Question

A capacitor of capacity $C_1$ is charged upto potential $V$ volt and then connected in parallel to an uncharged capacitor of capacity $C_2$. The final potential difference across each capacitor will be :

• A. $\frac{C_{2}V}{C_1+C_2}$
• B. $\frac{C_{1}V}{C_1+C_2}$
• C. $(1+\frac{C_2}{C_1})V$
• D. $(1-\frac{C_2}{C_1})V$

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{C_{2}V}{C_1+C_2}$
C $(1+\frac{C_2}{C_1})V$

Firstly, $C_1$ is charged upto potential $V$.
Charge stored $q=C_{1}V$..........(1)
Now $C_1$ and $C_2$ are connected in parallel.
Let charge on $C_1$ is $q_1$ and charge on $C_2$ is $q_2$.
Then $V_A-V_B=\frac{q_1}{C_1}=\frac{q_2}{C_2}$
$q_1{C}_2=q_2{C}_1$..............(2)
Also, $q_1+q_2=q\Rightarrow q_2=q-q_1$
Hence from equation (2)
$q_1{C}_2=(q-q_1)C_1$
$or,q_1{C}_2=q{C}_1-q_1{C}_1$
$or,q_1(C_1+C_2)=C_{1}V.C_1$ [ From equation (1)]
$or,q_1=\frac{C_1^{2}V}{C_1+C_2}$
Hence, $V_A-V_B=\frac{q_1}{C_1}=\frac{\frac{C_1^{2}V}{C_1+C_2}}{C_1}=\frac{C_{1}V}{C_1+C_2}$