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Energy Dissipation by Inductor (RL circuit)

  1. Nov 2, 2014 #1
    1. The problem statement, all variables and given/known data
    ffs.png

    2. Relevant equations
    The voltage across an inductor is V(T) = L i'
    Basic current division and voltage division .
    Step function where u(T) : when t is smaller than 0 : u(t) = 0
    t >0 : u(t) = 1

    3. The attempt at a solution

    The voltage source has the equation 40 - 40u(t).
    When t <0 , this means that the value of the VS is 40V.

    What happens to the current iL through the inductor...? Does the inductor acts as a short-circuit or open-circuit?
    Assuming an Inductor acts as a short-circuit under these conditions, iL= 40V/40ohms =1 Amps


    This is my i(0).
    Therefore Rth = 20 ohms
    t = L/R = 10 / 20 = 1/2

    So iL(t) is = 1A(e^(-2t))


    When t > 0 , Vin becomes 0 V so the whole circuit becomes an RL circuit which can be simplified using equivalent resistance equation and the final circuit becomes a 20 ohms resistor in series with the inductor.

    The energy dissipated by an Inductor is E = (L*i^2) /2 and since we know the initial energy stored in the inductor is (10H( 1A)^2) /2 = 5 Joules,

    It must be dissipated by the resistor. So the total energy dissipated by the circuit is 5 J?

    The main problem here is with me understand how an inductor behave in this kind of situation (DC I presume). How do we find the initial current and current at infinity? Kind of confused...
     
  2. jcsd
  3. Nov 2, 2014 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    You have it right. Inductor current starts at 1A and exponentially falls to zero on a time constant involving L and 20 ohms. The stored energy is all lost, so 5J looks correct.

    When voltages are steady, the inductor acts like a short circuit.

    You might find it instructive to work out what the voltage across the inductor would look like after t=0. Sketch the graph.
     
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