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In the circuit shown in figure $C_{1} = 2 C_{2} .$ Switch S is closed at time t=0. Let $i_{1} a n d i_{2}$ be the currents flowing through $C_{1} a n d C_{2}$ at any time t, then the ratio $i_{1} / i_{2}$

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is constant

increases with increase in time t

decrease with increase in time t

first increases then decreases

answer is B.

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Detailed Solution

Here, $I_{1} = \frac{V}{R} e^{\frac{- t}{R C_{1}}} a n d I_{2} = \frac{V}{R} e^{\frac{- t}{R C_{2}}}$

$\frac{I_{1}}{I_{2}} = \frac{e^{\frac{- t}{R C_{1}}}}{e^{\frac{- t}{R C_{2}}}} = e^{\frac{- t}{R} (\frac{1}{C_{1}} + \frac{1}{C_{2}})} = e^{\frac{- t (C_{1} + C_{2})}{R C_{1} C_{2}}}$

Hence option 2 is correct

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