Chapter 33, Problem 60PQ

An AC source of angular frequency ω is connected to a resistor R and a capacitor C in series. The maximum current measured is I_max. While the same maximum emf is maintained, the angular frequency is changed to ω/3. The measured current is now I/2. Determine the ratio of the capacitive reactance to the resistance at the initial frequency ω.

To determine

The ratio of the capacitive reactance to the resistance at the initial frequency.

Answer to Problem 60PQ

The ratio of the capacitive reactance to the resistance at the initial frequency is $0.77$.

Explanation of Solution

Write the expression to calculate the potential drop across the capacitor.

    $V_{\text{C}}=I_{\text{rms}}{X}_{\text{C}}$                                                                                                                (I)

Here, $V_{\text{C}}$ is the potential drop across the capacitor, $X_{\text{C}}$ is the capacitive reactance  and $I_{\text{rms}}$ is the value of current.

Write the expression for capacitive reactance.

    $X_{\text{C}}=\frac{1}{\omega C}$                                                                                                                 (II)

Here, $\omega$ is the angular frequency and $C$ is the capacitance of the capacitor.

Write the expression for the potential difference across the resistor.

    $V_{\text{R}}=I_{\text{rms}}R$                                                                                                                (III)

Here, $V_{\text{R}}$ is the potential difference across the resistor and $R$ is the resistance of the resistor.

Write the expression for the emf across the $RC$ combination.

    $\epsilon =\sqrt{V_{\text{R}}{}^2+V_{\text{C}}{}^2}$                                                                                                          (IV)

Here, $\epsilon$ is the emf across the $RC$ combination.

Substitute $I_{\text{rms}}R$ for $V_{\text{R}}$, $I_{\text{rms}}{X}_{\text{C}}$ for $V_{\text{C}}$ and $\frac{1}{\omega C}$ for $X_{\text{C}}$ in equation (IV).

    $\epsilon =\sqrt{{\left(I_{\text{rms}}R\right)}^2+{\left(I_{\text{rms}}\left(\frac{1}{\omega C}\right)\right)}^2}$                                                                                  (V)

Rearrange equation (V) to get the expression for $I_{\text{rms}}$.

    $\begin{array}{c}\epsilon =\sqrt{{\left(I_{\text{rms}}R\right)}^2+{\left(I_{\text{rms}}\left(\frac{1}{\omega C}\right)\right)}^2}\\ I_{\text{rms}}=\frac{\epsilon }{\sqrt{{\left(I_{\text{rms}}R\right)}^2+{\left(I_{\text{rms}}\left(\frac{1}{\omega C}\right)\right)}^2}}\end{array}$                                                                             (VI)

The relation for the maximum current is,

    $I_{\max }=\sqrt{2}{I}_{\text{rms}}$                                                                                                        (VII)

Substitute $\frac{\epsilon }{\sqrt{{\left(I_{\text{rms}}R\right)}^2+{\left(I_{\text{rms}}\left(\frac{1}{\omega C}\right)\right)}^2}}$ for $I_{\text{rms}}$ in equation (VII).

    $I_{\max }=\sqrt{2}\left(\frac{\epsilon }{\sqrt{{\left(I_{\text{rms}}R\right)}^2+{\left(I_{\text{rms}}\left(\frac{1}{\omega C}\right)\right)}^2}}\right)$                                                                (VIII)

Substitute $\frac{\omega }{3}$ for $\omega$ and $\frac{I_{\max }}{2}$ for $I_{\max }$ in equation (VIII).

    $\begin{array}{c}\frac{I_{\max }}{2}=\sqrt{2}\left(\frac{\epsilon }{\sqrt{{\left(I_{\text{rms}}R\right)}^2+{\left(I_{\text{rms}}\left(\frac{3}{\omega C}\right)\right)}^2}}\right)\\ I_{\max }=2\sqrt{2}\left(\frac{\epsilon }{\sqrt{{\left(I_{\text{rms}}R\right)}^2+{\left(I_{\text{rms}}\left(\frac{3}{\omega C}\right)\right)}^2}}\right)\end{array}$                                                                (IX)

Equate equation (VIII) and (XI)

    $\begin{array}{c}2\sqrt{2}\left(\frac{\epsilon }{\sqrt{{\left(R\right)}^2+{\left(\frac{3}{\omega C}\right)}^2}}\right)=\sqrt{2}\left(\frac{\epsilon }{\sqrt{{\left(R\right)}^2+{\left(\frac{1}{\omega C}\right)}^2}}\right)\\ 2\left(\frac{\epsilon }{\sqrt{{\left(R\right)}^2+{\left(\frac{3}{\omega C}\right)}^2}}\right)=\left(\frac{\epsilon }{\sqrt{{\left(R\right)}^2+{\left(\left(\frac{3}{\omega C}\right)\right)}^2}}\right)\end{array}$                                                   (X)

Square both side of the equation and substitute $X_{\text{C}}$ for $\frac{1}{\omega C}$ in equation (X) to solve for ratio $\frac{X_{\text{C}}}{R}$.

    $\begin{array}{c}{\left(2\left(\frac{\epsilon }{\sqrt{{\left(R\right)}^2+{\left(I_{\text{rms}}\left(\frac{3}{\omega C}\right)\right)}^2}}\right)\right)}^2={\left(\frac{\epsilon }{\sqrt{{\left(R\right)}^2+{\left(\frac{1}{\omega C}\right)}^2}}\right)}^2\\ 4\left(\frac{1}{{\left(R\right)}^2+{\left(\frac{3}{\omega C}\right)}^2}\right)=\left(\frac{1}{{\left(R\right)}^2+{\left(\frac{1}{\omega C}\right)}^2}\right)\\ 4\left({\left(R\right)}^2+{\left(X_{\text{C}}\right)}^2\right)={\left(R\right)}^2+9{\left(X_{\text{C}}\right)}^2\\ 4R^2-1R^2=9X_{\text{C}}{}^2-4X_{\text{C}}{}^2\end{array}$

Sole further.

    $\begin{array}{c}3R^2=5X_{\text{C}}{}^2\\ \frac{X_{\text{C}}}{R}={\left(\frac{3}{5}\right)}^{\frac{1}{2}}\\ =0.77\end{array}$

Conclusion:

Therefore, the ratio of the capacitive reactance to the resistance at the initial frequency is $0.77$