Question

The load shown in Figure 2.22 consists of a resistance R in parallel with a 2.6. 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 kVA at 0.8 power factor leading. (a) Find the values of R and X (b) Determine the supply voltage V. 8.4 +j11.2 2 120020 VRTX

Answer 1

Given power factor cos theta = O B V_rms at load = 1200 angla 0 degree V_rms Apparent power = V_rms I_rms 30 times 10^3 = 1200 I_rms I_rms = 25 cos theta = 0.8 theta = 36.86 (Impedance angle) I = 25 angle - 36.86 true power = V I cos theta = 1200 times 25 times 0.8 = 24,000 W V^2_rms/R = 24000 R = (1200)^2/24,000 = 60 Ohm Reactive power = squareroot (AP)^2 - (TYP)^2 = squareroot (30 times 10^3)^2 - (24 times 10^3)^2 = -18000 W 18000 = (1200)^2/X rightarrow X = 80 Ohm \r\n V_rms = 1200 1200 = V 60?? - j 80/8.4 + j11.2 + (60// - j 80) I_rms = V/8.4 + j11.2 + (60// - j 80) 60// - j 80 = -j 4800/60 - j 80 = 38.4 - j 28.8 25 angle - 36.86 = V/8.4 + j 11.2 + 38.4 - j 28.8 V = 25 times 50 angle - 36.86 - 20.6 V = 1250 angle - 57.47 degree.