Chapter 33, Problem 59AP

(a)

To determine

The maximum voltage $\Delta V_{\max }$.

Answer to Problem 59AP

The maximum voltage $\Delta V_{\max }$ is $22.4\text{V}$.

Explanation of Solution

Write the expression to calculate the maximum voltage using the given phasor diagram.

    $\Delta V_{\max }=\sqrt{\Delta V_R{}^2+{\left(\Delta V_{\text{L}}-\Delta V_{\text{C}}\right)}^2}$                                                                                 (I)

Here, $\Delta V_{\max }$ is the maximum voltage, $\Delta V_{\text{R}}$ is the voltage through the resistor, $\Delta V_{\text{L}}$ is the voltage through the inductor and $V_{\text{C}}$ is the voltage through the capacitor

Conclusion:

Substitute $25.0\text{V}$ for $\Delta V_{\text{L}}$, $15.0\text{V}$ for $\Delta V_{\text{C}}$, and $20.0\text{V}$ for $\Delta V_{\text{R}}$ in equation (I) to solve for $\Delta V_{\max }$

    $\begin{array}{c}\Delta V_{\max }=\sqrt{{\left(20.0\text{V}\right)}^2+{\left(25.0\text{V}-15.0{\text{V}}_{\text{C}}\right)}^2}\\ =\sqrt{500\text{V}}\\ =22.4\text{V}\end{array}$

Therefore, the maximum voltage $\Delta V_{\max }$ is $22.4\text{V}$.

(b)

To determine

The phase angle $\varphi$ of the circuit.

Answer to Problem 59AP

The phase angle $\varphi$ of the circuit is $26.6°$.

Explanation of Solution

Write the expression to calculate the phase angle using the given phasor diagram.

    $\varphi ={\tan }^{-1}\left(\frac{\Delta V_{\text{L}}-\Delta V_{\text{C}}}{\Delta V_{\text{R}}}\right)$                                                                                             (II)

Here, $\varphi$ is the phase angle.

Conclusion:

Substitute $25.0\text{V}$ for $\Delta V_{\text{L}}$, $15.0\text{V}$ for $\Delta V_{\text{C}}$, and $20.0\text{V}$ for $\Delta V_{\text{R}}$ in equation (I) to solve for $\varphi$.

    $\begin{array}{c}\varphi ={\tan }^{-1}\left(\frac{25.0\text{V}-15.0\text{V}}{20.0\text{V}}\right)\\ ={\tan }^{-1}\left(0.5\right)\\ =26.6°\end{array}$

Therefore, the phase angle $\varphi$ of the circuit is $26.6°$.

(c)

To determine

The maximum current in the circuit.

Answer to Problem 59AP

The maximum current in the circuit is $0.267\text{A}$.

Explanation of Solution

Write the expression to calculate the maximum current in the circuit.

    $I_{\max }=\frac{\Delta V_{\text{R}}}{R}$                                                                                                              (III)

Here, $I_{\max }$ is the maximum current in the circuit and $R$ is the resistance of the resistor.

Conclusion:

Substitute $20.0\text{V}$ for $\Delta V_{\text{R}}$ and $75.0\Omega$ for $R$ in equation (III) to solve for $I_{\max }$.

    $\begin{array}{c}{I}_{\max }=\frac{20.0\text{V}}{75.0\Omega }\\ =0.267\text{A}\end{array}$

Therefore, the maximum current in the circuit is $0.267\text{A}$.

(d)

To determine

The impedance of the circuit.

Answer to Problem 59AP

The impedance of the circuit is $83.9\Omega$.

Explanation of Solution

Write the expression to calculate the impedance of the circuit.

    $Z=\frac{\Delta V_{\max }}{I_{\max }}$                                                                                                               (IV)

Here, $Z$ is the impedance of the circuit.

Conclusion:

Substitute $0.267\text{A}$ for $I_{\max }$ and $22.4\text{V}$ for $\Delta V_{\max }$ in equation (IV) to solve for $Z$.

    $\begin{array}{c}Z=\frac{22.4\text{V}}{0.267\text{A}}\\ =83.9\Omega \end{array}$

Therefore, the impedance of the circuit is $83.9\Omega$.

(e)

To determine

The capacitance of the circuit.

Answer to Problem 59AP

The capacitance of the circuit is $47.2\mu \text{F}$.

Explanation of Solution

Write the expression to calculate the capacitance.

    $I_{\max }=2\pi fC\left(\Delta V_{\text{C}}\right)$                                                                                                   (V)

Here, $C$ is the capacitance and $f$ is the frequency.

Conclusion:

Substitute $0.267\text{A}$ for $I_{\max }$, $15.0\text{V}$ for $\Delta V_{\text{C}}$ and $60.0\text{Hz}$ for $f$ in equation (V) to solve for $C$.

    $\begin{array}{c}\left(0.267\text{A}\right)=2\pi \left(60.0\text{Hz}\right)C15.0\text{V}\\ C=\frac{\left(0.267\text{A}\right)}{2\pi \left(60.0\text{Hz}\right)15.0\text{V}}\\ =47.2\times {10}^{-6}\text{F}\times \frac{{10}^6\mu \text{F}}{1\text{F}}\\ =47.2\mu \text{F}\end{array}$

Therefore, the capacitance of the circuit is $47.2\mu \text{F}$.

(f)

To determine

The inductance of the circuit.

Answer to Problem 59AP

The inductance of the circuit is $0.249\text{H}$.

Explanation of Solution

Write the expression to calculate the capacitance.

    $I_{\max }=\frac{\left(\Delta V_{\text{L}}\right)}{2\pi fL}$                                                                                                           (VI)

Here, $L$ is the inductance and $f$ is the frequency.

Conclusion:

Substitute $0.267\text{A}$ for $I_{\max }$, $25.0\text{V}$ for $\Delta V_{\text{L}}$ and $60.0\text{Hz}$ for $f$ in equation (VI) to solve for $L$.

    $\begin{array}{c}0.267\text{A}=\frac{\left(25.0\text{V}\right)}{2\pi \left(60.0\text{Hz}\right)L}\\ L=\frac{\left(25.0\text{V}\right)}{2\pi \left(60.0\text{Hz}\right)\left(0.267\text{A}\right)}\\ =0.249\text{H}\end{array}$

Therefore, the inductance of the circuit is $0.249\text{H}$.

(g)

To determine

The average power delivered to the circuit.

Answer to Problem 59AP

The average power delivered to the circuit is $2.67\text{W}$.

Explanation of Solution

Write the expression to calculate the average power.

    $P_{\text{avg}}=I_{\text{rms}}{}^{2}R$                                                                                                           (VII)

Here, $P_{\text{avg}}$ is the average power, $I_{\text{rms}}$ is the rms value of current.

Write the expression for rms value of current.

    $I_{\text{rms}}=\frac{I_{\max }}{\sqrt{2}}$                                                                                                            (VIII)

Substitute $\frac{I_{\max }}{\sqrt{2}}$ for $I_{\text{rms}}$ in equation (VII)

    $P_{\text{avg}}={\left(\frac{I_{\max }}{\sqrt{2}}\right)}^{2}R$                                                                                                      (IX)

Conclusion:

Substitute $0.267\text{A}$ for $I_{\max }$ and $75.0\Omega$ for $R$ in equation (IX) to solve for $P_{\text{avg}}$.

    $\begin{array}{c}{P}_{\text{avg}}={\left(\frac{0.267\text{A}}{\sqrt{2}}\right)}^2\left(75.0\Omega \right)\\ =2.67\text{W}\end{array}$

Therefore, the average power delivered to the circuit is $2.67\text{W}$.