Question

2.6. The load shown in Figure 2.22 consists of a resistance R in parallel with a capacitor of reactance X. The load is fed from a single-phase supply through a line of impedance 8.4 + j 11.2 Ω. The rms voltage at the load terminal is 120020° V rms, and the load is taking 30 k VA at 0.8 power factor leading. (a) Find the values of R and X (b) Determine the supply voltage V. 8.4+511.2 Ω 120020° vsh 수 ._jX

Answer 1

From the given circuit Load R & X are in parallel - R times (-j x)/R - j X = -j Rx/R - jX S = 30 kVA cos phi = 0.5 pf lead s = p - j Q = 30 k VA cos phi = p/s P = 30 k times 0.8 = 24000 W = 24 kW Q = squareroot s^2 - p^2 = squareroot 30^2 - 24^2 = 18 kHz Q = 18 KVAR S = (24 - j 18) KVA S = VI I = S/V = 30 times 10^3/1200 angle 0 degree = 25 A cos phi = 0.8 phi = 36.86 degree I = 25 squareroot 36.86 degree A V = IZ 1200 angle 0 degree = (25 angle 36.86) (-j RX/R - jx) 1200 (R - jX) = (20 + j 15) (-j RX) 1200 (R - jX) = -j 20 RX + 15 RX 1200 R - j 1200 X