two wattmeter method lab experiment

Two Wattmeter Method of Power Measurement

Two Wattmeter Method can be employed to measure the power in a 3 phase, three-wire star or delta connected the balanced or unbalanced load.

In two wattmeter method, the current coils of the wattmeter are connected with any two lines, say R and Y and the potential coil of each wattmeter is joined on the same line, the third line i.e. B as shown below in figure (A):

The total instantaneous power absorbed by the three loads Z 1 , Z 2 and Z 3 , is equal to the sum of the powers measured by the two wattmeters, W 1 and W 2 .

Measurement of Power by Two Wattmeter Method in Star Connection

Measurement of power by two wattmeter method in delta connection.

Considering the above figure (A) in which Two Wattmeter W 1 and W 2 are connected, the instantaneous current through the current coil of Wattmeter, W 1 is given by the equation shown below:

The instantaneous potential difference across the potential coil of Wattmeter, W 1 is given as:

Instantaneous power measured by the Wattmeter, W 1 is

The instantaneous current through the current coil of Wattmeter, W 2 is given by the equation:

The instantaneous potential difference across the potential coil of Wattmeter, W 2 is given as:

Instantaneous power measured by the Wattmeter, W 2 is:

Therefore, the total power measured by the two wattmeters W 1 and W 2 will be obtained by adding the equation (1) and (2).

Considering the delta connected circuit shown in the figure below:

Instantaneous power measured by the Wattmeter, W 1 will be:

Therefore, the instantaneous power measured by the wattmeter, W 1 will be given as:

The instantaneous current through the current coil of the Wattmeter, W 2 is given as:

The instantaneous potential difference across the potential coil of wattmeter, W 2 :

Therefore, the instantaneous power measured by Wattmeter, W 2 will be:

Hence, to obtain the total power measured by the two wattmeter the two equations, i.e. equation (3) and (4) has to be added.

The power measured by the Two Wattmeter at any instant is the instantaneous power absorbed by the three loads connected in three phases. In fact, this power is the average power drawn by the load since the Wattmeter reads the average power because of the inertia of their moving system.

Related terms:

• Measurement of Three Phase Power: Three Wattmeter Method
• Two Wattmeter Method – Balanced Load Condition
• Reactive Power Measurement
• Low Power Factor Wattmeter
• Potier Triangle or Zero Power Factor Method

19 thoughts on “Two Wattmeter Method of Power Measurement”

What type of process we can run a single phase motor in reverse mode?

The motor can run in reverse mode by reversing the line connections of either the main winding or the starting winding.

Advantage of using two wattmeter method

1. Two wattmeter method measures the three phase power. 2. Useful for both the balanced and unbalanced loads.

1.use in balanced and unbalanced load 2.no need to connect neutral wire

Great Article.

Awesome 🙂 🙂

Very Informative. Please add more information regarding different types of connections made with watt meter in three phase systems…..

will there be any difference in result if we use 2wattmeter method vs 3 wattmeter method on same 11kV circuit?with same load?

I think it is “Instantaneous voltage measured by the Wattmeter, W1 will be

W sub 1 = e sub RB”

explain relation between single phase power and three phase power (show the equations)

3 phase 4 wire system power measure by……..

Is there any serious errors in measurement of power through watt metre of 1 and 2 it there any how to minimise it

While measuring power in a 3 phase load by 2 wattmeter methodbthe reading of 2 wattmeter are equal nd opposite why

Does the two wattmeter method of power measurement applicable for a 3 phase, 4 phase unbalanced load.if not how to measure the total power absorbed by three unequal load

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• Laboratory Practice
• Safety Rules

ECE 494 - Electrical Engineering Laboratory III

Experiment 1: three phase power measurements.

• To demonstrate the line and phase relations in 3-phase balanced networks.
• To study and demonstrate the two wattmeter method of measuring the power in 3-phase networks.
• Two digital multi-meters from the stockroom.
• One Set of banana cables and power quality meter Fluke 43B from the stockroom.
• HMRL-3 Resistor Load Cart.
• One three-phase variac.
• One small black box wattmeter (Murata AC power Meter)
• One switch box from cabinet or simple phase circuit breaker.
• Richard Dorf, Introduction to Electric Circuits, Ch. 11, 9th edition, John Wiley & Sons, Inc., 2013.
• D. Johnson, J. Johnson, J. Hilburn, Electric Circuit Analysis, Ch. 9, 10, 3rd edition, Prentice Hall, N. J., 1997.
• Turan Gönen, Electrical Machines with MATLAB , pp. 17-41, 2nd Edition, CRC Press, Boca Raton, Fla, 2012.

