Two capacitors $\mathrm{C}_1=3 \mu \mathrm{F}$ and $\mathrm{C}_2=2 \mu \mathrm{F}$ are connected in series…
Two capacitors $\mathrm{C}_1=3 \mu \mathrm{F}$ and $\mathrm{C}_2=2 \mu \mathrm{F}$ are connected in series across d.c. source of $100 \mathrm{~V}$. The ratio of the potential across $C_2$ to $C_1$ is
$2: 3$
$3:2$
$6:5$
$5:6$
Solution
$\begin{array}{ll}
\therefore \quad & \mathrm{C}_1=3 \mu \mathrm{F} \text { and } \mathrm{C}_2=2 \mu \mathrm{F} \\
& \mathrm{C}_{\text {series }}=\frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}=\frac{6}{5} \mu \mathrm{F} \\
& \text { Also, } \mathrm{Q}=\mathrm{CV} \\
\mathrm{Q} & =\mathrm{C}_{\text {series }} \times \mathrm{V} \\
& =\frac{6}{5} \times 100=120 \mu \mathrm{C}
\end{array}$
$\mathrm{Q}$ will be the same across both the capacitors as they are in series.
$\therefore \quad$ Potential across capacitors,
$\begin{aligned}
& \mathrm{V}_1=\frac{\mathrm{Q}}{\mathrm{C}_1}=\frac{120}{3}=40 \mathrm{~V} \\
& \mathrm{~V}_2=\frac{\mathrm{Q}}{\mathrm{C}_2}=\frac{120}{2}=60 \mathrm{~V} \\
& \therefore \quad \mathrm{V}_2: \mathrm{V}_1=60: 40=3: 2
\end{aligned}$