Question

3. A balanced three-phase delta connected load with 15+j10 ohms load on each balanced, wye connected source with Van branch is supplied by a -7.6240 KV, while the transmission line impedances are 1+7j on each line. Calculate the line currents (currents on the transmission lines) la, 1b, and 1c. (Hint: convert the Δ configured load into a Y configured load

Answer 1

v_ab = 7.62 < 0 degree kv when balanced load present the voltages and currents are in phase sequence with 120 degree part each other when Delta connected load converted into y connection z_A = z_1 z_3/z_1 z_2 + z_3 when z_1 = z_2 = z_3 = z z_A = z^2/3z = z/3 so y connected load per phase impendance z_y = 15 + j 10/3 = 5 + j 10/3 Ohm Now when we are consider single phase representation, because balanced system with a phase, i_a = v_an/1 + j7 + 5 + 10/3 j = 7.62 < 0 degree times 10^3/6 + j 10.333 = 7.62 < 0 degree times 10^3/11.9486 < 59.85 i_a = 0.6377 < - 59.85k Amps = The line currents having 120 degree a part, with same magnitude i_b = 0.6377 < - 59.85 + 120 = 0.6377 < 60.15 k Amp i_c = 0.6377 < - 59.85 + 240 = 0.6377 < 180.15 k Amp In star connection i_ph = i_L