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+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:e3360cc86683c7a5706db81b033b8664f6c6abe93730acff7b5ab6cc3ddee4ce"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3 : Application of Distribution Transformers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1  Page No : 118"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "S = 25.*(10**3);         #Rating of the transformer in VA\n",
+      "#Values in per unit\n",
+      "Rt = 0.014;         #Resismath.tance of Transformer\n",
+      "Xt = 0.012;         #Reacmath.tance of transformer\n",
+      "Vh = 7200;         #High Voltage End in V\n",
+      "Vx = 120;         # Low Voltage End in V\n",
+      "Rb = (Vh**2)/S;         #Base Value of Resismath.tance\n",
+      "#Accroding to Lloyd's Formula\n",
+      "\n",
+      "# Calculations\n",
+      "Zhx12 = (1.5*Rt)+(1j*1.2*Xt);         #Impedance referred to HV side when the winding x2x3 is shorted\n",
+      "\n",
+      "n = Vh/Vx;         #Turns Ratio\n",
+      "\n",
+      "Zhx13 = Rt+(1j*Xt);         #Use of Entire low voltage winding\n",
+      "\n",
+      "#Impedances of the required terms in pu\n",
+      "A = (2*Zhx13)-Zhx12;\n",
+      "B = ((2*Zhx12)-(2*Zhx13))/(n**2);\n",
+      "C = B;\n",
+      "\n",
+      "#Angle of Impedances\n",
+      "ta = math.degrees(math.atan(A.imag/A.real));\n",
+      "tb = math.degrees(math.atan(B.imag/B.real));\n",
+      "\n",
+      "# Results\n",
+      "print 'The Circuit impedances on the high voltage side is %g/_%g ohm'%(abs(A*Rb),ta)\n",
+      "print 'Each of the Circuit impedances on the low voltage side is %g/_%g ohm'%(abs(B*Rb),tb)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Circuit impedances on the high voltage side is 24.6366/_53.9017 ohm\n",
+        "Each of the Circuit impedances on the low voltage side is 0.0085248/_18.9246 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2  Page No : 119"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import exp\n",
+      "\n",
+      "# Variables\n",
+      "#Impedances from the previous example\n",
+      "Zh = 24.6437*exp(1j*53.9*math.pi/180);\n",
+      "Zl = 8.525*(10**-3)*exp(1j*18.9*math.pi/180);\n",
+      "#Voltages\n",
+      "Vh = 7200.;         #High End\n",
+      "Vx = 120.;         # Low End\n",
+      "S = 25.*1000;         #Transformer Rating in VA\n",
+      "N = Vh/Vx;         #Turns Ratio\n",
+      "\n",
+      "# Calculations\n",
+      "#R of service drop is zero         #Line to Neutral Currents\n",
+      "IfLVn = Vx/(Zl+((1/(N**2))*Zh));         #Secondary Fault Current\n",
+      "IfHVn = IfLVn/N;         #Primary Fault Current\n",
+      "\n",
+      "#R of service drop is zero         #Line to Line Currents\n",
+      "Nl = Vh/(2*Vx);         #New Truns Ratio\n",
+      "IfLVl = 2*Vx/((2*Zl)+((1/(Nl**2))*Zh));         #Secondary Fault Current\n",
+      "IfHVl = IfLVl/Nl;         #Primary Fault Current\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The Magnitude of Line to Neutral Fault Currentson HV and LV when R of service drop\\\n",
+      " is zero are %g A and %g A respectively'%(abs(IfHVn),abs(IfLVn))\n",
+      "print 'b) The Magnitude of Line to Line Fault Currentson HV and LV when R of service drop is zero are\\\n",
+      " %g A and %g A respectively'%(abs(IfHVl),abs(IfLVl))\n",
+      "print 'c) The Minimum Allowable interrupting capacity for circuit breaker isconnected to the LV is %g A'%(abs(IfLVn))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The Magnitude of Line to Neutral Fault Currentson HV and LV when R of service drop is zero are 136.353 A and 8181.2 A respectively\n",
+        "b) The Magnitude of Line to Line Fault Currentson HV and LV when R of service drop is zero are 188.283 A and 5648.5 A respectively\n",
