Measurement of three-phase power

A three-phase motor draws power that no single meter can honestly read unless you have access to the neutral wire — and in most real installations, you do not. Electrical engineers solve this with a trick that looks almost too simple: plant just two wattmeters in any two of the three supply lines, reference both of their voltage coils to the third, untouched line, and add the two readings. That sum is the total three-phase power, exactly, with no correction factor, whether the load is star or delta, balanced or unbalanced. Subtract the two readings instead of adding them and a second gift appears — the load's power factor angle drops out, with no separate phase measurement needed at all. This article builds the two-wattmeter method for a balanced three-phase load: why $W_1+W_2$ gives total power, why $\tan\theta = \sqrt3\,(W_1-W_2)/(W_1+W_2)$ gives the power factor, and the special readings — one wattmeter reading zero, or even negative — that examiners love to test.

The core insight

Two wattmeters with current coils in any two supply lines and voltage coils referenced to the third line always give total three-phase power as $W_1+W_2$ and the power factor angle as $\tan\theta=\sqrt3\,(W_1-W_2)/(W_1+W_2)$ , with no neutral connection required.

The core idea

A wattmeter has a current coil (in series with a line, sensing current) and a pressure coil (across two lines, sensing voltage); its reading is proportional to $VI\cos(\text{angle between them})$ . The two-wattmeter method places the current coils of two wattmeters in any two of the three lines R, Y, B, and both pressure coils from their own line back to the third, unused line. A common choice: Wattmeter 1's current coil in R, its voltage coil across R and B; Wattmeter 2's current coil in Y, its voltage coil across Y and B. Working through the phasor relationships for a balanced load (line voltage leads or lags phase voltage by $30^\circ$ ; load current lags or leads phase voltage by $\theta$ ), the two readings turn out to be:

W_1 = V_L I_L\cos(30^\circ-\theta), \qquad W_2 = V_L I_L\cos(30^\circ+\theta)

where: $V_L$ is the line voltage, $I_L$ is the line current, and $\theta$ is the load's power factor angle. Adding the two expressions with the cosine sum-to-product identity collapses them into the familiar balanced three-phase power formula, $W_1+W_2 = \sqrt3\,V_LI_L\cos\theta = P$ — the total active power falls out of just two meter readings, with no neutral wire and no correction term.

Figure

Two wattmeter connection diagram with current coils in lines R and Y and voltage coils of both meters referenced to line B, feeding a balanced three-phase load

Power factor angle. Subtracting the two readings instead isolates $\theta$ : $W_1-W_2 = V_LI_L\sin\theta$ . Dividing by the sum $W_1+W_2=\sqrt3 V_LI_L\cos\theta$ gives $\tan\theta=\sqrt3\,(W_1-W_2)/(W_1+W_2)$ , so power factor $\cos\theta$ is found from the two readings alone, with no separate phase-angle measurement. Useful special cases: at unity power factor, $W_1=W_2$ ; at $\theta=60^\circ$ lagging, $W_2=0$ ; beyond $\theta=60^\circ$ , $W_2$ turns negative and must be subtracted (in practice, by reversing that meter's current or pressure-coil connection and reading it as negative).

Why a single wattmeter is not enough. A lone wattmeter, with its current coil in one line and its pressure coil reading that line's voltage to neutral, measures the power of only one phase. Multiplying that single reading by three gives the correct total power only when the load is balanced and a low-impedance neutral point is actually available to tap. Real three-phase supplies and industrial loads frequently offer no such convenient neutral connection, so the single-wattmeter-times-three shortcut is unreliable outside a lab bench with a deliberately brought-out neutral. The two-wattmeter connection sidesteps the problem entirely: it never touches the neutral, so it is the method taught, tested, and used in practice for both star and delta loads.

Derivation: total power and power factor from two wattmeter readings

Derivation ATotal power and power factor angle from two wattmeter readings

The goal is to show, starting only from the two individual wattmeter reading formulas $W_1=V_LI_L\cos(30^\circ-\theta)$ and $W_2=V_LI_L\cos(30^\circ+\theta)$ already established for the balanced two-wattmeter connection, that their sum reproduces the standard total-power formula $P=\sqrt3\,V_LI_L\cos\theta$ exactly, and that their difference isolates the power-factor angle as $\tan\theta=\sqrt3\,(W_1-W_2)/(W_1+W_2)$ , using nothing beyond the cosine addition and subtraction identities.

Step 1. Expand both cosines using $\cos(A\mp B)=\cos A\cos B\pm\sin A\sin B$ with $A=30^\circ$ , $B=\theta$ : $W_1 = V_LI_L(\cos30^\circ\cos\theta+\sin30^\circ\sin\theta)$ and $W_2 = V_LI_L(\cos30^\circ\cos\theta-\sin30^\circ\sin\theta)$ .

Step 2. Add the two expansions. The $\sin30^\circ\sin\theta$ terms are equal and opposite, so they cancel, leaving $W_1+W_2 = 2V_LI_L\cos30^\circ\cos\theta$ . Since $\cos30^\circ=\sqrt3/2$ , this becomes $W_1+W_2 = 2V_LI_L\cdot\tfrac{\sqrt3}{2}\cos\theta = \sqrt3\,V_LI_L\cos\theta = P$ , exactly the balanced three-phase power formula already known from phasor analysis.

