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RL natural response

Natural response of an RL circuit. Written by Willy McAllister.
We investigate the natural response of a resistor and inductor circuit. This discussion parallels the analysis of RC circuits.
This $\text{RL}$ circuit is fairly common. It appears any time a coiled wire is involved in a circuit, such as when you drive a mechanical relay to cause a physical motion (a relay contains a coil used as an electromagnet). Inductors are found in almost every power supply and in many filters. All wires and circuit board traces have a small self-inductance, which can be important in very fast circuits.
This is a circuit where we have to account for time. To develop a precise understanding requires concepts from calculus. We use derivatives to describe how the $\text{RL}$ circuit behaves.

What we're building to

For an resistor-inductor circuit, if the inductor has an initial current ${\text{I}}_{0}$, the current will diminish exponentially according to:
$i\left(t\right)={\text{I}}_{0}\phantom{\rule{0.167em}{0ex}}{e}^{\phantom{\rule{-0.167em}{0ex}}-\text{R}t/\text{L}}$
Where ${\text{I}}_{0}$ is the current at time $t=0$. This is called the natural response.
The time constant for an $\text{RL}$ circuit is $\tau =\frac{\text{L}}{\text{R}}$.
The time constant is a measure of the steepness of the exponential. It has units of seconds.
The natural response of a circuit is what the circuit does when there are no external influences (no energy coming in). It is the most basic behavior of the circuit. When placed within a larger circuit, the natural response plays an essential part of the overall behavior.

Setting up the RL natural response

To get the $\text{RL}$ circuit to do something, we call on an outside helper to add some energy and then step back and leave it alone, while we watch what happens.
On the right side of this schematic we have inductor $\text{L}$, and resistor $\text{R}$. This is the circuit we want to study. On the left side is our "outside helper," consisting of a current source, $\text{I}$, resistor $\text{R}0$, and a switch in the closed position.
Assuming the switch has been closed for a long time, the blue loop shows how the current flows in this circuit:
How do we know all the current flows through just the inductor and no current flows in either resistor? The inductor equation tells us so:
$v=\text{L}\phantom{\rule{0.167em}{0ex}}\frac{di}{dt}$
The current from the source is constant, unchanging with time.
That means the change in current with time is $\frac{di}{dt}=0$.
If we put this value into the inductor equation we get $v=\text{L}\cdot 0=0$. The voltage across the inductor (and therefore both resistors) is $0$. Ohm's Law tells us resistors with $0$ volts have $0$ current.
When current through an inductor is constant, we say: The inductor "looks like" a short circuit because it has $0$ volts across its terminals, just like an ideal wire.

Initial conditions

Now there is a current flowing through our inductor. We throw open the switch at time $t=0$ and figure out the initial conditions.
The open switch disconnects the helper circuit $\left(\text{I},\text{R}0\right)$ from the $\text{RL}$ section. On the helper side, current $\text{I}$ starts flowing through $\text{R}0$, (the helper circuit has done its job and we won't pay attention to this part from now on). On the $\text{RL}$ side, the current flowing in $\text{L}$ instantly flips over to start flowing through $\text{R}$:

Summary of initial conditions

The instant before the switch opens, $t={0}^{-}$, the inductor has a current we'll call ${\text{I}}_{0}$, with $0$ volts across the inductor and resistor.
An instant later, at $t={0}^{+}$, the switch is open and current ${\text{I}}_{0}$ still flows in $\text{L}$ and now flows in $\text{R}$.
The current in the inductor does not, and in fact, cannot change instantaneously. So the current flowing in the inductor right after the switch opens is equal to the current just before the switch opens.
For all time in the past, up to $t={0}^{+}$, the current in the inductor is ${\text{I}}_{0}$:

RL natural response - intuitive description

Let's reason through what happens next. We want to figure out $i$ and $v$ as functions of time.
We stated above that ${\text{I}}_{0}$ is flowing in the inductor right after the switch is thrown open. What happens to the voltage?
The current in the resistor just jumped from $0$ to ${\text{I}}_{0}$, so the voltage instantly jumps up to $v\left({0}^{+}\right)={\text{I}}_{0}\phantom{\rule{0.167em}{0ex}}\text{R}$.
Now we know both the current and voltage just after the switch opens. Let's think next about where this circuit finally ends up, after a long time passes.
A resistor (unlike an ideal inductor or capacitor), dissipates energy as heat. That heat comes from the energy stored in the inductor's magnetic field (the only energy source in our natural response circuit). If we wait a long time, all the energy that started in the inductor will eventually be transformed into heat by the resistor. When all the energy is gone, $i$ will be $0$, and $v$ will be $0$. That's the final state of our circuit.
$i\left(t\right)$ and $v\left(t\right)$ now look like this with the final response filled in:

What happens in between?

