A wattmeter comprises of 2 coils. Those are:

**Potential or Pressure coil**

This possesses high resistance

**Current coil**

This possesses low resistance

**Methods**

There are mainly 3 methods with the help of which it is easy to measure power in 3 phase load. Those are:

- One wattmeter method
- Two wattmeter method
- Three wattmeter method

As per this diagram, the current coil is linked to a sequence with the line that carries the current. Across 2 points, a pressure coil is linked where its complete potential difference is measured.

The readings as per wattmeter state the proportionality of resultant of current via its current coil. Now, in accordance withit, we can easily seeitspotential difference across cosine of angle in between current and voltage and its pressure coil.

**Three-Wattmeters Method**

This figure depicts a star connected load.

This figurine depicts a delta connected load.

As per both these figures, we can clearly see that 3 wattmeters are linked with each other (in 3 phases). These phases are related to load whose connection can be differentiated as either delta connection or as a star connection.

Inside a wattmeter, 2 elements work in a synchronized pattern making its working complete.

- The pressure coil helps in the measurement of phase voltage of a certain phase.
- And the current is carried by a current coil that too in a single phase.

These 2 in combination makes sure that measurement of every individual wattmeter relays the amount of power inside one phase. In load, this power in totality is explained via the algebraic sum of readings regarding 3 wattmeters.

**Limitations involved in this method**

During the application of this method, there are certain difficult aspects that one had to face. Amongst those, 2 are highly important has to be kept in consideration.

- Under ordinary circumstance or prospect when it is about delta connected load, practical aspect suggests that phase of load should not be broken.
- The usual requirement for connection mainly revolves around neutral point. But when it is about star connected load, in this case, it is generally not possible to achieve that.

There isnâ€™t any specific rule that states that power measurement, the necessary requirement should revolve around 3 wattmeters. For the same purpose, we can use 2 wattmeters too.

**Two-wattmeters method**

Here are 2 figures.

This figure depicts a star connected load.

This figure depicts a delta connected load.

These figurines are based on 2 wattmeter arrangement. As per this method, in these 2 wattmeters, the insertion of current coils is done in any of the 2 lines. Its potential or pressure coils are connected to third line.

**Proof**

The provided 3 loads efficiently absorb the sum of instantaneous powers. Its proof can be found by the following equations and explanations.

Let us consider those 3 loads as L3, L2 and L1.

Indication of instantaneous power is relayed via W1 and W2.

In this instance, let us assume the load to be star connected load.

(Instead of this load, one can also assume to opt for delta connected load. And if delta connected load is not available, its easy and best replacement can be equivalent star connected load.)

The most important thing that we have to keep in consideration is the voltage direction when it passes via circuits. This has to be the same when we are trying to establish the readings related to this 2 wattmeter.

Now, with the help of Kirchhoffâ€™s point law, its equation will stand as,

Here, the power which is absorbed by the load depicted by L1 is denoted by P1. Similarly for load L2, the power is depicted by P2 and for load L3, the depiction of power is via P3.

Power in totality that is absorbed is equivalent to the sum of W2 and W1.

This proof can be only considered true when the load is either unbalanced or balanced.

Now in this case, it is a generalized sense that in a star connected load presence of neutral connection is impossible. This is because its arrangement is in a 3 phase and 4 wire format.

Now, if we assume that it has a neutral connection, we have to make sure that there is no flow of neutral current. And for this, the connection has to be exactly balanced. On failure to this, the equation as per Kirchhoffâ€™s point law will be,

iN + iB + iY + iR is equivalent to zero.

This derivation is obtained in accordance with instantaneous readings. Because of the inertia of wattmeter in a moving system is unable to follow swiftly the frequent changes that take place in that cycle. Therefore, the resultant showcases average power. So, the sum of total power regarding W2 and W1 is,

**Two wattmeters method related to balanced load**

When 2 wattmeters are used, we can find the total power consumption. This power is consumed by balanced load. In case the supposition of inductive load is explained via vector diagram. This assumption is for star connected load which is in a balanced state.

This is the vector figure.

Now, let us assume the entire problem in r.m.s. value. This is an alternate choice over instantaneous values.

As per this value, a star connected system has 3 phase voltages. Let us name those 3 r.m.s. values as E_{B}, E_{Y}, E_{R}.

B, Y and R represent the initials of primary colors, blue, yellow and red.

For current, the names for their r.m.s. values are I_{B}, I_{Y}, I_{R}.

The assumptions for these currents and voltages are sinusoidal or follow a curved path. The representations for both are highlighted via vectors. As per this, we can see the currents lagging behind phase voltages by É¸.

**For Pressure**

In accordance to the above figure, current which passes through the wattmeter, denoted by W1 is equal to IR. So as per vector representation, across the pressure coil of the wattmeter, its potential difference would be,

W1 = ERB

ER â€“ EB = ERB

So,

W1 â€“ ER â€“ EB

With the assistance and by compounding EB and ER in a reverse movement, ERBâ€™s value is obtained. After careful observation it is clearly understood that the phase difference between IR and ERB is equivalent to (30Â° â€“ É¸).

