Question

The Q factor of a series LCR circuit with L = 2 H, C=32$\mu$F and R=10$\Omega$is:

• A. 15
• B. 20
• C. 25
• D. 30

Answer 1

Answer: (C)

25

Here, $L=2H,C=32\mu F=32\times {10}^{-6}F,R=10\Omega$
$\therefore$ Resonance frequency,
${\omega }_r=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{2\times 32\times {10}^{-6}}}=\frac{{10}^3}{8}rad{s}^{-1}$

Quality factor = $\frac{{\omega }_{r}L}{R}=\frac{{10}^3\times 2}{8\times 10}=25$