb new file mode 100644 index 00000000..d7462a9e --- /dev/null +++ b/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter6.ipynb @@ -0,0 +1,483 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6: Measuring instruments" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1: pg 235" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The shunt resistance required in (ohm) = 0.163\n" + ] + } + ], + "source": [ + "#pg 235\n", + "#calculate the shunt resistance\n", + "# Given data\n", + "Rm = 8.;# in ohm\n", + "Im = 20.;# in mA\n", + "Im = Im * 10**-3;# in A\n", + "I = 1.;# in A\n", + "#calculations\n", + "# Multiplying factor\n", + "N = I/Im;\n", + "# Shunt resistance\n", + "Rsh = Rm/(N-1);# in ohm\n", + "#results\n", + "print \"The shunt resistance required in (ohm) = \",round(Rsh,3)\n", + " " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2: pg 235" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The multiplying factor is 241.0\n" + ] + } + ], + "source": [ + "#pg 235\n", + "#calculate the multiplying factor\n", + "# Given data\n", + "Rm = 6;# in ohm\n", + "Rsh = 0.025;# in ohm\n", + "#calculations\n", + "N = 1 + (Rm/Rsh);# multiplying factor\n", + "#results\n", + "print \"The multiplying factor is\",N\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3: pg 235" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance to be connected in parallel in (ohm) = 0.0761\n", + "The resistance to be connected in series in (ohm) = 661.67\n" + ] + } + ], + "source": [ + "#pg 235\n", + "#calculate the resistances to be connected in parallel and series\n", + "# Given data\n", + "Rm = 5.;# in ohm\n", + "Im = 15.;# in mA\n", + "Im = Im * 10**-3;# in A\n", + "I = 1.;# in A\n", + "#calculations\n", + "N = I/Im;# multiplying factor\n", + "Rsh = Rm/(N-1);# in ohm\n", + "print \"The resistance to be connected in parallel in (ohm) = \",round(Rsh,4)\n", + "V = 10;# in V\n", + "Rs = (V/Im)-Rm;# in ohm\n", + "print \"The resistance to be connected in series in (ohm) = \",round(Rs,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4: pg 236" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current range of instrument in A is 50.0\n" + ] + } + ], + "source": [ + "#pg 236\n", + "#calculate the current range of the instrument\n", + "# Given data\n", + "V=250.;# full scale voltage reading in V\n", + "Rm = 2.;# in ohm\n", + "Rsh = 2.;# in m ohm\n", + "Rsh = Rsh * 10**-3;# in ohm\n", + "R = 5000.;# in ohm\n", + "#calculations\n", + "Im = V/(Rm+R);# in A\n", + "Ish = (Im*Rm)/Rsh;# in A\n", + "# Current range of instrument\n", + "I = Im+Ish;# in A\n", + "#results\n", + "print \"The current range of instrument in A is\",round(I)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5: pg 236" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The percentage error in (percentage) is 249.38\n", + "The answer is a bit different from textbook due to rounding off error\n" + ] + } + ], + "source": [ + "#pg 236\n", + "#calculate the percentage error\n", + "import math\n", + "from math import acos,pi,cos\n", + "# Given data\n", + "V = 230.;# in V\n", + "I = 35.;# in A\n", + "N = 200.;\n", + "t = 64.;# in sec\n", + "kwh = 500.;\n", + "#calculations\n", + "phi= acos(0.8);# in radians\n", + "Er = N/kwh;# in kWh\n", + "Et = V*I*cos(phi)*t;# in Joules\n", + "Et = Et/3600.;# in W hour\n", + "Et = Et * 10**-3;# in kWh\n", + "# percentage error\n", + "PerError = ((Er-Et)/Et)*100;# in %\n", + "#results\n", + "print \"The percentage error in (percentage) is\",round(PerError,2)\n", + "print 'The answer is a bit different from textbook due to rounding off error'\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6: pg 237" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The percentage error in (percentage) is 3.22\n", + "The answer is a bit different from textbook due to rounding off error\n" + ] + } + ], + "source": [ + "#pg 237\n", + "#calculate the percentage error\n", + "from math import acos,cos,pi\n", + "# Given data\n", + "I = 50.;# in A\n", + "V = 230.;# in V\n", + "N = 61.;\n", + "t = 37.;# in sec\n", + "KWh = 500.;\n", + "#calculations\n", + "phi= acos(1);# in radians\n", + "Er = N/KWh;# in kWh\n", + "Et = V*I*cos(phi)*t;# in Joules\n", + "Et = Et/3600.;# in Wh\n", + "Et = Et*10**-3;# in kWh\n", + "# Percentage error\n", + "PerError = ((Er-Et)/Et)*100;# in %\n", + "#results\n", + "print \"The percentage error in (percentage) is \",round(PerError,2)\n", + "print 'The answer is a bit different from textbook due to rounding off error'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7: pg 237" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The series resistance in ohm is 24997.5\n" + ] + } + ], + "source": [ + "#pg 237\n", + "#calculate the series resistance\n", + "# Given data\n", + "Im = 20.;# in mA\n", + "Im = Im * 10**-3;# in A\n", + "Vm = 50.;# in mV\n", + "Vm = Vm * 10**-3;# in V\n", + "V = 500.;# in V\n", + "#calculations\n", + "Rm = Vm/Im;# in ohm\n", + "Rs = (V/Im)-Rm;# in ohm\n", + "#results\n", + "print \"The series resistance in ohm is\",Rs\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8: pg 238" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of Rs in (ohm) is 9950.0\n", + "The value of Rsh in (ohm) is 0.505\n" + ] + } + ], + "source": [ + "#pg 238\n", + "#calculate the values of resistances\n", + "# Given data\n", + "Rm = 50;# in ohm\n", + "Im = 10;# in mA\n", + "Im = Im * 10**-3;# in A\n", + "V = 100;# in V\n", + "#calculations\n", + "Rs = (V/Im)-Rm;# in ohm\n", + "print \"The value of Rs in (ohm) is\",Rs\n", + "N = 1/Im;\n", + "Rsh = Rm/(N-1);# in ohm\n", + "print \"The value of Rsh in (ohm) is\",round(Rsh,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9: