Question

One application of `L-R-C` series circuit is in high pass or low pass filter, which out either the low or high frequency components of a signal. A has pass filter is shown in figure where the output voltage is taken across the `L-R` where `L-R` combination represents and inductive coil that also has resistance due to the large length of the wire in the coil.

Find the ratio `V_("out") //V_(s)` as a function of the angular frequency `omega` of the source
A. `sqrt((R^(2) + omega L^(2))/(R^(2) + (omega L - (1)/(omega C))^(2)))`
B. `sqrt((R^(2) + omega^(2) L^(2))/(R^(2) + (omega L - (1)/(omega C))^(2)))`
C. `sqrt((R^(2) + omega^(2) L)/(R^(2) + (omega C - (1)/(omega L))^(2)))`
D. 1

Answer 1

Correct Answer - B
`V_("out") = (V_(s))/(sqrt(R^(2) + (omega L - (1)/(omega C))^(2))) xx sqrt(R^(2) + omega^(2) L^(2))`
`(V_("out"))/(V_(s)) = (sqrt(R^(2) + omega^(2) L^(2)))/(sqrt(R^(2) + (omega L - (1)/(omega C))^(2))) = sqrt((R^(2) + omega^(2) L^(2))/(R^(2) + (omega L - (1)/(omega C))^(2)))`