Discussion
In the given circuit, the average power dissipated in 6Ω resistor is :
Answer: B
The average power is 12 W.
The source has a peak value of 12 V, and average power in a resistor depends on the rms value, \(P = V_{rms}^{2}/R\). For a pure resistor the frequency (377 rad/s, that is 60 Hz) does not matter.
\[ \begin{aligned} V_{rms} &= \frac{12}{\sqrt{2}} \\ P &= \frac{V_{rms}^{2}}{R} = \frac{144/2}{6} = 12\ \text{W} \end{aligned} \]
Using the peak value in place of the rms value gives 144/6 = 24 W, which is twice the true average. 4 W, 14 W and 16 W do not follow from the circuit.
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