Obtain the resonant frequency and Q-factor of a series LCR circuit with L = 3.0H, C = 27μF, and R = 7.4Ω. It is desired to improve the sharpness of the resonance of the circuit by reducing its ‘full width at half maximum’ by a factor of 2. Suggest a suitable way.
Exercise
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Inductance, $L = 3.0H$
Capacitance, $C = 27\mu F = 27 \times 10^{-6}F$
Resistance, $\text{R}=7.4\Omega$
At resonance, angular frequency of the source for the given LCR series circuit is given as:
$\omega_{\text{r}}=\frac{1}{\sqrt{\text{LC}}}$
$=\frac{1}{\sqrt{3\times27\times10^{-6}}}=\frac{10^3}{9}=111.11\text{rad }\text{s}^{-1}$
Q-factor of the series:
$\text{Q}=\frac{\omega_{\text{r}}\text{L}}{\text{R}}$
$=\frac{111.11\times3}{7.4}=45.0446$
To improve the sharpness of the resonance by reducing its 'full width at half maximum'
by a factor of 2 without changing $\omega_{\text{r}},$ we need to reduce R to half i.e.,
$\text{Resistance}=\frac{\text{R}}{2}=\frac{7.4}{2}=3.7\Omega.$
Capacitance, $C = 27\mu F = 27 \times 10^{-6}F$
Resistance, $\text{R}=7.4\Omega$
At resonance, angular frequency of the source for the given LCR series circuit is given as:
$\omega_{\text{r}}=\frac{1}{\sqrt{\text{LC}}}$
$=\frac{1}{\sqrt{3\times27\times10^{-6}}}=\frac{10^3}{9}=111.11\text{rad }\text{s}^{-1}$
Q-factor of the series:
$\text{Q}=\frac{\omega_{\text{r}}\text{L}}{\text{R}}$
$=\frac{111.11\times3}{7.4}=45.0446$
To improve the sharpness of the resonance by reducing its 'full width at half maximum'
by a factor of 2 without changing $\omega_{\text{r}},$ we need to reduce R to half i.e.,
$\text{Resistance}=\frac{\text{R}}{2}=\frac{7.4}{2}=3.7\Omega.$
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