I am constructing a simple circuit to convert AC power from the wall to a 10V DC power across a load. I am fairly new at this though, so I have a question about how much power a capacitor can handle. I will be using a couple of 100μF capacitors. These are just your basic capacitors, purchased at your local university for college labs.

Using the OrCad tools, I have determined that these capacitors will have about 20 V across them, and will be handling around 10 W. I cannot find datasheets that will tell me the ratings of what they can handle. Does this seem too high? Am I going to blow my capacitors?

  • 1
    \$\begingroup\$ Can you post a circuit of what you're going to build? It's the current through the capacitor that counts and that depends on how the circuit is configured. \$\endgroup\$
    – PeterJ
    Feb 7, 2014 at 3:36
  • \$\begingroup\$ My current does not exceed 25mA. \$\endgroup\$
    – Riley
    Feb 7, 2014 at 3:38

2 Answers 2


The basic formula for a capacitor is: -

Q = \$C\cdot V\$ and this can be differentiated to \$\dfrac{dQ}{dt} = C\cdot\dfrac{dV}{dt}\$

rate of change of charge, \$\dfrac{dQ}{dt}\$, is current therefore: -

I = \$C\cdot\dfrac{dV}{dt}\$

Power\$^1\$ is I (current) x V (volts) therefore power taken by a capacitor is \$C\cdot V\dfrac{dV}{dt}\$.

But V multiplied by its differential will always average to zero therefore a capacitor doesn't get warm theoretically. However, due to losses in the capacitor (seen as either an equivalent series resistor or parallel resistor), it will get warm if pushed hard enough.

This is usually due to dielectric losses or the effective series resistance (ESR) of the capacitor.

If the current is 25mA and the power source is 50 or 60Hz then the capacitor is unlikely to get measurably warm.

\$^1\$ If you integrate this formula with respect to time, power becomes energy and you get the hopefully recognizable formula seen below: -

\$E = \dfrac{C\cdot V^2}{2}\$


A capacitor doesn't dissipate much power as heat. Check your capacitor's ESR and use the P=I^2*R equation to calculate it. As far as I understand you mean 10 W is the voltage and current product. But they are vectors with about 90 degrees between them, so their scalar product is low.


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