We investigate the natural response of a resistor inductor $(\text{RL})$ circuit. This derivation is similar to the RC natural response.

RL natural response circuit

An inductor’s $i$-$v$ equation is $v = \text L \,di/dt$. The voltage depends on how current is changing from moment to moment. We use concepts from calculus, specifically derivatives, to handle this dependence on time.

Written by Willy McAllister.


Contents


Where we’re headed

For a resistor-inductor $(\text{RL})$ circuit with an initial current $\text I_0$, the current diminishes exponentially,

$i(t) = \text I_0\,e^{-\text Rt/\text L}$

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 $\tau = \dfrac{\text L}{\text R}$.


The natural response of a circuit is what the circuit does “naturally” when it has some internal energy and we allow it to dissipate. It is the most basic behavior of a circuit.

Why do we study the $\text{RL}$ natural response? Because it appears any time a wire is involved in a circuit. If the wire is formed into a coil we call it an inductor. Even if it is not coiled up, every wire and every trace on a circuit board has a small inductance that might be important. Gold bond wires in an integrated circuit have enough inductance to be important in very fast circuits. There is always inductance around and always resistance nearby.

Intuitive description

To get the $\text{RL}$ circuit to do something, we call on an outside helper to create a current in the inductor. Then we disconnect the external circuit, step back, and watch what happens.

RL natural response circuit switch closed

On the right side 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,” made of current source $\text I$, resistor $\text R0$, and a switch in the closed position.

Assume the switch has been closed for a long time. The blue loop shows how the current flows in this circuit,

RL natural response circuit two current paths

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,

$v = \text{L}\,\dfrac{di}{dt}$

The current from the source is constant. The derivative of any constant is always $0$.

In derivative notation we say, $\dfrac{di}{dt} = \dfrac{d}{dt}\text I_0 = 0$.

If we put this value into the inductor equation we get,

$v = \text L\,\dfrac{di}{dt}=\text L \cdot 0$

$v = 0$

The voltage across the inductor (and therefore both resistors) is $0$. Ohm’s Law tells us a resistor with $0$ volts has $0$ current. Therefore, all of $\text I_0$ flows through the inductor.

When current in an inductor is constant (also known as DC or zero frequency) it has $0$ volts across its ends, just like an ideal wire. The inductor “looks like” a short circuit.

Summary: Before the switch opens,
Current in the inductor is $\text I_0$.
Voltage across the inductor and resistor is $0$ volts.

Initial conditions

We know what’s going on when the switch is closed. The next step is to find the $i$’s and $v$’s just after the switch opens. These are called the initial conditions. Opening the switch is an abrupt change to the circuit. What happens to current and voltage? Do they stay the same? Do they jump to some new value? We have to figure it out.

The switch opens at time $t = 0$. The $\text I$ and $\text R0$ helper circuit disconnects from the $\text{RL}$ section. We end up with two separate paths for current,

Switch opens, current loops

On the helper side, current from the source flows through $\text R0$.
On the $\text{RL}$ side, the current in $\text L$ flows through $\text R$.

The helper circuit has done its job and we won’t pay attention to it from now on.

Why do we need R0?

After the switch is thrown open, $\text R0$ provides a path for the current from the current source. It is not nice to ask a current source to drive an open circuit. The ideal current source creates an infinite voltage if it tries to drive current into an open circuit. If you simulate this circuit without $\text R0$ the simulator will be very unhappy. (For a similar reason, you should never short out an ideal voltage source, to avoid infinite current.)

Initial current

An instant after the switch opens, at $t=0^+$, what happens to the inductor current? Does it stop? Does it continue? Does it change?

Here is the most interesting property of an inductor,

The current in an inductor cannot change instantaneously.

why?

Current pushed into an inductor causes energy to be stored in a magnetic field surrounding the inductor. If the current stops, the stored energy does not vanish or go somewhere else in zero time. Stored magnetic energy returns to the circuit by continuing to push the inductor current.

Over a short time interval (like the tiny interval from before to after a switch opens) the current in an inductor is continuous—it does not suddenly jump to a different value. As the time interval approaches zero, the inductor current before and after the switch event is the same. $\text I_\text L(0^+) = \text I_\text L(0^-)$.

This behavior is predicted by the inductor equation,

$v = \text L\,\dfrac{di}{dt}$

If the current changes instantly, that implies a finite change of current, $di = \Delta i$, happening in zero time, $dt = 0$. The derivative of current is $di/dt = \Delta i/0$, or infinite. The inductor equation predicts there will be an infinite voltage across the inductor. Infinite voltage does not happen. There has to be some finite time interval $\Delta t$ to allow the energy to dissipate.

mechanical analogy

The energy stored in an inductor’s magnetic field is similar to momentum stored in a mass. If you try to stop a moving mass, its momentum does not dissipate instantly. You can’t stop a moving mass in an instant. We say: Momentum tends to sustain the motion of the mass.

