Question

In the circuit shown, initially there is no charge on capacitors and keys $S_1$ and $S_2$ are open. The values of the capacitors are $C_{1}=10\,\mu F, C_{2}=30\,\mu F$ and $C_{3}=C_{4}=80\,\mu F.$
Which of the statement(s) is/are correct?

• A. At time $t = 0$, the key $S_1$ is closed, the instantaneous current in the closed circuit will be 25 mA
• B. 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 4 V
• C. 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 $30\, \Omega$ resistor (between points P and Q) will be 0.2 A (round off to $1^{st}$ decimal place)
• D. 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 10 V

Answer 1

Answer: (A), (B)

When switch $S_1$ is closed, so the equivalent circuit at $t = 0$ is
$\therefore I=\frac{5}{200}=25\,mA$
At steady state for $S_1$ closed is
$C_{eq} = 8\, \mu F$ and $Q = 40\, \mu C$
$\therefore V(C_1) = 4$ volt; $V(C_3) = V(C_4) = 0.5$ volt
At steady state potential difference between 'P' and 'Q' is 4 volt.
Now when switch $S_2$ is closed (Then equivalent circuit)
$\therefore I_{PQ}=\frac{6\times2}{151}=0.08\,A$
Final answer (A, C)