Chapter 21, Problem 46P

(a)

To determine

The frequency of the RC circuit.

Answer to Problem 46P

The frequency of the series RC circuit is $24\text{Hz}$.

Explanation of Solution

The resistor with resistance of $3.3\text{k}\Omega$ is connected in series with a capacitor of capacitance $2.0\text{}\text{ μF}$. Both resistor and capacitor has the same rms voltage.

Write the expression for voltage drop across the resistor

  $V_{\text{R}}=IR$                                                                     (I)

Here, $V_{\text{R}}$ is the voltage drop across the resistor, $I$ is the current and $R$ is the resistance.

Write the expression for voltage drop across the capacitor

  $V_{\text{C}}=I{X}_{\text{C}}$                                                                     (II)

Here, $V_{\text{C}}$ is the voltage drop across the capacitor, $I$ is the current and $X_{\text{C}}$ is the reactance of the capacitor.

Write the expression for capacitive reactance

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

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

Substitute (III) in  (II)

  $V_{\text{C}}=\frac{I}{2\pi fC}$                                                                      (IV)

Equate (I) and (IV) since resistor and capacitor has the same rms voltage

  $IR=\frac{I}{2\pi fC}$                                                                      (V)

Rearrange for $f$

  $f=\frac{1}{2\pi RC}$                                                                      (VI)

Substitute $3.3\text{k}\Omega$ for $R$ and $2.0\text{}\text{ μF}$ for $C$ in (VI) to find $f$

  $\begin{array}{c}f=\frac{1}{2\pi \left(3.3\text{k}\Omega \right)\left(2.0\text{}\text{ μF}\right)}\\ =\frac{1}{2\pi \left(3.3\times {10}^3\Omega \right)\left(2.0\times {10}^{-6}\text{}\text{ F}\right)}\cdot \frac{1\Omega }{1{\text{sF}}^{-1}}\\ =24\text{Hz}\end{array}$

Thus, the frequency of the series RC circuit is $24\text{Hz}$.

(b)

To determine

The rms voltage of the resistor and capacitor as fraction of the source voltage.

Answer to Problem 46P

The rms voltage of the resistor and capacitor as fraction of the source voltage is $\frac{1}{\sqrt{2}}$.

Explanation of Solution

The phasor diagram for the series RC circuit is given below:

Write the expression for source voltage using the phasor diagram

  ${\epsilon }_{\text{m}}^2=V_{\text{R}}^2+V_{\text{C}}^2$                                                                     (VII)

Here, ${\epsilon }_m$ is the source voltage.

Substitute $V_{\text{R}}^2$ for $V_{\text{C}}^2$ in (VII)

  ${\epsilon }_{\text{m}}^2=2V_{\text{R}}^2$

Rearrange

  $\frac{V_{\text{R}}^2}{{\epsilon }_{\text{m}}^2}=\frac{1}{2}$                                                                       (VIII)

Take square root on both sides

  $\frac{V_{\text{R}}}{{\epsilon }_{\text{m}}}=\frac{1}{\sqrt{2}}$                                                                     (IX)

Since both resistor and capacitor has the same rms voltage.

  $\frac{V_{\text{R}}}{{\epsilon }_{\text{m}}}=\frac{V_{\text{C}}}{{\epsilon }_{\text{m}}}=\frac{1}{\sqrt{2}}$

Thus, the rms voltage of resistor and capacitor is not half the rms voltage of the source.

(c)

To determine

The phase angle between the source voltage and the current.

Answer to Problem 46P

The phase angle between the voltage and current is $45°$.

Explanation of Solution

Write the expression for phase angle

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

Here, $\varphi$ is the phase angle and $Z$ is the impedance.

Rearrange

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

Write the expression for impedance

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

Substitute $Z$ in (XIII).

  $\varphi ={\cos }^{-1}\left(\frac{R}{\sqrt{R^2+X_{\text{C}}^2}}\right)$

Multiply both numerator and denominator by $I$ .

  $\varphi ={\cos }^{-1}\left(\frac{IR}{\sqrt{{\left(IR\right)}^2+{\left(I{X}_{\text{C}}\right)}^2}}\right)$                                         (XIV)

Substitute (I) and (II) in (XIV).

  $\varphi ={\cos }^{-1}\left(\frac{V_{\text{R}}}{\sqrt{V_{\text{R}}^2+V_{\text{C}}^2}}\right)$                                                    (XV)

Substitute (VII) in (XV).

  $\varphi ={\cos }^{-1}\left(\frac{V_{\text{R}}}{{\epsilon }_m}\right)$                                                            (XIV)

Substitute (IX) in (XV).

  $\varphi ={\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)=45°$                                                   (XIV)

Thus, the phase angle between the voltage and current is $45°$ i.e. current leads voltage by $45°$.

(d)

To determine

The impedance of the series RC circuit.

Answer to Problem 46P

The impedance is $4.7\text{k}\Omega$.

Explanation of Solution

Rearrange (XI)

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

Substitute $3.3\text{k}\Omega$ for $R$ and $45°$ for $\varphi$ in (XV).

  $\begin{array}{c}Z=\frac{3.3\text{k}\Omega }{\cos 45°}\\ =4.7\text{k}\Omega \end{array}$

Thus, the impedance is $4.7\text{k}\Omega$.