Power dissipated in the inductor

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SUMMARY

The discussion centers on the power dissipated in an RL AC circuit with an AC source defined as vst = 220sin(40000t-45°), a 60mH inductor, and a 3kΩ resistor. It is established that the power dissipated in the inductor is zero due to the phase difference between voltage and current. The average power formula P_{avg} = V_{rms}I_{rms}cos(θ) is applicable, where θ is the phase angle, and in this case, the inductor causes the current to lag the voltage by 90 degrees, resulting in zero power dissipation.

PREREQUISITES
  • Understanding of RL AC circuits
  • Familiarity with phasor representation of voltage and current
  • Knowledge of the average power formula P_{avg} = V_{rms}I_{rms}cos(θ)
  • Basic calculus for evaluating integrals
NEXT STEPS
  • Study the concept of inductive reactance in AC circuits
  • Learn how to derive voltage and current functions for inductors over time
  • Explore the implications of phase differences in AC power calculations
  • Investigate the role of impedance in RL circuits
USEFUL FOR

Electrical engineering students, educators teaching AC circuit theory, and professionals analyzing power dissipation in inductive components.

noppawit
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In RL, AC circuit, an AC source vst = 220sin(40000t-45°) is connected to a 60mH inductor and a 3kΩ resistor.

Show that the power dissipated in the inductor is zero.



The current in this circuit is 0.057A as I calclulated. I know that inductor has no charging role in this circuit. But how can I write, describe this by equation?

Thank you.
 
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You need to find the power over one cycle, (or an integral number if you choose).

P = \int v(t)i(t)dt
 
Would you please make it simpler? I have not learned this before. I learned that

P_{avg} = V_{rms}I_{rms}cos(\theta) = V_{rms}I_{rms}\frac{R}{Z}

only.
 
You need to calculate both the current through, and the voltage across the inductor as a functions of time.

Inductors, being what they are, you should find that they are 90 degrees out of phase.

Since this is homework, you need to show some work. Generally, the more work the more help.
 

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