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Alternating Current — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsAlternating Current5 Marks
Question
Draw the graph showing the variation of alternating current with frequency in a L-C-R series resonant circuit and derive the expression for band width.
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Answer

Variation of current with frequency (an-gular frequency) in a series L-C-R resonant cir-cuit is shown in adjoining figure.
It is clear that for frequencies greater than or less than $\omega_0$, the values of current are less than the maximum value (I), where $\omega_0$ is resonant angular frequency.
Expression for Band Width : In series LCR resonant circuit at resonant angular frequency $\omega_0=1 / \sqrt{ LC }$ the amplitude of current is maximum. The quickness with which the current falls from its resonant value $\left( V _{ o } /\right.$R) with change in applied frequency is known as sharpness of resonance.
"It is measured by the ratio of resonant frequency $\omega_0$ to the diffe-rence of two frequencies $\omega_1$ and $\omega_2$ at which the current falls to $1 / \sqrt{2}$ of the resonant value.
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∴ Sharpness of resonance $=\frac{\omega_0}{\omega_2-\omega_1}$ Here $\omega_1$ and $\omega_2$ known as half power frequencies, because at these frequencies the power in the circuit reduces to half of its maximum value. The difference of half power frequencies ($\omega_1$ - $\omega_2$) is known as 'band width'. The smaller is the band width, the sharper is the resonance.
Expression for Band Width : The value of impedance of series LCR circuit at resonant frequency is Z, therefore at $\omega_1$ and $\omega_2$ it will be $\sqrt{2} R$ because at these frequencies current falls to $1 / \sqrt{2}$ of the resonant value.
$\therefore Z=\sqrt{2} R \Rightarrow \sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}=\sqrt{2} \cdot R$
On squaring both sides,
$R ^2+\left(\omega L -\frac{1}{\omega C }\right)^2=2 R ^2$
or $\left(\omega L -\frac{1}{\omega C }\right)^2= R ^2$
or $\left(\omega L -\frac{1}{\omega C }\right)= \pm R$
Thus, if $\omega_2$ > $\omega_1$ then it may be concluded that
$\omega_1 L-\frac{1}{\omega_1 C}=-R$ ...(1)
and $\omega_2 L-\frac{1}{\omega_2 C}=+R$ ...(2)
On adding equations (1) and (2) get
$\left(\omega_1-\omega_2\right) L-\frac{1}{C}\left(\frac{\omega_1-\omega_2}{\omega_1 \omega_2}\right)=0$
$\Rightarrow \quad \omega_2 \omega_1=\frac{1}{ LC }$ ...(3)
On subtracting eq. (1) from (2), we get
$\left(\omega_2-\omega_1\right) L+\frac{1}{C}\left(\frac{\omega_2-\omega_1}{\omega_1 \omega_2}\right)=2 R$
or $\left(\omega_2-\omega_1\right)\left[L+\frac{1}{C \omega_1 \omega_2}\right]=2 R$ ...(4)
On substituting the value of $\omega_1 \omega_2$ from equation (3) in equation (4),
$\left(\omega_2-\omega_1\right)\left[ L +\frac{1}{ C \times \frac{1}{ LC }}\right]=2 R$
$\Rightarrow \quad\left(\omega_2-\omega_1\right)= R / L$
i.e., Band width $=( R / L )$

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