# A question about the energy stored in a capacitor.

1. May 19, 2013

### NoddingDog

If we imagine a simple circuit with a battery and a capacitor with negligible internal resistance, the capacitor is charged up to a point where the voltage it stores is equal to the voltage of the battery. The battery has supplied an energy equal to V*I*T (voltage times current times time). However the capacitor now has a stored energy of ½*Q*V. The battery has supplied Q*V joules of energy but the capacitor is storing only ½ of that amount.

How is this possible? I know it can’t be resistance as it always a half difference in energy between the battery and the capacitor and resistance would have a varying affect depending on the circuit. Also the problem still occurs when the circuit has no resistance.

I’m wondering if my A-Level studies don’t explain in enough detail the theory about capacitors and that’s why I’m not understanding this.

Would appreciate any feasible explanations.

2. May 19, 2013

### Staff: Mentor

You are writing the energy supplied by the battery as V*I*T, which is only correct if the current I is constant throughout. It's not; the current flow starts at I when the capacitor is completely uncharged, but falls to zero when the capacitor is completely charged. Assuming an ideal capacitor and constant non-zero resistance in the battery and wire (if you don't make the latter assumption you'll get infinite current and a completely non-physical situation) the current falls linearly from its initial value to zero, with an average value of 1/2 the peak.

So the energy out of the battery is not V*I*T, it's half of that.

3. May 19, 2013

### Popper

You appear to have assumed that I = Q/T and therefore the charge on the capacitor is Q = IT and then assumed that U = QV that U = (IT)V = VIT. Is that correct? If so then you've made an error. I does not equal Q/T, it equals I = dQ/dT. You made the incorrect assumption that the current was a constant while it was charging in the manner in which you described, i.e. connecting the capacitor directly to the voltage. The current is not constant and it is the equation I = dQ/dT which holds and not I = Q/T.

You can eliminate these kinds of errors by paying closer attention to the meaning of the equations and how things are actually done in real life rather than simply juggling variables around.

Sorry Nugatory. Seems that you posted your response while I was writing mine.

4. May 19, 2013

### technician

This is a well known 'paradox' in physics and the explanation has been discussed many times here.
You are correct to recognise the energy from the battery = QV but the energy stored on the capacitor is 0.5QV
The 'missing' 0.5QV is lost in resistance of connecting wires, sparking at the switch and electromagnetic radiation due to the changing current as the capacitor charges.

5. May 19, 2013

### Staff: Mentor

Actually, the energy dissipated in the resistance for a charging series RC circuit is independent of the actual value of the resistance. For a series RC circuit with initial voltage 0 across the capacitor and with a constant supply voltage v_s after t=0 we have the differential equation:
$$C \frac{dv_c}{dt}+\frac{v_c-v_s}{R}=0$$
Which has the solution:
$$v_c=v_s(1-e^{-\frac{t}{RC}})$$

So the energy dissipated in the resistor is given by:
$$\int_0^{\infty} i_r v_r \, dt = \int_0^{\infty} \frac{v_c-v_s}{R} (v_c-v_s) \, dt = C \frac{v_s^2}{2}$$
which is not a function of R, and is also exactly equal to the "missing" energy.

So your assumption that it cannot be the resistance is mistaken. Even in the limit as the resistance goes to 0 the energy dissipated in the resistance is constant. Furthermore, even with superconductive wire, with 0 Ohmic resistance, there is still radiation resistance.

Last edited: May 19, 2013