pg 238" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The percentage error in (percentage) is 2.08\n" + ] + } + ], + "source": [ + "#pg 238\n", + "#calculate the percentage error\n", + "# Given data\n", + "from math import acos,cos\n", + "I = 40.;# in A\n", + "V = 230.;# in V\n", + "N = 600.;\n", + "t = 46.;# in sec\n", + "#calculations\n", + "phi= acos(1);# in radians\n", + "P = V*I*cos(phi);# in W\n", + "P = P * 10**-3;# in kW\n", + "# 1 kWh = 500 revolution \n", + "P = P * 500.;# in revolution\n", + "T = (3600./t)*60;# in revolution\n", + "# Percentage error\n", + "PerError = ((T-P)/P)*100;# in %\n", + "#results\n", + "print \"The percentage error in (percentage) is\",round(PerError,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10: pg 238" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The percentage error in (percentage) is 4.167\n" + ] + } + ], + "source": [ + "#pg 238\n", + "#calculate the percentage error\n", + "# Given data\n", + "N = 100.;\n", + "I = 20.;# in A\n", + "V = 210.;# in V\n", + "pf = 0.8;# in lad\n", + "Er = 350.;# in rev\n", + "a = 3.36;# assumed\n", + "#calculations\n", + "Et = (a*3600.)/3600;# in kWh\n", + "# 1 kWh = 100;# revolution\n", + "Et = Et*N;# revolution\n", + "# Percentage error\n", + "PerError = ((Er-Et)/Et)*100;# in %\n", + "#results\n", + "print \"The percentage error in (percentage) is\",round(PerError,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11: pg 239" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The percentage error in (percentage) is 3.22\n" + ] + } + ], + "source": [ + "#pg 239\n", + "#calculate the percentage error\n", + "# Given data\n", + "I = 5.;# in A\n", + "V = 230.;# in V\n", + "N = 61.;# number of revolution\n", + "t = 37.;# in sec\n", + "# speed of the disc\n", + "discSpeed= 500.;# in rev/kWh\n", + "#calculations\n", + "Er = N/discSpeed;\n", + "Et = (V*I*t)/(3600*100);\n", + "# percentage error\n", + "PerError = ((Er-Et)/Et)*100;# in %\n", + "#results\n", + "print \"The percentage error in (percentage) is\",round(PerError,2)\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter8.ipynb b/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter8.ipynb new file mode 100644 index 00000000..4011c1cd --- /dev/null +++ b/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter8.ipynb @@ -0,0 +1,759 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8: Magnetic Circuits" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1: pg 267" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current in A is 2.0\n" + ] + } + ], + "source": [ + "#pg 267\n", + "#calculate the current\n", + "# Given data\n", + "from math import pi\n", + "a = 3;# in cm^2\n", + "a = a * 10**-4;# in m^2\n", + "d = 20;# in cm\n", + "N = 500;\n", + "phi = 0.5*10**-3;# in Wb\n", + "miu_r = 833.33;\n", + "#calculations\n", + "miu_o = 4*pi*10**-7;\n", + "l = pi*d;# in cm \n", + "l = l * 10**-2;# in m\n", + "S = l/(miu_o*miu_r*a);# in AT/Wb\n", + "# Calculation of the current with the help of flux\n", + "# Formula phi = (m*m*f)/S = (N*I)/S;\n", + "I = (phi*S)/N;# in A\n", + "#results\n", + "print \"The current in A is\",round(I)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2: pg 268" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The coil mmf in AT is 750.0\n", + "The field strength in AT/m is 1875.0\n", + "Total flux in Wb is 0.00106\n", + "The reluctance of the ring in AT/Wb is 707355.3\n", + "Permeance of the ring in Wb/AT is 1.4e-06\n" + ] + } + ], + "source": [ + "#pg 268\n", + "#calculate the reluctance, flux, field strength, coil mmf and permeance\n", + "# Given data\n", + "from math import pi\n", + "N = 300.;\n", + "miu_r = 900.;\n", + "l = 40.;# in cm\n", + "a = 5.;# in cm**2\n", + "R = 100.;# in ohm\n", + "V = 250.;# in V\n", + "#calculations\n", + "miu_o = 4*pi*10**-7;\n", + "I = V/R;# in A\n", + "mmf = N*I;# in AT\n", + "print \"The coil mmf in AT is\",mmf\n", + "H = (N*I)/(l*10**-2);# in AT/m\n", + "print \"The field strength in AT/m is\",H\n", + "B = miu_o*miu_r*H;# in Wb/m**2\n", + "phi = B*a*10**-4;# in Wb\n", + "print \"Total flux in Wb is\",round(phi,5)\n", + "S = mmf/phi;# in AT/Wb\n", + "print \"The reluctance of the ring in AT/Wb is\",round(S,2)\n", + "# Permeance is recipocal of reluctance\n", + "Permeance = 1/S;# in Wb/AT\n", + "print '%s %.1e' %(\"Permeance of the ring in Wb/AT is\",Permeance)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3: pg 269" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The amphere turns for the gap in AT is 4138.0\n" + ] + } + ], + "source": [ + "#pg 269\n", + "#calculate the amphere turns\n", + "# Given data\n", + "from math import pi\n", + "Ig = 4;# in mm\n", + "Ig = Ig * 10**-3;# in m\n", + "B = 1.3;# in Wb/m**2\n", + "miu_r = 1;\n", + "#calculations\n", + "miu_o = 4*pi*10**-7;\n", + "H = B/(miu_o*miu_r);# in AT/m\n", + "Hg = H;# in AT/m\n", + "# Ampere turn required for air gap\n", + "AT = Hg*Ig;# AT for air gap in AT\n", + "#results\n", + "print \"The amphere turns for the gap in AT is\",round(AT)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4: pg 269" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total flux in the coil in m/Wb is 0.462\n" + ] + } + ], + "source": [ + "#pg 269\n", + "#calculate the total flux\n", + "# Given data\n", + "from math import pi\n", + "N = 500.;\n", + "R = 4.;# in ohm\n", + "d = 0.25;# in m\n", + "a = 700.;# in mm^2\n", + "a = a*10**-6;# in m^2\n", + "V = 6.;# in V\n", + "miu_r = 550.;\n", + "#calculations\n", + "miu_o = 4*pi*10**-7;\n", + "# Evaluation of current by ohm's law\n", + "I = V/R;# in A\n", + "l = pi*d;# in m\n", + "H = (N*I)/l;# in