RL Circuit Step Response: First-Order ODE

Hello! Welcome to the next lesson in our radioelectronics course.

In the last two lessons, we thoroughly analyzed the RC circuit, first deriving its governing differential equation and then solving it to understand its transient behavior. We saw how a physical system can be translated into a mathematical model, and how solving that model allows us to predict the system's response.

Today, we shift our focus to the inductor, the other primary energy-storage element in electronics. We will apply the exact same systematic approach to a series RL circuit.

Introduction

Approximate time to complete: 30-45 minutes.

This lesson is dedicated to the learning outcome: Deriving the First-Order ODE for an RL Circuit with a Step Input. Our goal is to translate a simple circuit containing a resistor (R), an inductor (L), and a voltage source into a first-order ordinary differential equation (ODE) that describes the current flowing through it.

Recap from previous lessons:

Kirchhoff's Voltage Law (KVL): The algebraic sum of the voltage drops around any closed loop in a circuit is zero.
Resistor V-I Relationship: $v_{R} (t) = i (t) R$ (Ohm's Law).
Capacitor V-I Relationship: The key relationship was for current, $i_{C} (t) = C \frac{d v _{C} ( t )}{d t}$ , which led us to an ODE in terms of voltage $v_{C} (t)$ .

Today, we will introduce the corresponding V-I relationship for an inductor and see how it naturally leads us to an ODE in terms of current, $i (t)$ .

1. The Series RL Circuit and its Components

Let's consider a simple series circuit consisting of a DC voltage source ( $V_{s}$ ), a switch that closes at time $t = 0$ , a resistor ( $R$ ), and an inductor ( $L$ ). Before the switch closes, no current is flowing.

The key new element is the inductor. While a capacitor stores energy in an electric field and opposes changes in voltage, an inductor stores energy in a magnetic field and opposes changes in current. This property is described by its fundamental voltage-current relationship:

$v_{L} (t) = L \frac{d i ( t )}{d t}$

This equation states that the voltage across an inductor ( $v_{L}$ ) is proportional to the rate of change of the current flowing through it. If the current is constant, $\frac{d i}{d t} = 0$ and the voltage across the inductor is zero—it behaves like a simple wire (a short circuit). If the current tries to change rapidly, the inductor generates a "back EMF" or voltage to oppose that change.

2. Deriving the ODE with Kirchhoff's Voltage Law

We will use the same tool as we did for the RC circuit: KVL. The process is straightforward and is clearly explained in the following video.

Please watch the first two minutes of this video, which walks through the entire derivation.

Let's formalize the steps shown in the video (from 0:00 to 2:00):

Step 1: Apply KVL to the circuit loop.
After the switch closes at $t = 0$ , we can write the KVL equation for the loop:
$V_{s} - v_{R} (t) - v_{L} (t) = 0$
This simply states that the source voltage $V_{s}$ is equal to the sum of the voltage drops across the resistor and the inductor.

Step 2: Substitute the V-I relationships for R and L.
Now, we substitute the specific voltage drop expressions for each component:

For the resistor: $v_{R} (t) = i (t) R$
For the inductor: $v_{L} (t) = L \frac{d i ( t )}{d t}$

Plugging these into our KVL equation gives:
$V_{s} - i (t) R - L \frac{d i ( t )}{d t} = 0$

Step 3: Rearrange into Standard Form.
To make this look like a standard linear first-order ODE, we rearrange the terms, typically placing the derivative term first:
$L \frac{d i ( t )}{d t} + R i (t) = V_{s}$
This is the first-order ordinary differential equation for the series RL circuit. It perfectly captures the dynamics: the source voltage $V_{s}$ drives a response that is balanced by the voltage drops from both the resistance $R i (t)$ and the inductor's opposition to current change $L \frac{d i}{d t}$ .

3. Comparing the RL and RC Circuit Equations

It's incredibly instructive to place the ODE for the RL circuit alongside the one we derived for the RC circuit. To make the comparison direct, let's divide the RL equation by $L$ to isolate the $\frac{d i}{d t}$ term, just as we did for the RC circuit.

Feature	RC Circuit	RL Circuit
Variable of Interest	Voltage, $v_{C} (t)$	Current, $i (t)$
Governing Equation	$RC \frac{d v _{C}}{d t} + v_{C} = V_{s}$	$L \frac{d i}{d t} + R i = V_{s}$
Standard ODE Form	$\frac{d v _{C}}{d t} + \frac{1}{RC} v_{C} = \frac{V _{s}}{RC}$	$\frac{d i}{d t} + \frac{R}{L} i = \frac{V _{s}}{L}$
Time Constant ( $τ$ )	$τ = RC$	$τ = L / R$

Notice the beautiful symmetry. Both equations have the exact same mathematical structure:
$\frac{d y ( t )}{d t} + \frac{1}{τ} y (t) = K$
This is not a coincidence! It reflects a fundamental characteristic of all single-energy-storage-element linear circuits. Because the underlying math is the same, we can predict that the solution for the RL circuit will also be an exponential function, just like the RC circuit.

4. Thinking Ahead: Initial and Final Conditions

Before we solve this ODE in the next lesson, let's use our circuit intuition to determine the initial and final states of the current. This will be crucial for finding the specific solution.

Initial Condition (at $t = 0^{+}$ ): The defining property of an inductor is that it resists changes in current. Therefore, the current through an inductor cannot change instantaneously. If the current was zero before the switch was closed ( $t = 0^{-}$ ), it must still be zero at the instant right after the switch is closed ( $t = 0^{+}$ ).
$i (0^{+}) = 0$
Final Condition (as $t \to \infty$ ): After a very long time, the circuit reaches a steady state. The current becomes constant, so its rate of change is zero ( $\frac{d i}{d t} = 0$ ). Looking at the inductor's V-I relation, $v_{L} = L \frac{d i}{d t} = 0$ . A zero voltage drop means the inductor behaves like a perfect wire (a short circuit). The only component limiting the current is the resistor.
$i (\infty) = \frac{V _{s}}{R}$

These two values, along with the time constant $τ = L / R$ , are all we will need to find the complete solution in our next session.

Conclusion

In this lesson, we successfully built the mathematical model for a series RL circuit. By applying KVL and the fundamental V-I relationship for an inductor, we derived its governing first-order ODE.

Key Takeaways:

The voltage across an inductor is proportional to the rate of change of current: $v_{L} (t) = L \frac{d i ( t )}{d t}$ .
Applying KVL to a series RL circuit yields the first-order ODE: $L \frac{d i}{d t} + R i = V_{s}$ .
The standard form of this equation is $\frac{d i}{d t} + \frac{R}{L} i = \frac{V _{s}}{L}$ , revealing the circuit's time constant to be $τ = L / R$ .
The RL and RC circuit ODEs share the same mathematical structure, which implies their solutions will be similar.

Preview of the Next Lesson:

We have our equation. The next logical step is to solve it. In the upcoming lesson, "Solving the First-Order RL Circuit ODE for its Complete Response," we will use the exact same techniques we learned for the RC circuit (both the formal solution and the engineer's shortcut) to find the function $i (t)$ that describes how current builds up in the inductor over time.