# Deriving the equation for capacitor discharge

• Sturk200
In summary: The change in voltage across the capacitor is independent of the direction of the current. It only depends on the charge on the capacitor, not the current through it. It is +q/c in both cases.
Sturk200
We have a circuit consisting of a charged capacitor and a resistor. First of all, we use Kirchhoff's loop rule to express this circuit in mathematical form. Now one teacher tells me Kirchhoff's loop rule yields iR+q/c=0. Another teacher tells me that the rule yields iR-q/c=0. Is there any way to reconcile these two interpretations? They seem to contradict.

Carrying out the derivation, if we use the first equation, we simply replace i=dq/dt. If we use the second equation we have to replace i=-dq/dt. Which of these routes is justified? I can't imagine that they are both valid.

You have to pick a direction for the current and the charge. The two equations simply use different conventions. The best way to see is to look at the two drawings.

I don't see how changing the current convention solves this. Say we have a square circuit with a capacitor on the top and a resistor on the bottom. The right side of the capacitor is charged positive.

(1) If we say that current flows in the direction of positive charge, we get the following. Start at the negative terminal of the capacitor and cross the capacitor to the positive terminal: the change in voltage is +q/c. The change of voltage across the resistor in the same direction is then -ir. Using Kirchhoff's rule, we get q/c-ir=0.

(2) If we say that current flows in the direction of negative charge, then the change in voltage from the negative to positive terminal of the capacitor is -q/c, the change in voltage across the resistor in the same direction is +ir, and Kirchhoff gives us ir-q/c=0=q/c-ir.

We never get ir+q/c=0. Am I missing something?

Sturk200 said:
(2) If we say that current flows in the direction of negative charge, then the change in voltage from the negative to positive terminal of the capacitor is -q/c, the change in voltage across the resistor in the same direction is +ir, and Kirchhoff gives us ir-q/c=0=q/c-ir.
The change in voltage from the negative terminal to the positive terminal of the capacitor is independent of the direction of the current. It only depends on the charge on the capacitor, not the current through it. It is +q/c in both cases. Only the sign of the voltage across the resistor changes with a change in the direction of the current.

I always use the passive sign convention. Define your currents so that current enters into the positive side of a passive device. If you do it consistently this way, even if you mis-assign your positive terminals of some devices, the signs will work out in the end.

DaleSpam said:
The change in voltage from the negative terminal to the positive terminal of the capacitor is independent of the direction of the current.

If the change in voltage across the capacitor is independent of the direction of the current, then so is the change in voltage across the resistor. We must choose whether to define voltage in terms of the potential energy of a positive charge or that of a negative charge. If we choose it to be for a positive charge, then the change in voltage from the positive to negative terminal of the capacitor is negative. Now if we continue to traverse the circuit, we see that the change in voltage across the resistor is opposite in sign to that across the capacitor -- this is true regardless of the direction of current, because it depends not on our current conventions but on our voltage conventions.

Suppose we did somehow end up with ir+q/c=0 using Kirchhoff's rule, that is suppose we somehow did convince ourselves that the change in voltage across each element of the circuit is indeed the same sign (which I still can't justify to myself), then what can we do in the next step of the derivation? We are supposed to replace i=-dq/dt. I was told that the justification for this step is that the charge separation on the capacitor is decreasing when it is discharging, so in our expression for the current the dq term must be negative (the amount of charge passing a given cross section of wire is decreasing). This would be the case regardless of how we choose our conventions in the first step, so how do we end up with the right derivation if we choose our conventions such that ir+q/c=0? It would seem that we would be forced to say q/c=dq/dt, which gives an incorrect result when we solve the differential equation.

EDIT: Sorry, I meant in the last sentence to say q/c=Rdq/dt. Still, the result would not come out right.

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Sturk200 said:
If the change in voltage across the capacitor is independent of the direction of the current, then so is the change in voltage across the resistor.
No, it isn't.

For a capacitor v=q/c

For a resistor v=ir

The current shows up in the resistor equation, but not in the capacitor equation. One depends on i, the other does not. So changing the sign convention of the current changes the sign of the voltage across the resistor but not the capacitor.

Sturk200 said:
It would seem that we would be forced to say q/c=dq/dt, which gives an incorrect result when we solve the differential equation.
All you get is that your function for i in one case will be the negative of your function for i in the other case, resulting in the same physical current. The choice of convention doesn't matter in the end.

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DaleSpam said:
All you get is that your function for i in one case will be the negative of your function for i in the other case, resulting in the same physical current. The choice of convention doesn't matter in the end.

If ir+q/c=0, then -(dq/dt)r+q/c=0. Solving this differential equation would give q=q0et/rc. This equation is incorrect. The exponent should be negative. With a positive exponent the equation implies that the charge is increasing infinitely. This violation of charge conservation, I think, is a direct consequence of the claim that the changes in voltage across both elements of the circuit are the same sign.

