Chapter 33, Problem 63PQ

A series RLC circuit driven by a source with an amplitude of 120.0 V and a frequency of 50.0 Hz has an inductance of 787 mH, a resistance of 267 Ω, and a capacitance of 45.7 µF. a. What are the maximum current and the phase angle between the current and the source emf in this circuit? b. What are the maximum potential difference across the inductor and the phase angle between this potential difference and the current in the circuit? c. What are the maximum potential difference across the resistor and the phase angle between this potential difference and the current in this circuit? d. What are the maximum potential difference across the capacitor and the phase angle between this potential difference and the current in this circuit?

To determine

The maximum current and the phase angle.

Answer to Problem 63PQ

The maximum current in the circuit is $0.374\text{A}$ and the phase angle is $33.6°$.

Explanation of Solution

Write the expression to calculate the inductive reactance.

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

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

Write the expression to calculate the capacitive reactance.

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

Here, the capacitance of the capacitor is $C$.

Write the expression for the impedance.

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

Here, $Z$ is the impedance, $X_{\text{L}}$ is the inductive reactance and $X_{\text{C}}$ is the capacitive reactance.

Write the expression to calculate the maximum current

    $I_{\max }=\frac{{\epsilon }_{\max }}{Z}$                                                                                        (IV)

Here, ${\epsilon }_{\max }$ is the maximum voltage and $I_{\max }$ is the maximum current.

Write the expression to calculate the phase angle.

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

Here, $\varphi$ is the phase angle.

Conclusion:

Convert form $\text{mH}$ to $\text{H}$.

    $\begin{array}{c}787\text{mH}=787\text{mH}\left(\frac{{10}^{-3}\text{H}}{1\text{mH}}\right)\\ =787\times {10}^{-3}\text{H}\end{array}$

Substitute $787\times {10}^{-3}\text{H}$ for $L$ and $50.0\text{Hz}$ for $f$ in equation (I) to calculate the value of $X_{\text{L}}$.

    $\begin{array}{c}{X}_{\text{L}}=2\pi \left(50.0\text{Hz}\right)\left(787\times {10}^{-3}\text{H}\right)\\ =\text{247}\text{.24}\Omega \end{array}$

Convert into Farad.

    $\begin{array}{c}45.7\mu \text{F}=45.7\mu \text{F}\left(\frac{{10}^{-6}\text{F}}{1\mu \text{F}}\right)\\ =45.7\times {10}^{-6}\text{F}\end{array}$

Substitute $45.7\times {10}^{-6}\text{F}$ for $C$ and $50.0\text{Hz}$ for $f$ in equation (II) to calculate the value of $X_{\text{C}}$.

    $\begin{array}{c}{X}_{\text{C}}=\frac{1}{2\text{π}\left(\text{50}\text{.0}\text{Hz}\right)\left(\text{45}\text{.7}\times {\text{10}}^{-6}\text{F}\right)}\\ =69.65\Omega \end{array}$

Substitute $\text{247}\text{.24}\Omega$ for $X_{\text{L}}$, $267\Omega$ for $R$ and $69.65\Omega$ for $X_{\text{C}}$ in the equation (III) to calculate the value of $Z$.

    $\begin{array}{c}Z=\sqrt{{\left(267\Omega \right)}^2+{\left(\text{247}\text{.24}\Omega -69.65\Omega \right)}^2}\\ =\sqrt{71289+31538}\Omega \\ =321\Omega \end{array}$

Substitute, $120.0\text{V}$ for ${\epsilon }_{\max }$ and $321\Omega$ for $Z$ in the equation (IV) to calculate the value of $I_{\max }$.

    $\begin{array}{c}{I}_{\max }=\frac{120\text{V}}{321\Omega }\\ =0374\text{A}\end{array}$

Substitute $\text{247}\text{.24}\Omega$ for $X_{\text{L}}$, $267\Omega$ for $R$ and $69.65\Omega$ for $X_{\text{C}}$ in equation (V)  to calculate the value of $\varphi$.

    $\begin{array}{c}\tan \varphi =\frac{\left(\text{247}\text{.24}\Omega -69.65\Omega \right)}{267\Omega }\\ \varphi ={\tan }^{-1}\left(0.665\right)\\ =33.6°\end{array}$

Therefore, the maximum current in the circuit is $0.374\text{A}$. The phase angle is $33.6°$.

To determine

The maximum potential difference and the phase angle between the potential difference across the inductor and the current.

Answer to Problem 63PQ

The maximum potential difference across inductor is $92.5\text{V}$ and the phase angle between the potential difference across the inductor and the current in the circuit is $90°$.

Explanation of Solution

Write the expression for potential difference across $L$.

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

Conclusion:

Substitute $0.374\text{A}$ for $I_{\max }$ and $\text{247}\text{.24}\Omega$ for $X_{\text{L}}$ in the equation (VI) to calculate the value of $V_{\max }$.

    $\begin{array}{c}{V}_{\max }=\left(0.374\text{A}\right)\left(\text{247}\text{.24}\Omega \right)\\ =92.47\text{V}\\ \approx 92.5\text{V}\end{array}$

Since, the potential difference across the inductor and the current in the circuit is perpendicular to each other. Hence the phase angle between the potential difference across the inductor and the current in the circuit is $90°$

Thus, the maximum potential difference across inductor is $92.5\text{V}$.

The phase angle between the potential difference across the inductor and the current in the circuit is $90°$.

To determine

The maximum potential difference and the phase angle between the potential difference across the resistor and the current.

Answer to Problem 63PQ

The maximum potential difference across resistor is $99.9\text{V}$. The phase angle between the potential difference across the resistor and the current in the circuit is $0°$.

Explanation of Solution

Write the expression for potential difference across $R$.

    $V_{\max }=I_{\max }R$                                                                                                   (VII)

Conclusion:

Substitute $0.374\text{A}$ for $I_{\max }$ and $\text{267}\Omega$ for $R$ in equation (VII) to calculate the value of $V_{\max }$.

    $\begin{array}{c}{V}_{\max }=\left(0.374\text{A}\right)\left(\text{267}\Omega \right)\\ =99.9\text{V}\end{array}$

Since, the potential difference across the resistor and the current in the circuit is in the same phase. Hence the phase angle between the potential difference across the resistor and the current in the circuit is $0°$.

Therefore, the maximum potential difference across resistor is $99.9\text{V}$. The phase angle between the potential difference across the resistor and the current in the circuit is $0°$.

To determine

The maximum potential difference and the phase angle between the potential difference across the capacitor and the current.

Answer to Problem 63PQ

The maximum potential difference across capacitor is $26.1\text{V}$. The phase angle between the potential difference across the resistor and the current in the circuit is $-90.0°$.

Explanation of Solution

Write the expression for potential difference across $C$.

    $V_{\max }=I_{\max }\times X_{\text{C}}$                                                                                         (VII)

Conclusion:

Substitute $0.374\text{A}$ for $I_{\max }$ and $69.65\Omega$ for $X_{\text{C}}$ in equation (VII) to calculate the value of $V_{\max }$.

    $\begin{array}{c}{V}_{\max }=0.374\text{A}\left(69.65\Omega \right)\\ =26.05\text{V}\end{array}$

The phase angle between the potential difference across the resistor and the current in the circuit is $-90.0°$.

Therefore, the maximum potential difference across capacitor is $26.1\text{V}$. The phase angle between the potential difference across the resistor and the current in the circuit is $-90.0°$.