Chapter 33, Problem 33.73AP

(a)

To determine

The inductive reactance in the circuit.

Answer to Problem 33.73AP

The inductive reactance in the circuit is $78.5\Omega$.

Explanation of Solution

Given info: The value of resistance is $150\Omega$, value of inductance is $0.250\text{H}$, value of capacitance is $2.00\text{μF}$, frequency is $50.0\text{Hz}$ and source with $210\text{V}$.

Formula to calculate the inductive reactance of the circuit is,

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

Here,

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

$f$ is the frequency of the source.

$L$ is the inductance of the inductor.

Substitute $50.0\text{Hz}$ for $f$ and $0.250\text{H}$ for $L$ to find $X_{\text{L}}$.

$\begin{array}{c}{X}_{\text{L}}=2\pi \times 50.0\text{Hz}\times 0.250\text{H}\\ =78.5\Omega \end{array}$

Conclusion:

Therefore, the inductive reactance in the circuit is $78.5\Omega$.

(b)

To determine

The capacitive reactance in the circuit.

Answer to Problem 33.73AP

The capacitive reactance in the circuit is $1.59\text{kΩ}$.

Explanation of Solution

Given info: The value of resistance is $150\Omega$, value of inductance is $0.250\text{H}$, value of capacitance is $2.00\text{μF}$, frequency is $50.0\text{Hz}$ and source with $210\text{V}$.

Formula to calculate the inductive reactance of the circuit is,

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

Here,

$X_{\text{C}}$ is the inductive reactance of the circuit.

$f$ is the frequency of the source.

$C$ is the capacitance of the capacitor.

Substitute $50.0\text{Hz}$ for $f$ and $2.00\text{μF}$ for $C$ to find $X_{\text{C}}$.

$\begin{array}{c}{X}_{\text{C}}=\frac{1}{2\pi \times 50.0\text{Hz}\times 2.00\text{μF}\times \frac{{10}^{-6}\text{F}}{1\text{μF}}}\\ =1.59\times {10}^3\Omega \times \frac{{10}^{-3}\text{kΩ}}{1\Omega }\\ =1.59\text{kΩ}\end{array}$

Conclusion:

Therefore, the capacitive reactance in the circuit is $1.59\text{kΩ}$.

(c)

To determine

The impedance in the circuit.

Answer to Problem 33.73AP

The impedance in the circuit is $1.52\text{kΩ}$.

Explanation of Solution

Given info: The value of resistance is $150\Omega$, value of inductance is $0.250\text{H}$, value of capacitance is $2.00\text{μF}$, frequency is $50.0\text{Hz}$ and source with $210\text{V}$.

Formula to calculate the impedance circuit is,

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

Here,

$Z$ is the impedance in the circuit.

$R$ is the resistance in the circuit.

Substitute $150\Omega$ for $R$, $78.5\Omega$ for $X_{\text{L}}$ and $1.59\text{kΩ}$ for $X_{\text{C}}$ to find $Z$.

$\begin{array}{c}Z=\sqrt{{\left(150\Omega \right)}^2+{\left(78.5\Omega -1.59\text{kΩ}\times \frac{{10}^3\text{kΩ}}{1\Omega }\right)}^2}\\ =\sqrt{2307132.25{\Omega }^2}\\ =1.52\times {10}^3\text{Ω}\times \frac{{10}^{-3}\text{kΩ}}{1\Omega }\\ =1.52\text{kΩ}\end{array}$

Conclusion:

Therefore, the impedance in the circuit is $1.52\text{kΩ}$.

(d)

To determine

The maximum current in the circuit.

Answer to Problem 33.73AP

The maximum current in the circuit is $138\text{mA}$.

Explanation of Solution

Given info: The value of resistance is $150\Omega$, value of inductance is $0.250\text{H}$, value of capacitance is $2.00\text{μF}$, frequency is $50.0\text{Hz}$ and source with $210\text{V}$.

Formula to calculate the maximum current in the circuit is,

$I_{\max }=\frac{\Delta V_{\max }}{Z}$

Here,

$I_{\max }$ is the maximum current in the circuit.

