Question

A resonance circuit having inductance and resistance $2\times {10}^{–4}H$ and $6.28\Omega$ respectively oscillates at $10MHz$ frequency. The value of the quality factor of this resonator is________. [$\pi =3.14$]

Answer 1

Step 1: Given data:

Resistance of the circuit,$R=6.28\Omega$

The inductance of the circuit, $L=2\times {10}^{-4}Henry$

Resonating frequency, $f=10MHz={10}^{7}Hz$

Step 2: Finding the value of the quality factor of this resonator:

We know that the quality factor $Q$ is given by

$$\begin{gathered} Q=XL/R \\ =\omega L/R \end{gathered}$$

where, $\omega =2\pi f$

$$\begin{gathered} \Rightarrow Q=\frac{2\pi fL}{R} \\ \Rightarrow Q=\frac{2\pi \times {10}^7\times 2\times {10}^{-4}}{6.28} \\ \Rightarrow Q=2000 \end{gathered}$$

Thus, the value of the quality factor of this resonator is $2000$