Obtain the resonant frequency ωr of a series LCR circuit with L = 2.0H, C = 32μF and R = 10Ω. What is the Q-value of this circuit?
Exercise
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Inductance, $L = 2.0H$
Capacitance, $C = 32\mu F = 32 \times 10^{-6}F$
Resistance, $\text{R}=10\Omega$
Resonant frequency is given by the relation,
$\omega_{\text{r}}=\frac{1}{\sqrt{\text{LC}}}$
$=\frac{1}{\sqrt{2\times32\times10^{-6}}}=\frac{1}{8\times10^{-3}}=125\text{s}^{-1}$
Now, Q-value of the circuit is given as:
$\text{Q}=\frac{1}{\text{R}}\sqrt{\frac{\text{L}}{\text{C}}}$
$=\frac{1}{10}\sqrt{\frac{2}{32\times10^{-6}}}=\frac{1}{10}\times\frac{1}{4\times10^{-3}}=25$
Hence, the Q-Value of this circuit is 25.
Capacitance, $C = 32\mu F = 32 \times 10^{-6}F$
Resistance, $\text{R}=10\Omega$
Resonant frequency is given by the relation,
$\omega_{\text{r}}=\frac{1}{\sqrt{\text{LC}}}$
$=\frac{1}{\sqrt{2\times32\times10^{-6}}}=\frac{1}{8\times10^{-3}}=125\text{s}^{-1}$
Now, Q-value of the circuit is given as:
$\text{Q}=\frac{1}{\text{R}}\sqrt{\frac{\text{L}}{\text{C}}}$
$=\frac{1}{10}\sqrt{\frac{2}{32\times10^{-6}}}=\frac{1}{10}\times\frac{1}{4\times10^{-3}}=25$
Hence, the Q-Value of this circuit is 25.
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