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  4. A series LCR circuit has R = 5 Ω, L = 40 mH and C = 1 μ F, the bandwidth of the circuit is

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Q. A series $LCR$ circuit has $R = 5 \,\Omega$, $L = 40\,mH$ and $C = 1\,\mu F$, the bandwidth of the circuit is

Alternating Current

Solution:

Resonant angular frequency $\omega_{r}=\frac{1}{\sqrt{LC}} \,...\left(i\right)$
Quality factor $Q=\frac{Resonant\, angular\, frequency}{Bandwidth}$
Bandwidth $=\upsilon_{2}-\upsilon_{1}=\frac{\upsilon_{r}}{Q} \,...\left(ii\right)$
where $\upsilon_{r} =$ resonant frequency $=\frac{1}{2\pi\sqrt{LC}}$
Also, $Q=\frac{\omega_{r}L}{R}$
$\therefore \upsilon_{2}-\upsilon_{1}=\frac{\upsilon_{r}R}{2\pi\upsilon_{r}L}=\frac{R}{2\pi L}$ (Using (i) and (ii))
$\upsilon_{2}-\upsilon_{1}=\frac{R}{2\pi L}$
$=\frac{5}{2\pi\times40\times10^{-3}}$
$=20\,Hz$