Question

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

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

Answer 1

Answer: (D)

$L=30MH$
$R=8\Omega$
resonant freq $=50Hz$
$\omega =2\pi f=100\pi$
Quantity factor $=Q=\frac{1}{R}\sqrt{\frac{L}{C}}$
$=\frac{1}{R}\sqrt{\frac{L}{C}}⟶(1)$
At resonance condition.
$\Rightarrow (100\pi {)}^2=\frac{1}{30\times 10^{-3}C}$
$\Rightarrow C=\frac{1}{30\times 10^{-3}\times (100\pi {)}^2}$
Now in (1) -
$Q=\frac{1}{8}\sqrt{\frac{30\times 10^{-3}\times 30\times 10^{-3}\times (100\pi {)}^2}{1}}$
$=\frac{1}{8}\sqrt{900\times 10^{-6}\times 10^4{\pi }^2}$
$=\frac{1}{8}\times 30\times 10^{-1}\times \pi$
$=\frac{3\pi }{8}$
$=1.1775$
$\approx 1.18$.