RC time constant

Simple RC Circuits

RC time constant

Consider a capacitor initially charged to V_init, and connected to a resistor:

Here we will apply the Kirchoff Voltage Law (described in the next chapter) for this loop (this is similar to the Loop Law we used in Hydraulic systems): V_r+V_c=0 or V_r=-V_c

Because the same current runs through each element (current is a through variable): i_r = i_c.

One of the homework problems which may be assigned will let you explore this discharge numerically. Here, we'll do it analytically.

Using the constitutive law of the capacitor and resistor

V_c = 1/C q_c
V_r = R i_r

(5)

and the relation between charge and current, q_c' = i_c , together with the voltage and current relations next to the figure above, we can obtain a differential equation of motion

q_c' = - (1/RC) q_c

(6)

If you would rather see it in terms of current, take its time derivative:

i_c' = - (1/RC) i_c

(7)

Or in terms of voltage, using (5)

V_c' = - (1/RC) V_c

(8)

All of these equations of motion have an easy solution, e.g.

V_c = A e ^{- t /RC}

(9)

which you should check. You might wonder how we guessed this solution - as you will learn as you study differential equations further in the next course, a differential equation can often be recognized as belonging to a particular class of equations which has a particular type of solution. If you look at this equation, you see that it states that the derivative of something is equal to a constant times itself. And what is the derivative of itself??? A simple exponential. The exact form of the exponential can be obtained by trial and error, or by using constants and matching terms such that the original diff eq and initial/boundary conditions are satisfied. In (9), we have already obtained the constant in the exponential itself such that the differential equation is satisfied, but A can be any constant and (9) holds true. However for this solution to be valid for the problem at hand, we must choose A such that we match the stated initial condition that V_c = V_init at t=0. For that to be so, A must be V_init, so the specific solution is

V_c = V_init e ^{- t /RC}

(10)

You can see that the voltage (and the current, and the charge) all decay away with time. The decay time constant is RC. This is a result worth remembering. Whenever there's a capacitor and a resistor around, there is a time period t=RC which is characteristic of the action.

Charging up a capacitor

That was discharing a capacitor, but what happens if the capacitor is initially discharged, then placed in a circuit with a resistor and a battery? Here is such a circuit. Note that there is a switch in the circuit above the battery which will be closed at time t=0. At t=0, V_C=0.

To analyze this circuit, we will apply the Kirchoff Voltage Law that we will see in more detail in the next section. The polarities of my meters has been assigned consistently with our convention:

V_R+V_C - V_B = 0

Because the same current runs through each element (current is a through variable): i_R= i_C= i

Our constitutive laws of the two elements are:

V_C = 1/C q_C
V_R = R i_R

Again, starting with our current/charge relation for the capacitor,

q_C' = i_C = V_R/R = ( V_B- V_C)/R
q_C' = V_B/R - (1/RC) q_C

Since we are not so interested in charge per se, but more so in the voltage changing across the capacitor, we can rewrite this diffeq using the constitutive law of the capacitor again:

(11)

Again, we have a differential equation in which the derivative of something is equal to a constant times itself, plus another constant. We anticipate an exponential solution for this and the following is the solution for (11):

(12)

Again, you should check this solution to convince yourself that it indeed satisfies the diffeq (11) and the initial condition, V_C(t=0)=0.

So, what does this equation mean? Let's plot it, using V_B=9 volts, R=2 ohms, C=10F:

This is the voltage in the capacitor as a function of time. It charges up exponentially until the capacitor reaches the voltage of the battery, 9 volts. After this point in time, nothing happens!

So, if after a certain time, V_C=constant, what does that mean? This is the steady-state condition, meaning no more changes with time. Thus in steady state, V'_C= 0. Since the constitutive law of the capacitor states that V_C= 1/C q_C, which we can also write as V'_C= 1/C i_C, therefore in steady state i_C=0! In fact, if we plot the current in the system as a function of time, we see that it starts off at some initial value, then decays to zero at the same time as the voltage in the capacitor reaches 9 volts:

From these two examples, we see that RC circuits have a time constant associated with them which is R*C (convince yourself that the units of ohms*farads really equals seconds!) and that such circuits could effectively be used as timers. We also see that the capacitor allows a current through only until it has reached an equilibrium voltage, at which point the current stops.