**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 _{L}I_{L}cosφ**

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_{R}X V_{RB}X cos∠I_{R}&V_{RB}

W2 = I_{Y}X V_{YB}X 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_{RB}cos(30° – φ ) i.e. W1 = I_{L} V_{L}cos(30° – φ )

> W2 = I_{Y}V_{YB}cos(30° + φ) i.e. W2 = I_{L}V_{L}cos(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**