Question

8.25 A three-phase balanced \(Y\) -connected load with series impedances of \((8+j 24) \Omega\) per phase and mutual impedance between any two phases of \(j 4 \Omega\) is supplied by a three-phase unbalanced source with line-to-neutral voltages of \(V_{a n}=200 \angle 25^{\circ}, V_{b n}=100 \angle-155^{\circ}\), \(V_{c n}=80 \angle 100^{\circ} \mathrm{V}\). The load and source neutrals are both solidly grounded. Determine: (a) the load sequence impedance matrix, (b) the symmetrical components of the line-toneutral voltages, (c) the symmetrical components of the load currents, and (d) the load currents.

Answer 1

Given, Z_s = 8 + j 24 Ohm Z_m = j4 Ohm V_an = 200 angle 25 degree, V_bn = 100 angle - 155 degree, V_cn = 80 angle 100 degree V (a) Load sequence impedance matrix = [Z_s + 2Z_m 0 0 0 Z_s - Z_m 0 0 0 Z_s - Z_m] = [8 + j24 + 2 (j4) 0 0 0 8 + j24 - j4 0 0 0 8 + j24 - j4] = [8 + j32 0 0 0 8 + j20 0 0 0 8 + j20] (b) Symmetrical components of voltages: [V_a0 V_a1 V_a2] = 1/3 [1 1 1 1 a a^2 1 a^2 a] [V_an V_bn V_cn] where a = 1 angle 120 degree \r\n V_a_0 = V_an + V_bn + V_cn/3 V_a_0 = 200 angle 25 + 100 angle - 155 + 80 angle 100/3 Rightarrow V_a_0 = 143.32168 angle 57.6288/3 V_a_0 = 47.773 angle 57.6268 degree V_a_1 = V_an + a V_bn + a^2 V_cn/3 = 200 angle 25 + 1 angle 120 times 100 angle - 155 + 1 angle 240 times 80 angle 100/3 * V_a_1 = 112.7840 angle - 0.03312 degree V_a_2 = V_an + a^2 V_bn + a V_c_n/3 V_a_2 = 200 angle 25 + 1 angle 240 times 100 angle - 155 + 1 angle 120 times 80 angle 100/3 * V_a_2 = 61.62310 angle 45.8824 degree