PW-B4 · Power and power factor

Measuring power: reading three phases with two wattmeters

Three-phase power cannot be measured with a voltmeter and an ammeter alone; the phase must be taken into account, yet two single-phase wattmeters are enough. Drag the current's phase yourself, see how the two readings split apart, and learn why their sum is always the total active power.

Drag the current's phase and watch the two meters

Dragging the current arrow on the left changes the phase difference φ from the voltage, and the two bars on the right are wattmeters W1, W2. The bars move separately, but their sum (gold) is always the total active power P = √3 V_L I_L cosφ, and once φ passes 60° W1 flips negative and turns red.

Drag the current arrow to change the phase φ between 0° and 85°. The numbers on the bars take V_L I_L as 1.

The two wattmeters and their sum (V_L I_L = 1)

W1 + W2 = 1.39 = P

W1 ≈ 0.39   W2 ≈ 0.99   cosφ ≈ 0.80

Why two and not three

By Blondel's theorem, the power delivered over n wires can be measured completely with n−1 wattmeters. Take one wire as the voltage reference and the wattmeter on that wire would always read 0, so it can be removed. A three-phase 3-wire system has n=3, so two wattmeters are enough: the two current coils go into two lines, and the two voltage coils are connected between their own line and the remaining third line.

The sum is always the total active power

Because of the geometry of the connection, the phase difference between the line voltage and line current each wattmeter reads is φ plus or minus 30°. W1 = V_L I_L cos(30°+φ), W2 = V_L I_L cos(30°−φ). Adding them, cos(30°+φ)+cos(30°−φ) = 2 cos30° cosφ = √3 cosφ, so the sum is exactly √3 V_L I_L cosφ, the total active power seen in A3. Even though 30° and φ are tangled in each reading, only P is left cleanly in the sum.

ObserveP = W1 + W2

The total active power is the sum of the two readings.

Choosecos(30°+φ) + cos(30°−φ) = 2 cos30° ?

cos(30°+φ)+cos(30°−φ) is 2 cos30° times cosφ.

Fill in2 cos30° = ?

2 cos30° is √3.

On your ownW1 + W2 = ? = P

So the sum is √3 V_L I_L cosφ = P.

The difference is reactive, the negative is a signal

Subtracting them, cos(30°−φ)−cos(30°+φ) = 2 sin30° sinφ = sinφ, so √3(W2−W1) = √3 V_L I_L sinφ = Q. So the two readings alone give active P, reactive Q and, from them, the power factor. When the power factor is below 0.5, 30°+φ passes 90°, the cosine goes negative and the W1 needle turns backward; then reversing the voltage terminals and subtracting the reading makes the sum come out right. A negative reading is not a fault but a sign of a low power factor.

Back to the first screen

Dragging the current to grow the phase φ made W1 and W2 move separately, and once φ passed 60°, W1 flipped negative. Even so, their sum (gold) stayed at √3 V_L I_L cosφ at every moment, the total active power of A3. The 30° that comes from the connection enters one reading added and the other subtracted, so it cancels cleanly in the sum, which is why three-phase power can be read in full without a meter on the third line.

Two-wattmeter method: the power of a three-phase 3-wire system is measured with 2 single-phase wattmeters (Blondel's theorem: n wires → n−1 meters). Each meter reads V_L I_L cos(30°∓φ), the sum W1+W2 = √3 V_L I_L cosφ = P (total active power), and the difference √3(W2−W1) = Q. Below a power factor of 0.5, W1 goes negative, and the two readings together give active power, reactive power and power factor.

The next step

Here section B (power and power factor) ends. If so far we looked at power from the load side, the next section C follows the path of that power from the power plant to the load. Its first unit (PW-C1) models the transmission line with resistance and reactance, starting from the point that a line is not a mere wire but a path with impedance, and lays the groundwork for where voltage drop and loss arise.