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Energy when charging a capacitor

Curricula: A-level, AP2, APC, Tle, Sek II

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Aim

Measure how much energy the source delivers when a capacitor is charged and how much of it the capacitor stores.

The bench

A 6 V source, a two-way switch, a 10 Ω resistor and a 0.47 F capacitor (\(\tau\) ≈ 5 s) — a supercapacitor: an assembly of cells with low internal resistance rated for 6 V and above; a single supercapacitor cell is rated for 2.7 or 5.5 V, and 6 V is too much for it. The first wattmeter is at the output of the source, the second on the capacitor. The switch opens in position "2", the capacitor is discharged.

Procedure

  1. Press ▶■ on both wattmeters.
  2. Move the switch to "1" and wait about 30 s, until the power on both instruments has dropped almost to zero. Record the energy \(E\) on both: the energy delivered by the source and the energy stored in the capacitor.
  3. Compare both energies with \(C \cdot U^2\) and \(C \cdot U^2 / 2\).
  4. Move the switch to "2" and watch the second wattmeter.
Expected result

During charging the source delivers about 16.5 J (\(C \cdot U^2\) = 0.47 · 6² ≈ 16.9 J), and the capacitor stores about 8.5 J — half of \(C \cdot U^2\) (\(C \cdot U^2 / 2\) ≈ 8.5 J). By its own wattmeter the source delivers slightly less than \(C \cdot U^2\): the wattmeter is placed after the internal resistance of the source, and the instrument does not see the roughly 0.4 J of heat in it. The other half turns into heat in the resistor, and this share does not depend on its resistance. During discharge the power on the second wattmeter is negative — energy flows from the capacitor into the circuit — and its counter decreases almost to zero: the stored energy goes into heat in the resistor.

Questions

  1. Why is as much energy released in the resistor during charging as the capacitor stores?
  2. Will the share of energy stored by the capacitor change if a 100 Ω resistor is used? What will change?
  3. What does a negative power on a wattmeter mean?
Answers
  1. The source moves the whole charge \(q = C \cdot U\) at its full voltage \(U\) and does the work \(q \cdot U = C \cdot U^2\) ≈ 16.9 J. The capacitor receives this charge while its voltage grows from 0 to \(U\), on average at half the voltage, so it stores only \(C \cdot U^2 / 2\) ≈ 8.5 J. The other half turns into heat, almost all of it in the resistor and about 0.4 J in the internal resistance of the source.
  2. No: the capacitor still stores half, about 8.5 J, because the calculation above does not contain the resistance. Charging becomes about ten times slower, \(\tau = R \cdot C\) = 100 Ω · 0.47 F = 47 s, so it takes several minutes, and the current and the power at the start are about ten times smaller.
  3. Energy flows through the instrument in the opposite direction: the part of the circuit it measures gives energy out. During discharge the capacitor works as a source and its 8.5 J go into heat in the resistor, so the counter runs down. A wattmeter with one of its circuits connected the other way round also shows a negative power.