Lesson 10 of 15
RL Step Response
The RL Circuit: Step Response
An inductor resists changes in current — the dual of a capacitor, which resists changes in voltage.
When a voltage Vs is suddenly applied to a series RL circuit:
Vs ---[R]---[L]--- GND
The current builds up exponentially:
I(t) = (Vs / R) · (1 − e^(−t / τ))
Where τ = L / R is the time constant (seconds).
The Duality with RC
| RC circuit | RL circuit |
|---|---|
| Voltage V(t) builds up | Current I(t) builds up |
| Final value: Vs (open cap) | Final value: Vs/R (short inductor) |
| τ = R·C | τ = L/R |
| Capacitor blocks DC (steady state) | Inductor passes DC (steady state) |
Time Constant τ = L/R
At t = τ: I = (Vs/R) · (1 − 1/e) ≈ 63.2% of final current
| t | I(t) |
|---|---|
| 0 | 0 |
| τ | 63.2% of Vs/R |
| 5τ | 99.3% of Vs/R |
Physical Intuition
The inductor's voltage is V_L = L · dI/dt. Initially, current is zero and V_L = Vs (all voltage across inductor). As current builds up, the voltage drop across R increases, reducing V_L, which slows the rate of change of current.
Applications
- Motor drivers (inductors are motor windings)
- Switching power supplies
- RF filters and chokes
- Flyback converters
Your Task
Implement double rl_step(double vs, double r, double l, double t) that returns the current I(t) for a series RL circuit with step voltage Vs.
Use #include <math.h> for exp.
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