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Consider the situation shown in diagram. Both capacitors are having same capacitance C. Capacitors are initially uncharged. Switch K is closed at time $t = 0$ . After a long time capacitors are fully charged. List-I represents value of r and List-II represents charge passing through r during complete charging process. Now match the followings:
List-I List-II
(P) r = 0 (1) $\frac{C V}{6}$
(Q) r = R (2) $\frac{C V}{5}$
(R) r = 2R (3) $\frac{C V}{4}$
(S) r = 3R (4) $\frac{C V}{3}$

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P-3; Q- 4; R-2; S-1

P-2; Q-4; R-3; S-1

P-4; Q-3; R-2; S-1

P-4; Q-2; R-3; S-1

answer is A.

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Detailed Solution

Our target is $\int_{0}^{\infty} x d t$

$\int_{0}^{\infty} I d t = 2 C V$ Now Let $\int_{0}^{\infty} x d t$
= q and $\int_{0}^{\infty} I_{1} d t = q_{1}$
Now applying loop rule
$2 I_{1} R = r x + R (I - I_{1}) \Rightarrow 3 I_{1} = \frac{r}{R} x + I 3 \int_{0}^{\infty} I_{1} d t = \frac{r}{R} \int_{0}^{\infty} x d t + \int_{0}^{\infty} I d t \Rightarrow 3 q = \frac{r}{R} q + 2 C V \dots \dots . \dots . . (i) a l s o \int_{0}^{\infty} (I_{1} + x) d t = C V ∴ q_{1} + q = C V \dots \dots \dots \dots \dots . (i i)$

Solving (i) and (ii)

$q = \frac{C V}{3 + r / R}$

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