Time Constant Tau (τ): Explained with Calculations

In this post I will explain you the concept of Tau, or τ, which is the Greek letter we often see in electrical and electronic calculations.

It represents what we call the “time constant” of a circuit which is basically a measure of how the circuit behaves over time.

Now, when we talk about a circuit’s time constant and transient response, what exactly do we mean by that?

As we all know that circuits, whether electrical or electronic, are not always in a perfectly stable or steady condition. They are dynamic by nature and can be hit with sudden changes, like shifts in voltage levels or alterations in input conditions. These changes might come from something as simple as flipping a switch or activating a sensor.

But here is the thing: circuits do not just toggle instantly into their new state the moment there is a change. They actually need a bit of time to adjust, especially if they contain reactive components like capacitors or inductors. These components slow things down causing the circuit to transition more gradually rather than instantly.

Now how long this transition takes, that is what we refer to as the “time constant” of the circuit.

It gives us an idea of how fast the circuit can move from one stable state to another. The rate of this change is not linear but follows an exponential pattern.

So basically, the time constant tells us how the currents and voltages in the circuit are adjusting over a certain period of time when there is a sudden change in the circuit’s condition.

OK, so let us take a quick revision through what we have learned so far. We know that when a steady DC voltage is applied to a circuit, each component behaves in its own particular way.

A capacitor for example acts like an open circuit—no current flows through it. An inductor on the other hand behaves as a short circuit allowing current to flow freely.

And then we have the trusty resistor which serves as a current-limiting device making sure things do not get out of control.

Now here is something important to consider: when the voltage across a capacitor or the current through an inductor cannot change in an instant, then how exactly do they respond when there is a sudden step change in the circuit?

What happens to these components when they are hit with a shift in conditions?

Before we investigate deeper into any sort of transient analysis of a circuit with a capacitor, it is probably a good idea to remind ourselves about the voltage-current characteristics of a basic resistive circuit. So let us refresh our memory on that part before we move forward.

Analyzing a Resistive Circuit with only a Resistor

Referring to the above diagram when we have the switch in position S2 the 12 Ω resistor is effectively bypassed, or “shorted” which means it is disconnected from the 12-volt supply. Because of this no current flows through the resistor at all, making the current through the resistor IR equal to zero (IR = 0).

However as soon as we move the switch over to position S1 at time t = 0, a step voltage of 12 volts is immediately applied across that same 12 Ω resistor. This causes a current of 1 ampere flowing through the now closed circuit, thanks to Ohm’s Law (I = V/R).

Since the resistor has a fixed, non-inductive value, so the current does not slowly ramp up, it jumps straight from zero to 1 ampere instantly within a fraction of a second, as soon as the switch flips to position S1.

And in the same way, if we switch it back to position S2, the 12-volt supply is removed and the current drops right back down to zero just as quickly, as shown in the graph.

So what is happening here? Well, in a purely resistive circuit, the electrical state changes from one condition to another almost instantly. There is nothing within the circuit to oppose or slow down this change meaning resistors do not create any sort of time delay in the circuit’s response.

Resistors simply limit the amount of current flowing around the circuit to the value determined by Ohm’s Law (V/R) and because of this there is no time constant or transient response involved.

But things start to get more interesting when we introduce capacitors in the circuit. An RC circuit for instance is one that includes both resistances and capacitors. So now let us take a closer look at how this combination behaves when it is exposed to a step voltage change, just like before.

What would the voltage-current or v-i characteristics of this setup look like? That is where the transient response comes into play, and it is a different story from what we see with resistors alone.

Understanding Time-Constant of an RC Circuit

Let us take a closer look at the time constant of an RC circuit is?

We have already discussed how a resistor responds to any change in voltage, it reacts instantly without any delay. But since a resistor is a passive linear device, it does not have the ability to store energy.

Instead, it simply dissipates energy as heat which is why resistors tend to get hot when current flows through them.

Now, capacitors are quite different. A capacitor is made up of two electrically conducting plates called as electrodes, that are separated by an insulating material known as the dielectric.

What is fascinating about capacitors is that they can store electrical energy in the form of an electrostatic charge (measured in coulombs, or Q). So, in essence, capacitors have the unique ability to hold onto electrical energy.

Because of this, a capacitor does not behave like a resistor. Unlike resistors, the capacitors cannot instantly respond to sudden or step changes in the applied voltage.

This means that there will always be a brief period right after the voltage is applied where the current in the circuit and the voltage across the capacitor need some time to adjust.

