- #1
ukamle
- 12
- 0
We know that energy stored in capacitor=
[tex]\int \frac{q}{C} dq = \frac{q^2}{2C} = qV/2 [/tex]
But work done by battery = qV
Where does the other qV/2 go ?
Assume NO resistance in circuit
The potential of the battery does not change with time
[tex]\int \frac{q}{C} dq = \frac{q^2}{2C} = qV/2 [/tex]
But work done by battery = qV
Where does the other qV/2 go ?
Assume NO resistance in circuit
The potential of the battery does not change with time