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

Question

A balanced three-phase source with the following instantaneous phase voltages
van = 2500 cos(!t)
vbn = 2500 cos(!t ¡ 120±)
vcn = 2500 cos(!t ¡ 240±)
supplies a balanced Y-connected load of impedance Z = 2506 36:87± ­ per phase.
(a) Using MATLAB, plot the instantaneous powers pa, pb, pc and their sum versus
!t over a range of 0 : 0:05 : 2¼ on the same graph. Comment on the nature of the
instantaneous power in each phase and the total three-phase real power.
(b) Use (2.44) to verify the total power obtained in part (a)..

Explanation / Answer

We use the following commands E1 = input('Source # 1 Voltage Mag. = '); a1 = input('Source # 1 Phase Angle = '); E2 = input('Source # 2 Voltage Mag. = '); a2 = input('Source # 2 Phase Angle = '); R = input('Line Resistance = '); X = input('Line Reactance = '); Z = R + j*X; % Line impedance E1 = (0.75*E1:1:E1)'; % Change E1 form 75% to 100% E1 a1r = a1*pi/180; % Convert degree to radian k = length(E1); E2 = ones(k,1)*E2;%create col. Array of same length for E2 a2r = a2*pi/180; % Convert degree to radian V1=E1.*cos(a1r) + j*E1.*sin(a1r); V2=E2.*cos(a2r) + j*E2.*sin(a2r); I12 = (V1 - V2)./Z; I21=-I12; S1= V1.*conj(I12); P1 = real(S1); Q1 = imag(S1); S2= V2.*conj(I21); P2 = real(S2); Q2 = imag(S2); SL= S1+S2; PL = real(SL); QL = imag(SL); Result1=[E1, Q1, Q2, QL]; disp(' E1 Q-1 Q-2 Q-L ') disp(Result1) plot(E1, Q1, E1, Q2, E1, QL), grid xlabel(' Source #1 Voltage Magnitude') ylabel(' Q, var') text(112.5, -180, 'Q2') text(112.5, 5,'QL'), text(112.5, 197, 'Q1') The result is 16 CONTENTS Source # 1 Voltage Mag. = 120 Source # 1 Phase Angle = -5 Source # 2 Voltage Mag. = 100 Source # 2 Phase Angle = 0 Line Resistance = 1 Line Reactance = 7

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.