Question

A $1\mu \text{F}$ capacitor is connected in the circuit shown below. The emf of the cell is $3\text{V}$ and internal resistance is $0.5\Omega$. The resistors $R_1$ and $R_2$ have values $4\Omega$ and $1\Omega$ respectively. The charge on the capacitor in steady state must be

• A. $1\mu \text{C}$
• B. $2\mu \text{C}$
• C. $1.33\mu \text{C}$
• D. zero

Answer 1

The correct option is B $2\mu \text{C}$

In steady state, current in the branch containing the capacitor is zero and hence, emf $E=3\text{V}$ is shared between internal resistance $(r=0.5\Omega )$ and $R_2$ in the ratio of their resistance.

Voltage across $R_2$ is $\frac{E{R}_2}{R_2+r}=2\text{Volts}$ = Voltage across capcitor.

$\therefore Q=C{V}_C=1\mu \text{F}\times 2\text{V}=2\mu \text{C}$