(I substituted i=-dq/dt because the charge on the capacitor is decreasing with time.)

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OK, so the STANDARD convention is the passive sign convention which was mentioned by EM Guy earlier. See the attached drawing for the definitions of currents and voltages used below. Applying KVL you get
##v_c+v_r=0##
One of the nice things about the passive sign convention is that all of the usual formulas for passive components are based on it, so this gives us
##q/c + ir =0##
So the formula that you don't see how to ever get is actually the correct formula under the standard convention.

Now, if we want to solve this equation then, again, all of the standard formulas assume the passive sign convention so:
##i=dq/dt=\dot q##
##q/c + r \dot q = 0##
##\dot q = -q/(rc)##
##q = q_0 e^{-t/rc}##

Sturk200 said:
(I substituted i=-dq/dt because the charge on the capacitor is decreasing with time.)
This is not correct reasoning (although it has a 50% chance of being right randomly).

The correct way to see if ##i=\dot q## or if ##i=-\dot q## is to look at your circuit drawing and see if a positive ##i## increases ##q## or decreases ##q##. It has nothing to do with whether or not the charge on the capacitor is increasing with time or not. For all you know, the charge on the capacitor could be negative, so it is increasing with time.

Bottom line. Draw your circuit and label everything. Then simply write your equations consistent with your drawing and labeling.

#### Attachments

• Scan_Pic0018.jpg
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That picture doesn't make sense. The current in that circuit is being driven by the charge on the capacitor. The positive terminal pushes a positive current from it in a counterclockwise direction. The left side of the resistor should be at a higher potential than the right side.

Check out this handout: http://web.mit.edu/sahughes/www/8.022/lec09.pdf. I think the way it's done here makes sense, but I don't see how it can be reconciled with the equation ir+q/c=0.

It makes perfect sense. It is the standard convention.

See: https://en.wikipedia.org/wiki/Passive_sign_convention

Note how, in my drawing, if you look at each component you see the current going into the positive terminal, just as shown in the picture on the wiki page.

Sturk200 said:
The left side of the resistor should be at a higher potential than the right side.
That depends on if ##q_0## is positive or negative.

DaleSpam said:
It is the standard convention.

I learned that the standard convention is to draw the current in the direction that positive charge would flow. In the circuit you drew positive charge would flow in the counterclockwise direction due to the configuration of charges on the capacitor.

DaleSpam said:
That depends on if q0q_0 is positive or negative.

What is this q0? Any capacitor has both negative and positive charges. On the standard view, when a capacitor is discharging current flows away from the side with positive charge and into the side with negative charge.

Sturk200 said:
What is this q0?
The initial charge on the capacitor, i.e. ##q(0)=q_0##.

Sturk200 said:
On the standard view, when a capacitor is discharging current flows away from the side with positive charge and into the side with negative charge.
That is the physics and it is true regardless of what sign convention you adopt for your circuit.

DaleSpam said:
That is the physics

The physics is that electrons flow from negative to positive. The convention is that current flows from positive to negative.

I think the difference between the approach which begins with ir+q/c=0 and the approach that begins with ir-q/c=0 is that the first approach uses the passive sign convention, like em_guy said, while the second approach sticks more closely to the actual current flow in the circuit. Both work, but the first approach requires a bit of falsification. Strictly speaking, it is not true that ir+q/c=0 in this circuit.

Sturk200 said:
Check out this handout: http://web.mit.edu/sahughes/www/8.022/lec09.pdf. I think the way it's done here makes sense, but I don't see how it can be reconciled with the equation ir+q/c=0.
I looked at the link. The problem is that in the handout none of the charges, currents, or voltages is labeled. I can infer what he is doing, but he doesn't make it clear. What he is doing is using the active sign convention for the capacitor and the passive sign convention for the resistor. But he never writes it down explicitly, which is bad form. Always label your voltages and currents in your circuit.

My personal preference is to use KCL rather than KVL. Then you never even need to label the currents. Plus, KCL works for some circuits where KVL does not work.

Sturk200 said:
I think the difference between the approach which begins with ir+q/c=0 and the approach that begins with ir-q/c=0 is that the first approach uses the passive sign convention, like em_guy said, while the second approach sticks more closely to the actual current flow in the circuit. Both work, but the first approach requires a bit of falsification. Strictly speaking, it is not true that ir+q/c=0 in this circuit.
I wouldn't say "falsification" I would say that in the passive sign convention you tolerate bad guesses about the sign of voltages and currents. You just recognize that if you guess wrong you will simply get a negative number instead of a positive number. Nothing is false, it is just negative. It is true that ir+q/c=0, given the definitions of i and q as drawn. It is just that some or all of those quantities may be negative.

