Question

In the given circuit, a charge of +80 c is given to the left plate of 4 f capacitor. Then in steady state, energy stored inside 2 f capacitor will be

Answer 1

Answer:

Then in steady state, energy stored inside 2 f capacitor will be +48 micro C

Explanation:

According to the problem ,

a charge of +80 c is given to the left plate of 4 f capacitor.

So, charge on capacitor of 3 μf

Q= 80 * 3/(2+3) = 48μC

Hence in steady state, energy stored inside 2 f capacitor will be +48 micro C