91Ó°ÊÓ

1. Impedance of the curve, $Z=500\text{Ω}$
2. On horizontal scale, the value of angular frequency is set at ${\mathrm{Ó\neg }}_{ds}=300\text{rad/s}$

The impedance of a circuit is the effective resistance of an electric circuit or component to alternating current, arising from the combined effects of ohmic resistance and capacitive reactance. We need to use the information from the graph and impedance formula of the circuit to find the resistance and capacitance of the coil.

The impedance of the circuit for the driving frequency,

$Z=\sqrt{R^2+{X_c}^2}$ …(¾±)

The reactance of a capacitor,

$X_C=\frac{1}{{\mathrm{Ó\neg }}_{d}C}$ …(¾±¾±)

The value of the impedance of equation (i) using equation (ii) becomes:

$Z=\sqrt{R^2+\frac{1}{{\left({\mathrm{Ó\neg }}_{d}C\right)}^2}}$

At asymptote $Z=R,$the second term in the square-root is zero. Here, from the above equation the value of the resistance in the coil is found to be:

$$\begin{gathered} Z=R \\ R=500\text{Ω} \end{gathered}$$

Hence, the value of the resistance is $500\text{Ω}$.

From the given graph of capacitive reactance, considering any one point, that is

So, consider the point with

$$\begin{gathered} {\mathrm{Ó\neg }}_d=50\text{rad/s}, \\ {X}_C=500\text{Ω} \end{gathered}$$

Thus, we can calculate the capacitance C using equation (ii) as follows:

$$\begin{gathered} C={\left({\mathrm{Ó\neg }}_d{X}_C\right)}^{-1} \\ ={\left(\left(\text{50}\text{rad/s}\right)\left(500\text{Ω}\right)\right)}^{-1} \\ =40\text{μ¹ó} \end{gathered}$$

Let us consider another point:

$$\begin{gathered} {\mathrm{Ó\neg }}_d=250\text{rad/s,} \\ {X}_C=100\text{Ω} \end{gathered}$$

Thus, we can calculate the capacitance C using equation (ii) as follows:

$$\begin{gathered} C={\left({\mathrm{Ó\neg }}_d{X}_C\right)}^{-1} \\ ={\left(\left(250\text{rad/s}\right)\left(100\text{Ω}\right)\right)}^{-1} \\ =40\text{μ¹ó} \end{gathered}$$

Hence, the capacitance of the capacitor is $40\text{μ¹ó}$.