Question

In a series LCR circuit having L = 30 mH, R = 8 $\Omega$ and the resonant frequency is 50 Hz. The quality factor of the circuit is:

• A. 0.118
• B. 11.8
• C. 118
• D. 1.18

Answer 1

The correct option is D 1.18

$L=30MH$

$R=8\Omega$

resonant freq $=50Hz$

$\omega =2\pi f=100\pi$

Quantity factor $=Q=\sqrt[\frac{1}{R}]{\frac{1}{R}\frac{L}{C}}$

$=\frac{1}{R}\sqrt{\frac{L}{C}}⟶\left(1\right)$

At resonance condition.

$\Rightarrow {\left(100\pi \right)}^2=\frac{1}{30\times {10}^{-3}C}$

$\Rightarrow C=\frac{1}{30\times {10}^{-3}\times {\left(100\pi \right)}^2}$

Now in (1) -

$Q=\frac{1}{8}\sqrt{\frac{30\times {10}^{-3}\times 30\times {10}^{-3}\times {\left(100\pi \right)}^2}{1}}$

$=\frac{1}{8}\sqrt{900\times {10}^{-6}\times {10}^4{\pi }^2}$

$=\frac{1}{8}\times 30\times {10}^{-1}\times \pi$

$=\frac{3\pi }{8}$

$=1.1775$

$\approx 1.18$.