Complete Response of First-Order RL Circuits

Hello! Welcome back to our course on radioelectronics.

In the last lesson, we successfully derived the first-order ordinary differential equation (ODE) that models the behavior of a series RL circuit. We saw how its structure, $\frac{d i}{d t} + \frac{R}{L} i = \frac{V _{s}}{L}$ , is mathematically identical to the RC circuit's ODE, which hints that its solution will also be an exponential function.

Today, we will solve this equation to find the complete expression for the current, $i (t)$ .

Introduction

Approximate time to complete: 45 minutes.

This lesson focuses on the learning outcome: Solving the First-Order RL Circuit ODE for its Complete Response. We will take the ODE we derived previously and solve it to find the precise function describing how current builds up in the circuit over time after a DC voltage is applied.

Recap from the previous lesson:

RL Circuit ODE: $L \frac{d i}{d t} + R i = V_{s}$
Standard Form: $\frac{d i}{d t} + \frac{R}{L} i = \frac{V _{s}}{L}$
Time Constant: $τ = L / R$
Initial Condition: The current is zero just after the switch closes, $i (0^{+}) = 0$ .
Final Condition: The current reaches a steady value after a long time, $i (\infty) = V_{s} / R$ .

We will use these facts to find our solution, first through a formal mathematical approach and then with a powerful intuitive shortcut.

1. Formal Solution using an Integrating Factor

The ODE for the RL circuit is a linear first-order differential equation. As you know from your calculus background, a standard method for solving such equations is the integrating factor method. This approach systematically transforms the equation into a form that can be easily integrated.

The following video provides a clear, step-by-step walkthrough of this exact process.

Please watch the following segment. The presenter will start with the ODE we derived, put it in standard form, find the integrating factor, and solve for the general solution for the current $i (t)$ .

Now, watch this second segment where the initial condition, $i (0) = I_{0}$ , is used to find the specific constant of integration, giving us the complete response. In our case, since the current starts at zero, you can set $I_{0} = 0$ .

Let's summarize the key steps from the video (2:00 to 7:04) to formalize our understanding.

Standard Form: The first step is to get the ODE into the standard form $y^{'} + P (x) y = Q (x)$ . For our circuit, this is:
$\frac{d i}{d t} + (\frac{R}{L}) i = \frac{V _{s}}{L}$
Integrating Factor: The integrating factor $μ (t)$ is calculated as $e^{\int P (t) d t}$ .
$μ (t) = e^{\int (R / L) d t} = e^{(R / L) t}$
Multiply and Integrate: Multiplying the standard form ODE by $μ (t)$ makes the left side the result of a product rule, $\frac{d}{d t} (μ (t) \cdot i (t))$ .
$\frac{d}{d t} (i (t) \cdot e^{(R / L) t}) = \frac{V _{s}}{L} e^{(R / L) t}$
Integrating both sides with respect to $t$ yields:
$i (t) \cdot e^{(R / L) t} = \int \frac{V _{s}}{L} e^{(R / L) t} d t = \frac{V _{s}}{L} \cdot \frac{L}{R} e^{(R / L) t} + C$
$i (t) \cdot e^{(R / L) t} = \frac{V _{s}}{R} e^{(R / L) t} + C$
General Solution: To find the general solution for $i (t)$ , we divide by the integrating factor:
$i (t) = \frac{V _{s}}{R} + C e^{- (R / L) t}$
Complete Solution: Now, we apply our initial condition, $i (0) = 0$ , to solve for the constant $C$ .
$0 = \frac{V _{s}}{R} + C e^{0} ⟹ C = - \frac{V _{s}}{R}$
Substituting this back into the general solution gives us the complete response:
$i (t) = \frac{V _{s}}{R} - \frac{V _{s}}{R} e^{- (R / L) t} = \frac{V _{s}}{R} (1 - e^{- t / (L / R)})$

2. Transient vs. Steady-State Response

This final equation is very descriptive. It's composed of two distinct parts:

Steady-State Response ( $i_{ss}$ ): This is the final value the current settles to as $t \to \infty$ .
$i_{ss} = \frac{V _{s}}{R}$
This is the constant part of the solution and is determined by the external voltage source and the resistance.
Transient Response ( $i_{t r}$ ): This is the part of the solution that decays to zero over time.
$i_{t r} (t) = - \frac{V _{s}}{R} e^{- t / (L / R)}$
This is the circuit's temporary, natural response as it transitions from its initial state to its final state. Its duration is governed by the time constant $τ = L / R$ .

