To determine Find the power factor of other load. Explanation Given data: Two 480 V ( rms ) loads are connected in parallel. The average power of total load ( P T ) is 40800 W . Power factor of the total load ( pf T ) is 0.8 lagging . The average power of the one of the load ( P 1 ) is 20 kVA . Power factor of the one of the load ( pf 1 ) is 0.96 leading . Formula used: Write the expression for apparent power of the total load as follows: S T = S 1 + S 2 (1) Here, S 1 is the apparent power of the one of the load and S 2 is the apparent power of the other load. Write the expression for apparent power as follows: S = P + j Q (2) Here, P is the real power, and Q is the reactive power. Write the expression for real power as follows: P = S cos ( θ v − θ i ) (3) Here, ( θ v − θ i ) is the power factor angle. Write the expression for reactive power as follows: Q = S sin ( θ v − θ i ) (4) Calculation: Modify the expression in Equation (2) for apparent power of total load as follows: S T = P T + j Q T As power factor of the total load is lagging, the positive sign is used for reactive power. Use the expression in Equation (4) and rewrite the expression as follows: S T = P T + j S T sin [ ( θ v − θ i ) T ] = P T + j P T cos ( θ v − θ i ) T sin [ ( θ v − θ i ) T ] S T = P T + j P T cos ( θ v − θ i ) T 1 − [ cos ( θ v − θ i ) T ] 2 (5) Substitute 40800 W for P T and 0

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Chapter 10, Problem 21P

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