Chapter 2, Problem 2.18QAP

Interpretation Introduction

(a)

Interpretation:

The capacitive reactance, the impedance and the phase angle $\varphi$ for the series RC circuit should be calculated.

Concept introduction:

The capacitive reactance (X_c) is a property of a capacitor that is analogous to the resistance of a resistor.

${\text{X}}_{\text{c}}\text{=}\frac{{\text{V}}_{\text{p}}}{{\text{I}}_{\text{p}}}=\frac{1}{\omega \text{C}}$

$\omega$ = frequency

C = capacitance

The impedance (Z) is given by following equation:

$\text{Z = }\sqrt{{\text{R}}^2+{\text{X}}^2{}_{\text{C}}}$

X_c = capacitive reactance

R = resistance

The phase angle ( $\varphi$ ) is given by following equation:

$\varphi =\arctan \frac{{\text{X}}_{\text{C}}}{\text{R}}$

X_c = capacitive reactance

R = resistance

Answer to Problem 2.18QAP

${\text{X}}_{\text{c}}=30.3\times {10}^6\Omega$

$\text{Z = 30}\text{.3}\times {\text{10}}^6\Omega$

$\varphi ={89.94}^0$

Explanation of Solution

Given information:

Frequency = 1 Hz

R = 30000 Ω

C = 0.033 µF

$\begin{array}{l}{\text{X}}_{\text{c}}\text{=}\frac{{\text{V}}_{\text{p}}}{{\text{I}}_{\text{p}}}=\frac{1}{\omega \text{C}}\\ {\text{X}}_{\text{c}}=\frac{1}{\omega \text{C}}=\frac{1}{1\times 0.033\times {\text{10}}^{-6}}\Omega =30.3\times {10}^6\Omega \end{array}$

$\begin{array}{l}\text{Z = }\sqrt{{\text{R}}^2+{\text{X}}^2{}_{\text{C}}}\\ \text{Z = }\sqrt{{(30000)}^2+{(30.3\times {10}^6)}^2}\\ \text{Z = }\sqrt{9\times {10}^8+9.1809\times {10}^{14}}\\ \text{Z = }\sqrt{9.180909\times {10}^{14}}\\ \text{Z = 30}\text{.3}\times {\text{10}}^6\Omega \end{array}$

$\begin{array}{l}\varphi =\arctan \frac{{\text{X}}_{\text{C}}}{\text{R}}\\ \varphi =\arctan \frac{30.3\times {10}^6}{30000}=\arctan 1010={89.94}^0\end{array}$

Interpretation Introduction

(b)

Interpretation:

The capacitive reactance, the impedance and the phase angle $\varphi$ for the series RC circuit should be calculated.

Concept introduction:

The capacitive reactance (X_c) is a property of a capacitor that is analogous to the resistance of a resistor.

${\text{X}}_{\text{c}}\text{=}\frac{{\text{V}}_{\text{p}}}{{\text{I}}_{\text{p}}}=\frac{1}{\omega \text{C}}$

$\omega$ = frequency

C = capacitance

The impedance (Z) is given by following equation:

$\text{Z = }\sqrt{{\text{R}}^2+{\text{X}}^2{}_{\text{C}}}$

X_c = capacitive reactance

R = resistance

The phase angle ( $\varphi$ ) is given by following equation:

$\varphi =\arctan \frac{{\text{X}}_{\text{C}}}{\text{R}}$

X_c = capacitive reactance

R = resistance

Answer to Problem 2.18QAP

${\text{X}}_{\text{c}}=30.3\times {10}^3\Omega$

$\text{Z = }42638\Omega$

$\varphi ={45.3}^0$

Explanation of Solution

Given information:

Frequency = 10³ Hz

R = 30000 Ω

C = 0.033 µF

$\begin{array}{l}{\text{X}}_{\text{c}}\text{=}\frac{{\text{V}}_{\text{p}}}{{\text{I}}_{\text{p}}}=\frac{1}{\omega \text{C}}\\ {\text{X}}_{\text{c}}=\frac{1}{\omega \text{C}}=\frac{1}{{10}^3\times 0.033\times {\text{10}}^{-6}}\Omega =30.3\times {10}^3\Omega \end{array}$

