# Capacitor Current–voltage relations conceptual doubt

I understand that a capacitor stores energy on the form of voltage. However, looking at the equations for the current and voltage on the capacitor, we have the following:  I was thinking here that if I connect the capacitor to a DC battery, it is going to charge, right? However, as the voltage is a constant, the current should be 0 according to the second equation. Therefore, according to the first one, the integral value is gonna equal 0, right? If it equals 0, then V(t) = V(t0). Assuming that V(t0) = 0, the voltage is always going to be 0, even though the capacitor charges when you apply a DC voltage to it.

Could someone please explain to me how to link this concept to the mathematical equations?

Thank you very much.

• Shall I assume practical batteries and capacitors with internal resistance ? Feb 20 '20 at 3:16
• Let me recommend that you read Chapter 18, circa page 732, of Matter & Interactions, 4th edition, by Chabay & Sherwood. You really need to visualize the surface charges and the fact that you are talking about a battery, a switch, and a capacitor. Even with a vacuum dielectric for the capacitor (just metal plates separated by a gap), the surface charges are arranged in a particular way at steady state before closing the switch. Then you need to imagine what happens right at the point of closing the switch and how those charges go through a transient event to re-arrange themselves.
– jonk
Feb 20 '20 at 6:02

In a perfect world, the current would be zero yes because once the capacitor "fills up" its charge, it'll behave like an open circuit and thus the current would be zero and your voltage V(t) would be equal to V(t0). Think of that as a taller water tank emptying itself out into another water tank. There will be equilibrium between the two water tanks. (forgive my drawing, it's not abiding by the laws of physics ◠‿◠ )

However, in the real world, capacitors take time to charge and that voltage is seen as an exponential growth until it plateaus completely. This exponential equation is where you take the derivative. The reason why it's exponential is because there is some resistivity inside the capacitor. This is also known as ESR.

If V(t0)=0 when you connect a DC voltage (V) to the capacitor, there it will see a voltage difference of V-V(t0)= V, and in practice, the capacitor will charge to DC voltage V in a minimal amount of time. There will be some current for a small amount of time, and once the capacitor charged to voltage V than current will be zero.

In theory, i.e. in the case of ideal capacitor and ideal voltage source, without resistances or inductivities or radiation of HF or other losses involved, the capacitor is strictly parallel to the source, which is a contradiction, since the initial value/voltage of the differential equation would be defined to be V_source and V_capacitor at the same time t0.

A capacitor is an ideal voltage source at t0. If it is an empty capacitor or the stored voltage is different from the voltage source, it would be like connecting 2 ideal voltage sources with different voltages in parallel (at time t0). Each of these 2 ideal voltage sources define another voltage resulting in the contradiction at t0.

Only in theory it is a contradiction to connect an ideal voltage source to an ideal capacitor if both voltages are different. In reality, there are always resistors, inductivities, waves, radiation etc. involved avoiding this contradiction.

I was thinking here that if I connect the capacitor to a DC battery, it is going to charge, right?

We know that: -

$$I = C\dfrac{dv}{dt}$$

And, at the instant the battery is applied to the capacitor, the rate of change of voltage has to be infinity therefore, the current into the capacitor is also infinity.

But, in practice there are series resistances and series inductances that MUST limit the infinities to practical real-world values.

However, as the voltage is a constant

It's only constant once things have settled down.