A/m\n", + "# Evaluation of flux density\n", + "B = miu_o*miu_r*H;# in T\n", + "# Evaluation of total flux\n", + "phi = B*a;# in Wb\n", + "phi= phi*10**3;# in mWb\n", + "#results\n", + "print \"The total flux in the coil in m/Wb is\",phi\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5: pg 269" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The mmf in AT is 1400.0\n", + "The total reluctance in AT/Wb is 23071142.12\n", + "The flux in Wb is 6.0682e-05\n", + "The flux density of the ring in (Wb/m^2) = 0.7726\n" + ] + } + ], + "source": [ + "#pg 269\n", + "#calculate the flux, density, reluctance, mmf\n", + "# Given data\n", + "from math import pi\n", + "d_r = 8;# diameter of ring in cm\n", + "d_r = d_r*10**-2;# in m\n", + "d_i = 1;# diameter of iron in cm\n", + "d_i = d_i * 10**-2;# in m\n", + "Permeability = 900;\n", + "gap = 2;# in mm\n", + "gap = gap * 10**-3;# in m\n", + "N = 400;\n", + "I = 3.5;# in A\n", + "#calculations\n", + "l_i = (pi*d_r)-gap;# length of iron in m\n", + "a = (pi/4)*(d_i**2);# in m**2\n", + "mmf = N*I;# in AT\n", + "print \"The mmf in AT is\",mmf\n", + "miu_o = 4*pi*10**-7;\n", + "miu_r = 900;\n", + "Si = l_i/(miu_o*miu_r*a);# in AT/Wb\n", + "miu_r = 1;\n", + "Sg = gap/(miu_o*miu_r*a);# in AT/Wb\n", + "S_T = Si+Sg;# in AT/Wb\n", + "print \"The total reluctance in AT/Wb is\",round(S_T,3)\n", + "phi = mmf/S_T;# in Wb\n", + "print '%s %.4e' %(\"The flux in Wb is\",phi)\n", + "# phi = B*a;\n", + "B = phi/a;# in Wb/m**2\n", + "print \"The flux density of the ring in (Wb/m^2) = \",round(B,4)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6: pg 271" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The reluctance of the magnetic circuit in AT/Wb is 7.958e+05\n", + "The inductance of the coil in H is 1.257\n", + "Note: In the book the calculated value of L is correct but at last they print its value wrong\n" + ] + } + ], + "source": [ + "#pg 271\n", + "#calculate the reluctance and inductance\n", + "# Given data\n", + "from math import pi\n", + "miu_r = 1400;\n", + "l = 70;# in cm\n", + "l = l * 10**-2;# in m\n", + "a = 5;# in cm**2\n", + "a = a * 10**-4;# in m**2\n", + "N = 1000;\n", + "#calculations\n", + "miu_o = 4*pi*10**-7;\n", + "S = l/(miu_o*miu_r*a);# in AT/Wb\n", + "# Calculation of inductance of the coil\n", + "L = (N**2)/S;# in H\n", + "#results\n", + "print '%s %.3e' %(\"The reluctance of the magnetic circuit in AT/Wb is\",S)\n", + "print \"The inductance of the coil in H is\",round(L,3)\n", + "\n", + "print 'Note: In the book the calculated value of L is correct but at last they print its value wrong'\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7: pg 271" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current in A is 12.223\n" + ] + } + ], + "source": [ + "#pg 271\n", + "#calculate the current\n", + "# Given data\n", + "from math import pi\n", + "l1 = 25.;# in cm\n", + "l1 = l1 * 10**-2;# in m\n", + "miu_o = 4*pi*10**-7;\n", + "miu_r = 750;\n", + "a1 = 2.5*2.5*10**-4;# in m\n", + "S1 = l1/(miu_o*miu_r*a1);# in AT/Wb\n", + "l2 = 40;# in cm\n", + "l2 = l2 * 10**-2;# in m\n", + "S2 = l2/(miu_o*miu_r*a1);# in AT/Wb\n", + "phi2 = 2.5*10**-3;# in Wb\n", + "N = 500;\n", + "#calculations\n", + "# mmf = phi1*S1 = phi2*S2;\n", + "phi1 = (phi2*S2)/S1;# in Wb\n", + "phi = phi1+phi2;# in Wb\n", + "# Sum of mmf required for AEFB\n", + "S_AEFB = S2;# in AT/Wb\n", + "mmfforAEFB = S_AEFB*phi;# mmf for AEFB in AT\n", + "totalmmf = mmfforAEFB+(phi1*S1);# total mmf in AT\n", + "# N*I = totalmmf;\n", + "# Calculation of current\n", + "I = totalmmf/N;# in A\n", + "#results\n", + "print \"The current in A is\",round(I,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8: pg 272" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current in A is 4.713\n" + ] + } + ], + "source": [ + "#pg 272\n", + "#calculate the current\n", + "# Given data\n", + "from math import pi\n", + "a = 16*10**-4;# in m**2\n", + "lg = 2*10**-3;# in m\n", + "N = 1000;\n", + "phi = 4*10**-3;# in Wb\n", + "miu_r = 2000;\n", + "miu_o = 4*pi*10**-7;\n", + "l=25.;# length of magnetic in cm\n", + "w= 20.;# in cm (width)\n", + "t= 4.;# in cm (thickness)\n", + "#calculations\n", + "li= ((w-t)*t/2+(l-t)*t/2-0.2);# in cm\n", + "li= li*10**-2;# in m\n", + "S_T= 1/(miu_o*a)*(li/miu_r+lg)\n", + "# Calculation of current with the help of flux\n", + "# phi = mmf/S_T = N*I/S_T;\n", + "I = (phi*S_T)/N;# in A\n", + "#results\n", + "print \"The current in A is\",round(I,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9: pg 273" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total flux in the ring in muWb is 462.0\n" + ] + } + ], + "source": [ + "#pg 273\n", + "#calculate the total flux\n", + "# Given data\n", + "from math import pi\n", + "N = 500.;\n", + "R = 4.;# in ohm\n", + "d_mean = 0.25;# in m\n", + "a = 700;# in mm^2\n", + "a = a * 10**-6;# in m\n", + "V = 6;# in V\n", + "miu_r = 550;\n", + "miu_o = 4*pi*10**-7;\n", + "#calculations\n", + "l_i = pi*d_mean;# in m\n", + "S = l_i/(miu_o*miu_r*a);# in AT/Wb\n", + "I = V/R;# in A\n", + "# Calculation of mmf\n", + "mmf = N*I;# in AT\n", + "# total flux\n", + "phi = mmf/S;# in Wb \n", + "phi = phi * 10**6;# in muWb\n", + "#results\n", + "print \"The total flux in the ring in muWb is\",phi\n", + "\n", + "# Note: In the book the value of flux calculated correct in muWb but at last they print only in Wb, so the answer in the book is wrong.