Chapter 21, Problem 94P

(a)

To determine

The frequency of the RC circuit.

Answer to Problem 94P

The capacitive reactance are $2.65\text{ k}\Omega$ and $21.2\Omega$

Explanation of Solution

The capacitance of capacitor is $5.00\text{}\text{ μF}$, the frequencies of the capacitor are $12.0\text{Hz}$ and $1.50\text{kHz}$.

Write the expression for capacitive reactance

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

Here, $X_{\text{C}}$ is the reactance of the capacitor, $f$ is the frequency and $C$ is the capacitance.

Substitute $12.0\text{Hz}$ for $f$ and $5.00\text{}\text{ μF}$ for $C$ in (I) to find $X_{\text{C1}}$. Here, $X_{\text{C1}}$ is the reactance of the capacitor with frequency $12.0\text{Hz}$.

  $\begin{array}{c}{X}_{\text{C1}}=\frac{1}{2\pi \left(12.0\text{Hz}\right)\left(5.00\text{}\text{ μF}\right)}\\ =\frac{1}{2\pi \left(12.0\text{Hz}\right)\left(5.00\times {10}^{-6}\text{ F}\right)}\cdot \frac{1\Omega }{1{\text{sF}}^{-1}}\\ =2.65\times {10}^3\Omega \\ =2.65\text{ k}\Omega \end{array}$

Thus, the capacitive reactance is $2.65\text{ k}\Omega$.

Substitute $1.50\text{kHz}$ for $f$ and $5.00\text{}\text{ μF}$ for $C$ in (I) to find $X_{\text{C2}}$. Here, $X_{\text{C2}}$ is the reactance of the capacitor with frequency $1.50\text{kHz}$.

  $\begin{array}{c}{X}_{\text{C2}}=\frac{1}{2\pi \left(1.50\text{kHz}\right)\left(5.00\text{}\text{ μF}\right)}\\ =\frac{1}{2\pi \left(1.50\times {10}^3\text{ Hz}\right)\left(5.00\times {10}^{-6}\text{ F}\right)}\cdot \frac{1\Omega }{1{\text{sF}}^{-1}}\\ =21.2\Omega \end{array}$

Thus, the capacitive reactance is $21.2\Omega$.

(b)

To determine

The impedance of the series circuit.

Answer to Problem 94P

The impedance of the series circuit is $3.32\text{k}\Omega$ and $2.00\text{k}\Omega$ at the given frequencies.

Explanation of Solution

The capacitor is connected in series with a resistor of resistance, $2.00\text{k}\Omega$.

Write the expression for impedance

  $Z=\sqrt{R^2+X_{\text{C}}^2}$                                                                    (II)

Here, $Z$ is the impedance.

Substitute $2.00\text{k}\Omega$ for $R$ and $2.65\text{ k}\Omega$ for $X_{\text{C1}}$ in (II) to find $Z_1$. Here $Z_1$ is the impedance for frequency $12.0\text{Hz}$.

  $\begin{array}{c}{Z}_1=\sqrt{{\left(2.00\text{k}\Omega \right)}^2+{\left(2.65\text{ k}\Omega \right)}^2}\\ =3.32\text{k}\Omega \end{array}$

Thus, the impedance is $3.32\text{k}\Omega$.

Substitute $2.00\text{k}\Omega$ for $R$ and $21.2\Omega$ for $X_{\text{C2}}$ in (II) to find $Z_2$. Here $Z_2$ is the impedance for the frequency $1.50\text{kHz}$.

  $\begin{array}{c}{Z}_2=\sqrt{{\left(2.00\text{k}\Omega \right)}^2+{\left(21.2\Omega \right)}^2}\\ =\sqrt{{\left(2000\Omega \right)}^2+{\left(21.2\Omega \right)}^2}\\ =2000.11\Omega \\ =2.00\text{k}\Omega \end{array}$

Thus, the impedance is $2.00\text{k}\Omega$.

(c)

To determine

The maximum current in the series circuit at the peak voltage.

Answer to Problem 94P

The maximum current for the given frequencies are $602\text{μA}$ and $1.00\text{mA}$.

Explanation of Solution

The peak voltage of the source is $2.00\text{V}$.

Write the expression for maximum current

  $I=\frac{{\epsilon }_{\text{m}}}{Z}$                                                                      (III)

Here, $I$ is the maximum current and ${\epsilon }_{\text{m}}$ is the peak voltage.

Substitute $3.32\text{k}\Omega$ for $Z$ and $2.00\text{V}$ for ${\epsilon }_{\text{m}}$ in (III) to find $I_1$. Here, $I_1$ is the maximum current at frequency $12.0\text{Hz}$

  $\begin{array}{c}{I}_1=\frac{2.00\text{V}}{3.32\text{k}\Omega }\\ =\frac{2.00\text{V}}{3.32\times {10}^3\Omega }\cdot \frac{1\text{}\Omega A}{1\text{V}}\\ =602\times {10}^{-6}\text{A}\\ =602\text{μA}\end{array}$

Thus, the maximum current is $602\text{μA}$.

Substitute $2.00\text{k}\Omega$ for $Z$ and $2.00\text{V}$ for ${\epsilon }_{\text{m}}$ in (III) to find $I_2$. Here, $I_2$ is the maximum current at frequency $1.50\text{kHz}$

  $\begin{array}{c}{I}_2=\frac{2.00\text{V}}{2.00\text{k}\Omega }\\ =\frac{2.00\text{V}}{2.00\times {10}^3\Omega }\cdot \frac{1\text{}\Omega A}{1\text{V}}\\ =1.00\times {10}^{-3}\text{A}\\ =1.00\text{mA}\end{array}$

Thus, the maximum current is $1.00\text{mA}$.

(d)

To determine

The phase angle between the voltage and the current.

Answer to Problem 94P

The current leads the voltage by $53.0°$ and $0.608°$ at the given frequencies.

Explanation of Solution

Initially, there is no charge on the capacitor when the ac supply is turned on. Thus, $V_C$ is zero and the current is maximum. Thus, the current leads the voltage.

Write the expression for phase angle

  $\cos \varphi =\frac{R}{Z}$                                                                     (IV)

Here, $\varphi$ is the phase angle.

Rearrange

  $\varphi ={\cos }^{-1}\left(\frac{R}{Z}\right)$                                                                    (V)

Substitute $3.32\text{k}\Omega$ for $Z$ and $2.00\text{k}\Omega$ for $R$ in (V) to find ${\varphi }_1$. Here, ${\varphi }_1$ is the phase angel at frequency $12.0\text{Hz}$

  $\begin{array}{c}{\varphi }_1={\cos }^{-1}\left(\frac{2.00\text{k}\Omega }{3.32\text{k}\Omega }\right)\\ =53.0°\end{array}$

Thus, the phase angle is $53.0°$.

Substitute $2000.11\Omega$ for $Z$ and $2.00\text{k}\Omega$ for $R$ in (V) to find ${\varphi }_2$. Here, ${\varphi }_2$ is the phase angel at frequency $1.50\text{kHz}$

  $\begin{array}{c}{\varphi }_2={\cos }^{-1}\left(\frac{2.00\text{k}\Omega }{2000.11\Omega }\right)\\ ={\cos }^{-1}\left(\frac{2000\Omega }{2000.11\Omega }\right)\\ =0.608°\end{array}$

Thus, the phase angle is $0.608°$.