Question

Question (4 (A)In Fig. 4a, letI43s A rms, and find the average power being supplied: (a) by the source: (b) to the 20a resistor: (c) to the load. Find the apparent power being supplied: (d) by the source: (e) to the 20 Ω resistor: (f) to the load. (g) what is the load PF? , 10.ror Ann-G) 2ο Ω_ 20 Load FIGURE 4.a 4.b, find the average power absorbed by (a) the source: (b) h of the two resistors: (e) each of the two inductances: (d) the mutual inductance (B) In the circuit shown in Fig.4 50 Ω зн 2 H 5 H 100 rads FIGURE 4.b

Answer 1

V = 4 < 35 degree times 20 Rightarrow V = 80 < 35 degree (A < 35 degree Vertical-bar times 20 = (7.1 < 18.84 Vertical-bar times z_load) Rightarrow z = 6.6 + 19.097 = 112.26 < 53.84 degree a) P_avg = I_rms V_rms cos Theta Angle b/w V and I = 10/squareroot 2 times 80/squareroot 2 cos (35) = 400 VA times cos 35 P_avg = 327.66 VA \r\n b) P_avg = V_rms I_rms cos Theta = 80/squareroot 2 times 4/squareroot 2 cos 0 degree P_avg = 160 VA c) P_avg = 80/squareroot 2 times 7.1/squareroot 2 cos 53.84 = 284 VA times cos 53.84 P_avg = 167.57 VA d) complex power S = V_rams I_rms^* = (80/squareroot 2 < 35) (10/squareroot 2 < 0 degree) S = 400 < 35 degree Apparent power: -Vertical-bar S Vertical-bar = 400 VA e) S = (80/squareroot 2 < 35 degree) (4/squareroot 2 < 35 degree) = 160 < 70 degree Therefore Vertical-bar S Vertical-bar = 160 VA