# RC circuit discharging capacitor, trouble understanding If we apply loop rule while charging the circuit we get:

$$\varepsilon -iR-\frac{q}{C}=0$$

The way I understand, we follow trough the circuit just before the battery, as we move trough the circuit there is a increase in potential so $+\varepsilon$. Then we go trough the resistor, since we are moving in the direction of the current, $-iR$. In the capacitor, there is a drop in potential, so $-\frac{q}{C}$.

The problem starts for me in the part of discharging the capacitor. Now the battery acts like part of a wire, $\varepsilon=0$. The top plate of the capacitor is at a higher potential. Current will flow from the top plate to the circuit in the direction of the resistor (opposite of what it was). Now loop rule would stand for (starting from just before the down plate):

$$\frac{q}{C}-iR=0$$

Because capacitor's top plate is at a higher potential, there is a increase in voltage. Loop rule is applied in the same direction of the current, so $-iR$.

But the equation resulted in the loop rule application when discharging a capacitor is given by:

$$\frac{q}{C}+iR=0$$

Why? What is the mistake in my reasoning? Please, help me understand. Excuse my bad english. Happy 2016.

• Does it depend on your definition of the direction of current? It is different during charging and discharging. I have to admit I find your question difficult to follow... Jan 2, 2016 at 3:44

You might have omitted part of the presentation. The way I think of it, during discharge we have as you say:$$\frac{q}{C}-iR=0$$ But there is more to the analysis

$q$ in these equations specifically refers to the charge on the capacitor. Usually when we think of the relationship between current and charge we're referring to the charge in the wire, but not so here.

In the discharging case, take the positive sense of current to be counterclockwise (as you have done). Consider what is happening to the charge on the capacitor as current flows in the positive direction: it gets smaller as the capacitor discharges. So for discharging we have $$i=-\frac{\mathrm{d}q}{\mathrm{d}t}$$

resulting in $$\frac{q}{C}+R\frac{\mathrm{d}q}{\mathrm{d}t}=0$$ If you consider what happens when charging, taking the positive sense of current to be clockwise, you see that the charge on the capacitor increases when the current it positive and charging the capacitor.$$i=\frac{\mathrm{d}q}{\mathrm{d}t}$$ Perhaps that's the source of your difficulty.

When you first apply Kirchhoff's second law you do it properly and you obtain \begin{equation} \sum_k V_k = \epsilon - iR - \frac{q}{C} \end{equation}

When you apply it for the second loop you have two options, to apply it clockwise or counterclockwise.

• Clockwise \begin{equation} iR - \frac{q}{C} = 0 \end{equation}

-Counterclockwise \begin{equation} -iR + \frac{q}{C} = 0 \end{equation}

As you see, you forgot to change the voltage sign for the capacitor. In a voltage source (charged capacitor) usually the sign is positive when your loop direction goes from the negative plate to the positive one and negative the other way around.