\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10: pg 274" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current required in A is 0.7958\n", + "The coil inductance in H is 0.3142\n" + ] + } + ], + "source": [ + "#pg 274\n", + "#calculate the current and coil inductance\n", + "# Given data\n", + "from math import pi\n", + "N = 1000.;\n", + "a = 5;# in cm**2\n", + "a = a * 10**-4;# in m**2\n", + "l_g = 2;# in mm\n", + "l_g = l_g * 10**-3;# in m\n", + "B = 0.5;# in T\n", + "#calculations\n", + "#miu_r= inf;\n", + "phi = B*a;# in Wb\n", + "miu_o = 4*pi*10**-7;\n", + "S = l_g/(miu_o*a);# in AT/Wb\n", + "# Calculation of current with the help of flux\n", + "# phi = mmf/S = N*I/S;\n", + "I = (phi*S)/N;# in A\n", + "# Evaluation of coil inductance\n", + "L = (N**2)/S;# in H\n", + "#results\n", + "print \"The current required in A is\",round(I,4)\n", + "print \"The coil inductance in H is\",round(L,4)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11: pg 274" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The amphere turns in AT is 4138.03\n" + ] + } + ], + "source": [ + "#pg 274\n", + "#calculate the amphere turns\n", + "# Given data\n", + "from math import pi\n", + "l_g = 4.;# in mm\n", + "l_g = l_g * 10**-3;# in m\n", + "Bg = 1.3;# in Wb/m**2\n", + "#calculations\n", + "miu_o = 4*pi*10**-7;\n", + "Hg = Bg/miu_o;\n", + "# Ampere turns for the gap\n", + "AT = Hg*l_g;# in AT\n", + "#results\n", + "print \"The amphere turns in AT is\",round(AT,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12: pg 274" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The mmf required in AT is 1492.0\n" + ] + } + ], + "source": [ + "#pg 274\n", + "#calculate the mmf required\n", + "# Given data\n", + "from math import pi\n", + "phi = 0.015;# in Wb\n", + "l_g = 2.5;# in mm\n", + "l_g = l_g * 10**-3;# in m\n", + "a = 200;# in cm**2\n", + "a = a * 10**-4;# in m**2\n", + "miu_o = 4*pi*10**-7;\n", + "# Calculation of reluctance of air gap\n", + "Sg = l_g/(miu_o*a);# in AT/Wb\n", + "mmf = phi*Sg;# in AT\n", + "#results\n", + "print \"The mmf required in AT is\",round(mmf)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13: pg 275" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The reluctance of magnetic circuit in AT/Wb is 255056.0\n", + "The reluctance of air gap in AT/Wb is 2652582.0\n", + "Total reluctance in AT/Wb is 2907638.0\n", + "The total flux in mWb is 0.27514\n", + "The flux density of air gap in Wb/m^2 is 0.2293\n" + ] + } + ], + "source": [ + "#pg 275\n", + "#calculate the reluctance, flux \n", + "# Given data\n", + "from math import pi\n", + "a = 12;# in cm**2\n", + "a = a * 10**-4;# in m**2\n", + "l_i = 50;# in cm\n", + "l_i = l_i * 10**-2;# in m\n", + "l_g = 0.4;# in cm\n", + "l_g = l_g * 10**-2;# in m\n", + "N = 2*400;\n", + "I = 1;# in A\n", + "miu_r = 1300;\n", + "#calculations\n", + "miu_o = 4*pi*10**-7;\n", + "Si = l_i/(miu_o*miu_r*a);# in AT/Wb\n", + "print \"The reluctance of magnetic circuit in AT/Wb is\",round(Si)\n", + "miu_r = 1;\n", + "Sg = l_g/(miu_o*miu_r*a);# in AT/Wb\n", + "print \"The reluctance of air gap in AT/Wb is\",round(Sg)\n", + "S_T = Si+Sg;# in AT/Wb\n", + "print \"Total reluctance in AT/Wb is\",round(S_T)\n", + "mmf = N*I;# in AT\n", + "phi_T = mmf/S_T;# in Wb\n", + "phi_T= phi_T*10**3;# in mWb\n", + "print \"The total flux in mWb is\",round(phi_T,5)\n", + "phi_T= phi_T*10**-3;# in Wb\n", + "# phi_T =B*a;\n", + "B = (phi_T)/a;# in Wb/m**2\n", + "print \"The flux density of air gap in Wb/m^2 is\",round(B,4)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14: pg 276" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current required in A is 2.7224\n" + ] + } + ], + "source": [ + "#pg 276\n", + "#calculate the current required\n", + "# Given data\n", + "from math import pi\n", + "l = 30.;# in cm\n", + "d = 2.;# in cm\n", + "N = 500.;\n", + "phi = 0.5;# in mWb\n", + "Airgap = 1;# in mm\n", + "miu_r = 4000;\n", + "#calculations\n", + "miu_o = 4*pi*10**-7;\n", + "Ac = (pi/4)*(d**2);# in cm^2\n", + "Ac = Ac * 10**-4;# in m^2\n", + "l_i = (l*10**-2)-(Airgap*10**-3);# in m\n", + "l_g = 1;# in mm\n", + "l_g = l_g * 10**-3;# in m\n", + "Si = l_i/(miu_r*miu_o*Ac);# in AT/Wb\n", + "Sg = l_g/(miu_o*Ac);# in AT/Wb\n", + "S =Si+Sg;# in AT/Wb\n", + "# phi = mmf/S = N*I/S;\n", + "I = (phi*10**-3*S)/N;# in A\n", + "#results\n", + "print \"The current required in A is\",round(I,4)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15: pg 276" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The inductance of the coil in H is 0.3593\n" + ] + } + ], + "source": [ + "#pg 276\n", + "#calculate the inductance\n", + "# Given data\n", + "from math import pi\n", + "l = 40;# in cm\n", + "l = l * 10**-2;# in m\n", + "a = 4;# in cm**2\n", + "a = a * 10**-4;# in m**2\n", + "miu_r = 1000;\n", + "miu_o = 4*pi*10**-7;\n", + "l_g = 1;# in mm\n", + "l_g = l_g * 10**-3;# in m\n", + "N = 1000;\n", + "#calculations\n", + "l_i = l-l_g;# in m\n", + "Si = l_i/(miu_r*miu_o*a);# in AT/Wb\n", + "Sg = l_g/(miu_o*a);# in AT/Wb\n", + "S = Si+Sg;# in AT/Wb\n", + "# The inductance of the coil \n", + "L = (N**2)/S;# in H\n", + "#results\n", + "print \"The inductance of the coil in H is\",round(L,4)\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter9.ipynb b/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter9.ipynb new file mode 100644 index 00000000..7415c45f --- /dev/null +++ b/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/Chapter9.ipynb @@ -0,0 +1,1090 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9: Single Phase Transformer" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1: pg 305" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The numbers of primary turns is 860.0\n", + "The secondary full load current in A is 200.0\n", + "The primary full load current in A is 20.0\n" + ] + } + ], + "source": [ + "# Exa 9.1\n", + "#pg 305\n", + "#calculate the number of turns, load current\n", + "# Given data\n", + "V1 = 3000.;# in V\n", + "V2 = 300.;# in V\n", + "N2 = 86.;# in Turns\n", + "Rating = 60.