Now we fill in what happens in the time interval between $t\left({0}^{+}\right)$ and "a long time from now." For now, I'm going to guess there is some smooth connecting curve that joins the two segments of the curve. I would guess the rate of change could be higher near the beginning, when the current is high and there's a higher rate of power dissipation in the resistor. Using this intuition, I can sketch in predicted curves for current and voltage.
This turns out to be a pretty good guess at the natural response of an $\text{RL}$ circuit. Just using our intuition we figured out the starting and ending state, and made an estimate of what the current and voltage look like during the transition between start and end. We are not exactly sure how fast the curves come down, or how long "a long time" really is.
Next, we develop a precise solution, which requires us to use some calculus.

Formal derivation of the $\text{RL}$‍  natural response

We want to derive the $\text{RL}$ natural response, $i$ and $v$ as a function of time. This derivation follows the same steps as the RC natural response.
We assume an initial current of ${\text{I}}_{0}$ flows in $\text{L}$.

Model the components

The two components are modeled by their characteristic $i$-$v$ equations.
The resistor is described by Ohm's Law:
${v}_{\text{R}}=i\phantom{\rule{0.167em}{0ex}}\text{R}$
The inductor is described by the inductor $i$-$v$ equation:
${v}_{\text{L}}=\text{L}\phantom{\rule{0.167em}{0ex}}\frac{di}{dt}$

Model the circuit

We can write Kirchhoff's Voltage Law starting at the top left corner of the schematic and going around counterclockwise:
${v}_{\text{L}}+{v}_{\text{R}}=0$
$\text{L}\phantom{\rule{0.167em}{0ex}}\frac{di}{dt}+i\phantom{\rule{0.167em}{0ex}}\text{R}=0$
This is the differential equation modeling the circuit.
From here on, we refer to ${v}_{\text{R}}$ as just $v$.

