# How can I graph an RC circuit?

##### 1 Answer

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**.

#DeltaV = 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 **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

#epsilon = q/C + IR#

Multiply by

#epsilonC = q + (dq)/(dt)RC#

#-(dq)/(dt)RC= q - epsilonC#

Separate the variables such that

#int_(0)^(q) 1/(q - epsilonC)dq = int_(0)^(t) -1/(RC)dt#

#ln|q - epsilonC| - ln|-epsilonC| = -t/(RC)#

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

#ln|(q - epsilonC)/(-epsilonC)| = -t/(RC)#

and then exponentiate both sides. Also note that

#(q - epsilonC)/(-epsilonC) = e^"-t/RC"#

#q - epsilonC = -epsilonCe^"-t/RC"#

#color(blue)(q(t) = epsilonC(1 - e^"-t/RC"))#

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

Then, at **the charge approaches**

Or, if you want to graph the current **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.

#color(blue)(I = (dq)/(dt)) = d/(dt)[epsilonC - epsilonCe^"-t/RC"]#

#= -epsiloncancel(C)*-1/(Rcancel(C))e^"-t/RC"#

#= color(blue)(epsilon/R e^"-t/RC")#

This one you can indeed see is exponential decay from the **the current approaches**

where