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A coil having both inductance and resistance is connected across a 200 V AC supply and draws a current of 10 A. The power consumed by the circuit is 1500 W. Determine the wattless (reactive) component of the current.
- 6.6 A
- 7.5 A
- 5.0 A
- 8.0 A
Correct answer: 6.6 A
Solution
The power factor is cos(phi) = P/(V*I) = 1500/(200*10) = 0.75. Then sin(phi) = sqrt(1 - 0.75²) = sqrt(1 - 0.5625) = sqrt(0.4375) = 0.661. The wattless (reactive) current is I*sin(phi) = 10*0.661 = 6.6 A.
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