+        "c) The Minimum Allowable interrupting capacity for circuit breaker isconnected to the LV is 8181.2 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3  Page No : 121"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from sympy import solve,Symbol\n",
+      "from numpy import exp\n",
+      "\n",
+      "# Variables\n",
+      "Vx = 120.;         #Low End Voltage\n",
+      "#When Service drop is Zero\n",
+      "IfLVn = 8181.7*exp(-1*1j*34.3*math.pi/180);         #Line to Neutral Vault Current\n",
+      "IfLVl = 5649*exp(-1*1j*40.6*math.pi/180);         #Line to Line Fault Current\n",
+      "\n",
+      "Ral4 = 2.58;         #        #4 AWG Aluminium Conductor Resismath.tance per mile\n",
+      "Ralinf = 1.03;         #        #1/0 AWG Aluminium Conductor Resismath.tance per mile\n",
+      "\n",
+      "# Calculations\n",
+      "#Impedances when Service drop is zero, suffix l denotes line to line\n",
+      "#Suffix n denotes line to neutral\n",
+      "Zl0 = (2*Vx)/IfLVl;\n",
+      "Zn0 = (Vx)/IfLVn;\n",
+      "\n",
+      "#When there is R service drop\n",
+      "#Magnitudes of Line to Line and Line to Earth fault currents are equal\n",
+      "\n",
+      "R = Symbol('R');         #Variable Value\n",
+      "#Effective Impedances\n",
+      "Zl = Zl0+(2*R);\n",
+      "Zn = Zn0+(2*R);\n",
+      "#Fault Currents\n",
+      "Ifl = 2*Vx/Zl;\n",
+      "Ifn = Vx/Zn;\n",
+      "#print Ifl[1]\n",
+      "#Magnitudes of Currents\n",
+      "MIfl = abs(240.)/abs(Ifl.subs(R,3));\n",
+      "MIfn = abs(Ifn.subs(R,2))/abs(Ifn.subs(R,3));\n",
+      "DI = MIfl-MIfn;\n",
+      "X = - 1.5781966 + 240*R   #DI.subs(R,2);         #Polynomial Equation to find 'R'\n",
+      "R = solve(X)[0];         #Numerical Value\n",
+      "\n",
+      "#The Magnitude of R found is Wrong in the Textbook\n",
+      "\n",
+      "#Length of service drop cable\n",
+      "SDL4 = R/Ral4;\n",
+      "SDLinf = R/Ralinf;\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The Value of Service drop in the Cable is %g ohm'%(R)\n",
+      "print 'b The Length of service drop cable for:'\n",
+      "print 'i)         #4 AWG Conductor is %g miles'%(SDL4) \n",
+      "print 'ii)         #1/0 AWG Conductor is %g miles'%(SDLinf) \n",
+      "\n",
+      "#Length is printed in Miles\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The Value of Service drop in the Cable is 0.00657582 ohm\n",
+        "b The Length of service drop cable for:\n",
+        "i)         #4 AWG Conductor is 0.00254877 miles\n",
+        "ii)         #1/0 AWG Conductor is 0.00638429 miles\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4  Page No : 122"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "#Transformer Ratings in kVA\n",
+      "Sr1 = 250.; \n",
+      "Sr2 = 500.;\n",
+      "\n",
+      "#percentage impedances\n",
+      "Zr1 = 2.4;\n",
+      "Zr2 = 3.1;\n",
+      "\n",
+      "# Calculations\n",
+      "#Ratio of Maximum Loads\n",
+      "R = Sr1*Zr2/(Sr2*Zr1);\n",
+      "\n",
+      "#If 500 kVA is chosen as the full load transformer, Transformer 1 becomes overloaded\n",
+      "SL1 = Sr1;         #To Avoid OverLoading of transformer 1\n",
+      "\n",
+      "SL2 = SL1/R;         #Maximum Load on transformer 2\n",
+      "\n",
+      "Tl = SL1+SL2;         #Total Load without overloading\n",
+      "\n",
+      "# Results\n",
+      "print 'The Maximum Load Carried without overloading any of the transformer is %g kVA'%(Tl)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Maximum Load Carried without overloading any of the transformer is 637.097 kVA\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.5  Page No : 127"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "#Considering Van as reference voltage\n",
