Measurement of power and wattmeters

Why it matters

Power, not volt-amperes, is what a load converts into heat, motion or light, and what efficiency and loss tests depend on. On AC the product V × I overstates power whenever current and voltage are out of phase, so a wattmeter — or the two-wattmeter method on three-phase supplies — is needed. Transformer and motor tests, energy audits and every three-phase lab use these methods.

Key ideas

Power quantities. For sinusoidal V and I with phase angle φ:

• Active power P = VI cos φ (W).
• Reactive power Q = VI sin φ (var).
• Apparent power S = VI (VA), with S² = P² + Q².
• Power factor = cos φ = P/S.

Electrodynamometer wattmeter.

• Fixed current coil (CC), a few thick turns, in series with the load.
• Moving pressure (potential) coil (PC), many fine turns with a large series non-inductive resistance, connected across the supply.
• The instantaneous torque is proportional to v·i, and the moving system responds to its average, so the deflection is proportional to VI cos φ, the true power, for any waveform (to the extent the PC current stays in phase with V). It works on DC and AC and has a nearly uniform scale.
• Air damping, spring control.
• Wattmeter ranges are set by current and voltage ratings as well as the power scale; an instrument can be overloaded at low power factor without the pointer going off scale, so always watch an ammeter and voltmeter too.

Errors in a wattmeter.

• Connection (loss) error: with the PC on the supply side it also measures the CC's I²R loss; with the PC on the load side the CC also carries the PC current and the meter reads the PC loss V²/Rp. Use the first for high current/low voltage loads and the second otherwise, or fit a compensating coil — a winding in series with the PC, wound with the CC, that cancels the field of the PC current.
• Pressure-coil inductance error: the PC current lags V by β = tan⁻¹(ωL/Rp). On lagging loads the meter reads high, and the error grows sharply at low power factor. It is reduced by a capacitor across part of the series resistance.
• Eddy currents in metal parts, stray fields (shielding), and temperature.
• Low power factor (LPF) wattmeters have a compensated PC, a low-inductance PC circuit and a strong torque so that full-scale deflection corresponds to a power factor of 0.1 or 0.2.

Three-phase power.

• Blondel's theorem: in an n-wire system, n − 1 wattmeters give the total power, each with its CC in one line and its PC between that line and the common line.
• Three-wattmeter method (4-wire star): total power is the sum of the three readings.
• Two-wattmeter method (any 3-wire load, balanced or not, star or delta): P = W1 + W2.
  • For a balanced load with sinusoidal supply, W1 = VL·IL·cos(30° − φ) and W2 = VL·IL·cos(30° + φ). The readings also give φ.
  • At unity pf both are equal; at pf = 0.5 one reads zero; below 0.5 one reads negative — reverse its PC (or CC) to get an up-scale reading and subtract it.
• One-wattmeter method works only for balanced loads (PC between a line and a star point; total = 3 × reading).
• Reactive power with one wattmeter: CC in line 1, PC across lines 2–3; reading = VL·IL·sin φ, total Q = √3 × reading (balanced load).

Formulas

P = V·I·cos φ ; Q = V·I·sin φ ; S = V·I ; S² = P² + Q² V, I = RMS values (V, A); φ = angle by which I lags V.

True power = Reading × cos φ / (cos β · cos(φ − β)) ; β = tan⁻¹(ωL/Rp) L = PC inductance (H); Rp = total PC circuit resistance (Ω). For lagging load.

P = √3·VL·IL·cos φ (balanced three-phase) W1 = VL·IL·cos(30° − φ) ; W2 = VL·IL·cos(30° + φ) (two-wattmeter, balanced) P = W1 + W2 ; Q = √3·(W1 − W2) ; tan φ = √3·(W1 − W2) / (W1 + W2) VL = line voltage (V); IL = line current (A); W1 = the larger reading on a lagging load.

Worked examples

Example 1 — pressure-coil inductance error. A wattmeter's pressure-coil circuit has Rp = 1000 Ω and L = 20 mH. It measures a load of 230 V, 10 A at power factor 0.1 lagging (50 Hz). Find the reading and the percentage error.

1. β = tan⁻¹(2π × 50 × 0.020 / 1000) = tan⁻¹(0.006283) = 0.360°.
2. φ = cos⁻¹(0.1) = 84.26°; φ − β = 83.90°.
3. Correction factor cos β·cos(φ − β)/cos φ = 1.0625.
4. True power = 230 × 10 × 0.1 = 230 W; reading = 230 × 1.0625 = 244.4 W.

Answer: the meter reads about 244 W, 6.25 % high — a small β gives a big error at low power factor, which is why LPF wattmeters are compensated.

Example 2 — two-wattmeter method (GATE level). A balanced three-phase load on a 400 V supply gives W1 = 2000 W and W2 = 1000 W. Find (a) total power, (b) power factor, (c) line current, (d) the readings if the same current flowed at pf 0.4 lagging.

1. P = 2000 + 1000 = 3000 W.
2. tan φ = √3 × (2000 − 1000)/3000 = 0.5774 → φ = 30°, pf = 0.866 lagging.
3. IL = P/(√3·VL·cos φ) = 3000/(1.732 × 400 × 0.866) = 5.0 A.
4. At pf 0.4 with IL = 5 A: φ = 66.42°. W1 = 400 × 5 × cos(−36.42°) = 1609 W; W2 = 400 × 5 × cos(96.42°) = −224 W.

Answer: (a) 3 kW; (b) 0.866 lagging; (c) 5.0 A; (d) W1 ≈ 1609 W, W2 ≈ −224 W (total 1386 W = √3 × 400 × 5 × 0.4).

Common mistakes

• Taking V × I as power on AC; it is apparent power in VA.
• Adding the two wattmeter readings when one has been reversed: the reversed reading must be subtracted.
• Using the balanced-load readings W1 = VL·IL·cos(30° ∓ φ) for an unbalanced load; only P = W1 + W2 holds in general.
• Using the two-wattmeter method on a 4-wire system with neutral current; Blondel's theorem needs three meters.
• Ignoring the current and voltage ratings of a wattmeter — at low pf the CC can be overloaded while the power reading is small.
• Forgetting that PC inductance makes a lagging-load reading high (and a leading-load reading low).

For GATE IN

Expect NAT questions on two-wattmeter readings (power, power factor, negative reading at pf below 0.5), reactive power from one wattmeter, and error due to PC inductance or connection. MCQs test Blondel's theorem, compensating coils and LPF wattmeters. Practise converting between W1, W2, φ and line current quickly, and watch for leading loads, where W1 and W2 swap roles.

Quick check

1. Two-wattmeter readings are equal and positive. What is the power factor?
2. One reading is zero. Power factor?
3. With a balanced load, W1 = 5 kW and W2 = 1 kW. Power factor?
4. A one-wattmeter reactive-power connection reads 1.2 kW. Total reactive power? Answers: 1. unity; 2. 0.5; 3. tan φ = √3 × 4/6 = 1.155, φ = 49.1°, pf = 0.655; 4. √3 × 1.2 = 2.08 kvar.