{ "metadata": { "name": "", "signature": "sha256:0910cfded5ae40d0ea2b9576604d8007ec8328215058593fcec218762063deff" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 14: A.C. Distribution" ] }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 14.1, Page Number: 359" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "import math\n", "\n", "#Variable declaration:\n", "l = 300 #line length(m)\n", "magI1 = 100 #current at load 1(A)\n", "pf1 = 0.707 #power factor at load 1\n", "l1 = 200 #line length till load 1(m)\n", "magI2 = 200 #current drawn at load 2(A)\n", "pf2 = 0.8 #power factor at laod 2\n", "l2 = 300 #line length till load 2(m)\n", "R = 0.2 #total resistance of line(ohm/km)\n", "X = 0.1 #total reactance of the line(ohm/km)\n", "\n", "#Calculation:\n", "phy1 = math.acos(pf1)\n", "phy2 = math.acos(pf2)\n", "Zac = (R+X*1j)*200/1000 #Impedance of section AC(ohm)\n", "Zcb = (R+X*1j)*100/1000 #Impedance of section CB(ohm)\n", "#Taking voltage at the far end B as the reference vector, we have\n", "I2 = magI2*(pf2-1j*math.sin(phy2)) #Load current at point B(A)\n", "I1 = magI1*(pf1-1j*math.sin(phy1)) #Load current at point C(A)\n", "Icb = I2 #A\n", "Iac = I1+I2 #A\n", "Vcb = Icb*Zcb #V\n", "Vac = Iac*Zac #V\n", "V = Vac+Vcb #Voltage drop in the distributor(V)\n", "\n", "\n", "\n", "#Result:\n", "print \"Voltage drop in the distributor is\",round(abs(V),2),\"V\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Voltage drop in the distributor is 17.85 V\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 14.2, Page Number: 359" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "import math\n", "\n", "#Variable declaration:\n", "l = 2000 #line length(m)\n", "magI1 = 80 #current at load 1(A)\n", "pf1 = 0.9 #power factor at load 1\n", "l1 = 1000 #line length till load 1(m)\n", "magI2 = 120 #current drawn at load 2(A)\n", "pf2 = 0.8 #power factor at laod 2\n", "l2 = 2000 #line length till load 2(m)\n", "R = 0.05 #total resistance of line(ohm/km)\n", "X = 0.1 #total reactance of the line(ohm/km)\n", "magVb = 230 #voltage maintained at point B(V)\n", "\n", "\n", "#Calculation:\n", "phy1 = math.acos(pf1)\n", "phy2 = math.acos(pf2)\n", "Zac = (R+X*1j)*1000/1000 #Impedance of section AC(ohm)\n", "Zcb = (R+X*1j)*1000/1000 #Impedance of section CB(ohm)\n", "#Taking voltage at the far end B as the reference vector, we have\n", "I2 = magI2*(pf2-1j*math.sin(phy2)) #Load current at point B(A)\n", "I1 = magI1*(pf1-1j*math.sin(phy1)) #Load current at point C(A)\n", "Icb = I2 #A\n", "Iac = I1+I2 #A\n", "Vcb = Icb*Zcb #V\n", "Vac = Iac*Zac #V\n", "V = Vac+Vcb+magVb*(1+0j) #Voltage drop in the distributor(V)\n", "theta = math.atan(V.imag/V.real)\n", "\n", "\n", "\n", "#Result:\n", "print \"(i) Voltage drop in the distributor is\",round(abs(V),2),\"V\"\n", "print \"(ii)The phase difference between Va and Vb is \",round(math.degrees(theta),2),\"degrees\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Voltage drop in the distributor is 261.67 V\n", "(ii)The phase difference between Va and Vb is 3.83 degrees\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 14.3, Page Number: 360" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "import math\n", "\n", "#Variable declaration:\n", "magI1 = 100 #current at load 1(A)\n", "pf1 = 0.6 #power factor at load 1\n", "magI2 = 100 #current drawn at load 2(A)\n", "pf2 = 0.8 #power factor at laod 2\n", "R = 0.1 #total resistance of line(ohm/km)\n", "X = 0.15 #total reactance of the line(ohm/km)\n", "magVb = 200 #voltage maintained at point B(V)\n", "\n", "\n", "#Calculation:\n", "phy1 = math.acos(pf1)\n", "phy2 = math.acos(pf2)\n", "Zam = (R+X*1j) #Impedance of section AM(ohm)\n", "Zmb = (R+X*1j) #Impedance of section MB(ohm)\n", "#Taking voltage at the far end B as the reference vector, we have\n", "I2 = magI2*(pf2-1j*math.sin(phy2)) #Load current at point B(A)\n", "Imb = I2 \n", "Vb = magVb*(1+0j) #V\n", "Vmb = Imb*Zmb #V\n", "Vm = Vb+Vmb #V\n", "alpha = math.atan(Vm.imag/Vm.real) #V\n", "#The load current I1 has a lagging p.f. of 0\u00b76 w.r.t. VM. It lags \n", "#behind Vm by an angle phy1.\n", "#Phase angle between I1 and Vb\n", "phy11 = phy1-alpha\n", "I1 = magI1*(math.cos(phy11)-math.sin(phy11)*1j) #A\n", "Iam = I1+I2 #A\n", "Vam = Iam*Zam #V\n", "Va = Vm+Vam #V\n", "theta = math.atan(Va.imag/Va.real)\n", "\n", "\n", "\n", "#Result:\n", "print \"(i)Voltage at mid-point is\",round(abs(Vm),1),\"V\"\n", "print \"(ii) Sending end voltage Va is\",round(abs(Va),2),\"V\"\n", "print \"(iii)The phase difference between Va and Vb is \",round(math.degrees(theta),2),\"degrees\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i)Voltage at mid-point is 217.1 V\n", "(ii) Sending end voltage Va is 252.33 V\n", "(iii)The phase difference between Va and Vb is 3.07 degrees\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 14.4, Page Number: 362" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "import math\n", "\n", "#Variable declaration:\n", "Zab = 1+1j #ohm\n", "Zbc = 1+2j #ohm\n", "Zac = 1+3j #ohm\n", "Ib = 20 #load current at B(A)\n", "pfb = 0.8 #power factor at A\n", "Ic = 15 #load current at C(A)\n", "pfc = 0.6 #power factor at B\n", "\n", "\n", "#Calculation:\n", "Iab = Ib*(pfb-1j*math.sin(math.acos(pfb))) #Current in section AB(A)\n", "Iac = Ic*(pfc-1j*math.sin(math.acos(pfc))) #Current in section AB(A)\n", "Vab = Iab*Zab #Voltage drop in section AB(V)\n", "Vac = Iac*Zac #Voltage drop in section AC(V)\n", "#point B is at higher potential than point C. The p.d. between B and C\n", "#is Thevenin\u2019s equivalent circuit e.m.f. Eo i.e.\n", "Eo = Vac-Vab #volt\n", "Zo = Zab+Zac #Thevenin\u2019s equivalent impedance(ohm)\n", "Ibc = Eo/(Zo+Zbc) #A\n", "Iab1 = Iab+Ibc #A\n", "Iac1 = Iac-Ibc #A\n", "Ia = Iab+Iac #Current fed at A(A)\n", "\n", "\n", "\n", "#Result:\n", "print \"The total current fed at A is\",Ia,\"A\"\n", "print \"\\nCurrent in AB is\",Iab1.real+round(Iab1.imag,2)*1j,\"A\"\n", "print \"\\nCurrent in BC is\",Ibc.real+round(Ibc.imag,2)*1j,\"A\"\n", "print \"\\nCurrent in AC is\",Iac1.real+round(Iac1.imag,2)*1j,\"A\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The total current fed at A is (25-24j) A\n", "\n", "Current in AB is (18.6-13.53j) A\n", "\n", "Current in BC is (2.6-1.53j) A\n", "\n", "Current in AC is (6.4-10.47j) A\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 14.5, Page Number: 363" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "import math\n", "\n", "\n", "#Variable declaration:\n", "l = 1000 #line length(m)\n", "magI1 = 5 #current at load 1(A)\n", "pf1 = 0.8 #power factor at load 1\n", "l1 = 600 #line length till load 1(m)\n", "pf2 = 0.85 #power factor at motor load B\n", "Po = 10 #power output at B(H.P)\n", "n = 0.9 #efficiency\n", "l2 = 400 #line length till load 2(m)\n", "R = 1 #total resistance of line(ohm/km)\n", "X = 0.5 #total reactance of the line(ohm/km)\n", "magVb = 400 #voltage maintained at point B(V)\n", "\n", "\n", "#Calculation:\n", "Zac = (R+X*1j)*l1/l #ohm\n", "Zcb = (R+X*1j)*l2/l #ohm\n", "magVbp = magVb/3**0.5 #volt per phase\n", "Vbp = magVbp*(1+0j) #V\n", "magIb = Po*746/(3**0.5*magVb*pf2*n) #Line current at B(A)\n", "magI2p = magIb\n", "I2p = magI2p*(pf2-1j*math.sin(math.atan(pf2))) #A\n", "I1p = magI1*(pf1-1j*math.sin(math.atan(pf1))) #A\n", "Iac = I1p+I2p #Current in section AC(A)\n", "Icb = I2p #Current in section CB(A)\n", "Vcb = Icb*Zcb #V\n", "Vac = Iac*Zac #V\n", "Va = Vbp+Vcb+Vac #V\n", "\n", "\n", "#Result:\n", "print \"Line voltage at A is\",round(abs(Va)*3**0.5),\"V\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Line voltage at A is 434.0 V\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 14.6, Page Number: 364" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "import math\n", "from sympy import *\n", "\n", "#Variable declaration:\n", "magVa = 11000 #volt\n", "magIb = 50 #load current at B(A)\n", "pf2 = 0.8 #power factor(lagging)\n", "magIc = 120 #load current at C(A)\n", "pf3 = 1.0 #power factor(lagging)\n", "magId = 70 #load current at D(A)\n", "pf4 = 0.866 #power factor(lagging)\n", "Zab = 1+0.6j #ohm\n", "Zbc = 1.2+0.9j #ohm\n", "Zcd = 0.8+0.5j #ohm\n", "Zda = 3+2j #ohm\n", "\n", "\n", "#Calculation:\n", "#Let current in section AB be (x + j*y).\n", "x,y = symbols('x,y')\n", "Iab = x+1j*y\n", "x,y = symbols('x,y')\n", "Ibc = Iab-magIb*(pf2-math.sin(math.atan(pf2))) #A\n", "Icd = (x-40+1j*(y+30))-(120+0j) #A\n", "Ida = ((x-160)+1j*(y+30))-(70*(0.866-0.5j)) #A\n", "Vab = Iab*Zab #Drop in section AB(V)\n", "Vbc = Ibc*Zbc #Drop in section BC(V)\n", "Vcd = Icd*Zcd #Drop in section CD(V)\n", "Vda = Ida*Zda #Drop in section DA(V)\n", "# Vab+Vbc+Vcd+Vda = 0\n", "# As the real and imaginary parts have to be separately zero,\n", "# 6*x-4*y-1009.8 = 0\n", "# 4*x+6*y-302.2 = 0\n", "x1 = solve(6*x-4*y-1009.8,x)[0]\n", "y1 = round(solve(4*x1+6*y-302.2,y)[0],1)\n", "x11 = round(solve(6*x-4*y1-1009.8,x)[0],1)\n", "#now putting the values of x11 and y1 in above equationa,\n", "Iab1 = x11+1j*y1 #A\n", "Ibc1 = (x11-40)+1j*(y1+30) #A\n", "Icd1 = (x11-40+1j*(y1+30))-(120+0j) #A\n", "Ida1 = ((x11-160)+1j*(y1+30))-(70*(0.866-0.5j)) #A\n", "magVap = round(magVa/3**0.5) #Voltage at supply end A(V)\n", "Vb = magVap*(1+0j)-Iab1*Zab #Voltage at station B(V/phase)\n", "Vc = Vb-Ibc1*Zbc #Voltage at station C(V/phase)\n", "Vd = Vc-Icd1*Zcd #Voltage at station D(V/phase)\n", "\n", "\n", "#Result:\n", "print \"Current in section AB is\",Iab1,\"A\"\n", "print \"Current in section BC is\",Ibc1,\"A\"\n", "print \"Current in section CD is\",Icd1,\"A\"\n", "print \"Current in section DA is\",Ida1,\"A\"\n", "print \"Voltage at A is\",magVap*(1+0j),\"V/phase\"\n", "print \"Voltage at B is\",Vb,\"V/phase\"\n", "print \"Voltage at C is\",Vc,\"V/phase\"\n", "print \"Voltage at D is\",Vd,\"V/phase\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Current in section AB is (139.8-42.8j) A\n", "Current in section BC is (99.8-12.8j) A\n", "Current in section CD is (-20.2-12.8j) A\n", "Current in section DA is (-80.82+22.2j) A\n", "Voltage at A is (6351+0j) V/phase\n", "Voltage at B is (6185.52-41.08j) V/phase\n", "Voltage at C is (6054.24-115.54j) V/phase\n", "Voltage at D is (6064-95.2j) V/phase\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 14.7, Page Number: 368" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "import math\n", "\n", "\n", "#Variable declaration:\n", "Pr = 10 #load connected to line R(kW)\n", "Py = 8 #load connected to line Y(kW)\n", "Pb = 5 #load connected to line B(kW)\n", "Vl = 400 #line voltage(V)\n", "#the loads are non-reactive.