Why does an uncharged capacitor short out a resistor in a basic DC circuit?

I’ve really only dealt with two laws thus far concerning circuits and capacitors: $$V=IR$$ $$Q=CV$$

The first obviously being Ohm’s law and the second the equation for a capacitor’s charge. I’ve seen many circuits with uncharged capacitors put in parallel with a resistor, and the solution always says the uncharged capacitor shorts out the resistor—all the current goes through the capacitor, not the resistor. This applies for the instant when the switch is closed, at least. But I don’t understand why that is. When the switch closes, I assume that the voltage across the resistor is 0, seeing as its charge is 0. But a capacitor doesn’t have a ‘resistance,’ does it? What would be its analog, and what equation would relate that quantity to current and voltage? Does Ohm’s law even apply here? Why, mathematically, is I infinitely large for an uncharged capacitor (i.e. short circuit)?

• The second equation is valid only for a DC steady state. For transients you need some more complex calc. – Eugene Sh. Apr 26 '16 at 14:13
• If a voltage supply is switched on at t=0, and the capacitor is initially uncharged, the initial voltage across the capacitor is zero. So if there's a resistor in parallel, it sees a short circuit at t=0. – Chu Apr 26 '16 at 14:16
• @EugeneSh. I just want to know the situation at the instant of the switch’s closing. Would that still be too complicated? – lightweaver Apr 26 '16 at 14:16
• @Chu Yes. That is the basis of my question, and my question is asking why, mathematically using equations that hopefully aren’t beyond the scope of my understanding. – lightweaver Apr 26 '16 at 14:17
• It is not too complicated. Just a bit more math. Look at capacitor charging/discharging equations. – Eugene Sh. Apr 26 '16 at 14:17

You can look at it this way (since you said you want an answer based on math):

The equation for the current through a capacitor is the following:

$I_C = C \dfrac{dV}{dt}$

So at $t=0$, when the switch is closed, there's a voltage across the capacitor and $\frac{dV}{dt}$ is really large (theoretically infinite) because there is a change in the voltage in technically 'no time,' therefore the slope $(\frac{dV}{dt})$ approaches infinity at $t=0$. Since $\frac{dV}{dt}$ is in theory infinity so is the current through the capacitor (they are directly proportional). That explains why it looks as a short at time zero.

As you probably know differentiation does not exist at sharp edges and this is exactly what is happening from a mathematical point of view. Hope it helps!

• Since I=dq/dt, that equation is the derivative of the second equation in the question. – Scott Seidman Apr 26 '16 at 15:24
• Of course differentiation exists at sharp edges. – Scott Seidman Apr 26 '16 at 15:57
• The slope doesn't exist at a corner because the limit of the function does not exist (the limit when you approach from the right side isn't the same as when you approach from the left side, therefore the function isn't differentiable at that point) – Big6 Apr 26 '16 at 16:11
• math.stackexchange.com/questions/515411/… – Scott Seidman Apr 26 '16 at 20:05
• en.wikipedia.org/wiki/… – Big6 Apr 26 '16 at 20:32

A capacitor does have a "resistance"; but since a capacitor is fundamentally different than a resistor, it is not considered this way.

A resistor has a static resistance. It doesn't matter at what time it is measured, or what voltage is applied - the resistance stays the same.

A capacitor has a static capacitance. It DOES matter at what time it is measured, AND what voltage is applied - as it's "resistance" will be different!

The moment the switch is thrown, the capacitor seems like a short circuit (low resistance) because there is no charge on it's plates. How can "no charge" make large currents flow? Because of the fact that there is no charge yet, this imposes the flow of electrons. It is like an empty battery with zero internal resistance - if it is empty, then it will absorb every single bit of energy that can be put into it. So initially, a capacitor seems like a short or low resistance value, until it starts charging.

As the capacitor charges, it starts to behave less like a short. So one could say that it's "resistance" starts increasing (as an analogy.) Up to the point where it is completely full and refuses to take any more electricity - then it would seem like a very high resistance.

But note this is assuming the voltage is constant. If a capacitor is "charged" to say, 5v, then the voltage is suddenly changed to 10v, then the capacitor will react in exactly the same way as it did for the transition from 0v to 5v. (Initially a "short", then gradually behaving less so.) This is where Sixto's answer is spot-on - the rate-of-change determines the current, which is proportional. Instant voltage change = instant current change.

Now another interesting part is, this "stored charge on it's plates" is potential energy, meaning it can be extracted and used elsewhere. So for instance, charging a small capacitor to 3v, then placing a white LED across it's terminals, will cause the capacitor to discharge it's stored charge in reverse - through the LED - causing it to light for a short time.

The length of time it can power the LED is directly related to it's capacitance value: $C=\frac{Q}{V}$ The larger the capacitor physically (the more Q potential), the more capacitance, and thus, the more ability to absorb and release electrons for any given voltage.

Ohm's law always applies to DC - always - that's why it's called a law. But this is not DC... charge varies with time, volts vary, amps vary... so this is the AC domain.

Ohm's law does not apply to an ideal capacitor because it's non-ohmic. Capacitors have a quantity called reactance, which measures its opposition to alternating current. Reactance+Resistance=Impedance (all of this is on Wikipedia).

If you want an intuitive model, its current is infinite for an infinitesimal time because it stores energy and that store fills when power is connected. How long does it take to fill? No time, because there's zero series resistance. Current is the rate of filling. To store a finite nonzero amount of energy in zero time requires an infinite fill rate.

(Readers who understand calculus and object to the arithmetic of infinity times zero should fill in "tends to zero" and "tends to infinity" as appropriate).