Chapter 2, Problem 2.6QAP

Interpretation Introduction

(a)

Interpretation:

The power dissipated between the points 1 and 2 is to be stated.

Concept introduction:

The resistive electrical circuit is the circuit which contains the voltage source and the resistor. The analysis of the circuit is done to evaluate the equivalent resistance, current through the circuit and power consumed by the circuit.

The power in dissipated in the electrical circuit is defined by $I^{2}R$, where the term $I$ is the current flowing through the circuit and the $R$ is the equivalent resistance of the electrical circuit across the source.

Answer to Problem 2.6QAP

The power dissipated between the points 1 and 2 is $0.0514\text{W}$.

Explanation of Solution

The below figure represents the electrical circuit.

Write the expression for the equivalent resistance between the point 1 and 2.

$R_{12}=\frac{R_B{R}_C}{R_B+R_C}$ ....... (I)

Write the expression for the equivalent resistance between the point 2 and 3.

$R_{23}=\frac{R_D{R}_E}{R_D+R_E}$ ....... (II)

Write the expression for the total equivalent resistance.

$R=R_A+R_{12}+R_{23}$ ....... (III)

Write the expression for the current flowing through the circuit.

$I=\frac{V}{R}$ ....... (IV)

Here, $V$ is the voltage of the voltage source.

Write the expression for the power dissipated between the points 1 and 2.

$P_{12}=I^2{R}_{12}$ ....... (V)

Substitute $3.0\text{kΩ}$ for $R_B$ and $4.0\text{kΩ}$ for $R_C$ in the Equation (I).

$\begin{array}{l}{R}_{12}=\frac{\left(3.0\text{kΩ}\right)\left(4.0\text{kΩ}\right)}{\left(3.0\text{kΩ}\right)+\left(4.0\text{kΩ}\right)}\\ R_{12}=1.714\text{kΩ}\end{array}$

Substitute $2.0\text{kΩ}$ for $R_D$ and $1.0\text{kΩ}$ for $R_E$ in the Equation (II).

$\begin{array}{l}{R}_{23}=\frac{\left(2.0\text{kΩ}\right)\left(1.0\text{kΩ}\right)}{\left(2.0\text{kΩ}\right)+\left(1.0\text{kΩ}\right)}\\ R_{23}=0.667\text{kΩ}\end{array}$

Substitute $1.714\text{kΩ}$ for $R_{12}$, $0.667\text{kΩ}$ for $R_{23}$ and $2.0\text{kΩ}$ for $R_A$ in the Equation (III).

$\begin{array}{l}R=\left(2.0\text{kΩ}\right)+\left(1.714\text{kΩ}\right)+\left(0.667\text{kΩ}\right)\\ R=4.381\text{kΩ}\end{array}$

Substitute $4.381\text{kΩ}$ for $R$ and $24\text{V}$ for the $V$ in the Equation (IV).

$\begin{array}{l}I=\frac{24\text{V}}{\left(4.381\text{kΩ}\right)\left(\frac{{10}^3\text{Ω}}{1\text{kΩ}}\right)}\\ I=5.478\times {10}^{-3}\text{A}\end{array}$

Substitute $5.478\times {10}^{-3}\text{A}$ for $I$ and $1.714\text{kΩ}$ for $R_{12}$ in the Equation (V).

$\begin{array}{l}{P}_{12}={\left(5.478\times {10}^{-3}\text{A}\right)}^2\left(1.714\text{kΩ}\right)\left(\frac{{10}^3\text{Ω}}{1\text{kΩ}}\right)\\ P_{12}=0.0514\text{W}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The current drawn from the source is to be stated.

Concept introduction:

The resistive electrical circuit is the circuit which contains the voltage source and the resistor. The analysis of the circuit is to evaluating the equivalent resistance, current through the circuit and power consumed by the circuit.

The current drown the from the source is define as $I=\frac{V}{R}$, where $V$ represents the voltage of the source and $R$ represents the equivalent resistance of the circuit.

Answer to Problem 2.6QAP

The current drown from the source is $5.478\text{mA}$.

Explanation of Solution

Write the expression for the equivalent resistance between the point 1 and 2.

$R_{12}=\frac{R_B{R}_C}{R_B+R_C}$ ....... (I)

Write the expression for the equivalent resistance between the point 2 and 3.

$R_{23}=\frac{R_D{R}_E}{R_D+R_E}$ ....... (II)

Write the expression for the total equivalent resistance.

$R=R_A+R_{12}+R_{23}$ ....... (III)

Write the expression for the current flowing through the circuit.

$I=\frac{V}{R}$ ....... (IV)

Here, $V$ is the voltage of the voltage source.

