Two-Wattmeter Method | Power Analysis

Two-Wattmeter Method — Three-Phase Power Measurement

For any three-phase, three-wire load — balanced or not, star or delta — the total power can be measured with just two wattmeters. Each wattmeter carries one line current in its current coil and the corresponding line-to-line voltage across its potential coil; the third line is the common reference. The algebraic sum of the two readings gives the total real power, and for a balanced load the difference yields the reactive power and hence the power factor.

Power Analysis · Three-Phase Circuits · 5 solved problems

fKey Relations (balanced load)

\[ \begin{aligned} W_1 &= V_L I_L \cos\!\left(\theta + 30^{\circ}\right), \quad W_2 = V_L I_L \cos\!\left(\theta - 30^{\circ}\right) \\[2pt] P_T &= W_1 + W_2 = \sqrt{3}\,V_L I_L \cos\theta \\[2pt] Q_T &= \sqrt{3}\,\left(W_2 - W_1\right) = \sqrt{3}\,V_L I_L \sin\theta \\[2pt] \tan\theta &= \sqrt{3}\,\frac{W_2 - W_1}{W_2 + W_1} \end{aligned} \]

Here \(\theta\) is the impedance (power-factor) angle. A lagging load makes \(W_2 > W_1\); when the power factor falls below \(0.5\) \((\theta > 60^{\circ})\), the smaller reading \(W_1\) becomes negative and its meter must be reversed to read.

Standard two-wattmeter connection for a three-wire, three-phase load. \(W_1\) measures \(V_{AB}I_A\)-type product, \(W_2\) measures \(V_{CB}I_C\); line B is the common potential-coil reference.

Demonstrative VideoWalkthrough

Problem 14-Wire Power (Three Wattmeters)

Find the total power absorbed by the unbalanced four-wire load with the phase voltages and line currents given below.

Four-wire three-phase load for Problem 1

\[\begin{aligned} \mathbf{V}_{AN} &= 100\angle 0^{\circ}, \quad \mathbf{V}_{BN} = 100 \angle 120^{\circ}, \quad \mathbf{V}_{CN} = 100 \angle -120^{\circ}~\mathrm{V} \\ \mathbf{I}_{a} &= 6.67 \angle 0^{\circ}, \quad \mathbf{I}_{b} = 8.94\angle 93.44^{\circ}, \quad \mathbf{I}_{c} = 10 \angle -66.87^{\circ}~\mathrm{A} \end{aligned}\]

Solution

Note on method. Because the neutral \(N\) is accessible (a four-wire system), each phase power is measured directly with its own wattmeter — the three-wattmeter method. The two-wattmeter method applies to three-wire systems where no neutral is brought out.

Each wattmeter reads the real part of the phase voltage times the conjugate of its line current:

\[ \begin{aligned} P_1 &= \operatorname{Re}\!\left(\mathbf{V}_{AN}\mathbf{I}_a^{*}\right) = V_{AN} I_a \cos\!\left(\theta_{V_{AN}}-\theta_{I_a}\right) \\ &= 100 \times 6.67 \times \cos\!\left(0^{\circ}-0^{\circ}\right) = 667~\mathrm{W} \end{aligned} \]

For phase B:

\[ \begin{aligned} P_2 &= \operatorname{Re}\!\left(\mathbf{V}_{BN}\mathbf{I}_b^{*}\right) = V_{BN} I_b \cos\!\left(\theta_{V_{BN}}-\theta_{I_b}\right) \\ &= 100 \times 8.94 \times \cos\!\left(120^{\circ}-93.44^{\circ}\right) = 800~\mathrm{W} \end{aligned} \]

For phase C:

\[ \begin{aligned} P_3 &= \operatorname{Re}\!\left(\mathbf{V}_{CN}\mathbf{I}_c^{*}\right) = V_{CN} I_c \cos\!\left(\theta_{V_{CN}}-\theta_{I_c}\right) \\ &= 100 \times 10 \times \cos\!\left(-120^{\circ}+66.87^{\circ}\right) = 600~\mathrm{W} \end{aligned} \]

The total absorbed power is the sum of the three phase powers:

\[ P_T = P_1 + P_2 + P_3 = 667 + 800 + 600 = 2067~\mathrm{W} \]

Answer\(P_T = 2067~\mathrm{W}\)

Problem 2Readings → pf & Impedance

The two-wattmeter method gives readings \(P_1 = 1560~\mathrm{W}\) and \(P_2 = 2100~\mathrm{W}\) when connected to a delta-connected load. If the line voltage is 220 V, calculate:

the per-phase average power,
the per-phase reactive power,
the power factor, and
the phase impedance.

