This is a theoretical question. Let's say that we have two black boxes, each with two terminals. One contain an infinity (or arbitrarily) large capacitor in series with a 1 ohm resistor and the other contains just a 1 ohm resistor.

Is there a way to tell which is which using finite finite resources (e.g. upper bound on experiment time, upper bound on voltage sources, upper bound on equipment accuracy, etc)?

My thinking is that since the capacitor is arbitrarily large, it will not get charge to a noticeable voltage in a given finite time and thus will be identical to a short (zero voltage regardless of finite current).

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    \$\begingroup\$ No it isn't, a capacitor blocks DC current. \$\endgroup\$
    – lucas92
    Jan 30, 2017 at 20:37
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    \$\begingroup\$ If you're happy with an infinite capacitor, then it's probably also sensible to talk about infinite time, or infinite current. Either of those would allow you to tell the difference. \$\endgroup\$
    – Jack B
    Jan 30, 2017 at 20:42
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    \$\begingroup\$ @JackB - He does have the series 1 ohm resistor in the RC box. You would never see infinite current with that unless you also started to consider infinite applied voltage. \$\endgroup\$ Jan 30, 2017 at 20:50
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    \$\begingroup\$ @MichaelKaras I would say that infinite current is what tells you the difference, and infinite voltage is just a side effect of infinite current through a resistor. One would also need infinite charge, infinite energy, and (to avoid hadronisation) infinite volume. And probably others too. That's the problem with infinities, once you start you can't stop until everything is infinite or zero. \$\endgroup\$
    – Jack B
    Jan 30, 2017 at 21:06
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    \$\begingroup\$ One difference is that, after a period of charging, the box containing the capacitor could be discharged. The other box couldn't. \$\endgroup\$ Jan 31, 2017 at 5:11

7 Answers 7


Yes, your analysis is correct for a infinite capacitor.

However, anything less than that can be detected in arbitrarily short time. The problem is that the size of the signal to notice the difference gets smaller as the time to run the experiment gets smaller. Larger current makes the effect larger in the same amount of time.

Let's say your current is limited to 1 A and you have a 12 bit A/D in a 3.3 V microcontroller. Let's see how large a capacitor this could detect. The voltage change of a cap as a result of some Amps for some seconds is:

  V = A s / F

Where A is the current in Amps, s is the time the current is applied in seconds, and F is the capacitance in Farads. Flippping this around to solve for the capacitance yields:

  F = A s / V

The minimum voltage change we can detect is (3.3 V)/4095 = 806 µV. Plugging in our particulars, we get:

  F = A s / V = (1 A)(1 s)/(806 µV) = 1.2 kF

That's a very large capacitor. If you can supply 5 A and wait 2 seconds, then you can detect a 10x larger capacitor. Or conversely, be able to measure 1.2 kF to 1 part in 10.

Yet another way to look at this is to apply a constant voltage for a fixed time, then see how much the open-circuit voltage went up afterwards. The voltage on the capacitor will rise exponentially, asymptotically approaching the fixed voltage being applied. Again let's say we can measure down to 1 part in 4095 of the applied voltage. That comes out to 0.000244 time constants. If that's how long 1 second is, then the time constant must be 4096 seconds. With a 1 Ω resistor, that means the cap is 4.1 kF.

Note that cheap $20 voltmeters can measure much smaller voltages than a 12 bit A/D running from 3.3 V.

Basically, it takes a unrealistically large capacitor to not be detectable via rather simple means.

  • 1
    \$\begingroup\$ Ever play with an ultracap? 2.7kF, max 3v, 500mA input (or you exceed a breakdown threshold). \$\endgroup\$
    – Joshua
    Jan 31, 2017 at 3:39
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    \$\begingroup\$ @Joshua This is easy to detect by exceeding the max current. \$\endgroup\$ Jan 31, 2017 at 9:24
  • \$\begingroup\$ @DmitryGrigoryev: Yeah, if you want to burn up your circuit. \$\endgroup\$
    – Joshua
    Jan 31, 2017 at 16:44
  • \$\begingroup\$ @Joshua Nobody told the test for an infinite cap had to be non-destructive. \$\endgroup\$ Jan 31, 2017 at 19:32

Yes, not only can you say that an infinite capacitor acts like a short, but a wire IS an infinite capacitor.

First, recall the formula for a parallel-plate capacitor:

$$C=\frac{\epsilon A}{d}$$

Grab your nearest monomolecular blade (1 Å) and stretch it a bit so it's even thinner. Now, cut your solid wire into two pieces, creating parallel surfaces separated by the thickness of your wire, so the parallel plate capacitor formula applies.

Observe that as the thickness of the blade approaches zero, eventually it passes between the atoms of the wire without interacting with any. So the continuous wire before and the capacitor after are physically indistinguishable.

