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NEET PHYSICSALTERNATING CURRENTHard

Question

For very high frequencies, the effective impedance of the circuit (shown in the figure) will be:

A

4 Ω\Omega

B

6 Ω\Omega

C

1 Ω\Omega

D

3 Ω\Omega

Step-by-Step Solution

In AC circuits, reactance depends on frequency. According to the sources, inductive reactance is XL=ωLX_L = \omega L and capacitive reactance is XC=1/(ωC)X_C = 1/(\omega C) . At very high frequencies (ω→∞\omega \to \infty), XLX_L becomes infinitely large, causing inductors to act as open circuits. Meanwhile, XCX_C approaches zero, causing capacitors to act as short circuits (wires). For the circuit provided in the NEET 2023 (Manipur) exam, these conditions effectively remove the branch containing the inductor and treat the capacitor branch as a purely resistive path. The resulting combination of remaining resistors yields a total effective impedance of 3 Ω\Omega.

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 CURRENTfrequencieseffectiveimpedancecircuitfigure

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