Chapter 33, Problem 60AP

(a)

To determine

The maximum current and the phase relative to the applied voltage.

Answer to Problem 60AP

The maximum current is $0.2\text{A}$ and the phase relative to the applied voltage is $36.9°$.

Explanation of Solution

Write the expression to calculate the inductive reactance.

    $X_{\text{L}}=2\pi fL$                                                                                                                 (I)

Here, $X_{\text{L}}$ is the inductive reactance, $f$ is the frequency and $L$ is the inductance.

Write the expression to calculate the capacitive reactance.

    $X_{\text{C}}=\frac{1}{2\pi fC}$                                                                                                              (II)

Here, $X_{\text{C}}$ is the capacitive reactance, $f$ is the frequency and $C$ is the capacitance.

Write the expression to calculate the impedance.

    $Z=\sqrt{R^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2}$                                                                                              (III)

Here, $Z$ is the impedance and $R$ is the resistance

Write the expression to calculate the maximum current.

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

Here, $I_{\max }$ is the maximum current and $\Delta V_{\max }$ is the maximum applied voltage.

Write the expression to calculate the phase angle.

    $\varphi ={\tan }^{-1}\left(\frac{X_{\text{L}}-X_{\text{C}}}{R}\right)$                                                                                                (V)

Here, $\varphi$ is the phase angle.

Conclusion:

Substitute $663\text{mH}$ for $L$ and $60.0\text{Hz}$ for $f$ in equation (I) to solve for $X_{\text{L}}$.

    $\begin{array}{c}{X}_{\text{L}}=2\pi \left(60.0\text{Hz}\right)\left(663\text{mH}\times \frac{{10}^{-3}\text{H}}{1\text{H}}\right)\\ =250\Omega \end{array}$

Substitute $26.5\mu \text{F}$ for $C$ and $60.0\text{Hz}$ for $f$ in equation (II) to solve for $X_{\text{C}}$.

    $\begin{array}{c}{X}_{\text{C}}=\frac{1}{2\pi \left(60.0\text{Hz}\right)\left(26.5\mu \text{F}\times \frac{{10}^{-6}\text{F}}{1\mu \text{F}}\right)}\\ =100.1\Omega \end{array}$

Substitute $250\Omega$ for $X_{\text{L}}$, $100.1\Omega$ for $X_{\text{C}}$ and $200\Omega$ for $R$ in equation (III) to solve for $Z$.

    $\begin{array}{c}Z=\sqrt{{\left(200\Omega \right)}^2+{\left(250\Omega -100.1\Omega \right)}^2}\\ =250\Omega \end{array}$

Substitute $250\Omega$ fore $Z$ and $50.0\text{V}$ for $\Delta V_{\max }$ in equation (IV) to solve for $I_{\max }$.

    $\begin{array}{c}{I}_{\max }=\frac{50.0\text{V}}{250\Omega }\\ =0.2\text{A}\end{array}$

Substitute $250\Omega$ for $X_{\text{L}}$, $100.1\Omega$ for $X_{\text{C}}$ and $200\Omega$ for $R$ in equation (V) to solve for $\varphi$.

    $\begin{array}{c}\varphi ={\tan }^{-1}\left(\frac{250\Omega -100.1\Omega }{200\Omega }\right)\\ ={\tan }^{-1}\left(0.7495\right)\\ =36.9°\end{array}$

Therefore, the maximum current is $0.2\text{A}$ and the phase relative to the applied voltage is $36.9°$.

(b)

To determine

The maximum voltage across the resistor and its phase relative to the current.

Answer to Problem 60AP

The maximum voltage across the resistor is $40\text{V}$ and its phase relative to the current is $0$.

Explanation of Solution

Write the expression to calculate the voltage across the resistor.

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

Here, $\Delta V_{\text{R}}$ is the maximum voltage across the resistor.

Conclusion:

Substitute $0.2\text{A}$ for $I_{\max }$ and $200\Omega$ for $R$ in equation (VI) to solve for $\Delta V_{\text{R}}$.

    $\begin{array}{c}\Delta V_{\text{R}}=\left(0.2\text{A}\right)\left(200\Omega \right)\\ =40\text{V}\end{array}$

The phase difference between the voltage and current is $0$.

Therefore, the maximum voltage across the resistor is $40\text{V}$ and its phase relative to the current is $0$.

(c)

To determine

The maximum voltage across the capacitor and its phase relative to the current.

Answer to Problem 60AP

The maximum voltage across the capacitor is $20.02\text{V}$ and voltage lags behind the current by $90°$.

Explanation of Solution

Write the expression to calculate the voltage across the capacitor.

    $\Delta V_{\text{C}}=I_{\max }{X}_{\text{C}}$                                                                                                        (VII)

Here, $\Delta V_{\text{C}}$ is the maximum voltage across the capacitor.

Conclusion:

Substitute $0.2\text{A}$ for $I_{\max }$ and $100.1\Omega$ for $X_{\text{C}}$ in equation (VII) to solve for $\Delta V_{\text{C}}$.

    $\begin{array}{c}\Delta V_{\text{C}}=\left(0.2\text{A}\right)\left(100.1\Omega \right)\\ =20.02\text{V}\end{array}$

The voltage lags behind the current by $90°$.

Therefore, the maximum voltage across the capacitor is $20.02\text{V}$ and voltage lags behind the current by $90°$.

(d)

To determine

The maximum voltage across the inductor and its phase relative to the current.

Answer to Problem 60AP

The maximum voltage across the inductor is $50\text{V}$ and voltage leads the current by $90°$.

Explanation of Solution

Write the expression to calculate the voltage across the inductor.

    $\Delta V_{\text{L}}=I_{\max }{X}_{\text{L}}$                                                                                                       (VII)

Here, $\Delta V_{\text{L}}$ is the maximum voltage across the inductor.

Conclusion:

Substitute $0.2\text{A}$ for $I_{\max }$ and $250\Omega$ for $X_{\text{L}}$ in equation (VIII) to solve for $\Delta V_{\text{L}}$.

    $\begin{array}{c}\Delta V_{\text{L}}=\left(0.2\text{A}\right)\left(250\Omega \right)\\ =50\text{V}\end{array}$

The voltage leads the current by $90°$

Therefore, the maximum voltage across the inductor is $50\text{V}$ and voltage leads the current by $90°$.