Chapter 33, Problem 24P

(a)

To determine

The maximum voltage between points $a$ and $b$.

Answer to Problem 24P

The maximum voltage between points $a$ and $b$ is $146.40\text{V}$.

Explanation of Solution

Consider the phasor diagram as shown in the figure 1.

Figure (1)

Write the expression for the maximum current in the circuit.

  $I_{\max }=\frac{\Delta V_{\max }}{\sqrt{R^2+{\left(2\pi fL-\frac{1}{2\pi fC}\right)}^2}}$                                                                             (I)

Here, $I_{\max }$ is the maximum current in the circuit, $\Delta V_{\max }$ is the maximum potential difference, $R$ is the resistance of resistor, $L$ is the inductance of inductor, $f$ is the frequency and $C$ is the capacitance of capacitor.

Write the expression for the maximum voltage between points $a$ and $b$.

  ${\left(\Delta V_{\text{ab}}\right)}_{\max }=I_{\max }R$                                                                                                   (II)

Here, ${\left(\Delta V_{\text{ab}}\right)}_{\max }$ is the maximum voltage across point $a$ and $b$.

Conclusion:

Substitute $150V$ for $\Delta V_{\max }$, $40.0\Omega$ for $R$, $50.0\text{Hz}$ for $f$, $185\text{mH}$ for $L$ and $65.0\mu \text{F}$ for $C$ in equation (I) to find $I_{\max }$.

  $\begin{array}{c}{I}_{\max }=\frac{150\text{V}}{\sqrt{{\left(40.0\Omega \right)}^2+{\left[2\pi \left(50.0\text{Hz}\right)\left(185\text{mH}\right)-\frac{1}{2\pi \left(50.0\text{Hz}\right)\left(185\text{mH}\right)\left(65.0\mu \text{F}\right)}\right]}^2}}\\ =\frac{150\text{V}}{\sqrt{{\left(40.0\Omega \right)}^2+{\left[2\pi \left(50.0\text{Hz}\right)\left(185\times {10}^{-3}\text{H}\right)-\frac{1}{2\pi \left(50.0\text{Hz}\right)\left(65.0\times {10}^{-6}\text{F}\right)}\right]}^2}}\\ =\frac{150}{41.03}\\ =3.66\text{A}\end{array}$

Substitute $3.66\text{A}$ for $I_{\max }$ and $40.0\Omega$ for $R$ in equation (II) to find ${\left(\Delta V_{\text{ab}}\right)}_{\max }$.

  $\begin{array}{c}{\left(\Delta V_{\text{ab}}\right)}_{\max }=\left(3.66\text{A}\right)\left(40.0\Omega \right)\\ =146.40\text{V}\end{array}$

Therefore, the maximum voltage between points $a$ and $b$ is $146.40\text{V}$.

(b)

To determine

The maximum voltage between points $b$ and $c$.

Answer to Problem 24P

The maximum voltage between points $b$ and $c$ is $212.72\text{V}$.

Explanation of Solution

Write the expression for the maximum voltage between points $b$ and $c$.

  $\begin{array}{l}{\left(\Delta V_{\text{bc}}\right)}_{\max }=I_{\max }{X}_{\text{L}}\\ {\left(\Delta V_{\text{bc}}\right)}_{\max }=I_{\max }\left(2\pi fL\right)\end{array}$                                                                                     (III)

Here, ${\left(\Delta V_{\text{bc}}\right)}_{\max }$ is the maximum voltage across point $b$ and $c$.

Conclusion:

Substitute $3.66\text{A}$ for $I_{\max }$, $50.0\text{Hz}$ for $f$ and $185\text{mH}$ for $L$ in equation (III) to find ${\left(\Delta V_{\text{bc}}\right)}_{\max }$.

  $\begin{array}{c}{\left(\Delta V_{\text{bc}}\right)}_{\max }=\left(3.66\text{A}\right)2\pi \left(50.0\text{Hz}\right)\left(185\text{mH}\right)\\ =\left(3.66\text{A}\right)2\pi \left(50.0\text{Hz}\right)\left(185\times {10}^{-3}\text{H}\right)\\ =212.72\text{V}\end{array}$

Therefore, the maximum voltage between points $b$ and $c$ is $212.72\text{V}$.

(c)

To determine

The maximum voltage between points $c$ and $d$.

Answer to Problem 24P

The maximum voltage between points $c$ and $d$ is $179.23\text{V}$.

Explanation of Solution

Write the expression for the maximum voltage between points $c$ and $d$.

  $\begin{array}{l}{\left(\Delta V_{\text{cd}}\right)}_{\max }=I_{\max }{X}_{\text{C}}\\ {\left(\Delta V_{\text{cd}}\right)}_{\max }=I_{\max }\left(\frac{1}{2\pi fC}\right)\end{array}$                                                                                     (IV)

Here, ${\left(\Delta V_{\text{cd}}\right)}_{\max }$ is the maximum voltage across point $c$ and $d$.

Conclusion:

Substitute $3.66\text{A}$ for $I_{\max }$, $50.0\text{Hz}$ for $f$ and $65.0\mu \text{F}$ for $C$ in equation (IV) to find ${\left(\Delta V_{\text{cd}}\right)}_{\max }$.

  $\begin{array}{c}{\left(\Delta V_{\text{cd}}\right)}_{\max }=\left(3.66\text{A}\right)\frac{1}{2\pi \left(50.0\text{Hz}\right)\left(65.0\mu \text{F}\right)}\\ =\left(3.66\text{A}\right)\frac{1}{2\pi \left(50.0\text{Hz}\right)\left(65.0\times {10}^{-6}\text{F}\right)}\\ =179.23\text{V}\end{array}$

Therefore, the maximum voltage between points $b$ and $c$ is $179.23\text{V}$.

(d)

To determine

The maximum voltage between points $b$ and $d$.

Answer to Problem 24P

The maximum voltage between points $b$ and $d$ is $33.49\text{V}$.

Explanation of Solution

Write the expression for the maximum voltage between points $b$ and $d$.

  ${\left(\Delta V_{\text{bd}}\right)}_{\max }={\left(\Delta V_{\text{bc}}\right)}_{\max }-{\left(\Delta V_{\text{cd}}\right)}_{\max }$                                                                           (V)

Here, ${\left(\Delta V_{\text{bd}}\right)}_{\max }$ is the maximum voltage across point $b$ and $d$.

Conclusion:

Substitute $212.72\text{V}$ for ${\left(\Delta V_{\text{bc}}\right)}_{\max }$ and $179.23\text{V}$ for ${\left(\Delta V_{\text{cd}}\right)}_{\max }$ in equation (V) to find ${\left(\Delta V_{\text{bd}}\right)}_{\max }$.

  $\begin{array}{c}{\left(\Delta V_{\text{bd}}\right)}_{\max }=\left(212.72\text{V}\right)-\left(179.23\text{V}\right)\\ =33.49\text{V}\end{array}$

Therefore, the maximum voltage between points $b$ and $c$ is $33.49\text{V}$.