A balanced three-phase source with the following instantaneous phase voltages v_

Question

A balanced three-phase source with the following instantaneous phase voltages v_an(t) = 2500 cos (omega t) v_bn (t) = 2500 cos(omega r -120 degree) v_cn (t) = 2500cos(omega t - 240 degree) supplies a balanced Y-connected load of impedance Z = 250 angle 36.87 degree per phase. Using Matlab, plot instantaneous powers p_a, p_b, p_c, and their sum versus omega t over a range of 0:0.05: 2 pi on the same graph. Comment on the nature of the instantaneous power in each phase and the three-phase real power. Include the code. Use P_active =3V_an I_a cos theta or P_active = squareroot 3V _ab I_a cos theta to verify the total power obtained in a.

Explanation / Answer

a.

PA = 2500*10*cos(wt)*cos(wt – iA)

PB = 2500*10*cos(wt - 1200)*cos(wt – 1200 – iB )

PC = 2500*10*cos(wt - 2400)*cos(wt - 2400 – iC)

PTOT = PA + PB + PC

b.

PACTIVE = 3 VAN IAN cos () where is phase difference between voltage and current for the given phase.

MATLAB CODE FOR THE ABOVE

clc;
close all;
clear all;

wt = 0:0.05:2*pi;

Va = 2500*cos(wt);
Vb = 2500*cos(wt -(120*pi/180));
Vc = 2500*cos(wt -(240*pi/180));

Vph1 = 2500*exp(j*0*pi/180);
Vph2 = 2500*exp(j*-120*pi/180);
Vph3 = 2500*exp(j*-240*pi/180);
Z = 250*exp(j*+36.87*pi/180);

IaA = Vph1/Z;
ThetaA = angle(IaA);

IbA = Vph2/Z;
ThetaB = angle(IbA);

IcC = Vph3/Z;
ThetaC = angle(IcC);

%INSTANTANEOUS POWER
Pa = 2500*10*cos(wt).*cos(wt-ThetaA);
Pb = 2500*10*cos(wt -(120*pi/180)).*cos(wt -(120*pi/180) -ThetaB);
Pc = 2500*10*cos(wt -(240*pi/180)).*cos(wt -(240*pi/180) -ThetaC);
Threephasepower = Pa + Pb + Pc;
%subplot(3,1,1);
plot(wt,Pa,'r',wt,Pb,'m',wt,Pc,'y',wt,Threephasepower,'c');
legend('Pa','Pb','Pc','Threephase');

%TOTAL POWER IN PHASE A
PtotA = 3*2500*10*cos( (0*pi/180) -ThetaA);
%PtotB = 2500*10*cos((-120*pi/180) -ThetaB);
%PtotC = 2500*10*cos((-240*pi/180) -ThetaC);

display(PtotA);
%display(PtotB);
%display(PtotC);

% subplot(3,1,2);
% plot(wt,Pb);
%
% subplot(3,1,3);
% plot(wt,Pc);

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