Charging and discharging a capacitor

What you are trying to do

Follow the pd across a discharging capacitor with a voltmeter and stopclock, and use a log-linear plot to find the time constant RC, and from it the capacitance.

Apparatus

• A large-value capacitor, for example 470 μF (note its polarity if it is electrolytic)
• A resistor chosen to make RC tens of seconds, for example 100 kΩ
• A 6 V battery pack and a two-way switch
• A digital voltmeter of high input resistance, connected across the capacitor
• A stopclock, or a data logger with a voltage sensor in place of both meters

Variables

• Independent: time since the discharge began
• Dependent: the pd across the capacitor
• Control: the resistance, the capacitance and the starting pd — the same components and the same fully charged start each run

Method

1. Choose components so the time constant suits a stopclock: 100 kΩ with 470 μF gives RC of 47 s, slow enough to read by eye. A small RC would be over before your first reading.
2. Charge the capacitor by flicking the switch to the battery side; the pd rises to the full 6.0 V in well under a second because the charging path has almost no resistance.
3. Flick the switch to the discharge side and start the clock at the same instant; this synchronisation is the single biggest source of timing error, so practise the flick-and-start before recording.
4. Read the pd at fixed intervals, every ten seconds here, until it has fallen to well under half its starting value. Repeat the whole run at least twice and average the readings at each time.

Analysis

1. The model to test is exponential decay, V = V₀e^-t/RC. Taking logarithms straightens it: ln V = ln V₀ − t/RC, which is a straight line in t.

1. Plot ln V against t. A straight line confirms the exponential; a curve would refute it. The gradient equals −1/RC, so RC is minus one over the gradient.
2. Divide RC by the known resistance to obtain the capacitance, and compare with the value printed on the component.

A worked set of readings

One discharge run with the circuit above, read every ten seconds:

The ln V column falls by the same amount every ten seconds, which is the exponential showing itself. The gradient is (0.515 − 1.792)/60 = −0.0213 s⁻¹, so RC = 1/0.0213 = 47 s. With R = 100 kΩ the capacitance is C = 47/1.00 × 10⁵ = 4.70 × 10⁻⁴ F, which is 470 μF: the label value, recovered from a stopclock and a voltmeter.

Where the uncertainty comes from

• Starting the clock: Human reaction is about 0.2 s, which is why RC is chosen at tens of seconds: 0.2 s in 47 s is under half a per cent.
• The voltmeter itself: The meter is a resistance in parallel with the capacitor and discharges it too. A digital meter's input resistance of order 10 MΩ beside 100 kΩ changes the effective R by about one per cent; an old analogue meter would wreck the experiment.
• Component tolerance: Electrolytic capacitors are commonly quoted at twenty per cent tolerance. Your measured RC is likely more accurate than the label; that is a feature, not an error.
• The late tail: Once V falls to a few tenths of a volt, the meter's last digit is a big fraction of the reading. Stop before the tail, or weight the early points; the log plot makes the scatter at the bottom end visible.

What earns the marks

Safety