Question

An LCR series a.c, circuit is connected to an a.c,source of variable angular frequency $\omega$ and a fixed voltage amplitude. The current amplitude in the circuit is found to be the same for two angular frequencies ${\omega }_1$ and ${\omega }_2$. The resonance angular frequency of this circuit must be

• A. $[{\omega }_1{\omega }_2{]}^{1/2}$
• B. $\frac{{\omega }_1+{\omega }_2}{2}$
• C. $\frac{{\omega }_1{\omega }_2}{{\omega }_1+{\omega }_2}$
• D. None

Answer 1

The correct option is A $[{\omega }_1{\omega }_2{]}^{1/2}$

$I=\frac{I_{max}}{\sqrt{2}}$ at $\omega ={\omega }_1$ and $\omega ={\omega }_2$

It is possible only when $X=\pm R$

$X=$ Reactance of circuit as $Z=R+jx$

at $\omega ={\omega }_1;X=-R$

$X=-X_L-X_C=-R$

${\omega }_{1}L-\frac{1}{{\omega }_{1}C}=-R$ (equation $01$)

at $\omega ={\omega }_2;X=R$

${\omega }_{2}L-\frac{1}{{\omega }_{2}C}=R$ (equation $02$)

Add $01$ and $02$

$({\omega }_1+{\omega }_2)L-\frac{1}{C}[\frac{1}{{\omega }_1}+\frac{1}{{\omega }_2}]=0$

$({\omega }_1+{\omega }_2)L=\frac{1}{C}\left(\frac{{\omega }_1+{\omega }_2}{{\omega }_1{\omega }_2}\right)$

$\Rightarrow {\omega }_1{\omega }_2=(LC{)}^{-1}$

$\Rightarrow {\omega }_1{\omega }_2=\frac{1}{LC}$

And as we know, ${\omega }_o=\frac{1}{\sqrt{LC}}$

$\because {\omega }_o^2={\omega }_1{\omega }_2$

$\Rightarrow {\omega }_o=\sqrt{{\omega }_1{\omega }_2}$