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

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

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

The key $S_{1}$ is kept closed for long time such that capacitors are fully charged. Now key $S_{2}$ is kept 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).

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.

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.

answer is A, B.

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Detailed Solution

When switch $S_{1}$ is closed, $I = \frac{5}{200} = 25 m A$
At steady state for $S_{1}$ closed is
$C_{e q} = 8 μ F a n d Q = 40 μ C ∴ V (C_{1}) = 4 v o l t; V (C_{3}) = V (C_{4}) = 0.5 v o l t$
At steady state potential difference between ‘P’ and ‘Q’ is 4 volt.
Now when switch $S_{2}$ is closed
$∴ I_{P Q} = \frac{6 \times 2}{151} = 0.08 A$