Time Constant of RC Circuit

If you've ever seen an automatic paper cutter, you have probably wondered how the people operating these things don't ever lose a finger or a hand. Surprisingly, the answer to your question is found in the time constant of RC circuits! This makes it possible for the machine operator to flick the "on" switch and then remove their hands from the paper well before the paper cutter actually starts cutting. Keep reading to learn more about how this time delay is created by the time constant in RC circuits.

Definition of the Time Constant in an RC Circuit

To understand what the time constant of an RC circuit is, we first need to make sure we know what an RC circuit is.

Like all other electric circuits, every RC circuit you will encounter has a total resistance \(R\) and a total capacitance \(C\). Now we can define what the time constant in such a circuit is.

Let's check that the units work out. We know that capacitance is charge \(Q\) divided by voltage \(V\), and we know that resistance is voltage divided by current \(I\). Thus, the units of capacitance are \(\mathrm{\tfrac{C}{V}}\) and the units of resistance are \(\mathrm{\tfrac{V}{A}}\). Therefore, the units of the time constant are

\[\mathrm{\frac{C}{V}}\mathrm{\frac{V}{A}}=\mathrm{\frac{C}{A}}=\mathrm{\frac{A\,s}{A}}=\mathrm{s}.\]

We see that indeed the units of the time constant are units of time!

Finding the Time Constant of an RC Circuit

To find the time constant of a specific RC circuit, we need to find the circuit's equivalent total resistance and capacitance. Let's recap how we find these.

To find the equivalent total resistance \(R\) of \(n\) resistors \(R_1,\dots,R_n\) that are connected in series, we just add up their individual resistances:

\[R=\sum_{i=1}^n R_i.\]

To find the equivalent total resistance \(R\) of \(n\) resistors \(R_1,\dots,R_n\) that are connected in parallel, we take the inverse of the sum of the inverses:

\[R=\left(\sum_{i=1}^n\frac{1}{R_i}\right)^{-1}.\]

To find the equivalent total capacitance \(C\) of \(n\) capacitors \(C_1,\dots,C_n\) that are connected in series, we take the inverse of the sum of the inverses:

\[C=\left(\sum_{i=1}^n\frac{1}{C_i}\right)^{-1}.\]

To find the equivalent total capacitance \(C\) of \(n\) capacitors \(C_1,\dots,C_n\) that are connected in parallel, we just add up their individual capacitances:

\[C=\sum_{i=1}^n C_i.\]

When you can simplify circuits with these rules, substituting multiple resistors and capacitors for only one resistor and one capacitor, you have the key to finding the time constant! This is because after the simplification, you have the two magic values for \(R\) and \(C\), the equivalent total resistance and capacitance, so you can just multiply these values to get the time constant according to

\[\tau=RC.\]

Derivation of the Time Constant of an RC Circuit

To see where this time constant comes from, we look at the simplest possible circuit containing resistors and capacitors, namely a circuit containing only one resistor and only one capacitor (so no battery!), seen in the figure below.

Fig. 1 - A simple circuit containing only a capacitor and a resistor.

Let's say we start out with some nonzero voltage \(V_0\) over the capacitor with capacitance \(C\). This means that there is some charge \(Q_0\) on either side of the capacitor, and these two sides are connected to each other by the circuit containing the resistor with resistance \(R\). Thus, there will be a current from one side to the other side to the capacitor, caused by the voltage over it. This current will change the charges \(Q\) on either side of the capacitor, so it will also change the voltage! That means we want to look at the voltage \(V\) over the capacitor and the charge \(Q\) on either side of it as a function of time. The voltage over a capacitor is given by

\[V=\frac{Q}{C},\]

so the current \(I\) through the circuit is given by

\[I=\frac{V}{R}=\frac{Q}{RC}.\]

But the current is the change in charge over time, so it is actually equal to the time derivative of the charge \(Q\) on either side of the capacitor! It is important to note that the net charge on either side of the capacitor decreases with the (positive) current, so there is a minus sign in our equation:

\[\frac{\mathrm{d}Q}{\mathrm{d}t}=-I=-\frac{Q}{RC}.\]

This is a differential equation for \(Q\) as a function of time that you don't have to be able to solve, so we just state the solution here:

\[Q(t)=Q_0\mathrm{e}^{-\tfrac{t}{RC}}.\]

There we have it! The factor \(RC\) just tells us how fast this process of charge balancing of the capacitor goes. After a time of \(t=\tau=RC\), the charge on either side of the capacitor is

\[Q(\tau)=\frac{1}{\mathrm{e}}Q_0,\]

and from the equation, we see that in general after every time duration \(\tau\), the charge decreased with a factor of \(\mathrm{e}\).

With this charge decrease, according to \(V=\tfrac{Q}{C}\), the voltage over the capacitor also decreases with a factor of \(\mathrm{e}\) every time duration \(\tau\). While the resistance stays constant, the current \(I=\tfrac{V}{C}\) also experiences the same decrease. Thus, the properties of the whole circuit (charge on either side of the capacitor, current through the circuit, and voltage over the capacitor) change with a factor of \(\mathrm{e}\) every time duration \(\tau\)!

Time Constant of an RC Circuit with Battery

Fig. 2 - The same circuit but now it contains a battery that supplies a voltage.

But what about if there is a battery in the circuit, like most circuits? Well, then we can start with a capacitor with zero charge on either side: this is a capacitor over which there is no voltage. If we connect it to a battery, the voltage will transport charges to the capacitor so that a voltage over the capacitor is created over time. This voltage \(V\) will look like this over time:

\[V(t)=V_0\left(1-\mathrm{e}^{-\tfrac{t}{RC}}\right).\]

We see the same exponential dependence in this formula, but now it goes the other way: the voltage over the capacitor grows.

At \(t=0\,\mathrm{s}\), we have \(V(0\,\mathrm{s})=0\,\mathrm{V}\) as expected. There is no resistance from any charges on the capacitor, so at the start, the capacitor behaves as a "bare wire" with zero resistance. Only after the start, when charge builds on the capacitor, does it become apparent to the circuit that it is actually a capacitor! It becomes more and more difficult to add charge to the capacitor as the charge on it, and thus the electric force against the current, grows.

After a long time (a large multiple of the time constant \(\tau\)), the exponential approaches zero, and the voltage over the capacitor approaches \(V(\infty)=V_0\). The constant voltage over the capacitor also means that the charge on the plate is constant, so there is no current flowing in and out of the capacitor. That means that the capacitor behaves as a resistor with infinite resistance.

• After turning on the battery, the capacitor behaves like a bare wire with zero resistance.
• After a long time, the capacitor behaves as if it is a resistor with infinite resistance.

Time Constant of an RC Circuit from a Graph

This all means that we should be able to determine the time constant of an RC circuit if we have a graph of either the voltage over the capacitor, the charge on either side of the capacitor, or the total current through the circuit with respect to time.

Significance of the Time Constant in an RC Circuit

The fact that there is a characteristic time constant in an RC circuit is very useful. As you can see from the formulae and the graphs, there is basically a time delay in voltage over the capacitor. This time delay can be used to get a time delay in voltage over any parallel connection. This way, you can create a time delay between turning a switch and turning on a machine. This is especially useful in high-risk industries where delays can avoid injuries.