+      "SL3phi = 200*(10**3);         #Load to be powered\n",
+      "pf3 = 0.8;         #Power Factor of three phase load\n",
+      "t3 = math.acos(pf3);         #Power FActor Angle for three phase load\n",
+      "pf1 = 0.9;         #Power Factor of math.single phase load\n",
+      "t1 = math.acos(pf1);         #Power Factor angle of math.single phase load\n",
+      "SL1 = 80.*(10**3);         #Single Phase Light Load\n",
+      "Vll = 240.;         #Secondary Voltage\n",
+      "#Rating of Single Phase Transformers between individual lines\n",
+      "Sbc = 100.*(10**3);\n",
+      "Sab = 75.*(10**3);\n",
+      "Sca = Sab;\n",
+      "#Angles of Three phase voltages \n",
+      "ta = 0.;\n",
+      "tb = -120.;\n",
+      "tc = 120.;\n",
+      "#Angles of three  line currents\n",
+      "tai = ta-t3;\n",
+      "tbi = tb-t3;\n",
+      "tci = tc-t3;\n",
+      "\n",
+      "# Calculations\n",
+      "I = SL3phi/(math.sqrt(3)*Vll);         #Magnitude of Current\n",
+      "#3 Phase Line Currents\n",
+      "Ia3 = I*exp(1j*math.pi*tai/180);\n",
+      "Ib3 = I*exp(1j*math.pi*tbi/180);\n",
+      "Ic3 = I*exp(1j*math.pi*tci/180);\n",
+      "\n",
+      "MIbc = SL1/Vll;         #Magnitude Single Phase Current\n",
+      "\n",
+      "tbc = -90;         #Lagging Van         #Angle of Vbc\n",
+      "tbci = tbc-t1;         #Angle of Current Ibc\n",
+      "Ibc = MIbc*exp(1j*math.pi*tbci/180);\n",
+      "\n",
+      "#Load Currents\n",
+      "Ia = Ia3;\n",
+      "Ta = math.degrees(math.atan(Ia.imag/Ia.real));\n",
+      "Ib = Ib3+Ibc;\n",
+      "Tb = math.degrees(math.atan(Ib.imag/Ib.real));\n",
+      "Ic = Ic3-Ibc;         #Current is wrong in the textbook\n",
+      "Tc = math.degrees(math.atan(Ic.imag/Ic.real));\n",
+      "\n",
+      "#Current Flowing in the secondary winding of the transformers 1,2 and 3\n",
+      "Iac = ((Ic/Sbc)-(Ia/Sab))/((1/Sab)+(1/Sbc)+(1/Sca));\n",
+      "T1 = math.degrees(math.atan(Iac.imag/Iac.real));         #Angle of the above current\n",
+      "Iba = ((Ia/Sca)-(Ib/Sbc))/((1/Sab)+(1/Sbc)+(1/Sca));\n",
+      "T2 = math.degrees(math.atan(Iba.imag/Iba.real));         #Angle of the above current\n",
+      "Icb = ((Ib/Sab)-(Ic/Sca))/((1/Sab)+(1/Sbc)+(1/Sca));\n",
+      "T3 = math.degrees(math.atan(Icb.imag/Icb.real));         #Angle of the above current\n",
+      "\n",
+      "#Kilovoltampere Load on each transformer\n",
+      "SLab = Vll*abs(Iba)/1000;\n",
+      "SLbc = Vll*abs(Icb)/1000;\n",
+      "SLca = Vll*abs(Iac)/1000;\n",
+      "\n",
+      "Vlls = Vll;         #Secondary Voltage\n",
+      "Vllp = 7620;         #Primary Voltage\n",
+      "n = Vllp/Vlls;         #Turns Ratio\n",
+      "\n",
+      "#Primary Currents of the transformer\n",
+      "IAC = Iac/n;\n",
+      "IBA = Iba/n;\n",
+      "ICB = Icb/n;\n",
+      "\n",
+      "#Primary Current in each each phase wire\n",
+      "IA = IAC-IBA;\n",
+      "TA = math.degrees(math.atan(IA.imag/IA.real));        #Angle of the above current\n",
+      "IB = IBA-ICB;\n",
+      "TB = math.degrees(math.atan(IB.imag/IB.real));        #Angle of the above current\n",
+      "IC = ICB-IAC;\n",
+      "TC = math.degrees(math.atan(IC.imag/IC.real));        #Angle of the above current\n",
+      "\n",
+      "# Results\n",
+      "print 'a The Line Currents flowing in secondary phase wire :'\n",
+      "print 'A phase is %g/_%g A'%(abs(Ia),Ta)\n",
+      "print 'B phase is %g/_%g A'%(abs(Ib),Tb)\n",
+      "print 'C phase is %g/_%g A'%(abs(Ic),Tc)\n",
+      "print 'b The Current flowing in secondary winding of each transformer:'\n",
+      "print 'AC is %g/_%g A'%(abs(Iac),T1)\n",
+      "print 'AB is %g/_%g A'%(abs(Iba),T2)\n",
+      "print 'BC is %g/_%g A'%(abs(Icb),T3)\n",
+      "print 'c The Load on Each Transformer is:'\n",