So, the reading as per wattmeter W1 will be,

IR x ERB x cos (30Â° â€“ É¸) (equation 1)

**For Current**

Now in similar way, when current passes through wattmeter W2, it is denoted via IY.

So as per vector representation, across the pressure coil of the wattmeter, its potential difference would be,

W2 = EYB

EY â€“ EBI = EYB

So,

W2 = EY â€“ EBI

After compounding EB and EY in a reverse movement, EYBâ€™s value is obtained. . After careful observation it is clearly understood that the phase difference between IY and EYB is equivalent to (30Â° + É¸).

So, the reading as per wattmeter W2 will be,

IY x EYB x cos (30Â° + É¸) (equation 2)

Since the load is found to be in a balanced position, line current will be,

IL = IY = IR

Similarly, for line voltage it will be,

EL = EYB = ERB

Therefore the equation will be,

Its total power will be,

P = W2 + W1

So, for 2 wattmeters, the sum of their readings that would relay the total power consumption with regards to 3 phase load. It is important that we must consider the fact that we have assumed the phase order of RYB for establishing the readings for this 2 wattmeter. Also we need to consider the fact that if this sequence or order changes or reverses, its resultant will also get reversed.

Variations in case of wattmeter readings

The above factors highlights lagging power factor from whom the equations will be,

It is evident from the equations of wattmeters that these separate readings are dependent on both its power factor and load.

There are 4 cases that will help us understand this variation.

In the first case,

#1

When the value of É¸ is equal to 0, it can be stated that the load is resistive. As per this, it can be also assumed that the value of power factor is considered to be unity.

So, it can be stated in equational form that,

W2 = W1 = IL x EL x cos 30Â°

According to this, it is evident that readings for every wattmeter will be both opposite and equal. This reading is an up-scale reading.

Now for the second case,

#2

When the value of É¸ is equal to 60Â°, it can be stated that the load is lagging. As per this, it can be also assumed that the value of power factor is considered to be 0.5.

So, it can be stated in equational form that,

W2 = IL x EL x cos 60Â° + 30Â°

Therefore, W2 = 0

**Two-wattmeters method**

**Â **

Here are 2 figures.

This figure depicts a star connected load.

This figure depicts a delta connected load.

These figurines are based on 2 wattmeter arrangement. As per this method, in these 2 wattmeters, the insertion of current coils is done in any of the 2 lines. Its potential or pressure coils are connected to third line.

**Proof**

The provided 3 loads efficiently absorb the sum of instantaneous powers. Its proof can be found by the following equations and explanations.

Let us consider those 3 loads as L3, L2, and L1.

An indication of instantaneous power is relayed via W1 and W2.

In this instance, let us assume the load to be star connected load.

(Instead of this load, one can also assume to opt for delta connected load. And if delta connected load is not available, its easy and best replacement can be equivalent star connected load.)

The most important thing that we have to keep in consideration is the voltage direction when it passes via circuits. This has to be the same when we are trying to establish the readings related to this 2 wattmeter.

Now, with the help of Kirchhoffâ€™s point law, its equation will stand as,

Here, the power which is absorbed by the load depicted by L1 is denoted by P1. Similarly, for load L2, the power is depicted by P2, and for load L3, the depiction of power is via P3.

Power in totality that is absorbed is equivalent to the sum of W2 and W1.

This proof can be only considered true when the load is either unbalanced or balanced.

Now, in this case, it is a generalized sense that in a star connected load presence of a neutral connection is impossible. This is because its arrangement is in a 3 phase and 4 wire format.

Now, if we assume that it has a neutral connection, we have to make sure that there is no flow of neutral current. And for this, the connection has to be exactly balanced. On failure to this, the equation as per Kirchhoffâ€™s point law will be,

iN + iB + iY + iR is equivalent to zero.

This derivation is obtained in accordance with instantaneous readings. Because of the inertia of wattmeter in a moving system, it is unable to follow swiftly the frequent changes that take place in that cycle. Therefore, the resultant showcases average power. So, the sum of total power regarding W2 and W1 is,

**Two wattmeters method related to balanced load**

When 2 wattmeters are used, we can find the total power consumption. This power is consumed by balanced load. In case the supposition of inductive load is explained via vector diagram. This assumption is for star connected load which is in a balanced state.

This is the vector figure.

Now, let us assume the entire problem in r.m.s. value. This is an alternate choice over instantaneous values.

As per this value, a star connected system has 3 phase voltages. Let us name that 3 r.m.s. values as E_{B}, E_{Y}, E_{R}.

B, Y, and R represent the initials of primary colors, blue, yellow and red.

For current, the names for their r.m.s. values are I_{B}, I_{Y}, I_{R}.

The assumptions for these currents and voltages are sinusoidal or follow a curved path. The representations for both are highlighted via vectors. As per this, we can see the currents lagging behind phase voltages by É¸.

**For Pressure**

In accordance with the above figure, current which passes through the wattmeter, denoted by W1 is equal to IR. So as per vector representation, across the pressure coil of the wattmeter, its potential difference would be,

W1 = ERB

ER â€“ EB = ERB

So,

W1 â€“ ER â€“ EB

With the assistance and by compounding EB and ER in a reverse movement, ERBâ€™s value is obtained. After careful observation, it is clearly understood that the phase difference between IR and ERB is equivalent to (30Â° â€“ É¸).