*10**3;# in VA\n", + "#calculations\n", + "K = V2/V1;\n", + "#Transformer ratio, N2/N1 = K;\n", + "N1 = N2/K;# in turns\n", + "I2 = Rating/V2;# in A\n", + "I1 = Rating/V1;# in A\n", + "#results\n", + "print \"The numbers of primary turns is\",N1\n", + "print \"The secondary full load current in A is\",I2\n", + "print \"The primary full load current in A is\",I1\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2: pg 306" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum flux density in Wb/m^2 is 0.75\n" + ] + } + ], + "source": [ + "# Exa 9.2\n", + "#pg 306\n", + "#calculate the max flux density\n", + "# Given data\n", + "E1 = 3000.;# in V\n", + "E2 = 200.;# in V\n", + "f = 50.;# in Hz\n", + "a = 150.;# in cm**2\n", + "N2 = 80.;# turns\n", + "#calculations\n", + "#Formula E2 = 4.44*phi_m*f*N2;\n", + "phi_m = E2/(4.44*f*N2);# in Wb\n", + "Bm = phi_m/(a*10**-4);# in Wb/m**2\n", + "#results\n", + "print \"The maximum flux density in Wb/m^2 is\",round(Bm,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3: pg 306" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The primary full load current in A is 8.33\n", + "The secondary full load current in A is 104.2\n", + "The secondary emf in V is 240.0\n", + "The maximum core flux in mWb is 27.0\n" + ] + } + ], + "source": [ + "# Exa 9.3\n", + "#pg 306\n", + "#calculate the load current, emf and core flux\n", + "# Given data\n", + "N1 = 500.;\n", + "N2 = 40.;\n", + "E1 = 3000.;# in V\n", + "f = 50.;# in Hz\n", + "Rating = 25*10**3;# in VA\n", + "#calculations\n", + "K = N2/N1;\n", + "I1 = Rating/E1;# in A\n", + "print \"The primary full load current in A is\",round(I1,2)\n", + "I2 = I1/K;# in A\n", + "print \"The secondary full load current in A is\",round(I2,1)\n", + "# K = E2/E1;\n", + "E2 = K*E1;# in V\n", + "print \"The secondary emf in V is\",round(E2)\n", + "# e.m.f equation of the transformer, E1 = 4.44*phi_m*f*N1;\n", + "phi_m = E1/(4.44*f*N1);# in Wb\n", + "phi_m = phi_m*10**3;# in mWb\n", + "print \"The maximum core flux in mWb is\",round(phi_m)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4: pg 307" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The active component of current in A is 1.522\n", + "The magnetizing component of current in A is 14.923\n" + ] + } + ], + "source": [ + "# Exa 9.4\n", + "#pg 307\n", + "#calculate the current\n", + "# Given data\n", + "from math import acos,cos,sin\n", + "Rating = 25.;# in KVA\n", + "f = 50.;# in Hz\n", + "Io = 15.;# in A\n", + "Wo = 350.;# in W\n", + "Vo = 230.;# in V\n", + "#calculations\n", + "# No load power factor\n", + "phi_o = acos(Wo/(Vo*Io));\n", + "# active component of current \n", + "Ic = Io*cos(phi_o);# in A\n", + "print \"The active component of current in A is\",round(Ic,3)\n", + "# magnetizing component of current \n", + "Im = Io*sin(phi_o);# in A\n", + "print \"The magnetizing component of current in A is\",round(Im,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5: pg 307" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Equivalent resistance reffered to primary in ohm is 3.55\n", + "Equivalent resistance reffered to secondary in ohm is 0.008875\n", + "Equivalent reactance reffered to primary in ohm is 5.6\n", + "Equivalent reactance reffered to secondary in ohm is 0.014\n", + "Equivalent impedance reffered to primary in ohm is : 6.6304\n", + "Equivalent impedance reffered to secondary in ohm is : 0.01658\n" + ] + } + ], + "source": [ + "# Exa 9.5\n", + "#pg 307\n", + "#calculate the resistance and reactance\n", + "# Given data\n", + "V1 = 2200.;# in V\n", + "V2 = 110.;# in V\n", + "R1 = 1.75;# in ohm\n", + "R2 = 0.0045;# in ohm\n", + "X1 = 2.6;# in ohm\n", + "X2 = 0.0075;# in ohm\n", + "#calculations\n", + "K = V2/V1;\n", + "#R1e = R1+R_2 = R1 + (R2/(K**2));\n", + "R1e = R1 + (R2/(K**2));# in ohm\n", + "print \"Equivalent resistance reffered to primary in ohm is\",R1e\n", + "# R2e = R2+R_1 = R2+((K**2)*R1);\n", + "R2e = R2+((K**2)*R1);# in ohm\n", + "print \"Equivalent resistance reffered to secondary in ohm is\",R2e\n", + "#X1e = X1+X_2 = X1+(X2/(K**2));\n", + "X1e = X1+(X2/(K**2));# in ohm\n", + "print \"Equivalent reactance reffered to primary in ohm is\",X1e\n", + "# X2e = X2+X_1 = X2 + ((K**2)*X1);\n", + "X2e = X2 + ((K**2)*X1);# in ohm\n", + "print \"Equivalent reactance reffered to secondary in ohm is\",X2e\n", + "Z1e= R1e+1j*X1e;# in ohm\n", + "Z2e= R2e+1j*X2e;# in ohm\n", + "print \"Equivalent impedance reffered to primary in ohm is : \",round(abs(Z1e),4)\n", + "print \"Equivalent impedance reffered to secondary in ohm is : \",round(abs(Z2e),5)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6: pg 308" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The equivalent impedance of the transformer reffered to primary in ohm is 2.05\n", + "The total copper loss in W is 1136.36\n" + ] + } + ], + "source": [ + "# Exa 9.6\n", + "#pg 308\n", + "#calculate the impedance and copper loss\n", + "from math import sqrt\n", + "# Given data\n", + "V1 = 2200.;# in V\n", + "V2 = 440.;# in V\n", + "R1 = 0.3;# in ohm\n", + "R2 = 0.01;# in ohm\n", + "X1 = 1.1;# in ohm\n", + "X2 = 0.035;# in ohm\n", + "#calculations\n", + "K = V2/V1;\n", + "Rating = 100;# in KVA\n", + "I1 = (Rating*10**3)/V1;# in A\n", + "I2 = (Rating*10**3)/V2;# in A\n", + "R1e = R1 + (R2/(K**2));# in ohm\n", + "X1e = X1+(X2/(K**2));# in ohm\n", + "Z1e = sqrt( (R1e**2) + (X1e**2) );# in ohm\n", + "# Total copper loss\n", + "totalcopperloss = (I1**2)*R1e;# in W\n", + "#results\n", + "print \"The