In the circuit shown, $R = \sqrt{\frac{L}{C}}$ . Switch S is closed at t = 0. The current through C and L would be equal after a time t equal to

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ln (2)

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ln (2)

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Solution

The correct option is B RC $ln (2)$
$i_{L} = \frac{ε}{R} (1 - e^{- R t / L})$
$i_{C} = \frac{ε}{R} e^{- t / R C}$
$i_{L} = i_{C} \Rightarrow \frac{ε}{R} e^{- t / R C} = \frac{ε}{R} (1 - e^{- R t / L})$
Substituting $R^{2} C$ in place of $L$
$\frac{ε}{R} e^{- t / R C} = \frac{ε}{R} (1 - e^{- t / R C})$
$\Rightarrow e^{\frac{- t}{R C}} = \frac{1}{2}$
$\Rightarrow t = R C ln 2$

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