Chapter 35, Problem 28SP

A 10.0-µF capacitor is in series with a 40.0-Ω resistance, and the combination is connected to a 110-V, 60.0-Hz line. Calculate (a) the capacitive reactance, (b) the impedance of the circuit, (c) the current in the circuit, (d) the phase angle between current and supply voltage, and (e) the power factor for the circuit.

To determine

The capacitive reactance of the $10.0\text{ μF}$ capacitor connected with $40.0\Omega$ resistance when the combination is connected to a $110\text{ V,}\text{60 Hz}$ line.

Answer to Problem 28SP

Solution:

$266\Omega$

Explanation of Solution

Given data:

The value of capacitance of the capacitor is $10.0\text{ μF}$.

The frequency of applied voltage is $60.0\text{ Hz}$.

Formula used:

The expression of capacitive reactance is expressed as,

$X_C=\frac{1}{2\pi fC}$

Here, $f$ is the frequency of the power supply, and $C$ is the capacitance of the capacitor.

Explanation:

Consider the expression of capacitive reactance,

$X_C=\frac{1}{2\pi fC}$

Substitute $60.0\text{ Hz}$ for $f$ and $10.0\text{ μF}$ for $C$

$\begin{array}{c}{X}_C=\frac{1}{2\pi \left(60.0\text{ Hz}\right)\left(10.0\text{ μF}\right)}\\ =\frac{1}{2\pi \left(60.0\text{ Hz}\right)\left(10.0\text{ μF}\left(\frac{{10}^{-6}\text{ F}}{\text{μF}}\right)\right)}\\ \simeq 266\Omega \end{array}$

Conclusion:

The capacitive reactance of the capacitor is $266\Omega$.

To determine

The impedance of the RC circuit having $10.0\text{ μF}$ capacitor connected with $40.0\Omega$ resistance when the combination is connected to a $110\text{ V,}\text{60 Hz}$ line.

Answer to Problem 28SP

Solution:

$269\Omega$

Explanation of Solution

Given data:

The value of capacitive reactance of the capacitor is $266\Omega$.

The resistance of the RC circuit is $40.0\Omega$.

Formula used:

The impedance of the capacitor is expressed as,

$Z=\sqrt{R^2+\left(X_C^2\right)}$

Here, $X_C$ is the capacitive reactance, and $R$ is the resistance of the resistor.

Explanation:

Consider the expression of impedance of the RC circuit.

$Z=\sqrt{R^2+\left(X_C^2\right)}$

Substitute $266\Omega$ for $X_C$ and $40.0\Omega$ for $R$

$\begin{array}{c}Z=\sqrt{{\left(266\Omega \right)}^2+{\left(40.0\Omega \right)}^2}\\ =\sqrt{70756+1600}\\ \approx 269\Omega \end{array}$

Conclusion:

The impedance of the RC circuit is $269\Omega$.

To determine

The current in the RC circuit having $10.0\text{ μF}$ capacitor connected with $40.0\Omega$ resistance when the combination is connected to a $110\text{ V,}\text{60 Hz}$ line.

Answer to Problem 28SP

Solution:

$0.409\text{ A}$

Explanation of Solution

Given data:

The rms value of voltage is $110\text{ V}$.

The impedance of the RC circuit is $269\Omega$.

Formula used:

Write the expression of Ohm’s law for ac circuit.

$V_{rms}=I_{rms}Z$

Here, $V_{rms}$ is the rms value of voltage, $I_{rms}$ is the rms value of current, and $Z$ is the impedance of thecircuit.

Explanation:

Consider the expressionof Ohm’s law.

$V_{rms}=I_{rms}Z$

Rearrange for $I_{rms}$.

$I_{rms}=\frac{V_{rms}}{Z}$

Substitute $110\text{ V}$ for $V_{rms}$ and $269\Omega$ for $Z$

$\begin{array}{c}{I}_{rms}=\frac{110\text{ V}}{269\Omega }\\ \approx 0.409\text{ A}\end{array}$

Conclusion:

The current through thecircuit is $0.409\text{ A}$.

To determine

The phase angle between current and supply voltage when the $10.0\text{ μF}$ capacitor connected with $40.0\Omega$ resistance, and the combination is connected to a $110\text{ V,}\text{60 Hz}$ line.

Answer to Problem 28SP

Solution:

Voltage lags by $81.4°$

Explanation of Solution

Given data:

The resistance of the circuit is $40.0\Omega$.

The value of capacitive reactance of the inductor is $266\Omega$.

Formula used:

The expression for phase angle between voltage and current in RL circuit is expressed as,

$\tan \varphi =\frac{X_C}{R}$

Here, $X_C$ is the capacitive reactance, and $R$ is the resistance of the capacitor.

Explanation:

Consider the expressionfor phase angle between voltage and current.

$\tan \varphi =\frac{X_C}{R}$

Solve for $\varphi$.

$\varphi ={\tan }^{-1}\left(\frac{X_C}{R}\right)$

Substitute $266\Omega$ for $X_C$ and $40.0\Omega$ for $R$

$\begin{array}{c}\varphi ={\tan }^{-1}\left(\frac{266\Omega }{40.0\Omega }\right)\\ =81.4°\end{array}$

The circuit is capacitive in nature; hence voltage is lagging in nature.

Conclusion:

The phase angle between current and supply voltage is $81.4°$ lagging.

To determine

The power factor of the circuit when the $10.0\text{ μF}$ capacitor connected with $40.0\Omega$ resistance, and the combination is connected to a $110\text{ V,}\text{60 Hz}$ line.

Answer to Problem 28SP

Solution:

$0.149$

Explanation of Solution

Given data:

The phase angle between current and supply voltage is $81.4°$.

Formula used:

The expression of the power factor is expressed as,

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

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

Explanation:

Consider the expression for power factor.

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

Substitute $81.4°$ for $\varphi$

$\begin{array}{c}\text{Power factor}=\cos \left(81.4°\right)\\ =0.149\end{array}$

Conclusion:

The power factor of the circuit is $0.149$.