equivalent impedance of the transformer reffered to primary in ohm is\",round(Z1e,2)\n", + "print \"The total copper loss in W is\",round(totalcopperloss,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7: pg 309" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The efficiency in percent is 95.24\n" + ] + } + ], + "source": [ + "# Exa 9.7\n", + "#pg 309\n", + "#calculate the efficiency\n", + "from math import cos,acos\n", + "# Given data\n", + "Rating = 150000.;# in VA\n", + "phi= acos(0.8);# in radians\n", + "Pcu = 1600.;# in W\n", + "Pi = 1400.;# in W\n", + "n = 1/4.;\n", + "#calculations\n", + "# Total loss of 25% load\n", + "totalloss = Pi + (n**2)*Pcu;# in W\n", + "# efficiency of transformer of 25% load\n", + "Eta = n*Rating*cos(phi)/(n*Rating*cos(phi)+Pi+n**2*Pcu)*100;# in %\n", + "#results\n", + "print \"The efficiency in percent is\",round(Eta,2)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8: pg 309" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The efficiency of full load power in percent is 97.09\n", + "The efficiency of half load power in percent is 96.53\n" + ] + } + ], + "source": [ + "# Exa 9.8\n", + "#pg 309\n", + "#calculate the efficiency\n", + "# Given data\n", + "from math import acos\n", + "Rating = 25.;# in KVA\n", + "V1 = 2000.;# in V\n", + "V2 = 200.;# in V\n", + "Pi = 350.;# in W\n", + "Pi = Pi * 10**-3;# in kW\n", + "Pcu = 400.;# in W\n", + "Pcu = Pcu * 10**-3;# in kW\n", + "#calculations\n", + "phi= acos(1);# in radians\n", + "output = Rating;\n", + "losses = Pi+Pcu;\n", + "Eta = (output/(output + losses))*100;# %Eta in %\n", + "print \"The efficiency of full load power in percent is\",round(Eta,2)\n", + "# For half load\n", + "output = Rating/2;# in kW\n", + "h = 1.;\n", + "Pcu = Pcu*((h/2)**2);# in kW\n", + "losses = Pi+Pcu;\n", + "# efficiency of half load power \n", + "Eta = (output/(output+losses))*100;# in %\n", + "print \"The efficiency of half load power in percent is\",round(Eta,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9: pg 310" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The efficiency at full load in percent is 98.14\n", + "The maximum efficiency in kVA is 237.17\n", + "The maximum efficiency in percent is 98.138\n" + ] + } + ], + "source": [ + "# Exa 9.9\n", + "#pg 310\n", + "#calculate the efficiency\n", + "from math import acos,cos,sqrt\n", + "# Given data\n", + "Rating = 250*10**3;# in VA\n", + "Pi = 1.8;# in kW\n", + "Pi = Pi * 10**3;# in W\n", + "Pcu_f1 = 2000;# in W\n", + "phi= acos(0.8);# in radians\n", + "Eta = ((Rating*cos(phi))/((Rating*cos(phi))+Pi+Pcu_f1))*100;# %Eta in %\n", + "print \"The efficiency at full load in percent is\",round(Eta,2)\n", + "# The maximum efficiency \n", + "Eta_max = Rating * sqrt(Pi/Pcu_f1 );# in VA\n", + "Eta_max = Eta_max *10**-3;# in kVA\n", + "print \"The maximum efficiency in kVA is\",round(Eta_max,2)\n", + "Eta_max = Eta_max *10**3;# in VA\n", + "Pcu = Pi;# in W\n", + "Eta_max1 = ((Eta_max*cos(phi))/((Eta_max*cos(phi)) + Pi+Pcu ))*100;# in %\n", + "print \"The maximum efficiency in percent is\",round(Eta_max1,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10: pg 311" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The iron loss in W is : 18518.519\n", + "The full load copper loss in watt 37037.04\n" + ] + } + ], + "source": [ + "# Exa 9.10\n", + "#pg 311\n", + "#calculate the iron loss, full load copper loss\n", + "# Given data\n", + "from math import acos,cos\n", + "phi= acos(1);# in radians\n", + "Pout = 500.;# in kW\n", + "Pout = Pout*10**3;# in W\n", + "Eta = 90.;# in %\n", + "n=1/2.;\n", + "#calculations\n", + "# For full load, Eta= Pout*100/(Pout+Pi+Pcu_f1) or Pi+Pcu_f1= (Pout*100-Eta*Pout)/Eta (i)\n", + "# For half load, Eta= n*Pout*100/(n*Pout+Pi+n**2*Pcu_f1) or Pi+n**2*Pcu_f1= (n*Pout*100-n*Eta*Pout)/Eta (ii)\n", + "# From eq(i) and (ii)\n", + "Pcu_fl= ((n*Pout*100-n*Eta*Pout)/Eta-(Pout*100-Eta*Pout)/Eta)/(n**2-1)\n", + "Pi=(Pout*100-Eta*Pout)/Eta-Pcu_fl\n", + "#results\n", + "print \"The iron loss in W is : \",round(Pi,3)\n", + "print \"The full load copper loss in watt\",round(Pcu_fl,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11: pg 311" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The magnetizing component of no load current in A is 9.68\n", + "The iron loss in W is 1000.0\n", + "The maximum value of flux in mWb is 3.6\n" + ] + } + ], + "source": [ + "# Exa 9.11\n", + "#pg 311\n", + "#calculate the flux, iron loass and load current\n", + "# Given data\n", + "from math import acos,sin,cos\n", + "Io = 10;# in A\n", + "phi_o= acos(0.25);# in radians\n", + "V1 = 400.;# in V\n", + "f = 50.;# in Hz\n", + "N1 =500.;\n", + "Im = Io*sin(phi_o);# in A\n", + "print \"The magnetizing component of no load current in A is\",round(Im,2)\n", + "Pi = V1*Io*cos(phi_o);# in W\n", + "print \"The iron loss in W is\",Pi\n", + "E1 = V1;# in V\n", + "#E1 v= 4.44*f*phi_m*N1;\n", + "phi_m = E1/(4.44*f*N1);# in Wb\n", + "phi_m=phi_m*10**3;# in mWb\n", + "print \"The maximum value of flux in mWb is\",round(phi_m,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12: pg 312" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The equivalent resistance to primary side in ohm is 5.0\n", + "The equivalent reactance to primary side in ohm is 6.5\n", + "The equivalent impedance to primary side in ohm is 8.2\n", + "Total copper loss in W is 1125.0\n" + ] + } + ], + "source": [ + "# Exa 9.12\n", + "#pg 312\n", + "#calculate the equivalent resistance\n", + "# Given data\n", + "from math import sqrt\n", + "Rating = 30.