Question

A $3\mu F$ and $4\mu F$ capacitors are charged to 6 V. After being disconnected from the battery, they are reconnected in such a manner that plates of opposite polarities are connected. The final potential difference is

• A. $\frac{6}{7}v$
• B. $\frac{7}{6}v$
• C. $6v$
• D. Zero

Answer 1

The correct option is A $\frac{6}{7}v$

Before connecting :
Charge on $3\mu F=3\times 6=18\mu C$
Charge on $4\mu F=4\times 6=24\mu C$
After connecting:
Charge magnitude on positive or
Negative plate =$(24-18)=6\mu C$
Net capacitance = $3+4=7\mu C$
Common potential =$\frac{6}{7}volt.$