Question:
A resonance circuit having inductance and resistance $2 \times 10^{-4} \mathrm{H}$ and $6.28 \Omega$
respectively oscillates at $10 \mathrm{MHz}$ frequency. The value of quality factor of this
resonator is $[\pi=3.14]$
Solution:
$(2000)$
Given : $\mathrm{R}=6.28 \Omega$
$\mathrm{L}=2 \times 10^{-4}$ Henry
we know that quality factor $Q$ is given by
$\Rightarrow Q=\frac{X_{L}}{R}=\frac{\omega L}{R}$
also, $\omega=2 \pi f$, so
$\Rightarrow Q=\frac{2 \pi f L}{R}$
$\Rightarrow Q=\frac{2 \pi \times 10 \times 10^{6} \times 2 \times 10^{-4}}{6.28}=2000$
$Q=2000$