Question

1-10 The power distribution characteristics of a factory's two departments A and B, in Fig. P1-10,are as shown in the following table. Each of the loads consists of resis- tors and inductors in series. Calculate and complete the table's missing information. Complex Average PowerPower Inductance Current Department (kVA) (w) 50/- 18° 20/25.84 B 20/25.84 Total 208 V, rms 60 Hz FIG.PI-10

Does 208 Vrms 60 Hz become v = 208sin(377t) V?

If this is correct where else am I going wrong?  

Answer 1

The power distribution of a factory two departments A & B are Department complex power Average power inductance current A B 20 25.84 degree 50 -18 degree so th circuit diagram is here I_A = 50 18&degree V = 208 Vms The complex power A = s_A cos theta_A = 9.89A B = 20 cos 25.84 degree kw \r\n saber te reductance I_A = 39.56 + 31.285 Ohm WC_A = 1.285 => L = 3.408 H current I_B can be founded by s_B = v_B I_B* I_B* = 96.15 25.04 degree then alpha_B = v/I_B = 1.946 + j09428 ohm L_B = 25 mH Current I_A = 50 18 degree A I_B = 96.15 25.84 degree A