1.

In a series RLC circuit R = 10 kΩ, L = 0.5 H and C = \(\frac{1}{250}\) μF. Calculate the frequency when the circuit is in a state of resonance.(a) 4.089 × 10^4 Hz(b) \(\frac{11111.1}{π}\)Hz(c) 4.089π × 10^4 Hz(d) \(\frac{1}{π}\) × 10^4 HzI got this question in a job interview.This interesting question is from Problems of Series Resonance Involving Frequency Response of Series Resonant Circuit topic in portion Resonance & Magnetically Coupled Circuit of Network Theory

Answer»

The CORRECT choice is (b) \(\FRAC{11111.1}{π}\)Hz

Easiest explanation: Resonance frequency = \(\frac{1}{2π\sqrt{LC}}\)

= \(\frac{1}{2π\sqrt{\frac{1}{250} × 10^{-6} × 0.5}} = \frac{11111.1}{π}\)Hz.



Discussion

No Comment Found

Related InterviewSolutions