Question

The frequency at which the impedance of the circuit becomes maximum is

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

Answer 1

Answer: (C)

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

Magnitude of admittance $Y=\frac{1}{Z}$ is given by

$|Y|=\frac{[R^2{X}_c^2+(R^2+X_L^2-X_L{X}_c{)}^2]}{X_c(R^2+X_L^2)}\frac{1}{2}$

Now, $|Z|$ is maximum or $|Y|$ is minimum at $\omega ={\omega }_0$ such that

$R^2+X_L^2-X_L{X}_C=0$

where $X_L=w_{o}L$ and $X_C=\frac{1}{w_{o}C}$

Or $R^2+{\omega }_0^2{L}^2={\omega }_{0}L\times \frac{1}{{\omega }_{0}C}=\frac{L}{C}$

or ${\omega }_0^2=\frac{1}{LC}-\frac{R^2}{L^2}$

or ${\omega }_0=\sqrt{\frac{1}{LC}-\frac{R^2}{L^2}}$

Hence resonance frequency $f_0=\frac{w_o}{2\pi }=\frac{1}{2\pi }\sqrt{\frac{1}{LC}-\frac{R^2}{L^2}}$