# Two Wattmeter Method – Measurement of Three Phase Power

There are three methods which are used for the measurement of three phase power in three phase circuits. The three methods are:

- Three Wattmeter Method

- Two Wattmeter Method

- Single Wattmeter Method

In this post, we will discuss the Two Wattmeter Method for power measurement.

**Two Wattmeter Method:**

In two wattmeter method, a three phase balanced voltage is to a balanced three phase load where the current in each phase is assumed lagging by an angle of Ø behind the corresponding phase voltage.

The schematic diagram for the measurement of three phase power using two wattmeter method is shown below.

From the figure, it is obvious that current through the Current Coil (CC) of Wattmeter W_{1} = I_{R} , current though Current Coil of wattmeter W_{2} = I_{B }whereas the potential difference seen by the Pressure Coil (PC) of wattmeter W_{1}= V_{RB} (Line Voltage) and potential difference seen by Pressure Coil of wattmeter W_{2} = V_{BY}. The phasor diagram of the above circuit is drawn by taking VR as reference phasor as shown below.

From the above phasor diagram,

Angle between the current I_{R}and voltage V_{RB} = (30° – Ø)

Angle between current I_{Y}and voltage V_{YB} = (30° + Ø)

Therefore, Active power measured by wattmeter W_{1} = V_{RB}I_{R} Cos (30° – Ø)

Similarly, Active power measured by wattmeter W_{2} = V_{YB}I_{Y}Cos(30° + Ø)

As the load is balanced, therefore magnitude of line voltage will be same irrespective of phase taken i.e. V_{RY}, V_{YB} and V_{RB} all will have same magnitude. Also for Star / Y connection line current and phase current are equal, say I_{R }= I_{Y} = I_{B} = I

Let V_{RY} = V_{YB} = V_{RB} = V_{L}

Therefore,

W_{1} = V_{RB}I_{R}Cos (30° – Ø)

= V_{L}ICos(30° – Ø)

In the same manner,

W_{2} = V_{L}ICos(30° + Ø)

Hence, total power measured by wattmeters for the balanced three phase load is given as,

W = W_{1} + W_{2}

= V_{L}I×Cos(30° – Ø) + V_{L}I×Cos(30° + Ø)

= V_{L}I [Cos(30° – Ø) + Cos(30° + Ø)]

= 2V_{L}I×Cos30°CosØ ……………….[ CosC + CosD = 2Cos(C+D)/2×Cos(C-D)/2 ]

=√3V_{L}ICosØ

Therefore, total power measured by wattmeters W = √3V_{L}ICosØ

Now, suppose you are asked to find the power factor of the load when individual power measured by the wattmeters are given, then we should proceed as

W_{1} + W_{2} = √3V_{L}ICosØ ……………………………..(1)

Similarly,

W_{1} – W_{2}= V_{L}I×Cos(30° – Ø) + V_{L}I×Cos(30° + Ø)

= V_{L}I [Cos(30° – Ø) + Cos(30° + Ø)]

= 2V_{L}I×Sin30°SinØ ………[ CosC – CosD = 2Sin(C+D)/2×Sin(D-C)/2 ]

= V_{L}ISinØ

Hence,

W_{1} – W_{2 }= V_{L}ISinØ ………………………………(2)

Dividing equation (2) by equation (1),

(W_{1} – W_{2}) / (W_{1} + W_{2}) = V_{L}ISinØ / √3V_{L}ICosØ

(W_{1} – W_{2}) / (W_{1} + W_{2}) = (tanØ) /√3

Hence,

tanØ = √3[(W
_{1} – W_{2}) / (W_{1} + W_{2})] |

From the above equation, we can find the value of Ø and hence the power factor Cos Ø of the load.

Hope you understand the method of measurement of three phase power using two wattmeter method. Now we will consider three cases and will observe the how the individual wattmeter measures the power in each case.

**Case1: When the power factor of load is unity.**

As the power factor of load is unity, hence Ø = 0

Therefore,

Power measured by first wattmeter W_{1}= V_{L}I Cos(30° – 0)

= V_{L}I Cos30°

= 0.866 V_{L}I

Power measured by second wattmeter W_{2}= V_{L}I Cos(30° + 0)

= V_{L}I Cos30°

= 0.866 V_{L}I

**Thus we see that, when the power factor of load is unity then both the wattmeter reads the same value.**

**Case2: When power factor of load is 0.5 lagging.**

As power factor is 0.5 hence CosØ = 0.5 i.e. Ø = 60°

Therefore,

Power measured by first wattmeter W_{1}= V_{L}I Cos(30° – 60°)

= V_{L}I Cos30° ……[Cos (-Ɵ) = CosƟ ]

= 0.866 V_{L}I

Power measured by second wattmeter W_{2}= V_{L}I Cos(30° + 60°)

= V_{L}I Cos90°

= 0

**Thus we see that, when power factor of load is 0.5 lagging then power is only measured by first wattmeter and reading of second wattmeter is ZERO. **

**Case3: When power factor of load is zero.**

As power factor of load is zero, hence CosØ = 0 i.e. Ø = 90°

Therefore,

Power measured by first wattmeter W_{1}= V_{L}I Cos(30° – 90°)

= V_{L}I Cos60°

= 0.5 V_{L}I

Power measured by second wattmeter W_{2}= V_{L}I Cos(30° + 90°)

= –V_{L}I Cos60°

= -0.5 V_{L}I

Thus we see that, when power factor of load is zero then one wattmeter reads +ive while second wattmeter reads –ive. As second wattmeter reads –ive hence wattmeter won’t read anything practically, therefore for second wattmeter we need to interchange the leads of either Pressure Coil or Current Coil so that second wattmeter may read value. As we have interchanged the connection of leads of either PC or CC, hence second wattmeter will read +ive but while calculating the total power measured we must take the reading of second wattmeter as –ive.

**It shall be noted that when ****60° <Ø < 90°, reading of one wattmeter will be positive while the reading of second wattmeter will be negative.**

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this is really helpful.

Good

thanks a lot

This is easily understandable

Good

great note! Simple but clear!

Thank you Tracey! Please share the post if you like it.

Super

Superb explanation

Calculation of case_2 is wrong. You have written cos30= 0.5 but in actual value is 0.866. please correct same. Ty

Thank you very much for pointing calculation mistake. Please check, the same has been rectified.

Nice

What are the advantages of this method?

Whenever we are saying that 2 watt meter method and 3 wattmeter method power is same.

so how can prove for

Case1. Unbalance current in all theree phase

case2. For 0.5 leg power factor total power with 3 wattmeter=2wattmeter ??

Simply calculate power measured by each wattmeter in case of 3 wattmeter using phase voltage and current and add them up. Now calculate the same using 2 wattmeter method. Check if result is same.

Is current is same in two wattmeter

If load is balanced, then current will be same. For detail, check calculation part.

Tq its helpful for me

too good…..

Thank you! Please share.

W2 = IB I think it should be IY instead of IB

plz reply