Question

A 500 uH inductor, 80/T12 pF capacitor and a 628 resistor are
connected to form a series RLC circuit. Calculate the resonant
frequency and Q-factor of this circuit at resonance.​

Answer 1

Answer:

The quality factor relates the maximum or peak energy stored in the circuit (the reactance) to the energy dissipated (the resistance) during each cycle of oscillation meaning that it is a ratio of resonant frequency to bandwidth and the higher the circuit Q, the smaller the bandwidth, Q = ƒr /BW.

Answer 2

Answer: The required resonant frequency is 2500 KHz and the Q-Factor is 12.5

Explanation:

It is given that,

L = 500 × 10⁻⁶ H ; C = $\frac{80}{\pi ^{2} }$ × 10⁻¹² F ; R = 628 Ω

(i) The required resonant frequency is :

$f_{r} = \frac{1}{2\pi \sqrt{LC} }$

   $= \frac{1}{2\pi \sqrt{500 * 10^{-6}*\frac{80}{\pi ^{2} } * 10^{-12} } }$

$f_{r} = \frac{1}{2\sqrt{40,000 * 10^{-18} } }$

    = $\frac{10,000 * 10^{3} }{4}$

$f_{r}$ = 2500 KHz

(ii) The required Q - Factor is:

= (ω, L) ÷ R = $\frac{2 * 3.14 * 2500 * 10x^{3} * 500 * 10^{-6} }{628}$

= Q = 12.5