Solve the circuit

The previous equation is a first-order ordinary differential equation (ODE).
We are now going to walk through the solution of an ODE. One way is to make an informed guess at a solution, and try it out. That's what we are going to do here, as we did with the analysis of the RC natural response.
To solve the differential equation we dream up a function for current, $i\left(t\right)$ and plug the function into the differential equation and see if it turns out true.
$\text{L}\phantom{\rule{0.167em}{0ex}}\frac{di}{dt}+i\phantom{\rule{0.167em}{0ex}}\text{R}=0\phantom{\rule{2em}{0ex}}$ (differential equation)
Just as we did with the $\text{RC}$ circuit, we try an exponential function with some adjustable parameters, $K$ and $s$.
$i\left(t\right)=K\phantom{\rule{0.167em}{0ex}}{e}^{st}$
• $t$ is time
• $i\left(t\right)$ is current as a function of time
• $K$ and $s$ are constants we have to figure out
• $K$ is an amplitude term that scales current up or down
• $s$ must have units of $1/t$ so we get a unit-less exponent.
Substitute our proposed solution into the differential equation to see if it works:
$\text{L}\frac{d}{dt}\left(K{e}^{st}\right)+\text{R}\left(K{e}^{st}\right)=0$
Let's compute the derivative found in the first term:
$\frac{d}{dt}\left(K{e}^{st}\right)=sK{e}^{st}$
Plug the derivative back into the differential equation:
$s\text{L}\phantom{\rule{0.167em}{0ex}}K{e}^{st}+\text{R}\phantom{\rule{0.167em}{0ex}}K{e}^{st}=0$
Now we can factor out the common $K{e}^{st}$ term.
$\left(s\text{L}+\text{R}\right)\phantom{\rule{0.167em}{0ex}}K{e}^{st}=0$
This equation describes our specific circuit, with the proposed $i\left(t\right)$.
Now we work out the two constants, $K$ and $s$, to see if we can make the equation true.
We could set $K=0$ to get a solution. But that's pretty boring. You put nothing in, and get nothing out.
We could make ${e}^{st}=0$ to get another solution. This is boring, too. If we make $s$ a negative number and let $t$ go to $+\mathrm{\infty }$, it means we sit around forever and wait for the current to die out to zero. Snooze.
The third way we can make the equation true is to set $s\text{L}+\text{R}=0$. This becomes interesting. This is true if:
$s=-\frac{\text{R}}{\text{L}}$
This solves for $s$, and makes our function for current look like this:
$i\left(t\right)=K\phantom{\rule{0.167em}{0ex}}{e}^{\phantom{\rule{-0.167em}{0ex}}-\text{R}t/\text{L}}$
The last step is to figure out $K$, the amplitude factor. We do this using the initial conditions. The inductor had a known current at the instant the switch was flipped. To find $K$ we plug in everything we know about $t={0}^{+}$. The current was $i\left({0}^{+}\right)={\text{I}}_{0}$.
$i\left(0\right)={\text{I}}_{0}=K\phantom{\rule{0.167em}{0ex}}{e}^{\phantom{\rule{-0.167em}{0ex}}-\text{R}\cdot 0/\text{L}}$
${\text{I}}_{0}=K\phantom{\rule{0.167em}{0ex}}{e}^{\phantom{\rule{-0.167em}{0ex}}0}$
$K={\text{I}}_{0}$
All done! We found a function and two constants that made the differential equation true. We've solved the current for all time after the switch is opened.
The general solution for the natural response of an $\text{RL}$ circuit is,
$i\left(t\right)={\text{I}}_{0}\phantom{\rule{0.167em}{0ex}}{e}^{\phantom{\rule{-0.167em}{0ex}}-\text{R}t/\text{L}}$
We can get voltage $v\left(t\right)$ straight from Ohm's Law:
$v\left(t\right)=\text{R}\cdot i\left(t\right)$
$v\left(t\right)=\text{R}\phantom{\rule{0.167em}{0ex}}{\text{I}}_{0}\phantom{\rule{0.167em}{0ex}}{e}^{\phantom{\rule{-0.167em}{0ex}}-\text{R}t/\text{L}}$

The $\text{RL}$‍  natural response looks like this

These graphs show the shape of the $\text{RL}$ natural response. For $t\le 0$ the current is ${\text{I}}_{0}$. After $t=0$ the current falls on an exponential curve until it becomes $0$. The rate of change (the slope) is greatest at the beginning, when the current is highest. The ratio $\text{R}/\text{L}$ determines the steepness of the exponential response.
The voltage across the inductor is flat at $0$ for $t\le 0$, and makes a sharp jump at $t=0$ when the current starts to change. The peak voltage depends on the starting current ${\text{I}}_{0}$ and the resistance, $\text{R}$ (and, oddly, does not depend on the value of the inductor $\text{L}$). Voltage follows a similar exponential curve downward until it fades to $0$.
Compare these computed graphs to the ones we sketched earlier. The sketches have the right shape.

Time Constant of a resistor-inductor combination

An exponent has to be a plain number, it can't have dimensions. That means the quotient $\text{R}/\text{L}$ has to have units of $1/\text{time}$, so it can cancel out $t$. That means $\text{L}/\text{R}$ has units of $\text{seconds}$.
$\text{L}/\text{R}$ is called the time constant of a resistor-inductor combination. It has the same properties as the corresponding product $\text{R}\cdot \text{C}$ in the resistor-capacitor circuit. We use the Greek letter $\tau$ (tau) as the symbol for time constant. For a resistor-inductor pair:
$\tau =\frac{\text{L}}{\text{R}}$
The time constant for an inductor and resistor gets longer with a larger inductor, and shorter with a larger resistor (in contrast to the $\text{RC}$ time constant, which gets longer with both bigger $\text{C}$ and $\text{R}$.)
Using $\tau$, we can write the natural response equation like this:
$i\left(t\right)={\text{I}}_{0}\phantom{\rule{0.167em}{0ex}}{e}^{-t/\tau }$
When $t$ is equal to the time constant, the exponent of $e$ becomes $-1$, and the exponential term is equal to $1/e$, or about $0.37$. The time constant determines how fast the exponential curve comes down to zero. After $1$ time constant has passed, the current is down to $37\mathrm{%}$ of its initial value.

