Why is cap voltage negative in KVL for discharge circuit?

Question

I was just trying to derive the equation for a capacitor discharging through a resistor, and I've run into in a problem.

If I set up my KVL, then I would say $iR = V_c$ (where $i$ is instantaneous current). Then you just replace $i$ with $dQ/dt$ and it's a separable DE. Easy. Except you get this: $$ \frac{dQ}{dt}R=\frac{Q}{C}\\ \int_{t_0}^t{\frac{dt}{RC}}=\int_{Q_0}^Q{\frac{dQ}{Q}}\\ Q=Q_0e^{\frac{\Delta t}{RC}}\\ V_c=V_0e^{\frac{\Delta t}{RC}} $$

This is the wrong sign for the exponent on $e$.

I looked it up, and in all other derivations, the original KVL is written as $-V_c-iR=0$.

But why should this be the case (besides the fact that it works)?

Answer 1

This is a common question. The issue is that the "Q" in $i = dQ/dt$ is not the same as the $Q$ that represents the charge on the capacitor. The variable $Q$ in use here is simply the charge on the capacitor. No problem. When the capacitor discharges the quantity of charge that is introduced into the circuit after a time $\delta t$ has elapsed is $$\delta q = Q_0-Q(\delta t)$$ a positive quantity.

However, the charge on the capacitor has decreased: $$\delta Q = Q_f-Q_i = Q(\delta t) - Q_0 = -\delta q$$

That is: the current increases when the charge on the capacitor decreases. One book I saw recently gets it right at first, and then muddles it later. Watch out.

Answer 2

I recently answered a similar question here.

The ideal capacitor equation

$$i_C = C\frac{dv_C}{dt}$$

assumes the passive sign convention which means that the reference direction for $i_C$ is into the positive labelled terminal.

When you write

$$iR = v_C$$

it is necessarily the case that

$$i_C = - i$$

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To see this, assume that both positive labelled terminals are connected together (so that both negative labelled terminals are connected together).

Now, by the passive sign convention, the capacitor current $i_C$ is into the positive labelled terminal of the capacitor while the resistor current $i_R$ is into the positive labelled terminal of the resistor.

Since only the resistor and capacitor are connected there, it follows from KCL that

$$i_C + i_R = 0 \rightarrow i_C = - i_R$$

Since you've chosen $i$ to be in the direction of $i_R$, it follows that your second equation should be

$$\frac{dQ}{dt} = -i$$

which leads to the correct differential equation

$$\frac{dQ}{dt} + \frac{1}{RC}Q = 0 $$

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The quick summary is this:

• when the current $i$ in your first equation is positive, the charge $Q$ is decreasing, i.e., $\frac{dQ}{dt}$ is negative.