The time constant of an LR circuit is 40 ms. The circuit is connected at t = 0 and the steady-state current is found to be 2.0 A. Find the current at (a) t = 10 ms (b) t = 20 ms, (c) t = 100 ms and (d) t = 1 s.

Sol. τ = 40 ms i0 = 2 A a) t = 10 ms i = i0 (1 – e^–t/τ) = 2(1 – e^–10/40) = 2(1 – e^–1/4) = 2(1 – 0.7788) = 2(0.2211)A = 0.4422 A = 0.44 A b) t = 20 ms i = i0 (1 – e^–t/τ) = 2(1 – e^–20/40) = 2(1 – e^–1/2) = 2(1 – 0.606) = 0.7869 A = 0.79 A c) t = 100 ms i = i0 (1 – e^–t/τ) = 2(1 – e^–100/40) = 2(1 – e^–10/4) = 2(1 – 0.082) = 1.835 A =1.8 A d) t = 1 s i = i0 (1 – e^–t/τ) = 2(1 – e^-1/40x10^-3) = 2(1 – e^-10/40) = 2(1 – e^–25) = 2 × 1 = 2 A