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

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In summary, the conversation discusses various equations and concepts related to power and efficiency, such as apparent power, real power, power factor, and joule loss. The equations (V^2/R) / (VI) = (VI) / (RI^2) = V / RI and (V^2 / R) * PF = VA are mentioned, as well as the question of what causes the magnetic field and the potential typo of "IR^2" instead of "V^2/R". The discussion also touches on the relationship between V^2/R and electrical power input, and how this can be affected by a power factor less than 1.
  • #1
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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  • #2
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.
 

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

1. Why is (v^2/r) greater than or equal to V*i in this equation?

The equation (v^2/r) >= V*i represents the relationship between the centripetal acceleration (v^2/r) and the tangential velocity (V) multiplied by the angular velocity (i). This is because the centripetal acceleration is equal to the tangential velocity squared divided by the radius of the circular motion. Therefore, in order for the object to maintain a circular path, the centripetal acceleration must be greater than or equal to the tangential velocity multiplied by the angular velocity.

2. What is the significance of the radius in this equation?

The radius in this equation represents the distance from the center of the circular path to the object. It plays a crucial role in determining the amount of centripetal acceleration needed for the object to maintain its circular motion. As the radius decreases, the centripetal acceleration must increase in order to keep the object moving in a circle at a constant speed.

3. How does the tangential velocity affect the overall equation?

The tangential velocity (V) in this equation represents the linear speed of the object along its circular path. As the tangential velocity increases, the amount of centripetal acceleration needed to maintain the circular motion also increases. This is because the object is moving faster and therefore requires a greater force to keep it in a circular path.

4. Why is the angular velocity (i) included in this equation?

The angular velocity (i) in this equation represents how quickly the object is rotating around the center of the circular path. It is essential in determining the overall velocity of the object and plays a crucial role in the amount of centripetal acceleration needed. As the angular velocity increases, the tangential velocity also increases, resulting in a greater need for centripetal acceleration.

5. What are the practical applications of this equation?

This equation is commonly used in the fields of physics and engineering to understand and calculate the forces required for objects to maintain circular motion. It is also used in designing and analyzing circular motion systems, such as roller coasters, centrifuges, and satellite orbits.

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