Answer:
C = 59.17 nF
Q = 2.6
Explanation:
given data
frequencies = 40k Hz
frequencies = 90k Hz
solution
we take here R, L C take in series
so cut off frequency is express as
Wc1 = [tex]-\frac{R}{2L}+ \sqrt{(\frac{R}{2L})^2+\frac{1}{LC}}[/tex] = 40000
wc2 = [tex]\frac{R}{2L}+ \sqrt{(\frac{R}{2L})^2+\frac{1}{LC}}[/tex] = 90000
so here
wc2 - wc1 will be
wc2 - wc1 = 90000 - 40000 = 50000
so
[tex]\frac{R}{L}[/tex] = 50000
we consider here R is 500
so L = [tex]\frac{500}{50000}[/tex]
L = 10 m H
and here total cut off frequency is
total cut off frequency = 40000 + 90000 = 130000
so capacitance will be
capacitance C is = [tex]\frac{1}{10\times 10^{-3}\times 130000^2}[/tex]
so C = 59.17 nF
quality factor Q will be
Q = [tex]\frac{130000}{50000}[/tex]
Q = 2.6