Step 3. Subtract instead. The $\cos30^\circ\cos\theta$ terms now cancel, leaving $W_1-W_2 = 2V_LI_L\sin30^\circ\sin\theta$ . Since $\sin30^\circ=1/2$ , this simplifies to $W_1-W_2 = V_LI_L\sin\theta$ .

Step 4. Divide the Step 3 result by the Step 2 result: $\dfrac{W_1-W_2}{W_1+W_2}=\dfrac{V_LI_L\sin\theta}{\sqrt3\,V_LI_L\cos\theta}=\dfrac{\sin\theta}{\sqrt3\cos\theta}=\dfrac{\tan\theta}{\sqrt3}$ . Rearranging gives $\tan\theta=\sqrt3\,\dfrac{W_1-W_2}{W_1+W_2}$ , a formula that needs only the two meter readings, no separate measurement of voltage, current magnitude, or phase.

Worked example

Two wattmeters connected to measure the power input to a balanced three-phase inductive load read $W_1=5000\ \text{W}$ and $W_2=2000\ \text{W}$ . Find the total power and the load power factor.

Step 1. Total power is the direct sum: $P = W_1+W_2 = 5000+2000 = 7000\ \text{W} = 7\ \text{kW}$ .

Step 2. Power factor angle from the difference formula: $\tan\theta = \sqrt3\,\dfrac{W_1-W_2}{W_1+W_2} = 1.732\times\dfrac{5000-2000}{7000}=1.732\times\dfrac{3000}{7000}=1.732\times0.4286=0.7423$ , so $\theta = \tan^{-1}(0.7423) = 36.6^\circ$ .

Step 3. Power factor: $\cos\theta = \cos(36.6^\circ) = \mathbf{0.803\ \text{lagging}}$ (lagging because the load is inductive), with total power $P=\mathbf{7\ \text{kW}}$ .

Competitive stretch

A balanced three-phase motor draws a line current of $20\ \text{A}$ from a $400\ \text{V}$ line at a lagging power factor of $0.6$ ; predict the two wattmeter readings. First, $\theta = \cos^{-1}(0.6) = 53.13^\circ$ . Then $W_1 = V_LI_L\cos(30^\circ-\theta) = 400\times20\times\cos(-23.13^\circ)=8000\times0.9191=7353\ \text{W}$ and $W_2 = 8000\times\cos(83.13^\circ)=8000\times0.1191=953\ \text{W}$ . Check: $P=\sqrt3\times400\times20\times0.6=8313.6\ \text{W}$ , and $W_1+W_2=7353+953=8306\ \text{W}\approx8313.6\ \text{W}$ (small rounding gap). Since $\theta=53.13^\circ<60^\circ$ , both readings stay positive, consistent with $W_2=953\ \text{W}$ being small but not zero.

Common traps

Forgetting that beyond $\theta=60^\circ$ (very low lagging power factor) one wattmeter reading goes negative and must be subtracted, not recorded as a small positive number, in $P=W_1+W_2$ .
Misremembering the power-factor-angle formula's sign or $\sqrt3$ factor — it is always $\tan\theta=\sqrt3\,(W_1-W_2)/(W_1+W_2)$ with $W_1$ the larger reading under the standard connection convention.
Assuming the two-wattmeter method needs a balanced load — it works for any three-phase load, balanced or unbalanced, for total power; the neat $\tan\theta$ formula is specifically the balanced-load simplification.
Believing a "one wattmeter per phase, three total" scheme with each voltage coil across a neutral is the standard method — the syllabus method uses exactly two wattmeters and three lines, with no neutral connection at all.
Applying the $\tan\theta$ formula to a load you already know is unbalanced — the sum $W_1+W_2$ still gives correct total power there, but the neat power-factor shortcut silently gives a meaningless number.

Ask the tutor: want to work through a case where one wattmeter reads negative, or practice going both directions between wattmeter readings and power factor?

Why do both wattmeter voltage coils connect back to the same third line?

That choice is what makes the algebra collapse so neatly. Referencing both pressure coils to the untouched third line gives each wattmeter a line-to-line voltage that is exactly $30^\circ$ away from the corresponding phase voltage, so each reading becomes $V_LI_L\cos(30^\circ\mp\theta)$ . Any other reference would break this clean $30^\circ$ offset and the simple sum-gives-total-power result would no longer hold in this tidy form. The third line is not measured directly; it only supplies the reference point.

Does the two-wattmeter method still work if the three-phase load is unbalanced?

Yes, for total power. The sum $W_1+W_2$ gives the correct total active power for any three-phase, three-wire load, balanced or not, because the underlying derivation only assumes conservation of power across the three lines, not equal phase currents. What breaks for an unbalanced load is the tidy $\tan\theta=\sqrt3\,(W_1-W_2)/(W_1+W_2)$ formula, since that shortcut was derived specifically using the balanced-load phasor symmetry; an unbalanced load needs a fuller phasor analysis to extract a meaningful power factor.

How do you physically get a negative wattmeter reading in the lab?

A wattmeter's pointer tries to move backward past zero when the angle between the current-coil current and pressure-coil voltage exceeds $90^\circ$ , which happens for one of the two meters once the load's power factor angle passes $60^\circ$ lagging. Since the pointer cannot indicate a negative number on its normal scale, the standard fix is to reverse the connections of either the current coil or the pressure coil on that one meter (not both), which flips the sign of its torque and lets it read a positive deflection — that reading is then recorded as negative when computing $P=W_1+W_2$ .