Question

2. (25 points) For the given ideal transformer in Fig. (a), where /10 Ω, and ZL 50 Ω, calculate the average power delivered to Zulf connections primary and the secondary are added, as shown in Fig.(b), and all other qualities recalculate the average power delivered to ZL between the are the same. Zy (b) Without electrical isolation (a) With electrical isolation

Answer 1

Given diagram shown below: - rightarrow The above diagram is transfer equivalent circuit referred to primary. n = squareroot 2 v_s = 500 angle 0 degree V_peak z_1 = j 10 ohm z_L = 50 ohm Here, connect in the transformer. So, I_1 = 500/25 + j10 I = 18.56 angle -21.80 Amps. \r\n rightarrow average power declared to the load, avg power declared to load = I^2 R equation (1) where, I = 18.562 times 7 R = 25 Now, substitute the values in eqn (1) = (18.562)^2 times 25 therefore avg power load = 8611.25 wat Now, when winding are directly canned: - \r\n voltage across the primary winding v_1 = 500 - j 10 (I_1) equation (a) where I_1 = 18.56 angle -21.80 Now, substitute the values in eqn (a) v_1 = 500 - (310) times 18.56 angle -21.80 v_1 = 464.242 angle -21 v now, v_L = v_1 * n equation (b) where, v_1 = 464.242 angle -21 v n