- Year not given · 8 marks
Single-phase electrodynamometer wattmeter
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.
Describe the constructional details and working of a single-phase electrodynamometer type of wattmeter. Also derive the expression for deflection for ac operation.
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
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.
✓ Answer checked by Claude Opus 5.5 (high effort)
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