# Math of capacitor current/voltage when connecting capacitor to battery

I think I'm missing something fundamental in the math surrounding capacitors.

Suppose I have a 1 F capacitor that is not charged. The voltage across its terminals is 0 V.

Now suppose I connect a 1 V battery to it. The voltage across its terminals is therefore 1 V.

The current to it is given by:

I(t) = C dV(t)/dt


But the voltage has jumped in a non-continuous manner from 0V to 1V.

Therefore dV(t)/dt is "infinite" (or undefined) at the time t where I connected the battery. It jumped from 0V to 1V "instantaneously". This implies the current is infinite at this point.

What am I missing here?

• Boah, I should know this, university is not too far back for me... However, the only thing I can come up right now is that the resistance from the battery to the capacitor will not be zero, so that will slow down the charging process (given you do not look at an idealized setup). But yes, you will get an huge current spike also in reallife. – jwsc Jul 13 '15 at 13:20
• A 1V battery is not just a 1V voltage source. (Wiring and capacitors have resistance too, but the battery is probably the dominant resistance source in most such circuits. However many real 1F capacitors have resistances in the 50 ohm region) – Brian Drummond Jul 13 '15 at 14:45

What you are missing is that you are working with ideal components. You are assuming that the capacitor and the battery have no resistance. In that case, yes, current would be infinite and the time to charge tha capacitor would be infinitely short.

We don't, however, have ideal components to work with. The battery will have an internal resistance (NOT a discrete part, but caused by the physics and chemistry of the battery.) The capacitor also has an internal resistance (again, this is inherent to the materials.)

These resistances prevent infinite currents from flowing, and prevent an instantaneous change in voltage.

You need to use equations that take the effects of the resitances into consideration. Specifically you need to look into the RC time constant.

The RC time constant (T) is the product of the resistance (in Ohms) and the capacitance (in Farads.) T = RC

T is the number of seconds it will take the capacitor to charge to 63 percent of the target voltage - in your case, 0.63 Volts.

If the capacitor and the battery together have 1 Ohm of internal resistance, then T would be 1 Second. Taking t as 1 second, dV(t)/dt would then be 0.63 which leads to I(t) being 0.63Amperes, at which point the capacitor is only charged to 0.63 volts.

I've probably screwed up something in there, but the point is that infinite currents only occur in ideal circuits with no resistance.

You only missed the point that you assume that the internal capacitor resistance is zero, which means it is an ideal capacitor, which does not exists. And you are assuming that the power source does have a zero internal resitance, which makes it an ideal power source, which does not exists. But, there are out there capacitors with very low internal resistance. And power sorces which have also very low internal resistance too. So, when you connect such power source to such capacitor, the current does not tend to infinite, but to a very high value for a very short time. And this is an issue. For example, consider the main AC voltage you have at home. You design a power supply with and rectifier, connected to some capacitors, in order to have a DC power signal (it is very common in switch mode power supplies), when you plug the circuit to the mains plug, the initial current tend to a very high value. This is a problem, it can burn the protection fuse, and damage the capacitors. It is common to put a PTC resistor in series, which increases the internal resistance if the current increases, in order to "hold" the amount of current charges the capacitors.

So, your math is completely right.