written 9.4 years ago by
teamques10
★ 70k
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modified 9.4 years ago
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$R=10Ω, \\
L=0.014H, \\
C=100μF \\
1) F_r=\dfrac{1}{2 \pi \sqrt{LC}}=134.51 Hz \\
2) \text{Quality Factor}=Q=\dfrac 1R \sqrt{\dfrac LC}=1.1832 \\
3) \text{Bandwidth} \Delta f=\dfrac{R}{2 \pi L}=113.6821 Hz \\
4) f_1=f_r-\dfrac{\Delta f}{2}=77.6689 Hz \\
5) f_2= \Delta f+f_1=191.351 Hz$
$\boxed{\text{Resonant Frequency} (Fr)=134.51Hz \\
\text{Quality Power(Q)}=1.1832 \\
\text{Bandwidth}(\Delta F)=113.6821Hz \\
\text{Lower cutoff frequency}(F1)=77.6689Hz \\
\text{Upper cutoff frequency}(F2)=191.351Hz}$