A great example of a moving mass is a bicycle wheel. Pick up the front wheel of your bicycle off the ground and give it a good hard spin. Now grab it with your hand and try to stop it instantly. Your hand gets quite a jolt and the wheel does not stop right away. There is a lot of energy stored in a spinning wheel.

The inductor current just before the switch opens,

Current before switch opens

equals the inductor current just after the switch opens,

Current after switch opens

$i(0^+) = i(0^-) = \text I_0$

This is the initial condition for the $\text{RL}$ current.

Initial voltage

What happens to voltage when the switch opens? All of a sudden there is a current forced to flow in $\text R$—so the voltage jumps abruptly up to $v = \text I_0\,\text R$.

RL initial Io RL initial Vo

Inductor current can’t change suddenly, but its voltage is allowed to make sudden jumps.

Inductor voltage can change in an instant

An inductor has no problem if its voltage changes instantaneously. If you force the voltage to jump from $1$ to $10\,\text V$ the $i$-$v$ equation says the slope of current, $di/dt$, will increase by a factor of $10$. Inductor current can change slope; it just can’t jump abruptly to a new value.

In the mechanical analogy, the voltage is analogous to the force. Nothing stops you from suddenly changing how hard you push on your bicycle wheel as you spin it up.

We reasoned our way through the switch event.

What happens after a long time?

Before we look at what happens after the switch event, let’s first think about what happens in the long run. If we wait a long time, magnetic energy in the inductor emerges as current, which eventually is transformed into heat by the resistor. When all the energy has turned into heat, $i$ will be $0$, and $v$ will be $0$. This is the final state of our circuit.

$i(t)$ and $v(t)$ look like this with the long-time response added,

Current after a long time Voltage after a long time

What happens in between?

RL circuit

Now we fill in what happens between $t=0$ and “a long time from now.” Let’s make a guess. There is probably some sort of smooth curve joining the two segments. I would guess the rate of change could be higher near the beginning when the current is high. That means the power dissipation in the resistor is high so energy dissipates faster. With this intuition, we sketch in predicted curves for current and voltage,

Current sketched in Voltage sketched in

This will turn out to be a pretty good guess for the $\text{RL}$ natural response. Just with our intuition we figured out where it starts and ends, and we estimated what the current and voltage look like during the transition. We are not exactly sure how fast the curves come down, or how long “a long time” really is.

Formal derivation

We want to derive the $\text{LC}$ natural response, $\blueD i$ and $\goldD v$ as a function of time. This derivation follows the same steps as the RC natural response.

RL basic circuit

We assume $\text I_0$ is the initial current flowing in the inductor.

Model the components

$\text R$ and $\text L$ can be modeled by their characteristic $i$-$v$ equations.

The resistor is described by Ohm’s Law,

$v_\text R = i \,\text R$

The inductor is described by the inductor $i$-$v$ equation,

$v_\text L = \text L \,\dfrac{di}{dt}$

RL with voltage labels

passive sign convention

The resistor and inductor voltages are oriented using the sign convention for passive components. The current direction is indicated by the blue arrow, and the two voltages are oriented so the current flows into the positive voltage terminals of $\text R$ and $\text L$.

We are careful about this because it determines signs in equations coming up soon.

Model the circuit

Once we have a model for each component, we create a model of how they are connected with Kirchhoff’s Voltage Law. Let’s start at the top left corner of the schematic and go around counterclockwise,

$v_\text L + v_\text R = 0$

$\text L \,\dfrac{di}{dt} + i\,\text R = 0$

This is a differential equation. It models the circuit.

From here on, we refer to $v_\text R$ as just $v$.

Solve the circuit

$\text L \,\dfrac{di}{dt} + i\,\text R = 0$

This equation is called a first-order ordinary differential equation (ODE). What does this jargon mean?

  • It is a differential equation because it contains derivatives.
  • It is first-order because the highest derivative is a first derivative $({di} / {dt})$.
  • It is ordinary because there is just a single independent variable $(t)$, (as opposed to partial derivatives of multiple variables).

Now we go through the solution of an ODE step by step.

Our goal is to discover a function $i(t$) that, when plugged in, makes the ODE true. One way to come up with an $i(t)$ is to make an informed guess and try it out. We will guess a solution, just like we did with the RC natural response.

Are there other ways?

This ODE turns out to be a separable differential equation. A way to solve this separable equation is in the appendix at the end of this article. When you use this method, there is no guessing involved.

Sal goes into depth in a sequence of videos on separable differential equations.

He also has videos about guessing a solution to solve second-order differential equations. Second-order equations come up in electrical circuits when we get to $\text{LC}$ and $\text{RLC}$ circuits.

