I'm using a large capacitor to buffer the load requirement of a solenoid (solar/battery operated setup, with solenoid kicking in a few times a day). Someone mentioned that if I use a large-enough capacitor, I need to put a resistor in series with it to moderate the "capacitive load" (a term I found out later has little to do with this situation).

So - how much current (theoretically) does a 3300uF electrolytic cap draw in a 12V circuit when it's first energized? (I realize the value drops off over time as the cap charges.)

  • \$\begingroup\$ It has 2 components, when initially turned ON, inrush current exists, which depends on ESR of your cap and dV/dT of turn ON. after that transient event, capacitor slowly charges. Charging time constant will be RC, How much series resistor you will kepp based on that it will vary. we can assume 5RC time to completely charge the capacitor. as far as i know, Q=CV, it's only charge that is important, Current varies based on your Series resistor initially, as capacitor approches completely charged state, current slowly decreases, when capacitor completely charges , current will be Zero. Charge mate \$\endgroup\$
    – user19579
    Commented Aug 29, 2013 at 3:23

1 Answer 1


The current when charging a capacitor is not based on voltage (like with a resistive load); instead it's based on the rate of change in voltage over time, or ΔV/Δt (or dV/dt).

The formula for finding the current while charging a capacitor is:

$$I = C\frac{dV}{dt}$$

The problem is this doesn't take into account internal resistance (or a series current-limiting resistor if you include one) or if the capacitor already has some charge.

You have to account for the continually changing charge being applied to the capacitor. In other words, at the very beginning, it looks like a short circuit to your power supply (barring resistance, again). Thus, whatever maximum current your power supply can handle is the theoretical max current. As the capacitor charges, this current decreases exponentially, until the capacitor reaches max charge Q.

The formula for this is:

$$I = \frac{V_b}{R}e^{-t/RC}$$

Where \$V_b\$ is the source voltage, R is resistance, t is time and RC is the time constant (product of resistance and capacitance).

Let's say you don't use a current-limiting resistor and your power supply has an internal resistance of 4Ω:

$$I = \frac{12}{4}e^{-0/0.0132}$$

At time 0 s, the current is 3A. If we figure for, say, 1 ms later:

$$I = \frac{12}{4}e^{-0.001/0.0132}$$

Now the current is ~1 A.

So, how long will it take to charge the capacitor? If you take the time constant, RC (the 0.0132 in the exponent) as a value in seconds, there's a rule of thumb that a capacitor will be charged in 5 times this duration:

$$5\cdot0.0132 = 0.066s$$

The initial current (or the current during some portion of this duration) is referred to as the inrush current. You may want to reduce it by adding a series current-limiting resistor to protect your power supply.

  • \$\begingroup\$ this is helpful, thanks. how do i size the resistor? should i just use the I=12/R e^(-t/RC) formula and pick a target current? \$\endgroup\$
    – kolosy
    Commented Aug 29, 2013 at 4:25
  • 2
    \$\begingroup\$ You'd want to pick a value that puts some upper limit on the initial current, something your circuit and power supply can safely handle. There are other ways to limit inrush current, some of which may or may not apply to your device. Look around on the site here, there are a few questions about just that. \$\endgroup\$
    – JYelton
    Commented Aug 29, 2013 at 4:30
  • 1
    \$\begingroup\$ The INITIAL current is Vb/R, where Vb is the supply voltage and R is the series resistance (including ESR of the capacitor, which will generally be small). At any given instant, the instantaneous current is given by (Vb - Vc)/R, where Vb and R are as above, and Vc is the already-charged voltage on the capacitor. To get the full story, you have to solve the differential equation, which is where the exponential factor comes from. \$\endgroup\$ Commented Aug 29, 2013 at 5:05
  • \$\begingroup\$ @John Thanks for the clarification, I'll edit accordingly. \$\endgroup\$
    – JYelton
    Commented Aug 29, 2013 at 5:07
  • 1
    \$\begingroup\$ @MisforMary I recommend posting a new question and reference this one. Then you can provide details about how you are measuring and calculating. It would be more likely to help than a comment exchange here. :) \$\endgroup\$
    – JYelton
    Commented Aug 10, 2017 at 16:30

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