Question

A capacitor of capacitance $C_1$ charged at a certain potential v. It is connected with another uncharged capacitor $C_2$. What is the final p.d. of this new system.

• A. $\frac{C_{2}V}{C_1{C}_2}$
• B. $\frac{C_{1}V}{C_1{C}_2}$
• C. $(1+\frac{C_2}{C_1})$
• D. $D(1-\frac{C_2}{C_1})V$

Answer 1

The correct option is B $\frac{C_{1}V}{C_1{C}_2}$

Charge on first capacitors, $q_1=c_{1}v$

Charge on second capacitor, $q_2=c$

when they are connected in parallel the total charge, $q=q_1+q_2$

$\therefore q_1=c_{1}v$

and capacitor, $c=c_1+c_2$

Let $V$ be the common potential difference across each capacitors, then $q=cv$

$\therefore v^{\prime }=\frac{q}{c}=\frac{c_{1}v}{c_1+c_2}$