# When to use $e^{-t/RC}$ - Capacitor discharge and charge

Respected Team Members,

I am learning how to calculate the amount of time it takes to charge/discharge a capacitor.

The Formula given in the text book is $$\ V_{f}=V_{s}(1-e^{-t/RC})$$

But we are also required to use $$\ V_{f}=V_{s}(e^{-t/RC})$$

Can you please guide me when to use which ?

I have requested some fellow students and also my professor, and the answer is somewhat ambiguous. For example if t=0 and charging or discharging use the former, if t is a higher value say 40s and counting down then use the latter. Can you please guide me.

Thank you.

• Don't just memorize an answer. Investigate the math behind this problem. Assume a value for starting voltage, and see what happens for each equation when t=0. Then, choose small values of t and see what happens. Then choose successively larger values for t, and see what happens. My point? Take the time to learn the concept behind the math, and also learn a way to diagnose problems whereby you can answer some of your own questions. – David White Oct 18 '19 at 18:19

They are both in essence the same formula.

The claim is that you have a capacitor with a voltage $$V(t)$$ as a function of time, starting at some value $$V_0$$, that is in series with a resistor and in parallel with a battery at some voltage $$V_1$$.

The physics of this system is that the current through the capacitor must charge the capacitor and must also induce a voltage drop along the resistor, so that Kirchoff's voltage law says that,$$V_1 - \frac{Q}{C} - I~R = 0,$$ or taking a derivative with respect to time, $$\frac{I}{C} + R~\frac{dI}{dt} = 0.$$ This is an extremely common form of equation, one where the rate-of-change of a quantity, in this case the current, is linear in the value of that quantity. You see it for example if you are losing weight, where when you are heavier your body needs to burn more calories to simply exist and when you lose weight your body goes into a hibernation mode where it burns fewer calories to simply exist. You see it when a sink or bathtub is draining through a clog and the water is being forced through that clog by pressure but the lower the water level in the sink, the lower the pressure forcing water out through the clog. On the flip side, when you see the rate growing, you have the phenomenon of compound interest where the amount of money you earn monthly by saving money is proportional to the amount of money in your savings account. And it is solved by exponential growth or decay,$$I(t) = \alpha_1~\exp\left(\frac{-t}{RC}\right).$$ Note that the equations do not say what this arbitrary parameter $$\alpha_1$$ is, that is determined by what we call boundary conditions—you choose it to match whatever that current was at $$t=0$$, say, because $$I(0)=\alpha_1.$$ Or you can choose a different time $$\tau$$ if you compute $$I(\tau)~e^{\tau/(RC)}.$$ But the dynamics of the system do not care how exactly it started out; they tell you for any starting state how this circuit will continue.

In turn this means that the voltage $$Q/C$$ on the capacitor is $$V(t) = \frac QC = V_1 - IR = V_1 - \alpha_1~R~\exp\left(\frac{-t}{RC}\right).$$ The battery voltage $$V_1$$ specifies where the capacitor eventually “wants to go,” the arbitrary initial constant $$\alpha_1$$ specifies something about where it has “started.”

One of your equations corresponds to the case where there is no battery in the loop, which you can think of as $$V_1 = 0.$$ It also corresponds to the case where $$\alpha_1 R = -V_\text s$$ for some $$V_\text s.$$ In other words the battery capacitor has been charged initially to the value $$V_\text s$$ but it wants to go to zero because there is no battery in the loop, so it exponentially decays from $$V_\text s$$ down to $$0$$.

In your other case, you have a battery with voltage $$V_1 = V_\text s$$ in parallel but the capacitor is starting out with absolutely zero charge on it $$V(0) = 0$$, and it then exponentially decays upwards to $$V_\text s.$$

You can also consider situations that do not match either of these two. For example, maybe your capacitor starts with voltage $$2V_1$$ and decays down to voltage $$V_1.$$ That requires a new choice of $$\alpha_1,$$ namely $$V(t) = V_1 + V_1 e^{-t/(RC)}.$$All of these are valid, it just depends on what the battery voltage $$V_1$$ is and what the initial condition voltage $$V_0$$ is. You can always write this as $$V(t) = V_1 + (V_0 - V_1) e^{-t/(RC)}.$$

You can easily determine which equation to use based on how you expect the circuit to behave as well as the shape of each of your functions.

The first equation starts at $$0$$ for $$t=0$$ and then grows towards a maximum value of $$V_s$$.

The second equation starts at a maximum of $$V_s$$ and then decays towards $$0$$.

So, just look at the circuit element in question and ask yourself what it is doing. For example, in a charging RC circuit the capacitor starts with no potential difference and then grows to a maximum potential difference. Therefore, the first equation describes a charging capacitor.

For a discharging capacitor the capacitor starts at a maximum potential difference and decays to no potential difference. Therefore, the second equation describes a discharging capacitor.

The main point though is don't rely on memorization. Rely on reasoning like what I outline above. By knowing how the equations and circuits behave, you can easily determine how to tackle the problem at hand.