How Does V^2/R Relate to Electrical Power and Efficiency?

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SUMMARY

The discussion centers on the relationship between V^2/R, electrical power, and efficiency in electrical systems. Key equations established include V^2/R = VA^2/W and the implications of power factor (PF) on efficiency calculations. The participants clarify that apparent power (VA) is always greater than or equal to real power (W), leading to the conclusion that V^2/R must also be greater than or equal to electrical power input. The conversation highlights the significance of understanding these relationships for accurate efficiency assessments in electrical circuits.

PREREQUISITES
  • Understanding of electrical power concepts, including apparent power (VA) and real power (W).
  • Familiarity with power factor (PF) and its implications on efficiency.
  • Knowledge of basic electrical circuit principles, including Ohm's Law and Joule's Law.
  • Ability to interpret and manipulate electrical equations involving voltage (V), current (I), and resistance (R).
NEXT STEPS
  • Research the derivation and applications of the equation V^2/R in electrical engineering.
  • Learn about the implications of power factor on energy efficiency in AC circuits.
  • Explore the relationship between inductance, current, and magnetic fields in electrical systems.
  • Study efficiency calculations in electrical systems, focusing on minimizing joule loss.
USEFUL FOR

Electrical engineers, students in electrical engineering, and professionals involved in optimizing power efficiency in electrical systems will benefit from this discussion.

kmarinas86
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(V^2/R) / (VI) = (VI) / (RI^2) = V / RI

VA = VI = Apparent power
W = RI^2 = Real power
PF = RI/V = Power factor

(V^2/R) / VA = VA / W = V / RI = 1 / PF

(V^2/R) = VA^2 / W

VA>=W

Therefore:

V^2/R >= VA

!

(V^2 / R) * PF = VA
VA * PF = W

!

What is V^2/R? If you have an large inductor, you could have V^2/R with very little current! What causes the magnetic field?

Is it "IR^2"? I know that equals joule loss. (real power=joule loss)
Efficiency = (? - Joule loss) / ?

Is it "VI"? I know that equals the electrical power input. (electrical power input>=joule loss)
Efficiency = (VI - ? - Joule loss) / VI

Is it "V^2/R"? But this is greater than or equal to electrical power input! (V/R>=I) What does this mean when PF<1 ?
Efficiency = ?
 
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Your definitions appear to have some typos and issues mixed in. Can you post links to the sources of the equations you are using? Especially the equations you are using to calculate the Power Factor. Thanks.
 

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