The Two-Wattmeter Method

Start from the instantaneous power crossing the boundary into the load. With all voltages measured from an arbitrary reference point \(o\):

This is exact — it is just the definition of power flow, with no assumption yet made about the load.

Apply KCL to the whole load. Charge cannot accumulate inside it, so

One current is redundant. This is the entire content of the theorem.

Choose the reference to be conductor \(n\) itself, so \(v_{no} = 0\):

The \(n\)th term vanishes because its voltage is zero by construction, not because its current is. There are now \(n-1\) terms, each a voltage times a current — exactly what one wattmeter measures.

Averaging gives the readings:

What the proof did not assume:

Only KCL and the definition of average power were used, which is why the result is so robust.

The practical cases:

Note the four-wire case needs three — the neutral breaks the constraint that made two sufficient, because the three line currents no longer sum to zero.

Set up. Positive sequence, phase voltages referred to the neutral, load impedance \(|\mathbf{Z}|\angle\theta\):

The two line voltages, using Set 24, Problem 3:

Note \(\mathbf{V}_{cb} = -\mathbf{V}_{bc}\), which is why the second meter is connected C-to-B rather than B-to-C — reversing it would make \(W_2\) read negative for every load.

Each wattmeter reads \(\operatorname{Re}\left(\mathbf{V}\mathbf{I}^{*}\right)\) — Set 23, Problem 17:

Where the 30° comes from. Each meter compares a line voltage with a line current. But the impedance angle \(\theta\) relates the phase voltage to that current, and the line voltage leads the phase voltage by 30°:

One meter picks up \(+30°\) and the other \(-30°\) because their line voltages sit on opposite sides of their currents. The shift derived in Set 24 for its own sake is exactly what makes the method work.

Sum and difference by the standard identities:

Numerical confirmation. Taking \(V_{ph} = 100\ \text{V}\), \(I_L = 10\ \text{A}\), \(\theta = 36.87°\), computing directly from the phasors gives \(W_1 = 680.4\ \text{W}\) and \(W_2 = 1719.6\ \text{W}\), matching the formulas exactly; their sum is 2400.0 W, and \(\sqrt3(W_2-W_1) = 1800.0\ \text{var}\), both agreeing with \(\sqrt3V_LI_L\cos\theta\) and \(\sqrt3V_LI_L\sin\theta\).

Why three wattmeters here. The neutral is accessible, so \(n = 4\) and Blondel requires \(n-1 = 3\). Each meter measures one phase directly, referred to the neutral — the two-wattmeter method does not apply.

Note also that \(\mathbf{V}_{BN}\) leads and \(\mathbf{V}_{CN}\) lags, so this is negative (acb) sequence. That makes no difference to the arithmetic, since the currents are given rather than derived.

Each reading is the real part of voltage times conjugate current:

The total:

Note the three power factors differ:

Three different loads on three phases — which is exactly the situation a four-wire system exists to accommodate, and exactly why each phase must be metered separately.

The neutral current, for completeness:

Non-zero because the load is unbalanced, and it is precisely this that destroys the constraint \(\sum i_k = 0\) over the three lines and so forbids the two-wattmeter method.

A shortcut when the neutral is available. If only the total is wanted and the load impedances are known, \(P_T = \sum|\mathbf{I}_k|^2R_k\) avoids all the angles. Here the angles were given rather than the impedances, so the direct route was the shorter one.

The impedance angle is the power-factor angle:

Line current — star, so it equals the phase current:

The two readings:

\(W_2 > W_1\), confirming a lagging load — consistent with the \(+j6\) in the impedance. Note how unequal they are: a 2.5:1 ratio at a perfectly ordinary 0.8 power factor.

Total real and reactive power:

Check both against Set 24's formulas:

The second is the more convincing check, since it uses the reactance directly and involves no wattmeter theory at all.

Why prediction is worth practising. Working forward from a known load to the expected readings is how a measurement is verified — if the instruments disagree with the prediction, either the connection is wrong or the load is not what it was assumed to be. Problem 16 lists the connection faults this catches.

Total power is the algebraic sum:

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

Why the formula needs no voltage or current. Both readings contain the factor \(V_LI_L\), which cancels in the ratio:

So the power factor is obtained from the two readings alone — no voltmeter, no ammeter, and no knowledge of the load's connection.

The reactive power follows too:

Confirming \(\text{pf} = P/S = 1380/1594 = 0.866\) ✓ — the two routes to the power factor agree.

A memorable special case. Here \(W_2 = 2W_1\) exactly, and the answer came out at exactly 30°. That is worth remembering as a calibration point:

Problem 9 develops the ratio into a general method.

Total and per-phase real power:

Total and per-phase reactive power:

Dividing by three is legitimate here because the load is balanced — each phase carries an equal share. Problem 12 shows what happens when it is not.