*10**3;# in VA\n", + "V1 = 2000.;# in V\n", + "V2 = 200.;# in V\n", + "f = 50.;# in Hz\n", + "R1 = 3.5;# in ohm\n", + "X1 = 4.5;# in ohm\n", + "R2 = 0.015;# in ohm\n", + "X2 = 0.02;# in ohm\n", + "#calculations\n", + "K = V2/V1;\n", + "R1e = R1 + (R2/(K**2));# in ohm\n", + "print \"The equivalent resistance to primary side in ohm is\",R1e\n", + "X1e = X1 + (X2/(K**2));# in ohm\n", + "print \"The equivalent reactance to primary side in ohm is\",X1e\n", + "Z1e = sqrt( (R1e**2) + (X1e**2) );# in ohm\n", + "print \"The equivalent impedance to primary side in ohm is\",round(Z1e,1)\n", + "I1 = Rating/V1;# in A\n", + "# Total copper loss in transformer\n", + "Pcu_total = (I1**2)*R1e;# in W\n", + "print \"Total copper loss in W is\",Pcu_total\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13: pg 313" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The full load secondary voltage in V is 377.65\n" + ] + } + ], + "source": [ + "# Exa 9.13\n", + "#pg 313\n", + "#calculate the full load secondary voltage\n", + "from math import cos,sin,acos\n", + "# Given data\n", + "Rating = 10;# in KVA\n", + "phi= acos(0.8)\n", + "V1 = 2000.;# in V\n", + "V2 = 400.;# in V\n", + "R1 = 5.5;# in ohm\n", + "X1 = 12;# in ohm\n", + "R2 = 0.2;# in ohm\n", + "X2 = 0.45;# in ohm\n", + "K = V2/V1;\n", + "#R1e = R1 + R_2 = R1 + (R2/(K**2));\n", + "R1e = R1 + (R2/(K**2));# in ohm\n", + "#X1e = X1 + X_ = X1 + (X2/(K**2));\n", + "X1e = X1 + (X2/(K**2));# in ohm\n", + "I2 = (Rating*10**3)/V2;# in A\n", + "R2e = (K**2)*R1e;# in ohm\n", + "X2e = (K**2)*X1e;# in ohm\n", + "Vdrop = I2 * ( (R2e*cos(phi)) + (X2e*sin(phi)) );# voltage drop in V\n", + "#E2 = V2 +Vd;\n", + "E2 = V2;# in V\n", + "# The full load secondary voltage \n", + "V2 = E2-Vdrop;# in V\n", + "#results\n", + "print \"The full load secondary voltage in V is\",V2\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14: pg 313" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Full load efficiency in percent is 96.77\n", + "The percentage of the full load in percent is 70.71\n" + ] + } + ], + "source": [ + "# Exa 9.14\n", + "#pg 313\n", + "#calculate the efficiency\n", + "from math import cos,acos,sqrt\n", + "# Given data\n", + "Rating = 40*10**3;# in VA\n", + "Pi = 400.;# in W\n", + "Pcu_f1 = 800.;# in W\n", + "#calculations\n", + "phi= acos(0.9);# in radians\n", + "Eta_f1 = ((Rating*cos(phi))/( (Rating*cos(phi)) + Pi + Pcu_f1 ))*100;# in %\n", + "print \"Full load efficiency in percent is\",round(Eta_f1,2)\n", + "# percentage of the full load\n", + "Eta_max = Rating*sqrt( Pi/Pcu_f1);# in KVA\n", + "Eta_max = Eta_max/Rating*100;# in %\n", + "print \"The percentage of the full load in percent is\",round(Eta_max,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15: pg 314" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The full load efficiency in percent is 97.56\n" + ] + } + ], + "source": [ + "#pg 314\n", + "# Exa 9.15\n", + "#calculate the full load efficiency\n", + "from math import cos,acos\n", + "# Given data\n", + "Rating = 8*10**3;# in VA\n", + "phi= acos(0.8);# in radians\n", + "V1 = 400.;# in V\n", + "V2 = 100.;# in V\n", + "f = 50.;# in Hz\n", + "Pi = 60.;# in W\n", + "Wo = Pi;# in W\n", + "Pcu = 100.;# in W\n", + "#results\n", + "# The full load efficiency \n", + "Eta_f1 = ((Rating*cos(phi))/((Rating*cos(phi)) + Pi + Pcu))*100;# in %\n", + "#results\n", + "print \"The full load efficiency in percent is\",round(Eta_f1,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16: pg 314" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The full load efficiency in percent is 96.0\n" + ] + } + ], + "source": [ + "#pg 314\n", + "# Exa 9.16\n", + "# Given data\n", + "from math import acos,cos\n", + "Rating = 10*10**3;# in VA\n", + "phi= acos(0.8);# in radians\n", + "V1 = 500.;# in V\n", + "V2 = 250.;# in V\n", + "Pi = 200.;# in W\n", + "Pcu = 300.;# in W\n", + "Isc = 30.;# in A\n", + "#calculations\n", + "I1 = Rating/V1;# in A\n", + "# Pcu/(Pcu(f1)) = (Isc**2)/(I1**2);\n", + "Pcu_f1 = Pcu * ((I1**2)/(Isc**2));# in W\n", + "# The efficiency at full load\n", + "Eta_f1 = Rating*cos(phi)/(Rating*cos(phi) + Pi + Pcu_f1)*100;# in %\n", + "#results\n", + "print \"The full load efficiency in percent is\",Eta_f1\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17: pg 315" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum efficiency of transformer in percent is 97.683\n" + ] + } + ], + "source": [ + "#pg 315\n", + "# Exa 9.17\n", + "#calculate the max efficiency\n", + "# Given data\n", + "from math import acos,cos,sqrt\n", + "Rating = 20*10**3;# in VA\n", + "phi= acos(0.8);# in radians\n", + "V1 = 2000.;# in V\n", + "V2 = 200.;# in V\n", + "Pi = 120.;# in W\n", + "Pcu = 300.;# in W\n", + "#calculations\n", + "Eta_max = Rating*(sqrt( Pi/Pcu ));# in VA\n", + "Pcu = Pi;# in W\n", + "# The maximum efficiency of transformer \n", + "Eta_max = ((Eta_max*cos(phi))/( Eta_max*cos(phi) + (2*Pi) ))*100;# in %\n", + "#results\n", + "print \"The maximum efficiency of transformer in percent is\",round(Eta_max,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18: pg 315" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The equivalent parameters referred to secondary side are : \n", + "The value of R_2e is : 0.041 ohm\n", + "The value of X_2e is : 0.248 ohm\n", + "The value of R''c is : 14.0 ohm\n", + "The value of X''m is : 3.92 ohm\n" + ] + } + ], + "source": [ + "#pg 315\n", + "# Exa 9.18\n", + "#calculate the resistances\n", + "# Given data\n", + "Turnratio = 5;\n", + "R1 = 0.5;# in ohm\n", + "R2 = 0.021;# in ohm\n", + "X1 = 3.2;# in ohm\n", + "X2 = 0.12;# in ohm\n", + "Rc = 350.;# in ohm\n", + "Xm = 98.;# in ohm\n", + "N1 = 5.;\n", + "N2 = 1.;\n", + "#calculations\n", + "K = N2/N1;\n", + "# Evaluation of the