Question

An alternating source with a variable frequency, a capacitor with capacitance C, and a resistor with resistance Rare connected in series. Figure gives the impedance Zof the circuit versus the driving angular frequency${\mathit{\omega }}_{\mathbf{d}}$; the curve reaches an asymptote of $\mathbf{500}\mathbf{}\mathbf{\text{Ω}}$, and the horizontal scale is set by${\left({\omega }_d\right)}_{\mathbf{s}}\mathbf{=}\mathbf{300}\mathbf{}\mathbf{\text{rad/s}}$. The figure also gives the reactance ${\mathit{X}}_{\mathbf{C}}$for the capacitor versus${\mathit{\omega }}_{\mathbf{d}}$. What are (a) What is R ?(b) What is C?

Short answer

1. The value of resistance is $500\text{Ω}$.
2. The value of capacitance is $40\text{μF}$.

Step-by-step solution

The given data

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

Understanding the concept of impedance versus angular frequency graph

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,

$Z=\sqrt{R^2+{X_c}^2}$ …(i)

The reactance of a capacitor,

$X_C=\frac{1}{{\omega }_{d}C}$ …(ii)

a) Calculation of the resistance

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

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

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 $500\text{Ω}$.

b) Calculation of the capacitance

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

So, consider the point with

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

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

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

Let us consider another point:

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

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

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

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