Back to Directory
NEET PHYSICSMedium

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=∣XL−XC∣=∣25.12−53.08∣=27.96 ΩZ = |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 .
Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started