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Suppose we have a capacitor of capacitance $C$ and a resistor of resistance $R$ connected in series on both sides. This circuit is called an RC circuit. Let $Q$ be the charge on the capacitor, and let $I$ be the current. Suppose that at time $t=0$, the capacitor is fully charged, i.e. $Q=Q_0$ at $t=0$. Define $V_C$ and $V_R$ to be the potential differences across the capacitor and resistor, respectively. Since the capacitor is discharging, the current actually represents how quickly the charge on the capacitor is decreasing, not increasing. Thus, $I=-\frac{dQ}{dt}$.

Let us first build an intuition for the situation. At time $t=0$, since $Q=Q_0$, we would expect the current to be at its maximum, so that the capacitor can start to discharge. However, as time goes on, we expect the current to decrease as the capacitor approaches zero charge. Ultimately, after a very long time (we can say as $t\rightarrow\infty$), it seems that $I$ and $Q$ should both approach $0$, since the charge will leave the capacitor, and thus the rate at which the charge changes should also settle down to zero.

To analyze the RC circuit mathematically, we use Kirchhoff's Loop Rule, which states that the total potential difference is zero over a complete loop: $$\frac{Q}{C}-IR=0$$ $$\frac{dQ}{dt}=-\frac{Q}{RC}$$

This is a standard differential equation, which we can solve quite easily: $$\frac{dQ}{Q}=-\frac{dt}{RC}$$ Since $Q=Q_0$ at $t=0$, we may integrate as follows: $$\int_{Q_0}^{Q}\frac{dQ}{Q}=\int_{0}^{t}-\frac{dt}{RC}$$ $$\ln\frac{Q}{Q_0}=-\frac{t}{RC}$$ $$\frac{Q}{Q_0}=e^{-t/RC}$$ $$Q=Q_0 e^{-t/RC}$$

This answer is intuitively pleasing, since it agrees with our earlier notions that $Q=Q_0$ at $t=0$ and $Q\rightarrow 0$ as $t\rightarrow\infty$. Now that we have formally worked through the calculations, we can be assured that our intuition is correct.

To find the current, we can simply differentiate: $$I=-\frac{dQ}{dt}=\frac{Q_0}{RC}e^{-t/RC}$$

This also agrees with our intuition, since $I$ starts out at its maximum value and tends toward zero as the capacitor discharges. The fact that $I$ is positive and $I=-\frac{dQ}{dt}$ is an easy check to show that the charge is indeed decreasing and not increasing.

Note the constant $RC$. This is called the time constant of the circuit, and it is often denoted $\tau$. You are encouraged to check for yourself that $\tau=RC$ indeed has dimension of time. The time constant tells us how quickly or slowly the capacitor discharges.

It's great that we have solved for all of the variables at any time $t$. However, the special times $t=0$ and $t=\infty$ still deserve some special attention.

Note that, at $t=0$, we have: $$Q=Q_0$$ $$V_C=\frac{Q_0}{C}$$ $$V_R=V_C=\frac{Q_0}{C}$$ $$I=\frac{V_R}{R}=\frac{Q_0}{RC}$$

Also note that, as $t\rightarrow\infty$, we have: $$Q=0$$ $$V_C=0$$ $$V_R=0$$ $$I=0$$

It is worthwhile to keep these special cases fresh in your mind, but of course it is more valuable to just remember how to derive these relations.