Concept:
Power in a three-phase circuit is,
P=√3VLILcosϕ
Two wattmeter method:
The connection diagram using wattmeters as shown below.
W1=IRVRBcos(IR ∧VRB)
W2=IYVYBcos(IY∧VYB)
From the phasor diagram,
IR∧VRB=30−ϕ
IY∧VYB=30+ϕ
W1=IRVRBcos(30−ϕ)
⇒W1=VLILcos(30−ϕ)
W2=IYVYBcos(30+ϕ)
⇒W2=VLILcos(30+ϕ)
W1+W2=VLIL[cos(30−ϕ)+cos(30+ϕ)]
=√3VLILcosϕ
⇒ Total three-phase power =;W1+W2=√3VLILcosϕ
Total three-phase power is the sum of two wattmeters.
W1=VLILcos(30−ϕ)
W2=VLILcos(30+ϕ)
W1−W2=√3VphIphsinϕ
√3(W1−W2)=3VphIphsinϕ
Reactive power =√3;(W1−W2)
Reactive power is equal to √3 times the difference between the readings of the two wattmeters.
W1+W2=3VphIphcosϕ
√3(W1−W2)=3VphIphcosϕ
⇒ϕ=tan−1(W1+W2√3(W1−W2))
Power factor =cos;ϕ
cosϕ=cos[tan−1(W1+W2√3(W1−W2))]
Calculation:
Reactive power =√3;(W1−W2)
= √3 (20 - 10) = 17.32 kVar