Question

Find out the condition that LCR circuit show an oscillatory decay and find out the relaxation time

Answer 1

Answer:

Electrical resonance occurs in an AC circuit when the two reactances which are opposite and equal cancel each other out as XL = XC and the point on the graph at which this happens is were the two reactance curves cross each other. In a series resonant circuit, the resonant frequency, ƒr point can be calculated as follows.

We can see then that at resonance, the two reactances cancel each other out thereby making a series LC combination act as a short circuit with the only opposition to current flow in a series resonance circuit being the resistance, R. In complex form, the resonant frequency is the frequency at which the total impedance of a series RLC circuit becomes purely “real”, that is no imaginary impedance’s exist. This is because at resonance they are cancelled out. So the total impedance of the series circuit becomes just the value of the resistance and therefore: Z = R.

XL = XC

$2\pi \: f l \: = \frac{1}{2\pi \: f \: c}$

$f \: = \: \frac{1}{2\pi \: \sqrt{lc} }$

So relaxation time is,

$t \: = \frac{1}{f} = 2\pi lc$