Chapter 21, Problem 92P

(a)

To determine

The impedance of the inductor in the circuit.

Answer to Problem 92P

The impedance of the inductor in the circuit is $33.8\Omega$.

Explanation of Solution

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

  $Z=\sqrt{R^2+X_L^2}$                                                                                  (I)

Here, $Z$ is the impedance, $R$ is the resistance, and $X_L$ is the inductive reactance.

Write the expression for inductive reactance.

  $X_L=\omega L$                                                                                           (II)

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

Conclusion:

Substitute equation (II) in the equation (I).

  $Z=\sqrt{R^2+{\omega }^2{L}^2}$                                                                                (III)

Write the equation for the induced emf of the ac source.

  $\epsilon \left(t\right)=\left(286\text{V}\right)\sin \left[\left(390\text{rad/s}\right)t\right]$

Here, the angular frequency of the ac mains is $390\text{rad/s}$.

Substitute $390\text{rad/s}$ for $\omega$, $30.0\Omega$ for $R$, and $40.0\text{mH}$ for $L$ in equation (III) to find $Z$.

  $\begin{array}{c}Z=\sqrt{{\left(30.0\Omega \right)}^2+{\left(390\text{rad/s}\right)}^2{\left[\left(40.0\text{mH}\right)\left(\frac{{10}^{-3}\text{H}}{1\text{mH}}\right)\right]}^2}\\ =33.8\Omega \end{array}$

Therefore, the impedance of the inductor in the circuit is $33.8\Omega$.

(b)

To determine

The peak and rms voltage across the inductor.

Answer to Problem 92P

The peak voltage across the inductor is $286\text{V}$ and rms voltage across the inductor is $202\text{V}$.

Explanation of Solution

The peak voltage across the inductor including its internal resistance is equal to the peak source voltage,

  $V_{\text{L}}={\epsilon }_{\text{m}}$                                                                                         (IV)

Here, $V_{\text{L}}$ is the voltage across the inductor and ${\epsilon }_{\text{m}}$ is the peak source voltage.

Write the expression for rms voltage across the inductor.

  $V_{\text{rms}}=\frac{V_{\text{L}}}{\sqrt{2}}$                                                                                       (V)

Here, $V_{\text{rms}}$ is the rms voltage.

Conclusion:

Substitute $286\text{V}$ for ${\epsilon }_{\text{m}}$ in equation (IV) to find $V_{\text{L}}$.

  $V_{\text{L}}=286\text{V}$

Substitute $286\text{V}$ for $V_{\text{L}}$ in the equation (V) to find $V_{\text{rms}}$

  $\begin{array}{c}{V}_{\text{rms}}=\frac{286\text{V}}{\sqrt{2}}\\ =202.3\text{V}\\ =202\text{V}\end{array}$

Therefore, the peak voltage across the inductor is $286\text{V}$ and rms voltage across the inductor is $202\text{V}$.

(c)

To determine

The peak current in the circuit.

Answer to Problem 92P

The peak current in the circuit is $8.46\text{A}$.

Explanation of Solution

Write the expression for peak current.

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

Here, $I$ is the peak current.

Conclusion:

Substitute $286\text{V}$ for ${\epsilon }_{\text{m}}$ and $33.8\Omega$ for $Z$ in equation (VI) to find $I$.

  $\begin{array}{c}I=\frac{286\text{V}}{33.8\Omega }\\ =8.46\text{A}\end{array}$

Therefore, the peak current in the circuit is $8.46\text{A}$.

(d)

To determine

The average power dissipated in the circuit.

Answer to Problem 92P

The average power dissipated in the circuit is $1.07\text{kW}$.

Explanation of Solution

Write the expression average power dissipated in the circuit.

  $P_{\text{avg}}=I_{\text{rms}}{\epsilon }_{\text{rms}}\cos \varphi$                                                                             (VII)

Here, $P_{\text{avg}}$ is the average power dissipated in the circuit, $I_{\text{rms}}$ is the rms current, $\varphi$ is the phase angle, and ${\epsilon }_{\text{rms}}$ is the rms voltage.

Write the expression for the phase angle.

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

Write the expression for the rms current.

  $I_{\text{rms}}=\frac{I}{\sqrt{2}}$                                                                                                (IX)

Conclusion:

Substitute the equation (VIII) and (IX) in the equation (VII).

  $P_{\text{avg}}=\frac{I}{\sqrt{2}}{\epsilon }_{\text{rms}}\frac{R}{Z}$

Substitute $8.46\text{A}$ for $I$, $202\text{V}$ for ${\epsilon }_{\text{rms}}$, $30.0\Omega$ for $R$, and $33.8\Omega$ for $Z$ to find $P_{\text{avg}}$.

  $\begin{array}{c}{P}_{\text{avg}}=\frac{8.46\text{A}}{\sqrt{2}}\left(202\text{V}\right)\frac{30.0\Omega }{33.8\Omega }\\ =1072.8\text{W}\\ =1.07\times {10}^3\text{W}\left(\frac{{10}^{-3}\text{kW}}{1\text{W}}\right)\\ =1.07\text{kW}\end{array}$

The average power dissipated in the circuit is $1.07\text{kW}$.

(e)

To determine

Derive the expression for the current through the inductor.

Answer to Problem 92P

The expression for the current through the inductor is

$i\left(t\right)=\left(8.46\text{A}\right)\sin \left[\left(390\text{rad/s}\right)t-0.480\text{rad}\right]$.

Explanation of Solution

Write the expression for current through the inductor lags the voltage across it.

  $i\left(t\right)=I_{\text{peak}}\sin \left(\omega t-\varphi \right)$ (X)

Rewrite the expression for the phase angle from part (d) in the equation (VIII).

  $\varphi ={\cos }^{-1}\left(\frac{R}{Z}\right)$ (XI)

Conclusion:

Substitute $30.0\Omega$ for $R$, and $33.8\Omega$ for $Z$ to find $\varphi$ in equation (XI).

  $\begin{array}{c}\varphi ={\cos }^{-1}\left(\frac{30.0\Omega }{33.8\Omega }\right)\\ =0.480\text{rad}\end{array}$

Substitute $8.46\text{A}$ for $I$, $390\text{rad/s}$ for $\omega$, and $0.480\text{rad}$ for $\varphi$ in equation (X) to find $i\left(t\right)$.

  $i\left(t\right)=\left(8.46\text{A}\right)\sin \left[\left(390\text{rad/s}\right)t-0.480\text{rad}\right]$

Therefore, the expression for the current through the inductor is

$i\left(t\right)=\left(8.46\text{A}\right)\sin \left[\left(390\text{rad/s}\right)t-0.480\text{rad}\right]$.