JEE Main 2023 — Alternating Current Question with Solution

The Q factor (quality factor) in a series RLC circuit can be given by the formula:

$Q=\frac{X_L}{R}=\frac{\omega L}{R}$

where ($X_L$) is the inductive reactance, (R) is the resistance, (L) is the inductance, and ($\omega$) is the angular frequency.

In a series RLC circuit at resonance, the resonant frequency (f) is given by

$f=\frac{1}{2\pi \sqrt{LC}}$

or equivalently, the angular frequency ($\omega$) at resonance is

$\omega =\frac{1}{\sqrt{LC}}$

Substituting (L = 1H) and ($C=6.25\mu F=6.25\times 10^{-6}F$) into the equation for (\omega) gives

$\omega =\frac{1}{\sqrt{1H\times 6.25\times 10^{-6}F}}$ = 400 rad/s

Substituting ($\omega =400rad/s$), (L = 1H), and ($R=100\Omega$) into the equation for the Q factor gives

$Q=\frac{\omega L}{R}=\frac{400rad/s\times 1H}{100\Omega }=4$

So, the Q factor of the circuit is 4.