$\begin{array}{l}\text{Z = }\sqrt{{\text{R}}^2+{\text{X}}^2{}_{\text{C}}}\\ \text{Z = }\sqrt{{(30000)}^2+{(30.3\times {10}^3)}^2}\\ \text{Z = }\sqrt{9\times {10}^8+9.1809\times {10}^8}\\ \text{Z = }\sqrt{18.18\times {10}^8}\\ \text{Z = 42638 }\Omega \end{array}$

$\begin{array}{l}\varphi =\arctan \frac{{\text{X}}_{\text{C}}}{\text{R}}\\ \varphi =\arctan \frac{30.3\times {10}^3}{30000}=\arctan 1.01={45.285}^0\end{array}$

Interpretation Introduction

(c)

Interpretation:

The capacitive reactance, the impedance and the phase angle $\varphi$ for the series RC circuit should be calculated.

Concept introduction:

The capacitive reactance (X_c) is a property of a capacitor that is analogous to the resistance of a resistor.

${\text{X}}_{\text{c}}\text{=}\frac{{\text{V}}_{\text{p}}}{{\text{I}}_{\text{p}}}=\frac{1}{\omega \text{C}}$

$\omega$ = frequency

C = capacitance

The impedance (Z) is given by following equation:

$\text{Z = }\sqrt{{\text{R}}^2+{\text{X}}^2{}_{\text{C}}}$

X_c = capacitive reactance

R = resistance

The phase angle ( $\varphi$ ) is given by following equation:

$\varphi =\arctan \frac{{\text{X}}_{\text{C}}}{\text{R}}$

X_c = capacitive reactance

R = resistance

Answer to Problem 2.18QAP

${\text{X}}_{\text{c}}=303.03\Omega$

$\text{Z = }30000\Omega$

$\varphi ={0.58}^0$

Explanation of Solution

Given information:

Frequency = 10⁶ Hz

R = 30000 Ω

C = 0.0033 µF

$\begin{array}{l}{\text{X}}_{\text{c}}\text{=}\frac{{\text{V}}_{\text{p}}}{{\text{I}}_{\text{p}}}=\frac{1}{\omega \text{C}}\\ {\text{X}}_{\text{c}}=\frac{1}{\omega \text{C}}=\frac{1}{{10}^6\times 0.0033\times {\text{10}}^{-6}}\Omega =303.03\Omega \end{array}$

$\begin{array}{l}\text{Z = }\sqrt{{\text{R}}^2+{\text{X}}^2{}_{\text{C}}}\\ \text{Z = }\sqrt{{(30000)}^2+{(303.03)}^2}\\ \text{Z = }\sqrt{9\times {10}^8+91827.2}\\ \text{Z = }\sqrt{9.00\times {10}^8}\\ \text{Z = 30000 }\Omega \end{array}$

$\begin{array}{l}\varphi =\arctan \frac{{\text{X}}_{\text{C}}}{\text{R}}\\ \varphi =\arctan \frac{303.03}{30000}=\arctan 0.010101={0.58}^0\end{array}$

Interpretation Introduction

(d)

Interpretation:

The capacitive reactance, the impedance and the phase angle $\varphi$ for the series RC circuit should be calculated.

Concept introduction:

The capacitive reactance (X_c) is a property of a capacitor that is analogous to the resistance of a resistor.

${\text{X}}_{\text{c}}\text{=}\frac{{\text{V}}_{\text{p}}}{{\text{I}}_{\text{p}}}=\frac{1}{\omega \text{C}}$

$\omega$ = frequency

C = capacitance

The impedance (Z) is given by following equation;

$\text{Z = }\sqrt{{\text{R}}^2+{\text{X}}^2{}_{\text{C}}}$

X_c = capacitive reactance

R = resistance

The phase angle ( $\varphi$ ) is given by following equation;

$\varphi =\arctan \frac{{\text{X}}_{\text{C}}}{\text{R}}$

X_c = capacitive reactance

R = resistance

Answer to Problem 2.18QAP

${\text{X}}_{\text{c}}=30.3\times {10}^7\Omega$

$\text{Z = }30.3\times {10}^7\Omega$

$\varphi ={89.9999}^0={90}^0$

Explanation of Solution

Given information:

Frequency = 1Hz

R = 300 Ω

C = 0.0033 µF

$\begin{array}{l}{\text{X}}_{\text{c}}\text{=}\frac{{\text{V}}_{\text{p}}}{{\text{I}}_{\text{p}}}=\frac{1}{\omega \text{C}}\\ {\text{X}}_{\text{c}}=\frac{1}{\omega \text{C}}=\frac{1}{1\times 0.0033\times {\text{10}}^{-6}}\Omega =30.3\times {10}^7\Omega \end{array}$

$\begin{array}{l}\text{Z = }\sqrt{{\text{R}}^2+{\text{X}}^2{}_{\text{C}}}\\ \text{Z = }\sqrt{{(300)}^2+{(30.3\times {10}^7)}^2}\\ \text{Z = }\sqrt{9\times {10}^4+9.1809\times {10}^{16}}\\ \text{Z = }\sqrt{9.1809\times {10}^{16}}\\ \text{Z = 30}\text{.3}\times {\text{10}}^7\Omega \end{array}$

$\begin{array}{l}\varphi =\arctan \frac{{\text{X}}_{\text{C}}}{\text{R}}\\ \varphi =\arctan \frac{30.3\times {10}^7}{300}=\arctan 1010000={89.9999}^0={90}^0\end{array}$

Interpretation Introduction

(e)

Interpretation:

The capacitive reactance, the impedance and the phase angle $\varphi$ for the series RC circuit should be calculated.

Concept introduction:

The capacitive reactance (X_c) is a property of a capacitor that is analogous to the resistance of a resistor.

${\text{X}}_{\text{c}}\text{=}\frac{{\text{V}}_{\text{p}}}{{\text{I}}_{\text{p}}}=\frac{1}{\omega \text{C}}$

$\omega$ = frequency

C = capacitance

The impedance (Z) is given by following equation:

$\text{Z = }\sqrt{{\text{R}}^2+{\text{X}}^2{}_{\text{C}}}$

X_c = capacitive reactance

R = resistance

The phase angle ( $\varphi$ ) is given by following equation:

$\varphi =\arctan \frac{{\text{X}}_{\text{C}}}{\text{R}}$

X_c = capacitive reactance

R = resistance

Answer to Problem 2.18QAP

${\text{X}}_{\text{c}}=30.3\times {10}^4\Omega$

$\text{Z = }30.3\times {10}^4\Omega$

$\varphi ={89.94}^0$

Explanation of Solution

Given information:

Frequency = 10³ Hz

R = 300 Ω

C = 0.0033 µF

$\begin{array}{l}{\text{X}}_{\text{c}}\text{=}\frac{{\text{V}}_{\text{p}}}{{\text{I}}_{\text{p}}}=\frac{1}{\omega \text{C}}\\ {\text{X}}_{\text{c}}=\frac{1}{\omega \text{C}}=\frac{1}{{10}^3\times 0.0033\times {\text{10}}^{-6}}\Omega =30.3\times {10}^4\Omega \end{array}$

$\begin{array}{l}\text{Z = }\sqrt{{\text{R}}^2+{\text{X}}^2{}_{\text{C}}}\\ \text{Z = }\sqrt{{(300)}^2+{(30.3\times {10}^4)}^2}\\ \text{Z = }\sqrt{9\times {10}^4+9.1809\times {10}^{10}}\\ \text{Z = }\sqrt{9.1809\times {10}^{10}}\\ \text{Z = 30}\text{.3}\times {\text{10}}^4\Omega \end{array}$

$\begin{array}{l}\varphi =\arctan \frac{{\text{X}}_{\text{C}}}{\text{R}}\\ \varphi =\arctan \frac{30.3\times {10}^4}{300}=\arctan 1010={89.94}^0\end{array}$

Interpretation Introduction

(f)

Interpretation:

The capacitive reactance, the impedance and the phase angle $\varphi$ for the series RC circuit should be calculated.

Concept introduction:

The capacitive reactance (X_c) is a property of a capacitor that is analogous to the resistance of a resistor.