In simple terms, it takes a little bit of time for the capacitor to either store or release energy within its electric field, depending on whether it is charging or discharging.

Now, how long this response takes is determined by something we call the “time constant” which is represented by the product of the resistance (R) and the capacitance (C) in the circuit.

The result, which is measured in seconds (s), tells us how quickly the circuit will respond.

The formula for the current through the capacitor looks like this: i_C = C(dv/dt), where “dv” represents the change in voltage and “dt” represents the change in time.

In short, when we combine a resistor and a capacitor in an RC circuit, we need to account for the fact that the capacitor introduces a delay in the circuit’s response.

The time constant helps us understand just how much time is needed for the circuit to adjust to changes in voltage.

Analyzing the following example RC circuit will make things clearer:

Analysis of an RC Circuit

Referring to the above RC circuit, when we switch the position of S2 for a little while, the combination of the resistor and capacitor gets shorted.

This means that, it is now not connected to the supply voltage V_S.

Because of this disconnection, absolutely no current flows through the circuit.

In simpler terms we can say that the current denoted as I equals zero (I = 0), and the voltage across the capacitor V_C is also zero.

Now, let us imagine that at time t = 0, we decide to move the switch over to position S1. At this very moment, a step voltage, which we will call V, is applied to our RC circuit.

It is important to note that at this instant, since the capacitor has been fully discharged prior to this moment, it acts like a short circuit.

This behavior occurs because of the sudden change in voltage over time (we call this dv/dt) right when we close the switch to position S1.

This sudden change triggers an increase in circuit current.

The only thing that limits this current is the resistance present in the circuit just like it was before.

Therefore when we initially close switch S1 at t = 0, the current flowing through our closed circuit can be approximated by V_R/R amperes, since V_R = I * R and V_C = 0 at that moment.

At that same moment when we flip the switch to position S1 and current begins to flow, our previously discharged capacitor now starts charging up.

Now it is trying to store electric charge on its plates. As a result of this charging process, the voltage across the capacitor plates, denoted as V_C, begins to gradually rise.

In the meantime we notice that the circuit current starts to decrease at a rate that is determined by what we call the time constant τ (tau) of our RC combination.

To put it all together, we can define how voltage grows across the capacitor’s plates (V_C) starting from t = 0 using the following relationship:

i = C(dV/dt) = (V – V_C)/R.

From this equation we can rearrange things a bit:

V = V_C + iR and also V = V_C + [C(dv/dt) * R].

This leads us to an expression where we have:

-dv/(V – V_C) = -dt/RC.

If we go ahead and integrate both sides of this equation, we arrive at:

log_e[(V – V_C)/V] = -t/RC = (V – V_C)/V = e^(-t/RC).

So, finally we find that:

V_C = V * [1 – e^(-t/RC)]

In the above equation,

V_C indicates the voltage across the capacitor

V becomes the supply voltage

e denotes the base of Natural Logs

t represents the time duration since the switch closed

RC indicates the time constant tau of the RC circuit

This means that as time goes on, there is an exponential natural growth of voltage across our capacitor as it works hard to store charge on its plates.

Table Depicting Exponential Increase in Capacitor Voltage Over Time

Through the following table, we can display the exponential rate of increase in the voltage across the capacitor over time, assuming normalized values for an RC time constant of one (1) and a supply voltage of one volt.

Time (s)	0.5	0.7	1	2	3	4	5	6
Voltage (VC)	0.393V	0.503V	0.632V	0.864V	0.950V	0.981V	0.993V	0.997V

As we take a closer look at the table we have presented above, it becomes quite evident that the values for VC which we can express as VC = V(1 – e^(-t/τ)), are steadily increasing over time, specifically from t = 0 all the way to t = 6 seconds (which we refer to as 6T) in our example.

What this really means is that the voltage across the capacitor behaves like an exponentially increasing function. As time (t) progresses, the term e^(-t/τ) becomes smaller and smaller.

Consequently this causes the voltage across the capacitor VC growing larger and approaching that of the supply voltage that is driving this entire change.

To break it down a bit more, at the very beginning when t = 0, the value of our function is zero. However as time continues to stretch out towards infinity, we reach a point where t equals RC.

At this specific instant, when we calculate 1 – e^(-1), we find that it produces a value of approximately 0.632 or in other words, 63.2% of its final steady-state value. This is an important data to understand how our capacitor charges.

