Question

There are $n$ identical capacitors which are connected in parallel to a potential difference $V.$ These capacitors are then reconnected in series. The potential difference between the extreme ends is

• A. $nV$
• B. $0$
• C. $(n-1)V$
• D. $(n-2)V$

Answer 1

The correct option is A $nV$

Initially the capacitors are connected in parallel:


So, the voltage across each capacitor will be $V$.

Now, when they are connected in series like this:


Assuming, $V_A<V_B$

$V_A+nV=V_B$

$\Rightarrow V_B-V_A=nV$

Hence, option (b) is the correct answer.

$\text{Alternate solution:}$

If $C$ is the capacitance of the each capacitor, then the charge on each capacitor is, $q=CV$

In series, the total potential difference is the sum of potential across each capacitor. so,

$V_{tot}=V_1+V_2+V_3...+V_n$

$\Rightarrow V_{tot}={\left(\frac{q}{C}\right)}_1+{\left(\frac{q}{C}\right)}_2+{\left(\frac{q}{C}\right)}_3+...+{\left(\frac{q}{C}\right)}_n$

$\Rightarrow V_{tot}=n\left(\frac{q}{C}\right)$

As, $q=CV$,

$\Rightarrow V_{tot}=n\left(\frac{CV}{C}\right)$

$\Rightarrow V_{tot}=nV$

Hence, option (b) is the correct answer.

Key concept: Voltage across series and parallel arrangement of capacitors.