Capacitor + Resistor in Circuit

[justadream]

Let's say you have a battery, a resistor, and a capacitor all in series.

Does the resistor affect the capacitor in any way?

I know the charge on the capacitor is Q = CV.

Since the resistor doesn't affect V or C, doesn't that mean that the charge on the capacitor remains the same?

So what does change when you add a capacitor? The battery being depleted faster?

---

Members don't see this ad.

 

[Cawolf]

This is commonly referred to as an RC circuit and behaves differently depending on whether it is charging or discharging.

When charging the current starts high and decays to zero as the capacitor is charged. So the current will flow until the capacitor is charged. Initially when the charge on capacitor is zero it looks like a simple circuit with I = V/R. When the potential difference across the capacitor is the same as the battery - it is charged and current stops flowing.

When the battery is switched off the charge on the capacitor decays exponentially with a formula that can be derived.

Q(t) = Q(0)e^(-t/RC)

So you can see that there is exponential decay of charge that is inversely proportional to the resistance of the circuit and the capacitance of the capacitor.

I guess I did not really address your question, but maybe that helped.

 

[justadream]

@Cawolf

"When charging the current starts high and decays to zero as the capacitor is charged. So the current will flow until the capacitor is charged. Initially when the charge on capacitor is zero it looks like a simple circuit with I = V/R. When the potential difference across the capacitor is the same as the battery - it is charged and current stops flowing."

Here, are you talking about a circuit with just a capacitor (or a circuit with both a capacitor and a resistor)?

In other words, I am asking - in the process of charging, does the resistor affect anything?

I guess in the process of discharging the capacitor the resistor would affect the current?

 

[milski]

It affects the current through the circuit and how fast the capacitor is charged and discharged. Larger resistance will lead to longer time to charge. Cawolf explained the initial and final condition quite well. What happens while the capacitor is charging: as the charge increases, the potential across the capacitor increases as well (from V(t)=Q(t)/C). As a result, the potential across the resistor drops, since the sum of Vc+Vr=Vbat. The current in the resistor is Ir=Vr/R and since Vr decreases over time, so does I. You need calculus to derive the exact formula but Cawolf already put it up there - note that R is part of it.

 

[justadream]

@milski

Does it take longer to discharge as well?

Also, when charging, since V decreases, I must decrease?

What about when discharging?

 

[Cawolf]

@justadream

It applies to any circuit. The current will just be proportional to the resistance of the circuit if the voltage is fixed (ie a battery).

In the process of charging, the resistance does not effect the capacitor other than it's rate of charging - it effects the magnitude of the current through the circuit. So having a larger resistance will decrease the I and therefore the rate of charge being stored in the capacitor.

@milski filled in anything else I would've said.

 

[justadream]

@Cawolf

When discharging, does the voltage decrease or stay constant? Also, does having a resistor increase time for discharge?

 

[Cawolf]

If you look at the formula I posted you can see that the resistance is in there.

A larger resistance will have e raised to a smaller negative power - or a larger number overall.

This shows how a greater resistance will result in a longer time to discharge.

The voltage is the potential difference across the plates of the capacitor V = Q/C. Since the charge decreases exponentially, so does the potential.

 

[milski]

@milski

Does it take longer to discharge as well?

Also, when charging, since V decreases, I must decrease?

What about when discharging?

Yes, a higher R will lead to longer discharge time. Same logic applies as before. Let's say you replace the battery with a switch or a direct connection. The initial potential across R is the potential of the capacitor and the current will start high. As the capacitor discharges, the potential decreases and so does the current.

As a summary - the value of R will affect the time for which the capacitor charges/discharges to/from a given charge. It will also affect the rate at which the current decreases over time - higher R will start with lower initial current. What R does not affect is the final charge of the capacitor and correspondingly the final potential across it.

Also note - in reality, there is no circuit with only battery and a capacitor and without a resistor. There is always some sort of resistance. Otherwise the initial current will be infinite and that's generally.. bad. 🙂

 

[Cawolf]

Haha yah - and even without a technical resistor, all wire has some resistance.