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Charging and discharging a capacitor with a voltmeter

Curricula: A-level RP9, Tle (dipôle RC), AP2, APC, Sek II

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Aim

Record the charging and discharging curves of a capacitor, and determine the RC time constant and the capacitance of two capacitors connected in parallel.

The bench

A 9 V source, a two-way switch, a 10 kΩ resistor and a 1000 µF capacitor (\(\tau\) = 10 s), and a voltmeter across the capacitor. Jaw "1" of the switch is connected to the positive terminal of the source, jaw "2" to the negative terminal, and the common terminal to the resistor. In position "1" the capacitor charges through the resistor, in position "2" it discharges through the same resistor. The blade is in jaw "2", and the capacitor is discharged.

Procedure

  1. Throw the blade into jaw "1": drag the handle or select position "1" in the settings panel. Using the simulation clock in the top panel, record the voltage every 5 s for one minute.
  2. Throw the blade into jaw "2" and record the voltage during the discharge in the same way.
  3. Plot \(U(t)\) and \(\ln U(t)\) for the discharge; find \(RC\) from the slope.
  4. Drag a second capacitor from the elements panel, place it above the first one and set it to 1000 µF in the settings panel. Connect it with cables to the same two buses as the first one — in parallel.
  5. Throw the blade into jaw "1" again and use the clock to find the time it takes the voltage to reach 5.7 V (63% of 9 V). Find the capacitance of the pair from \(\tau = R \cdot C\).
Expected result

Charging: after 10 s (\(\tau\)) the voltage reaches about 5.7 V (63% of 9 V), and after 50 s almost 9 V. Discharging: in 10 s the voltage falls to 37% of its initial value — from 9 V to 3.3 V. The \(\ln U(t)\) graph for the discharge is a straight line with a slope of \(-1/RC\) ≈ −0.1 s⁻¹. With two capacitors in parallel, 5.7 V is reached in 20 s: \(\tau\) is twice as large, and the capacitance of the pair is 2000 µF — capacitances add when connected in parallel.

Questions

  1. How will the charging time change if a 20 kΩ resistor is used?
  2. Why is the charging current greatest at the first moment?
  3. Why does the current through the resistor flow in the opposite direction during discharge? Answer using the circuit: where does the charge flow from and to?
  4. What would \(\tau\) be if two such capacitors were connected in series? Why does the capacitance decrease in that case?
Answers
  1. \(\tau = R \cdot C\) doubles, to 20 kΩ · 1000 µF = 20 s: the voltage reaches 5.7 V after 20 s instead of 10 s, and the capacitor is almost fully charged after about 100 s instead of 50 s.
  2. At the first moment the capacitor is uncharged and has no voltage across it, so the whole 9 V is across the resistor and the current is 9 V / 10 kΩ = 0.9 mA. As the capacitor charges, its voltage opposes the source, the voltage across the resistor falls, and the current falls with it.
  3. During charging, charge flows from the positive terminal of the source through jaw "1", the common terminal and the resistor onto the capacitor plate that faces the resistor, and this plate becomes positive. In position "2" the source is out of the loop: the resistor, the common terminal and jaw "2" connect that plate to the negative terminal, where the other plate is also connected, so the charge flows back from the positive plate through the resistor towards the switch, opposite to the charging current.
  4. In series \(1 / C = 1 / C_1 + 1 / C_2\), so two 1000 µF capacitors give 500 µF, and \(\tau\) = 10 kΩ · 500 µF = 5 s. Both capacitors carry the same charge, and the source voltage is split between them, so a given charge needs twice the voltage, and \(C = Q / U\) is halved.