Chapter 33, Problem 59PQ

(a)

To determine

The total resistance of the circuit.

Answer to Problem 59PQ

The total resistance is $8.41\Omega$.

Explanation of Solution

Write the expression to calculate the voltage across the $\text{AC}$ generator.

    $\epsilon \left(t\right)=V_{\text{L}}\left(t\right)+V_{\text{R}}\left(t\right)+V_{\text{C}}\left(t\right)$                                                                                  (I)

Here, $\epsilon$ is the total voltage across the AC generator, $V_{\text{L}}$ the voltage across the inductor, $V_{\text{C}}$ is the voltage across the capacitor and $V_{\text{R}}$ is the voltage across the resistor.

Write the expression for voltage across the generator

    $\epsilon \left(t\right)={\epsilon }_{\max }\sin \left(\omega t+\varphi \right)$

Here, ${\epsilon }_{\max }$ is the maximum voltage, $\omega$ is the angular frequency and $\varphi$ is the phase angle.

Write the expression for voltage across the inductor.

    $V_{\text{L}}\left(t\right)=X_{\text{L}}{I}_{\max }\sin \left(\omega t+\frac{\pi }{2}\right)$

Here, $I_{\max }$ is the maximum current in the inductor and $X_{\text{L}}$ is the inductive reactance.

Write the expression for voltage across the capacitor.

    $V_{\text{C}}\left(t\right)=-X_{\text{C}}{I}_{\max }\sin \left(\omega t+\frac{\pi }{2}\right)$

Here, the capacitive reactance is $X_C$

Write the expression for the voltage across the resistor.

    $V_{\text{R}}\left(t\right)=R{I}_{\max }\sin \left(\omega t\right)$

Here, the resistance of the resistor is $R$.

Substitute $R{I}_{\max }\sin \left(\omega t\right)$ for $V_{\text{R}}\left(t\right)$, $X_{\text{C}}{I}_{\max }\sin \left(\omega t+\frac{\pi }{2}\right)$ for $V_{\text{C}}\left(t\right)$, $X_{\text{L}}{I}_{\max }\sin \left(\omega t+\frac{\pi }{2}\right)$ for $V_{\text{L}}\left(t\right)$ and ${\epsilon }_{\max }\sin \left(\omega t+\varphi \right)$ for $\epsilon \left(t\right)$ in equation (I)

    ${\epsilon }_{\max }\sin \left(\omega t+\varphi \right)=X_{\text{L}}{I}_{\max }\sin \left(\omega t+\frac{\pi }{2}\right)+R{I}_{\max }\sin \left(\omega t\right)+X_{\text{C}}{I}_{\max }\sin \left(\omega t+\frac{\pi }{2}\right)$        (II)

Now, substitute $0$ for $t$ in equation (II)

    $\begin{array}{c}{\epsilon }_{\max }\sin \left(\omega \times 0+\varphi \right)=\left(\begin{array}{c}{X}_{\text{L}}{I}_{\max }\sin \left(\omega \times 0+\frac{\pi }{2}\right)\\ +R{I}_{\max }\sin \left(\omega \times 0\right)-X_{\text{C}}{I}_{\max }\sin \left(\omega \times 0+\frac{\pi }{2}\right)\end{array}\right)\\ {\epsilon }_{\max }\sin \left(\varphi \right)=X_{\text{L}}{I}_{\max }\sin \left(\frac{\pi }{2}\right)+R{I}_{\max }\sin \left(0\right)-X_{\text{C}}{I}_{\max }\sin \left(\frac{\pi }{2}\right)\\ {\epsilon }_{\max }\sin \left(\varphi \right)=\left(X_{\text{L}}-X_{\text{C}}\right)I_{\max }\end{array}$                         (III)

Substitute $\sqrt{2}{I}_{\text{rms}}$ for $I_{\max }$ and $\sqrt{2}{\epsilon }_{\text{rms}}$ for ${\epsilon }_{\max }$ in the equation (III).

    $\begin{array}{c}\sqrt{2}{\epsilon }_{\text{rms}}\sin \varphi =\sqrt{2}{I}_{\text{rms}}\left(X_{\text{L}}-X_{\text{C}}\right)I_{\text{rms}}\times \sqrt{2}\\ \left(X_{\text{L}}-X_{\text{C}}\right)=\frac{{\epsilon }_{\text{rms}}\sin \varphi }{I_{\text{rms}}}\end{array}$                (IV)

Write the expression to calculate the phase angle.

    $\tan \varphi =\left(\frac{X_{\text{L}}-X_{\text{C}}}{R}\right)$                        (V)

Conclusion:

Substitute $-23.5°$ for $\varphi$, $110\text{V}$ for ${\epsilon }_{\text{rms}}$ and $12\text{A}$ for $I_{\text{rms}}$ in equation (IV) to calculate the value of $\left(X_{\text{L}}-X_{\text{C}}\right)$

    $\begin{array}{c}\left(X_{\text{L}}-X_{\text{C}}\right)=\frac{110\text{V}\sin \left(-23.5°\right)}{12\text{A}}\\ =-3.66\Omega \end{array}$

Substitute $\left(-23.5°\right)$ for $\varphi$ and $-3.66\Omega$ for $X_{\text{L}}-X_{\text{C}}$ in equation (V) to calculate the value of $R$.

    $\begin{array}{c}\tan \left(-23.5°\right)=\left(\frac{-3.66\Omega }{R}\right)\\ R=\frac{-3.66\Omega }{-0.4348}\\ =8.41\Omega \end{array}$

Therefore, the total resistance in the circuit is $8.41\Omega$.

(b)

To determine

The net reactance $\left(X_{\text{L}}-X_{\text{C}}\right)$.

Answer to Problem 59PQ

The net reactance $\left(X_{\text{L}}-X_{\text{C}}\right)$ is $-3.66\Omega$.

Explanation of Solution

Conclusion:

Substitute $-23.5°$ for $\varphi$, $110\text{V}$ for ${\epsilon }_{\text{rms}}$ and $12\text{A}$ for $I_{\text{rms}}$ in equation (IV) to calculate the value of $\left(X_{\text{L}}-X_{\text{C}}\right)$

    $\begin{array}{c}\left(X_{\text{L}}-X_{\text{C}}\right)=\frac{110\text{V}\sin \left(-23.5°\right)}{12\text{A}}\\ =-3.66\Omega \end{array}$

Therefore, the net reactance $\left(X_{\text{L}}-X_{\text{C}}\right)$ is $-3.66\Omega$.