Chapter 33, Problem 33.66AP

A capacitor, a coil, and two resistors of equal resistance are arranged in an AC circuit as shown in Figure P33.66 (page 1028). An AC source provides an emf of ΔV_rms = 20.0 V at a frequency of 60.0 Hz. When the double throw switch S is open as shown in the figure, the rms current is 183 mA. When the switch is closed in position a, the rms current is 298 mA. When the switch is closed in position b, the rms current is 137 mA. Determine the values of (a) R, (b) C, and (c) L. (d) Is more than one set of values possible? Explain.

To determine

 The value of resistance.

Answer to Problem 33.66AP

 The value of resistance is $\text{99}\text{.6}\Omega$.

Explanation of Solution

Given info: The value of the source emf is $20.0\text{V}$ and frequency of the source emf is $60.0\text{Hz}$. The current through the circuit when switch is open is $183\text{mA}$, the current through the circuit when switch is at a position is $298\text{mA}$ and the current through the circuit when the switch is at b position is $137\text{mA}$.

The expression for inductive reactance is,

$X_{\text{L}}=2\pi fL$

Here,

$f$ is the source frequency.

$L$ is the inductance.

The expression for capacitive reactance is,

$X_{\text{C}}=\frac{1}{2\pi fC}$

Here,

$C$ is the capacitance.

The expression for the impedance of the circuit is.

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

Here,

$R_{\text{eq}}$ is the equivalent resistance of the circuit.

The expression of the impedance in terms of voltage and current is,

$Z=\frac{\Delta V_{\text{rms}}}{I_{\text{rms}}}$

Here,

$\Delta V_{\text{rms}}$ is the source voltage.

$I_{\text{rms}}$ is the current.

The figure below depicts the circuit when the switch is in open condition.

Figure (1)

From figure(1), for the throw switch in open condition:

Substitute $R$ for $R_{\text{eq}}$ and $\frac{\Delta V_{\text{rms}}}{I_{\text{rms}}}$ for $Z$ in the expression of the impedance of the circuit.

$\begin{array}{c}\frac{\Delta V_{\text{rms}}}{I_{\text{rms}}}=\sqrt{R^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2}\\ R^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2={\left(\frac{\Delta V_{\text{rms}}}{I_{\text{rms}}}\right)}^2\end{array}$

Substitute $20.0\text{V}$ for $\Delta V_{\text{rms}}$ and $183\text{mA}$ for $I_{\text{rms}}$ in the above expression.

$\begin{array}{c}{R}^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2={\left(\frac{20.0\text{V}}{183\text{mA}}\right)}^2\\ R^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2=11944.22{\Omega }^2\end{array}$ (1)

For the throw switch at a position:

The two resistances are in parallel.

The expression of the equivalent resistance is.

$\begin{array}{c}{R}_{\text{eq}}={\left(\frac{1}{R}+\frac{1}{R}\right)}^{-1}\\ =\frac{R}{2}\end{array}$

Substitute $\frac{R}{2}$ for $R_{\text{eq}}$ and $\frac{\Delta V_{\text{rms}}}{I_{\text{rms}}}$ for $Z$ in the expression of the impedance of the circuit.

$\begin{array}{c}\frac{\Delta V_{\text{rms}}}{I_{\text{rms}}}=\sqrt{{\frac{R}{4}}^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2}\\ {\frac{R}{4}}^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2={\left(\frac{\Delta V_{\text{rms}}}{I_{\text{rms}}}\right)}^2\end{array}$

Substitute $20.0\text{V}$ for $\Delta V_{\text{rms}}$ and $298\text{mA}$ for $I_{\text{rms}}$ in the above expression.

$\begin{array}{c}{\frac{R}{4}}^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2={\left(\frac{20.0\text{V}}{298\text{mA}}\right)}^2\\ {\frac{R}{4}}^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2=4504.30{\Omega }^2\end{array}$ (2)

Subtract equation (2) from equation (1).

$\begin{array}{c}\left(R^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2\right)-\left({\frac{R}{4}}^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2\right)=11944.22{\Omega }^2-4504.30{\Omega }^2\\ \frac{3}{4}{R}^2=7439.92{\Omega }^2\\ R=\text{99}\text{.6}\Omega \end{array}$

Conclusion:

Therefore, the value of resistance is $\text{99}\text{.6}\Omega$.

To determine

 The value of the capacitance.

Answer to Problem 33.66AP

 The value of the capacitance is $24.8\text{μF}$.

Explanation of Solution

Given info: The value of the source emf is $20.0\text{V}$ and frequency of the source emf is $60.0\text{Hz}$. The current through the circuit when switch is open is $183\text{mA}$, the current through the circuit when switch is at a position is $298\text{mA}$ and the current through the circuit when the switch is at b position is $137\text{mA}$.

The figure below depicts the circuit when switch is at position b.

Figure (2)

From figure (2), for the throw switch at b position:

When the switch is at position b the inductor gets short circuited.