${\text{X}}_{\text{c}}\text{=}\frac{{\text{V}}_{\text{p}}}{{\text{I}}_{\text{p}}}=\frac{1}{\omega \text{C}}$

$\omega$ = frequency

C = capacitance

The impedance (Z) is given by following equation;

$\text{Z = }\sqrt{{\text{R}}^2+{\text{X}}^2{}_{\text{C}}}$

X_c = capacitive reactance

R = resistance

The phase angle ( $\varphi$ ) is given by following equation;

$\varphi =\arctan \frac{{\text{X}}_{\text{C}}}{\text{R}}$

X_c = capacitive reactance

R = resistance

Answer to Problem 2.18QAP

${\text{X}}_{\text{c}}=303.03\Omega$

$\text{Z = }426.4\Omega$

$\varphi ={45.29}^0$

Explanation of Solution

Given information:

Frequency = 10⁶ Hz

R = 300 Ω

C = 0.0033 µF

$\begin{array}{l}{\text{X}}_{\text{c}}\text{=}\frac{{\text{V}}_{\text{p}}}{{\text{I}}_{\text{p}}}=\frac{1}{\omega \text{C}}\\ {\text{X}}_{\text{c}}=\frac{1}{\omega \text{C}}=\frac{1}{{10}^6\times 0.0033\times {\text{10}}^{-6}}\Omega =303.03\Omega \end{array}$

$\begin{array}{l}\text{Z = }\sqrt{{\text{R}}^2+{\text{X}}^2{}_{\text{C}}}\\ \text{Z = }\sqrt{{(300)}^2+{(303.03)}^2}\\ \text{Z = }\sqrt{9\times {10}^4+91827.2}\\ \text{Z = }\sqrt{181827.2}\\ \text{Z = 426}\text{.4 }\Omega \end{array}$

$\begin{array}{l}\varphi =\arctan \frac{{\text{X}}_{\text{C}}}{\text{R}}\\ \varphi =\arctan \frac{303.03}{300}=\arctan 1.0101={45.29}^0\end{array}$

Interpretation Introduction

(g)

Interpretation:

The capacitive reactance, the impedance and the phase angle $\varphi$ for the series RC circuit should be calculated.

Concept introduction:

The capacitive reactance (X_c) is a property of a capacitor that is analogous to the resistance of a resistor.

${\text{X}}_{\text{c}}\text{=}\frac{{\text{V}}_{\text{p}}}{{\text{I}}_{\text{p}}}=\frac{1}{\omega \text{C}}$

$\omega$ = frequency

C = capacitance

The impedance (Z) is given by following equation;

$\text{Z = }\sqrt{{\text{R}}^2+{\text{X}}^2{}_{\text{C}}}$

X_c = capacitive reactance

R = resistance

The phase angle ( $\varphi$ ) is given by following equation;

$\varphi =\arctan \frac{{\text{X}}_{\text{C}}}{\text{R}}$

X_c = capacitive reactance

R = resistance

Answer to Problem 2.18QAP

${\text{X}}_{\text{c}}=30.3\times {10}^5\Omega$

$\text{Z = }30.3\times {10}^5\Omega$

$\varphi ={89.94}^0$

Explanation of Solution

Given information:

Frequency = 1 Hz

R = 3000 Ω

C = 0.33 µF

$\begin{array}{l}{\text{X}}_{\text{c}}\text{=}\frac{{\text{V}}_{\text{p}}}{{\text{I}}_{\text{p}}}=\frac{1}{\omega \text{C}}\\ {\text{X}}_{\text{c}}=\frac{1}{\omega \text{C}}=\frac{1}{1\times 0.33\times {\text{10}}^{-6}}\Omega =30.3\times {10}^5\Omega \end{array}$

$\begin{array}{l}\text{Z = }\sqrt{{\text{R}}^2+{\text{X}}^2{}_{\text{C}}}\\ \text{Z = }\sqrt{{(3000)}^2+{(30.3\times {10}^5)}^2}\\ \text{Z = }\sqrt{9\times {10}^6+9.1809\times {10}^{12}}\\ \text{Z = }\sqrt{9.1809\times {10}^{12}}\\ \text{Z = 30}\text{.3}\times {\text{10}}^5\Omega \end{array}$

$\begin{array}{l}\varphi =\arctan \frac{{\text{X}}_{\text{C}}}{\text{R}}\\ \varphi =\arctan \frac{30.3\times {10}^5}{3000}=\arctan 1010={89.94}^0\end{array}$

Interpretation Introduction

(h)

Interpretation:

The capacitive reactance, the impedance and the phase angle $\varphi$ for the series RC circuit should be calculated.

Concept introduction:

The capacitive reactance (X_c) is a property of a capacitor that is analogous to the resistance of a resistor.

