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In a series RLC resonance circuit, the resonance frequency f0 = 700 kHz. The resistor R = 10 Ohm. The specified bandwidth (BW) should be 10 kHz. Please calculate the quality factor (Q) and the values of inductor and capacitor.

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onyebuchinnaji

Answer:

  • quality factor (Q) = 69.99
  • inductor = 1.591 x 10⁻⁴ H
  • capacitor = 3.248 x 10⁻¹⁰ F

Explanation:

Given;

resonance frequency (F₀) = 700 kHz

resistor, R =  10 Ohm

bandwidth (BW) = 10 kHz

bandwidth (BW)  [tex]= \frac{R}{2\pi L}[/tex]

[tex]BW = \frac{R}{2\pi L}[/tex]

make L (inductor) the subject of the formula

[tex]L = \frac{R}{2\pi *BW} = \frac{10}{2\pi *10,000} =1.591 *10^{-4} \ H = \ 0.1591\ mH[/tex]

[tex]F_o =\frac{1}{2\pi\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi F_o} \\\\LC = \frac{1}{4\pi ^2F_o^2}= \frac{1}{4\pi ^2(700,000)^2} = 5.168*10^{-14}[/tex]

make C (capacitor)  the subject of the formula

[tex]C = \frac{5.168*10^{-14}}{1.591*10^{-4}} = 3.248*10^{-10} \ F = \ 3.248*10^{-4} \ \mu F[/tex]

quality factor (Q) [tex]= \frac{1}{R} \sqrt{\frac{L}{C}} \ = \frac{1}{10} \sqrt{\frac{1.591*10^{-4}}{3.248*10^{-10}}}=69.99[/tex]

quality factor (Q) =  69.99

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