equivalent parameters referred to secondary side \n", + "R2e = R2 + ((K**2)*R1);# in ohm\n", + "print \"The equivalent parameters referred to secondary side are : \"\n", + "print \"The value of R_2e is : \",R2e,\" ohm\"\n", + "X2e = X2 + ((K**2)*X1);# in ohm\n", + "print \"The value of X_2e is : \",X2e,\" ohm\"\n", + "R_c = (K**2)*Rc;# in ohm\n", + "print \"The value of R''c is : \",R_c,\" ohm\"\n", + "X_m = (K**2)*Xm;# in ohm\n", + "print \"The value of X''m is : \",X_m,\" ohm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19: pg 315" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of R''o is : 24.2 ohm\n", + "The value of X''o is : 5.0 ohm\n", + "The value of R1e is : 36.3073 ohm\n", + "The value of Z1e is : 55.0055 ohm\n", + "The value of X1e is : 41.3206 ohm\n", + "The value of R2e is : 0.0145 ohm\n", + "The value of X2e is : 0.0165 ohm\n", + "The value of Z2e is : 0.022 ohm\n" + ] + } + ], + "source": [ + "#pg 315\n", + "# Exa 9.19\n", + "#calculate the resistances\n", + "# Given data\n", + "from math import acos,cos,sin,sqrt\n", + "Rating = 100.*10**3;# in VA\n", + "V1 = 11000.;# in V\n", + "V2 = 220.;# in V\n", + "Wo = 2*10**3;# in W\n", + "Vo = 220.;# in V\n", + "Io = 45.;# in A\n", + "#calculations\n", + "phi_o = acos(Wo/(Vo*Io));\n", + "I_c = Io*cos(phi_o);# in A\n", + "I_m = Io*sin(phi_o);# in A\n", + "Ro= V2/I_c;# in ohm\n", + "Xo= V2/I_m;# in ohm\n", + "Wsc= 3*10**3;# in W\n", + "Vsc= 500.;# in V\n", + "Isc= 9.09;# in A\n", + "R1e= Wsc/Isc**2;# in ohm\n", + "Z1e= Vsc/Isc;# in ohm\n", + "X1e= sqrt(Z1e**2-R1e**2);# in ohm\n", + "K= V2/V1;\n", + "R2e= K**2*R1e;# in ohm\n", + "X2e= K**2*X1e;# in ohm\n", + "Z2e= K**2*Z1e;# in ohm\n", + "#results\n", + "print \"The value of R''o is : \",Ro,\" ohm\"\n", + "print \"The value of X''o is : \",round(Xo),\" ohm\"\n", + "print \"The value of R1e is : \",round(R1e,4),\" ohm\"\n", + "print \"The value of Z1e is : \",round(Z1e,4),\" ohm\"\n", + "print \"The value of X1e is : \",round(X1e,4),\" ohm\"\n", + "print \"The value of R2e is : \",round(R2e,4),\" ohm\"\n", + "print \"The value of X2e is : \",round(X2e,4),\" ohm\"\n", + "print \"The value of Z2e is : \",round(Z2e,4),\" ohm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20: pg 316" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The efficiency of the transformer in percent is 96.59\n" + ] + } + ], + "source": [ + "#pg 316\n", + "# Exa 9.20\n", + "#calculate the efficiency\n", + "# Given data\n", + "from math import acos,cos\n", + "V1 = 250.;# in V\n", + "V2 = 500.;# in V\n", + "Pcu = 100.;# in W\n", + "Pi = 80.;# in W\n", + "V = V2;# in V\n", + "A = 12.;# in A\n", + "#calculations\n", + "phi= acos(0.85);# in radians\n", + "# The efficiency of the transformer \n", + "Eta = ((V*A*cos(phi))/( V*A*cos(phi) + Pi+Pcu ))*100;# in %\n", + "print \"The efficiency of the transformer in percent is\",round(Eta,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21: pg 317" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The iron loss on full load and half load remain same in W which are : 1012.0226\n", + "The copper loss on full load in W is : 2972.993\n", + "The copper loss on half load in W is : 743.248\n" + ] + } + ], + "source": [ + "#pg 317\n", + "# Exa 9.21\n", + "#calculate the iron and copper loss\n", + "# Given data\n", + "from math import acos,cos,sin\n", + "VA = 400.*10**3;# in Mean\n", + "Eta_fl = 98.77/100;# in %\n", + "phi1= acos(0.8);# in radians\n", + "phi2= acos(1);# in radians\n", + "Eta_hl = 99.13/100;# in %\n", + "n = 1/2.;\n", + "#calculations\n", + "#For full load, Eta_f1 = ((VA*cosd(phi1))/( VA*cosd(phi1) + Pi + Pcu_f1 )) or Pi+Pcu_f1 = VA*cosd(phi1)*(1-Eta_fl)/(Eta_f1) (i)\n", + "#For half load, Eta_hl = n*VA*cosd(phi2)/(n*VA*cosd(phi2)+Pi+n**2*Pcu_f1) or Pi+n**2*Pcu_f1 = n*VA*cosd(phi2)*( 1-Eta_hl)/Eta_hl (ii)\n", + "# From eq(i) and (ii)\n", + "Pcu_fl=(n*VA*cos(phi2)*( 1-Eta_hl)/Eta_hl-VA*cos(phi1)*(1-Eta_fl)/(Eta_fl))/(n**2-1);# in W\n", + "Pi=VA*cos(phi1)*(1-Eta_fl)/(Eta_fl)-Pcu_fl;# in W\n", + "# The copper loss on half load \n", + "C_loss_half_load=n**2*Pcu_fl;# in W \n", + "#results\n", + "print \"The iron loss on full load and half load remain same in W which are : \",round(Pi,4)\n", + "print \"The copper loss on full load in W is : \",round(Pcu_fl,3)\n", + "print \"The copper loss on half load in W is : \",round(C_loss_half_load,3)\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 22: pg 317" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The efficiency of a transformer in percent is 98.3913\n" + ] + } + ], + "source": [ + "#pg 317\n", + "# Exa 9.22\n", + "#calculate the efficiency\n", + "# Given data\n", + "from math import cos,acos\n", + "VA = 100.*10**3;# in VA\n", + "Eta_max = 98.40/100;# in %\n", + "Eta_max1 = 90./100;# in %\n", + "phi= acos(1);# in radians\n", + "#Eta_max = (Eta_max1*VA*cos(phi)/(Eta_max1*VA*cos(phi) + 2*Pi);\n", + "Pi = (Eta_max1*VA*cos(phi)/Eta_max - Eta_max1*VA*cos(phi))/2;# in W\n", + "Pcu = Pi;# in W\n", + "n = 0.9;\n", + "# Pcu_fl/Pcu = (VA/(0.9*VA) )**2;\n", + "Pcu_fl = Pcu*(VA/(0.9*VA) )**2;# in W\n", + "Eta_fl = ( (VA*cos(phi))/( (VA*cos(phi)) + Pi + Pcu_fl ) )*100;# in %\n", + "#results\n", + "print \"The efficiency of a transformer in percent is\",round(Eta_fl,4)\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/screenshots/chap1.png b/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/screenshots/chap1.png Binary files differnew file mode 100644 index 00000000..2a86ad69 --- /dev/null +++ b/Introduction_to_Electrical_Engineering_by_Er._J.P._Navani_&_Er._Sonal_Sapra/screenshots/chap1.png diff --git 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