Chapter 33, Problem 40P

(a)

To determine

The rms current in the circuit.

Answer to Problem 40P

The rms current in the circuit is $\underset{\_}{5.42\text{A}}$.

Explanation of Solution

Write expression for rms current.

    $I_{\text{rms}}=\frac{V_{\text{rms}}}{Z}$                                                                                                                 (I)

Here, $I_{\text{rms}}$ is the rms current, $V_{\text{rms}}$ is the rms voltage and $Z$ is the impedance.

Write expression for $Z$.

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

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

Write expression for $X_{\text{C}}$.

    $X_{\text{C}}=\frac{1}{\omega C}$                                                                                                               (III)

Here, $\omega$ is the angular frequency and $C$ is the capacitance.

Write expression for $X_{\text{L}}$.

    $X_{\text{L}}=\omega L$                                                                                                                (IV)

Here, $L$ is inductance.

Write expression for $\omega$.

    $\omega =2\pi f$                                                                                                                 (V)

Here, $f$ is the frequency.

Conclusion:

Substitute $60.0\text{-Hz}$  for $f$ in equation (V) to calculate $\omega$.

    $\begin{array}{c}\omega =2\pi \left(60.0\text{-Hz}\right)\\ =377\text{rad}/\text{s}\end{array}$

Substitute $377\text{rad}/\text{s}$ for $\omega$ and $25.0\text{-mH}$ for $L$ is equation (IV) to calculate $X_{\text{L}}$.

    $\begin{array}{c}{X}_{\text{L}}=\left(377\text{rad}/\text{s}\right)\left(25.0\text{-mH}\right)\left(\frac{{10}^{\text{-3}}\text{H}}{1\text{mH}}\right)\\ =9.43\Omega \end{array}$

Substitute $9.43\Omega$ for $X_{\text{L}}$, $0$ for $X_{\text{C}}$ and $20.0\text{-}\Omega$ for $R$ in equation(II) to calculate $Z$.

    $\begin{array}{c}Z=\sqrt{{20}^2+{\left(9.43-0\right)}^2}\\ =22.1\Omega \end{array}$

Substitute $22.1\Omega$ for $Z$ and $120\text{-V}$ for $V_{\text{rms}}$ in equation (I), to calculate $I_{\text{rms}}$.

    $\begin{array}{c}{I}_{\text{rms}}=\frac{120\text{-V}}{22.1\Omega }\\ =5.42\text{A}\end{array}$

Therefore, the rms current in the circuit is $\underset{\_}{5.42\text{A}}$.

(b)

To determine

The power factor in the circuit.

Answer to Problem 40P

The power factor of the circuit is $\underset{\_}{0.9}$.

Explanation of Solution

Write expression for power factor in the circuit.

    $\text{Power}\text{factor}=\cos \varphi$                                                                                              (VI)

Here, $\varphi$ is the phase angle.

Write expression for phase angle.

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

Conclusion:

Substitute $9.43\text{H}$ for $X_{\text{L}}$, $0$ for $X_{\text{C}}$ and $20.0\text{-}\Omega$ for $R$ in equation (VII), to calculate $\varphi$.

    $\begin{array}{c}\varphi ={\tan }^{-1}\frac{9.43\text{H}-0}{20.0\text{-}\Omega }\\ =25.24°\end{array}$

Substitute $25.24°$ for $\varphi$ in equation (VI), to calculate power factor.

    $\begin{array}{c}\text{Power}\text{factor}=\cos \left(25.24°\right)\\ =0.9\end{array}$

Therefore, the power factor of the circuit is $\underset{\_}{0.9}$.

(c)

To determine

The capacitor that must be added in series to make the power factor equal to 1.

Answer to Problem 40P

The capacitor that must be added in series to make the power factor equal to 1 is $\underset{\_}{2.81\times {10}^{\text{-4}}\text{F}}$.

Explanation of Solution

To make the power factor equals to $1$ $X_{\text{C}}$ and $X_{\text{L}}$ should be same.

Substitute $9.43\Omega$ for $X_{\text{C}}$ and $377\text{rad}/\text{s}$ for $\omega$ in equation (III) to calculate $C$.

    $\begin{array}{c}9.43\Omega =\frac{1}{\left(377\text{rad}/\text{s}\right)C}\\ C=2.81\times {10}^{\text{-4}}\text{F}\end{array}$

Therefore, the capacitor that must be added in series to make the power factor equal to 1 is $\underset{\_}{2.81\times {10}^{\text{-4}}\text{F}}$.

(d)

To determine

The reduction in the supply voltage.

Answer to Problem 40P

The reduction in the supply voltage is $\underset{\_}{108.2\text{V}}$.

Explanation of Solution

Write expression for average power in terms for rms volage.

    $P_{\text{avg}}=\frac{\Delta V_{\text{rms}}{}^2}{R}$                                                                                                      (VIII)

Here, $P_{\text{avg}}$ is the average power delivered and $\Delta V_{\text{rms}}$ is the rms voltage.

Write expression for power in terms of power factor.

    $P_{\text{avg}}=I_{\text{rms}}{V}_{\text{rms}}\cos \varphi$                                                                                                (IX)

Conclusion:

Equate equation (VIII) and equation (IX).

    $\frac{\Delta V_{\text{rms}}{}^2}{R}=I_{\text{rms}}{V}_{\text{rms}}\cos \varphi$

Substitute $5.42\text{A}$ for $I_{\text{rms}}$, $120\text{-V}$ for $V_{\text{rms}}$, $0.9$ for $\cos \varphi$ and  $20.0\text{-}\Omega$ for $R$ in above equation, to calculate $\Delta V_{\text{rms}}$.

    $\begin{array}{c}\frac{\Delta V_{\text{rms}}{}^2}{20.0\text{-}\Omega }=\left(5.42\text{A}\right)\left(120\text{-V}\right)\left(0.9\right)\\ \Delta V_{\text{rms}}{}^2=11707.2\\ \Delta V_{\text{rms}}=108.2\text{V}\end{array}$

Therefore, The reduction in the supply voltage if the power remains same is $\underset{\_}{108.2\text{V}}$.