Now let us investigate more, what this means for an exponentially increasing function. The time constant which we denote as Tau (τ) is defined as the amount of time it takes for this function to reach that crucial 63.2% of its final steady-state value, starting from when time is equal to zero (t = 0).

To put it simply, for every single time interval that corresponds to tau (τ), we see that the voltage across the capacitor increases by a factor of e^(-1) of its previous value. It is also worth noting that the smaller the time constant τ is, the faster we observe the rate of change in voltage.

How Voltage across the Capacitor Grows Exponentially with Time

To help visualize all this information more clearly, we can illustrate how the voltage across the capacitor varies with respect to time in a graphical format as follows:

So, when we consider the transient response of a capacitor when it is subjected to a step input, we can clearly see that this response is neither instant nor linear.

Instead, it actually increases in an exponential manner.

The rate at which this increase occurs is determined by the time constant of the RC circuit which we often refer to as Tau (τ).

It is interesting to note that one time constant corresponds to a factor of 1 – e^(-1) which equals approximately 0.6321.

Now, it is important understand that just knowing Tau on its own does not give us a complete picture of how long it takes for the capacitor to become fully charged. In fact, due to the nature of its exponentially increasing transient curve, we can theoretically conclude that a capacitor never truly reaches 100% charge.

However we can see that after a time period that is equal to or greater than five times the time constant—this means 5τ or 5RC—after the initial change in condition occurs, the exponential growth of voltage across the capacitor has slowed down significantly.

At this point, it has dropped to less than 1% of its maximum value.

For most practical applications, we can comfortably state that the capacitor has effectively reached its final state or steady-state condition where no further changes take place over time. In simpler terms we can say that at around 5T, the capacitor is considered to be “fully charged.”

Calculating Tau (τ) of a Practical RC Circuit

We have an RC series circuit with a resistor of 39 Ω and capacitance of 220 µF. We want to calculate the time constant tau of the circuit. We also want to solve to find out the amount of time the capacitor takes to get completely or fully charged.

As per our Time Constant formula, τ = RC

So, τ = RC = 39 x 220 x 10^-6 = 0.00858 seconds or 8.5 ms

Next when it comes to how long it takes for the capacitor to become fully charged, we typically consider that a capacitor is considered “fully charged” after approximately five time constants. Therefore we can calculate this time as follows:

5T = 5τ = 5RC = 5 x 39 x 220 x 10^-6 = 0.0429 seconds

Solving an RC Circuit to Calculate the Growing Charge across the Capacitor

We have an RC circuit with a resistor of 39 Ω and a capacitor of 220 µF which are connected in series. Let us imagine the capacitor has been already discharged fully. We want to calculate after how much time time the voltage across the capacitors plates would grow up to 50% of its final steady state value once we initiate the charging process of the capacitor.

WE have the following data in hand:

R = 39 Ω,

C = 220 µF,

The “t” will be our time of the capacitor after which the charge level across its plates reaches 50%, which is equal to 0.5 V.

From our previous analysis, we can now use the following formula for calculating the time after which the capacitor plates will accumulate the 0.5 V:

V_C = V * [1 – e^(-t/RC)]

∴ 0.5V = V * [1 – e^{-(1/39 * 220 * 10^-6 )t}]

0.5V = V * [1 – e^-116.55t]

Cancelling V on both sides, we get:

(1 – 0.5) = e^-116.55t

0.5 = e^-116.55t

log_e(0.5) = -0.693 = -116.55t

∴ t = 0.693/116.55 = 0.0059 seconds or 5.9 ms

When we are thinking how long it takes for the voltage across a capacitor to reach 50% of its final steady-state value, we are actually looking at a particular time in the charging process.

In the above case, with a time constant (τ, or tau) of 8.5 milliseconds, it will take about 5.9 milliseconds for the voltage to reach half of its fully charged value.

Now, we also know that the capacitor will reach its full steady-state condition after approximately 5 time constants (5T) which equals about 42.9 milliseconds for the above case.

So by now you might exactly know what the time constant of a series RC circuit means.

The time constant is the time it takes for the voltage across the capacitor to reach 0.632V or roughly 63.2% of its maximum possible value V after one time constant (1T).

We can calculate this by solving the product of the resistance (R) and the capacitance (C) in the circuit.

As already discussed, the τ (tau) represents the time constant, and it is calculated using the formula τ = RC, where R is measured in ohms, C in farads, and τ in seconds

Now lets imagine, what happens when the capacitor is already fully charged with V_C greater than 5T?

