InterviewSolution
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InterviewSolution
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A series LCR circuit has `R=5 Omega, L=40 mH` and `C=1mu F`, the bandwidth of the circuit isA. 10 HzB. 20 HzC. 30 HzD. 40 Hz |
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Answer» Correct Answer - B 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` |
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