Question

In the circuit shown there is steady state with the switch closed. The switch is opened at $t=0$. Choose the correct option(s). (Given : $\epsilon =24\text{V},C_1=3\text{F}$ and $C_2=2\text{F})$

• A. The voltage across $C_1$ before the switch is open is $12\text{V}$.
• B. The voltage across $C_1$ after the switch is open for a long time is $12\text{V}$.
• C. The voltage across $C_2$ after the switch is open for a long time is $24\text{V}$.
• D. The voltage across $C_2$ before the switch is open is $8\text{V}$.

Answer 1

Answer: (A), (C), (D)

The correct options are
A The voltage across $C_1$ before the switch is open is $12\text{V}$.
C The voltage across $C_2$ after the switch is open for a long time is $24\text{V}$.
D The voltage across $C_2$ before the switch is open is $8\text{V}$.

In steady state before opening the switch,
$R_{eq}=18\Omega$
$\epsilon =24\text{V}$
$i=\frac{\epsilon }{R_{eq}}=\frac{24}{18}=\frac{4}{3}\text{A}$
Potential difference across capacitor $C_1$ is
$V_1=\frac{4}{3}\times 9=12\text{V}$
Potential difference across capacitor $C_2$ is
$V_2=\frac{4}{3}\times 6=8\text{V}$
After opening the switch and waiting for a long time,
$V_2=V_1=24\text{V}$