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NEET generalGeneralMedium

Question

An inductor 20 mH20 \text{ mH}, a capacitor 100 μF100 \text{ }\mu\text{F} and a resistor 50 Ω50 \text{ }\Omega are connected in series across a source of emf, V=10sin⁡314 tV = 10 \sin 314 \text{ t}. The power loss in the circuit is

1

2.74 W

2

0.43 W

3

0.79 W

4

1.13 W

Step-by-Step Solution

Pav=(VRMSZ)2RP_{av} = \left( \frac{V_{RMS}}{Z} \right)^2 R. Z=R2+(ωL−1ωC)2=56 ΩZ = \sqrt{R^2 + \left( \omega L - \frac{1}{\omega C} \right)^2} = 56 \text{ }\Omega. Therefore, Pav=(10(2)56)2×50=0.79 WP_{av} = \left( \frac{10}{(\sqrt{2}) 56} \right)^2 \times 50 = 0.79 \text{ W}.

Exam Context & Concepts Covered

This question aligns with the NEET general syllabus, specifically targeting concepts from General. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

generalinductorcapacitormutextfresistorconnected

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