Chapter 21, Problem 37P

(a)

To determine

The voltages $V_{\text{L}},V_{\text{C}},V_{\text{R}}$, and the phase angle.

Answer to Problem 37P

The values of the voltages are $\underset{\_}{V_{\text{L}}=3.4\text{V}}$, $\underset{\_}{V_{\text{C}}=9.2\text{V}}$, $\underset{\_}{V_{\text{R}}=6.9\text{V}}$, and the phase angle is $\underset{\_}{\varphi =-40°}$.

Explanation of Solution

Given that, in the series $RLC$ circuit, the capacitance is $0.20\text{mF}$, the inductance is $13\text{mH}$, the resistance is $10.0\Omega$, the voltage amplitude is $9.0\text{V}$, and the frequency is $60\text{Hz}$. The circuit is shown in Figure 1.

Write the expression for the phase angle in the series $RLC$ circuit.

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

Here, $\varphi$ is the phase angle, $X_{\text{L}}$ is the inductive reactance, $X_{\text{C}}$ is the capacitive reactance, and $R$ is the resistance.

Write the expression for the inductive reactance.

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

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

Write the expression for the capacitive reactance.

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

Here, $C$ is the capacitance.

Write the expression for the impedance of the series $RLC$ circuit.

$Z=\frac{R}{\cos \varphi }$ (IV)

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

Write the expression for the current in the series $RLC$ circuit.

$I=\frac{{\epsilon }_{\text{m}}}{Z}$ (V)

Here, $I$ is the current, and ${\epsilon }_{\text{m}}$ is the voltage amplitude.

Write the expression for the voltage across the inductor ($V_{\text{L}}$).

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

Write the expression for the voltage across the capacitor ($V_{\text{C}}$).

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

Write the expression for the voltage across the resistor ($V_{\text{R}}$).

$V_{\text{R}}=IR$ (VIII)

Conclusion:

Substitute $60\text{Hz}$ for $f$, and $13\text{mH}$ for $L$ in equation (II) to find $X_{\text{L}}$.

$\begin{array}{c}{X}_{\text{L}}=2\pi \left(60\text{Hz}\right)\left(13\text{mH}\right)\\ =2\pi \left(60\text{Hz}\right)\left(13\text{mH}\times \frac{1\text{H}}{1000\text{mH}}\right)\\ =4.901\Omega \end{array}$

Substitute $60\text{Hz}$ for $f$, and $0.20\text{mF}$ for $C$ in equation (III) to find $X_{\text{C}}$.

$\begin{array}{c}{X}_{\text{C}}=\frac{1}{2\pi \left(60\text{Hz}\right)\left(0.20\text{mF}\right)}\\ =\frac{1}{2\pi \left(60\text{Hz}\right)\left(0.20\text{mF}\times \frac{1\text{F}}{1000\text{mF}}\right)}\\ =13.26\Omega \end{array}$

Substitute $4.901\Omega$ for $X_{\text{L}}$, $13.26\Omega$ for $X_{\text{C}}$, and $10.0\Omega$ for $R$ in equation (I) to find $\varphi$.

$\begin{array}{c}\varphi ={\tan }^{-1}\frac{4.901\Omega -13.26\Omega }{10.0\Omega }\\ =-39.9°\\ \approx -40°\end{array}$

Substitute $-39.9°$ for $\varphi$, and $10.0\Omega$ for $R$ in equation (IV) to find $Z$.

$\begin{array}{c}Z=\frac{10.0\Omega }{\cos \left(-39.9°\right)}\\ =13.035\Omega \end{array}$

Substitute $13.035\Omega$ for $Z$, and $9.0\text{V}$ for ${\epsilon }_{\text{m}}$ in equation (V) to find $I$.

$\begin{array}{c}I=\frac{9.0\text{V}}{13.035\Omega }\\ =0.6904\text{A}\end{array}$

Substitute $0.6904\text{A}$ for $I$, and $4.901\Omega$ for $X_{\text{L}}$ in equation (VI) to find $V_{\text{L}}$.

$\begin{array}{c}{V}_{\text{L}}=\left(0.6904\text{A}\right)\left(4.901\Omega \right)\\ =3.4\text{V}\end{array}$

Substitute $0.6904\text{A}$ for $I$, and $13.26\Omega$ for $X_{\text{C}}$ in equation (VII) to find $V_{\text{C}}$.

$\begin{array}{c}{V}_{\text{L}}=\left(0.6904\text{A}\right)\left(13.26\Omega \right)\\ =9.2\text{V}\end{array}$

Substitute $0.6904\text{A}$ for $I$, and $10.0\Omega$ for $R$ in equation (VIII) to find $V_{\text{R}}$.

$\begin{array}{c}{V}_{\text{L}}=\left(0.6904\text{A}\right)\left(10.0\Omega \right)\\ =6.9\text{V}\end{array}$

Therefore, the values of the voltages are $\underset{\_}{V_{\text{L}}=3.4\text{V}}$, $\underset{\_}{V_{\text{C}}=9.2\text{V}}$, $\underset{\_}{V_{\text{R}}=6.9\text{V}}$, and the phase angle is $\underset{\_}{\varphi =-40°}$.

(b)

To determine

The phasor diagram of the voltages of the given $RLC$ circuit.

Answer to Problem 37P

The phasor diagram of the voltages of the given $RLC$ circuit is shown in Figure 2.

Explanation of Solution

For the given series $RLC$ circuit I Figure 1, the values of the voltages obtained as $V_{\text{L}}=3.4\text{V}$, $V_{\text{C}}=9.2\text{V}$, $V_{\text{R}}=6.9\text{V}$, and the phase angle is $\varphi =-40°$.

The difference of the voltages across the inductor and capacitor is obtained to be,

$\begin{array}{c}{V}_{\text{L}}-V_{\text{C}}=3.4\text{V}-9.2\text{V}\\ =-5.8\text{V}\end{array}$

Conclusion:

The phase difference between the voltages $V_{\text{L}}$ and $V_{\text{C}}$ is $\pi \text{rad}$, and the phase difference between $V_{\text{L}}$ and $V_{\text{R}}$ is $\pi /2\text{rad}$. The phasor diagram of the voltages of the given $RLC$ circuit is shown in Figure 2.

Therefore, the phasor diagram of the voltages of the given $RLC$ circuit is shown in Figure 2.