# How can I graph an RC circuit?

Dec 13, 2015

If you don't remember the concept, then if you can derive the formula, it will be very easy to graph.

With a basic RC circuit, you have:

So, going clockwise, you get a change in voltage for a closed circuit to follow Kirchoff's law.

$\Delta V = 0 = \epsilon - {V}_{C} - {V}_{R}$

where $\epsilon$ is the "electromotive force" (the voltage increase through the battery), ${V}_{C}$ is the voltage drop through the capacitor as it stores charge on the parallel plates, and ${V}_{R}$ is the voltage drop through the resistor.

I recall that the capacitance $C$ in Farads can be written as $\text{C/V}$, so $C = \frac{q}{V} _ C$, where $q$ is the charge in $\text{C}$. The voltage through the resistor can be written as ${V}_{R} = I R$ from Ohm's law.

Furthermore, the current can be written as the change in charge over time, since the capacitor is going to store charge over time $\left(\frac{\mathrm{dq}}{\mathrm{dt}}\right)$. So, we have so far:

$\epsilon = \frac{q}{C} + I R$

Multiply by $C$ and plug in $I = \frac{\mathrm{dq}}{\mathrm{dt}}$:

$\epsilon C = q + \frac{\mathrm{dq}}{\mathrm{dt}} R C$

$- \frac{\mathrm{dq}}{\mathrm{dt}} R C = q - \epsilon C$

Separate the variables such that $\frac{1}{q - \epsilon C}$ is on the same side as $\mathrm{dq}$ and $- \frac{1}{R C}$ is on the same side as $\mathrm{dt}$, then begin to take the integral.

${\int}_{0}^{q} \frac{1}{q - \epsilon C} \mathrm{dq} = {\int}_{0}^{t} - \frac{1}{R C} \mathrm{dt}$

$\ln | q - \epsilon C | - \ln | - \epsilon C | = - \frac{t}{R C}$

Using the properties of logarithms, you can turn the left side into a fraction:

$\ln | \frac{q - \epsilon C}{- \epsilon C} | = - \frac{t}{R C}$

and then exponentiate both sides. Also note that $q - \epsilon C$ will be negative since $q$, the current charge, is less than $\epsilon C$, the initial charge. So, the absolute values don't matter here.

$\frac{q - \epsilon C}{- \epsilon C} = {e}^{\text{-t/RC}}$

$q - \epsilon C = - \epsilon C {e}^{\text{-t/RC}}$

$\textcolor{b l u e}{q \left(t\right) = \epsilon C \left(1 - {e}^{\text{-t/RC}}\right)}$

So you can graph the charge with respect to time as it gets stored into the capacitor using this equation. All you have to do is note that it is the vertical reflection of an exponential decay, seen as $- {e}^{- u}$.

Then, at $t = 0$, $q = 0$, while at $t \to \infty$, the charge approaches $\setminus m a t h b f \left(\epsilon C\right)$.

Or, if you want to graph the current $I$ instead as the capacitor discharges the stored electrical charge, since that is supposed to be exponential decay and is easier to visualize, you can take the derivative.

$\textcolor{b l u e}{I = \frac{\mathrm{dq}}{\mathrm{dt}}} = \frac{d}{\mathrm{dt}} \left[\epsilon C - \epsilon C {e}^{\text{-t/RC}}\right]$

$= - \epsilon \cancel{C} \cdot - \frac{1}{R \cancel{C}} {e}^{\text{-t/RC}}$

$= \textcolor{b l u e}{\frac{\epsilon}{R} {e}^{\text{-t/RC}}}$

This one you can indeed see is exponential decay from the ${e}^{- u}$ equation. At $t = 0$, $I = \frac{\epsilon}{R}$, while as $t \to \infty$, the current approaches $\setminus m a t h b f \left(\text{0 A}\right)$.

where $V$ on the diagram is the same as $\epsilon$ and $i$ is the same as the current $I$.