Chapter 33, Problem 33.60AP

(a)

To determine

The current in the circuit and phase relative to the applied voltage.

Answer to Problem 33.60AP

The current in the circuit is $200\text{mA}$ and phase relative to the applied voltage is $36.8°$.

Explanation of Solution

Given info: The resistance of the circuit is $200\Omega$, the inductance of the circuit is $663\text{mH}$, the capacitance of the circuit is $26.5\text{μF}$, the input voltage is $50.0\text{V}$ and frequency of the system is $60.0\text{Hz}$.

Write the expression to calculate the inductive resistance of the circuit.

$X_{\text{L}}=2\pi fL$

Here,

$X_{\text{L}}$ is the inductive resistance of the circuit.

$f$ is the frequency of the circuit.

$L$ is the inductance of the circuit.

Substitute $663\text{mH}$ for $L$ and $60.0\text{Hz}$ for $f$ in above expression.

$\begin{array}{c}{X}_{\text{L}}=2\pi \times 60.0\text{Hz}\times 663\text{mH}\times \frac{1\text{H}}{1000\text{mH}}\\ =249.9\Omega \end{array}$

Thus, the inductive resistance of the system is $249.9\Omega$.

Write the expression to calculate the capacitive resistance of the circuit.

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

Here,

$X_{\text{C}}$ is the capacitive resistance of the system.

$f$ is the frequency of the system.

$C$ is the capacitance of the circuit.

Substitute $60.0\text{Hz}$ for $f$ and $26.5\text{μF}$ for $C$ in above expression.

$\begin{array}{c}{X}_{\text{C}}=\frac{1}{2\pi \times 60.0\text{Hz}\times 26.5\text{μF}\times \frac{1\text{F}}{{10}^6\text{μF}}}\\ =100.1\Omega \end{array}$

Thus, the capacitive resistance of the system is $100.1\Omega$.

Write the expression to calculate the impedance of the circuit.

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

Here,

$Z$ is the impedance of the circuit.

$R$ is the resistance of the circuit.

$X_{\text{L}}$ is the inductive resistance of the circuit.

$X_{\text{C}}$ is the capacitive resistance of the system.

Substitute $200\Omega$ for $R$, $249.9\Omega$ for $X_{\text{L}}$ and $100.1\Omega$ for $X_{\text{C}}$ in above expression.

$\begin{array}{c}Z=\sqrt{{\left(200\Omega \right)}^2+{\left(249.9\Omega -100.1\Omega \right)}^2}\\ =249.8\Omega \end{array}$

Thus, the impedance of the circuit is $249.8\Omega$.

Write the expression to calculate the current in the circuit.

$I=\frac{V}{Z}$

Here,

$I$ is the current in the circuit.

$V$ is the input voltage.

$Z$ is the impedance of the circuit.

Substitute $249.8\Omega$ for $Z$ and $50.0\text{V}$ for $V$ in above expression.

$\begin{array}{c}I=\frac{50.0\text{V}}{249.8\Omega }\\ =0.200\text{A}\times \frac{1000\text{mA}}{1\text{A}}\\ =200\text{mA}\end{array}$

Thus, the current in the circuit is $200\text{mA}$.

Write the expression to calculate the phase angle.

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

Here,

$\varphi$ is the phase angle.

$R$ is the resistance of the circuit.

$X_{\text{L}}$ is the inductive resistance of the circuit.

$X_{\text{C}}$ is the capacitive resistance of the system.

Substitute $200\Omega$ for $R$, $249.9\Omega$ for $X_{\text{L}}$ and $100.1\Omega$ for $X_{\text{C}}$ in above expression.

$\begin{array}{c}\varphi ={\tan }^{-1}\left(\frac{249.9\Omega -100.1\Omega }{200\Omega }\right)\\ =36.8°\end{array}$

Thus, the phase angle is $36.8°$

Conclusion:

Therefore, the current in the circuit is $200\text{mA}$ and phase relative to the applied voltage is $36.8°$.

