RL circuit current as a function of time

A series RL circuit with a constant EMF E obeys Kirchhoff's voltage law:

L di/dt + R i = E

Given R = 100 Ω, L = 0.5 H, E = 20 V, and (as usual) the current is zero at the instant of closing the switch, i(0) = 0.

Set up

0.5 di/dt + 100 i = 20 → di/dt + 200 i = 40

This is a linear equation with integrating factor e^(200t):

d/dt(i e^(200t)) = 40 e^(200t)

i e^(200t) = (40/200) e^(200t) + K = 0.2 e^(200t) + K

i = 0.2 + K e^(−200t)

Apply i(0) = 0

0 = 0.2 + K → K = −0.2

i(t) = 0.2(1 − e^(−200t)) amperes

Interpretation

• Steady-state current E/R = 20/100 = 0.2 A, reached as t → ∞.
• Time constant L/R = 0.5/100 = 0.005 s (5 ms); after about 5 time constants (25 ms) the current is within 1% of 0.2 A.