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If $I_{0}$ and $V_{0}$ are the peak current and voltage across the resistor in a series L-C-R circuit, then the power dissipated in the circuit is (Power factor = $\cos θ$ )

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$\frac{V_{0} I_{0}}{2}$

$\frac{V_{0} I_{0}}{\sqrt{2}}$

$V_{0} I_{0} \cos θ$

$\frac{V_{0} I_{0}}{2} \cos θ$

answer is D.

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Detailed Solution

$\begin{array}{rcl} ∵ P & = & V_{r m s} I_{r m s} \cos θ \\ = & (\frac{V_{0}}{\sqrt{2}}) (\frac{I_{0}}{\sqrt{2}}) \cos θ \\ = & \frac{V_{0} I_{0}}{2} \cos θ \end{array}$

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