Question

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 Hz

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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 1

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.