Question

An AC source of variable frequency is applied across a series L-C-R circuit. At a frequency double the resonance frequency, the impedance is $\sqrt{10}$ times the minimum impedance. The inductive reactance is

• A. $R$
• B. $2R$
• C. $3R$
• D. $4R$

Answer 1

The correct option is D $4R$

Let ${\omega }_0$ be the resonance frequency

minimum impedance is at ${\omega }_0$

$Z_{min}=X_R=R$

At $\omega =2{\omega }_0$,

$Z=\sqrt{10}{Z}_{min}$

$\sqrt{R^2+(\omega L-\frac{1}{\omega C}{)}^2}=\sqrt{10}\times R$

$R^2+(2{\omega }_{0}L-\frac{1}{2{\omega }_0\times C}{)}^2=10\times R^2$

because at ${\omega }_0$ , ${\omega }_0\times L=\frac{1}{{\omega }_{0}C}$ ,

$R^2+(2{\omega }_{0}L-\frac{{\omega }_{0}L}{2}{)}^2=10\times R^2$

$(\frac{3}{2}{\omega }_{0}L{)}^2=9\times R^2$

$\frac{3}{2}\times {\omega }_{0}L=3\times R$

$X_L=\omega L=2{\omega }_{0}L=4R$ (at $\omega =2{\omega }_0$)