An alternating source with a variable frequency, an inductor with inductance L, and a resistor with resistance Rare connected in series. Figure gives the impedance Zof the circuit versus the driving angular frequency ${\mathit{\omega }}_{\mathbf{d}}$, with the horizontal axis scale set by ${\mathbf{\left(}{\mathbf{\omega }}_{\mathbf{d}}\mathbf{\right)}}_{\mathbf{s}}\mathbf{=}\mathbf{}\mathbf{1600}\mathbf{}\mathbf{}\mathbf{\text{rad/s}}$. The figure also gives the reactance ${\mathit{X}}_{\mathbf{L}}$for the inductor versus ${\mathit{\omega }}_{\mathbf{d}}$. (a) What isR? (b) What isL?

1. Variation of impedance Z with driving angular frequency ${\omega }_d$
2. Variation of inductive reactance $X_L$with driving angular frequency ${\omega }_d$
3. Horizontal scale${\omega }_{ds}=1600\text{rad/s}$

An electrical circuit composed of a resistor and an inductor, connected in series, is known as anLRcircuit. We can use the formulae for inductive reactance and the impedance of the series circuit. By observing the given plot, we can get the user data required to solve the problem.

The inductive reactance of the inductor,

${\mathit{X}}_{\mathit{L}}\mathbf{=}{\mathit{\omega }}_{\mathit{d}}\mathit{L}$ ...(1)

The impedance of theLRcircuit for the driving frequency,

$\mathit{Z}\mathbf{=}\sqrt{{\mathit{R}}^{\mathbf{2}}\mathbf{+}{{\mathit{X}}_{\mathit{L}}}^{\mathbf{2}}}$ ...(2)

Here, Ris the resistance of the resistor and Lis the inductance of the inductor.

From the plot, we can observe that as ${\omega }_d\to 0$, we have$Z=40\text{Ω}$ and$X_L=0$

Thus, using these values in equation (2), we can get the resistance value as follows:

$$\begin{gathered} 40\mathrm{\Omega }=\sqrt{R^2+0} \\ R=40\text{Ω} \end{gathered}$$

Hence, the value of the resistance is $40\text{Ω}$.

To find the value of inductance, we have to find out the slope of the plot of inductive reactance$X_L$versus driving frequency ${\omega }_d$.

Thus, the slope of the plotof inductive reactance$X_L$versus driving frequency ${\omega }_d$,is given as:

$$\begin{gathered} L=\text{slope} \\ =\frac{X_L}{{\omega }_d} \\ =\frac{\left(120\text{Ω}-0\right)}{\left(2000\text{rad/s}-0\right)} \\ =0.06\text{H} \end{gathered}$$

So, the inductance of the inductor is role="math" localid="1662805509759" $60\text{mH}$.