How Is Thermal Energy Calculated in an Inductor?

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SUMMARY

The discussion focuses on calculating thermal energy in an inductor with an inductance of 13H, resistance of 160 ohms, and a current of 0.4 A. The stored energy in the magnetic field is calculated using the formula W = 1/2Li^2, resulting in 1.04 J. The thermal energy developed in the inductor is primarily due to Ohmic losses, which are calculated using the formula P = I^2R, where P represents power loss due to resistance.

PREREQUISITES
  • Understanding of inductance and its units (Henry)
  • Knowledge of Ohm's Law and resistance
  • Familiarity with energy storage in magnetic fields
  • Ability to perform basic electrical calculations
NEXT STEPS
  • Research "Ohmic losses in inductors" for a deeper understanding of thermal energy dissipation.
  • Learn about "power loss calculations in electrical components" to apply similar principles.
  • Explore "thermal management in inductors" to understand heat dissipation techniques.
  • Study "inductor applications in DC power supplies" to see practical implementations.
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Electrical engineers, students studying electromagnetism, and professionals working with DC power supplies will benefit from this discussion.

cmalle09
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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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