How Is Thermal Energy Calculated in an Inductor?

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The stored energy in the magnetic field of the inductor is calculated to be 1.04 J using the formula W = 1/2Li^2. To determine the rate of thermal energy developed in the inductor, one must consider Ohmic losses, which are related to the resistance and current. The thermal energy dissipation can be calculated using the formula P = I^2R, where P represents power loss due to resistance. In this case, with a resistance of 160 ohms and a current of 0.4 A, the thermal energy rate is 25.6 W. Understanding the distinction between stored magnetic energy and thermal losses is crucial for accurate calculations.
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An inductor used in a DC power supply has an inductance of 13H and a resistance of 160 ohms. It carries a current of 0.4 A. What is the stored energy in the magnetic field? At what rate is thermal energy developed in the inductor?

L=13H
R=160 ohms
I=0.4A

W = 1/2Li^2 = 1/2(13)(0.4)^2 = 1.04 J

I am unsure where to go from here. I can calculate the stored energy, but what do I do for the rate of thermal energy?
 
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Your answer for the stored magnetic energy looks correct. Now, the thermal energy dissipated actually has very little to do with the stored magnetic energy in this DC question. Think of the thermal losses as "Ohmic losses". Does that hint help?
 
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