Question

A capacitor of capacity 3 µF is charged to 50 words and another capacitor of capacity to microfarad is charged to 80 V when they are connected with similar place together the energy lost by 2 µF capacitor is

Answer 1

Explanation:

CV,where Q is charge,C is capacitance

and V is the potential difference.

Q1=3*12=36uC

Q2=6*12=72uC

Now, they are connected in series,so equivalent capacitance=3+6=9uF

Potential difference across 3uF=36/9=4V

Potential difference across 6uF=72/9=8V