To solve the differential equation we,

  • Dream up a function for current, $i(t)$.
  • Plug the function into the differential equation.
  • Solve for constants using the initial conditions.
  • If the constants are indeed constant (not a function of time), then $i(t)$ is a winner!

Just as we did with the $\text{RC}$ circuit, let’s guess an exponential function with some adjustable parameters, $K$ and $s$,

$i(t) = Ke^{\displaystyle st}$

  • $t$ is time
  • $i(t)$ 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 $1/t$, to make sure the exponent is dimensionless

Substitute our proposed solution into the differential equation,

$\text L \dfrac{d}{dt} (K e^{\displaystyle st}) + K e^{\displaystyle st}\,\text R = 0$

The first term includes a derivative we know how to do,

$\dfrac{d}{dt} (K e^{\displaystyle st}) = sKe^{\displaystyle st}$

reminder: derivative of an exponential

$\dfrac d {dt} \,e^{\displaystyle \alpha t} = \alpha \,e^{\displaystyle \alpha t}$

Plug the derivative back into the differential equation,

$s\text L \, K e^{\displaystyle st} + \text R \, K e^{\displaystyle st} = 0$

There is a common $Ke^{st}$ term we can factor out,

$(s\text L + \text R)\, K e^{\displaystyle st} = 0$

This is what the differential equation looks like with the proposed $i(t)$.

Now we work out the two constants, $K$ and $s$, to see if we can make the equation true. There are three different ways we can make the left side equal zero.

We could set $K=0$ to get a solution. You put nothing in and get nothing out. I’m bored already.

We could make $e^{st} = 0$ to get another solution. If we make $s$ a negative number and let $t$ go to $+\infty$, then $e^{st}$ goes to zero. It means we sit around forever and wait for the current to die out. 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,

$s = -\dfrac{\text R}{\text L}$

If we pick this value for $s$, it makes our function for current look like this,

$i(t) = Ke^{-\text Rt/\text L}$

There’s one more step before we can declare victory. We have to figure out $K$, the amplitude factor. We find $K$ using the initial conditions.

The inductor had a known current at the instant the switch was flipped. We figured out the initial current up above in the intuition section. We plug in what we know about $t=0^+$, namely, the current is $i(0^+) = \text I_0$.

$i(t) = Ke^{-\text Rt/\text L}$

$\text I_0 = Ke^{-(\text R\cdot 0)/\text L}$

$\text I_0 = Ke^0 = K \cdot 1$

$K = \text I_0$

Done! We found a function and two constants and the differential equation came true. The general solution for the natural response of an $\text{RL}$ circuit is,

$\boxed{i(t) = \text I_0\,e^{-\text Rt/\text L}}$

What about the voltage? The voltage $v(t)$ falls out of Ohm’s Law,

$v = \text R \, i$

$\boxed{v(t) = \text R\,\text I_0\,e^{-\text Rt/\text L}}$

I always like to see what the equations look like,

RL current RL voltage

Before the switch closes the current is flat at $\text I_0$. The switch closes at $t=0$ and the current falls on an exponential curve until it fades to $0$.

The voltage across the inductor is $0$ before the switch closes. It makes a sharp jump at $t=0$ as soon as the current starts to change. The peak voltage depends on the initial current $\text I_0$ and the resistance, $\text R$. Voltage follows a similar exponential curve until it fades to $0$.

Compare these computed graphs to the ones we sketched earlier. The sketches have the right shape.

Current sketched in Voltage sketched in

Time constant

An exponent has to be a plain vanilla number. It isn’t allowed to have dimensions. $\text{R}/\text L$ must have units of $1/\text{time}$, so it can cancel out $t$. The reciprocal, $\text L/\text R$, has units of $\text{seconds}$, something you might not have guessed.

$\text L/\text R$ 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,

$\tau = \dfrac{\text L}{\text R}$ seconds

We can rewrite the natural response equation like this,

$i(t) = \text I_0e^{-t/\tau}$

When $t$ is equal to the time constant, the exponent becomes $-1$. The exponential term is equal to $1/e = 1/1.569$, 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 $37\%$ of its initial value. If you wait $3$ to $5$ time constants, the natural response is pretty much over.

If you make the inductor bigger, the time constant gets longer.
If you make the resistor bigger, the time constant get shorter.

This is different than the $\text{RC}$ time constant, $\tau_{\text{RC}}=\text{RC}$, which gets longer with both larger $\text R$ and $\text C$. For the $\text{RL}$, when the resistor is large energy its dissipation is high and the natural response is snuffed out rapidly. When the resistor is small the inductor current passes more freely through the resistor so energy dissipates slowly and the current circulates round and round for a long time.

The RL time constant is not R times L.

