Question

What does meant by half power point frequencies
obtain an expression for bandwidth in a LCR series circuit
Shoe half power point frequencies in curve between alternating current and frequency

Answer 1

Half power point frequency: The input frequency at which the power dissipation in LCR circuit becomes half of maximum power.
Resonance condition:
Here
${\omega }_1={\omega }_0-{\Delta }_{\omega }$
${\omega }_2={\omega }_0+{\Delta }_{\omega }$
Band width
${\omega }_2-{\omega }_1=({\omega }_0+{\Delta }_{\omega })-({\omega }_0-{\Delta }_{\omega })$
${\omega }_2-{\omega }_1=2\Delta \omega$
The ${\omega }_1$ and${\omega }_2$ has frequency at which maximum current becomes $\frac{1}{\sqrt{2}}$ time of its value
$V_0=I_{0}Z$
$\frac{V_0}{Z}=\frac{I_0}{\sqrt{2}}$
$\frac{I_{0}R}{Z}=\frac{I_0}{\sqrt{2}}[V_0=I_{0}R]$
$\sqrt{2}R=Z$
$\sqrt{2}R=\left(\sqrt{R^2+(X_C-X_L{)}^2}\right)$
$2R^2=R^2+(\frac{1}{\omega C}-\omega L)$
$R^2={(\frac{1}{{\omega }_{1}C}-{\omega }_{1}L)}^2$
$R^2=\dfrac{1}{\omega_0 +\Delta \omega )C}-(\omega_0+\Delta
\omega)L$
$R=\frac{1}{{\omega }_{0}C(1+\frac{\Delta \omega }{{\omega }_0})-{\omega }_{0}L(1+\frac{\Delta \omega }{{\omega }_0})}$
$R={\omega }_0{(1+\frac{\Delta \omega }{{\omega }_0})}^{-1}-{\omega }_{0}L(1+\frac{\Delta \omega }{{\omega }_0})$
$R={\omega }_{0}L(1-\frac{\Delta \omega }{{\omega }_0})-{\omega }_{0}L(1+\frac{\Delta \omega }{{\omega }_0})$
$R={\omega }_{0}L[1-\frac{\Delta \omega }{{\omega }_0}-1-\frac{\Delta \omega }{{\omega }_0}]$
$R={\omega }_{0}L\left[\frac{2\Delta \omega }{{\omega }_0}\right]$
$R=L2\Delta \omega$
$\Delta \omega =\frac{R}{2L}$ band width