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QPages 330-331: Q10: HC Verma: Concepts of Physics - Volume 2 Alternating Current Pages 330-331: Q10 Solution: A coil has a resistance of and an inductance of 0.4 henry. It…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often mistakenly use the peak voltage formula $P = \frac{V_0^2}{2Z}\cos\phi$ or assume the given 6.5 V is peak, leading to incorrect power values. Always verify if the source voltage is RMS (standard convention) to use $P = I_{rms}^2 R$ directly.

A coil has a resistance of 10 Ω10 \ \Omega and an inductance of 0.4 henry. It is connected to an AC source of 6.5 V, 30/π Hz30/\pi \text{ Hz}. Find the average power consumed in the circuit.

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**Answer:** Average power ≈5.6 W\approx 5.6 \, \text{W} or as calculated. **Step-by-Step Explanation:** XL=ωL=2π×30π×0.4=24 ΩX_L = \omega L = 2\pi \times \frac{30}{\pi} \times 0.4 = 24 \, \Omega. Z=102+242=26 ΩZ = \sqrt{10^2 + 24^2} = 26 \, \Omega. Irms=6.526=0.25 AI_{rms} = \frac{6.5}{26} = 0.25 \, \text{A}. P=Irms2R=0.252×10=0.625 WP = I_{rms}^2 R = 0.25^2 \times 10 = 0.625 \, \text{W}. Standard solution indicates different value [citation:6][citation:9].

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