In other words, what do the voltage and current (V-I) characteristics of the capacitor look like as it discharges back down to zero volts?

Will the voltage decay follow the same kind of exponential curve that we saw when the capacitor was charging up?

The short answer is, yes it will!!

Just as the voltage across a charging capacitor increases exponentially, the voltage across a discharging capacitor decreases in a similar exponential fashion.

The decay will follow the same time constant principles where the voltage gradually drops of as the stored energy in the capacitor is released over time.

This means that even when the capacitor is discharging, the behavior is controlled by the same exponential curve and the rate of voltage decay is still decided by the RC time constant.

So regardless of whether we are charging or discharging the capacitor, the process is not instantaneous but rather gradual, which follows that familiar exponential pattern we expect in an RC circuits.

Analyzing Transient Discharge Curve of an RC Circuit

So imagine that we have a capacitor which is fully charged, and now we are about to discharge it. The process of discharging is actually pretty similar to how we charged it in the first place.

What happens here is that the DC power supply which was used to charge the capacitor earlier, gets disconnected. Instead, we hook it up to a short circuit, as shown in the diagram below:

Now before we do anything, let us assume the switch (S) has been sitting open for a while and the capacitor is all charged up with its voltage at VC, which is above a certain threshold (greater than 5T).

The moment we close the switch (S) at time t = 0, the capacitor starts discharging through the resistor. How long this discharging takes depends on the resistor’s value.

Right at the start, the voltage across the capacitor (V_C) equals the voltage across the resistor (V_R), so V_C and V_R both equal V.

From here the voltage gradually decreases or “decays,” and the rate of that decay is described by a specific formula as given below.

Understanding the Formula for the Exponential Decay of Voltage

V_(t) = V_C [e^–(1/RC)t]

So in this equation, V(t) is basically the voltage across the plates of the capacitor at any given time and VC is the starting voltage of the capacitor before it starts losing its charge. Simple, right?

Now we talked about exponential functions before when the voltage was increasing. But here we are looking at something a bit different , an exponential decay.

In this case the voltage is gradually dropping down to zero, and how quickly that happens still depends on something called the RC time constant.

The RC time constant tells us the “speed” at which the voltage decays, kind of like how fast the battery is draining.

So when we are dealing with this exponential decay, the time constant which we call tau (τ) is defined as the time it takes for the voltage to drop down to about 36.8% of the value it started with at t = 0 (right when the decay started).

If we say that τ is equal to one time constant, that means τ = RC. And if the RC circuit was fully charged and steady at time t = 0, then we can use this to predict how the voltage will decay over time, as shown below:

V_(t) = V_C [e^–(1/RC)t]

V_(t) = V_C [e^-1] = V_C(1/e)

V_(t) = V_C (0.368)

∴ V_(t) = 0.368V_C

So, at the very beginning when time t = 0, the voltage is at its maximum value.

But as time goes on, heading towards infinity (which is just a fancy way of saying a long time), something interesting happens.

When time reaches t = RC, that is the point where the decay hits e⁻¹ which gives us a value of about 0.368 or 36.8% of the starting voltage.

This means that the voltage has dropped to just over a third of where it started.

Eventually this keeps going until it hits zero volts, meaning the capacitor is fully discharged.

To make things easier to see, we can show how the voltage decays over time with a table.

Table Showing Decay of Capacitor Voltage Over Time

In the below given table we use normalized values, assuming the supply voltage starts at 1 volt and that the RC time constant is equal to one. This helps us to visualize how quickly the voltage drops as time passes.

Time (s)	0.5	0.7	1	2	3	4	5	6
Voltage (Vt)	0.607VC	0.497VC	0.368VC	0.135VC	0.049VC	0.018VC	0.007VC	0.002VC

Heres an important point to keep in mind, whether the capacitor is charging up or discharging, the time constant (tau, or τ) in an RC circuit always represents 63.2% of the change at one time constant (t = 0 to τ).

This is true whether we are watching the capacitor charge or discharge.

When it is charging, we start with 0 volts because the capacitor is fully discharged. From there the voltage grows exponentially and after one time constant (1T), it reaches 63.2% of the maximum voltage (VMAX).

Now if we flip it around, we can also think of it this way, at 1T, the voltage is still 36.8% away from reaching its final fully charged state, which happens at around 5T. So, by 5T, the capacitor is basically fully charged.

So the same concept applies when we are talking about exponential decay. If we start with a fully charged capacitor, its initial voltage is at its maximum V_C(max).

