Jitender Pal
Last Activity: 10 Years ago
Sol. L = 4.0 H, R = 10 Ω, E = 4 V
a) Time constant = τ = L/R = 4/10 = 0.4 s.
b) i = 0.63 i0
Now, 0.63 i0 = i0 (1 – e^–t/τ)
⇒ e–t/τ = 1 – 0.63 = 0.37
⇒ ℓ ne^–t/τ = In 0.37
⇒ –t/τ = –0.9942
⇒ t = 0.9942 x 0.4 = 0.3977 = 0.40 s.
c) i = i0 (1 – e^–t/τ)
⇒ 4/10 (1 – e^-0.4/0.4) = 0.4 x 0.6321 = 0.2528 A.
Power delivered = VI
= 4 x 0.2528 = 1.01 = 1ω.
d) Power dissipated in Joule heating =I^2R
= (0.2528)2 x 10 = 0.639 = 0.64 ω.