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What do you understand by "sharpness of resonance" in a series LCR circuit ? Find expression for Q-factor of the circuit. |
| Answer» <html><body><p></p>Solution :We <a href="https://interviewquestions.tuteehub.com/tag/know-534065" style="font-weight:bold;" target="_blank" title="Click to know more about KNOW">KNOW</a> that current amplitude in a series LCR circuit is maximum when `omega = omega_(0) = 1/sqrt(LC)`. The maximum value of current at resonance is `(I_(m))_(rms) = V_(m)/<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>`For any other value of `omega` , the current amplitude is less. Choose values of o for which current amplitude is `1/sqrt(2) (I_(m))_(max)` . In Fig. 7.34, these values of <a href="https://interviewquestions.tuteehub.com/tag/ware-1448850" style="font-weight:bold;" target="_blank" title="Click to know more about WARE">WARE</a> represented by <a href="https://interviewquestions.tuteehub.com/tag/points-1157347" style="font-weight:bold;" target="_blank" title="Click to know more about POINTS">POINTS</a> B and C having angular frequencies `omega_(1)` and `omega_(2)`such that `omega_(1) = omega_(2)`and `omega_(2) = omega_(0) + Deltaomega`.The quantity `omega_(0)/(2.Deltaomega)`is regarded as a measure of sharpness of resonance and is also known as the Q-factor of the circuit. Smaller the value of Aw, sharper is the resonance.<br/> It can be easily calcualted that `Deltaomega = R/(2L)` <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/U_LIK_SP_PHY_XII_C07_E10_010_S01.png" width="80%"/> <br/>`<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>` Q-factor representing the sharpness of resonance <br/> `Q = (omega_(0))/(2Deltaomega) = (omega_(0)L)/R` <br/> As, `omega_(0) =1/sqrt(LC)`, hence, `Q = 1/sqrt(LC).1/R = 1/Rsqrt(L/C)`</body></html> | |