Question

Show the work out but try to make simple as much as u can 0lz

Question 4 (5+5+5+5+5+5 30 marks) Given the AC circuit shown in Figure 4 a) Find the total impedance ZT seen by the voltage source; b) Determine the total current drawn from the voltage source. Express your (5 marks) answer in the form IT/0 Amps; c) Determine the average power delivered by the source d) Determine the reactive power absorbed by the total load; e) Determine the apparent power delivered by the source f Determine the power factor of the total load. (5 marks) (5 marks) (5 marks) (5 marks) (5 marks) 652 202 240 6 j2052 10S2 -j122 Figure 4: Circuit for Question 4

Answer 1

V = 240 angle 0 degree Impedance of first branch = -j 40 + 20 + 10 Z_1 = (30 - j 40) ohm = 50 angle -53.13 degree ohm Impedance of 2nd Branch Z_2 = 6 + j 20 - j 12 Z_2 = (6 + j 8) ohm = 10 angle 53.13 degree ohm Total Impedance Z_T = Z_1//Z_2 = Z_1 Z_2/Z_1 + Z_2 Z_T = 50 angle -53.13 times 10 angle 53.13/50 angle -53.13 + 10 angle 53.13 = 10.38 angle 41.63 degree ohm Z_T = (7.7586 + j 6.8965)ohm = 10.38 angle 41.63 degree ohm I_T = V/Z_T = 240 angle 0/10.38 angle 41.63 = 23.12 angle -41.63 degree A Average Power P_avg = V I cos phi P_avg = 240 times 23.12 times cos(41.63 degree) = 4147.45W Reactive power Q = V I sin phi Q = 240 times 23.12 times sin(41.63 degree) = 3686.165W Apparent power S = squareroot P^2 + Q^2 = 55488 VA {If v =

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