Three-phase balanced networks are used in the power industry for reasons of economy and performance. Three-phase generators and motors run smoothly, with no torque pulsations, unlike single phase machines. In addition balanced three phase systems may be operated as three wire or four wire systems, with much less copper needed for the power delivered as compared with three single phase systems.

At a power generating plant, the windings of a three phase machine are arranged to provide three voltages, each 120° apart in time and, in the common balanced system, usually all of the same magnitude. These three voltage sources may be connected in a wye (Y) or a delta (∆) configuration. Three phase loads may also be connected in wye or delta connections. The wye connection has a central node to which a neutral wire may be joined, but the delta connection is a three wire system without a node for a neutral or ground) connection.

To measure power in a 3-phase system, it would seem necessary to use three wattmeters, each connected to neutral for a common terminal, and each responding to a line-to-neutral voltage and a line current. One would then add up the powers indicated on each wattmeter. Analysis of such a circuit shows that one wattmeter is redundant, hence the two-wattmeter method of measuring 3-phase power was developed for three wire systems. This method is satisfactory even if the loads are unbalanced. It is necessary to connect the wattmeters taking into account the polarity of their coils. When the current enters the marked terminal of the current coil and the voltage positive is connected to the marked terminal of the voltage coil, the reading represents power absorbed. In that case the algebraic sum of the wattmeters determines the total load power. In reactive circuits it may be necessary to reverse the current coil of one wattmeter in order to get an upscale deflection. This reading is taken as negative when the total power is determined algebraically.

If a 3-phase system has four wires, it is necessary to use three wattmeters, unless it is known that the system is balanced and therefore no current is flowing in the neutral wire. For any balanced N wire system it is necessary to use N − 1 wattmeters to measure the total power.

• Assume a phase voltage of 120 V (a line voltage of 208V) in Figure 1.1 and that the three resistors have values of 800 Ω. Calculate the expected values of I 1 = I 2 = I 3 for the completely balanced circuit.
• Review the two power meter method for measuring three phase power. Determine how to connect the meters into the circuits of figures 1.1 and 1.2 to measure power supplied by the variac. The power meters we use will also read out the voltage and currents they are measuring but you will need to connect DMVs to measure the remaining phase voltage and line current as well as to measure the additional voltages and currents asked for in the lab description. (Line voltages: V AB , V BC , V CA . Phase Voltages: V AN , V BN , V CN . Powers: W 1 , W 2 . Line currents: I 1 , I 2 , I 3 . Phase Currents: I P1 , I P2 , I P3 . Neutral current: I N ) Print out these circuits and indicate on them where your watt meters and DVMs will connect.
• Under what conditions will one of the watt meters in a two-wattmeter measurement read negative powers with a balanced source feeding a balanced load?

Power Measurements on 3- φ Systems

• Set Load Rack switches so that all 3 resistances are nominally identical. Measure the resistor values before the experiment; their values should be closely matched.
• Connect the three-phase wye circuit as shown in figure 1.1. Connect in the power meters and the DVM's to allow measurement of the power flowing into the load, the line voltages ( V AB , V BC , and V CA ), the phase voltages across the resistors ( V AN , etc.), neutral current ( I N ) and the line currents. Note: It is important to monitor the current through power meters to ensure that it doesn't exceed the rated current. Low power can be observed when there are large voltages and large currents if there is a low power factor. Note that all measurements in this experiment are AC. Estimate all instrument readings for a source phase voltage of 120V (line voltage of 208V between phases). Select your meter scales accordingly.
• A voltage distribution panel is located on the side of the bench. Use a voltmeter to verify that the voltage is 208 volts between lines. Connect the three-phase variac to voltage distribution panel.
• Carefully adjust the variac output voltage to a phase voltage of 120V (208V line voltage).
• Without connecting the neutral switch at open position, measure and record all currents, voltages (line and phase) , and power it with different balanced loads of the resistor load cart. Record the results in table 1.1. Turn down the variac and shut the power off. Note: Measuring power requires the measurement of Voltage, Current, and the phase between them. The Fluke meter has a current clamp which is an inductive pick up that converts current to voltage for measurement by the instrument. The clamp has two scale settings and it is important to check that the meter is set to the same scale as the current clamp. The small black box meters should have their current “coil” connections in series with the circuit. For most measurements (all in this lab) you will want to short the input current connection to one of the voltage “coil” connections. These meters turn on when the voltage exceeds about 65 Volts. They do not read negative power (power flow from the load towards the source). If the meter shows current and voltage but no power then the direction of current through the device must be reversed. The watt meter reading should then be considered negative. Both Current and Voltage can be very high while there is still almost no power dissipated in the circuit when they are out of phase (low power factor). Hence it is important to always monitor the voltage, current, and power to ensure that none of them exceeds the ratings of the power meters.
• Move switch to closed position to connect an ammeter from the neutral of the resistor circuit to the neutral of the three-phase variac and observe the current flow. The current should be read on the 300 mA (or lower) scale.
• Measure all currents, voltages and power readings at same load settings of resistor load cart from step 5. Record all measurements in table 1.1. Turn down the variac and shut the power off.
• Connect the 3-phase circuit as shown in figure 1.2. Raise the line voltage to 120 Volts (phase voltage of 69.3V). Measure and record all currents, voltages and power readings at the same balanced load settings of resistor load cart from step 5. Note: There will not be enough ammeters for measuring all the phase currents and phase voltages at the same time. Measure the phase currents first, then reconnect to measure the phase voltages.
• Why do we use 208 V for the line voltage on the "wye" circuit but only 120 V for the line voltage on the "delta" load?
• Calculate the total load power in a wye(ү) and delta(Δ) configuration at each balance load from experiment, using the current and voltage data, by two different methods.
• Tabulate the total load power from the calculations from previous question and from the two- wattmeter measurement method. Discuss any differences.
• Verify phase and line voltage/current relationship of wye(ү) and delta(Δ) configuration circuits.

