Dynamometer wattmeter

Every voltmeter and ammeter in the lab tells you $V$ or $I$ separately — neither one, by itself, tells you the active power a load actually consumes, because power also depends on the phase angle $\theta$ between them. The dynamometer wattmeter solves this with a clever piece of mechanical arithmetic: it builds a moving-coil instrument whose deflecting torque is naturally proportional to the product of two currents, wires one coil to carry the load current and the other to carry a current proportional to the load voltage, and lets the torque itself compute $VI\cos\theta$ . This article works through the electrodynamometer principle, how the current coil and pressure coil combine inside a wattmeter, and the practical consequences of connection polarity — the direct hardware realisation of the active-power formula you met in AC circuit theory.

The core insight

A dynamometer wattmeter uses a fixed current coil (carrying load current $I$ ) and a moving pressure coil (carrying a current proportional to load voltage $V$ ) whose product-torque law makes the pointer deflection read active power $P = VI\cos\theta$ directly.

The core idea

Why a dynamometer instrument is different. A PMMC meter has a permanent magnet and one moving coil; a moving-iron meter has a fixed coil and an iron vane. A dynamometer (electrodynamometer) instrument replaces the permanent magnet with a second fixed coil, split into two halves, so the field itself is produced by a current rather than by magnetised steel. A moving coil, suspended between the two fixed-coil halves and restrained by control springs (which also carry current in and out), carries the second current. Both currents jointly set up the torque, so the torque law is a product rather than a square:

T_d \propto I_1 I_2 \cos\phi

where: $I_1$ is the (RMS) current in the fixed coil, $I_2$ is the (RMS) current in the moving coil, and $\phi$ is the phase angle between $I_1$ and $I_2$ . Contrast this with a PMMC or MI meter, where the torque is proportional to a single current (or its square) — a dynamometer's torque naturally multiplies two independent quantities together, and that is precisely the mathematical shape of electrical power.

Turning the two coils into a wattmeter. To measure power you need one coil carrying a signal proportional to load current and the other carrying a signal proportional to load voltage, with their relative phase preserved. A dynamometer wattmeter assigns the roles as follows:

The current coil (CC) is the fixed coil: a low-resistance winding of a few turns of heavy wire, connected in series with the load, so $I_1 = I$ , the actual load current.
The pressure coil (PC) (also called the voltage coil) is the moving coil: a high-resistance winding of many turns, wired with a series multiplier resistor and connected in parallel across the load. Because the PC branch is deliberately made almost purely resistive, its current $I_2$ is proportional to $V$ and stays essentially in phase with $V$ .

Since the PC current is in phase with $V$ and lags/leads $I$ by the load's own phase angle, the angle $\phi$ between the two coil currents in the torque law is the same $\theta$ that appears in the load's power factor $\cos\theta$ . Substituting $I_1 = I$ , $I_2 \propto V$ , and $\phi = \theta$ into the torque law gives a deflecting torque proportional to the load's active power itself — the instrument scale can therefore be calibrated directly in watts rather than in some intermediate mechanical unit.

Figure

A dynamometer wattmeter schematic showing a fixed current coil in series with the load and a moving pressure coil in parallel across the load, with the pointer torque proportional to the product of the two coil currents and the phase angle between them

Derivation: from the torque law to a direct power reading

Derivation ADeriving the wattmeter's

VI\cos\theta

reading from the dynamometer torque law

The goal of this derivation is to show, starting only from the general dynamometer torque law $T_d \propto I_1 I_2 \cos\phi$ and the two specific ways the wattmeter's current coil and pressure coil are wired into a circuit, that the resulting deflecting torque — and hence the calibrated pointer reading — works out to be exactly the load's active power $P = VI\cos\theta$ , with no separate multiplication step needed from the observer.

Step 1. Identify the fixed-coil current. The current coil is wired in series with the load, carrying the full load current. So $I_1 = I$ , the RMS load current, with whatever phase it naturally has relative to the supply voltage.

Step 2. Identify the moving-coil current. The pressure coil is wired in parallel across the load through a large multiplier resistance, making the PC branch almost purely resistive. A current in a purely resistive branch is in phase with the voltage across it, so $I_2 = kV$ for some constant $k$ set by the PC's resistance, and $I_2$ is in phase with $V$ , the load voltage.

Step 3. Identify the phase angle in the torque law. The angle $\phi$ between $I_1$ and $I_2$ in the torque law is the angle between the load current $I$ and the load voltage $V$ — because $I_2$ tracks $V$ 's phase exactly. That angle is precisely the load's own phase angle $\theta$ , the same $\theta$ that defines the power factor $\cos\theta$ .

Step 4. Substitute into the torque law. $T_d \propto I_1 I_2 \cos\phi = I \cdot (kV) \cdot \cos\theta = k(VI\cos\theta)$ . The deflecting torque is directly proportional to $VI\cos\theta$ , with all the other quantities (the constant $k$ , the spring constant, coil geometry) absorbed into a single calibration constant.

