ElectricalTransientsmediumCourseworkPE / Advanced

Current Rise in an R-L Circuit

A $20\ \text{V}$ source is switched onto a series circuit of $R = 100\ \Omega$ and $L = 0.50\ \text{H}$ at $t = 0$ , with zero initial current.

Find the current at $t = 10\ \text{ms}$ .

Given

V	20 V	Source voltage
R	100 ohm	Resistance
L	0.5 H	Inductance
t	10 ms	Elapsed time

Hint 1

The RL time constant is not the same expression as the RC one.

Hint 2

τ = L/R = 5 ms, so 10 ms is two time constants.

Hint 3

Current rises toward V/R: i = (V/R)(1 − e^(−t/τ)).

Worked solution — try the problem first

Time constant. For an R-L circuit it is $L/R$ , not $RL$ :

$\tau = \frac{L}{R} = \frac{0.50}{100} = 5.0\times10^{-3}\ \text{s} = 5\ \text{ms}$

Steady-state current. In the long run the inductor behaves as a short circuit:

$I_\infty = \frac{V}{R} = \frac{20}{100} = 0.20\ \text{A}$

Transient response:

$i(t) = I_\infty\left(1 - e^{-t/\tau}\right)$

At $t = 10$ ms $= 2\tau$ :

$i = 0.20\left(1 - e^{-2}\right) = 0.20(1 - 0.1353) = 0.20(0.8647)$

$\boxed{i = 0.173\ \text{A}}$

Contrast with RC. In an RC circuit $\tau = RC$ ; in an RL circuit $\tau = L/R$ . Increasing $R$ makes an RL circuit faster but an RC circuit slower — worth keeping straight, since the two are easily transposed.

Concepts:RL transientsTime constantInductor behaviour

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