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A series RLC circuit has following parameter values: $R=10\Omega$, L=0.014H, $C=100 \mu F$. Compute resonant frequency, quality factor, bandwidth, located cut off frequency.
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written 7.8 years ago by
teamques10
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• modified 7.8 years ago |
$R=10Ω, \\ L=0.014H, \\ C=100μF \\ 1) F_r=\dfrac{1}{2 \pi \sqrt{LC}}=134.51 Hz \\ 2) \text{Quality Factor}=Q=\dfrac 1R \sqrt{\dfrac LC}=1.1832 \\ 3) \text{Bandwidth} \Delta f=\dfrac{R}{2 \pi L}=113.6821 Hz \\ 4) f_1=f_r-\dfrac{\Delta f}{2}=77.6689 Hz \\ 5) f_2= \Delta f+f_1=191.351 Hz$
$\boxed{\text{Resonant Frequency} (Fr)=134.51Hz \\ \text{Quality Power(Q)}=1.1832 \\ \text{Bandwidth}(\Delta F)=113.6821Hz \\ \text{Lower cutoff frequency}(F1)=77.6689Hz \\ \text{Upper cutoff frequency}(F2)=191.351Hz}$
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