# Discharging capacitor in RC circuit

In lab, we built a circuit with a 4.7 micro-F capacitor hooked up to a DC source. In parallel to the capacitor was a 10megaOhm resistance voltmeter. We used the voltmeter to find the voltage drop across the resistor and find the RC constant (supposing unknown C). We would turn on the battery, charge up the capacitor, then turn off the battery and take watch/record the values in the voltmeter.
My question has to do with the following: Say we started at 1V and read measurements at 0.9, 0.8, 0.7 V etc, as the voltage decreased. Well, when the capacitor was charged at 1V, when we turned the battery off, the voltage would decrease, then start increasing back to 1, and then start steadily decreasing. I had not expected that increase in the voltage. I though it would decrease smoothly from 1 to 0.9 etc. Nobody could explain the temporary increase to me. Can somebody help me explain why the voltage increased and then decreased? Thanks.

Were you measuring the voltage with an analogue voltmeter?

Added: I think the type of voltmeter doesn't matter!

My idea:

Before cutting off the battery, you had a current in the battery-resistor loop. By cutting this current, you had a sudden change in the flux, which induces an emf in the voltmeter-resistor loop. The voltmeter shows this emf added to the capacitor voltage.
By adding a small capacitor parallel with the switch, this induced emf would be negligible.

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Philip Wood
Gold Member
Hassan2. Ingenious, though I'm surprised the measured voltage fell before the effects of the induced emf kicked in.

Couldn't reproduce the effect with a 50 mu F electrolytic capacitor and a 10 M ohm digital voltmeter. Voltage just fell steadily. And also when I connected the capacitor the wrong way round!

Would be interesting to know if the meter used originally was digital or analogue, whether the initial fall in voltage was very quick, and to what value it fell.

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The voltmeter was a digital multimeter. The initial fall in the voltage was very quick. No matter what voltage we started at, either 1V or 20V, it seemed to drop down to close to zero almost immediately, then climb back up to about 1V in about 10seconds, then begin it's steady decline back down.

Philip Wood
Gold Member
Sorry to be a bore, but I wonder exactly how you 'turned off the battery'. Did you physically disconnect at least one of the wires going to it?

It's an interesting phenomenon, and I wish I could reproduce it.

No, the battery has an on/off switch. It was not just my group where this was happening. It happened to all the groups. We tried first with a 2.2 micro-F capacitor, and since it discharged too fast, we switched to a 4.7 micro-F capacitor. I don't know if that affects anything.

Philip Wood
Gold Member
Fascinating! My hunch is that there's something odd about the switches in the battery units. Maybe they do more than just connect/disconnect one of the wires. Any chance of your seeing inside the unit? Or – easier – seeing what happens if you use an ordinary battery and connect it across the capacitor with a pair of wires, and then disconnect it by disconnecting one of the wires.

BTW Don't forget that your own resistance, say between one hand and the other, is likely to be less than 10 M ohm. If you touch two connections in the circuit with both hands or the fingers of one hand, you may well disturb the circuit.

Philip Wood,

According to my little experience, digital meters often mix up in sharp changes of voltage and takes a while to become steady. So, what we read may not be the true voltages. But why they behave like that, I it's about their input capacitor. Again , the only way for the meter to feel the sharp change is the flux linkage.

cosmogrl,

Sorry for bring up the topic again. Was that a battery or a power supply? Power supplies have internal capacitor and also output resistance.
If it was a power supply, then the phenomenon can be explained easily.

It was a power supply.

Philip Wood
Gold Member
See what happens if, instead of using the switch on the power supply, you manually disconnect a wire (or both wires) going from the power supply to the capacitor.

The figure shows a typical power supply circuit. Look at R2 and R3 of the circuit. The series of R2 and R3 is parallel with the 4.7 uF capacitor.

When you turn of the switch ( see the location of the switch in the figure), the 4.7uF and the output 10uF capacitors discharges through R1+R2=20K. The time constant is about (10+4.7)*.02=.294 seconds. That's why the voltage drops to zero almost immediately.

Then, the remaining voltage of the internal capacitor C1=2.2mF=2200 uF starts charging the 4.7 uF transistor through all possible resistors of the circuit. Then both capacitors discharge gradually.

The right way to measure the time constant is to disconnect the output. I think the time constant you measured is not the time constant of your parallel RC circuit.

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there was no external resistor. The voltmeter itself (the multimeter) has a resistance of 10Mohms. But, what do you mean disconnect the output? Do you mean remove one of the wires leading into/out of the power supply?

In your experiment, the switch must isolate your capacitor from the power supply. The on-off switch does not do so.
In the simplest way, just disconnect one of the wires which connect the power supply to the capacitor. This isolates the circuit from the power supply.