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.

Answer 1

Given:
Inductance, L = 4.0 H
Resistance, R = 10 Ω
Emf of the battery, E = 4 V

(a) Time constant
$\tau =\frac{L}{R}=\frac{4}{10}=0.4\mathrm{s}$

(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
⇒ $-\frac{t}{\tau }=-0.9942$
⇒ 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