1.

Read the following passage and then answer questions on the basis of your under standing of the passage and the related studied concepts. Passage: For an LCR series circuit driven with an alternating voltage of amplitude Vmand angular frequency oo, the current amplitude is given as I_(m) =V_(m)/z = V_(m)/sqrt(R^(2) + (X_(L)-X_(C))^(2)) =V_(m)/sqrt(R^(2) + (Iomega -1/(c omega))^(2)) If omega is varied then for a particular frequency omega_(0), X_( C) = X_(L) and then Z = R and hence, I_(m) = V_(m)/R is maximum. This frequency is called the resonant frequency. The resonant frequency omega_(0) = 1/sqrt(LC) = Resonance of a LCR series a.c. circuit is said to be sharping current amplitude Im falls rapidly on increasing/decreasing the angular frequency from its resonant value 0. Mathematically, sharpness of resonance is measured by the quality factor of the circuit, which is given as: Q = (omega L)/R = 1/R sqrt(L/C) (e) An alternating series LCR circuit consists of aninductance of 10 mH, a capacitance of 100 uF and a resistance of 5 Omega.Compute its resonance frequency w, as well as the Q-factor.

Answer»

Solution :Here, L = 10 mH = 0.01 H, C = 100 `muF = 100 XX 10^(-6) F` and R = `5 Omega`
`therefore` Resonant frequency `omega_(0) = 1/sqrt(LC) =1/sqrt(0.01 xx 100 xx 10^(-6)) = 100 s^(-1)`
and Q-factor `= (omega_(0)L)/R = (1000 xx 0.01)/5 =2`


Discussion

No Comment Found

Related InterviewSolutions