# A resonance circuit having inductance and resistance $2 imes 10^{-4} mathrm{H}$ and $6.28 Omega$

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

$\Rightarrow Q=\frac{2 \pi f L}{R}$

$\Rightarrow Q=\frac{2 \pi \times 10 \times 10^{6} \times 2 \times 10^{-4}}{6.28}=2000$

$Q=2000$