Question

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.

Answer 1

Given:-
Time constant of the given LR circuit, τ = 40 ms
Steady-state current in the circuit, i₀ = 2 A

(a) Current at time t = 10 ms:
i = i₀(1 − e^−t/τ)
  = 2(1 − e^−10/40)
  = 2(1 − e^−1/4)
  = 2(1 − 0.7788)
  = 0.4422 A
  = 0.44 A

(b) Current at time t = 20 ms:
i = i₀(1 − e^−t/τ)
  = 2(1 − e^−20/40)
  = 2(1 − e^−1/2)
  = 2(1 − 0.606)
  = 0.788 A
  = 0.79 A

(c) Current at t = 100 ms:
i = i₀(1 − e^−t/τ)
  = 2(1 − e^−100/40)
  = 2(1 − e^−10/4)
  = 2(1 − e^−5/2)
  = 2(1−0.082)
  =1.835 A
  = 1.8 A

(d) Current at t = 1 s:
i = i₀(1 − e^−t/τ)
  = 2(1 − e^−1000/40)
  = 2(1 − e^−100/4)
  = 2(1 − e⁻²⁵)
  = 2 × 1 A
  = 2 A