Example
When connected to a three-phase motor, two
wattmeters gave readings of 5 kW and—1 kW.
Ptotal = W1 + W2
Ptotal = (—1000) + 5000 = 4000
Ptotal = 4 kW
tan Φ = √3 x 
5 - (-1)



             5 + (-1)

 tan Φ = √3 x 
6



             4

 tan Φ = √3 x 
1.5 = 2.598



     Φ ...
So as, the power factor (pf) = Cos Φ

Cos Φ = Cos 68.9⁰ = 0.3599 = 0.36

Power factor = 0.36.
14.4.3 Example
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14.4.3 Example

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14.4.3 Example

  1. 1. Example When connected to a three-phase motor, two wattmeters gave readings of 5 kW and—1 kW.
  2. 2. Ptotal = W1 + W2 Ptotal = (—1000) + 5000 = 4000 Ptotal = 4 kW
  3. 3. tan Φ = √3 x 5 - (-1) 5 + (-1) tan Φ = √3 x 6 4 tan Φ = √3 x 1.5 = 2.598 Φ = 68.9⁰
  4. 4. So as, the power factor (pf) = Cos Φ Cos Φ = Cos 68.9⁰ = 0.3599 = 0.36 Power factor = 0.36.

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