A series LCR circuit has `R=5 Omega, L=40 mH` and `C=1mu F`, the bandwidth of the circuit is

Options

A. 10 Hz
B. 20 Hz
C. 30 Hz
D. 40 Hz

Answer

B

Explanation

Resonant angular frequency `omega_(r ) = (1)/(sqrt(LC))` …(i)
Quality factor `Q=("Resonant angular frequency")/("Bandwidth")`
Bandwidth `=upsilon_(2)-upsilon_(1)=(upsilon_(r ))/(Q)` ….(ii)
where `upsilon_(r )` = resonant frequency `=(1)/(2pi sqrt(LC))`
Q = quality factor.
Also, `Q = (omega_(r )L)/(R )`
`therefore upsilon_(2)-upsilon_(1)=(upsilon_(r )R)/(2pi upsilon_(r )L)=(R )/(2pi L)` (Using (i) and (ii))
`upsilon_(2)-upsilon_(1)=(R )/(2pi L)=(5)/(2pi xx 40xx10^(-3))=20 Hz`