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March 29, 2017

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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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