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

\[\tau = \frac{L}{R} = \frac{4}{10} = 0 . 4 s\]

(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