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This is for a science project. I have built my own electrolytic capacitor with aluminum foil and paper dielectric soaked in salt water solution. My parallel plate capacitor is simply aluminum foil with a paper dielectric.

While they both have voltage when measured with a multimeter, only the electrolytic capacitor has current. Why might only the electrolytic capacitor be measuring current?

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    \$\begingroup\$ Electrolysis... \$\endgroup\$
    – user16324
    Commented Sep 9, 2020 at 17:31
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    \$\begingroup\$ @gstudent that's because it works. \$\endgroup\$
    – Marcus Müller
    Commented Sep 9, 2020 at 17:33
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    \$\begingroup\$ What do you mean by current? 1) Only the electrolytic capacitor passes current through it continuously or that, 2) once charged to the same voltage, only the electrolytic capacitor provides current (while discharging) ? \$\endgroup\$
    – AJN
    Commented Sep 9, 2020 at 17:35
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    \$\begingroup\$ To specify the question: what is your precise measurement setup? Draw a schematic! Nothing less will do. \$\endgroup\$
    – Marcus Müller
    Commented Sep 9, 2020 at 17:36
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    \$\begingroup\$ @gstudent how are measuring the current? A plate capacitor, as Spehro says, should have low capacitance and hence expectedly will offer less current, and also, both should be discharged quicker than you can observe if you're shorting them with an amperemeter. This question is really lacking a precise measurement description, and if I know anything about science projects, is that they expect you to write down exactly what you observed by doing what. \$\endgroup\$
    – Marcus Müller
    Commented Sep 9, 2020 at 17:44

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With most off the shelf meters, the meter won't detect much current because the capacitor essentially gets shorted when the meter is placed across it.

With a capacitor the exponential timeconstant (~60% of the initial voltage) would be

$$ \tau = RC$$

With a meter in current mode most likely having a resistance of lower than 0.1Ω

$$ \tau = (0.1Ω)(1000uF)= 100us$$

So this means in about 100us most of the voltage fades away

enter image description here
Source: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/imgele/capdis.gif

The current could also be calculated with the equation above, but you must know how much voltage \$V_0\$ you charged the capacitor to.

The current at 100us and 5V would be less than 19mA, in another few hundred us (like 1000us) the current from a 1000uF cap would be less than 1mA which would be hard to see for many meters because of the short duration of the current.

The capacitor you constructed probably has a smaller value of capacitance (maybe lower than 1uf) so the time would be even shorter than 100us.

So, one way to overcome this would be to put a resistor in series with the capacitor to make the time constant longer and/or use a really good current meter that can measure. Another good capability to have would be graphing. Don't get me started on fitting exponential curves (big hint).

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  • \$\begingroup\$ EDITED: I'm wrong :-). He is talking about discharge current, as you say: |||| I'd assume he was talking about leakage current, which for an ecap will be much higher than for a "polymer" cap. If he uses eg an Ohm-meter on a high range he'll see finite resistance with an ecap whereas with the paper dielectric, if dry, the effective resistance may well be above his max Ohms measurement. \$\endgroup\$
    – Russell McMahon ♦
    Commented Sep 10, 2020 at 4:25

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