# Discharge in a DC RC circuit and Kirchhoff's Loop Rule

• FranzDiCoccio
In summary, the author is trying to understand how to solve a differential equation for a charging RC circuit, but is having difficulty understanding the signs of the current. He has three possible approaches, but believes that the simplest is to use Kirchhoff's laws and to be careful about the sign of the current.
FranzDiCoccio
Hi all,

I think this issue periodically resurfaces in PF. I have found a similar discussion in this closed post and possibly others. I'm posting this because I'd like to check my understanding, if anyone is available to provide some furtherinsight.

So I'm trying to gather a "overall" understanding of simple circuits, without resorting to advanced circuit analysis. This is meant for people that is able to analyze simple circuits using Kirchhoff laws and knows derivatives (but not yet integrals or differential equations).
The idea is that they should be able to write down the differential equation and at least check that some given function solves it (they are not yet able to properly solve a differential equation).

My problem is with signs in the discharging RC circuit. Of course, formally, the equation should be
$$\frac{q}{C}+IR=0$$
which is solved by a decreasing exponential function: the charge must drop exponentially to 0.

But what is the best way of obtaining that equation? I can think of three approaches
1. The charging of the RC circuit has been already discussed. One starts with the same equation, where the generator term is removed ${\cal E}=0$. This gives the "correct" signs. However, it seems that this approach is not compatible with KLR. There is nothing "compensating" the two voltage drops of the resistor and capacitor. But then again, perhaps it does not need to be: after all you suddenly removed an element from a circuit which previously satisfied KLR.
2. You insist on KLR. You start from the circuit containing only the charged capacitor and the resistor. It seems to me that this should result in the (apparently?) wrong signs: $q/C-IR=0$. Indeed the current should flow from the positively to the negatively charged plate of the capacitor, and the resistor produces a voltage drop in the direction of the current. I think this can be salvaged when connecting the current and the charge. Since the charge is decreasing, it should be $I=-\dot q$, as opposed to the naive choice $I=-\dot q$. That gives once again the expected decreasing exponential function.
3. You forget about Kirchhoff and reason in terms of energy. The capacitor stores an energy $E_C=q^2/C$ which is dissipated by the capacitor at a rate $P=I^2 R$. Since it is dissipation, the resulting equation is $\dot E_C=-I^2 R$, which is equivalent to the previous ones.

I guess what ultimately confuses me is the meaning of the sign of the current. The above considerations seem to work fine with a direct current, where somehow I can assume that the current is always positive. But what if I have to consistently extend this to alternate currents?
It seems to me that approach n° 3 works fine in all cases, provided the circuit is simple (one loop). I guess it's unfeasible with more complex circuits, e.g. with several loops. I guess that one should go with Kirchhoff rules, i.e. approach n° 2.

Suppose I have a complex circuit with R, L, C and (possibly a.c.) generators. Would KLR work? And if this is the case, how would one deal with the signs of the derivatives when setting up the differential equations for the charge/current in the circuit? Does this question even make sense?

I'm referring to approach n° 2, where one applies KLR and then has to be careful about the sign of $\dot q$. Perhaps this can be done algorithmically by assigning an arbitrary sign to the currents? Perhaps positive for anticlockwise and negative for clockwise currents?

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FranzDiCoccio said:
I can assume that the current is always positive. But what if I have to consistently extend this to alternate currents?

The DC equations apply instantaneously. Think of a DC voltage source that varies with time, but at each instant Kirchoff's laws apply. Therefore, instantaneously, AC is no different than DC. AC becomes different only when you try to calculate averages over one or more complete cycles rather than instantaneous values.

FranzDiCoccio said:
You insist on KLR
Always insist on Kirchoffs laws, they are the foundation of circuit theory. If they don’t work, then you need to use something much more difficult, like Maxwell’s equations or quantum electrodynamics.

FranzDiCoccio said:
Indeed the current should flow from the positively to the negatively charged plate of the capacitor, and the resistor produces a voltage drop in the direction of the current. I think this can be salvaged
I am not sure what you think needs to be salvaged. You simply need to draw the circuit, and label things. Then write Kirchoffs laws consistently with your drawings. The sign depends on the convention you adopt in your circuit diagram.

Dale said:
I am not sure what you think needs to be salvaged. You simply need to draw the circuit, and label things. Then write Kirchoffs laws consistently with your drawings. The sign depends on the convention you adopt in your circuit diagram.

What I mean here is that, if I get it right, KLR in a simple RC circuit gives $q/C-IR=0$. On the other hand, "removing" the generator term in the KLR for a charging circuit gives $q/C+IR=0$. This is what for instance my book by Halliday, Resnick and Walker does (without giving too much explanation).
These two equations are clearly different, and it does not seem to me that this is a matter of sign convention.

And, by the way, I'm not entirely clear on the right way of applying KLR to capacitors.
From what I know, as far as only resistors are present, you arbitrarily choose a direction for the current and whenever that goes through a resistor there is a voltage drop (hence a minus if you "follow" the current, and a plus if you go against it). Then you solve the linear system of equations for the currents and if one of your currents turns out to be negative that means that the direction you originally chose should be reversed. This is algorithmic, you do not need to worry too much about signs.

