- #1

- 261

- 26

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

[tex]\frac{q}{C}+IR=0[/tex]

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

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 [itex]\dot q[/itex]. Perhaps this can be done

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

[tex]\frac{q}{C}+IR=0[/tex]

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

- The
*charging*of the RC circuit has been already discussed. One starts with the same equation, where the generator term is removed [itex]{\cal E}=0[/itex]. 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. - 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: [itex]q/C-IR=0[/itex]. 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 [itex]I=-\dot q[/itex], as opposed to the naive choice [itex]I=-\dot q[/itex]. That gives once again the expected decreasing exponential function.
- You forget about Kirchhoff and reason in terms of energy. The capacitor stores an energy [itex]E_C=q^2/C[/itex] which is dissipated by the capacitor at a rate [itex]P=I^2 R[/itex]. Since it is dissipation, the resulting equation is [itex]\dot E_C=-I^2 R[/itex], 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 [itex]\dot q[/itex]. Perhaps this can be done

*algorithmically*by assigning an arbitrary sign to the currents? Perhaps positive for anticlockwise and negative for clockwise currents?
Last edited: