A single wattmeter can also do the average three-phase power measurement that is balanced so that P_{1} = P_{2} = P_{3}; the total power is three times the reading of that one wattmeter.

**Contents**show

*Make sure to read what is three-phase circuit first.*

After we learn about three phase circuit, we will learn:

- Balanced three phase voltage
- Balanced three phase power
- Unbalanced three phase power
- Three phase power measurement

## Three-Phase Power Measurement Explained in Detail

The **three-wattmeter method** of power measurement, shown in Figure.(1) will work regardless of whether the load is balanced or unbalanced, wye- or delta-connected.

Figure 1. Three-wattmeter method for measuring three-phase power. |

The three-wattmeter method is well suited for power measurement in a three-phase system where the power factor is constantly changing. The total average power is the algebraic sum of the three wattmeter readings,

(1) |

where P_{1}, P_{2}, and P_{3} correspond to the readings of wattmeters W_{1}, W_{2}, and W_{3}, respectively. Notice that the common or reference point *o* in Figure.(1) is selected arbitrarily. If the load is wye-connected, point *o* can be connected to the neutral point *n*.

For a delta-connected load, point *o* can be connected to any point. If point *o* is connected to point *b*, for example, the voltage coil in wattmeter W_{2} reads zero and P_{2} = 0, indicating that wattmeter W_{2} is not necessary. Thus, two wattmeters are sufficient to measure total power.

The two-wattmeter method is the most commonly used method for three-phase power measurement. The two wattmeters must be properly connected to any two phases, as shown typically in Figure.(2).

Notice that the current coil of each wattmeter measures the line current, while the respective voltage coil is connected between the line and the third line and measures the line voltage.

Figure 2. Two-wattmeter method for measuring three-phase power. |

Also, notice that the ± terminal of the voltage coil is connected to the line to which the corresponding current coil is connected.

Although the individual wattmeters no longer read the power taken by any particular phase, the algebraic sum of the two wattmeter readings equals the total average power absorbed by the load.

Regardless of whether it is wye- or delta-connected, balanced or unbalanced. The total real power is equal to the algebraic sum of the two wattmeter readings,

(2) |

We will show here that the method works for a balanced three-phase system. It is wise to read how does wattmeter work first.

Consider the balanced, wye-connected load in Figure.(3). Our objective is to apply the two-wattmeter method to find the average power absorbed by the load. Assume the source is in the abc sequence and the load impedance **Z**_{Y} = **Z**_{Y}∠θ.

Figure 3. The two-wattmeter method applied to a balanced wye load |

Due to the load impedance, each voltage coil leads its current coil by θ, so that the power factor is cosθ. We recall that each line voltage leads the corresponding phase voltage by 30◦.

V_{ab} is θ + 30 ◦, and the average power read by wattmeter W_{1} is

(3) |

Similarly, we can show that the average power read by wattmeter 2 is

(4) |

We now use the trigonometric identities

(5) |

to find the sum and the difference of the two wattmeter readings in Equations.(3) and (4):

(6) |

since 2 cos 30◦ = √3 shows that the sum of the wattmeter readings gives the total average power,

(7) |

Similarly,

(8) |

since 2 sin 30◦ = 1 shows that the difference of the wattmeter readings is proportional to the total reactive power, or

(9) |

From Equations.(7) and (9), the total apparent power can be obtained as

(10) |

Dividing Equations.(9) by (7) gives the tangent of the power factor angle as

(11) |

from which we can obtain the power factor as pf= cos θ. Thus, the two wattmeter method not only provides the total real and reactive powers, but it can also be used to compute the power factor.

From Equations.(7), (9), and (11), we conclude that:

1. If P2 = P1, the load is resistive.

2. If P2 > P1, the load is inductive.

3. If P2 < P1, the load is capacitive.

Although these results are derived from a balanced wye-connected load, they are equally valid for a balanced delta-connected load. However, the two-wattmeter method cannot be used for power measurement in a three-phase four-wire system unless the current through the neutral line is zero.

We use the three-wattmeter method to measure the real power in a three-phase four-wire system.

Read also : nodal and supernode ac circuits

## Measurement of Three-Phase Power Examples

1. Three wattmeters W_{1}, W_{2}, and W_{3} are connected, respectively, to phases *a*, *b*, and *c* to measure the total power absorbed by the unbalanced wye connected load in Figure.(4). (a) Predict the wattmeter readings. (b) Find the total power absorbed.

Figure 4 |

*Solution:*

Assume that the wattmeters are properly connected as in Figure.(5).

Figure 5 |

(a) The voltages

while

We calculate the wattmeter readings as follows:

(b) The total power absorbed is

We can find the power absorbed by the resistors in Figure.(5) and use that to check or confirm this result.

which is exactly the same thing.

2. The two-wattmeter method produces wattmeter readings P_{1} = 1560 W and P_{2} = 2100 W when connected to a delta-connected load.

If the line voltage is 220 V, calculate: (a) the per-phase average power, (b) the per phase reactive power, (c) the power factor, and (d) the phase impedance.

*Solution:*

We can apply the given results to the delta-connected load.

(a) The total real or average power is

The per-phase average power is then

(b) The total reactive power is

so that the per-phase reactive power is

(c) The power angle is

Hence, the power factor is

It is a leading pf because Q_{T} is positive or P_{2} > P_{1}.

(d) The phase impedance is **Z**_{P} = **Z**_{P}∠θ. We know that θ is the same as the pf angle; that is, θ = 14.57◦.

We recall that for a delta-connected load, V_{P} = V_{L} = 220 V.

Hence,

and

3. The three-phase balanced load in Figure.(2) has impedance per phase of **Z**_{Y} = 8 + j6 Ω. If the load is connected to 208-V lines, predict the readings of the wattmeters W_{1} and W_{2}. Find P_{T} and Q_{T}.

*Solution:*

The impedance per phase is

so that the pf angle is 36.87◦. Since the line voltage V_{L} = 208 V, the line current is

Then

Thus, wattmeter 1 reads 980.48 W, while wattmeter 2 reads 2478.1 W. Since P_{2} > P_{1}, the load is inductive.

This is evident from the load **Z**_{Y} itself.

Next,

and

Very much good

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