Question

LC tank circuit consists of an ideal capacitor C connected in parallel with a coil of inductance L having an internal resistance R. The resonant frequency of the tank circuit is

• A. $\frac{1}{2\pi \sqrt{LC}}$
• B. $\frac{1}{2\pi \sqrt{LC}}\sqrt{1-R^2\frac{C}{L}}$
• C. $\frac{1}{2\pi \sqrt{LC}}\sqrt{1-\frac{L}{R^{2}C}}$
• D. $\frac{1}{2\pi \sqrt{LC}}(1-R^2\frac{C}{L})$

Answer 1

The correct option is B $\frac{1}{2\pi \sqrt{LC}}\sqrt{1-R^2\frac{C}{L}}$


$Z_{eq}=\frac{\frac{1}{j\omega C}\times (j\omega L+R)}{\frac{1}{j\omega C}+j\omega L+R}$

$Z_{eq}=\frac{(\frac{R}{j\omega C}+\frac{L}{C})}{R+j(\omega L-\frac{1}{\omega C})}\times \frac{R-j(\omega L-\frac{1}{\omega C})}{R-j(\omega L-\frac{1}{\omega C})}$

Equating imaginary part to zero.
$Img=-\frac{R^2}{\omega C}-\frac{L}{C}(\omega L-\frac{1}{\omega C})=0$

$\frac{R^2}{\omega C}+\frac{L}{C}(\omega L-\frac{1}{\omega C})=0$

$\frac{C{R}^2+{\omega }^2{L}^{2}C-L}{\omega C^2}=0$

${\omega }^2=\frac{L-R^{2}C}{L^{2}C}$

$\omega =\frac{1}{\sqrt{LC}}\sqrt{1-\frac{R^{2}C}{L}}$

$f=\frac{1}{2\pi \sqrt{LC}}\sqrt{1-\frac{R^{2}C}{L}}$