---
title: "Single-phase electrodynamometer wattmeter"
subject: "Instrumentation I"
chapter: 7
key: "semester_4/instrumentation_i:electrodynamometer-wattmeter"
source: https://www.bctengineer.com/q/semester_4/instrumentation_i/electrodynamometer-wattmeter
site: BCT Engineering
---

# Single-phase electrodynamometer wattmeter

Instrumentation I · Chapter 7: Electrical equipment

- Asked in: year not given, 8 marks
- Video: [Electrodynamometer Wattmeter- construction and working](https://www.youtube.com/watch?v=Igf2nRb8p8M) — Anet Jose
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/instrumentation_i/electrodynamometer-wattmeter
- Whole chapter: https://www.bctengineer.com/md/semester_4/instrumentation_i/chapter-7

## Question

Describe the constructional details and working of a single-phase electrodynamometer type of wattmeter. Also derive the expression for deflection for ac operation.

## Short answer

The electrodynamometer wattmeter has two fixed current coils in series with the load and a moving voltage coil in series with a high resistance across the load. Deflecting torque is proportional to instantaneous product of currents; for AC, average torque Td = (VI cosφ / Rp)(dM/dθ). With spring control, deflection θ = (P/(K Rp))(dM/dθ), making deflection proportional to average power.

## Answer

The **electrodynamometer wattmeter** measures power in DC and single-phase AC circuits. It works on the principle that a current-carrying coil placed in the magnetic field of another current-carrying coil experiences a torque.

#### Construction
```text
  Fixed coils (current coil, 2 sections, thick wire, low R)
        ____         ____
       |    |  /|\  |    |
 load ->|CC  |  | | |  CC |-> to load
       |____| spindle|____|
              moving coil (pressure coil, thin wire)
              + series high resistance R
```
- **Fixed coils (current coils):** two identical coils of a few turns of thick wire, connected in series with the load and carrying the load current I. Using two coils gives a nearly uniform field.
- **Moving coil (pressure/voltage coil):** many turns of fine wire, pivoted between the fixed coils and connected across the load voltage in series with a high non-inductive resistance R; it carries a current ∝ V.
- **Control:** springs (which also lead the current into the moving coil); **damping:** air friction; a **pointer** and scale; and **shielding** with high-permeability material against stray fields.
- The scale is nearly uniform when properly designed.

#### Working
The load current flows in the fixed coils and produces a field proportional to it. The moving coil carries a current proportional to the voltage. The interaction produces a deflecting torque proportional to the product of the two currents, i.e. to the power.

#### Deflection for AC operation
Let the supply voltage be v = Vm sin ωt and the load current (through the fixed coils) be i1 = Im sin(ωt − φ), where φ is the load phase angle. The pressure-coil circuit is taken as purely resistive (its inductance is negligible), so its current is i2 = v/Rp = (Vm/Rp) sin ωt, where Rp is the total resistance of the pressure branch.

Instantaneous torque: Td(inst) = i1 i2 (dM/dθ), where M is the mutual inductance between the coils.

Mean torque over one cycle:

Td = (dM/dθ) · (1/T)∫₀ᵀ i1 i2 dt = (dM/dθ) · (Im Vm/ Rp) · ½ cos φ

With rms values I and V:

**Td = (V I cos φ / Rp) (dM/dθ)**

At equilibrium, the controlling torque of the spring, Tc = Kθ, equals Td:

**θ = (1/K) (V I cos φ / Rp) (dM/dθ) = (P/ (K Rp)) (dM/dθ)**

If dM/dθ is nearly constant over the working range, θ ∝ P = V I cos φ, so the deflection is proportional to the **average power** and the scale is uniform.

**Errors:** pressure-coil inductance, capacitance of the series resistor, eddy currents, stray fields and temperature. They are reduced by compensation (a capacitor across part of R) and by shielding.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In an electrodynamometer wattmeter, how are the fixed and moving coils built and connected?
   A. Fixed coils: many turns of fine wire across the load; moving coil: thick wire in series with the load
   B. Fixed coils: few turns of thick wire in series with the load; moving coil: many turns of fine wire with a high series resistance across the load
   C. Both coils are thick wire connected in series with the load
   D. Both coils are fine wire connected across the load
   Answer: B. The fixed coils are the current coils carrying load current. The moving pressure coil is fine wire with a high non-inductive resistance, across the load voltage. (Key point: Coil construction)

2. What does the deflecting torque of the wattmeter depend on?
   A. The sum of the two coil currents
   B. The product of the fixed-coil current and the moving-coil current, which represents power
   C. The square of the load voltage only
   D. The ratio of the load current to the voltage
   Answer: B. The moving-coil current is proportional to V and the fixed-coil current is the load current I, so the torque is proportional to the product, which is power. (Key point: Principle of torque)

3. With i1 = Im sin ωt and i2 = (Vm/Rp) sin(ωt − φ), why does the mean torque contain cos φ?
   A. The average of sin ωt · sin(ωt − φ) over a cycle is ½ cos φ
   B. The average of sin ωt · sin(ωt − φ) over a cycle is ½ sin φ
   C. The mutual inductance M varies as cos φ
   D. The spring torque is proportional to cos φ
   Answer: A. Averaging the product over one cycle gives ½ cos φ, so Td = (V I cos φ / Rp)(dM/dθ) with rms values. (Key point: Mean torque for AC)

4. Which is the deflection expression for AC operation, and what does it show if dM/dθ is nearly constant?
   A. θ = (K Rp / VI) (dM/dθ), so θ is inversely proportional to power
   B. θ = (1/K)(V I cos φ / Rp)(dM/dθ), so θ is proportional to average power and the scale is uniform
   C. θ = (1/K) V I sin φ (dM/dθ), so θ is proportional to reactive power
   D. θ = (1/K) V^2 (dM/dθ), so θ is proportional to voltage squared
   Answer: B. Setting Kθ equal to the mean deflecting torque gives θ proportional to P = VI cos φ, which gives a uniform scale. (Key point: Deflection expression)

## Related in Instrumentation I

- [Types of frequency meter and one working](https://www.bctengineer.com/question.md?key=semester_4%2Finstrumentation_i%3Afrequency-meter-types-working) — Instrumentation I, chapter 6

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