Just a small caution: Even though we call this an $\text{RL}$ circuit, the time constant is not the product $\text{RL}$. It’s $\tau_{\text{LC}} = \dfrac{\text L}{\text R}$.

For the $\text{RC}$ circuit the time constant is $\text{RC}$.

Do your best not to get trapped.

How do you remember the order of the quotient? Is $\text R$ on top, or is it $\text L$? I remember $\text L/\text R$ because “el over r” starts with hello! while “arr over el” sounds like a pirate.

Worked example

Let’s work through an example,

Worked example circuit

Problem 1
What is $\blueD i$ if the switch is closed?

$i = $________ $\text{mA}$

show answer

When the switch is closed, all the current from the current source flows up through the inductor. $i = 8 \,\text{mA}$.

Problem 2
What is $\goldD v$ if the switch is closed?

$i = $________ $\text{V}$

show answer

When the switch is closed, the derivative of the constant current, $di/dt = 0$.

The inductor equation tells us $v = \text L\,\dfrac{di}{dt} = 0 \,\text{V}$.

The switch is thrown open at $t=0$.

Worked example switch thrown open

Problem 3
What is $\blueD i$ in the inductor the instant after the switch opens?

$i = $ ________ $\text{mA}$

show answer

In the instant after the switch is thrown open, the current in the inductor is unchanged from what it was just before. $i = 8 \,\text{mA}$. The current in an inductor cannot change instantaneously.

Problem 4
What is the time constant, $\tau$?

$\tau = $________ $\text{seconds}$

show answer

The time constant for an $\text{RL}$ circuit is,

$\tau = \dfrac{\text L}{\text R}$

$\tau = \dfrac{16\,\mu\text H}{200\,\Omega} = \dfrac{16\times 10^{-6}}{200} = 8 \times 10^{-8}\,\text{seconds}$

$\tau = 80\,\text{ns}$

Problem 5
Write expressions for $i(t)$ and $v(t)$ after $t=0$.

$i(t) = $ ________________

$v(t) = $ ________________

show answer

$i(t) = \text I_0\,e^{-\text Rt/\text L} = \text I_0\,e^{-\text t/\tau}$

$\tau = \dfrac{\text L}{\text R} = \dfrac{16\times 10^{-6}}{200} = 80\,\text{ns}$

$i(t) = 0.008\,e^{-t/80\,\text{ns}}$

$ $

$v(t) = \text R \cdot \text I_0\,e^{-\text Rt/\text L} = \text R \cdot \text I_0\,e^{-\text t/\tau}$

$v(t) = 200 \cdot 0.008\,e^{-t/80\,\text{ns}}$

$v(t) = 1.6\,e^{-t/80\,\text{ns}}$

The natural response looks like this,

RL natural response current RL natural response voltage

Simulation model

Try this simulation model. Click on TRAN to perform a transient analysis. The current source has its own internal switch. It starts at $8\,\text{mA}$ and steps abruptly down to $0$ at $t=0$. Confirm for yourself the current is down to $37\%$ of its $t(0)$ value after one time constant.

Summary

The natural response of an $\text{RL}$ circuit is an exponential,

$i(t) = \text I_0\,e^{-\text Rt/\text L}$

where $\text I_0\,$ is the inductor current at time $t=0$.

The time constant for an $\text{RL}$ circuit is $\tau = \dfrac{\text L}{\text R}$ seconds.

The current can be expressed with $\tau$ notation as,

$i(t) = \text I_0\,e^{-\text t/\tau}$

Appendix - Separable differential equation

The differential equation we derived for the $\text{LC}$ circuit is,

$\text L \dfrac{di}{dt} + i\,\text R = 0$

This is a separable differential equation. A differential equation is separable if it is possible to sort all the $i$’s and $di$’s to one side of the equation and get all the $dt$’s on the other side. This happens in the second step below.

If you have covered this technique in your calculus studies, you can solve both the $\text{RL}$ and $\text{RC}$ first-order differential equations with this method, without guessing a solution.

$\begin{aligned} \text L \frac{di}{dt} &= -i\,\text R \\ \\ \text L \frac{di}{i} &= -\text R \,dt \\ \\ \int_0^{t} \text L \frac{di}{i} &= -\int_0^t \text R \,dt \\ \\ \text L\,(\ln i(t) - \ln i(0)) &= -\text R\,t \\ \\ \text L\,\ln(i(t)/\text I_0) &= -\text R\,t \\ \\ \ln(i(t)/\text I_0) &= -\text R\,t/\text L \\ \\ i(t)/\text I_0 &= e^{-\text Rt/\text L} \\ \\ i(t) &= \text I_0\,e^{-\text Rt/\text L} \end{aligned}$

This is the same result we came up with in the main article by guessing a solution. Sal has a sequence of videos where he solves this type of separable differential equation.