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 statements is/are correct?

• A. 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 $4V$.
• B. 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 instantancous current across $30\Omega$ resistor (between points $P$ and $Q$ ) will be $0.2A$ (round off to $1^{\text{st}}$ decimal place).
• C. At time $t=0,$ the key $S_1$ is closed, the instantaneous current in the closed circuit will be $25mA.$
• D. If key $S_1$ is kept closed for long time such that capacitors are fully charged, the voltage difference between point $P$ and $Q$ will be $10V$.

Answer 1

Answer: (A), (C)

At time $t=0,$ the key $S_1$ is closed, the instantaneous current in the closed circuit will be $25mA.$


For option C,: $S_1$ closed
When switch $S_1$ being closed, so the equivalent circuit at $t=0$ is given by,

Here, $R_{eq}=R_1+R_2+R_3{R}_{eq}=200\Omega$
Instataneous current will be given by $i=\frac{V}{R_{eq}}=\frac{5}{200}=25mA$

For part A,
At steady state for $S_1$,
$\frac{1}{Ceq}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{10}+\frac{1}{80}+\frac{1}{80}\Rightarrow C_{eq}=8\mu F$
Now, total charge consumption by the circuit at steady state condition,
$Q=C_{eq}V8\mu F\times 5=40\mu C$

Now, the voltage across $C_1=\frac{Q}{V}=\frac{40\mu C}{10\mu f}=4V$
& $V\left(C_3\right)=V\left(C_4\right)=0.5$ volt
Hence, option A is correct.

For option B & D,
At steady state potential difference between ${}^{\prime }{P}^{\prime }$ and ${}^{\prime }{Q}^{\prime }$ is 4 volt.
Now, when switch $S_2$ is closed (Then equivalent circuit)


so, $l_{pQ}=\frac{6\times 2}{151}=0.08A$
So, option B and D are wrong.