Answer:
C = 59.17 nF
Q = 2.6
Step-by-step 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 =
= 40000
wc2 =
= 90000
so here
wc2 - wc1 will be
wc2 - wc1 = 90000 - 40000 = 50000
so
= 50000
we consider here R is 500
so L =
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 =
so C = 59.17 nF
quality factor Q will be
Q =
Q = 2.6