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Energy stored in an inductor of an LR circuit 
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#1
Mar1812, 04:02 PM

P: 28

1. The problem statement, all variables and given/known data
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] U_{m}=[itex]\frac{1}{2}[/itex]LI^{2} L=[itex]\phi[/itex]_{m}/I 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
Mar1812, 04:21 PM

P: 1,506

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 


#3
Mar1812, 04:43 PM

P: 28

That would be the correct answer. Thank you! 


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