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NEET PHYSICSALTERNATING CURRENTEasy

Question

An inductor of inductance 2 mH is connected to a 220 V, 50 Hz AC source. Let the inductive reactance in the circuit be X1X_1. If a 220 V DC source replaces the AC source in the circuit, then the inductive reactance in the circuit is X2X_2. X1X_1 and X2X_2, respectively, are:

A

6.28 Ω\Omega, zero

B

6.28 Ω\Omega, infinity

C

0.628 Ω\Omega, zero

D

0.628 Ω\Omega, infinity

Step-by-Step Solution

According to the sources, inductive reactance (XLX_L) is given by the formula XL=ωL=2πfLX_L = \omega L = 2\pi f L . For the AC source, with f=50f = 50 Hz and L=2L = 2 mH (2×10−32 \times 10^{-3} H), the reactance X1=2×π×50×2×10−3=0.2π≈0.628 ΩX_1 = 2 \times \pi \times 50 \times 2 \times 10^{-3} = 0.2\pi \approx 0.628 \ \Omega. For a DC source, the frequency (ff) is zero, resulting in an inductive reactance X2=2π(0)L=0X_2 = 2\pi(0)L = 0 (zero) .

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 CURRENTinductorinductanceconnectedsourceinductive

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