Question

Five capacitors are connected as shown in the figure. Initially, all capacitors are uncharged and $S$ is open. When $S$ is closed, then the potential difference between points $\text{M}$ and $\text{N}$ is :
[assume steady state to be achieved after $S$ closes]

• A. $15\text{V}$
• B. $10\text{V}$
• C. $14\text{V}$
• D. $12\text{V}$

Answer 1

The correct option is D $12\text{V}$

After switch $S$ is closed, all the capacitor and both batteries form a closed loop. Let that higher emf battery supplies a charge $q$ in circuit.


Now applying $KVL$ in the circuit in anti-clockwise path,
(All the capacitor will have same $q$ charge )

$\Sigma \Delta V=0$

$\Rightarrow -\left(\frac{q}{4}\right)-\left(\frac{q}{2}\right)-\left(\frac{q}{4}\right)-7-\left(\frac{q}{6}\right)-\left(\frac{10q}{12}\right)+31=0$

$\Rightarrow \frac{3q+6q+3q+2q+10q}{12}=24$

$\Rightarrow 24q=24\times 12$

$\therefore q=\frac{24\times 12}{24}=12\mu \text{C}$

Now, considering the path $\text{M}$ to $\text{N}$ along direction of charge flow;

$V_M-\left(\frac{q}{4}\right)-7-\left(\frac{q}{6}\right)=V_N$

$\Rightarrow V_M-V_N=\frac{5q}{6}+7$

$\therefore V_M-V_N=\left(\frac{5\times 12}{12}\right)+7=12\text{V}$

$$\begin{array}{c}\text{Why this question?}\\ \text{Concept: When a higher emf battery is}\\ \text{present in a closed loop, we assume charge to be}\\ \text{supplied in circuit from higher emf battery.}\\ \\ \text{Tip: In order to avoid any confusion, its better to}\\ \text{apply KVL along direction of charge flow.}\end{array}$$