91Ó°ÊÓ

1. The alternating EMF source is connected in turn, to a resistor, a capacitor, and then an inductor.
2. The driving frequency is $f_d$.
3. The amplitude of the resulting current is$I$ .
4. Fig.31-22 of the three plots is given

Applying Ohm’s law that voltage is proportional to the product of current and resistance in the formula for angular frequency, we can find the relation between current and frequency, and using this relation, we can find which given plot corresponds to which given device.

Formulae:

The angular frequency of an oscillation, ${\mathit{Ó\neg }}_{\mathbf{d}}\mathbf{=}\mathbf{2}\mathit{Ï€}{\mathit{f}}_{\mathbf{d}}$ (i)

The alternating emf of a circuit, ${\mathit{Î\mu }}_{\mathbf{m}}\mathbf{=}{\mathit{V}}_{\mathbf{R}}\mathbf{=}{\mathit{V}}_{\mathbf{C}}\mathbf{=}{\mathit{V}}_{\mathbf{L}}$ (ii)

The current equation according to the Ohm’s law, ${\mathit{I}}_{\mathbf{R}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{R}}}{\mathbf{R}{\mathbf{\text{orX}}}_{\mathbf{c}}{\mathbf{\text{orX}}}_{\mathbf{L}}}$ (iii)

The reactance of the capacitor, ${\mathit{X}}_{\mathbf{C}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{Ó\neg }}_{\mathbf{d}}\mathbf{C}}$ (iv)

The reactance of the inductor, ${\mathit{X}}_{\mathbf{L}}\mathbf{=}{\mathit{Ó\neg }}_{\mathbf{d}}\mathit{L}$ (v)

Substituting the value of voltage in equation (ii) in equation (iii), we can get the current of the resistor equation as:

$I_R=\frac{{\mathrm{Î\mu }}_m}{R}$

where,${\mathrm{Î\mu }}_m$and$R$both are constant.

Thus, the current$I_R$is also constant.

Now from Fig.31-22, we can write that plot b corresponds to circuit with$R$.

For circuit with capacitor $C$:

Substituting the value of reactance from equation (iv) in equation (ii), we can get that the current through the capacitor is given as:

$$\begin{gathered} I_C=\frac{{\mathrm{Î\mu }}_m}{\left(\frac{1}{{\mathrm{Ó\neg }}_{d}C}\right)} \\ ={\mathrm{Î\mu }}_m{\mathrm{Ó\neg }}_d \end{gathered}$$

Now, substituting equation (i) in the above equation, we get the current equation as:

$I_C=\left(2\mathrm{Ï€}{\mathrm{Î\mu }}_{m}C\right)f_d$

Where$\left(2\mathrm{Ï€}{\mathrm{Î\mu }}_{m}C\right)$ is constant. Let,$k=\left(2\mathrm{Ï€}{\mathrm{Î\mu }}_{m}C\right)$ .

Thus, the value of the current now becomes

$I_C=k{f}_d$

Here, the current$I_C$is directly proportional todriving frequency $f_d$.

Therefore, from Fig.31-22, we can write that plot c corresponds to circuit with $C$.

For circuit with inductor $L$:

Substituting the value of reactance from equation (v) in equation (ii), we can get that the current through the capacitor is given as:

$I_L=\frac{{\mathrm{Î\mu }}_m}{\left({\mathrm{Ó\neg }}_{d}L\right)}$

Now, substituting equation (i) in the above equation, we get the current equation as:

$$\begin{gathered} I_L=\frac{{\mathrm{Î\mu }}_m}{\left(2\mathrm{Ï€}{f}_d\right)L} \\ =\frac{{\mathrm{Î\mu }}_m}{\left(2\mathrm{Ï€}L\right)}\left(\frac{1}{f_d}\right) \end{gathered}$$

where,$\frac{{\mathrm{Î\mu }}_m}{\left(2\mathrm{Ï€}L\right)}$is constant. Let,$k=\frac{{\mathrm{Î\mu }}_m}{\left(2\mathrm{Ï€}L\right)}$.

Thus, the current equation becomes:

$I_L=k\left(\frac{1}{f_d}\right)$

Therefore, the current $I_L$is inversely proportional to driving frequency$f_d$ and from Fig.31-22, we can write that plot a corresponds to circuit with $L$.