Question

A capacitor of capacity C${}_1$ charged up to V volt and then connected to an uncharged capacitor C${}_2$.Then final P.D. across each will be:

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

Answer 1

The correct option is B $\frac{C_{1}V}{C_1+C_2}$

The chaege on capacitor of capacity $C_1$=$C_{1}V$

and we know that charge is conserved therefore we can say that total charge of both the capacitors after connecting them will also =$C_{1}V$

$\therefore$ we can write that $C_1{V}_f+C_2{V}_f$=$C_{1}V$

$\Rightarrow$ $V_f=\frac{C_{1}V}{C_1+C_2}$ (here , $V_f=$ final potential difference accross the capacitors)