Question

Obtain the resonant frequency and Q – factor of a series LCR circuit with L = 3H, C = `27muF, " " R = 7.4Omega`. It is desired to improve the sharpness of resonance of circuit by reducing its full width at half maximum by a factor of 2. Suggest a suitable way.

Answer 1

` because " resonant frequency for LCR circuit is given by " v_(0) = (1)/(2pisqrt(lc))`
` = (1)/(2 xx 3.14 sqrt(3 xx 27 xx 10^(-6)))`
= 17.69 Hz
`" or "omega_(0) = 2piv_(0) = 111"rad"//s.`
`because " quality factor of resonance "`
`Q = (omega0)/(2Deltaw) = (omega_(o)L)/(R) = (1)/(R) sqrt((L)/(C))`
`therefore Q = (1)/(7.4)sqrt((3)/(27 xx 10^(-6))) = 45.0`
To improve sharpness of resonance circuit by a factor 2, without reducing `omega_(0)`, reduce R to half of its value i.e. R = `3.7 Omega`