Power factor:

Now use the delta connection. The phase voltage equals the line voltage:

The phase impedance:

Check via the phase powers directly:

Note the essential extra information. The wattmeter readings alone gave \(P_T\), \(Q_T\) and the power factor — but the impedance required two further facts: the line voltage, and the knowledge that the load is delta. Had it been star, the same readings would give

The measurement cannot distinguish the two, since Set 24, Problem 6 showed they are indistinguishable from the terminals.

The angle:

Already above 60°, which is the threshold Problem 8 identifies.

The readings, with \(V_LI_L = (415)(12) = 4980\ \text{VA}\):

Why \(W_1\) is negative. Its argument \(\theta+30°\) has passed 90°, so the cosine turns negative:

The voltage across that meter's pressure coil and the current in its current coil are more than 90° apart, so the average of their product is negative — the meter's element is returning power in the accounting sense, though nothing physically flows backwards.

The total is still the algebraic sum:

What to do in practice. A moving-coil wattmeter cannot deflect below zero — the pointer simply drives against its stop, which can damage the movement:

Reversing both coils restores the original negative reading, since two sign changes cancel — a classic laboratory error.

The diagnostic value. A negative reading is not a fault but information: it says immediately that the power factor is below 0.5, without any calculation. On a large motor that would prompt an investigation — a lightly loaded induction motor runs at very poor power factor, and correcting it (Set 24, Problem 17) would show up as the negative reading returning to positive.

Normalise by taking \(V_LI_L = 1000\ \text{VA}\), so the readings are directly comparable:

The four landmarks:

Notice \(W_2\) is not monotonic. It peaks at \(\theta = 30°\), where its argument \(\theta-30°\) is zero:

So the larger reading actually rises as the power factor falls from 1.0 to 0.866, then falls thereafter. Only \(W_1\) decreases throughout — which is why it is \(W_1\) that carries the diagnostic information.

The zero-power-factor case is a useful sanity check. With a purely reactive load the readings are equal and opposite, so

as it must be — no real power is consumed. Yet each meter shows a substantial deflection, and each carries full line current. A wattmeter reading zero total does not mean the instruments are idle, exactly as in Set 23, Problem 17.

A useful bound. Since \(|\cos| \le 1\), neither reading can exceed \(V_LI_L\), while the total reaches \(\sqrt3V_LI_L\) at unity power factor. So each meter must be rated for the full \(V_LI_L\) even though it never reads more than \(1/\sqrt3\) of the maximum total.

Start from the standard formula and divide numerator and denominator by \(W_2\):

Verify at three points:

Each row confirms the corresponding row of Problem 8's table.

Why the ratio form is convenient:

The third is the practically important one: if both wattmeters share a current transformer of uncertain ratio, the ratio method still gives the correct power factor even though neither reading is trustworthy in absolute terms.

A worked instance. Two meters read 35 and 87 scale divisions on identical instruments:

No scale factor was needed. To obtain \(P_T\) as well, the multiplier would then have to be applied — the ratio gives the power factor for free, but not the power.

A caution on the sign of \(r\). When \(W_1\) is negative, \(r\) is negative and the formula still holds — but the reading must be entered with its sign. Entering a reversed meter's reading as positive gives a badly wrong power factor, and is the commonest error in the low-power-factor case.

The total is unambiguous:

Addition is commutative, so the labelling does not matter for the power.

The power factor magnitude is also unambiguous. Taking the two assignments in turn:

Same magnitude, opposite sign — one is a lagging load and the other leading, and the readings alone cannot tell them apart.

The general rule follows from the derivation:

Swapping a leading load for a lagging one of the same power factor simply exchanges the two readings — which is exactly why the identity of the meters must be recorded.

Three ways to resolve it:

The third is a genuine field technique — Problem 17 works it through. It resolves the ambiguity by perturbing the system in a known direction.

The deeper point. This is the same ambiguity as Set 23, Problem 16: \(\cos\theta\) is an even function, so no measurement of a cosine can recover a sign. The wattmeter pair does encode the sign — in which meter reads higher — but that information is lost the moment the readings are written down as an unlabelled pair.

The difference, from Problem 2:

Why the \(\pm30°\) produces a sine. The identity

converts a difference of cosines into a product of sines. The sum did the reverse, giving \(2\cos\theta\cos30°\). So one combination extracts \(\cos\theta\) and the other \(\sin\theta\) — the two meters between them resolve the load's complex power into its rectangular components.

The single-wattmeter method for \(Q\). Connect one wattmeter with its current coil in line A but its pressure coil across \(\mathbf{V}_{bc}\) — the voltage between the other two lines:

One instrument, connected "across the other two lines", reads reactive power directly. This is the cross-connection or varmeter arrangement.

