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Question
A solenoid having inductance 4.0 H and resistance 10 Ω is connected to a 4.0 V battery at t = 0. Find (a) the time constant, (b) the time elapsed before the current reaches 0.63 of its steady-state value, (c) the power delivered by the battery at this instant and (d) the power dissipated in Joule heating at this instant.
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Solution
Given:-
Inductance, L = 4.0 H
Resistance, R = 10 Ω
Emf of the battery, E = 4 V
(a) Time constant
\[\tau = \frac{L}{R} = \frac{4}{10} = 0 . 4 s\]
(b) As the current reaches 0.63 of its steady-state value, i = 0.63 i0.
Now,
0.63 i0 = i0(1 − e−t/τ)
⇒ e−t/τ = 1 − 0.063 = 0.37
⇒ ln e−t/τ = ln 0.37
`rArr -t/tau=-0.9942`
⇒ t = 0.942 × 0.4
= 0.3977 = 0.4 s
(c) The current in the LR circuit at an instant is given by
i = i0(1 − e−t/τ)
\[= \frac{4}{10}(1 - e^{- 0 . 4/0 . 4} )\]
= 0.4 × 0.6321
= 0.2528 A
Power delivered, P = Vi
⇒ P = 4 × 0.2528
= 1.01 = 1 W
(d) Power dissipated in Joule heating, P' = i2R
⇒ P' = (0.2258)2 × 10
= 0.639 = 0.64 W
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