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.

#### Solution

Given:-

Time constant of the given LR circuit, τ = 40 ms

Steady-state current in the circuit, i_{0} = 2 A

(a) Current at time t = 10 ms:

i = i_{0}(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 = i_{0}(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 = i_{0}(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 = i_{0}(1 − e^{−t}^{/τ})

= 2(1 − e^{−1000}^{/40})

= 2(1 − e^{−100}^{/4})

= 2(1 − e^{−25})

= 2 × 1 A

= 2 A