Chapter 21, Problem 93P

(a)

To determine

The reactance of the capacitor and the inductor.

Answer to Problem 93P

The reactance of the capacitor is $20\Omega$ and the reactance of the inductor is $35\Omega$.

Explanation of Solution

Write an expression for the capacitive reactance.

    $X_C=\frac{1}{\omega C}$                                                                                                      (I)

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

Write an expression for the inductive reactance.

    $X_L=\omega L$                                                                                                    (II)

Here, $X_L$ is the inductive reactance and $L$ is the inductance.

Conclusion:

Substitute $1.0\times {10}^3\text{rad/s}$ for $\omega$ and $50.0\text{μF}$ for $C$ in equation (I) to find $X_C$.

    $\begin{array}{c}{X}_C=\frac{1}{\left(1.0\times {10}^3\text{rad/s}\right)\left(50.0\text{μF}\right)}\\ =\frac{1}{\left(1.0\times {10}^3\text{rad/s}\right)\left(\left(50.0\text{μF}\right)\left(\frac{{10}^{-6}\text{F}}{1\text{μF}}\right)\right)}\\ =20\Omega \end{array}$

Substitute $1.0\times {10}^3\text{rad/s}$ for $\omega$ and $35.0\text{mH}$ for $L$ in equation (II) to find $X_L$.

    $\begin{array}{c}{X}_C=\frac{1}{\left(1.0\times {10}^3\text{rad/s}\right)\left(35.0\text{mH}\right)}\\ =\frac{1}{\left(1.0\times {10}^3\text{rad/s}\right)\left(\left(35.0\text{mH}\right)\left(\frac{{10}^{-3}\text{H}}{1\text{mH}}\right)\right)}\\ =35\Omega \end{array}$

Thus, the reactance of the capacitor is $20\Omega$ and the reactance of the inductor is $35\Omega$.

(b)

To determine

The impedance.

Answer to Problem 93P

The impedance is $25\Omega$.

Explanation of Solution

Write an expression for the impedance.

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

Here, $Z$ is the impedance and $R$ is the resistance.

Conclusion:

Substitute $20.0\Omega$ for $R$, $35\Omega$ for $X_L$ and $20\Omega$ for $X_C$ in equation (III) to find $Z$.

    $\begin{array}{c}Z=\sqrt{{\left(20.0\Omega \right)}^2+{\left(35\Omega -20\Omega \right)}^2}\\ =\sqrt{{\left(20.0\Omega \right)}^2+{\left(15\Omega \right)}^2}\\ =25\Omega \end{array}$

Thus, the impedance is $25\Omega$.

(c)

To determine

The rms current.

Answer to Problem 93P

The rms current is $4.0\text{A}$.

Explanation of Solution

Write an expression for the rms current.

    $I_{rms}=\frac{{\epsilon }_{rms}}{Z}$                                                                                      (IV)

Here, $I_{rms}$ is the rms current and ${\epsilon }_{rms}$ is the rms voltage.

Conclusion:

Substitute $100\text{V}$ for ${\epsilon }_{rms}$ and $25\Omega$ for $Z$ in equation (IV) to find $I_{rms}$.

    $\begin{array}{c}{I}_{rms}=\frac{100\text{V}}{25\Omega }\\ =4.0\text{A}\end{array}$

Thus, the rms current is $4.0\text{A}$.

(d)

To determine

The current amplitude.

Answer to Problem 93P

The current amplitude is $5.7\text{A}$.

Explanation of Solution

Write an expression for the current amplitude.

    $I=\sqrt{2}{I}_{rms}$                                                                                                 (V)

Here, $I$ is the current amplitude.

Conclusion:

Substitute $4.0\text{A}$ for $I_{rms}$ in equation (V) to find $I$.

    $\begin{array}{c}I=\sqrt{2}\left(4.0\text{A}\right)\\ =5.7\text{A}\end{array}$

Thus, the current amplitude is $5.7\text{A}$.

(e)

To determine

The phase angle.

