
Two capacitors $C_1$ and $C_2$ are connected in parallel to a battery. Charge-time graph is shown below for…

- $\mathrm{C}_2 \gt \mathrm{C}_1, \mathrm{U}_2 \lt \mathrm{U}_1$
- $\mathrm{C}_1 \gt \mathrm{C}_2, \mathrm{U}_1 \gt \mathrm{U}_2$
- $\mathrm{C}_1 \gt \mathrm{C}_2, \mathrm{U}_1 \lt \mathrm{U}_2$
- $\mathrm{C}_2 \gt \mathrm{C}_1, \mathrm{U}_2 \gt \mathrm{U}_1$
Solution
$q=C V \text { and } U=\frac{1}{2} C V^2$
Since $C_1$ and $C_2$ are connected in parallel, $V_1=V_2$.
Also from graph $q_1 < q_2$
$\begin{aligned}
& \Rightarrow \quad C_1 V_1 < C_2 V_2 \\ & \text { i.e. } C_1 < C_2 \text { or } C_2>C_1 \\ & \qquad \frac{U_1}{U_2}=\frac{C_1 V_1^2}{C_2 V_2^2}=\frac{C_1}{C_2} < 1
\end{aligned}$
or $U_1 < U_2$ or $U_2>U_1$
Asked in: JEE Main 2025 (28 Jan Shift 1)