When a d.c. source $V_{0}$ volt is connected across an inductor, power dissipated is P. When an a.c. source having peak value $V_{0}$ is connected across the same inductor, average power consumed is P/4. Then power factor of the inductor is

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1/4

$\frac{1}{\sqrt{3}}$

1/2

$\frac{1}{\sqrt{2}}$

answer is C.

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Detailed Solution

$\frac{V_{0}^{2}}{R} = P a n d (\frac{V_{0}}{\sqrt{2}}) (\frac{V_{0}}{\sqrt{2} Z}) . \frac{R}{Z} = P / 4$

$\begin{array}{l} ∴ \frac{V_{0}^{2} R}{2 Z^{2}} = \frac{V_{0}^{2}}{4 R} \\ \Rightarrow Z^{2} = 2 R^{2} \Rightarrow R^{2} + X_{L}^{2} = 2 R^{2} \Rightarrow X_{L} = R \end{array}$

$∴ \cos ϕ = \frac{R}{Z} = \frac{R}{\sqrt{R^{2} + R^{2}}} = \frac{1}{\sqrt{2}}$

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