Demonstrate that, two wattmeter’s are sufficient to measure power in a three phase balanced star-connected circuit

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 V_LI_Lcosφ

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 = I_RX V_RBX cos∠I_R&V_RB
W2 = I_YX V_YBX cos∠I_Y&V_YB

• To find the angle between ( I_R and V_RB) and (I_Y and V_YB) phasor diagram is drawn.
(Assuming power factor to be lagging) V̅_RB= V̅_R− V̅_b

And V̅_YB=V̅_Y−V̅_B
The angle between V_R and I_R = φ

The angle between V_Y and I_Y = φ
V_ph= V_R= V_Y=V_B and V_RB= V_YB= V_L
I_R= I_Y= I_L=I_PH(star)

From the vector diagram
The angle between V_RB and I_R = 30° – φ
The angle between V_YB and I_Y = 30° + φ
> W1 = I_R V_RBcos(30° – φ ) i.e. W1 = I_L V_Lcos(30° – φ )
> W2 = I_YV_YBcos(30° + φ) i.e. W2 = I_LV_Lcos(30° + φ)
>> W1 + W2 = I_L V_L [cos(30° – φ ) + cos(30° + φ) ]

= I_L V_L [ cos 30 cos φ + sin 30 sin φ + cos 30 cos φ – sin 30 sin φ]
= 2 I_L V_L cos 30 cos φ
= 2 I_L V_L cos φ

W1 + W2= √3 I_L V_L cos φ = Total 3 phase power