Question

Question 9 (26 marks) Figure 1 shows a single phase load (represented by an impedance ZD) connected to a single phase 50 Hz supply system represented by an ideal voltage source Vs behind a system impedance Zs Figure 1 (i) If Vs-230 volts, Zs=(0.1tj0.05) ohms and ZDF(4.940.95) ohms find the voltage across the load (13 marks) (ii) If a 121.3 uF compensation capacitor is connected in parallel with the load recalculate the (13 marks) and the real and reactive power absorbed by the load. voltage across the load and calculate the real and reactive power absorbed by the compensated load.

Answer 1

Z_s = (0.1 + j 0.05)ohm Z_D = (4.9 + j 0.95) ohm frequency �f� = 50 HZ w = 2pi f = 2 times pi times 50 w = 314.159 rad/sec. According to voltage division rule V_D = V_s times Z_D/Z_D + z_s V_D = 230 angleof 0 degree times 4.9 + j 0.95/0.1 + j 0.05 + 4.9 + j 0.95 v_D = 230 angleof 0 degree times 0.9788 angleof -0.3377 degree V_D = 225.124 angleof -0.3377 degree volts Current I_D = V_D/Z_D = 225.124 angleof -0.3377 degree/4.9 + j 0.95 I_D = 45.1038 angleof -11.31 degree Real power absorbed by the load P_D = v_D I_D cos phi P_D = 225.124 times 45.1038 times cos (11.31 - 0.3377) P_D = 9968.3267 watts Reactive power absorbed by the load Q_D = V_D I_D sin phi Q_D = 225.124 times 45.1038 X (11.31 - 0.3377) Q_D = 1932.645 VA r \r\n (ii) Z_s = (0.1 + j 0.05)ohm Z_D = (4.9