Q106 / 180 · AIIMS 2010 · Physics
$$
\text { What would be the voltage across } C_3 \text { ? }
$$
- Correct answerA$\frac{\left(C_1+C_2\right) V}{C_1+C_2+C_3}$
- B$\frac{C_1 V}{C_1+C_2+C_3}$
- C$\frac{C_2 V}{C_2+C_2+C_3}$
- D$\frac{C_3 V}{C_1+C_2+C_3}$
Solution
Capacitors $C_1$ and $C_2$ are in parallel, their equivalent capacitance is in series with capacitor $C_3$. Hence, the equivalent capacitance of the given circuit is
$$
C_{e q}=\frac{\left(C_1+C_2\right) C_3}{C_1+C_2+C_3}
$$
Charge on capacitor $C_3$ is
$$
\begin{aligned}
Q & =V C_{\mathrm{eq}} \\
& =\frac{V\left(C_1+C_2\right) C_3}{C_1+C_2+C_3}
\end{aligned}
$$
Voltage across $C_3$ is
$$
=\frac{Q}{C_3}=\frac{V\left(C_1+C_2\right)}{C_1+C_2+C_3}
$$
Hence, the correct answer is option: 'A'.