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Energy stored in an inductor of an LR circuit

  1. Mar 18, 2012 #1
    An LR circuit has a resistance R = 25 Ω, an inductance L = 5.4 mH, and a battery of EMF = 9.0 V. How much energy is stored in the inductance of this circuit when a steady current is achieved?

    2. Relevant equations
    [itex]\epsilon[/itex]= -d[itex]\phi[/itex]m/dt=-L[itex]\frac{dI}{dt}[/itex]

    3. The attempt at a solution
    According to the equations, to find the energy stored in the inductance of the circuit, I need to find current, but I don't know how. For the equation of emf, by a "steady" current, I suppose this means that dI/dt is equal to zero. I don't know how that helps, but it's as far as I got trying to understand this problem. Perhaps there is an equation that is necessary to solve this problem, but nothing comes to mind. Maybe... Ohm's law? But I doubt it as the potential difference across the circuit isn't known, and I don't think emf can be substituted for potential difference V even thought they have the same units (voltage).
  2. jcsd
  3. Mar 18, 2012 #2
    You are on the right track!!!!!!
    The steady current is simply given by I = V/R
    The inductance determines the RATE at which the current rises
  4. Mar 18, 2012 #3
    Oh. So I was. Supposing emf can be substituted into I=V/R as V, then the standing current is equal to 9V/25 ohms= 0.36A. Um= 0.5(5.4 x 10-3H)(0.36A)2= 3.4992 x 10-4J= 0.35 x 10-3J= 0.35 mJ

    That would be the correct answer. Thank you!

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