+      "print '1 : %g kVA'%(SLca)\n",
+      "print '2 : %g kVA'%(SLab)\n",
+      "print '3 : %g kVA'%(SLbc)\n",
+      "print 'd The Current flowing in primary winding of each transformer:'\n",
+      "print 'AC is %g/_%g A'%(abs(IAC),T1)\n",
+      "print 'AB is %g/_%g A'%(abs(IBA),T2)\n",
+      "print 'BC is %g/_%g A'%(abs(ICB),T3)\n",
+      "print 'e The Line Currents flowing in primary phase wire :'\n",
+      "print 'A phase is %g/_%g A'%(abs(IA),TA)\n",
+      "print 'B phase is %g/_%g A'%(abs(IB),TB)\n",
+      "print 'C phase is %g/_%g A'%(abs(IC),TC)\n",
+      "\n",
+      "#Ic is calculation is wrong, the author has added Ibc instead of subtracting, so if you change - into + in line 45, you get the answer as in the textbook\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The Line Currents flowing in secondary phase wire :\n",
+        "A phase is 481.125/_-0.643501 A\n",
+        "B phase is 787.293/_71.6504 A\n",
+        "C phase is 787.977/_-72.7823 A\n",
+        "b The Current flowing in secondary winding of each transformer:\n",
+        "AC is 316/_-40.9814 A\n",
+        "AB is 315.535/_39.7662 A\n",
+        "BC is 545.454/_89.442 A\n",
+        "c The Load on Each Transformer is:\n",
+        "1 : 75.84 kVA\n",
+        "2 : 75.7283 kVA\n",
+        "3 : 130.909 kVA\n",
+        "d The Current flowing in primary winding of each transformer:\n",
+        "AC is 9.95276/_-40.9814 A\n",
+        "AB is 9.9381/_39.7662 A\n",
+        "BC is 17.1796/_89.442 A\n",
+        "e The Line Currents flowing in primary phase wire :\n",
+        "A phase is 15.1536/_-0.643501 A\n",
+        "B phase is 24.7966/_71.6504 A\n",
+        "C phase is 24.8182/_-72.7823 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6  Page No : 140"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "#Considering Van as reference voltage\n",
+      "SL3phi = 100.*(10**3);         #Load to be powered\n",
+      "pf3 = 0.8;         #Power Factor of three phase load\n",
+      "t3 = math.acos(pf3);         #Power FActor Angle for three phase load\n",
+      "pf1 = 0.9;         #Power Factor of math.single phase load\n",
+      "t1 = math.acos(pf1);         #Power Factor angle of math.single phase load\n",
+      "SL1 = 50.*(10**3);         #Single Phase Light Load\n",
+      "Vll = 240.;         #Secondary Voltage\n",
+      "#Angles of Three phase voltages \n",
+      "ta = 0.;\n",
+      "tb = -120.;\n",
+      "tc = 120.;\n",
+      "#Angles of three  line currents\n",
+      "tai = ta-t3;\n",
+      "tbi = tb-t3;\n",
+      "tci = tc-t3;\n",
+      "\n",
+      "# Calculations\n",
+      "I = SL3phi/(math.sqrt(3)*Vll);         #Magnitude of Current\n",
+      "#3 Phase Line Currents\n",
+      "Ia3 = I*exp(1j*math.pi*tai/180);\n",
+      "Ib3 = I*exp(1j*math.pi*tbi/180);\n",
+      "Ic3 = I*exp(1j*math.pi*tci/180);\n",
+      "\n",
+      "MI1 = SL1/Vll;         #Magnitude Single Phase Current\n",
+      "\n",
+      "t1v = 30;         #Leading Van         #Angle of Vbc\n",
+      "t1i = t1v-t1;         #Angle of Current Ibc\n",
+      "I1 = MI1*exp(1j*math.pi*t1i/180);\n",
+      "\n",
+      "#Load Currents\n",
+      "Ia = Ia3+I1;\n",
+      "Ta = math.degrees(math.atan(Ia.imag/Ia.real));\n",
+      "Ib = Ib3-I1;\n",
+      "Tb = -180+(math.degrees(math.atan(Ib.imag/Ib.real)));\n",
+      "Ic = Ic3;         #Current is wrong in the textbook\n",
+      "Tc = math.degrees(math.atan(Ic.imag/Ic.real));\n",
+      "\n",
+      "#Current flowing in the secondary of the transformer\n",
+      "Iba = Ia;\n",
+      "T2 = math.degrees(math.atan(Iba.imag/Iba.real));         #Angle of the above current\n",
+      "Icb = Ic;\n",
+      "T3 = 180+(math.degrees(math.atan(Icb.imag/Icb.real)));         #Angle of the above current\n",
+      "\n",
