L-R Circuits

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Question

A capacitor of capacity $2 μ F$ is charged to a potential difference of 12 V.It is then connected across an inductor of inductance $6 μ H$ . What is the current (in A) in the circuit at a time when the potential difference across the capacitor is 6.0 V?

Moderate

Solution

From energy conservation : $\frac{1}{2} C V_{0}^{2} = \frac{1}{2} C V^{2} + \frac{1}{2} L I^{2}$
$\Rightarrow \frac{1}{2} \times 2 \times 10^{- 6} \times 12^{2} = \frac{1}{2} \times 2 \times 10^{- 6} \times 6^{2} + \frac{1}{2} \times 6 \times 10^{- 6} I^{2}$
$\Rightarrow I = 6 A$

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