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# 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. - Physics

Sum

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.

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#### Solution

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)
= 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)
= 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)
= 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)
= 2(1 − e−1000/40)
= 2(1 − e−100/4)
= 2(1 − e−25)
= 2 × 1 A
= 2 A

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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 16 Electromagnetic Induction
Q 74 | Page 312
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