# Signs in derivation of capacitor discharge differential equation

In deriving the discharge current for a capacitor I have seen two different approaches:

1. By Kirchhoff's law we have:

\begin{align} 0 &= I R + \frac{Q}{C}\\ \implies 0 &= \dot I R + \frac{1}{C} \dot Q\\ \implies RC \dot I &= -I \end{align}

Where in the last step the equation $I = \dot Q$ was used.

(Compare: Demtröder, Experimentalphysik - I am looking for an english reference too...)

1. We see the capacitor as an EMF, then:

\begin{align*} IR &= \frac{Q}{C} \\ \implies \dot IR &= \frac{1}{C} \dot Q \\ \implies RC\dot I &= -I \end{align*}

Where in the last step $I = -\dot Q$ was used since the charge on the capacitor is decreasing. (Compare for example: Jeans: The mathematical Theory of Electricity and Maganetism, page 321)

Why not $I = \dot Q$ in the second case? (But then the differential equation would be completely wrong).

For me there seems to be some "magic" sign conventions which I don't understand. So can somebody explain in detail where the signs come from in both cases and how they are connected.

I am feeling a detailed and careful explanation using the definitions of the quantities (for example $I = \int_A \vec{j} d\vec{A}$ which involves to choose the orientation of $d\vec{A}$ etc. but I don't see how to do it.

Edit:

To make it clearer. My question has two parts. One is about the actual conventions beeing used (which is addressed by Alpha-Centauri's answer), the other one is the physical reason where does the ambiguities come from physics and why are the conventions consistent in this case (I think you have to choose multiple normal vectors; for example you need to make a closed oriented surface around one capacitor plate to to define $I$, you need to define a normal vector of the cross section to define $I$ in the resistor. Both selections seem to be unrelated...). In particular how can I see in an early stage why mixing both conventions is physically wrong. Clearly it leads to the wrong differential equations which doesn't meet the experiment, but one should see this in an earlier stage when choosing degrees of freedom.

• May I suggest that you include a diagram of the circuit and your chosen convention for direction of current and charge; when you do, I think the sign convention will jump out at you. – Floris Mar 16 '15 at 19:47

By the passive sign convention, the current enters the positively labelled terminal.

For a capacitor, the voltage and current variables, assuming the passive sign convention, are related by

$$i_C = C \frac{d v_C}{dt}$$

Assuming the reference direction for the current is clockwise, KVL clockwise around the loop gives

$$v_R + v_C = 0 = \frac{d v_R}{dt} + \frac{d v_C}{dt} = \frac{d i}{dt} R + i\frac{1}{C}$$

Why not I=Q˙ in the second case?

If one treats the capacitor as a source and uses the active sign convention, the current exits the positive labelled terminal of the capacitor. In other words, the polarity of the capacitor voltage variable is reversed:

$$v'_C = - v_C$$

In that case, KVL clockwise around the loop gives

$$v_R - v'_C = 0$$

But now

$$\frac{d v'_C}{dt} = - \frac{d i}{dt} \frac{1}{C}$$

so the two cases lead to the same result.

Generally speaking, one should used the passive sign convention in which case, positive power (the product of the voltage across and current through) implies the circuit element is receiving power from the circuit while negative power implies the circuit element is supplying power to the circuit.

• How did I survive all these years without knowing this?!!! – Floris Mar 17 '15 at 1:24