$\text{RL}$‍  natural response example

Let's do an example together. For this circuit:
Problem 1
What is $i$ if the switch is closed?
$i=$
$\text{mA}$

Problem 2
What is $v$ if the switch is closed?
$v=$
$\text{V}$

The switch is thrown open at $t=0$.
Problem 3
What is $i$ in the inductor the instant after the switch is thrown open?
$i=$
$\text{mA}$

Problem 4
What is the time constant, $\tau$ ?
$\tau =$
$\text{seconds}$

Write expressions for $i\left(t\right)$ and $v\left(t\right)$ after $t=0$.
$i\left(t\right)=\mathrm{_}\mathrm{_}\mathrm{_}\mathrm{_}\phantom{\rule{0.167em}{0ex}},$ $v\left(t\right)=\mathrm{_}\mathrm{_}\mathrm{_}\mathrm{_}$
The natural response for the example circuit looks like this:

Summary

The natural response of an $\text{RL}$ circuit is an exponential:
$i\left(t\right)={\text{I}}_{0}\phantom{\rule{0.167em}{0ex}}{e}^{\phantom{\rule{-0.167em}{0ex}}-\text{R}t/\text{L}}$
Where ${\text{I}}_{0}$ is the current at time $t=0$.
The time constant for an $\text{RL}$ circuit is $\tau =\frac{\text{L}}{\text{R}}$.

Appendix - Solving a separable differential equation

As a reminder, the differential equation of the $\text{LC}$ circuit is:
$\text{L}\phantom{\rule{0.167em}{0ex}}\frac{di}{dt}+i\phantom{\rule{0.167em}{0ex}}\text{R}=0$
Here are the steps to solve this separable differential equation. If you have covered this technique in your calculus studies, you can solve both the $\text{RL}$ and $\text{RC}$ first-order differential equations without guessing a solution.
$\begin{array}{rl}\text{L}\frac{di}{dt}& =-i\phantom{\rule{0.167em}{0ex}}\text{R}\\ \\ \text{L}\frac{di}{i}& =-\text{R}\phantom{\rule{0.167em}{0ex}}dt\\ \\ {\int }_{0}^{t}\text{L}\frac{di}{i}& =-{\int }_{0}^{t}\text{R}\phantom{\rule{0.167em}{0ex}}dt\\ \\ \text{L}\phantom{\rule{0.167em}{0ex}}\left[\mathrm{ln}i\left(t\right)-\mathrm{ln}i\left(0\right)\right]& =-\text{R}\phantom{\rule{0.167em}{0ex}}t\\ \\ \text{L}\phantom{\rule{0.167em}{0ex}}\mathrm{ln}\left(i\left(t\right)/{\text{I}}_{0}\right)& =-\text{R}\phantom{\rule{0.167em}{0ex}}t\\ \\ \mathrm{ln}\left(i\left(t\right)/{\text{I}}_{0}\right)& =-\text{R}\phantom{\rule{0.167em}{0ex}}t/\text{L}\\ \\ i\left(t\right)/{\text{I}}_{0}& ={e}^{\phantom{\rule{-0.167em}{0ex}}-\text{R}t/\text{L}}\\ \\ i\left(t\right)& ={\text{I}}_{0}\phantom{\rule{0.167em}{0ex}}{e}^{\phantom{\rule{-0.167em}{0ex}}-\text{R}t/\text{L}}\end{array}$
This is the same result we came up with in the main article by guessing a solution.
Sal has a sequence of videos showing how to solve this type of separable differential equation.

Want to join the conversation?

• Is there any video/article related to RL step response?
• Why did we chose constant current source in RL natural response, but constant voltage source in RC natural response ? Can we do other way round ? What is reason for picking these sources in such manner ?
• For RL and RC natural response we want to start the circuit off with some initial energy. A resistor doesn't store energy, so that means we place the energy in either L or C. An inductor's energy is stored in its magnetic field, which comes back out of the inductor as a current (the magnetic field pushes charge through the wire). For a capacitor, its energy is stored in the form of charge. That stored charge causes a voltage to appear on the capacitor, as in q = C v, or v = q/C.

