3.A) Demonstrate that two wattmeters are sufficient to measure power in a three-phase balanced star-connected circuit with the help of a neat circuit diagram and phasor diagram.
Answer;–
If W1 and W2 are the two wattmeter readings then total power
W= W1 + W2 = three phase power = √3 VLILcosφ
Consider star-connected load and two-wattmeter connected as shown in fig.
Let us consider the RMS values of current and voltage to prove that sum of two
wattmeter gives the total power consumed by the three phase load.
W1 = IRX VRBX cos∠IR&VRB
W2 = IYX VYBX cos∠IY&VYB
• To find the angle between ( IR and VRB) and (IY and VYB) phasor diagram is drawn.
(Assuming power factor to be lagging) V̅RB= V̅R− V̅b
And V̅YB=V̅Y−V̅B
The angle between VR and IR = φ
The angle between VY and IY = φ
Vph= VR= VY=VB and VRB= VYB= VL
IR= IY= IL=IPH(star)
From the vector diagram
The angle between VRB and IR = 30° – φ
The angle between VYB and IY = 30° + φ
> W1 = IR VRBcos(30° – φ ) i.e. W1 = IL VLcos(30° – φ )
> W2 = IYVYBcos(30° + φ) i.e. W2 = ILVLcos(30° + φ)
>> W1 + W2 = IL VL [cos(30° – φ ) + cos(30° + φ) ]
= IL VL [ cos 30 cos φ + sin 30 sin φ + cos 30 cos φ – sin 30 sin φ]
= 2 IL VL cos 30 cos φ
= 2 IL VL cos φ
W1 + W2= √3 IL VL cos φ = Total 3 phase power
