A circuit consists of a permanent source of e.m.f. E, a resistor R and a capacitor C connected in series, the internal resistance of the source is negligibly small. At the moment t=0, the capacitance of the capacitor was abruptly (jumpwise) decreased by factor $η$ . Then

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The current in the circuit as a function of time is $\frac{(η - 1) E}{R} e^{- η t / R C}$

The current in the circuit as a function of time is $\frac{(η + 1) E}{R} e^{- η t / R C}$

The current in the circuit just after the capacitance changes $\frac{(η - 1) E}{R}$

The current in the circuit just after the capacitance changes is $\frac{(η + 1) E}{R}$

answer is A, C.

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Detailed Solution

In the closed circuit, $\frac{q}{C / η} - E = i R$
Or, $i R = \frac{η q}{C} - E ...... (1)$
We differentiate this equation with respect to time (considering that un our case q decreases). $d q / d t = - i$
$R \frac{d i}{d t} = \frac{- η}{C} i o r, \frac{d i}{i} = - \frac{η}{C} d t$
Integration of this equation gives
In $\frac{i}{i_{0}} = - \frac{η t}{R C} o r, i = i_{0} e^{- η t / R C} ...... (2)$
Where $i_{0}$ is determined by condition (1). Indeed, we can write $R i_{0} = η q_{0} / C - E,$
$= (η - 1) E / R .$ hence $i = \frac{(η - 1) E}{R} e^{- η t / R C}$