A wattmeter is an instrument used to measure the electrical power supplied in a circuit. It consists of a voltage coil connected across the load and a current coil connected in series with the load. So that deflection of the pointer is proportional to the magnetic field produced by the two coils, which is nothing but proportional to power.

#### A wattmeter can measure power in single-phase circuits directly. But in order to measure 3-phase power, one or more wattmeters are used. The various methods available for the measurement of power in 3-phase circuits are,- One wattmeter method
- Two wattmeter mehtod
- Three wattmeter method

In this article let us see about the measurement of power by one wattmeter method.

## One Wattmeter Method of Power Measurement :

This method is used only for a balanced load. When a 3-phase system is connected to a balanced 3-phase load, then the entire power consumption of the load can be determined by using a single wattmeter, whose arrangement is shown below.

The wattmeter current coil is connected in series with any one of the lines and hence carries the full line current. The voltage coil is connected in such a way that, one end to the line in which the current coil is placed, the other end is connected to the remaining two lines through an SPDT switch, and hence the line voltage will be impressed on it.

Here an SPDT switch is used to change the voltage coil connection between two phases. Hence, the sum of readings of the wattmeter with two different positions of the switch will give the total power consumed by the load. Let the phase voltages of the phases R, Y, and B be V_{R}, V_{Y}, and V_{B} respectively the phase/line currents of the phases R, Y, and B be I_{R}, I_{Y}, and I_{B} respectively.

#### As the system is a balanced one, the impedances of the phases must be equal.The phasor diagram of the balanced 3-phase system is shown below.When SPDT switch is at position 1, the wattmeter reading indicates,*W*_{1} = V_{RY} I_{R} cos (30° + Φ)*= √3VI cos (30° + Φ)*When SPDT switch is thrown to position 2, the wattmeter reading indicates,*W*_{2} = V_{RB} I_{R} cos (30° - Φ)*= √3VI cos (30° - Φ)*Total power i.e., the sum of the readings of two wattmeters is,

*W*

_{1}= V_{RY}I_{R}cos (30° + Φ)*= √3VI cos (30° + Φ)*

*W*

_{2}= V_{RB}I_{R}cos (30° - Φ)*= √3VI cos (30° - Φ)*