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  4. The Q factor of a series LCR circuit with L = 2 H, C = 32 μ F and R= 10 Ω is

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

Alternating Current

Solution:

Here, $L = 2\,H$,
$C= 32 \,\mu F = 32 \times 10^{-6}\,F$,
$R = 10\,\Omega$
$\therefore $ Resonance frequency,
$\omega_{r}=\frac{1}{\sqrt{LC}}$
$=\frac{1}{\sqrt{2\times32\times10^{-6}}}$
$=\frac{10^3}{8}\,rad\,s^{-1}$
Quality factor $\frac{\omega_{r}\,L}{R}$
$=\frac{10^{3}\times2}{8\times10}$
$=25$