A 24-V battery is connected in series with a resistor and an inductor, with R = 5.6 Ω and L = 2.0 H, respectively.

Question

(a) Find the energy stored in the inductor when the current reaches its maximum value.


(b) Find the energy stored in the inductor one time constant after the switch is closed.

Answer 1

V = iR + L di/dt

i = a(1-e^-kt)
for large t
i =24/5.6 = a
so
a = 4.29
i = 4.29(1-e^-kt)
di/dt = 4.29 k e^-kt
24 = 24-24e^-kt + 2(4.29)k e^-kt
24 = 2(4.29) k
k = 24/(2*4.29) = R/L
so
i = 4.29(1-e^-(Rt/L))
current is max at great t
i max = 4.29 - 0
energy = (1/2) L i^2 =(1/2)(2)4.29^2
= 18.4 Joules

one time constant T =L/R and e^-(Rt/L) = 1/e = .368
i = 4.29 (1-.368) = 4.29*.632 = 2.71 amps
energy = (1/2)(2)2.71^2 = 7.35 Joules