Question

A $100V,AC$ source of frequency $500Hz$ is connected to an LCR circuit with $L=8.1mH,$ $C=12.5\mu F,R=10\Omega$ 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, $Q=\frac{\text{ voltage across }L\text{ or }C}{\text{ applied voltage }(=\text{ voltage actross }R)}$
$=\frac{\left({\omega }_{r}L\right)i}{Ri}=\frac{{\omega }_{r}L}{R}$
or $=\frac{\left(1/{\omega }_{r}C\right)i}{Ri}=\frac{1}{RC{\omega }_r}$
Using, ${\omega }_r=\frac{1}{\sqrt{LC}}$, we get
$Q=\frac{L}{R}\cdot \frac{1}{\sqrt{LC}}=\frac{1}{R}\sqrt{\frac{L}{C}}$
or $=\frac{1}{R}\sqrt{\frac{L}{C}}$
Here, $L=8.1mH,C=12.5\mu F,R=10\Omega ,f=500Hz$
$\therefore Q=\frac{{\omega }_{r}L}{R}=\frac{2\pi fL}{R}$
$=\frac{2\times \pi \times 500\times 8.1\times 10^{-3}}{10}=\frac{8.1\pi }{10}$
$=2.5434$