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An inductor of inductance ‘L’ in series with resistance R is connected to an ideal battery of emf $V_{0}$ at t = 0. The maximum power developed across the inductor during growth of current is

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$\frac{V_{0}^{2}}{3 R}$

$\frac{V_{0}^{2}}{R}$

$\frac{V_{0}^{2}}{4 R}$

Zero

answer is B.

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Detailed Solution

$i^{2} R + P = V_{0} i$

$\Rightarrow i^{2} R - V_{0} i + P = 0$

$For i to be real, V_{0}^{2} \geq 4 PR$

$∴ P_{\max} = \frac{V_{0}^{2}}{4 R}$

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