$\Delta V_{\max }$ is the maximum source voltage.

Substitute $1.52\text{kΩ}$ for $Z$ and $210\text{V}$ for $\Delta V_{\max }$ to find $I_{\max }$.

$\begin{array}{c}{I}_{\max }=\frac{210\text{V}}{1.52\text{kΩ}}\\ =0.138\text{A}\times \frac{{10}^{-3}\text{mA}}{1\text{A}}\\ =138\text{mA}\end{array}$

Conclusion:

Therefore, the maximum current in the circuit is $138\text{mA}$.

(e)

To determine

The phase angle between the current and the source voltage.

Answer to Problem 33.73AP

The phase angle between the current and the source voltage is $-84.3°$.

Explanation of Solution

Given info: The value of resistance is $150\Omega$, value of inductance is $0.250\text{H}$, value of capacitance is $2.00\text{μF}$, frequency is $50.0\text{Hz}$ and source with $210\text{V}$.

Formula to calculate the phase angle is,

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

Here,

$\varphi$ is the phase angle between the current and the source voltage.

Substitute $150\Omega$ for $R$, $78.5\Omega$ for $X_{\text{L}}$ and $1.59\text{kΩ}$ for $X_{\text{C}}$ to find $\varphi$.

$\begin{array}{c}\varphi ={\tan }^{-1}\left(\frac{78.5\Omega -1.59\text{kΩ}}{150\Omega }\right)\\ ={\tan }^{-1}\left(-10.07\right)\\ =-84.3°\end{array}$

Conclusion:

Therefore, the phase angle between the current and the source voltage is $-84.3°$.

(f)

To determine

The power factor for the circuit.

Answer to Problem 33.73AP

The power factor for the circuit is $0.0987$.

Explanation of Solution

Given info: The value of resistance is $150\Omega$, value of inductance is $0.250\text{H}$, value of capacitance is $2.00\text{μF}$, frequency is $50.0\text{Hz}$ and source with $210\text{V}$.

Formula to calculate the power factor for the circuit is,

$\text{power}\text{factor}=\cos \varphi$

Substitute $-84.3°$ for $\varphi$ to find $\text{power}\text{factor}$.

$\begin{array}{c}\text{power}\text{factor}=\cos \left(-84.3°\right)\\ =0.0987\end{array}$

Conclusion:

Therefore, the power factor for the circuit is $0.0987$.

(g)

To determine

The power input to the circuit.

Answer to Problem 33.73AP

The power input to the circuit is $1.43\text{W}$.

Explanation of Solution

Given info: The value of resistance is $150\Omega$, value of inductance is $0.250\text{H}$, value of capacitance is $2.00\text{μF}$, frequency is $50.0\text{Hz}$ and source with $210\text{V}$.

Formula to calculate the power input to the circuit is,

$P=\frac{{\left(\Delta V_{\text{rms}}\right)}^2}{Z}\cos \varphi$ (1)

Write the expression for the rms voltage.

$\Delta V_{\text{rms}}=\frac{\Delta V_{\max }}{\sqrt{2}}$

Here,

$\Delta V_{\text{rms}}$ is the rms voltage to the circuit.

Replace $\left(\Delta V_{\text{max}}/\sqrt{2}\right)$ by $\Delta V_{\text{rms}}$ in equation (1).

$P=\frac{{\left(\Delta V_{\text{max}}/\sqrt{2}\right)}^2}{Z}\cos \varphi$

Substitute $210\text{V}$ for $\Delta V_{\text{max}}$, $1.52\text{kΩ}$ for $Z$ and $0.0987$ for $\cos \varphi$ to find $P$.

$\begin{array}{c}P=\left(\frac{\frac{210\text{V}}{\sqrt{2}}}{1.52\text{kΩ}\times \frac{{10}^3\text{Ω}}{1\text{kΩ}}}\right)\cdot \left(0.0987\right)\\ =\text{W}\times 0.0987\\ =1.43\text{W}\end{array}$

Conclusion:

Therefore, the power input to the circuit is $1.43\text{W}$.