Difficulty: Easy
Correct Answer: True
Explanation:
Introduction / Context:AC power in reactive circuits splits into real power (converted to heat or useful work) and reactive power (alternately stored and released by fields). An RL circuit includes both a resistor and an inductor, so it naturally involves both components.
Given Data / Assumptions:
Concept / Approach:
Instantaneous power fluctuates over a cycle, but average real power over one cycle is P = V_rms * I_rms * cos(phi). The reactive component is Q = V_rms * I_rms * sin(phi), associated with energy stored in and returned from the inductor's magnetic field. Apparent power is S = V_rms * I_rms, with S^2 = P^2 + Q^2.
Step-by-Step Solution:
Compute XL = 2 * pi * f * L.Find the impedance magnitude: |Z| = sqrt(R^2 + XL^2); current I_rms = V_rms / |Z|.Determine phase: cos(phi) = R / |Z| and sin(phi) = XL / |Z|.Evaluate powers: P = V_rms * I_rms * cos(phi) and Q = V_rms * I_rms * sin(phi).Verification / Alternative check:
Power factor PF = cos(phi) is less than 1 whenever XL > 0, proving that some of the apparent power is reactive. Measuring input and load waveforms with a wattmeter and a VAR meter confirms nonzero P and Q.
Why Other Options Are Wrong:
Common Pitfalls:
Assuming P + Q addition without using RMS and phase relationships. Always calculate with RMS quantities and the phase angle between voltage and current.
Final Answer:
True
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