In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance $C$ will be
JEE MAIN 2017, Medium
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In steady state, flow of current through capacitor will be zero.
Current through the circuit,
$i=\frac{E}{r+r_{2}}$
Potential difference through capacitor
$V_{c}=\frac{Q}{C}=E-i r=E-\left(\frac{E}{r+r_{2}}\right) r$
$\therefore \quad Q=C E \frac{r_{2}}{r+r_{2}}$
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