\n", "\n", "#Calculation:\n", "Vp = round(Vl/3**0.5) #phase voltage(V)\n", "Ir = Pr*1000/Vp #A\n", "Iy = Py*1000/Vp #A\n", "Ib = Pb*1000/Vp #A\n", "\n", "#Resolving the three currents along x-axis and y-axis, we have,\n", "Ih = Iy*math.cos(math.pi/6)-Ib*math.cos(math.pi/6) #Resultant horizontal component(A)\n", "Iv = Ir-Iy*math.cos(math.pi/3)-Ib*math.cos(math.pi/3) #Resultant vertical component(A)\n", "In = (Ih**2+Iv**2)**0.5 #current in neutral wire(A)\n", "\n", "\n", "#Result:\n", "print \"(i) Ir =\",round(Ir,1),\"A\"\n", "print \" Iy =\",round(Iy,1),\"A\"\n", "print \" Ib =\",round(Ib,2),\"A\"\n", "print \"(ii) Current in neutral wire is\",round(In,1),\"A\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Ir = 43.3 A\n", " Iy = 34.6 A\n", " Ib = 21.65 A\n", "(ii) Current in neutral wire is 18.9 A\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 14.8, Page Number: 369" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "import math\n", "\n", "#Variable declaration:\n", "Vl = 400 #line voltage(V)\n", "Vp = 230 #voltage across lamp(V)\n", "I1 = 70 #current in load RN(A)\n", "I2 = 84 #current in load YN(A)\n", "I3 = 33 #current in load BN(A)\n", "Im = 200 #current taken by the motor(A)\n", "pf = 0.2 #power factor(lagging)\n", "\n", "#Calculation:\n", "#Lamp load alone:\n", "Ih = I2*math.cos(math.pi/6)-I3*math.cos(math.pi/6) #Resultant H-component(A)\n", "Iv = I1-I3*math.cos(math.pi/3)-84*math.cos(math.pi/3) #Resultant V-component(A)\n", "In = math.sqrt(Ih**2+Iv**2) #Neutral current(A)\n", "\n", "\n", "#Both lamp load and motor load:\n", "Ir = Im*pf #Active component of motor current(A)\n", "Ix = Im*math.sin(math.acos(pf)) #Reactive component of motor current(A)\n", "IR = ((Ir+I1)**2+Ix**2)**0.5 #A\n", "IY = ((Ir+I2)**2+Ix**2)**0.5 #A\n", "IB = ((Ir+I3)**2+Ix**2)**0.5 #A\n", "P = Vp*(I1+I2+I3)*1 #Watt #( cos phy_L = 1)\n", "Pm = 3**0.5*Vl*Im*pf #Power supplied to motor(W)\n", "\n", "\n", "\n", "#Result:\n", "print \"Lamp load alone: neutral curent is\",round(In,2),\"A\"\n", "print \"\\nWhen Both lamp load and motor load is present:\"\n", "print \"The current components are:\"\n", "print \"Neutral current is\",round(In,2),\"A\"\n", "print \"IR =\",round(IR,1),\"A;\\tIY =\",round(IY,0),\"A;\\tIB =\",round(IB,2),\"A\"\n", "print \"Power supplied to the lamp is\",P,\"W\"\n", "print \"Power supplied to the motor is\",round(Pm),\"W\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Lamp load alone: neutral curent is 45.64 A\n", "\n", "When Both lamp load and motor load is present:\n", "The current components are:\n", "Neutral current is 45.64 A\n", "IR = 224.7 A;\tIY = 232.0 A;\tIB = 209.11 A\n", "Power supplied to the lamp is 43010 W\n", "Power supplied to the motor is 27713.0 W\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 14.9, Page Number: 370" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "import math\n", "\n", "#Variable declaration:\n", "Prn = 20 #kW\n", "pf1 = 1 #power factor of loaf RN\n", "kVAyn = 28.75 #kVA of load YN\n", "kVAbn = 28.75 #kVA of load BN\n", "pf2 = 0.866 #power factor of laod YN & BN each.(lagging)\n", "Vl = 400 #line voltage(V)\n", "Vp = 230 #phase voltage(V)\n", "\n", "\n", "#Calculation:\n", "phy1 = math.acos(pf1)\n", "phy2 = math.acos(pf2)\n", "phy3 = phy2\n", "Ir = Prn*1000/Vp #A\n", "Iy = kVAyn*1000/Vp #A\n", "Ib = kVAbn*1000/Vp #A\n", "Ih = Ir-Iy*math.cos(phy2)-Ib*math.cos(phy2) #A\n", "Iv = 0+Iy*math.sin(phy2)-Iy*math.sin(phy3) #A\n", "In = math.sqrt(Ih**2+Iv**2) #A\n", "#When load from B to N removed.:\n", "#When the load from B to N is removed, the various line currents are:\n", "#Ir in phase with Vrn; Iy lagging by 30 deg.; Ib = 0.