Asked by Help
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.
(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.
Answers
Answered by
Damon
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
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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