Reading of a dynamometer wattmeter connected in an AC circuit: determine what the instrument indicates under normal operation.

Difficulty: Easy

Correct Answer: average power (true power)

Explanation:


Introduction / Context:
An electrodynamometer wattmeter has a current coil in series with the load and a pressure coil across it. The interaction torque is proportional to the product of instantaneous voltage and current averaged over a cycle. The question asks which power quantity the pointer shows in AC.


Given Data / Assumptions:

  • AC steady state; frequency is constant.
  • Proper connections and instrument calibration.


Concept / Approach:

Average (real) power in an AC circuit is P = V_rms * I_rms * cosφ. The dynamometer’s average torque is proportional to this P, so its steady reading corresponds to true power. Apparent power S = V_rms * I_rms differs by the power factor and cannot be read directly from a single wattmeter.


Step-by-Step Solution:

Instantaneous torque ∝ v(t) * i(t).Average over a cycle → pointer deflection ∝ average of v(t) * i(t).Therefore, reading equals true power in watts.


Verification / Alternative check:

For a pure inductor (cosφ ≈ 0), the wattmeter reads nearly zero, confirming it measures real—not apparent—power.


Why Other Options Are Wrong:

  • Apparent/peak/instantaneous/rms power are different quantities; the instrument is not designed to indicate these directly.


Common Pitfalls:

  • Equating wattmeter reading to VA without accounting for power factor.


Final Answer:

average power (true power)

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