In the circuit shown, initially there is no charge on capacitors and keys S1 and S2 are open. The values of the capacitors are $C_{1} = 10 μ F, C_{2} = 30 μ F$ and $C_{3} = C_{4} = 80 μ F$ .

Which of the statement(s) is/are correct?

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If key $S_{1}$ is kept closed for long time such that capacitors are fully charged, the voltage across the capacitor $C_{1}$ will be 4V.

At time t=0, the key $S_{1}$ is closed, the instantaneous current in the closed circuit will be 25mA.

If key $S_{1}$ is kept closed for long time such that capacitors are fully charged, the voltage difference between points P and Q will be 10V.

The key $S_{1}$ is kept closed for long time such that capacitors are fully charged. Now key $S_{2}$ is closed, at this time, the instantaneous current across $30 Ω$ resistor (between points P and Q) will be 0.2A (round off to 1st decimal place)

answer is A, B.

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Detailed Solution

$C_{net} = \frac{40 \times 10}{40 + 10} = 8 μ F \Rightarrow Q = 40 μ C$

V across $C_{1} = \frac{40 μ C}{10 μ F} = 4 V$ (at steady state)

V across $C_{3} = \frac{40 μ C}{80 μ F} = \frac{1}{2} V$ (at steady state)

V across $C_{4} = \frac{40 μ C}{80 μ F} = \frac{1}{2} V$ (at steady state)
Just after S1 is closed, $i = \frac{5}{30 + 70 + 100} = 25 m A$
Long time after S1 is closed, $V_{P Q} = {V_{C}}_{_{1}} = 4 V$