In a "Snap Circuits" project ("Leaky Capacitor"), the instructions have me put a 470 uF polarized capacitor in backwards with the negative side towards the batteries. This is to demonstrate that the capacitor will leak current when installed backwards. (The green LED stays dimly lit after the capacitor is fully charged.) Everything I read on-line says this will damage the capacitor and that it might explode. Is this experiment really dangerous to the capacitor or to the experimenter? Thanks!


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Don't feel pushed into doing this experiment - there is nothing useful to be gained here as far as I can tell. Stay safe. \$\endgroup\$
    – Andy aka
    Feb 26, 2022 at 15:47
  • \$\begingroup\$ 6 V is not enough to give you a shock or electrocute you. If the capacitor "explodes" it will not be very forceful. Just wear glasses so no flying pieces hit you in the eye. But I also tend to agree with Andy that it is not a very interesting experiment. You could just take note of the fact that electrolytic caps should not be hooked up backwards and move on to the next experiment. \$\endgroup\$
    – user57037
    Feb 26, 2022 at 17:24
  • \$\begingroup\$ negative side towards the batteries is not necessarily backward \$\endgroup\$
    – jsotola
    Feb 26, 2022 at 18:58

2 Answers 2


In that circuit the current through the capacitor will be limited by the diode and the \$100\Omega\$ resistor. A typical aluminum electrolytic capacitor shouldn't explode in that setup, and I suspect you'd have to leave the circuit on for days to damage the cap.

When electrolytic capacitors explode it's because they're connected to a power supply that's capable of supplying a lot of current. The reverse current flow heats things up, and because the electrolyte has water in it, there's a steam explosion.

I don't go around burning up caps on purpose so I don't have much experience, but if your 6V battery supply used D cells and your \$470\mu\mathrm F\$ cap were of appropriately low voltage (i.e. 10 or 16V) then you might be able to explode it -- or it may just go "fizz", with or without a puff of smoke, and stop being a capacitor.


The probability of negative events is highly accelerated if the reverse voltage is >= 10% of rated voltage and the severity depends on the both the max power limited series R and max energy that can be stored in cap, triggered by an abrupt short from breakdown. It also depends if the Capacitor is degraded or new and has low ESR.

Since one green Vf = 3 V offsets one 3 V battery, the comparison is then second 3 V battery relative to 10% of the Capacitor rating.

If Vcap rating >= 30 V, it is more likely to produce Green light with Ir > 3% of CV = 3% of 470 x 30 V ~ 0.4 mA

  • computing a max power transfer limit of 3V^2/100ohm= 90 mW is unlikely to cause much heat, but as LED gets brighter with a lower rated voltage then there is a strong likelihood of thermal avalanche breakdown and state change of abrupt toxic outgassing.
  • but if a breakdown occurs with a low ESR cap rated near ESR*C<=10us or ESR<=10/470 = 20 mohms and now with up to 3V charge squared over 20 mohms, Pd= 3^2/0.02 = 450 watts of discharge caused by a voltage breakdown and short circuit in the insulation. poof****

Is this experiment really dangerous to the capacitor or to the experimenter? Thanks!

If the test kit supplies the capacitor, then I believe it would have been selected to be safe with -3V and this cct. being both low voltage (-3V) and low current (30 mA max) the risk of damage to the capacitor is low. The LED brightness changes abruptly on connection to indicate the charging. Reversing the direction of the capacitor with a charged potential of 3V will give even a brighter initial burst but only last for T=RC= 100*470u = 4.7 ms which is just a flash.

Using the resistance mode of a DMM can perform the same test with low voltage but using a much higher series resistance and much lower limited current, so the rate of change of values will be an indication of capacitance. If the leakage resistance is < 1 Mohm at 1V then it may settle on a steady value but leakage current is normally done at rated current.

The purpose of the test is to demonstrate visually that all electrolytic capacitor leak with voltage applied proportional to their capacitance and significantly more in reverse voltage but safely at <=10% rated.


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