Question

(30) In the circuit with a voltage source and two loads below, find the following: 2. The complex power of each load a. b. The total complex power for the two loads - note units of power in the figure c. Sketch the triangle for the total complex power e total current from the source as a phasor assuming the line voltage at the load is 20820 VRMS The value of the parallel capacitance in parallel with the two loads needed to correct the power factor of the total load to 0.95. Assume the source frequency is 60 Hz. e. S-32.0 KVA 208 0° VRMS 0.8 pf lagging P 44.3 KW 0.7 pf lagging

Answer 1

Complex power of each load a) four load s_1 complex power S = VI P = VI cos theta [cos theta = -7 pf log 9] implies 4I cos theta/VI = P/S S_1 = P/cos theta = 44.3/0.7 = 63.28 KVA S_1 = 63.28 KVA angleof cos^-1 0.7 lagging = 63.28 KVA angleof 45.57 lagging for load S_2 complex power S = 32 KVA angleof cos^-1 0.8 = 32 KVA angleof 36.87 Let voltage Across each branch is V angleof 0 degree current through load S_1 = I_1 VI_1* = S_1 V angleof 0 I_1* = 63.28 KVA angleof 95.57 I_! = (63.28/V angleof -95) KA current through load S_2 = I_2 VI_2* = S_2 = 32 KVA angleof 36.87 I_2 = (32/V angleof -36.87) KA \r\n By applying KCL at A I = I_1 + I_2 implies I = 63.28 angleof -45.57/V + 32/V angleof -36.87 0 = 1/V [63.28 angleof -45.57 + 32 angleof -36.87] = 95.035 angleof -42.65/V