Discussion Questions

• Discuss any differences or similarities for the data obtained for the Y connection with or without neutral connection.
• Would the results be affected if wattmeter 2 were placed to measure the line current B-B’ and both wattmeter potential coils were brought to line C, instead of line B.
• Show a diagram for using only one wattmeter to measure the power in one phase of a balanced three-phase load.

• Engineering & Technology
• Electrical Engineering

6. POWER MEASUREMENT BY TWO WATTMETER METHOD

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Three Phase Power Measurement            

   The connection diagram for the measurement of power in three phase power measurement circuit using two wattmeter's method is shown in figure 1. This is irrespective of the circuit connection star or delta. The circuit may be taken as balanced or unbalanced one, balanced type being only a special case. Please note the connection of two wattmeter's. The current coil of the wattmeter's 1 and 2 in series with R and B phase with the pressure voltage coils being connected across R-Y and B-Y respectively. Y is the third phase in which no current coil is connected.

   If star connected circuit is taken as an example the total instantaneous power consumed in the circuit is, 

$$W= I_{RN}*V_{RN} + I_{YN}*V_{YN} + I_{BN}*V_{BN} ...(1)$$

    Each of the terms in the above expression equation (1) is the instantaneous power consumed by the phases. From the connection diagram, the circuit in and the voltages across the respective (current, pressure or voltage) coils in the wattmeter, W 1 are I RN   and $$V_{RY} = V_{RN} - V_{YN}$$. So, the instantaneous power measured by the wattmeter W 1 is $$W_1 = I_{RN}*V_{RY}$$.

Similarly the instantaneous power measured by the wattmeter W 2 is . $$W_2 = I_{BN}*V_{BY} = I_{BN}* (V_{BN} - V_{YN})$$

  Some of the two readings as given above is,

$$ = I_{RN} V_{RN} + I_{BN} V_{BN} - V_{YN} ( I_{RN} + I_{BN} ) ....(2)$$

$$and \ I_{RN} + I_{BN} + I_{YN} = 0$$

$$applying \ in \ equation \ (2),$$

$$W_1 + W_2 = I_{RN} V_{RN} + I_{BN} V_{BN} + V_{YN} I_{YN} .....(3)$$

Equation (1) is compared with equation (3)  to give the total instantaneous power consumed in the circuit . They are found to be same. The phasor diagram of three phase balanced star connected circuit is shown in figure 2.

Electrical Concepts

Tricky but Easy Electrical Engineering!

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

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 ]

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 = 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  

W 1 – W 2 = V L ISinØ ………………………………(2)

Dividing equation (2) by equation (1),

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

Power measured by first wattmeter W 1 = V L I Cos (30° – 0)

                                                               = V L I Cos 30°

                                                               = 0.866 V L I

Power measured by second wattmeter W 2 = V L I Cos (30° + 0)

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°

Power measured by first wattmeter W 1 = V L I Cos (30° – 60°)

                                                               = V L I Cos 30° ……[Cos (-Ɵ) = CosƟ ]

Power measured by second wattmeter W 2 = V L I Cos (30° + 60°)

                                                                     = V L I Cos 90°

                                                                     = 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°

Power measured by first wattmeter W 1 = V L I Cos (30° – 90°)

                                                               = V L I Cos 60°

                                                               = 0.5 V L I

Power measured by second wattmeter W 2 = V L I Cos (30° + 90°)

                                                               = – V L I Cos 60°

                                                               = -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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21 thoughts on “Two Wattmeter Method – Measurement of Three Phase Power”

this is really helpful.

thanks a lot

This is easily understandable

great note! Simple but clear!

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

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.

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

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