A $1\,\mu F$ capacitor is connected in the circuit shown below. The emf of the cell is $3\ volts$ and internal resistance is $0.5\ ohms$ . The resistors $R_1$ and $R_2$ have values $4\ ohms$ and $1\ ohm$ respectively. The charge on the capacitor in steady state must be.......$\mu C$
Medium
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In steady state current in the branch containing the capacitor is zero and hence
$\mathrm{emf}$ $\mathrm{E}$ is shared between
$\mathrm{r} $ and $ \mathrm{R}_{2}=$ in the ratio of their resistance voltage across $\mathrm{R}_{2}$ is $\frac{\mathrm{ER}_{2}}{\mathrm{R}_{2}+\mathrm{r}}=2 \mathrm{\,Volts}$
$=$ Voltage across capacitor.
$\therefore $ $\mathrm{Q}=\mathrm{CV}_{\mathrm{c}}=2 \mu \mathrm{C}$
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