What is the rule with capacitors? I can assign a sign to the voltage if I know the sign of the charge on the plates, which is kind of easy in a simple one-loop circuit. But what is the "algorithmic" procedure? Also what is the way of dealing with the relation between charge and current? Because the problem is now a set of differential equations.
The minimal example is with the initial equation I wrote: $q/C-IR=0$. If you naively assume $I=\dot q$ you get the wrong answer (increasing exponential). So you somehow need to have an idea of what's happening, which is easy for simple circuits only, I guess.

I hope I clarified my doubt (and btw thanks a lot for your time).

FranzDiCoccio said:
So you somehow need to have an idea of what's happening, which is easy for simple circuits only, I guess.
No.
Here's how you can write the KVL for a loop:
Independent source voltage= sum of potential drops across the components.

Treat the component voltages as potential drops.

If you have an RC charging circuit containing a battery of emf V,
you can write the KVL as
V=Rdq/dt+q/C.
Here, V would be your 'independent' source voltage.

In a discharging RC circuit, there is no independent voltage source, which is why this circuit is also called 'source free' circuit.
Hence, the above KVL equation modifies to
0=Rdq/dt+q/C.

FranzDiCoccio said:
What is the rule with capacitors? I can assign a sign to the voltage if I know the sign of the charge on the plates, which is kind of easy in a simple one-loop circuit. But what is the "algorithmic" procedure?
Just use the passive sign convention, the same as with a resistor.

https://www.khanacademy.org/science...s-topic/circuit-elements/a/ee-sign-convention

Under the passive sign convention ##i=C \frac{d}{dt} v##. It can be done completely algorighmically as long as you are consistent.

If for some reason you must use the active sign convention then you would flip the sign of the voltage in the capacitor equation, but I try to avoid that to avoid confusion

cnh1995 said:
No.
Here's how you can write the KVL for a loop:
Independent source voltage= sum of potential drops across the components.

Treat the component voltages as potential drops.

Ok... Would this work in any case?
What I'm trying to say here is: as far as I have to analyze only the charging or discharging RC circuit everyithing is pretty clear to me, and I do not even need an algorithmic procedure: there is only one circuit to be analyzed.
What you suggest would work in an arbitrarily complex circuit, with many loops, different voltage sources, some charged capacitor, some uncharged ones?

I mean, the KLR I was taught seems more complex than that, even where only resistors are present.
In that case you choose an arbitrary direction for the current in a loop and one direction for analyzing the loop. Then the rhs is zero (overall voltage drop, including sources) but you need to choose the sign for each voltage in the lhs. The sign is + if you go through a source from - to + and if you go through a resistor against the direction you originally chose for the current.

Is this unnecessarily complex?

FranzDiCoccio said:
I mean, the KLR I was taught seems more complex than that, even where only resistors are present.
In general, I prefer the node voltage approach instead of the loop current approach. However, if you just consistently use the passive sign convention then you will be fine either way.

cnh1995
FranzDiCoccio said:
Ok... Would this work in any case?
What I'm trying to say here is: as far as I have to analyze only the charging or discharging RC circuit everyithing is pretty clear to me, and I do not even need an algorithmic procedure: there is only one circuit to be analyzed.
It works in any case.
But like Dale, I too prefer node voltage method when solving more complex circuits.

Dale
Hi Dale and cnh1995

thanks a lot for your help! I'll look into what you suggest ASAP.
Franz

## 1. What is discharge in a DC RC circuit?

Discharge in a DC RC circuit refers to the process of releasing stored electrical energy in a capacitor through a resistor. When a capacitor is charged, one plate of the capacitor accumulates positive charge while the other plate accumulates negative charge. The capacitor will discharge when connected to a circuit, allowing the electrons to flow from one plate to the other through a resistor.

## 2. How does Kirchhoff's Loop Rule apply to a DC RC circuit?

Kirchhoff's Loop Rule, also known as Kirchhoff's Voltage Law, states that the sum of the voltage drops in a closed loop must equal the sum of the voltage sources in that loop. In a DC RC circuit, this means that the voltage drop across the capacitor, which is equal to the voltage across the resistor, must equal the voltage of the battery or power source.

## 3. What is the formula for calculating the time constant in a DC RC circuit?

The time constant, denoted by the symbol τ (tau), is calculated by multiplying the resistance (R) in ohms by the capacitance (C) in farads. This can be represented by the formula τ = RC.

## 4. How does the value of the resistor affect the discharge time in a DC RC circuit?

The larger the resistance value, the longer it will take for the capacitor to discharge. This is because a higher resistance means there is more opposition to the flow of electrons, resulting in a slower discharge rate.

## 5. Can Kirchhoff's Loop Rule be applied to circuits with multiple capacitors and resistors?

Yes, Kirchhoff's Loop Rule applies to any closed loop in a circuit, regardless of the number of components. The rule still states that the sum of the voltage drops must equal the sum of the voltage sources. However, the calculations may be more complex in circuits with multiple components.

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