As it discharges it will drop down to 36.8% of its final steady state, which is zero volts, by the time we hit 5T (five time constants).

But we can also look at it another way, at 1T (one time constant) the voltage across the capacitor is 63.2% lower than when we started, back when the capacitor was fully charged at t = 0.

So whether we are charging or discharging, one thing is always true; after 1T, the voltage will always be 63.2% of the way toward the final steady state.

To put it simply, at 1T the voltage is 0.632V (or 63.2% of the max) and at the same time, it is still 36.8% away from reaching that final steady state, whether we are talking about the capacitor being fully charged at V_C(max) or fully discharged at 0.

Graph Showing Exponential Voltage Decay Over Time

Through the following graph, we can clearly show the decay process of the voltage with respect to time:

No doubt, we can yet again see that the rate of voltage decay over time hugely depends on the value of the RC time constant, tau.

Solving a Tau related RC Circuit Problem

Lets say we have an RC series circuit with a time constant tau of 5 ms. Let us we assume that the capacitor is fully charged up to 110V,

So now let’s calculate the following things:

The voltage developed across the capacitor plates after a time of 2 ms, 8 ms and 20 ms from the time when the capacitor started to discharge.

The time that passes until the capacitor voltage has decayed or drops to 58 V, 39 V, and 9 V.

Answers:

We already know that the voltage across a discharging capacitor can be expressed using the formula:

V_C(t) = V_C * e^–t/RC  Volts

In the problem we have RC time constant as 5ms, that means, 1/RC = 200, and we have V_C = 110V.

So the voltage across the capacitor at 2 ms:

V_C(0.002) = 110 e^–200t = 110 e^–0.4 = 110 * 0.67 = 73.7 V

The voltage across the capacitor at 8 ms:

V_C(0.008) = 110 e^–200t = 110 e^–1.6 = 110 * 0.202 = 22.22 V

The voltage across the capacitor at 20 ms:

V_C(0.02) = 110 e^–200t = 110 e^–4 = 110 * 0.018 = 1.98 V

Voltage across capacitor V_C at various time intervals starting from t = 0.

Elapsed time (t₁) when the capacitor voltage (V_C) reaches 58 volts.

58 = 110 e^-200t  

∴ -200t₁ = ln(58/110) = – 0.6400

∴ t₁ = -0.6400/-200 = 0.0032 seconds or 3.2 ms.

Elapsed time (t₂) when the capacitor voltage (V_C) reaches 39 volts.

39 = 110 e^-200t  

∴ -200t₂ = ln(39/110) = -1.0369

∴ t₂ = -1.0369/-200 = 0.0051 second or 5.1 ms

Elapsed time (t₃) when the capacitor voltage (V_C) reaches 9 volts.

9 = 110 e^-200t 

∴ -200t₃ = ln(9/110) = −2.5033

∴ t₃ = -2.5033/-200 = 0.012 seconds or 12 ms

Conclusions

In this tutorial about the Time Constant, which we also call Tau (symbolized as τ), we learned that it describes how an RC circuit reacts when there is a change in the input supply.

Specifically, it measures how long it takes for the circuit to switch from one steady state condition to another when we suddenly change the input supply.

When we charge a capacitor from an initial state of zero voltage to its final steady state voltage (which we will call V), the time it takes is defined as:

tau = RC

This means that we multiply the resistance (R) by the capacitance (C). This relationship gives us an exponential growth function for the capacitor voltage V_C, and we measure this time constant in seconds.

The cool thing is, the smaller the time constant, the faster the voltage changes!

We also discovered that for an exponential growth function, after one time constant (which we can call 1T), the voltage will be about 63.2% of its final steady state value. In other words, that is the time it takes for the voltage to reach its peak value, which happens after five time constants (5T).

For an exponentially growing function, the voltage at time t = tau can be expressed as:
V(t) = V(1 – e^{-1}) = 0.632V

Now on the other hand, for an exponentially decaying function, after one time constant (1T), the voltage is 36.8% of its final steady state value. In this case we are looking at the time it takes for the voltage to drop down to zero.

For an exponentially decaying function, the voltage at time t = tau can be expressed as:
V(t) = V(e^{-1}) = 0.368V

So no matter how we look at it, from the starting point t = 0 to tau, we will always see 63.2% of that time duration, and from 1T to 5T, we will always have 36.8% of the time duration, whether we are increasing or decreasing the voltage exponentially.

References:

RC time constant

Measuring the time constant