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NEET PHYSICSALTERNATING CURRENTMedium

Question

A circuit containing an 80 mH inductor and a 60 µF capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible. The current amplitude is:

A

11.63 A

B

8.22 A

C

10.0 A

D

9.21 A

Step-by-Step Solution

To find the current amplitude (imi_m), we follow these steps based on formulas in the sources:

  1. Calculate Reactances: The inductive reactance is X_L = \omega L = 2\pi fL = 2 \times 3.14 \times 50 \times 0.08 pprox 25.12 \, \Omega . The capacitive reactance is X_C = 1/(\omega C) = 1/(2\pi fC) = 1/(2 \times 3.14 \times 50 \times 60 \times 10^{-6}) pprox 53.08 \, \Omega .
  2. Calculate Impedance: For an LC circuit with negligible resistance, impedance Z=XLXC=25.1253.08=27.96ΩZ = |X_L - X_C| = |25.12 - 53.08| = 27.96 \, \Omega.
  3. Calculate RMS Current: I_{rms} = V_{rms} / Z = 230 / 27.96 pprox 8.226 A .
  4. Determine Current Amplitude: The peak current is i_m = \sqrt{2} \times I_{rms} = 1.414 \times 8.226 pprox 11.63 A .

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 CURRENTcircuitcontaininginductorcapacitorseries

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