Also $$\lim_{d \rightarrow 0} C \rightarrow \infty$$

  • 1
    \$\begingroup\$ But isn’t there a requirement for the two surfaces to be isolated from each other (to some degree and voltage)? \$\endgroup\$
    – Michael
    Jan 31, 2017 at 16:46
  • \$\begingroup\$ @Michael: They are isolated. By separation \$d\$. Which is smaller than microscopically small. \$\endgroup\$
    – Ben Voigt
    Jan 31, 2017 at 17:06
  • \$\begingroup\$ But what's the breakdown voltage? \$\endgroup\$
    – Joshua
    Jan 31, 2017 at 20:16
  • \$\begingroup\$ @Joshua: Zero, same as it is for any infinite capacitor. \$\endgroup\$
    – Ben Voigt
    Jan 31, 2017 at 20:58

In your theoretical world of infinite capacitance, I suppose the step input would be indistinguishable between the R and the series RC. That is only the case because of the infinite capacitance: anything less would gain charge slowly and start to block the DC.

This happens since when solving for the initial current through a capacitor you ignore the capacitor and I = V/R. Since the capacitor would never charge up at all, being infinitely large, this would remain the current.


You are correct that for an arbitrarily large capacitor (approaching infinite farads of capacitance) and for a finite limit on testing strategy you would not be able to tell the differences between the two black boxes.

If you begin to bring the capacitor value back into realistic and implementable sizes (so you can at least carry these black boxes around) and remove any restrictions on your test time and equipment accuracy you would begin to see the current into the series RC box begin to diminish as the capacitor starts to charge.


If you are dealing with DC, a capacitor is always a open circuit.
If you are on transient domain (ie: calculating the circuit reaction to a key switching), the capacitor is an short until it is fully loaded. Then it will work as an open circuit like the DC model.
If you are dealing with AC, a very large capacitor (a capacitor with theoretical infinite capacitance) is an short circuit.
So, answering your question, Feed the black boxes with a 1V ideal DC Voltage source and wait.
After a very long time (infinite time), the box with the resistor alone will be draining 1A while the one with the capacitor in series will be draining no current.

  • \$\begingroup\$ Note that for a sufficiently large capacitor and a sufficiently short timescale, a capacitor behaves indistinguishably from a wire under both AC and DC. \$\endgroup\$
    – Mark
    Jan 31, 2017 at 1:28
  • \$\begingroup\$ @Mark The notion of timescale makes no sense in DC analysis. \$\endgroup\$ Jan 31, 2017 at 9:30
  • \$\begingroup\$ @Mark, using a hydraulic metaphor, a very large pool is undistinguishable from a sink until it is full. \$\endgroup\$ Jan 31, 2017 at 13:49
  • \$\begingroup\$ The hydraulic metaphor is interesting. Do capacitors become heavier when the charge? Do they displace more air? Any other 'physical' affect? \$\endgroup\$ Feb 2, 2017 at 16:05
  • \$\begingroup\$ @user1139880 I don't think capacitors would get heavier. Despite the fact that electrons have mass, one plate will be holding positive charges (electron absences) while the other will be holding negative charges (electron surpluses), thou I clearly remember calculating the Earth's capacitance in a Physics class assuming a second plate at infinity... \$\endgroup\$ Feb 2, 2017 at 16:21

The way to tell is to apply your voltage for a small given time, then disconnect the voltage source and short the terminals through a current measuring device. Since the capacitor stores energy and the wire doesn't the one that indicates a current when shorted is the capacitor.

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That and the whole... "I can't lift the infinite capacitor box" thing.

BTW: All capacitors present themselves as almost a short when power is first applied to them. This can introduce a troublesome inrush current when powering up a circuit-board or system if there is a large capacitor farm. This is one of the reasons you often see a small inductor added to the in-line before the capacitors.


It depends... Are we talking AC or DC? In AC a cap is a timed short. In DC it is an open. Now, if you had to identify what might be in the black box just hook up a VOM in Ohms mode across the pins. The one with resistance ONLY will read the resistance immediately and it will read the same value in either direction the leads are hooked up. Now the other black box will show a slow change in the resistance - Especially, if you use an analog meter and when you quickly swap the leads you will see a rapid discharge followed by a slow increase in resistance. This was once called capacitive action. TC = RxC and a full charge should be 5 x TC.

  • \$\begingroup\$ He kind of mentions infinitely large cap. Maybe you are familiar with how an Ohm meter or capacitive sensor work but both send a small test current and measure the delta-V \$\endgroup\$
    – jbord39
    Jan 30, 2017 at 21:48
  • \$\begingroup\$ Exactly jBord... That is what is charging the Cap. So what is your point? \$\endgroup\$
    – Barry
    Jan 31, 2017 at 17:32
  • \$\begingroup\$ so an infinitely large cap will never charge up or down from a small trickle current and the delta-v will be unmeasurable ... the TC is infinity \$\endgroup\$
    – jbord39
    Jan 31, 2017 at 17:38

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