+      "#Load on Each Transformer\n",
+      "SLba = Vll*abs(Iba)/1000;\n",
+      "SLcb = Vll*abs(Icb)/1000;\n",
+      "\n",
+      "Vlls = Vll;         #Secondary Voltage\n",
+      "Vllp = 7620;         #Primary Voltage\n",
+      "n = Vllp/Vlls;         #Turns Ratio\n",
+      "\n",
+      "#Primary Currents of the transformer\n",
+      "IA = Iba/n;\n",
+      "TA = math.degrees(math.atan(IA.imag/IA.real));        #Angle of the above current\n",
+      "IB = Icb/n;\n",
+      "TB = T3;        #Angle of the above current\n",
+      "IN = IA+IB;         #Neutral Current\n",
+      "TN = math.degrees(math.atan(IN.imag/IN.real));        #Angle of the above current\n",
+      "\n",
+      "# Results\n",
+      "print 'a The Line Currents flowing in secondary phase wire :'\n",
+      "print 'A phase is %g/_%g A'%(abs(Ia),Ta)\n",
+      "print 'B phase is %g/_%g A'%(abs(Ib),Tb)\n",
+      "print 'C phase is %g/_%g A'%(abs(Ic),Tc)\n",
+      "print 'b The Current flowing in secondary winding of each transformer:'\n",
+      "print 'AB is %g/_%g A'%(abs(Iba),T2)\n",
+      "print 'BC is %g/_%g A'%(abs(Icb),T3)\n",
+      "print 'c The Load on Each Transformer is:'\n",
+      "print '1 : %g kVA'%(SLba)\n",
+      "print '2 : %g kVA'%(SLcb)\n",
+      "print 'd The Line Currents flowing in primary phase wire :'\n",
+      "print 'AB is %g/_%g A'%(abs(IA),TA)\n",
+      "print 'CB is %g/_%g A'%(abs(IB),TB)\n",
+      "print 'The Neutral Current is %g/_%g'%(abs(IN),TN)\n",
+      "\n",
+      "#Note the mistake in the Textbook for the calulation for Neutral Current\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The Line Currents flowing in secondary phase wire :\n",
+        "A phase is 433.486/_13.3432 A\n",
+        "B phase is 433.874/_-134.453 A\n",
+        "C phase is 240.563/_-60.6435 A\n",
+        "b The Current flowing in secondary winding of each transformer:\n",
+        "AB is 433.486/_13.3432 A\n",
+        "BC is 240.563/_119.356 A\n",
+        "c The Load on Each Transformer is:\n",
+        "1 : 104.037 kVA\n",
+        "2 : 57.735 kVA\n",
+        "d The Line Currents flowing in primary phase wire :\n",
+        "AB is 13.6531/_13.3432 A\n",
+        "CB is 7.57678/_119.356 A\n",
+        "The Neutral Current is 13.6653/_45.5475\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.8  Page No : 152"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "Vll = 480.;         #Line to Line Voltage\n",
+      "Vln = 277.;         #Line to neutral Voltage\n",
+      "\n",
+      "# Calculations\n",
+      "#From the Phasor Diagram from the result file\n",
+      "Vab = Vll*exp(1j*0);         #Vab is taken as reference\n",
+      "Vabh = 50*Vab/100;\n",
+      "VAB = 4160.;\n",
+      "VABh = 50*VAB/100; \n",
+      "VH1H2o = math.sqrt((VAB**2)-(VABh**2));\n",
+      "VH1H2t = (VABh);\n",
+      "Vx1x2o = 1*math.sqrt((Vab**2)-(Vabh**2))/3;\n",
+      "Vx2x3o = 2*math.sqrt((Vab**2)-(Vabh**2))/3;\n",
+      "VH2H3t = (VABh);\n",
+      "Vx1x2t = Vabh;\n",
+      "Vx2x3t = Vabh;\n",
+      "\n",
+      "# Results\n",
+      "print 'a The Phasor diagram is shown in the result file attached to the code'\n",
+      "print 'b) Vab is %g/_%g V'%(abs(Vab),Vab.imag/Vab.real)\n",
+      "print 'c The Magnitudes of the following rated winding voltages'\n",
+      "print 'i) The Voltage VH1H2 on transformer 1 : %g V'%(VH1H2o)\n",
+      "print 'ii) The Voltage Vx1x2 on transformer 1 : %g V'%(Vx1x2o)\n",
+      "print 'iii) The Voltage Vx2x3 on transformer 1 : %g V'%(Vx2x3o)\n",
+      "print 'iv) The Voltage VH1H2 on transformer 2 : %g V'%(VH1H2t)\n",
+      "print 'v) The Voltage VH2H3 on transformer 2 : %g V'%(VH2H3t)\n",
+      "print 'vi) The Voltage Vx1x2 on transformer 2 : %g V'%(Vx1x2t)\n",
+      "print 'vii) The Voltage Vx1x2 on transformer 2 : %g V'%(Vx2x3t)\n",
