Page 317 - Electrician - TT (Volume 1)

ELECTRICIAN - CITS

P + P = 4.5 + 3 = 7.5 KW
1 2
P P = 4.5 3 = 1.5 KW
2
1
tan f =

f = tan1 0.3464 = 1906’

Power factor Cos 19°6’ = 0.95
Example 2: Two wattmeters connected to measure the power input to a balanced three-phase circuit indicate 4.5
KW and 3 KW respectively. The latter reading is obtained after reversing the connection of the voltage coil of that
wattmeter. Find the power factor of the circuit.
Soultion
tan f =

f = tan—1 8.66 = 83°.27’
since power factor (Cos 83o 27’) = 0.114.
Example 3: The reading on the two wattmeters connected to measure the power input to the three-phase,
balanced load are 600W and 300W respectively.
Calculate the total power input and power factor of the load.
Solution
Total power = P = P + P 2
1
T
P = 600W.
1
P = 300W.
2
P = 600 + 300 = 900
T

f = tan10.5774 = 30o

Power factor = Cos 30o = 0.866.
Assignment
Two wattmeters connected to measure the power input to a balanced, three-phase load indicate 25KW and 5KW
respectively.
Find the power factor of the circuit when (i) both readings are positive and (ii) the latter reading is obtained after
reversing the connections of the pressure coil of the wattmeter.

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CITS : Power - Electrician & Wireman - Lesson 50-53