(b)

To determine

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

Answer to Problem 33.60AP

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

Explanation of Solution

Given info: The resistance of the circuit is $200\Omega$, the inductance of the circuit is $663\text{mH}$, the capacitance of the circuit is $26.5\text{μF}$, the input voltage is $50.0\text{V}$ and frequency of the system is $60.0\text{Hz}$.

Write the expression to calculate the voltage across resistor.

$V_{\text{R}}=IR$

Here,

$V_{\text{R}}$ is the voltage across resistor.

$I$ is the current in the circuit.

$R$ is the resistance of the circuit.

Substitute $200\text{mA}$ for $I$ and $200\Omega$ for $R$ in above expression.

$\begin{array}{c}{V}_R=200\text{mA}\times \frac{1\text{A}}{1000\text{mA}}\times 200\Omega \\ =40\text{V}\end{array}$

Thus, the voltage across resistor is $40\text{V}$.

In case of resistance, the phase difference between the voltage and current across resistor is zero. Hence, the phase relative to current is $0$.

Conclusion:

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

(c)

To determine

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

Answer to Problem 33.60AP

The maximum voltage across capacitor is $20.02\text{V}$ and its phase relative to the current is $-90°$.

Explanation of Solution

Given info: The resistance of the circuit is $200\Omega$, the inductance of the circuit is $663\text{mH}$, the capacitance of the circuit is $26.5\text{μF}$, the input voltage is $50.0\text{V}$ and frequency of the system is $60.0\text{Hz}$.

Write the expression to calculate the voltage across capacitor.

$V_{\text{C}}=I{X}_{\text{C}}$

Here,

$V_{\text{C}}$ is the voltage across capacitor.

$I$ is the current in the circuit.

$X_{\text{C}}$ is the capacitive resistance of the system.

Substitute $200\text{mA}$ for $I$ and $100.1\Omega$ for $X_{\text{C}}$ in above expression.

$\begin{array}{c}{V}_{\text{C}}=200\text{mA}\times \frac{1\text{A}}{1000\text{A}}\times 100.1\Omega \\ =20.02\text{V}\end{array}$

Thus, the voltage across capacitor is $20.02\text{V}$.

In case of capacitor the current leads in the capacitor lags the voltage by $90°$. Hence, the phase relative to current is $-90°$.

Conclusion:

Therefore, the maximum voltage across capacitor is $20.02\text{V}$ and its phase relative to the current is $-90°$.

(d)

To determine

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

Answer to Problem 33.60AP

The maximum voltage across inductor is $49.98\text{V}$ and its phase relative to the current is $90°$.

Explanation of Solution

Given info: The resistance of the circuit is $200\Omega$, the inductance of the circuit is $663\text{mH}$, the capacitance of the circuit is $26.5\text{μF}$, the input voltage is $50.0\text{V}$ and frequency of the system is $60.0\text{Hz}$.

Write the expression to calculate the voltage across inductor.

$V_{\text{L}}=I{X}_{\text{L}}$

Here,

$V_{\text{L}}$ is the voltage across inductor.

$I$ is the current in the circuit.

$X_{\text{L}}$ is the inductive resistance of the system.

Substitute $200\text{mA}$ for $I$ and $249.9\Omega$ for $X_{\text{L}}$ in above expression.

$\begin{array}{c}{V}_{\text{L}}=200\text{mA}\times \frac{1\text{A}}{1000\text{A}}\times 249.9\Omega \\ =49.98\text{V}\end{array}$

Thus, the voltage across inductor is $49.98\text{V}$.

In case of capacitor the current leads in the inductor leads the voltage by $90°$. Hence, the phase relative to current is $90°$.

Conclusion:

Therefore, the maximum voltage across inductor is $49.98\text{V}$ and its phase relative to the current is $90°$.