91Ó°ÊÓ

Fractional half width is given by,

$\frac{\left({\mathrm{Ó\neg }}_{d2}-{\mathrm{Ó\neg }}_{d1}\right)}{\mathrm{Ó\neg }}=\frac{\mathrm{Δ}{\mathrm{Ó\neg }}_d}{\mathrm{Ó\neg }}$

Here, $\mathrm{Ó\neg }$is the angular frequency at resonance.

When driving angular frequency ${\mathit{Ó\neg }}_{\mathbf{d}}$is equal to natural angular frequency$\mathit{Ó\neg }$of the circuit, then the current amplitude is maximum because$\mathit{Z}\mathbf{=}\mathit{R}$.

The above condition is satisfied only when the capacitive reactance is exactly matched with inductive reactance, and the condition is called the resonance condition.

Formulas:

Inductive reactance,

${\mathrm{χ}}_L={\mathrm{Ó\neg }}_{d}L$

Capacitive reactance,

${\mathrm{χ}}_c=\frac{1}{{\mathrm{Ó\neg }}_{d}C}$

Impedance,

$Z=\sqrt{R^2+{\left({\mathrm{χ}}_L-{\mathrm{χ}}_C\right)}^2}$

The amplitude of current,

$I=\frac{{\mathrm{Î\mu }}_m}{Z}$

We have angular frequency ${\mathrm{Ó\neg }}_d$corresponding to maximum current amplitude as

localid="1663234080329" $$\begin{gathered} {\mathrm{Ó\neg }}_d=\sqrt{\frac{1}{LC}} \\ =\frac{1}{\sqrt{LC}} \end{gathered}$$

$\frac{1}{\sqrt{LC}}=\mathrm{Ó\neg }$ ...(1)

For a given amplitude of emf, current amplitude is given by

$I=\frac{{\mathrm{Î\mu }}_m}{Z}$

Substitute $\sqrt{R^2+{\left({\mathrm{χ}}_L-{\mathrm{χ}}_C\right)}^2}$ for Z in the above equation.

$I=\frac{{\mathrm{Î\mu }}_m}{\sqrt{R^2+{\left({\mathrm{χ}}_L-{\mathrm{χ}}_C\right)}^2}}$ ….. (2)

To find the angular frequency corresponding to half of the maximum current amplitude,you have

$I=\frac{{\mathrm{Î\mu }}_m}{2R}$

Substitute the above equation into equation (2).

$$\begin{gathered} \frac{{\mathrm{Î\mu }}_m}{2R}=\frac{{\mathrm{Î\mu }}_m}{\sqrt{R^2+{\left({\mathrm{χ}}_L-{\mathrm{χ}}_C\right)}^2}} \\ \frac{1}{2R}=\frac{1}{\sqrt{R^2+{\left({\mathrm{χ}}_L-{\mathrm{χ}}_C\right)}^2}} \end{gathered}$$

Squaring both sides, you get

$\frac{1}{4R^2}=\frac{1}{R^2+{\left({\mathrm{χ}}_L-{\mathrm{χ}}_C\right)}^2}$

Rearranging the terms in the above equation, you get

$$\begin{gathered} 4R^2=R^2+{\left({\mathrm{χ}}_L-{\mathrm{χ}}_C\right)}^2 \\ 3R^2={\left({\mathrm{χ}}_L-{\mathrm{χ}}_C\right)}^2 \end{gathered}$$

Taking the square root on both sides,

$$\begin{gathered} Â\pm \sqrt{3}·R={\mathrm{χ}}_L-{\mathrm{χ}}_C \\ =\left({\mathrm{Ó\neg }}_{d}L-\frac{1}{{\mathrm{Ó\neg }}_{d}C}\right) \end{gathered}$$

Simplifying the above equation further, you get a quadratic equation in${\mathrm{Ó\neg }}_d$.

$\left(LC\right){\mathrm{Ó\neg }}_d^2Â\pm \sqrt{3}\left(RC\right){\mathrm{Ó\neg }}_d-1=0$

The two roots of this equation will givelower angular frequency${\mathrm{Ó\neg }}_{d1}$and higher angular frequency${\mathrm{Ó\neg }}_{d2}$corresponding to half of the maximum current amplitude.

The solution is

${\mathrm{Ó\neg }}_d=\frac{-\left(Â\pm \sqrt{3}RC\right)Â\pm \sqrt{3{\left(RC\right)}^2+4LC}}{2LC}$

As negative frequency is not possible, you will neglect the negative sign in the second term. Hence, you have

${\mathrm{Ó\neg }}_d=\frac{-\left(Â\pm \sqrt{3}RC\right)+\sqrt{3{\left(RC\right)}^2+4LC}}{2LC}$

Now considering the positive sign, let${\mathrm{Ó\neg }}_d={\mathrm{Ó\neg }}_{d1}$. The corresponding root will givelower angular frequency${\mathrm{Ó\neg }}_{d1}$.

${\mathrm{Ó\neg }}_{d1}=\frac{-\left(\sqrt{3}RC\right)+\sqrt{3{\left(RC\right)}^2+4LC}}{2LC}$

Similarly, considering the negative sign, let ${\mathrm{Ó\neg }}_d={\mathrm{Ó\neg }}_{d2}$. The corresponding root will give higher angular frequency ${\mathrm{Ó\neg }}_{d2}$.

$$\begin{gathered} \mathrm{Δ}{\mathrm{Ó\neg }}_d={\mathrm{Ó\neg }}_{d2}-{\mathrm{Ó\neg }}_{d1} \\ =\frac{\left(+\sqrt{3}RC\right)+\sqrt{3{\left(RC\right)}^2+4LC}}{2LC}+\frac{\left(\sqrt{3}RC\right)-\sqrt{3{\left(RC\right)}^2+4LC}}{2LC} \\ \mathrm{Δ}{\mathrm{Ó\neg }}_d=\frac{\left(\sqrt{3}RC+\sqrt{3}RC\right)}{2LC} \\ =\frac{\sqrt{3}R}{L} \end{gathered}$$

Divide both side by angular frequency.

$\frac{\mathrm{Δ}{\mathrm{Ó\neg }}_d}{\mathrm{Ó\neg }}=\frac{\sqrt{3}R}{L\mathrm{Ó\neg }}$

From equation (1), substitute $\frac{1}{\sqrt{LC}}$for $\mathrm{Ó\neg }$in the above right hand side expression, hence you obtain

$$\begin{gathered} \frac{\mathrm{Δ}{\mathrm{Ó\neg }}_d}{\mathrm{Ó\neg }}=\frac{\sqrt{3}R}{L}×\sqrt{LC} \\ =R\sqrt{\frac{3C}{L}} \end{gathered}$$

From the above equation, notice that fractional width is directly proportional to resistance$R$.

Hence if$R$increases,thebell-shaped curve will become broader. Therefore,

$\frac{\mathrm{Δ}{\mathrm{Ó\neg }}_d}{\mathrm{Ó\neg }}=R\sqrt{\frac{3C}{L}}$