In the circuit shown, switch S is closed at $t = 0$ . Let $i_{1}$ and $i_{2}$ be the current at any finite time t, then the ratio $i_{1} / i_{2}$

is constant

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increases with time

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decreases with time

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first increases and then decreases

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Solution

The correct option is B increases with time
We know that the current in RC circuit at any time t is $i = \frac{V_{0}}{R} e^{- t / R C}$
here, $i_{1} = \frac{V}{2 R} e^{- t / (2 R) (3 C)}$ and $i_{2} = \frac{V}{R} e^{- t / R C}$

$∴ \frac{i_{1}}{i_{2}} = \frac{1}{2} . e^{t / R C (1 - 1 / 6)} = \frac{1}{2} . e^{5 t / 6 R C}$

Thus, we can see that this ratio is increasing with time.

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