+      "print 'd i NO ii NO iii YES'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The Phasor diagram is shown in the result file attached to the code\n",
+        "b) Vab is 480/_0 V\n",
+        "c The Magnitudes of the following rated winding voltages\n",
+        "i) The Voltage VH1H2 on transformer 1 : 3602.67 V\n",
+        "ii) The Voltage Vx1x2 on transformer 1 : 138.564 V\n",
+        "iii) The Voltage Vx2x3 on transformer 1 : 277.128 V\n",
+        "iv) The Voltage VH1H2 on transformer 2 : 2080 V\n",
+        "v) The Voltage VH2H3 on transformer 2 : 2080 V\n",
+        "vi) The Voltage Vx1x2 on transformer 2 : 240 V\n",
+        "vii) The Voltage Vx1x2 on transformer 2 : 240 V\n",
+        "d i NO ii NO iii YES\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "-c:16: ComplexWarning: Casting complex values to real discards the imaginary part\n",
+        "-c:17: ComplexWarning: Casting complex values to real discards the imaginary part\n",
+        "-c:31: ComplexWarning: Casting complex values to real discards the imaginary part\n",
+        "-c:32: ComplexWarning: Casting complex values to real discards the imaginary part\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.9  Page No : 154"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import exp\n",
+      "\n",
+      "# Variables\n",
+      "R = 2.77;         #Resismath.tance of the balanced load\n",
+      "#From Phasor Diagram in Result file\n",
+      "Vab = 480*exp(1j*0);         #Reference Voltage\n",
+      "MVn = abs(Vab)/math.sqrt(3);         #Magnitude of line to neutral voltages\n",
+      "#Angles of Three phase voltages \n",
+      "ta = -30.;\n",
+      "tb = -150.;\n",
+      "tc = 90.;\n",
+      "\n",
+      "# Calculations\n",
+      "#Angles of Winding according to the Line Currents\n",
+      "tx3x2 = 30;         #Leading\n",
+      "tx1x2 = -30;         #Lagging\n",
+      "\n",
+      "I = MVn/R;         #Magnitude of current\n",
+      "\n",
+      "#Low Voltage Current Phasors\n",
+      "Ia = I*exp(1j*math.pi*ta/180);\n",
+      "Ib = I*exp(1j*math.pi*tb/180);\n",
+      "Ic = I*exp(1j*math.pi*tc/180);\n",
+      "pfT = math.cos(math.radians(ta-ta));         #Angle of Ia is same as phase voltage         #Resismath.tance load\n",
+      "\n",
+      "# Results\n",
+      "print 'a The Low voltage current phasors are:'\n",
+      "print 'A is %g/_%g A'%(abs(Ia),ta)\n",
+      "print 'B is %g/_%g A'%(abs(Ib),tb)\n",
+      "print 'C is %g/_%g A'%(abs(Ic),tc)\n",
+      "print 'b The Phasor Diagram is the ''b'' diagram of in the result file'\n",
+      "print 'c) The Power Factor of the Transformer is %g'%(pfT)\n",
+      "print 'd) Power Factor as seen by winding x3x2 of transformer 2 is %g leading'%(math.cos(math.radians(tx3x2)))\n",
+      "print 'e) Power Factor as seen by winding x1x2 of transformer 2 is %g lagging'%(math.cos(math.radians(tx1x2)))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The Low voltage current phasors are:\n",
+        "A is 100.046/_-30 A\n",
+        "B is 100.046/_-150 A\n",
+        "C is 100.046/_90 A\n",
+        "b The Phasor Diagram is the b diagram of in the result file\n",
+        "c) The Power Factor of the Transformer is 1\n",
+        "d) Power Factor as seen by winding x3x2 of transformer 2 is 0.866025 leading\n",
+        "e) Power Factor as seen by winding x1x2 of transformer 2 is 0.866025 lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.10  Page No : 156"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import exp\n",
+      "\n",
+      "#'o' and 't' represent transformer one and two respectively\n",
+      "# Variables\n",
+      "#Objective is to find the Factor which has to be multiplied to get VA rating\n",
+      "Vll = 1.;         #Assumption Made\n",