Step 5. Read off the calibrated result. Since the torque — and hence the steady-state pointer deflection — is proportional to $VI\cos\theta$ , the manufacturer scales the dial directly in watts:

T_d \propto VI\cos\theta \quad\Rightarrow\quad \text{Wattmeter reading} = VI\cos\theta = P

where: $V$ is the RMS load voltage, $I$ is the RMS load current, $\theta$ is the phase angle between them, and $P$ is the active power delivered to the load. No separate multiplication by the observer is required — the instrument's mechanics perform the product internally.

Worked example

A dynamometer wattmeter measures the power drawn by a single-phase load with $V = 230\ \text{V}$ (RMS), $I = 8\ \text{A}$ (RMS), and lagging power factor $\cos\theta = 0.85$ .

Step 1. Identify what each coil carries: the current coil, in series with the load, carries $I = 8$ A; the pressure coil, in parallel across the load, carries a current proportional to $V = 230$ V and essentially in phase with it.

Step 2. Apply the calibrated reading formula directly, since the torque law has already reduced to $T_d \propto VI\cos\theta$ :

P = VI\cos\theta = 230 \times 8 \times 0.85 = 1564\ \text{W} = 1.564\ \text{kW}

Step 3. Check the sign/behaviour for a leading load of the same magnitude. If the load is reconnected so its phase angle becomes $-\theta$ (leading) instead of $+\theta$ (lagging), with $|\cos\theta| = 0.85$ unchanged, the reading is unaffected because cosine is an even function: $\cos(-\theta) = \cos(\theta)$ , so $P = 230 \times 8 \times 0.85 = 1564\ \text{W}$ again. The wattmeter reads 1564 W in both cases — it cannot by itself tell you whether the power factor is leading or lagging.

Step 4. Consider reversing the current-coil terminals. Swapping the CC's two leads reverses the sign of $I_1$ relative to $I_2$ in $T_d \propto I_1 I_2 \cos\phi$ , flipping the sign of the torque. The pointer tries to deflect backward against the mechanical stop instead of giving an upscale reading — a purely wiring-polarity effect, not a change in the load's actual power.

Competitive stretch

If the load's current instead rose to $I = 10\ \text{A}$ at the same $V$ and $\cos\theta$ , the reading follows the same formula: $P = 230 \times 10 \times 0.85 = 1955\ \text{W}$ . Since $P \propto I$ at fixed $V, \cos\theta$ , a 25% rise in current gives exactly a 25% rise in reading — a quick sanity check without redoing the full multiplication.

Common traps

Mixing up the current coil (low resistance, series with the load, carries $I$ ) with the pressure coil (high resistance, parallel across the load, carries a current proportional to $V$ ) — this inverts the whole torque argument.
Reading the wattmeter as if it measured $VI$ (apparent power) rather than $VI\cos\theta$ (active power); the two only coincide at unity power factor.
Expecting the wattmeter to reveal whether the power factor is leading or lagging — it cannot, because $\cos\theta$ is identical for $+\theta$ and $-\theta$ .
Ignoring polarity marks on the CC or PC terminals; a reversed coil connection flips the torque sign and drives the pointer backward instead of giving a valid upscale reading.

Ask the tutor: want to see how the wattmeter reading changes for a different load current or power factor, or check your own worked-example arithmetic — ask the AI tutor to walk through it with you.

Why does the pressure coil need to be almost purely resistive?

The torque law needs the pressure-coil current to be in phase with the load voltage $V$ , so that the phase angle between the two coil currents in $T_d \propto I_1I_2\cos\phi$ equals the load's own angle $\theta$ . If the PC branch had significant inductance, its current would lag $V$ by some extra angle, and the wattmeter would read $VI\cos(\theta + \text{error})$ instead of $VI\cos\theta$ — a systematic error. Making the PC branch almost purely resistive (large series multiplier resistor, low PC inductance) keeps that extra angle negligible.

Is a wattmeter reading exactly the same thing as multiplying a voltmeter reading by an ammeter reading?

Only at unity power factor. A voltmeter times an ammeter gives you $VI$ , the apparent power, ignoring phase entirely. A wattmeter's torque law inherently includes the $\cos\theta$ factor because the pressure-coil current tracks $V$ 's phase while the current-coil current tracks $I$ 's phase, so their product-with-phase gives $VI\cos\theta$ , the active power. Whenever the load is reactive ( $\theta \ne 0$ ), the wattmeter reading will be less than the plain $V \times I$ product.

What actually happens if I connect the current coil the wrong way round?

The current-coil current effectively reverses sign relative to the pressure-coil current, which flips the sign of $\cos\phi$ in $T_d \propto I_1I_2\cos\phi$ . The deflecting torque reverses direction, so instead of deflecting upscale the pointer is driven the wrong way and jams against the mechanical stop at the low end of the scale. The fix is simply to respect the manufacturer's polarity marks (often marked with a plus or dot) on the CC and PC terminals and reconnect accordingly, restoring a proper upscale reading.