Capacitor Energy in an RC Circuit

[t_n_p]

In an RC circuit, current flows from the battery until the capacitor is completely charged. Is the total energy supplied by the battery equal to the total energy stored by the capacitor? If not, where does the extra energy go?

I'm a total physics newbie, but taking an almost wild guess I think that some of the energy in the wires as heat due to resistance. Can anybody confirm/deny my theory?

Thanks

 

[Hootenanny]

All the information you could ever need; http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html#c2

 

[t_n_p]

I'm still stuggling to understand.
So U stands for stored energy in the capacitor and U=.5QV

but how can that be used to answer my question?

 

[Hootenanny]

Transporting differential charge dq to the plate of the capacitor requires work dU = Vdq. But as the voltage rises toward the battery voltage in the process of storing energy, each successive dq requires more work to force it onto the positive plate.

[...]

Note that the total energy stored QV/2 is exactly half of the energy QV which is supplied by the battery, independent of R!

So the capacitor stores half the energy supplied by the battery, the other half is the work done placing charge on the plates, irrespective of the resistance.

 

[Weimin]

Hmm, then how about the heat released from the resistance?

 

[t_n_p]

Hmm, then how about the heat released from the resistance?

Yeah, that's what I am wondering, or is it negligible?

 

[Hootenanny]

So the capacitor stores half the energy supplied by the battery, the other half is the work done placing charge on the plates, irrespective of the resistance.

Hmm, then how about the heat released from the resistance?

Perhaps it would be better if I put it this was. For any RC circuit with a finite resistance, half the energy supplied by the battery is dissapated as heat through the resistor. This is independent of the value of the resistance.

 

[t_n_p]

Perhaps it would be better if I put it this was. For any RC circuit with a finite resistance, half the energy supplied by the battery is dissapated as heat through the resistor. This is independent of the value of the resistance.

So of the energy supplied by a battery, half is stored in the capacitor and half is lost as heat through resistors?

 

[Weimin]

It seems that we have more than two halves:

- half is stored in the capacitor
- half is the work done placing charges on the plate
- half is lost on the resistor

And I also wonder why we did not take into account the energy of the electro-magnetic field: as the charges are gradually filled up the capacitor there is a variable current in the circuit. This induces a variable magnetic field and so on, finally we have emf propagating in the space, meaning that we are losing the energy. I am a bit confused on this.

 

[t_n_p]

It seems that we have more than two halves:

- half is stored in the capacitor
- half is the work done placing charges on the plate
- half is lost on the resistor

And I also wonder why we did not take into account the energy of the electro-magnetic field: as the charges are gradually filled up the capacitor there is a variable current in the circuit. This induces a variable magnetic field and so on, finally we have emf propagating in the space, meaning that we are losing the energy. I am a bit confused on this.

Way to confuse me...

 

[Hootenanny]

So of the energy supplied by a battery, half is stored in the capacitor and half is lost as heat through resistors?

Correct! This is always the case, irrespective of the values of R or C.

 

[t_n_p]

Correct! This is always the case, irrespective of the values of R or C.

Thanks, that's all I was after!