Question

A condenser of capacity $2\mu F$ is charged to a potential of 200V. It is now connected to an uncharged condenser of capacity $3\mu F$. The common potential is :

• A. 200 V
• B. 100 V
• C. 80 V
• D. 40 V

Answer 1

The correct option is C 80 V

After connecting to the second capacitor the Potential difference across both the capacitor will be same Therefore we know that

$V=\frac{q}{C}$

so,

$\frac{q_1}{C_1}=\frac{q_2}{C_2}=V$

or

$\frac{q_1}{2}=\frac{q_2}{3}=V$

By conservation of charge, $q_1+q_2=400$

Solving the above two equations give the charge on the both the capacitor

$q_1=160\mu C$

$q_2=240\mu C$

and the potential difference across both the capacitor is $V=\frac{q_2}{C_2}=80V$