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.

 

[Carl Pugh]

Homework ?

 

[seahs]

Homework ?

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

 

[uart]

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?

 

[uart]

A battery is not an ideal voltage source, but under some circumstances is a good approximation. The above analysis is based on an ideal voltage source. It is applicable to the case of a battery only to the extent that the battery can be modeled as an ideal voltage source.

 

[sophiecentaur]

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

Yes - the 'voltage source' receives charge from the Capacitor. But, as stated above, half the energy is lost in a 'simple' connecting circuit. However, using a suitable inductor can ensure (virtually) no energy loss.