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Power in an RL Circuit

  1. Mar 20, 2008 #1
    An RL circuit in which L = 9.00 H and R = 5.00 is connected to a 24.0 V battery at t = 0.
    (a) What energy is stored in the inductor when the current is 0.500 A?
    (b) At what rate is energy being stored in the inductor when I = 1.00 A?
    (c) What power is being delivered to the circuit by the battery when I = 0.500 A?

    I got the energy; that was pretty straightforward. Just U=1/2 LI^2.

    For the rate of energy, I tried solving for energy at 1.00 A, then solving for t when inductor is at 1A. To do that I used I=V/R(1-e^-Rt/L), then I divided energy by the time to find power but that didn't work. Any tips?
     
  2. jcsd
  3. Mar 21, 2008 #2
    Any power delivered by the battery not used up in the resistor is stored in the inductor
     
  4. Mar 21, 2008 #3
    Okay, I found the formula IE=RI^2+IL(dI/dt) and I used that to solve (b), which turned out to be 19 W, but that formula didn't work when I tried it on (c). I think what I have to do is find the total power in the system and subtract my answer from (b) from that. But I'm not sure. Any suggestions?
     
  5. Mar 21, 2008 #4
    c) seems suspiciously easy. The power that a battery delivers is just V*I. Using
    I = V/R(1-e^-Rt/L) and Power = RI^2+IL(dI/dt) works as well but seems rather complicated compared with just computing I*V
     
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