Discharge in a DC RC circuit and negative sign of current

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Hi everyone!
I'm trying to understand why when writing the differential equation of a discharging RC circuit [itex]V_C-Ri(t)=0 \to q(t)/C - Ri(t)=0[/itex] we replace [itex]i(t)[/itex] with [itex]-\frac{dq(t)}{dt}[/itex].
I read many threads but I don't understand the physics behind this. The usual answers I read are something like "by using the passive sign convention" or "because the current must be positive".
I'd like to solve this problem because it is driving me crazy!
 

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  • #2
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You need to sketch the circuit and label the variables.
 
  • #4
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So look at the way you have i and q labeled in your drawing. Does a positive i lead to an increase in q or a decrease?
 
  • #5
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Doesn't it lead to an increase of charge in the circuit and an equivalent decrease in the capacitor?
 
  • #6
sophiecentaur
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Doesn't it lead to an increase of charge in the circuit and an equivalent decrease in the capacitor?
Would you have the same problem with two plus two equal four? Once you step into the maths, the results can be believed because you can trust Maths ( better people than you or I can be relied on for true rigour). Don’t look for things just to make sense; use the tools.
 
  • #7
DaveE
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One of the trickiest parts of circuit analysis is keeping track of the signs of voltages and currents. There really isn't a better way of doing this than labeling your schematic with all of the pertinent voltage and current polarities and carefully following rules about adding things up in the equations you derive from those drawings.
This usually isn't about understand how circuits work, it's more like a book-keeping problem. Until you get used to this you will be struggling with sign errors over and over again. After you create your equations, add in a step where you explicitly check for these sort of sign errors. It often isn't emphasized in school, but adding in checks to verify you haven't made simple errors is a key part of engineering in practice.
 
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  • #8
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Doesn't it lead to an increase of charge in the circuit and an equivalent decrease in the capacitor?
Don’t forget that “charge” in a capacitor is actually “charge separation”. There is no net charge on any circuit component. So there is a decrease on the amount of positive charge on the top plate and also an identical decrease on the amount of negative charge on the bottom plate.

The resistor has no substantial charge separation. So it does not get either a net charge nor a charge separation.
 
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  • #9
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I can't disagree with you all, but I still don't understand the choice of a negative [itex] I(t) [/itex]. As sophiecentaur says we have to believe in maths but if so, why should we put a negative sign? A derivative is just a mathematical tool that expresses the change of something, and as such, it has an implicit sign depending on the change of [itex]q(t)[/itex] in [itex]t[/itex].
Maybe my problem is that I don't understand what [itex]q(t)[/itex] really means. I'm very sorry for being too stubborn, but I do really like physics and I want to understand every choice that is made to derive equations.
 
  • #10
DaveE
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These polarities can be quite arbitrary. The important thing is to keep track. In more complicated circuits, you may find it impossible to have an implicitly correct sign. For example, the current that flows from one capacitor into another. People do tend to have conventions about these assignments (like the passive sign convention, for example), but these tend to break down in practice.
 
  • #11
sophiecentaur
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The usual answers I read
How many 'usual' answers have you read? If you find a source confusing then read a few more. This is all such basic stuff and the equation you started with is all you need. It's Kirchoff's Current equation; current in minus current out at a node is zero. The reference direction for the current flow has to be the same for in and out, of course.
I still don't understand the choice of a negative I(t)
The signs are different because one is going out and one is going in. The direction is dictated by the PD across the Capacitor.
 
  • #12
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The signs are different because one is going out and one is going in. The direction is dictated by the PD across the Capacitor.
If I use KVL in the circuit I sketched, I get that the voltage across the capacitor must be the same as the voltage across the resistor [itex]V_C=V_R[/itex] hence [itex]V_C-V_R=0 \to q(t)/C - I(t) R=0[/itex]. Am I wrong?
How can I 'see' the other current you talk about and what causes it? Could you add it in my sketch, please?
 
  • #13
DaveE
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If I use KVL in the circuit I sketched, I get that the voltage across the capacitor must be the same as the voltage across the resistor [itex]V_C=V_R[/itex] hence [itex]V_C-V_R=0 \to q(t)/C - I(t) R=0[/itex]. Am I wrong?
How can I 'see' the other current you talk about and what causes it? Could you add it in my sketch, please?
Looks good to me. I think we are making this harder than it needs to be.
In my analysis of this circuit, there is only one voltage and only one current. I tend to skip the step Vc=Vr and Ic=Ir. You can arbitrarily assign voltage and current polarities, but then you have to be careful about defining the relation between V and I at each circuit element. Sometimes V=IR, sometimes V=-IR.
 
  • #14
DaveE
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OK, more on this.
The people than really do circuit analysis (like simulation software, for example) tend to assign voltages and currents (and their polarities) to the circuit meshes (or branches) without any regard to what the actual circuit components are, they are all just "some impedance"; the polarities are quite arbitrary. Then they define the relationship between voltage and current at each circuit element according to it's specific rules. That gives you a bunch of equations that your circuit simulation software will solve. When in doubt, do this for simple circuits; step by step keeping careful track of the polarities. Humans tend to invent their own conventions, and they will tell you you should do it their way, but that doesn't matter, you just need to do it carefully which ever way you choose.
For really complicated circuits, the humans use other techniques to simplify the meshes/nodes since we aren't great at solving problems with lots of equations. If that doesn't work you have to rely on linear algebra or computers.
 
  • #15
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For really complicated circuits, the humans use other techniques to simplify the meshes/nodes since we aren't great at solving problems with lots of equations. If that doesn't work you have to rely on linear algebra or computers.
Thank you very much for your time and for all you wrote! One last question, could you tell me the convention used in this case to make it work, please?
 
  • #16
sophiecentaur
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Thank you very much for your time and for all you wrote! One last question, could you tell me the convention used in this case to make it work, please?
Look in any text book and see how it's done. As @DaveE says, you seem to be over thinking this. If one terminal of the C is + then that defines the direction of the (conventional - of course) current. There's nothing more to it.
 
  • #17
DaveE
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Thank you very much for your time and for all you wrote! One last question, could you tell me the convention used in this case to make it work, please?
You will learn these as you continue in your studies. Basically you can look up Thevenin's theorem on the web, that is most of it.
The other more complex tool is to learn to do approximate solutions by ignoring features in the circuit that you don't care about; For example, splitting the problem into different models for different frequency ranges. If you only care about audio frequencies, you can (often, not always) ignore the stuff that happens above 10MHz. This, however, is more in the realm of practical engineering and is rarely covered in textbooks or tests.

One of my favorite quotes, from a former professor:
"Engineering is the art of approximation" - R.D. Middlebrook
 
  • #18
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I still don't understand the choice of a negative I(t) .
There is no real reason for it. Whoever drew the figure drew i as leaving the + terminal, so i decreases q. If they had instead drawn the figure so that i was entering the + terminal then i would increase q so the sign you are confused about would be positive.

There is no fundamental physics reason why it is negative. It is purely due to how the author chose to draw and label the current and voltage in the figure.
 
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