Answer to Problem 93P

The phase angle is $37°$.

Explanation of Solution

Write an expression for the phase angle.

    $\begin{array}{c}\cos \varphi =\frac{R}{Z}\\ \varphi ={\cos }^{-1}\left(\frac{R}{Z}\right)\end{array}$                                                                                     (VI)

Here, $\varphi$ is the phase angle.

Conclusion:

Substitute $20.0\Omega$ for $R$ and $25\Omega$ for $Z$ in equation (VI) to find $\varphi$.

    $\begin{array}{c}\varphi ={\cos }^{-1}\left(\frac{20.0\Omega }{25\Omega }\right)\\ ={\cos }^{-1}\left(0.8\right)\\ =37°\end{array}$

Thus, the phase angle is $37°$.

(f)

To determine

The rms voltage across each of the circuit elements.

Answer to Problem 93P

The rms voltage across the resistance is $80\text{V}$, the rms voltage across the inductor is $140\text{V}$, and the rms voltage across the capacitor is $80\text{V}$.

Explanation of Solution

Write an expression for the rms voltage across the resistance.

    $V_{R,rms}=I_{rms}R$                                                                                     (VII)

Here, $V_{R,rms}$ is the rms voltage across the resistance.

Write an expression for the rms voltage across the inductor.

    $V_{L,rms}=I_{rms}{X}_L$                                                                                      (VIII)

Here, $V_{L,rms}$ is the rms voltage across the inductor.

Write an expression for the rms voltage across the capacitor.

    $V_{C,rms}=I_{rms}{X}_C$                                                                                     (IX)

Here, $V_{C,rms}$ is the rms voltage across the capacitor.

Conclusion:

Substitute $4.0\text{A}$ for $I_{rms}$ and $20.0\Omega$ for $R$ in equation (VII) to find $V_{R,rms}$.

    $\begin{array}{c}{V}_{R,rms}=\left(4.0\text{A}\right)\left(20.0\Omega \right)\\ =80\text{V}\end{array}$

Substitute $4.0\text{A}$ for $I_{rms}$ and $35\Omega$ for $X_L$ in equation (VIII) to find $V_{L,rms}$.

    $\begin{array}{c}{V}_{L,rms}=\left(4.0\text{A}\right)\left(35\Omega \right)\\ =140\text{V}\end{array}$

Substitute $4.0\text{A}$ for $I_{rms}$ and $20\Omega$ for $X_C$ in equation (IX) to find $V_{C,rms}$.

    $\begin{array}{c}{V}_{C,rms}=\left(4.0\text{A}\right)\left(20\Omega \right)\\ =80\text{V}\end{array}$

Thus, the rms voltage across the resistance is $80\text{V}$, the rms voltage across the inductor is $140\text{V}$, and the rms voltage across the capacitor is $80\text{V}$.

(g)

To determine

If the current leads or lags the voltage.

Answer to Problem 93P

The current lags the voltage.

Explanation of Solution

From the values of the capacitive reactance and inductive reactance, the inductive reactance is greater than the capacitive reactance.

    $X_L>X_C$

Conclusion:

As the dominant element is the inductor, the current lags the voltage.

Thus, the current lags the voltage.

(h)

To determine

Sketch the phasor diagram.

Answer to Problem 93P

The phasor diagram is shown in figure 1.

Explanation of Solution

The voltages of the different circuit elements found in part (f) is used to draw the phasor diagram.

The rms voltage across the resistance $V_R$ is $80\text{V}$.

The rms voltage across the inductor $V_L$ is $140\text{V}$.

The rms voltage across the capacitor $V_C$ is $80\text{V}$.

  $\begin{array}{c}{V}_L-V_C=140\text{V}-80\text{V}\\ =\text{60}\text{V}\end{array}$

As the dominant element is the inductor, the current lags the voltage.

Conclusion:

The difference between the voltage across the inductor and the voltage across the capacitor is zero.

Sketch the phasor diagram.

Thus, the phasor diagram is shown in figure 1.