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A series LCR circuit has R=5 Omega, L=40...

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

A

10Hz

B

20Hz

C

30Hz

D

40Hz

Text Solution

Verified by Experts

The correct Answer is:
B
`Q = (omega_r L)/( R)`
` therefore v_2 -v_1 = ( R)/(2pi sqrt(LC)omega_r L) = ( R)/(2pi L)" " (because omega_r = (1)/(sqrt(LC)) )`
` v_2 - v_1 = ( R)/(2pi L) = (5)/(2pi xx 40 xx 10^(-3)) =20Hz`
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MTG GUIDE-ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS-NEET CAFE TOPICWISE PRATICE QUESTIONS (LC OSCILLATIONS)
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