The important thing about using a convention is that you just apply it consistently and you get the right answer, and you don't have to spend a lot of time thinking about the details.

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DaleSpam said:
Nothing is false, it is just negative.

I still can't resist thinking that this is an either/or. We have to assume that the MIT guy's justification for setting i=-dq/dt is valid. Moreover this same justification appears in a Yale lecture on Youtube (you can see I have been trying to get perspectives on this). If it's true that in the case of a discharging capacitor i=-dq/dt, then this is true regardless of our current convention. Therefore, if you accept the passive sign convention, then the only way to get the right solution to the differential equation is to ignore the physics -- i.e. to set i=dq/dt regardless of the fact that the charge is decreasing with time. Physically speaking, this approach involves a falsification. It is not merely altering the sign on a charge, but rather mistaking a decrease in charge for an increase. Plus the passive sign convention in the case of a discharging capacitor requires a complete butchering of the physics of the current flow, since you have to say that the positive terminal of the capacitor connects to the negative terminal of the resistor, which in fact is not the case, since in the actual circuit the capacitor is driving the current. (Note this is not just switching signs; in the actual circuit like signs attach to like, regardless of what they are).

Using the passive sign convention works mathematically, but it requires claims that are physically untrue.

DaleSpam said:
The important thing is that you just apply it consistently and you get the right answer, and you don't have to spend a lot of time thinking about the details.

True, but I like to understand the justifications.

It ends up as using nodes or loops to solve the circuit.

The summation of all current into or out of a node equals zero. ie what comes in must come out.
The summation of all voltages around a loop equals zero.

How we write it is dependant on whether we think we know the current direction, or voltage magnitude. Or if we do not know.
Consistancy is important throughout the whole analysis, especially when the node has greater than than 2 terminals.

for the circuit in discussion
One could write at a node:
Vc+Vr = 0
Vc - Vr = 0

Also, using currents,
Ic + Ir = 0 , or
Ic - Ir = 0

In either case the directions of currents and magnitudes of voltages sort themselves out.

Sturk200 said:
I still can't resist thinking that this is an either/or. We have to assume that the MIT guy's justification for setting i=-dq/dt is valid. Moreover this same justification appears in a Yale lecture on Youtube (you can see I have been trying to get perspectives on this).
So, what? I just pulled out four of my textbooks that cover capacitors and RC circuits. All four of them used the passive convention. It is by far more common than any other convention.

Nilsson and Riedel. "Electric Circuits, 5 ed." p 221.
Serway. "Physics for Scientists and Engineers, 3 ed." p 775.
Rizzoni. "Principles and Applications of Electrical Engineering". p 123.
Oppenheim and Wilsky. "Signals and Systems, 2 ed". p 240.

Perhaps some other textbooks do not use it, and perhaps even in these textbooks it uses a different convention in other problems, but certainly the passive sign convention is physically correct, commonly used, and well motivated.

Sturk200 said:
Therefore, if you accept the passive sign convention, then the only way to get the right solution to the differential equation is to ignore the physics -- i.e. to set i=dq/dt regardless of the fact that the charge is decreasing with time. ... Using the passive sign convention works mathematically, but it requires claims that are physically untrue.
Nonsense. There is nothing at all "physically untrue" about the passive sign convention. If you look at my analysis or the analysis in any of these sources then you will see that the passive sign convention is both physically and mathematically correct.

I am not claiming that other conventions are incorrect, but the use of the passive sign convention is clearly supported by standard EE practice and most textbooks. You may choose a different convention if you like, but to claim that there is something wrong with the passive sign convention is simply untrue.

Also, you are again using your incorrect reasoning about the relationship between i and dq/dt, as I explained in post 9.

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## 1. What is a capacitor?

A capacitor is an electronic component that stores electrical energy in the form of an electric field. It consists of two conductive plates separated by an insulating material called a dielectric.

## 2. What is capacitor discharge?

Capacitor discharge refers to the process of releasing the stored energy in a capacitor. This can happen when the capacitor is connected to a circuit or when it is discharged through a resistor.

## 3. What is the equation for capacitor discharge?

The equation for capacitor discharge is Q = Q0e-t/RC, where Q is the charge on the capacitor at time t, Q0 is the initial charge on the capacitor, R is the resistance in the circuit, and C is the capacitance of the capacitor.

## 4. How is the equation for capacitor discharge derived?

The equation for capacitor discharge is derived using the principles of Kirchhoff's voltage law and Ohm's law. By analyzing the circuit and applying these principles, we can come up with an equation that describes the behavior of a discharging capacitor.

## 5. What factors affect the discharge of a capacitor?

The discharge of a capacitor is affected by its capacitance, the resistance in the circuit, and the initial charge on the capacitor. It is also affected by the type of dielectric material used and the temperature of the capacitor.

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