A 24.0-V battery is connected in series with a resistor and an inductor, with R = 9.40 Ω and L = 6.40 H, respectively. (a) Find the energy stored in the inductor when the current reaches its maximum value. J (b) Find the energy stored in the inductor at an instant that is a time interval of one time constant after the switch

a) 20.81 J

b) 8.29 J

Explanation:

V = iR + L di/dt

where

i = a(1-e^-kt)

for large t

i = V/R

i = 24 / 9.4

i = 2.55 A

so

i = 2.55(1-e^-kt)

di/dt = 2.55 k e^-kt

24 = 24-24e^-kt + 6.4(2.55)k e^-kt

24 = 6.4(2.55) k

k = 24 / (6.4 * 2.55)

k = 24 / 16.32

k = 1.47 = R/L

so

i = 2.55(1-e^-(Rt/L))

current is maximum at great t

i max = 2.55 - 0

energy = (1/2) L i^2

E = (1/2)(6.4)2.55^2

E = 20.81 Joules

one time constant T = L/R and e^-(Rt/L) = 1/e = .368

i = 2.55 (1 - 0.368)

i = 2.55 * 0.632

i = 1.61 amps

energy = (1/2)(6.4)1.61^2

E = 8.29 Joules