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A circuit consists of a source of a constant emf , a resistance R and a capacitor with capacitance C connected in series. The internal resistance of the source is negligible. At a moment t= 0, the capacitance of the capacitor is abruptly decreased to $η$ -fold. Find the current flowing through the circuit as a function of time t.

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$\frac{E}{R} (η - 1) e^{- \frac{ηt}{CR}}$

$\frac{E}{R} (η - 1) (1 - e^{- \frac{t}{CR}})$

$\frac{E}{R} (η - 1) e^{\frac{t}{CR}}$

$\frac{E}{R} (η - 1) (1 - e^{- \frac{ηt}{CR}})$

answer is A.

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Detailed Solution

Let at any time charge in the capacitor be q and current at that instant i = $\frac{d q}{d t} .$

Now applying KVL , $\frac{d q}{d t} + \frac{q}{R \frac{C}{η}} - \frac{E}{R} = 0$

Solving ,q= $\frac{C E}{η} + C E (1 - \frac{1}{η}) e^{- \frac{η t}{R C}} \Rightarrow i = \frac{d q}{d t} = \frac{E}{R} (η - 1) e^{- \frac{η t}{R C}}$

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