Question

Calculate the relaxation time and quality factor
in a series LCR see arcuit when the charge is
driven by a dc emf given L = 1 mH, C = 5 uf and
R = 0.5ohm​

Answer 1

Answer:

The relaxation time is 2.5 micro-seconds

The quality factor (Q) is 2√200

Explanation:

We know that the relaxation time is the time required to charge the capacitor through resistor and also known as time constant denoted by tau.

Formula for relaxation time:

t = RC

By substituting the values we get

t = 0.5 × 5 × $10^{-6}$ = 2.5 micro-seconds

So, the relaxation time is derived as 2.5 micro-seconds

It is defined that the quality factor (Q) is the ratio of resonant frequency to the bandwidth.

Formula for quality factor (Q):

Q = $\frac{1}{R}\sqrt{\frac{L}{C}}$

By substituting the values we get

Q = $\frac{1}{0.5}\sqrt{\frac{1(10^{-3})}{5(10^{-6})}}$

Q = $2\sqrt{200}$

So, the quality factor (Q) is derived as 2√200.

Hope this answer may help you