n identical capacitors are joined in parallel and charged to a potential V . The charged capacitors are disconnected and then connected in series using insulating handles such that the positive plate of one is connected to the negative plate of the other. The potential difference across the free plates of the combination is

Question

n identical capacitors are joined in parallel and charged to a potential V . The charged capacitors are disconnected and then connected in series using insulating handles such that the positive plate of one is connected to the negative plate of the other. The potential difference across the free plates of the combination is (A)  V  (B)  V/n  (C)  nV  (D)  (n+1)V

Tardigrade Admin · Accepted Answer

When n identical capacitors are connected in parallel, each capacitor will charged to potential V and q=CV Now, these capacitors are connected in series ∴ Total potential difference across free plates of combination is V'=V1+V2+ dots Vn=nV

$V$

$V / n$

$nV$

$(n + 1) V$

When $n$ identical capacitors are connected in parallel, each capacitor will charged to potential
$V$ and $q = C V$
Now, these capacitors are connected in series
$∴$ Total potential difference across free plates of combination is
$V^{'} = V_{1} + V_{2} + \dots V_{n} = nV$

V

V/n

nV

(n+1)V

When n identical capacitors are connected in parallel, each capacitor will charged to potential V and q=CV Now, these capacitors are connected in series ∴ Total potential difference across free plates of combination is V′=V1​+V2​+…Vn​=nV

$V$

$V / n$

$nV$

$(n + 1) V$

When $n$ identical capacitors are connected in parallel, each capacitor will charged to potential
$V$ and $q = C V$
Now, these capacitors are connected in series
$∴$ Total potential difference across free plates of combination is
$V^{'} = V_{1} + V_{2} + \dots V_{n} = nV$