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A single-phase load draws 10 kW from a 416 V line at a power factor of 0.9 lagging. Calculate the complex power S=P+jQ.

User Karan Shaw
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Final answer:

In a single-phase AC circuit, the complex power S is given by the formula S = P + jQ, where P is the real power and Q is the reactive power. Given that the load draws 10 kW at a power factor of 0.9 lagging, we can calculate the real power as P = 10 kW * 0.9 = 9 kW. Since the load is purely resistive, the reactive power is zero, so Q = 0. Therefore, the complex power S = 9 kW + j0.

Step-by-step explanation:

In a single-phase AC circuit, the complex power S is given by the formula S = P + jQ, where P is the real power and Q is the reactive power.

Given that the load draws 10 kW at a power factor of 0.9 lagging, we can calculate the real power as P = 10 kW * 0.9 = 9 kW.

Since the load is purely resistive, the reactive power is zero, so Q = 0. Therefore, the complex power S = 9 kW + j0.

User Jared Beach
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