Question

In the given circuit, $R_1=10\Omega ,R_2=20\Omega ,$$C_1=1\mu \text{F},C_2=2\mu \text{F}$ and $E=6\text{V}$. Find the charge on each capacitor in the steady state.

• A. $2\mu \text{C},12\mu \text{C}$
• B. $6\mu \text{C},12\mu \text{C}$
• C. $4\mu \text{C},2\mu \text{C}$
• D. $2\mu \text{C},4\mu \text{C}$

Answer 1

The correct option is A $2\mu \text{C},12\mu \text{C}$

At steady state, capacitor is fully charged and no current flows through the capacitor.

Current in the outer loop is,

$I=\frac{E}{R_1+R_2}=\frac{6}{10+20}=\frac{1}{5}\text{A}$


$\because R_1$ and $C_1$ are in parallel, thus potential difference across them will be the same.

Charge on capacitor $C_1$,

$q_1=C_1{V}_1=C_{1}I{R}_1=1\times \frac{1}{5}\times 10=2\mu \text{C}$

Using $KVL$ to the open loop form $\text{A}$ to $\text{B}$ through $C_1$ and $R_2$ we get potential difference across $C_2$ as,

$V_2=\frac{q_1}{C_1}+I{R}_2=\frac{2}{1}+0.2\times 20=6\text{V}$

Charge on capacitor $C_2$,

$q_2=C_2{V}_2=2\times 2=12\mu \text{C}$

Hence, option $\left(a\right)$ is correct.