Question: In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance $\mathrm{C}$ will be :
$\mathrm{CE} \frac{\mathrm{r}_{2}}{\left(\mathrm{r}+\mathrm{r}_{2}\right)}$
$\mathrm{CE} \frac{r_{1}}{\left(r_{1}+r\right)}$
CE
$\mathrm{CE} \frac{\mathrm{r}_{1}}{\left(\mathrm{r}_{2}+\mathrm{r}\right)}$
Correct Option: 1
Solution:
