# Voltage across resistor in a broken circuit (capacitors)

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#1
Hi! a quick (hopefully!) question about voltages in a circuit where a capacitor can be charged (like in the picture).

So, I understand that V = Vc + Vr, because the battery, resistor and capacitor are all in series. And I know that when the switch is open (provided the capacitor is initially fully discharged), Vc = 0, since there is 0 charge on the capacitor plates.

So does this mean that Vr = V when the switch is open? This is the part that confuses me. Can a resistor connected to a battery have a voltage across it, even if no current is flowing through the resistor?

Or, is Vr actually 0 when the switch is open, but increases instantaneously to the maximum value V as soon as the switch is closed (when t = 0)?

I know that there is a pd across the terminals of a battery or cell when no current is flowing (i.e. from the definition of emf, right?), but I'm struggling to understand whether this applies to resistors as well.

Thanks! x
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2 years ago
#2
(Original post by levi ackerman)

So does this mean that Vr = V when the switch is open? This is the part that confuses me. Can a resistor connected to a battery have a voltage across it, even if no current is flowing through the resistor?
No.

There is no conduction path across the capacitor plates. When the capacitor has no charge at t=0, there is no p.d. across either the resistor or the capacitor. No current flows through the resistor so no p.d. can develop.

Or, is Vr actually 0 when the switch is open, but increases instantaneously to the maximum value V as soon as the switch is closed (when t = 0)?
Correct.

Because when the switch is closed at t=0, there is initially no electric field across the capacitor plates and therefore no opposition to charge flowing onto the capacitor. The capacitor initially appears as if it is a short circuit from the supply perspective and so all of the supply p.d. appears across the resistor at that instant.

As the charge on the capacitor builds, so does the p.d. across the plates, and from Kirchoff's voltage rule (sum of potential drops must equal the supply).

Because the charge building across the capacitor plates stores energy on the electric field between them, the electric field opposes the supply e.m.f. and immediately starts to limit the current flow.

The p.d. across the resistor will fall accordingly.
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#3
(Original post by uberteknik)
No.

There is no conduction path across the capacitor plates. When the capacitor has no charge at t=0, there is no p.d. across either the resistor or the capacitor. No current flows through the resistor so no p.d. can develop.

Correct.

Because when the switch is closed at t=0, there is initially no electric field across the capacitor plates and therefore no opposition to charge flowing onto the capacitor. The capacitor initially appears as if it is a short circuit from the supply perspective and so all of the supply p.d. appears across the resistor at that instant.

As the charge on the capacitor builds, so does the p.d. across the plates, and from Kirchoff's voltage rule (sum of potential drops must equal the supply), the p.d. across the resistor will fall accordingly.
awesome, thank you!
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