Solution

Total and per-phase real power:

\[ P_T = P_1 + P_2 = 1560 + 2100 = 3660~\mathrm{W}, \qquad P_p = \tfrac{1}{3}P_T = 1220~\mathrm{W} \]

Total and per-phase reactive power:

\[ Q_T = \sqrt{3}\,(P_2 - P_1) = \sqrt{3}\,(2100 - 1560) = 935.3~\mathrm{VAR}, \qquad Q_p = \tfrac{1}{3}Q_T = 311.77~\mathrm{VAR} \]

Power-factor angle and power factor:

\[ \theta = \tan^{-1}\frac{Q_T}{P_T} = \tan^{-1}\frac{935.3}{3660} = 14.33^{\circ}, \qquad \cos\theta = 0.9689~(\text{lagging}) \]

For a delta load the phase voltage equals the line voltage \((V_p = V_L = 220~\mathrm{V})\). From \(P_p = V_p I_p \cos\theta\):

\[ I_p = \frac{P_p}{V_p \cos\theta} = \frac{1220}{220 \times 0.9689} = 5.723~\mathrm{A} \]

The phase impedance magnitude and complex value:

\[ Z_p = \frac{V_p}{I_p} = \frac{220}{5.723} = 38.44~\Omega, \qquad \mathbf{Z}_p = 38.44\,\angle 14.33^{\circ}~\Omega \]

Answer\(P_p = 1220~\mathrm{W},\ Q_p = 311.77~\mathrm{VAR},\ \cos\theta = 0.969,\ \mathbf{Z}_p = 38.44\,\angle 14.33^{\circ}~\Omega\)

Problem 3Predict W₁ and W₂

A balanced three-phase load has impedance per phase \(\mathbf{Z}_Y = 8 + j6~\Omega\). If the load is connected to 208 V lines, predict the readings of wattmeters \(W_1\) and \(W_2\), and find \(P_T\) and \(Q_T\).

Solution

Express the per-phase impedance in polar form; the angle is the power-factor angle:

\[ \mathbf{Z}_Y = 8 + j6 = 10\,\angle 36.87^{\circ}~\Omega \;\Rightarrow\; \theta = 36.87^{\circ} \]

For the wye load the phase voltage is \(V_L/\sqrt{3}\), so the line current is:

\[ I_L = \frac{V_p}{|\mathbf{Z}_Y|} = \frac{208/\sqrt{3}}{10} = 12~\mathrm{A} \]

The two wattmeter readings:

\[ \begin{aligned} W_1 &= V_L I_L \cos\!\left(\theta + 30^{\circ}\right) = 208 \times 12 \times \cos\!\left(36.87^{\circ}+30^{\circ}\right) = 980.5~\mathrm{W} \\ W_2 &= V_L I_L \cos\!\left(\theta - 30^{\circ}\right) = 208 \times 12 \times \cos\!\left(36.87^{\circ}-30^{\circ}\right) = 2478.1~\mathrm{W} \end{aligned} \]

Since \(W_2 > W_1\), the load is inductive (lagging) — consistent with the \(+j6\) reactance in \(\mathbf{Z}_Y\).

Total real and reactive power:

\[ \begin{aligned} P_T &= W_1 + W_2 = 3.459~\mathrm{kW} \\ Q_T &= \sqrt{3}\,(W_2 - W_1) = \sqrt{3}\,(1497.6) = 2.594~\mathrm{kVAR} \end{aligned} \]

Answer\(W_1 = 980.5~\mathrm{W},\ W_2 = 2478.1~\mathrm{W},\ P_T = 3.459~\mathrm{kW},\ Q_T = 2.594~\mathrm{kVAR}\)

Additional Practice Problems

Two further worked examples: recovering the power factor directly from a pair of readings, and the important low-power-factor case where one wattmeter reads negative.

Problem 4Power Factor from Readings

Two wattmeters measuring the power of a balanced three-phase load read \(W_1 = 460~\mathrm{W}\) and \(W_2 = 920~\mathrm{W}\). Determine the total power and the power factor of the load (assume lagging).

Solution

The total real power is the sum of the readings:

\[ P_T = W_1 + W_2 = 460 + 920 = 1380~\mathrm{W} \]

The power-factor angle follows from the ratio of the difference to the sum:

\[ \tan\theta = \sqrt{3}\,\frac{W_2 - W_1}{W_2 + W_1} = \sqrt{3}\,\frac{460}{1380} = 0.5774 \;\Rightarrow\; \theta = 30^{\circ} \]

Hence the power factor:

\[ \cos\theta = \cos 30^{\circ} = 0.866~(\text{lagging}) \]

Answer\(P_T = 1380~\mathrm{W},\ \cos\theta = 0.866~\text{lagging}\)

Problem 5Low pf · Negative Reading

A balanced three-phase load operates at a power factor of \(0.4\) lagging with a line voltage of 415 V and a line current of 12 A. Find the two wattmeter readings and verify the total power. Comment on the sign of the readings.

Solution

The power-factor angle for a lagging load:

\[ \theta = \cos^{-1}(0.4) = 66.42^{\circ} \]

The two readings, with \(V_L I_L = 415 \times 12 = 4980\):

\[ \begin{aligned} W_1 &= V_L I_L \cos\!\left(\theta + 30^{\circ}\right) = 4980 \times \cos\!\left(96.42^{\circ}\right) = -557~\mathrm{W} \\ W_2 &= V_L I_L \cos\!\left(\theta - 30^{\circ}\right) = 4980 \times \cos\!\left(36.42^{\circ}\right) = 4007~\mathrm{W} \end{aligned} \]

Negative reading. Because the power factor is below \(0.5\) \((\theta > 60^{\circ})\), the argument of \(W_1\) exceeds \(90^{\circ}\) and the meter deflects backwards. In practice its connections are reversed and the reading is subtracted.

The total power is still the algebraic sum, which matches the direct formula:

\[ P_T = W_1 + W_2 = -557 + 4007 = 3450~\mathrm{W} = \sqrt{3}\,V_L I_L \cos\theta \]

Answer\(W_1 = -557~\mathrm{W},\ W_2 = 4007~\mathrm{W},\ P_T = 3450~\mathrm{W}\)