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NEET PHYSICSALTERNATING CURRENTMedium

Question

Power dissipated in an L-C-R series circuit connected to an AC source of emf ε is

A

ε²R / [R² + (L\omega - 1/C\omega )²]

B

ε²√[R² + (L\omega - 1/C\omega )²] / R

C

ε²[R² + (L\omega - 1/C\omega )²] / R

D

ε²R / √[R² + (L\omega - 1/C\omega )²]

Step-by-Step Solution

In an L-C-R series circuit, power is dissipated only in the resistor, as the average power over a cycle for pure inductors and capacitors is zero. The average power dissipated is given by P=I2RP = I^2 R, where II is the rms current . The rms current is given by I=ε/ZI = \varepsilon / Z, where ε\varepsilon is the rms emf and ZZ is the impedance. The impedance Z=R2+(XL−XC)2=R2+(Lω−1Cω)2Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + (L\omega - \frac{1}{C\omega})^2} . Substituting II into the power equation: P=(εZ)2R=ε2RZ2=ε2RR2+(Lω−1Cω)2P = \left( \frac{\varepsilon}{Z} \right)^2 R = \frac{\varepsilon^2 R}{Z^2} = \frac{\varepsilon^2 R}{R^2 + (L\omega - \frac{1}{C\omega})^2}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ALTERNATING CURRENT. 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.

PHYSICSALTERNATING CURRENTdissipatedseriescircuitconnectedsource

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