Chapter 33, Problem 66AP

(a)

To determine

The value of $R$.

Answer to Problem 66AP

The value of $R$ is $99.3\Omega$.

Explanation of Solution

Write the expression for the rms voltage when the switch $S$ is open.

    $\Delta V_{\text{rms}}=I_{\text{rms}}{}^{\prime }{Z}_1$                                                                                                            (I)

Here, $\Delta V_{\text{rms}}$ is the rms voltage, $I_{\text{rms}}{}^{\prime }$ is the rms value of current when switch $S$ is open and $Z$ is the impedance when the switch $S$ is open.

Write the expression for impedance when switch $S$ is open.

    $Z_1=\sqrt{\left(R^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2\right)}$                                                                                         (II)

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

Substitute $\sqrt{\left(R^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2\right)}$ for $Z_1$ in equation (I).

    $\Delta V_{\text{rms}}=I_{\text{rms}}{}^{\prime }\left(\sqrt{\left(R^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2\right)}\right)$                                                                        (III)

Write the expression for current when the switch is at position $a$.

    $\Delta V_{\text{rms}}=I_{{\text{rms}}_{\text{a}}}{Z}_{\text{a}}$                                                                                                         (IV)

Here, $I_{{\text{rms}}_{\text{a}}}$ is the current when the switch is at position $a$ and $Z_{\text{a}}$ is the impedance when the switch is at position $a$.

Write the expression for impedance when the switch is at position $a$.

    $Z_{\text{a}}=\sqrt{\left({\left(\frac{R}{2}\right)}^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2\right)}$

Substitute $\sqrt{\left({\left(\frac{R}{2}\right)}^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2\right)}$ for $Z_{\text{a}}$ in equation (IV).

    $\Delta V_{\text{rms}}=I_{{\text{rms}}_{\text{a}}}\left(\sqrt{\left({\left(\frac{R}{2}\right)}^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2\right)}\right)$                                                                   (V)

Conclusion:

Substitute $20.0\text{V}$ for $\Delta V_{\text{rms}}$ and $183\text{mA}$ for $I_{\text{rms}}{}^{\prime }$ in equation (III).

    $\begin{array}{c}20.0\text{V}=\left(183\text{mA}\times \frac{{10}^{-3}\text{A}}{1\text{mA}}\right)\left(\sqrt{\left(R^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2\right)}\right)\\ {\left(\sqrt{\left(R^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2\right)}\right)}^2={\left(\frac{20.0\text{V}}{\left(183\text{mA}\times \frac{{10}^{-3}\text{A}}{1\text{mA}}\right)}\right)}^2\\ R^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2=11.9\times {10}^3\end{array}$                  (VI)

Substitute $20.0\text{V}$ for $\Delta V_{\text{rms}}$ and $298\text{mA}$ for $I_{{\text{rms}}_{\text{a}}}$ in equation (V).

    $\begin{array}{c}20.0\text{V}=\left(298\text{mA}\times \frac{{10}^{-3}\text{A}}{1\text{mA}}\right)\left(\sqrt{\left({\left(\frac{R}{2}\right)}^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2\right)}\right)\\ {\left(\sqrt{\left({\left(\frac{R}{2}\right)}^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2\right)}\right)}^2={\left(\frac{20.0\text{V}}{\left(298\text{mA}\times \frac{{10}^{-3}\text{A}}{1\text{mA}}\right)}\right)}^2\\ {\left(\frac{R}{2}\right)}^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2=4.5\times {10}^3\end{array}$      (VII)

Subtract equation (VI) by (VII) to solve for $R$.

    $\begin{array}{c}{R}^2-{\left(\frac{R}{2}\right)}^2=\left(\left(11.9\times {10}^3\right)-\left(4.5\times {10}^3\right)\right)\Omega \\ 3R^2=4\left(7.4\times {10}^3\right)\Omega \\ R=\sqrt{\frac{29.6\times {10}^3}{3}}\Omega \\ R=99.3\Omega \end{array}$

Therefore, the value of $R$ is $99.3\Omega$

(b)

To determine

The value of $C$.

Answer to Problem 66AP

The value of $C$ is $24.7\mu \text{F}$.

Explanation of Solution

Write the expression for current when the switch is at position $b$.

    $\Delta V_{\text{rms}}=I_{{\text{rms}}_{\text{b}}}{Z}_{\text{b}}$                                                                                                    (VIII)

Here, $I_{{\text{rms}}_{\text{b}}}$ is the current when the switch is at position $b$ and $Z_{\text{b}}$ is the impedance when the switch is at position $b$.

Write the expression for impedance when the switch is at position $b$.

    $Z_{\text{a}}=\sqrt{\left(R^2+X_{\text{C}}{}^2\right)}$

Substitute $\sqrt{\left(R^2+X_{\text{C}}{}^2\right)}$ for $Z_{\text{b}}$ in equation (VIII).

    $\Delta V_{\text{rms}}=I_{{\text{rms}}_{\text{b}}}\left(\sqrt{\left(R^2+X_{\text{C}}{}^2\right)}\right)$                                                                                   (IX)

Write the expression to calculate the capacitance.

