Question

The fractional half-width $\mathit{\Delta }{\mathit{\omega }}_{\mathbf{d}}$of a resonance curve, such as the ones in Fig. 31-16, is the width of the curve at half the maximum value of I. Show that $\frac{{\mathbf{\Delta \omega }}_{\mathbf{d}}}{\mathbf{\omega }}\mathbf{=}\mathit{R}\mathbf{\times }\mathbf{\left(}\frac{\mathbf{3}\mathbf{c}}{\mathbf{I}}\mathbf{\right)}\frac{\mathbf{1}}{\mathbf{2}}$, where$\mathit{\omega }$ is the angular frequency at resonance. Note that the ratio$\frac{{\mathbf{\Delta \omega }}_{\mathbf{d}}}{\mathbf{\omega }}$ increases with R, as Fig. 31-16 shows.

Short answer

$\frac{\Delta {\omega }_d}{\omega }=R\sqrt{\frac{3C}{L}}$

Step-by-step solution

Listing the given quantities:

Fractional half width is given by,

$\frac{\left({\omega }_{d2}-{\omega }_{d1}\right)}{\omega }=\frac{\Delta {\omega }_d}{\omega }$

Here, $\omega$is the angular frequency at resonance.

Understanding the concepts of resonance:

When driving angular frequency ${\mathit{\omega }}_{\mathbf{d}}$is equal to natural angular frequency$\mathit{\omega }$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,

${\chi }_L={\omega }_{d}L$

Capacitive reactance,

${\chi }_c=\frac{1}{{\omega }_{d}C}$

Impedance,

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

The amplitude of current,

$I=\frac{{\epsilon }_m}{Z}$

Explanation:

We have angular frequency ${\omega }_d$corresponding to maximum current amplitude as

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

$\frac{1}{\sqrt{LC}}=\omega$ ...(1)

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

$I=\frac{{\epsilon }_m}{Z}$

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

$I=\frac{{\epsilon }_m}{\sqrt{R^2+{\left({\chi }_L-{\chi }_C\right)}^2}}$ ….. (2)

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

$I=\frac{{\epsilon }_m}{2R}$

Substitute the above equation into equation (2).

$$\begin{gathered} \frac{{\epsilon }_m}{2R}=\frac{{\epsilon }_m}{\sqrt{R^2+{\left({\chi }_L-{\chi }_C\right)}^2}} \\ \frac{1}{2R}=\frac{1}{\sqrt{R^2+{\left({\chi }_L-{\chi }_C\right)}^2}} \end{gathered}$$

Squaring both sides, you get

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

Rearranging the terms in the above equation, you get

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

Taking the square root on both sides,

$$\begin{gathered} \pm \sqrt{3}·R={\chi }_L-{\chi }_C \\ =\left({\omega }_{d}L-\frac{1}{{\omega }_{d}C}\right) \end{gathered}$$

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

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

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

The solution is

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

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

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

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

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

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

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

Divide both side by angular frequency.

$\frac{\Delta {\omega }_d}{\omega }=\frac{\sqrt{3}R}{L\omega }$

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

$$\begin{gathered} \frac{\Delta {\omega }_d}{\omega }=\frac{\sqrt{3}R}{L}\times \sqrt{LC} \\ =R\sqrt{\frac{3C}{L}} \end{gathered}$$

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,

$\frac{\Delta {\omega }_d}{\omega }=R\sqrt{\frac{3C}{L}}$