Why it works. \(\mathbf{V}_{bc}\) is in quadrature with \(\mathbf{V}_{an}\) — it lags by 90°:

Feeding a wattmeter with a voltage in quadrature to the one that defines \(\theta\) turns its cosine into a sine — which is the definition of reactive power. The three-phase system supplies that quadrature voltage for free, which a single-phase system cannot do without a phase-shifting network.

A worked check. For the load of Problem 4 — \(V_L = 208\), \(I_L = 12.01\), \(\theta = 36.87°\):

Matching the two-wattmeter result exactly — and note that 1499 is precisely \(W_2-W_1\) from that problem, as it must be.

Recover the currents from Set 24, Problem 15, where Millman's theorem gave a star-point displacement of \(28.99\angle89.04°\ \text{V}\):

These sum to zero, as a three-wire system requires.

The true powers, computed element by element:

Only phase B has reactance, so it alone contributes to \(Q\).

The wattmeter readings, with line B as the common reference:

Test the sum:

Exact. Blondel's theorem holds to the last digit, exactly as Problem 1 promised — no assumption of balance was ever made.

Now test the power-factor formula:

Wrong by a factor of more than two — and wrong in sign. The formula reports a strongly capacitive load where the true load is inductive.

Why the sum survives and the difference does not. The sum was derived from KCL alone. The difference was derived from the specific phasor relationships of a balanced load — equal current magnitudes, equal angles, and the exact \(\pm30°\) pattern:

What to do instead. For an unbalanced load, the two wattmeters give \(P_T\) and nothing more. Obtaining \(Q\) or a meaningful power factor requires either three-wattmeter measurement with an artificial neutral, or full phasor instrumentation — and even then, "the power factor" of an unbalanced load is a quantity needing careful definition, since each phase has its own.

Delta quantities from Set 24, Problem 11:

The readings use line quantities — which is the whole point:

Matching Set 24, Problem 11's answer of 27.56 kW exactly.

Why the connection is irrelevant to the meters. The wattmeters are connected to the lines, and they see only line voltages and line currents. From outside, a delta of \(\mathbf{Z}\) and a star of \(\mathbf{Z}/3\) are indistinguishable — Set 24, Problem 6 — so they must give identical readings:

The same line current, hence the same readings.

What the meters cannot tell you. Given the readings and the line voltage, the per-phase impedance follows only once the connection is known:

Same measurements, same power, same power factor — three-to-one difference in the deduced impedance. This is Problem 6's point restated with numbers.

Verify the power factor from the readings:

Recovering the impedance angle exactly, which confirms both the readings and the formula.

Blondel's theorem sets the floor at \(n-1\) instruments for a general load. A single wattmeter can only be enough if extra information — specifically, that the load is balanced — is supplied from outside the measurement.

aBalanced load with an accessible neutral. Measure one phase and multiply by three:

The simplest case, and valid only because balance guarantees the other two phases are identical.

bBalanced load, three-wire, artificial neutral. Create a neutral with two resistors equal to the wattmeter's pressure-coil resistance:

Three equal resistances in star place their common point at the true neutral potential — the balanced case of Set 24, Problem 15, where the Millman displacement vanishes. The meter then reads \(P_T/3\).

cThe two-reading method with one meter. Connect one wattmeter as \(W_1\), record the reading, then switch it to the \(W_2\) position and record again:

Valid only if the load is steady between the two readings — the measurements are no longer simultaneous. A fluctuating load makes this useless, which is the practical reason two meters are preferred.

dReactive power alone, by the cross-connection of Problem 11:

Summary of what each arrangement needs:

The two-wattmeter method is the only row requiring no assumption about the load, which is why it is the standard.

The three connections, each omitting one line from the pressure-coil references:

All three agree on the total, exactly as Problem 1's proof requires — the reference conductor was arbitrary throughout.

But the individual readings are permuted. The same pair of numbers appears in every row, in different orders. This matters because

depends on which is called \(W_1\). Rows 2 and 3 would give \(\theta = -36.87°\) — reporting a leading load — unless the labelling convention is applied consistently.

The convention that fixes it. With positive sequence and line B common, label so that

Equivalently: for a lagging load \(W_2 > W_1\) always. Since most industrial loads are lagging, assigning the larger reading to \(W_2\) is usually correct — but it assumes the answer, which is exactly the ambiguity of Problem 10.

Which line to choose in practice:

There is no electrical reason to prefer one, which is itself a useful fact: the choice can be made entirely on practical grounds.

The one thing that must not change is the pairing. Meter 1's current coil and its pressure coil must refer to the same line — pairing \(\mathbf{I}_A\) with \(\mathbf{V}_{CB}\) gives a reading with no useful interpretation, and the sum is then simply wrong. Problem 16 lists this among the connection faults.

