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NEET PHYSICSALTERNATING CURRENTMedium

Question

A coil of inductive reactance 31Ω31 \, \Omega has a resistance of 8Ω8 \, \Omega. It is placed in series with a condenser of capacitive reactance 25Ω25 \, \Omega. The combination is connected to an a.c. source of 110110 V. The power factor of the circuit is:

A

0.56

B

0.64

C

0.8

D

0.33

Step-by-Step Solution

The power factor (cosϕ\cos \phi) of a series LCR circuit is defined as the ratio of resistance (RR) to impedance (ZZ), given by cosϕ=RZ\cos \phi = \frac{R}{Z} . First, calculate the impedance ZZ. The formula for impedance in a series LCR circuit is Z=R2+(XLXC)2Z = \sqrt{R^2 + (X_L - X_C)^2} . Given: Resistance R=8ΩR = 8 \, \Omega Inductive Reactance XL=31ΩX_L = 31 \, \Omega Capacitive Reactance XC=25ΩX_C = 25 \, \Omega Calculate the net reactance: X=XLXC=3125=6ΩX = X_L - X_C = 31 - 25 = 6 \, \Omega. Calculate the impedance: Z=82+62=64+36=100=10ΩZ = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \, \Omega. Calculate the power factor: cosϕ=RZ=810=0.8\cos \phi = \frac{R}{Z} = \frac{8}{10} = 0.8.

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 CURRENTinductivereactanceresistanceplacedseries

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