Confused about the reactive power

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SUMMARY

The discussion centers on the calculation of reactive power (Q) and its relationship with apparent power (S) in electrical systems. The formulas Q = UI sin(a) and S = UI are established, but the exercise reveals that Q can also be expressed as Q = I² R sin²(a) when analyzing heat loss in power grids. The correct ratio for heat loss due to reactive power is Q/S = sin²(a), particularly when the power factor cos(a) = 0.85 is given. Understanding when to use each formula is crucial for accurate electrical calculations.

PREREQUISITES
  • Understanding of electrical power concepts, including reactive power and apparent power.
  • Familiarity with trigonometric functions, particularly sine and cosine.
  • Knowledge of Ohm's Law and its application in electrical circuits.
  • Basic grasp of power factor and its significance in electrical systems.
NEXT STEPS
  • Study the derivation and application of the power factor in AC circuits.
  • Learn about the implications of reactive power on energy efficiency in power systems.
  • Explore the relationship between current (I), resistance (R), and power loss in electrical grids.
  • Investigate advanced topics in electrical engineering, such as power quality and harmonics.
USEFUL FOR

Electrical engineers, power system analysts, and students studying electrical engineering concepts related to reactive power and energy loss in power grids.

Gavroy
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hey,
i am a little bit confused about the reactive power.
we defined it as
Q=UI sin(a)
and the apparent power as
S=UI

Now I have to solve this exercise:
Which percentage of heat loss in a power grid is caused by the reactive power, when the powerfactor is cos(a)=0,85?

Now I thought:
Q/S=sin(a) would be the solution

but instead of this it is said that one has to set:
Q= I² R sin²(a)
S= I² R

and therefore Q/S=sin²(a)

now my question is: when can I use Q=U I sin(a) and when do I have to use Q= I² R sin²(a)...can somebody explain this to me?
 
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Put it in the homework section
 

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