The correct connection, stated precisely:

The faults and their symptoms:

The second row is the trap: reversing both coils changes two signs, which cancel. A meter that appears correctly connected may have had both reversed by a previous user.

The prediction check. The most reliable safeguard is to predict the readings before connecting, as in Problem 4. Given an approximate load rating:

A sum that is out by \(\sqrt3\), or a ratio wildly different from expectation, localises the fault immediately.

A worked diagnostic. A load known to be about 5.5 kW at roughly 0.8 lagging is measured, giving readings of 1571 W and −3971 W:

But the magnitudes 1571 and 3971 are exactly those predicted in Problem 15. The diagnosis is a single reversed coil on the second meter; correcting the sign gives

Note that a genuine negative reading (Problem 7) also occurs, so a negative value is not by itself evidence of a fault — the test is whether the sum is plausible.

Distinguishing a true negative from a reversal:

Before correction, \(\theta = 53.13°\), \(V_LI_L = (400)(69.28) = 27\,712\ \text{VA}\):

After correction, \(\theta = 18.19°\), \(V_LI_L = (400)(43.76) = 17\,504\ \text{VA}\):

The comparison:

The three signatures of a successful correction:

The first is the crucial one. A capacitor consumes no real power, so if \(P_T\) changed, something other than the correction has altered — the load itself, or the connection.

Resolving Problem 10's ambiguity. This is the field technique promised there. Add a known capacitance and observe:

A definite answer from a single perturbation, using instruments already in place.

A warning about overcorrection. Continuing past unity power factor makes the load capacitive, and the readings begin to diverge again — with \(W_1\) now the larger. Since \(r\) passes through 1 and comes back, a single reading of \(r\) cannot distinguish 0.9 lagging from 0.9 leading; only the direction of travel can.

1Real power — the sum:

2Reactive power — \(\sqrt3\) times the difference:

Positive, and \(W_2 > W_1\), so the load is lagging.

3Power factor:

4Apparent power:

Check: \(P/S = 9.2/11.947 = 0.770\) ✓ agreeing with the power factor.

5Line current, from \(S = \sqrt3V_LI_L\):

6Correction to unity requires the capacitors to supply all of \(Q_T\):

Using Set 24, Problem 17's delta formula. At unity power factor the readings would become equal at 4.6 kW each, and the line current would fall to \(9200/(\sqrt3\times415) = 12.80\ \text{A}\) — a 23% reduction.

What could not be determined from these measurements:

The second is worth dwelling on. Every step after the sum assumed balance, and Problem 12 showed the consequences of that assumption failing.

aPressure-coil burden. The pressure coil draws current, and where that current flows depends on the connection:

Choose whichever error is smaller: the first connection for low currents, the second for high ones — precisely the ammeter–voltmeter trade-off of Set 1.

A numerical instance. With \(V = 400\ \text{V}\), \(R_p = 10\ \text{k}\Omega\), \(I = 20\ \text{A}\), \(R_c = 0.05\ \Omega\):

Comparable here; at 200 A the second becomes 2 kW and the first connection is clearly preferable.

bPressure-coil inductance. The pressure coil is not purely resistive, so its current lags the voltage slightly by an angle \(\beta\). The meter then reads

The error grows as the power factor worsens, and it is worst exactly where the two-wattmeter method is most needed. This is why low-power-factor wattmeters, with compensated pressure coils, exist as a separate instrument class.

cTransformer ratio and phase errors. At high voltage or current, instrument transformers are used, and each introduces both a ratio error and a small phase displacement — the latter again mattering most at low power factor.

Which errors the ratio method escapes:

The ratio is immune to anything scaling both readings equally — which is the commonest instrumentation error, since both meters typically share the same transformer set.

The practical conclusion. A two-wattmeter installation of modest accuracy will give a power factor considerably better than its power measurement, because the power factor depends only on the ratio. Reporting "9.2 kW ± 3% at a power factor of 0.770 ± 0.005" is a perfectly consistent statement, though it looks odd at first sight.

The six sets, in one line each:

The shape of the whole. One transform (Set 20) made two sets of transferred results almost free, because the transform was built to preserve Kirchhoff's laws and linearity. Genuine novelty appeared only where those two premises were insufficient:

Both are non-linear operations on the circuit variables — a maximisation and a multiplication — which is exactly why linearity did not carry them across.

The one idea running through everything is that phase, introduced in Set 20 as a bookkeeping device for time shifts, turns out to control energy:

The last is a small delight: a phase shift that seemed a mere nuisance in Set 24 became the mechanism of the measurement in Set 25.

The standing limitation. Everything assumed a single frequency in the steady state. Two omissions follow:

Each was met once and set aside. Each now becomes a subject in its own right.

What comes next. Treating \(\omega\) as a variable rather than a constant opens the second half of Part 3: