Power And Power Factor
An AC circuit draws an RMS current of 4 A from a 100 V RMS source while the phase angle between voltage and current is 60 degrees. What is the average power consumed by the circuit?
Select the correct option:
Solution
200 W
Average power in an AC circuit is delivered only by the in-phase component of current, so it equals the product of RMS voltage, RMS current and the power factor (\cos\phi), giving (P_{avg} = V_{rms} I_{rms}\cos\phi). The reactive part of the current carries energy back and forth without net dissipation. Substituting (V_{rms} = 100) V, (I_{rms} = 4) A and (\cos 60^\circ = 0.5) yields (P_{avg} = 100 \times 4 \times 0.5 = 200) W. The option 400 W is the apparent power (V_{rms}I_{rms}), which ignores the phase factor. The option 100 W mistakenly applies (\cos\phi) twice. The option 346 W uses (\sin 60^\circ), which would give the wattless reactive power instead. This is the NCERT distinction between true, apparent and reactive power, where true power is measured in watts, apparent power in volt-amperes, and reactive power in volt-amperes reactive. Improving the power factor toward unity, for instance by adding a correcting capacitor to an inductive load, reduces the current needed to deliver the same useful power and so cuts transmission losses. A plausibility check confirms the true power must be smaller than the apparent power whenever a phase difference exists, and is exactly half here because the power factor is 0.5, consistent with a moderately reactive load drawing substantial wattless current.
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About This Question
- Subject
- physics
- Chapter
- electromagnetic induction and alternating currents
- Topic
- power and power factor
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
200 W
Average power in an AC circuit is delivered only by the in-phase component of current, so it equals the product of RMS voltage, RMS current and the power factor (\cos\phi), giving (P_{avg} = V_{rms} I_{rms}\cos\phi). The reactive part of the current carries energy back and forth without net dissipation. Substituting (V_{rms} = 100) V, (I_{rms} = 4) A and (\cos 60^\circ = 0.5) yields (P_{avg} = 100 \times 4 \times 0.5 = 200) W. The option 400 W is the apparent power (V_{rms}I_{rms}), which ignores the phase factor. The option 100 W mistakenly applies (\cos\phi) twice. The option 346 W uses (\sin 60^\circ), which would give the wattless reactive power instead. This is the NCERT distinction between true, apparent and reactive power, where true power is measured in watts, apparent power in volt-amperes, and reactive power in volt-amperes reactive. Improving the power factor toward unity, for instance by adding a correcting capacitor to an inductive load, reduces the current needed to deliver the same useful power and so cuts transmission losses. A plausibility check confirms the true power must be smaller than the apparent power whenever a phase difference exists, and is exactly half here because the power factor is 0.5, consistent with a moderately reactive load drawing substantial wattless current.
This medium difficulty physics question is from the chapter electromagnetic induction and alternating currents, covering the topic of power and power factor. It appeared in the 2025 exam.
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