Question

Four capacitors of capacitance $C_1=1\mu \text{F},C_2=2\mu \text{F},C_3=3\mu \text{F}$ and $C_4=4\mu \text{F}$ are connected as shown in the figure. Find the potential differences across $C_3$ when switch $S$ is closed.

• A. $7\text{V}$
• B. $3\text{V}$
• C. $7.5\text{V}$
• D. $2.5\text{V}$

Answer 1

The correct option is B $3\text{V}$


$C_1$ & $C_2$ are in parallel, so the equivalent of these two capacitors is given as

$C^{\prime }=C_1+C_2=(1+2)=3\mu \text{F}$

Similarly, $C_3$ & $C_4$ are in parallel, so

$C^{\prime \prime }=C_3+C_4=3+4=7\mu \text{F}$


Since, $C^{\prime }$ & $C"$ are in series, so the charge on both capacitors in the circuit will be same.

Charge on $C^{\prime }=$ charge on $C^{\prime \prime }$

Let voltage across $C^{\prime }$ and $C^{\prime \prime }$ be $V^{\prime }$ and $V^{\prime \prime }$ respectively.

$\therefore C^{\prime }{V}^{\prime }=C^{\prime \prime }{V}^{\prime \prime }$

$V^{\prime }=\frac{C^{\prime \prime }}{C^{\prime }}{V}^{\prime \prime }=\frac{7}{3}{V}^{\prime \prime }$

Also, given that,

$V^{\prime }+V^{\prime \prime }=10\text{V}$

$\frac{7V^{\prime \prime }}{3}+V^{\prime \prime }=10$

$\therefore V^{\prime \prime }=3\text{V}$

Therefore, potential difference across $C_3$ & $C_4$ will be $3\text{volts}$.

So. the right option is $\left(b\right)$