Discussion
The current in the circuit lags the voltage by \(30^\circ\). If the input power is 400 W, supply voltage is \(v = 100\sin(377t + 10^\circ)\). The value of input current is:
Answer: B
The real power is \(P = V_{rms} I_{rms}\cos\phi\). The supply voltage has a peak of 100 V, so
\[ V_{rms} = \frac{100}{\sqrt{2}} = 70.7\ \text{V} \]
\[ \begin{aligned} I_{rms} = \frac{P}{V_{rms}\cos 30^\circ} \\ &= \frac{400}{70.7 \times 0.866} = 6.53\ \text{A} \end{aligned} \]
The current lags the voltage by \(30^\circ\), so its angle is \(-30^\circ\) relative to the voltage. The key gives 6.53 A at \(-30^\circ\), option B; strictly the phase with respect to the reference would be \(10^\circ - 30^\circ = -20^\circ\), but none of the options says that, so the angle is read as the lag. The values of 65.3 A come from forgetting the rms conversion by a factor of ten.
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