Chapter 21, Problem 44P

(a)

To determine

The voltage amplitude across the capacitor and resistor.

Answer to Problem 44P

The voltage amplitude across the capacitor is $0.97\text{V}$ and voltage across the resistor is $1.8\text{V}$.

Explanation of Solution

The figure 2 shows the $RC$ circuit for ac mains.

Write the expression for current generated in an ac source.

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

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

Write the expression for impedance $RC$ circuit.

  $Z=\sqrt{R^2+X_C^2}$                                                       (II)

Here, $R$ is the resistance and $X_C$ is the capacitive reactance.

Write the expression for capacitive reactance.

  $X_C=\frac{1}{\omega C}$                                                              (III)

Here, $\omega$ is the angular frequency and $C$ is the capacitance.

The expression for frequency of the ac source is.

  $\omega =2\pi f$                                                                (IV)

Here, $f$ is the frequency of the ac mains.

Conclusion:

Substitute the equation (II) and (III) in equation (I).

  $\begin{array}{c}I=\frac{{\epsilon }_{\text{m}}}{\sqrt{R^2+X_C^2}}\\ =\frac{{\epsilon }_{\text{m}}}{\sqrt{R^2+{\left(\frac{1}{\omega C}\right)}^2}}\\ =\frac{{\epsilon }_{\text{m}}}{\sqrt{R^2+\left(\frac{1}{{\omega }^2{C}^2}\right)}}\end{array}$                                                 (V)

Substitute equation (IV) in the equation (V).

  $I=\frac{{\epsilon }_{\text{m}}}{\sqrt{R^2+\left(\frac{1}{4{\pi }^2{f}^2{C}^2}\right)}}$

Substitute $2.0\text{V}$ for ${\epsilon }_{\text{m}}$, $5.00\text{k}\Omega$ for $R$, $120\text{Hz}$ for $f$, and $0.48\text{μF}$ for $C$ to find $I$.

  $\begin{array}{c}I=\frac{\left(2.0\text{V}\right)}{\sqrt{{\left[\left(5.00\text{k}\Omega \right)\left(\frac{{10}^3\Omega }{1\text{k}\Omega }\right)\right]}^2+\left(\frac{1}{4{\left(3.14\right)}^2{\left(120\text{Hz}\right)}^2{\left[\left(0.48\text{μF}\right)\left(\frac{{10}^{-6}\text{F}}{1\text{μF}}\right)\right]}^2}\right)}}\\ =0.35\times {10}^{-3}\text{A}\\ =3.5\times {10}^{-4}\text{A}\end{array}$

Write the equation for the voltage across the capacitor.

  $V_C=I{X}_C$                                                                (VI)

Substitute the equation (III) and (IV) in the equation (VI).

  $\begin{array}{c}{V}_C=\frac{I}{\omega C}\\ =\frac{I}{2\pi fC}\end{array}$

Substitute $3.5\times {10}^{-4}\text{A}$ for $I$, $120\text{Hz}$ for $f$, and $0.48\text{μF}$ for $C$ to find $V_C$.

  $\begin{array}{c}{V}_C=\frac{\left(3.5\times {10}^{-4}\text{A}\right)}{2\left(3.14\right)\left(120\text{Hz}\right)\left(0.48\text{μF}\right)\left(\frac{{10}^{-6}\text{F}}{1\text{μF}}\right)}\\ =0.0097\times {10}^2\text{V}\\ =0.97\text{V}\end{array}$

Write the equation for the voltage across the resistor.

  $V_R=IR$                                                                           (VII)

Substitute $3.5\times {10}^{-4}\text{A}$ for $I$ and $5.00\text{k}\Omega$ for $R$ to find $V_R$.

  $\begin{array}{c}{V}_R=\left(3.5\times {10}^{-4}\text{A}\right)\left(5.00\text{k}\Omega \right)\left(\frac{{10}^3\Omega }{1\text{k}\Omega }\right)\\ =17.5\times {10}^{-1}\text{V}\\ =1.8\text{V}\end{array}$

Therefore, the voltage across the capacitor is $0.97\text{V}$ and voltage across the resistor is $1.8\text{V}$.

(b)

To determine

If the amplitude of voltage and source voltage add together gives $2.0\text{V}$.

Answer to Problem 44P

No, the amplitude of voltage do not add to give the source voltage.

Explanation of Solution

Since the voltage amplitude do not add as scalars to give the amplitude of the source voltage because the voltage across the resistor and capacitor are $90°$ out of phase. Similar to the impedance the source voltage is the square root of the sum of squares is given by.

  ${\epsilon }_{\text{m}}=\sqrt{V_C^2+V_R^2}$

Conclusion:

Thus, the amplitude of voltage does not give the source voltage by adding as scalars of voltage across the resistance and capacitors.

(c)

To determine

Draw the phasor diagram for the voltages.

Answer to Problem 44P

The phasor diagram is drawn for the voltage across the resistance and capacitors.

Explanation of Solution

The phasor diagram is used to show the phase relationship between two or more sine waves having the same frequency. The voltage phasor diagram is drawn across the series RLC circuit such as voltage across resistor, voltage across inductor, and voltage across capacitor.

The figure 1 shows the phasor diagram.