A series ac circuit containing a resistor, inductor, and a capacitor has a peak voltage of $116 \mathrm {~V}$ and a peak current of $2.00 \mathrm {~A}$. If the current lags the voltage by $52.0 ^ { \circ }$, what is the average power of the circuit?
A) $296 \mathrm {~W}$
B) $71 \mathrm {~W}$
C) $148 \mathrm {~W}$
D) $142 \mathrm {~W}$

Question

Quizplus · Accepted Answer

The average power in an AC circuit is given by $P = V_{\text{rms}} I_{\text{rms}} \cos(\phi)$, where $V_{\text{rms}}$ is the root mean square voltage, $I_{\text{rms}}$ is the root mean square current, and $\phi$ is the phase difference between the voltage and the current. Given a peak voltage ($V_{\text{peak}}$) of 116 V and a peak current ($I_{\text{peak}}$) of 2.00 A, the rms values are $V_{\text{rms}} = V_{\text{peak}}/\sqrt{2}$ and $I_{\text{rms}} = I_{\text{peak}}/\sqrt{2}$. The phase difference ($\phi$) is $52.0^\circ$, so $\cos(\phi) = \cos(52.0^\circ)$. Plugging these values in, we get $P = (116/\sqrt{2}) \times (2.00/\sqrt{2}) \times \cos(52.0^\circ) \approx 71 \mathrm {~W}$.

A Series Ac Circuit Containing a Resistor, Inductor, and a Capacitor