Calculating average power using impedance and rms current

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SUMMARY

The average power (P) absorbed by a load with an impedance of 56 + j15 ohms and a root mean square (rms) current of 21 A is calculated using the formula P = Irms² * R. Converting the impedance to polar form results in a magnitude of 57.9741 ohms at an angle of 14.9951 degrees. The calculated average power is 25.5 kW. The discussion also raises a question regarding the power dissipated in the j15 ohm inductor and whether it should be included in the power dissipation calculation.

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  • Understanding of complex impedance in AC circuits
  • Familiarity with polar form representation of complex numbers
  • Knowledge of root mean square (rms) current calculations
  • Proficiency in power calculation formulas in electrical engineering
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pokie_panda
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Homework Statement



Calculate the average power (P) absorbed by the load shown below when its impedance is 56 +j15 ohms 21 A rms?
Answer to three significant figures

Homework Equations



P=Irms^2*R

The Attempt at a Solution



Converting 56 +j15 to polar form yields 57.9741 ohms @ 14.9951 degrees
So 21^2*57.9741 = 25.5kW
 
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pokie_panda said:

Homework Statement



Calculate the average power (P) absorbed by the load shown below when its impedance is 56 +j15 ohms 21 A rms?
Answer to three significant figures

Homework Equations



P=Irms^2*R


The Attempt at a Solution



Converting 56 +j15 to polar form yields 57.9741 ohms @ 14.9951 degrees
So 21^2*57.9741 = 25.5kW

"Load shown below"??

Anyway,
What is the power dissipated in the j15 ohm inductor?

Should you be including it in your power dissipation calculation? You had the right formula!
 

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