+      "#From the Phasor Diagram from the result file\n",
+      "Vab = Vll*exp(1j*0);         #Vab is taken as reference\n",
+      "Vabh = 50.*Vab/100;\n",
+      "Vx1x2o = 1*math.sqrt((Vab**2)-(Vabh**2))/3;\n",
+      "Vx2x3o = 2*math.sqrt((Vab**2)-(Vabh**2))/3;\n",
+      "Vx1x2t = Vabh;\n",
+      "Vx2x3t = Vabh;\n",
+      "\n",
+      "#Let I be unity\n",
+      "I = 1;\n",
+      "\n",
+      "# Calculations\n",
+      "#VA Ratings of the respective windings\n",
+      "Sx1x2o = Vx1x2o*I;\n",
+      "Sx2x3o = Vx2x3o*I;\n",
+      "Sx1x2t = Vx1x2t*I;\n",
+      "Sx2x3t = Vx2x3t*I;\n",
+      "\n",
+      "#Total VA rating of transformer\n",
+      "S1 = Sx1x2o+Sx2x3o;\n",
+      "S2 = Sx1x2t+Sx2x3t;\n",
+      "\n",
+      "#Ratio of total rating to maximum rating\n",
+      "Rt = (S1+S2)/(math.sqrt(3)*Vll*I);\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The voltampere raing of x1x2 of transformer 1 is %g*VI VA'%(Sx1x2o)\n",
+      "print 'b) The voltampere raing of x1x2 of transformer 1 is %g*VI VA'%(Sx2x3o)\n",
+      "print 'c) The Total Output from transformer 1 is %g*VI VA'%(S1)\n",
+      "print 'd) The voltampere raing of x1x2 of transformer 2 is %g*VI VA'%(Sx1x2t)\n",
+      "print 'e) The voltampere raing of x1x2 of transformer 2 is %g*VI VA'%(Sx2x3t)\n",
+      "print 'f) The Total Output from transformer 2 is %g*VI VA'%(S2)\n",
+      "print 'g) The Total Rating to the Maximum Continous Output is %g'%(Rt)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The voltampere raing of x1x2 of transformer 1 is 0.288675*VI VA\n",
+        "b) The voltampere raing of x1x2 of transformer 1 is 0.57735*VI VA\n",
+        "c) The Total Output from transformer 1 is 0.866025*VI VA\n",
+        "d) The voltampere raing of x1x2 of transformer 2 is 0.5*VI VA\n",
+        "e) The voltampere raing of x1x2 of transformer 2 is 0.5*VI VA\n",
+        "f) The Total Output from transformer 2 is 1*VI VA\n",
+        "g) The Total Rating to the Maximum Continous Output is 1.07735\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "-c:11: ComplexWarning: Casting complex values to real discards the imaginary part\n",
+        "-c:12: ComplexWarning: Casting complex values to real discards the imaginary part\n",
+        "-c:37: ComplexWarning: Casting complex values to real discards the imaginary part\n",
+        "-c:38: ComplexWarning: Casting complex values to real discards the imaginary part\n",
+        "-c:39: ComplexWarning: Casting complex values to real discards the imaginary part\n",
+        "-c:40: ComplexWarning: Casting complex values to real discards the imaginary part\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.11  Page No : 157"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import exp\n",
+      "\n",
+      "# Variables\n",
+      "#Per unit value\n",
+      "Zt = 0.01+(1j*0.03);         #Transformer impedance\n",
+      "\n",
+      "Vll = 240.;         #Secondary Voltage\n",
+      "\n",
+      "Sl = 90.;         #Lighting Load\n",
+      "pfl = 0.9;\n",
+      "tl = math.acos(pfl);\n",
+      "S = 25.;         #Balanced Load\n",
+      "pf = 0.8;\n",
+      "t = math.acos(pf);\n",
+      "\n",
+      "\n",
+      "# Calculations\n",
+      "def angle(y): \n",
+      "    return math.degrees(math.atan(y.imag/y.real))\n",
+      "\n",
+      "tab = 30;         #Phase angle of Vab\n",
+      "\n",
+      "Il = Sl*1000/Vll;         #Magnitude of Light Load\n",
+      "#Umath.sing the symmetrical - components theory\n",
+      "Ia1 = Il*exp(1j*math.pi*(tab-tl)/180);\n",
+      "Ta1 = angle(Ia1);         #Angle for the above current\n",
+      "Ib1 = -1*Ia1;\n",
+      "Ic1 = 0;         #Neutral Wire\n",
+      "#Angles of three line to line voltages\n",
+      "ta = 0;\n",
