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.

Theory

A capacitor stores charge. The charge \(Q\) on its plates is proportional to the voltage \(U_C\) between them, \(Q = C \cdot U_C\), where \(C\) is the capacitance, measured in farads. The voltage of a capacitor cannot jump: to change \(U_C\), charge has to flow onto the plates or off them, and the resistor \(R\) limits the current that carries it.

In position "1" the source \(U\) charges the capacitor through \(R\). The voltage across the resistor is \(U - U_C\), so the current is \(I = (U - U_C) / R\): the closer \(U_C\) comes to \(U\), the smaller the current, and the slower the capacitor charges. A quantity that changes at a rate proportional to the difference still left follows an exponential: \(U_C = U \cdot (1 - e^{-t / RC})\). In position "2" the source is out of the loop, and the capacitor discharges through the same resistor, starting from the voltage \(U_0\) it had at the moment of switching: \(U_C = U_0 \cdot e^{-t / RC}\).

The product \(\tau = R \cdot C\) is the time constant of the circuit: ohms times farads give seconds. In one \(\tau\) a charging capacitor covers \(1 - 1/e\), about 63%, of the way to \(U\), and the voltage of a discharging one falls to \(1/e\), about 37%, of \(U_0\). Every further \(\tau\) shrinks the difference still left by the same factor: after \(3\tau\) the change is about 95% complete, after \(5\tau\) about 99%, which is usually taken as fully charged or fully discharged.

Taking the logarithm of the discharge equation gives \(\ln U_C = \ln U_0 - t / RC\): the graph of \(\ln U_C\) against \(t\) is a straight line with a slope \(k = -1/RC\), so \(RC = -1/k\). The slope uses every reading of the discharge rather than one point of the curve, and it does not depend on the starting voltage. With \(R\) known, the time constant also gives the capacitance: \(C = \tau / R\).

The same curves, a thousand times faster, are seen on a screen in the lab RC circuit on the oscilloscope. A camera flash charges its capacitor over a few seconds and empties it through the flash tube in about a millisecond; timers, delays and the smoothing capacitor of a rectifier use the same charge and discharge. Where the energy of the source goes while the capacitor charges is the subject of the lab Energy when charging a capacitor.

The bench

The diagram in Theory keeps the left-to-right order of the bench; only the source is turned upright. A 9 V source \(U\), a two-way switch \(S\), a 10 kΩ resistor \(R\) and a 1000 µF capacitor \(C\) (\(\tau\) = 10 s), and a voltmeter across the capacitor. Terminal "1" of the switch is connected to the positive terminal of the source, terminal "2" to the negative terminal, and the common terminal to the resistor. \(C\) stands between two buses, one on each side of it, and the voltmeter is connected to them. The switch is in position "2", and the capacitor is discharged.

A tap on the lever moves it one step, from "2" to Open, where it touches neither terminal, and on to "1": to flip the switch, drag its lever all the way across or tap it twice. The simulation clock on the toolbar counts from the opening of the lab: write down its reading at the moment you flip the switch and count \(t\) from there. A capacitor taken from the element panel comes at 100 µF.

Procedure

1. Set the speed of the simulation clock on the toolbar to ×0.2: a minute of the experiment then takes five.
2. Flip the switch to position "1" and record the voltage every 5 s for one minute.
3. Flip the switch to position "2" and record the voltage during the discharge in the same way.
4. Plot \(U(t)\) for the charge and for the discharge. From the charging curve read \(\tau\) as the time the voltage takes to reach 5.7 V (63% of 9 V). Plot \(\ln U(t)\) for the discharge, with the readings in volts, draw the best straight line and find \(RC = -1/k\) from its slope \(k\). Compare both values with \(R \cdot C\).
5. Place a second capacitor above the first one, set it to 1000 µF and connect it with cables to the buses on both sides of the first one — in parallel.
6. Flip the switch to position "1" again and find the time it takes the voltage to reach 5.7 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), after 30 s about 8.5 V (95%), 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, so both methods give \(RC\) ≈ 10 s, equal to 10 kΩ · 1000 µF. 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. With a 20 kΩ resistor, when would the voltage reach 5.7 V, and what would the slope of the \(\ln U(t)\) graph of the discharge be?
2. From your charging readings, calculate the current through the resistor at \(t\) = 0 and at \(t = \tau\). Then draw the tangent to the charging curve at \(t\) = 0: at what time does it reach 9 V?
3. During discharge, does the current through the resistor flow the same way as during charging? Trace on the diagram where the charge flows 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 slope of the \(\ln U(t)\) graph halves, to −0.05 s⁻¹.
2. At \(t\) = 0 the capacitor is empty, the whole 9 V is across the resistor, and \(I\) = 9 V / 10 kΩ = 0.9 mA. At \(t = \tau\) the capacitor has about 5.7 V, the resistor 9 V − 5.7 V = 3.3 V, and \(I\) ≈ 0.33 mA. At the first moment the capacitor charges at \(I / C\) = 0.9 mA / 1000 µF = 0.9 V/s; at that rate it would reach 9 V in 10 s, so the tangent crosses 9 V at \(t = RC\).
3. No, it flows the opposite way. During charging, charge flows from the positive terminal of the source through terminal "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 terminal "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.
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.