The capacitor as an energy store

Curricula: AP2, A-level, Tle, Sek I/II

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Aim

Verify that a charged capacitor stores energy and can power a lamp.

The bench

A 6 V source, a switch (closed), a 10 Ω resistor and a 0.47 F capacitor in series (a capacitance like this is a supercapacitor: an assembly of cells with low internal resistance, rated 6 V and above); in parallel with the capacitor — a 6 V / 0.5 W lamp and a voltmeter. When the bench is opened, the capacitor is discharged and starts charging.

Procedure

1. Wait about 20 s until the voltmeter reading stops rising, and write it down.
2. Open the switch and watch the lamp and the voltmeter.
3. Calculate the stored energy \(W = C \cdot U^2 / 2\) from the voltage in step 1.

Expected result

After the bench is opened, the lamp brightens as the capacitor charges; after 15–20 s the voltage across it settles at about 5.2 V — part of the voltage drops across the 10 Ω resistor. After the switch is opened, the lamp does not go out at once but gradually dims over about half a minute: it is powered by the capacitor, and the voltage across it falls. The stored energy is about 0.47 · 5.2² / 2 ≈ 6.3 J.

Questions

1. Where does the energy come from when the switch is open?
2. How will the glow time change if you use a capacitor with twice the capacitance?

Answers

1. From the electric field of the charged capacitor. While the switch was closed, the source stored about 6.3 J in it; after the switch is opened, the capacitor discharges through the lamp, and its voltage falls as that energy turns into heat and light in the filament.
2. The glow time doubles, to about a minute. At the same 5.2 V the capacitor stores twice the energy, about 12.7 J, and the lamp draws the same current at each voltage, so the voltage falls half as fast; charging in step 1 also takes twice as long.