RC Circuit

The RC circuit is composed of a resistor, a switch and a capacitor. The switch controls flow from the battery to the resistor and capacitor that are wired in series.

1). Turn circuit “on” by switching S to the “a” position. 2). Write potential for each element 3). Write loop rule paying attention to signs.

$$EM{F}_R=ir=R\frac{dq}{dt}$$ $$EM{F}_C=\frac{I}{C}\int idt=\frac{q}{c}$$

Using the above two we can use the loop rule, as depicted by the purple rectangle in the figure above. This yields the below equation.

$$\mathcal{E}-iR-\frac{q}{c}=0$$

Rewriting the above equation in terms of charge instead of current yields the below:

$$\mathcal{E}-r\frac{dq}{dt}-\frac{q}{c}=0$$

Then we can integrate:

$${\int }_Q^{Q_f}\frac{dq}{{\mathcal{E}}_c-q}={\int }_{ti}^{tf}\frac{dt}{RC}$$

Letting $u={\mathcal{E}}_c-q$ so $dq=-du$

$$-{\int }_{{\mathcal{E}}_c-Qi}^{\mathcal{E}c-Qf}\frac{du}{u}=-\ln (-\frac{{\mathcal{E}}_c-Q_f}{{\mathcal{E}}_c-Q_i})=\frac{t_f-t_i}{RC}+-\frac{{\mathcal{E}}_c-Q_f}{{\mathcal{E}}_c-Q_i}=e^{-(t_f-t_i)/RC}$$

Initial conditions: Uncharged capacitor at $t=0$

$$\frac{{\mathcal{E}}_c-Q_f}{{\mathcal{E}}_c}=e^{-t/RC}$$ $$⟹Q(t)=\mathcal{E}C(1-e^{-t/RC})$$

All of the above was about turning the circuit “on”. Now what if we turn the circuit “off”?

We can redo the loop rule again with the switch in the B position after the capacitor is charged:

$$-ir-\frac{q}{c}=0$$

or

$$\frac{dq}{dt}R+\frac{q}{c}=0$$ $$\frac{dq}{q}=-\frac{I}{RC}dt$$ $$⟹\ln (Q(t)-\ln (Q(0)))=-\frac{t}{RC}$$ $$Q(t)=Q(0)e^{-t/RC}$$