Substitute $R$ for $R_{\text{eq}}$, $0$ for $X_{\text{L}}$  and $\frac{\Delta V_{\text{rms}}}{I_{\text{rms}}}$ for $Z$ in the expression of the impedance of the circuit.

$\frac{\Delta V_{\text{rms}}}{I_{\text{rms}}}=\sqrt{R^2+{\left(0-X_{\text{C}}\right)}^2}$

Rearrange the above expression for value of $X_{\text{C}}$

$X_{\text{C}}={\left({\left(\frac{\Delta V_{\text{rms}}}{I_{\text{rms}}}\right)}^2-R^2\right)}^{\frac{1}{2}}$

Substitute $20.0\text{V}$ for $\Delta V_{\text{rms}}$ and $137\text{mA}$ for $I_{\text{rms}}$ in the above expression.

$\begin{array}{c}{X}_{\text{C}}{}^2={\left({\left(\frac{20.0\text{V}}{137\text{mA}}\right)}^2-{\left(99.6\Omega \right)}^2\right)}^{\frac{1}{2}}\\ X_{\text{C}}=106.73\Omega \end{array}$

Substitute $\frac{1}{2\pi fC}$ for $X_{\text{C}}$ in the above expression.

$\frac{1}{2\pi fC}=106.73\Omega$

Rearrange the above equation for value of value of $C$.

$C=\frac{1}{2\pi f\left(106.73\Omega \right)}$

Substitute $60.0\text{Hz}$ for $f$ in the above expression.

$\begin{array}{c}C=\frac{1}{2\pi \left(60.0\text{Hz}\right)\left(106.73\Omega \right)}\\ =24.8\times {10}^{-6}\text{F}\\ \text{=}24.8\text{μF}\end{array}$

Conclusion:

Therefore, the value of the capacitance is $24.8\text{μF}$.

To determine

 The value of the inductance.

Answer to Problem 33.66AP

 The values of the inductance are $0.164\text{H}$, $0.402\text{H}$.

Explanation of Solution

Given info: The value of the source emf is $20.0\text{V}$ and frequency of the source emf is $60.0\text{Hz}$. The current through the circuit when switch is open is $183\text{mA}$, the current through the circuit when switch is at a position is $298\text{mA}$ and the current through the circuit when the switch is at b position is $137\text{mA}$.

Substitute $99.6\Omega$ for $R$ and $106.73\Omega$ for $X_{\text{C}}$ in the equation (1).

$\begin{array}{c}{R}^2+{\left(X_{\text{L}}-106.73\Omega \right)}^2=11944.22{\Omega }^2\\ X_{\text{L}}=106.73\Omega \pm \sqrt{11944.22{\Omega }^2-{\left(99.6\Omega \right)}^2}\\ X_{\text{L}}=106.73\Omega \pm 44.98\Omega \\ X_{\text{L}}=61.75\Omega ,151.71\Omega \end{array}$ (3)

For first value of the inductor,

Substitute $2\pi fL$ for $X_{\text{L}}$ in the above expression.

$\begin{array}{c}2\pi fL=61.75\Omega \\ L=\frac{61.75\Omega }{2\pi f}\end{array}$

Substitute $60.0\text{Hz}$ for f in the above expression.

$\begin{array}{c}L=\frac{61.75\Omega }{2\pi \left(60.0\text{Hz}\right)}\\ =0.164\text{H}\end{array}$

Thus, the first value of the inductor is $0.164\text{H}$

For second value of the inductor,

Substitute $2\pi fL$ for $X_{\text{L}}$ in equation (3).

$\begin{array}{c}2\pi fL=151.71\Omega \\ L=\frac{151.71\Omega }{2\pi f}\end{array}$

Substitute  for f in the above expression.

$\begin{array}{c}L=\frac{151.71\Omega }{2\pi \left(60.0\text{Hz}\right)}\\ =0.402\text{H}\end{array}$

Thus, the second value of the inductor is $0.402\text{H}$.

Conclusion:

Therefore, the values of the inductance are $0.164\text{H}$, $0.402\text{H}$.

To determine

 Whether more than one set of value possible.

Answer to Problem 33.66AP

 The resistance and capacitance has one set of values, the inductor has in two set of values.

Explanation of Solution

Given info: : The value of the source emf is $20.0\text{V}$ and frequency of the source emf is $60.0\text{Hz}$. The current through the circuit when switch is open is $183\text{mA}$, the current through the circuit when switch is at a position is $298\text{mA}$ and the current through the circuit when the switch is at b position is $137\text{mA}$.

From the calculation in part (a),(b) and (c),

The value of resistance is $\text{99}\text{.6}\Omega$.

The value of capacitance is $24.8\text{μF}$.

The values of the inductor are $0.164\text{H}$, $0.402\text{H}$.

Hence the resistance and the capacitance has one value while the inductor has two set of value in the circuit.

Conclusion:

Therefore, the resistance and capacitance has one set of values, the inductor has in two set of values.