+      "tb = -120;\n",
+      "tc = 120;\n",
+      "\n",
+      "Ib = S*1000/(math.sqrt(3)*Vll);         #Magnitude of balanced load\n",
+      "\n",
+      "#Currents in Three phase load\n",
+      "Ta2 = ta-t;\n",
+      "Ia2 = Ib*exp(1j*math.pi*Ta2/180);\n",
+      "Tb2 = tb-t;\n",
+      "Ib2 = Ib*exp(1j*math.pi*Tb2/180);\n",
+      "Tc2 = tc-t;\n",
+      "Ic2 = Ib*exp(1j*math.pi*Tc2/180);\n",
+      "\n",
+      "#Currents in phase wire\n",
+      "Ia = Ia1+Ia2;\n",
+      "Ta = angle(Ia);         #Angle corresponding to the above angle\n",
+      "Ib = Ib1+Ib2;\n",
+      "Tb = angle(Ib);         #Angle corresponding to the above angle\n",
+      "Ic = Ic1+Ic2;\n",
+      "Tc = angle(Ic);         #Angle corresponding to the above angle\n",
+      "\n",
+      "#Transformer Loads\n",
+      "ST1 = Vll*abs(Ia)/1000;\n",
+      "T1 = 100;         #From the above value of Load, this transformer is chosen to meet the specific characteristic\n",
+      "ST1pu = ST1/T1;         #Per unit Load\n",
+      "ST2 = Vll*abs(Ic)/1000;\n",
+      "T2 = 15;         #From the above value of Load, this transformer is chosen to meet the specific characteristic\n",
+      "ST2pu = ST2/T2;         #Per unit Load\n",
+      "\n",
+      "#Transformer Power Factors\n",
+      "pfT1 = math.cos(math.radians(tab-Ta));\n",
+      "pfT2 = math.cos(math.radians(90-Tc));         #Vcb makes angle of 90\n",
+      "\n",
+      "Vh = 7200;         #High End Voltage\n",
+      "n = Vh/Vll;         #Turns Ratio\n",
+      "\n",
+      "# The Primary Line Currents\n",
+      "IA = Ia/n;\n",
+      "IB = -1*Ic/n;\n",
+      "IN = -1*(IA+IB);\n",
+      "\n",
+      "Ibase = T1*1000/Vll;         #Base Current\n",
+      "Iapu = Ia/Ibase;\n",
+      "Icpu = Ic/Ibase;\n",
+      "\n",
+      "#Phase Voltages\n",
+      "Vab = Vll*exp(1j*math.pi*tab/180); \n",
+      "Vbc = Vll*exp(-1*1j*math.pi*90/180);\n",
+      "#Per Unit Voltages\n",
+      "VANpu = (Vab/Vll)+(Iapu*Zt);\n",
+      "VBNpu = (Vbc/Vll)-(Icpu*Zt);\n",
+      "\n",
+      "#Actual Voltages\n",
+      "VAN = VANpu*Vh;\n",
+      "VBN = VBNpu*Vh;\n",
+      "\n",
+      "# Results\n",
+      "print 'a The Phasor Currents:'\n",
+      "print 'Ia is %g/_%g A'%(abs(Ia),Ta)\n",
+      "print 'Ib is %g/_%g A'%(abs(Ib),180+Tb)\n",
+      "print 'Ic is %g/_%g A'%(abs(Ic),Tc)\n",
+      "print 'b) Suitable ratings of the transformers are %g kVA and %g kVA'%(T1,T2)\n",
+      "print 'c) The Per Unit kVA load on each transformer is %g pu and %g pu'%(ST1pu,ST2pu)\n",
+      "print 'd) The power factor of output of each transformer is %g and %g both lagging'%(pfT1,pfT2)\n",
+      "print 'e The phasor currents at the high voltage leads:'\n",
+      "print 'IA is %g/_%g A'%(abs(IA),Ta)\n",
+      "print 'IB is %g/_%g A'%(abs(IB),180+angle(IB))\n",
+      "print 'IN is %g/_%g A'%(abs(IN),angle(IN))\n",
+      "print 'f) VAN is %g/_%g V and VBN is %g/_%g V'%(abs(VAN),angle(VAN),abs(VBN),angle(VBN))\n",
+      "\n",
+      "#Highly Accuracy of Answers; Upto 5 decimal Places\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The Phasor Currents:\n",
+        "Ia is 428.052/_25.4972 A\n",
+        "Ib is 428.229/_213.552 A\n",
+        "Ic is 60.1407/_-60.6435 A\n",
+        "b) Suitable ratings of the transformers are 100 kVA and 15 kVA\n",
+        "c) The Per Unit kVA load on each transformer is 1.02732 pu and 0.96225 pu\n",
+        "d) The power factor of output of each transformer is 0.996914 and -0.871586 both lagging\n",
+        "e The phasor currents at the high voltage leads:\n",
+        "IA is 14.2684/_25.4972 A\n",
+        "IB is 2.00469/_119.356 A\n",
+        "IN is 14.5415/_17.5913 A\n",
+        "f) VAN is 7294.34/_31.6923 V and VBN is 7193.85/_-89.743 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file