${\text{X}}_{\text{c}}\text{=}\frac{{\text{V}}_{\text{p}}}{{\text{I}}_{\text{p}}}=\frac{1}{\omega \text{C}}$

$\omega$ = frequency

C = capacitance

The impedance (Z) is given by following equation:

$\text{Z = }\sqrt{{\text{R}}^2+{\text{X}}^2{}_{\text{C}}}$

X_c = capacitive reactance

R = resistance

The phase angle ( $\varphi$ ) is given by following equation:

$\varphi =\arctan \frac{{\text{X}}_{\text{C}}}{\text{R}}$

X_c = capacitive reactance

R = resistance

Answer to Problem 2.18QAP

${\text{X}}_{\text{c}}=30.3\times {10}^2\Omega$

$\text{Z = 4}\text{.26}\times {10}^3\Omega$

$\varphi ={45.29}^0$

Explanation of Solution

Given information:

Frequency = 1000 Hz

R = 3000 Ω

C = 0.33 µF

$\begin{array}{l}{\text{X}}_{\text{c}}\text{=}\frac{{\text{V}}_{\text{p}}}{{\text{I}}_{\text{p}}}=\frac{1}{\omega \text{C}}\\ {\text{X}}_{\text{c}}=\frac{1}{\omega \text{C}}=\frac{1}{1000\times 0.33\times {\text{10}}^{-6}}\Omega =30.3\times {10}^2\Omega \end{array}$

$\begin{array}{l}\text{Z = }\sqrt{{\text{R}}^2+{\text{X}}^2{}_{\text{C}}}\\ \text{Z = }\sqrt{{(3000)}^2+{(30.3\times {10}^2)}^2}\\ \text{Z = }\sqrt{9\times {10}^6+9180900}\\ \text{Z = }\sqrt{18180900}\\ \text{Z = 4}\text{.26}\times {\text{10}}^3\Omega \end{array}$

$\begin{array}{l}\varphi =\arctan \frac{{\text{X}}_{\text{C}}}{\text{R}}\\ \varphi =\arctan \frac{30.3\times {10}^2}{3000}=\arctan 1.01={45.29}^0\end{array}$

Interpretation Introduction

(i)

Interpretation:

The capacitive reactance, the impedance and the phase angle $\varphi$ for the series RC circuit should be calculated.

Concept introduction:

The capacitive reactance (X_c) is a property of a capacitor that is analogous to the resistance of a resistor.

${\text{X}}_{\text{c}}\text{=}\frac{{\text{V}}_{\text{p}}}{{\text{I}}_{\text{p}}}=\frac{1}{\omega \text{C}}$

$\omega$ = frequency

C = capacitance

The impedance (Z) is given by following equation:

$\text{Z = }\sqrt{{\text{R}}^2+{\text{X}}^2{}_{\text{C}}}$

X_c = capacitive reactance

R = resistance

The phase angle ( $\varphi$ ) is given by following equation:

$\varphi =\arctan \frac{{\text{X}}_{\text{C}}}{\text{R}}$

X_c = capacitive reactance

R = resistance

Answer to Problem 2.18QAP

${\text{X}}_{\text{c}}=3.03\Omega$

$\text{Z = }3000\Omega$

$\varphi ={0.058}^0$

Explanation of Solution

Given information:

Frequency = 10⁶ Hz

R = 3000 Ω

C = 0.33 µF

$\begin{array}{l}{\text{X}}_{\text{c}}\text{=}\frac{{\text{V}}_{\text{p}}}{{\text{I}}_{\text{p}}}=\frac{1}{\omega \text{C}}\\ {\text{X}}_{\text{c}}=\frac{1}{\omega \text{C}}=\frac{1}{{10}^6\times 0.33\times {\text{10}}^{-6}}\Omega =3.03\Omega \end{array}$

$\begin{array}{l}\text{Z = }\sqrt{{\text{R}}^2+{\text{X}}^2{}_{\text{C}}}\\ \text{Z = }\sqrt{{(3000)}^2+{(3.03)}^2}\\ \text{Z = }\sqrt{9\times {10}^6+91809}\\ \text{Z = }\sqrt{9\times {10}^6}\\ \text{Z = 3000 }\Omega \end{array}$

$\begin{array}{l}\varphi =\arctan \frac{{\text{X}}_{\text{C}}}{\text{R}}\\ \varphi =\arctan \frac{3.03}{3000}=\arctan 1.01\times {10}^{-3}={0.058}^0\end{array}$