    $C=\frac{1}{2\pi f{X}_{\text{C}}}$                                                                                                             (X)

Here, $C$ is the capacitance and $f$ is the frequency.

Conclusion:

Substitute $20.0\text{V}$ for $\Delta V_{\text{rms}}$, $99.3\Omega$ and $137\text{mA}$ for $I_{{\text{rms}}_{\text{b}}}$ in equation (IX) to solve for $X_{\text{C}}$.

    $\begin{array}{c}20.0\text{V}=\left(137\text{mA}\times \frac{{10}^{-3}\text{A}}{1\text{mA}}\right)\left(\sqrt{\left({\left(99.3\Omega \right)}^2+{\left(X_{\text{C}}\right)}^2\right)}\right)\\ {\left(\sqrt{\left({\left(99.3\Omega \right)}^2+{\left(X_{\text{C}}\right)}^2\right)}\right)}^2={\left(\frac{20.0\text{V}}{\left(137\text{mA}\times \frac{{10}^{-3}\text{A}}{1\text{mA}}\right)}\right)}^2\\ {\left(X_{\text{C}}\right)}^2=\left(21.3\times {10}^3-9.8\times {10}^3\right){\Omega }^2\\ X_{\text{C}}=107.2\Omega \end{array}$

Substitute $107.2\Omega$ for $X_{\text{C}}$ and $60.0\text{Hz}$ for $f$ in equation (X) to solve for $C$.

    $\begin{array}{c}C=\frac{1}{2\pi \left(60.0\text{Hz}\right)\left(107.2\Omega \right)}\\ =2.47\times {10}^{-5}\text{F}\times \frac{{\text{10}}^6\mu \text{F}}{1\text{F}}\\ =24.7\mu \text{F}\end{array}$

Therefore, the value of $C$ is $24.7\mu \text{F}$.

(c).

To determine

The value of $L$.

Answer to Problem 66AP

The value of $L$ is $404\text{mH}$.

Explanation of Solution

Write the expression to calculate the inductance.

    $L=2\pi f{X}_{\text{L}}$                                                                                                              (XI)

Here, $L$ is the inductance.

Conclusion:

Substitute $99.3\Omega$ for $R$, $107.2\Omega$ for $X_{\text{C}}$ in equation (VI) to solve for $X_{\text{L}}$.

    $\begin{array}{c}{\left(99.3\Omega \right)}^2+{\left(X_{\text{L}}-107.2\Omega \right)}^2=11.9\times {10}^3\\ {\left(X_{\text{L}}-107.2\Omega \right)}^2=11.9\times {10}^3-{\left(99.3\Omega \right)}^2\\ X_{\text{L}}-107.2\Omega =\sqrt{2039.51}\Omega \\ X_{\text{L}}=152.4\Omega \end{array}$

Substitute $152.4\Omega$ for $X_{\text{L}}$ and $60.0\text{Hz}$ for $f$ in equation (XI) to solve for $L$.

    $\begin{array}{c}\left(152.4\Omega \right)=2\pi \left(60.0\text{Hz}\right)L\\ L=\frac{\left(152.4\Omega \right)}{2\pi \left(60.0\text{Hz}\right)}\\ =0.404\text{H}\times \frac{{10}^3\text{mH}}{1\text{H}}\\ =404\text{mH}\end{array}$

Therefore, the value of $L$ is $404\text{mH}$.

(d).

To determine

Whether more than one set of values are possible.

Answer to Problem 66AP

The other possible value of $L$ is $164\text{mH}$ and no other possible values for $R$ and $C$.

Explanation of Solution

Conclusion:

Substitute $99.3\Omega$ for $R$, $107.2\Omega$ for $X_{\text{C}}$ in equation (VI) to solve for $X_{\text{L}}$.

    $\begin{array}{c}{\left(99.3\Omega \right)}^2+{\left(X_{\text{L}}-107.2\Omega \right)}^2=11.9\times {10}^3\\ {\left(X_{\text{L}}-107.2\Omega \right)}^2=11.9\times {10}^3-{\left(99.3\Omega \right)}^2\\ X_{\text{L}}-107.2\Omega =\sqrt{2039.51}\Omega \end{array}$

Take the negative root.

    $\begin{array}{c}{X}_{\text{L}}=\sqrt{2039.51}\Omega +107.2\Omega \\ =\left(-45.2+107.2\right)\Omega \\ =62\Omega \end{array}$

Substitute $62\Omega$ for $X_{\text{L}}$ and $60.0\text{Hz}$ for $f$ in equation (XI) to solve for $L$.

    $\begin{array}{c}\left(152.4\Omega \right)=2\pi \left(60.0\text{Hz}\right)L\\ L=\frac{\left(62\Omega \right)}{2\pi \left(60.0\text{Hz}\right)}\\ =0.164\text{H}\times \frac{{10}^3\text{mH}}{1\text{H}}\\ =164\text{mH}\end{array}$

Therefore, the other possible value of $L$ is $164\text{mH}$ and there are no other possible values for $R$ and $C$.