\n", "Ir1 = Ir\n", "Iy1 = Iy; Ib1 = 0 #A\n", "Ih1 = Ir1-Iy1*math.cos(math.pi/6) #A\n", "Iv1 = 0-Iy1*math.sin(math.pi/6) #A\n", "In1 = math.sqrt(Ih1**2+Iv1**2) #A\n", "\n", "\n", "\n", "#Result:\n", "print \"When no changes were made, the various currents are:\"\n", "print \"Ir =\",round(Ir,2),\"A;\\tIy =\",Iy,\"A;\\tIb =\",Ib,\"A;\\tIn =\",round(In,2),\"A\"\n", "print \"\\nWhen load from B to N removed, the various currents are:\"\n", "print \"Ir =\",round(Ir1,2),\"A;\\tIy =\",Iy1,\"A;\\tIb =\",Ib1,\"A;\\tIn =\",round(In1,2),\"A\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "When no changes were made, the various currents are:\n", "Ir = 86.96 A;\tIy = 125.0 A;\tIb = 125.0 A;\tIn = 129.54 A\n", "\n", "When load from B to N removed, the various currents are:\n", "Ir = 86.96 A;\tIy = 125.0 A;\tIb = 0 A;\tIn = 66.03 A\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 14.10, Page Number: 371" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "import cmath\n", "import math\n", "\n", "#Variable declaration:\n", "Vl = 400 #line voltage(V)\n", "Vp = 230 #phase voltage(V)\n", "Ir = 30 #load current at R-phase(A)\n", "pf1 = 0.866 #power factor for R-phase(lagging)\n", "Iy = 30 ##load current at Y-phase(A)\n", "pf2 = 0.866 #power factor for R-phase(lagging)\n", "Ib = 30 ##load current at R-phase(A)\n", "pf3 = 1.0 #power factor for R-phase(lagging)\n", "R = 0.2 #resistance of each line conductor(ohm)\n", "\n", "\n", "#Calculation:\n", "phy1 = math.acos(pf1)\n", "phy2 = math.acos(pf2)\n", "phy3 = math.acos(pf3)\n", "\n", "VR = Vp*(1+0j) #V\n", "VY = Vp*(math.cos(-2*math.pi/3)+math.sin(-2*math.pi/3)) #V\n", "VB = Vp*(math.cos(2*math.pi/3)+math.sin(-2*math.pi/3)) #V\n", "\n", "#the line currents can be expressed as :\n", "IR = cmath.rect(30,-math.pi/6) #A\n", "IY = cmath.rect(30,-math.pi/2) #A\n", "IB = cmath.rect(30,2*math.pi/3) #A\n", "IN = IR+IY+IB #A\n", "\n", "#Since, the area of X-section of neutral is half of any line conductor.\n", "Rn = 2*R #resistance of neutral(ohm)\n", "#ER = VR + Drop in R phase + Drop in neutral\n", "ER = VR+R*IR+IN*2*R #V\n", "\n", "\n", "#Result:\n", "print \"The supply end voltage for R phase is\",round(ER.real,3)+1j*round(ER.imag,3),\"V\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The supply end voltage for R phase is (239.588-10.608j) V\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 14.11, Page Number: 371" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "\n", "#Variable declaration:\n", "Vl = 400 #line voltage(V)\n", "Vp = 230 #phase voltage(A)\n", "Pln = 100 #load connected b/n LN(W)\n", "Pyn = 150 #load connected b/n YN(W)\n", "\n", "\n", "\n", "#Calculation:\n", "#before disconnecting the neutral wire,\n", "R1 = Vp**2/Pln #Resistance of lamp L1(ohm)\n", "R2 = Vp**2/Pyn #Resistance of lamp L2(ohm)\n", "\n", "#When the neutral wire is disconnected,\n", "EL = 400 #V\n", "I = EL/(R1+R2) #A\n", "V1 = I*R1 #Voltage across lamp L1(V)\n", "V2 = I*R2 #Voltage across lamp L2(V)\n", "\n", "\n", "\n", "#Result:\n", "print \"The voltage across the lamps are:\"\n", "print \"Lamp 1, Voltage =\",V1,\"V ;\\tLamp 2, voltage =\",V2,\"V\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The voltage across the lamps are:\n", "Lamp 1, Voltage = 240.0 V ;\tLamp 2, voltage = 160.0 V\n" ] } ], "prompt_number": 5 } ], "metadata": {} } ] }