AP Physics teacher here, preparing to teach my students about capacitors. We have some basic breadboards, capacitors, battery packs, and DC power supplies. My plan was to have them put three different capacitors in series and parallel and measure the voltage on each to figure out how these arrangements affect total capacitance.

My understanding of capacitors in series is that they have to have the same amount of charge on each plate. Using V = Q/C, this indicates that the smallest capacitors should have the highest voltage. But, whether I use a battery or DC regulated power supply, I keep seeing that the biggest capacitors use the most voltage. Why is this? Am I measuring it wrong, or am I misunderstanding the calculations?

  • $\begingroup$ Can you show actual readings for your set-up? Do you get constant voltage, or does it change with time? $\endgroup$ – Brick Oct 2 '18 at 0:34
  • $\begingroup$ @Brick Caps are about 100, 10, and .05 microfarads. The Voltage in series is about 5.3, .5, and. 05 respectively. But all of those voltage readings drop quickly, which I imagine is discharging through the voltmeter. If there is a different way to measure voltage, let me know. $\endgroup$ – Wesleyrm76 Oct 2 '18 at 1:12

All voltmeters have an internal resistance that is designed to be very high (like 100Mohm) so my guess is you end up with a voltage that is a comparison of the high dc resistance of the cap compared to the high dc resistance of the voltmeter. Maybe the meter shorts out the cap, you could test this by charging a cap to say 10v and then attaching it only to the voltmeter, it should say 10 but the charge may dissipate quickly. That's my first educated guess.


Am I measuring it wrong, or am I misunderstanding the calculations?

The answer by PhysicsDave gets close to what I believe is at the root of your measurements.

According to theory, an ideal voltmeter (with infinite input impedance) would measure the following voltages across the capacitors:

$$V_{.05} = V\frac{C_{eq}}{0.05\,\mathrm{\mu F}}$$

$$V_{10} = V\frac{C_{eq}}{10\,\mathrm{\mu F}}$$

$$V_{100} = V\frac{C_{eq}}{100\,\mathrm{\mu F}}$$


$$C_{eq} \equiv \frac{1}{\frac{1}{100} + \frac{1}{10} + \frac{1}{0.05}}\,\mathrm{\mu F} $$

However, a physical voltmeter has finite input resistance and so, there is a time constant associated with each capacitor. If the time constant is relatively short, you may significantly discharge the capacitor before the voltmeter reading stabilizes enough to record a value.

You can estimate the time constant for measuring each capacitor voltage as follows:

Assume, for example, an input impedance of $1\,\mathrm{M\Omega}$ and find that the time constant associated with the smallest capacitance is just

$$\tau_{.05} = 1\,\mathrm{M\Omega}\cdot 0.05\,\mathrm{\mu F} = 50 \,\mathrm{ms}$$

while the time constant for the largest capacitance is

$$\tau_{100} = 1\,\mathrm{M\Omega}\cdot 100\,\mathrm{\mu F} = 100 \,\mathrm{s}$$

So, it makes sense that you're measuring the largest voltage across the largest capacitance because you can take your time with the measurement and find that the voltage across the capacitor is falling very slowly as you watch the voltmeter reading.

But, for the smallest capacitance, you've essentially discharged the capacitor in about a 1/4 of a second so it's unlikely that you can get an accurate measurement of the voltage unless your meter has a very quick "peak hold" function.

  • $\begingroup$ So, is there an alternate way of measuring the voltage or charge on a capacitor? $\endgroup$ – Wesleyrm76 Oct 2 '18 at 2:29
  • $\begingroup$ Some meters are very high in resistance, you might get a reading for the bigger caps that lasts a while, try them in and taken out of the circuit. You could try AC voltage off of a transformer (wired the correct way), now it's called impedance at it varies with 1 over (frequency x capacitance) with an ac meter, its getting complicated. $\endgroup$ – PhysicsDave Oct 2 '18 at 3:07

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