The complete response is the sum of these two: $i (t) = i_{ss} + i_{t r} (t)$ .

The following video provides a worked example with numerical values and gives an excellent explanation of these two terms at the end.

Watch from 6:09 to 7:14 for a clear summary of the transient and steady-state components in the context of an RL circuit.

3. The Engineer's Shortcut: An Intuitive Solution

Just as with the RC circuit, now that we understand the form of the solution, we can use an intuitive method that bypasses the formal differential equation steps. This is a powerful tool for quickly analyzing first-order circuits.

The solution for any voltage or current $y (t)$ in a first-order circuit driven by a DC source can be written as:

$y (t) = y (\infty) + [y (0^{+}) - y (\infty)] e^{- t / τ}$

This formula reads: "The value at any time $t$ is the final value, plus the difference between the initial and final values, which decays exponentially over time."

Let's apply this to our RL circuit to find $i (t)$ . From our previous lesson, we already determined all the necessary components:

Final value: $i (\infty) = V_{s} / R$
Initial value: $i (0^{+}) = 0$
Time constant: $τ = L / R$

Substituting these directly into the formula gives:

$i (t) = \frac{V _{s}}{R} + [0 - \frac{V _{s}}{R}] e^{- t / (L / R)}$

Simplifying this expression, we get:

$i (t) = \frac{V _{s}}{R} (1 - e^{- t / (L / R)})$

This is the exact same result we obtained through the formal integrating factor method, but derived almost instantly from our physical understanding of the circuit's initial state, final state, and time constant.

4. Visualizing the Response

A graph of the current $i (t)$ helps solidify these concepts. The current starts at 0 and exponentially approaches its final, steady-state value of $V_{s} / R$ .

Graph of current vs. time in an RL circuit — *Current i(t) in a series RL circuit after a DC voltage is applied. The current rises exponentially from 0 to its final value of Vs/R. The time constant τ determines the rate of this rise.*

The time constant $τ = L / R$ is a crucial parameter. It represents the time it takes for the current to reach approximately 63.2% ( $1 - e^{- 1}$ ) of its final value. After five time constants ( $5 τ$ ), the transient response has decayed to less than 1% of its initial value, and for most practical purposes, the circuit is considered to be in its steady state.

Conclusion

In this lesson, we have fully solved the first-order ODE for the series RL circuit. We've seen how both a rigorous mathematical method and a practical, intuitive shortcut lead to the same solution, reinforcing the deep connection between the physics of the circuit and the mathematics that describes it.

Key Takeaways:

The complete response for current in a DC-driven series RL circuit is $i (t) = \frac{V _{s}}{R} (1 - e^{- t / τ})$ , where $τ = L / R$ .
This response is the sum of a steady-state part ( $V_{s} / R$ ) and a transient part ( $- \frac{V _{s}}{R} e^{- t / τ}$ ).
The time constant $τ = L / R$ dictates how quickly the current rises to its final value.
The intuitive formula $y (t) = y (\infty) + [y (0^{+}) - y (\infty)] e^{- t / τ}$ is a powerful tool for quickly solving any first-order linear circuit.

Preview of the Next Lesson:

We have now mastered the analysis of first-order circuits (RC and RL), which contain a single energy-storage element. What happens when we include both a resistor, an inductor, AND a capacitor in the same circuit? In our next lesson, "Deriving the Second-Order ODE for Series and Parallel RLC Circuits," we will see that these circuits are described by second-order ODEs, which opens the door to more complex and interesting behaviors like damping and natural oscillation—the very foundation of radio filters and tuners.