So in setting up to analyze natural response we pick the type of source that directly injects the energy into L or C.
• 1)could you explain this more "and, oddly, does not depend on the value of the inductor L" why does not the voltage depend on L ?

and Why is THERE not a forced response analysis like an RC circuit?
• The peak voltage value does not depend on the inductance of the inductor. This might seem non-intuitive. but it happens because 1. the math tells us so, and 2. The starting voltage depends only on the starting current I0 and the resistor value R. The inductor value has a substantial impact on the steepness of the response, but not its starting value.

I did not write up the forced step response of LR for two reasons. First, it is rare for this circuit to come up in a forced/stepped situation, and second, the analysis is an exact dual of the RC step response. The math follows the exact same steps at RC Step.
• The ideal current source produces an infinite voltage as it tries to drive current across the open circuit. (For a similar reason, you should avoid shorting an ideal voltage source, to avoid infinite current.)

Since The ideal current source produces an infinite voltage, so even the practical current source could provide a considerable level of voltage, then the R0 selection must could endure such high voltage?
• Hello Wayne,

By Ohm's law the resistor will only need to withstand I * R volts. That is 8 volts in this case.

Here is something fun for you. Let's take a 1 million volt power supply and place a 125 million Ohm resistor in series with it. This is a thought experiment – not something you would want to build. Anyway, my question to you is this. Could you tell the difference between this circuit and the 8 mA constant current source mentioned on this page?

Regards,

APD
(1 vote)
• Excuse my ignorance, but why is the KVL VL+VR=0 instead of VL-VR=0?
(1 vote)
• It is somewhat tricky to set up the KVL equation for this RL circuit (and the RC circuit as well). You have to be careful with the sign convention for passive components.

Look at the circuit in section "Model the Components". The starting point is to add the current arrow pointing up out of the inductor. That means it is flowing down through the resistor.

The second step is to add voltage labels to L and R, while respecting the sign convention for passive components. For the L, the current flowing up means the positive polarity of the voltage is at the bottom of the inductor. For the R the current is flowing down from the top, so the positive polarity of the resistor's voltage is at the top of the resistor. Admittedly, the voltage on the resistor looks "right" while the upside down voltage on the inductor seems a bit odd. That's not because it is wrong, but because it is unfamiliar.

The last step is to write KVL around the loop. I chose to start on the top node and circle down through the L and then up through the L. That results in two positive voltages being added together.
Write expressions for i(t)i(t)i, left parenthesis, t, right parenthesis and v(t)v(t)v, left parenthesis, t, right parenthesis after t=0t=0t, equals, 0.

I found that we used s in equations for i(t) and v(t)
s = R/L

But when the values were used, we don't get the answer we should from 200 ohms / 16 micro-Henrys. Instead, I think the time constant (T or Tao) was used to get 80 nano-seconds.

Should we use s times t in the exponential or the time constant?
We got s from the integration, so where does the time constant come in?
(1 vote)
• Natural frequency s = -R/L is closely related to the time constant tau = L/R.

We found a solution to the differential equation by solving for s in the equation sL + R = 0. This gave us s = -R/L and we were able to assemble a solution for i(t).

What are the units of s? If the exponent is -st, the units of s have to be 1/time. (exponents can't have units, so all the quantities in an exponent must cancel out.)

When you talk about the time constant you ask a slightly different question. Given the solution for i:

i = I_0 e^{-Rt/L}

what value of time makes the exponential term equal -1? Another way to ask the same thing is to ask, what value of time makes the exponential term equal to 1/e?

The answer is at t = L/R the exponent is -1. That special time is called the time constant and we use the Greek letter tau. What are the units of tau? Since tau is a particular value we chose for t, the units of tau are seconds.
• I am confused, if the volt is zero, then what causes the circuit to have 8 mA current? It was said the inductor equation tells us the volt is zero, then if volt is zero, then there should be no current anywhere.
(1 vote)
• The i-v equation for an inductor is v = L di/dt. That says the voltage across the inductor depends on how fast the current is changing (the di/dt term). If the current is not changing, the voltage is zero. The current could be 0 or +100 or -1000, but if it is not changing, the voltage is 0.

You can look at it like this... For the case where di/dt = 0 (the current is constant), an inductor acts exactly like an ideal wire. An ideal wire always has 0 volts between its ends, but it can certainly carry a current if it is connected to something else.

This is what is happening when the inductor is initially connected to the current source before the switch opens. The constant 8mA flows through the inductor. Because it is constant, the voltage across the inductor is 0.