So, the reading as per wattmeter W1 will be,

IR x ERB x cos (30Â° â€“ É¸) (equation 1)

**For Current**

Now in a similar way, when current passes through wattmeter W2, it is denoted via IY.

So as per vector representation, across the pressure coil of the wattmeter, its potential difference would be,

W2 = EYB

EY â€“ EBI = EYB

So,

W2 = EY â€“ EBI

After compounding EB and EY in a reverse movement, EYBâ€™s value is obtained. After careful observation, it is clearly understood that the phase difference between IY and EYB is equivalent to (30Â° + É¸).

So, the reading as per wattmeter W2 will be,

IY x EYB x cos (30Â° + É¸) (equation 2)

Since the load is found to be in a balanced position, line current will be,

IL = IY = IR

Similarly, for line voltage, it will be,

EL = EYB = ERB

Therefore the equation will be,

Its total power will be,

P = W2 + W1

So, for 2 wattmeters, the sum of their readings that would relay the total power consumption with regards to 3 phase load. It is important that we must consider the fact that we have assumed the phase order of RYB for establishing the readings for this 2 wattmeter. Also, we need to consider the fact that if this sequence or order changes or reverses, its resultant will also get reversed.

**Variations in case of wattmeter readings**

The above factors highlights lagging power factor from whom the equations will be,

It is evident from the equations of wattmeters that these separate readings are dependent on both its power factor and load.

There are 4 cases that will help us understand this variation.

In the first case,

**#1**

When the value of É¸ is equal to 0, it can be stated that the load is resistive. As per this, it can also be assumed that the value of power factor is considered to be unity.

So, it can be stated in equational form that,

W2 = W1 = IL x EL x cos 30Â°

According to this, it is evident that readings for every wattmeter will be both opposite and equal. This reading is an up-scale reading.

Now for the second case,

**#2**

When the value of É¸ is equal to 60Â°, it can be stated that the load is lagging. As per this, it can also be assumed that the value of power factor is considered to be 0.5.

So, it can be stated in equational form that,

W2 = IL x EL x cos 60Â° + 30Â°

Therefore, W2 = 0

So, the measurement of power is done via W1.

**#3**

As per this case, we can see that W1â€™s reading is positive whereas its counterpart, W2â€™s reading is just the opposite. And even for leading p.f., the required condition is just reversed. The reason for W1 to give a reading that is reversed or negative is due to its phase angle which is more than 90Â° between voltage and current.

W2 â€“ W1 = total power

Here, the reading for W2 is termed as backward or down scale. And in order to acquire its readings, it is highly essential that either its current coil or its pressure coil is reversed. This will give its reading in negative.

**#4**

In accordance to this 2 readings, both of their magnitudes are equivalent, but their signs are opposite.

Therefore, sum of W1 and W2 is zero.

The consideration and sign for lagging angles are taken as positive. But when power factor with respect to a wattmeter changes, its readings will be,

Therefore, via angles related to 2 wattmeter leads to readings which are interchanged. So, the equations when power is leading will be,

**Power factor in case the load is balanced**

In case the balanced load has a lagging power factor, and it is clearly showcased that current and voltage are sinusoidal, its equation will be,

After the value of tan É¸ is obtained, with the help of trigonometrical tables the value of cos É¸ can also be calculated.

One of the most important while taking this reading is the consideration for W2, which is obtained by reversing the significant pressure coil. So now, the equation will be,

The expression for power factor can also be represented in ratio format related to the readings of 2 wattmeters.

**Watt ratio curve**

Plotting of curve between cos É¸ and Î± is known as watt ratio curve. This curve will look like,

**With two wattmeters reactive volt amperes**

We are already aware that,

As per this, tangent of angle of lag which is between phase voltages and phases current of a circuit is equivalent to ratio of reactive power over true power. Therefore, the difference of reactive powerâ€™s reading with respect to 2 wattmeters is utilized for power measurement of 3 phase circuit.

**Mathematical proof with figure**

**One-wattmeter method**

In accordance with this method, in any certain one line, the current coil is linked, and its connection for pressure coil is alternately placed between the other 2 lines. Inside a balanced load, important 2 readings which are acquired are based on wattmeter method.

**Links of Previous Main Topic:-**

- Current Electricity Basic Concepts
- Introduction to Alternating Current
- Introduction Three Phase A C Circuits
- Advantages of Polyphase Systems

**Links of NextÂ Electrical Engineering Topics:-**

- Measurement of Power in 3 Phase Circuit
- Measurement of Reactive Volt Amperes
- Unbalanced Star Connected Load
- Unbalanced Delta Connected Load
- Power Factor Improvement
- Magnetic Field
- General Aspects
- Elementary Theory of Ideal Transformer
- General Aspects Polyphase Induction Motors
- Single Phase Motors
- Characteristics of D C Generators
- Measuring Instruments
- Power Supply System