Question

Obtain the overall impedance for the following cases: a. P 1000 W, pf 0.8 (leading), Vms 110 Volt b. S 5000 60° VA, V 120245° volt 1. 2. For the circuit shown in figure P-02, determine: a. The total load power factor, and power factor angle! b. The average power delivered by the source C. The complex power absorbed by the load 2 S2 16245° v + 10 C2 Figure P-02

Answer 1

a P = 1000N: pf = 0.8 load: V_rms = 10V rightarrow Vertical-bar V Vertical-bar Vertical-bar I Vertical-bar C_m theta = P: Now Vertical-bar V Vertical-bar / Vertical-bar I Vertical-bar = Vertical-bar Z Vertical-bar rightarrow Vertical-bar I Vertical-bar = Vertical-bar V Vertical-bar / Vertical-bar Z Vertical-bar Vertical-bar V Vertical-bar ^2/ Vertical-bar Z Vertical-bar C_m theta = P rightarrow Vertical-bar Z Vertical-bar = Vertical-bar V Vertical-bar ^2 C_m theta/p rightarrow Vertical-bar Z Vertical-bar = (110)^2 times (0.8)/1000 = 9.68 Ohm Now impedance = Vertical-bar Z Vertical-bar C_m theta - j Vertical-bar Z Vertical-bar sin theta {leading pt} = 9.68 (0.8) - j (9.68) sin (cm^- 1) 0.8 = 7.744 - j (9.68 times 0.6) = 7.744 - j 5.808 Ohm b S = 5000 angle 60 degree VA: V = 120 angle 48 degree V rightarrow VI* = S rightarrow V times (V*/Z*) = S rightarrow VV*/Z* = S rightarrow Vertical-bar V Vertical-bar ^2/S = Z* Therefore Z* = (120) (120)/(5000) angle 60 degree = 2.88 angle - 60 degree Therefore Z = 2.88 angle 60 degree \r\n (- j5 + 10) Ohm & (j6 + 8 Ohm) are parallel Therefore Z_L = (- j5 + 10) (j6 + 8)/10 + 8 + j6 - j5 = 30 - 40j + 60j + 80/18 + j1 = 110 + 20j/18 + j1 = 6.15 + 0.77j Ohm a Load pt Z_L 6.15 + 0.77 j Ohm Therefore Load power factor = C_m {tan^- 1 (0.77/6.15)}
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