Question

A $100\, V, AC$ source of frequency $500\, Hz$ is connected to an LCR circuit with $ L=8.1\,mH, $ $ C=12.5\mu F,R=10\,\Omega $ all connected in series as shown in figure. What is the quality factor of circuit?

• A. $ 2.02 $
• B. $ 2.5434 $
• C. $ 20.54 $
• D. $ 200.54 $

Answer 1

Answer: (B)

Quality factor defines sharpness of tuning at resonance.
The $Q-$ resonant circuit is denned as the ratio of the voltage developed across the inductance or capacitance at resonance to the impressed voltage, which is the voltage applied across $R$.
ie, $Q=\frac{\text { voltage across } L \text { or } C }{\text { applied voltage }(=\text { voltage actross } R )}$
$=\frac{\left(\omega_{r} L\right) i}{R i}=\frac{\omega_{r} L}{R}$
or $=\frac{\left(1 / \omega_{r} C\right) i}{R i}=\frac{1}{R C \omega_{r}}$
Using, $\omega_{r}=\frac{1}{\sqrt{L C}}$, we get
$Q=\frac{L}{R} \cdot \frac{1}{\sqrt{L C}}=\frac{1}{R} \sqrt{\frac{L}{C}}$
or $=\frac{1}{R} \sqrt{\frac{L}{C}}$
Here, $L=8.1\, m H, C=12.5 \,\mu \,F, R=10 \,\Omega,\,f=500\, H z$
$\therefore Q=\frac{\omega_{r} L}{R}=\frac{2 \pi f L}{R}$
$=\frac{2 \times \pi \times 500 \times 8.1 \times 10^{-3}}{10}=\frac{8.1 \pi}{10}$
$=2.5434$