Q81 / 120 · VITEEE 2006 · Physics
A potential difference of $300 \mathrm{~V}$ is applied to a combination of $2.0 \mu \mathrm{F}$ and $8.0 \mu \mathrm{F}$ capacitors connected in series. The charge on the $2.0 \mu \mathrm{F}$ capacitor is
- A$2.4 \times 10^{-4} \mathrm{C}$
- Correct answerB$4.8 \times 10^{4} \mathrm{C}$
- C$7.2 \times 10^{-4} \mathrm{C}$
- D$9.6 \times 10^{-4} \mathrm{C}$
Solution
$\mathrm{V}=300 \mathrm{~V}, \mathrm{C}_{1}=2.0 \mu \mathrm{F}, \mathrm{C}_{2}=8.0 \mu \mathrm{F}$,
Net capacitance, $\frac{1}{\mathrm{C}_{\mathrm{s}}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}$
$$
\begin{array}{l}
\Rightarrow \mathrm{C}_{\mathrm{s}}=\frac{\mathrm{C}_{1} \mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}} \\
\Rightarrow \mathrm{C}_{\mathrm{s}}=\frac{2 \times 8}{2+8}=\frac{16}{10}=1.6 \mu \mathrm{F}
\end{array}
$$
Now total charge,
$\mathrm{Q}=\mathrm{V}_{\mathrm{s}} \times \mathrm{C}_{\mathrm{s}}=300 \times 1.6 \times 10^{-6}=4.8 \times 10^{-4} \mathrm{C}$
$\because$ In series charge is same on capacitors
$\Rightarrow$ Charge on $2 \mu \mathrm{F}$ capacitor is $4.8 \times 10^{-4} \mathrm{C}$
Hence, the correct answer is option: 'B'.