# Voltage drop across capacitor in a DC circuit? (Simple question)

I'm trying to find the voltage drop across a capacitor in a DC circuit. We can assume that the capacitor is fully charged and in a steady state. Therefore, the capacitor is now an open circuit. So this is really an open circuit voltage problem.

My question concerns the following section of the circuit:

Node a -> resistor -> node b -> capacitor -> ground

Is the voltage at node b zero because no current flows through the capacitor $(V=0R)$? Or does the voltage at node b equal voltage at node a?

• The capacitor is not a resistor, so it doesn't care to follow Ohm's law.
– pipe
Commented Oct 13, 2016 at 12:35

The voltage across a resistor is proportional to the current thru it. When there is no current thru a resistor, the voltage across it is therefore zero. Since the voltage across the resistor is zero, the voltage of the two nodes at either end are at the same potential.

The voltage at node b is equal to the voltage at node a. No current is flowing through the resistor and therefore the voltage drop is zero.

The capacitor is an open circuit, therefore any voltage drop is possible. The equation V=0*R cannot be used since this equation is only valid for a resistor.

You are interpreting the results of your calculation incorrectly.

You wrote:

V=0*R

(Therefore Voltage is zero.)

And assumed this was the voltage at the node between the resistor and the capacitor. In fact, this is the voltage drop across the resistor. You are actually proving the voltage is the same with respect to ground on either side of the resistor. Which is what you would expect if using an fully charged ideal capacitor.

Being pedantic (but accurate) the capacitor (even an ideal one) never fully charges to the supply voltage so there would always be a very small current flowing and so a very small (and decreasing with time) voltage drop between node a and node b.

By following Ohm's Law, you can see that there is no voltage drop across the 50ohms resistor, therefore the voltage is the same at node a and node b.

But considering the fact that no electronic is EVER ideal, the capacitor will never charge fully. So there will be a small amount of current flowing through the resistor and there will be a voltage drop a node b.