PW-B4 · Power and power factor

Measuring power: reading three phases with two wattmeters

Three-phase power needs more than a voltmeter and ammeter; the phase must be seen too. Yet two single-phase wattmeters are enough. Drag the current phase to see how the two readings split, and learn why their sum is always the total active power.

Drag the current phase and watch the two meters

Drag the current arrow on the left to change its phase φ relative to the voltage. The two bars on the right are wattmeters W1 and W2. The bars move on their own, but their sum (gold) is always the total active power P = √3 V_L I_L cosφ. Past φ = 60 degrees, W1 flips negative.

Drag the current arrow to set the phase φ.

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 carried by n wires can be measured completely with n−1 wattmeters, because if one wire is taken as the voltage reference its wattmeter reads zero and can be removed. A three-phase three-wire system has n = 3, so two wattmeters suffice. The two current coils go in two lines, and the two voltage coils each tie to the remaining third line.

The sum is always the total active power

Because of the wiring geometry, each wattmeter sees the product of a line voltage and a line current at a phase shifted by 30 degrees: 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 from A3. The 30 degrees and φ are tangled in each reading, yet the sum leaves only P.

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 the two, cos(30°−φ)−cos(30°+φ) = 2 sin30° sinφ = sinφ, so √3(W2−W1) = √3 V_L I_L sinφ = Q. The two readings alone thus give active P, reactive Q and even the power factor together. When the power factor is below 0.5, 30°+φ passes 90 degrees and its cosine turns negative, so the W1 needle runs backward. You then reverse its voltage terminals and subtract that reading to keep the sum right — the negative is not a fault but a signal of low power factor.

Back to the first screen

Dragging the current to grow the phase φ moved W1 and W2 apart, and past 60 degrees W1 flipped negative. Yet their sum (gold) stayed √3 V_L I_L cosφ at every moment — exactly the total active power of A3. The 30-degree wiring angle enters both readings and complements their phases, so it vanishes cleanly in the sum. That is how three-phase power is read in full without a meter on the third line.

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

The next step

Section B (power and power factor) ends here. So far we looked at power at the load; in the coming section C we follow its journey from the plant to the load. The first unit (PW-C1) models a transmission line as resistance and reactance. Seeing the wire not as a plain conductor but as a path with impedance lays the ground for where voltage drop and loss arise.