Figure (i) shows two capacitors connected in series and connected by a battery. The graph (ii) shows the…
Figure (i) shows two capacitors connected in series and connected by a battery. The graph (ii) shows the variation of potential as one moves from left to right on the branch $\mathrm{AB}$ containing the capacitors. Then
$C_{1}=C_{2}$
$\mathrm{C}_{1} < \mathrm{C}_{2}$
$\mathrm{C}_{1}>\mathrm{C}_{2}$
$\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ cannot be compared
Solution
Since, potential difference across $\mathrm{C}_{2}$ is greater than $\mathrm{C}_{1}$. $\Rightarrow \mathrm{C}_{1}>\mathrm{C}_{2}\left[\because \mathrm{V}=\frac{\mathrm{q}}{\mathrm{C}}\right.$ and $\mathrm{q}$ is same in series
