An AC source vₛ = 200*sqrt(2) sin(omega t + 15 deg) is applied to a circuit, producing a current i = 2 sin(omega t + 45 deg). Find the average power consumed by the circuit.

100*sqrt(6) watt
400*sqrt(2) watt
200 watt
200*sqrt(2) watt

Correct answer: 100*sqrt(6) watt

Solution

Peak voltage = 200 sqrt2, so V_rms = 200. Peak current = 2, so I_rms = 2/sqrt2 = sqrt2. Phase difference phi = 45-15 = 30 deg, cos30 = sqrt3/2. P = 200 * sqrt2 * (sqrt3/2) = 100 sqrt6 W.

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