Chapter 33, Problem 66PQ

(a)

To determine

The phasor diagram if the capacitance is $13.7\mu \text{F}$.

Answer to Problem 66PQ

The phasor diagram of the circuit is shown below.

Explanation of Solution

Write the expression for the impedance.

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

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

Write the expression to calculate the inductive reactance.

    $X_{\text{L}}=\omega L$

Here, $\omega$ is the angular frequency and $L$ is the inductor.

Write the expression for the capacitive reactance.

  $X_{\text{C}}=\frac{1}{\omega C}$

Here, $C$ is the capacitor.

Substitute the $\omega L$ for $X_L$ and $\omega C$ for $X_C$ in the equation (I) to calculate $Z$.

    $Z=\sqrt{\left({\left(\omega L\right)}^2-{\left(\frac{1}{\omega C}\right)}^2\right)+R^2}$

Write the expression for maximum current in the circuit.

    $I_{\text{max}}=\frac{{\epsilon }_{\text{max}}}{Z}$                                                                     (II)

Here, ${\epsilon }_{\text{max}}$ is the maximum voltage.

Write the expression for phase angle.

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

Conclusion:

Substitute $377\text{rad/s}$ for $\omega$, $126\text{mH}$ for $L$, $13.7\mu \text{F}$ for $C$ and $325\Omega$ for $R$ in the above equation to calculate $Z$.

    $\begin{array}{c}Z=\sqrt{\begin{array}{l}\left({\left(377\text{rad/s}\times \left(126\text{mH}\times \frac{{10}^{-3}\text{H}}{\text{mH}}\right)\right)}^2-{\left(\frac{1}{377\text{rad/s}\times \left(13.7\mu \text{F}\times \frac{{10}^{-6}\text{F}}{\mu \text{F}}\right)}\right)}^2\right)\\ +{\left(325\Omega \right)}^2\end{array}}\\ =\sqrt{\left({\left(47.502\Omega \right)}^2-{\left(193.6\Omega \right)}^2\right)+{\left(325\Omega \right)}^2}\\ =356.33\Omega \end{array}$

Substitute $18.8\text{V}$ for ${\epsilon }_{\text{max}}$ and $356.33\Omega$ for $Z$ in equation (II) to calculate $I_{\text{max}}$.

    $\begin{array}{c}{I}_{\text{max}}=\frac{18.8\text{ V}}{356.33\Omega }\\ =0.0528\text{A}\end{array}$

Substitute $47.502\Omega$ for $X_{\text{L}}$, $325\Omega$ for $R$ and $193.6\Omega$ for $X_{\text{C}}$ in equation (III) to calculate $\varphi$.

    $\begin{array}{c}\tan \varphi =\frac{\left(47.502\Omega -193.6\Omega \right)}{325\Omega }\\ =\left(-0.449\right)\\ \varphi ={\tan }^{-1}\left(-0.449\right)\\ =-24.18°\end{array}$

The phase angle is $-24.18°$ if the capacitance is $13.7\mu \text{F}$.

The inductive reactance is less than the capacitive reactance. Hence, the phasor angle is negative.

The phasor diagram of the circuit is shown below.

Figure (1)

(b)

To determine

The phasor diagram if the capacitance is $137\text{mF}$.

Answer to Problem 66PQ

The phasor diagram of the circuit is shown below.

Explanation of Solution

Write the expression for the impedance.

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

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

Write the expression to calculate the inductive reactance.

  $X_{\text{L}}=\omega L$

Here, $\omega$ is the angular frequency and $L$ is the inductor.

Write the expression for the capacitive reactance.

  $X_{\text{C}}=\frac{1}{\omega C}$

Here, $C$ is the capacitor.

Substitute the $\omega L$ for $X_L$ and $\omega C$ for $X_C$ in the equation (IV) to calculate $Z$.

  $Z=\sqrt{\left({\left(\omega L\right)}^2-{\left(\frac{1}{\omega C}\right)}^2\right)+R^2}$

Write the expression for maximum current in the circuit.

    $I_{\text{max}}=\frac{{\epsilon }_{\text{max}}}{Z}$                                                                            (V)

Here, ${\epsilon }_{\text{max}}$ is the maximum voltage.

Write the expression for phase angle.

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

Conclusion:

Substitute $377\text{rad/s}$ for $\omega$, $126\text{mH}$ for $L$, $13.7\text{mF}$ for $C$ and $325\Omega$ for $R$ in the above equation to calculate $Z$.

    $\begin{array}{c}Z=\sqrt{\begin{array}{l}\left({\left(377\text{rad/s}\times \left(126\text{mH}\times \frac{{10}^{-3}\text{H}}{\text{mH}}\right)\right)}^2-{\left(\frac{1}{377\text{rad/s}\times \left(13.7\text{mF}\times \frac{{10}^{-3}\text{F}}{\text{mF}}\right)}\right)}^2\right)\\ +{\left(325\Omega \right)}^2\end{array}}\\ =\sqrt{\left({\left(47.502\Omega \right)}^2-{\left(0.1936\Omega \right)}^2\right)+{\left(325\Omega \right)}^2}\\ =328\Omega \end{array}$

Substitute $18.8\text{V}$ for ${\epsilon }_{\text{max}}$ and $328\Omega$ for $Z$ in equation (V) to calculate $I_{\text{max}}$.

    $\begin{array}{c}{I}_{\text{max}}=\frac{18.8\text{V}}{328\Omega }\\ =0.0573\text{A}\end{array}$

Substitute $47.502\Omega$ for $X_{\text{L}}$, $325\Omega$ for $R$ and $0.1936\Omega$ for $X_{\text{C}}$ in equation (VI) to calculate $\varphi$.

    $\begin{array}{c}\tan \varphi =\frac{\left(47.502\Omega -0.1936\Omega \right)}{325\Omega }\\ \tan \varphi =\left(0.1455\right)\\ \varphi ={\tan }^{-1}\left(0.1455\right)\\ =8.28°\end{array}$

The phase angle is $8.28°$ if the capacitance is $13.7\text{mF}$.

The inductive reactance is less than the capacitive reactance. Hence, the phasor angle is negative.

The phasor diagram of the circuit is shown below.