Physics!!!

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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.
(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.

• Physics!!! -

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

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