How does the power factor affect the reading of a wattmeter if voltage and current are unaltered?

Question

Examoogle · Accepted Answer

The correct answer is Option C: The reading increases with the power factor.. The reading of a wattmeter is directly proportional to the active power consumed by the circuit. Since active power is the product of voltage, current, and the power factor ($\cos \phi$), keeping voltage and current constant results in the wattmeter reading varying linearly with the power factor.

Step-by-Step Solution & Concepts:
The reading of a wattmeter is directly proportional to the active power consumed by the circuit. Since active power is the product of voltage, current, and the power factor ($\cos \phi$), keeping voltage and current constant results in the wattmeter reading varying linearly with the power factor.

Working Principle: A wattmeter consists of a pressure coil (connected across the load) and a current coil (connected in series with the load). The torque developed by the electrodynamometer movement is proportional to the instantaneous product of voltage and current, which, when averaged over a cycle, equals $P = VI \cos \phi$. As $\cos \phi$ increases (moving towards unity), the real power component for fixed $V$ and $I$ magnitudes increases, leading to a higher deflection.

Key Formulas: $P = V \cdot I \cdot \cos \phi$ — Where $P$ is active power, $V$ is RMS voltage, $I$ is RMS current, and $\cos \phi$ is the power factor.; $	au = K \cdot V \cdot I \cdot \cos \phi$ — Where $	au$ is the deflecting torque and $K$ is the instrument constant.

Key Points: Wattmeters measure true power, not apparent power.; The power factor represents the cosine of the phase angle between voltage and current.; For a constant $V$ and $I$, the reading is directly proportional to $\cos \phi$.; At $\cos \phi = 0$ (purely reactive load), the wattmeter reading will be zero.

Advantages: Measures real power effectively in AC circuits.; Applicable for both balanced and unbalanced loads.

Disadvantages: Sensitive to phase angle errors at low power factors.; Does not directly indicate the power factor value without external measurement.

Applications: Power measurement in industrial utility grids.; Efficiency testing of electrical motors and transformers.

Additional Info: Option B is incorrect because a decrease in power factor reduces the real power component, thus decreasing the reading.. Option D is incorrect because the wattmeter is specifically designed to account for phase displacement between $V$ and $I$.

Answer: C is correct — The reading is directly proportional to the power factor since $P = VI \cos \phi$.

Exam Tip: In exams, always distinguish between a 'Wattmeter' (measures active power, $P = VI \cos \phi$) and an 'Ammeter-Voltmeter' method (measures apparent power, $S = VI$) to avoid confusion regarding power factor dependency.

Examoogle · Answer

Option A: The reading increases with square of the power factor.

Examoogle · Answer

Option B: The reading decreases with the power factor.

Examoogle · Answer

Option D: The reading is independent of the power factor.

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