Reading of a dynamometer wattmeter connected in an AC circuit: determine what the instrument indicates under normal operation.
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AVA product (apparent power)
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Baverage power (true power)
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Cpeak power
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Dinstantaneous power
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Erms power
Answer
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)