# Discharging a high voltage capacitor using a low voltage source?

seahs
Suppose a capacitor is initially charged up to 2V, and then a 1V DC voltage source is used to discharge the capacitor to 1V, what is the energy consumption from the voltage source and what is the total energy loss?

Assume the capacitor is connected to the voltage source using a resistor and the capacitor is 1F.

## Answers and Replies

Carl Pugh
Homework ?

seahs
Homework ?

nope, just wondering what happens when the current flows into the voltage source

If Vb is the battery (voltage source) voltage and Vc is the initial capacitor voltage then :

The energy gained by the voltage source is Vb*Q = Vb C ( Vc - Vb )

The reduction in stored energy in the capacitor is 1/2 * C ( Vc^2 - Vb^2 )

The energy lost is the difference of the above, which equals 1/2 * C (Vc - Vb)^2.

Note that this energy loss is independent of the value of the resistor used to connect the two.

seahs
If Vb is the battery (voltage source) voltage and Vc is the initial capacitor voltage then :

The energy gained by the voltage source is Vb*Q = Vb C ( Vc - Vb )

The reduction in stored energy in the capacitor is 1/2 * C ( Vc^2 - Vb^2 )

The energy lost is the difference of the above, which equals 1/2 * C (Vc - Vb)^2.

Note that this energy loss is independent of the value of the resistor used to connect the two.

Can the voltage source gain energy? what if the voltage source is like a non-rechargeable battery?