Substitute $3.0\text{kΩ}$ for $R_B$ and $4.0\text{kΩ}$ for $R_C$ in the Equation (I).

$\begin{array}{l}{R}_{12}=\frac{\left(3.0\text{kΩ}\right)\left(4.0\text{kΩ}\right)}{\left(3.0\text{kΩ}\right)+\left(4.0\text{kΩ}\right)}\\ R_{12}=1.714\text{kΩ}\end{array}$

Substitute $2.0\text{kΩ}$ for $R_D$ and $1.0\text{kΩ}$ for $R_E$ in the Equation (II).

$\begin{array}{l}{R}_{23}=\frac{\left(2.0\text{kΩ}\right)\left(1.0\text{kΩ}\right)}{\left(2.0\text{kΩ}\right)+\left(1.0\text{kΩ}\right)}\\ R_{23}=0.667\text{kΩ}\end{array}$

Substitute $1.714\text{kΩ}$ for $R_{12}$, $0.667\text{kΩ}$ for $R_{23}$ and $2.0\text{kΩ}$ for $R_A$ in the Equation (III).

$\begin{array}{l}R=\left(2.0\text{kΩ}\right)+\left(1.714\text{kΩ}\right)+\left(0.667\text{kΩ}\right)\\ R=4.381\text{kΩ}\end{array}$

Substitute $4.381\text{kΩ}$ for $R$ and $24\text{V}$ for the $V$ in the Equation (IV).

$\begin{array}{l}I=\frac{24\text{V}}{\left(4.381\text{kΩ}\right)\left(\frac{{10}^3\text{Ω}}{1\text{kΩ}}\right)}\\ I=\left(5.478\times {10}^{-3}\text{A}\right)\left(\frac{1\text{mA}}{{10}^{-3}\text{A}}\right)\\ I=5.478\text{mA}\end{array}$

Interpretation Introduction

(c)

Interpretation:

The voltage drop across the resistor $R_A$ is to be stated.

Concept introduction:

The resistive electrical circuit is the circuit which contains the voltage source and the resistor. The analysis of the circuit is done to evaluate the equivalent resistance, current through the circuit and power consumed by the circuit.

The voltage drop across the resistor is defined as $V=IR$, where $I$ is the current through the resistor and the $R$ is the resistance of the resistor.

Answer to Problem 2.6QAP

The voltage drop across the resistance $R_A$ is $10.956\text{V}$.

Explanation of Solution

Write the expression for the equivalent resistance between the point 1 and 2.

$R_{12}=\frac{R_B{R}_C}{R_B+R_C}$ ....... (I)

Write the expression for the equivalent resistance between the point 2 and 3.

$R_{23}=\frac{R_D{R}_E}{R_D+R_E}$ ....... (II)

Write the expression for the total equivalent resistance.

$R=R_A+R_{12}+R_{23}$ ....... (III)

Write the expression for the current flowing through the circuit.

$I=\frac{V}{R}$ ....... (IV)

Here, $V$ is the voltage of the voltage source.

Write the expression for the voltage drop across the resistor $R_A$.

$V_A=I{R}_A$ ....... (V)

Substitute $3.0\text{kΩ}$ for $R_B$ and $4.0\text{kΩ}$ for $R_C$ in the Equation (I).

$\begin{array}{l}{R}_{12}=\frac{\left(3.0\text{kΩ}\right)\left(4.0\text{kΩ}\right)}{\left(3.0\text{kΩ}\right)+\left(4.0\text{kΩ}\right)}\\ R_{12}=1.714\text{kΩ}\end{array}$

Substitute $2.0\text{kΩ}$ for $R_D$ and $1.0\text{kΩ}$ for $R_E$ in the Equation (II).

$\begin{array}{l}{R}_{23}=\frac{\left(2.0\text{kΩ}\right)\left(1.0\text{kΩ}\right)}{\left(2.0\text{kΩ}\right)+\left(1.0\text{kΩ}\right)}\\ R_{23}=0.667\text{kΩ}\end{array}$

Substitute $1.714\text{kΩ}$ for $R_{12}$, $0.667\text{kΩ}$ for $R_{23}$ and $2.0\text{kΩ}$ for $R_A$ in the Equation (III).

$\begin{array}{l}R=\left(2.0\text{kΩ}\right)+\left(1.714\text{kΩ}\right)+\left(0.667\text{kΩ}\right)\\ R=4.381\text{kΩ}\end{array}$

Substitute $4.381\text{kΩ}$ for $R$ and $24\text{V}$ for the $V$ in the Equation (IV).

$\begin{array}{l}I=\frac{24\text{V}}{\left(4.381\text{kΩ}\right)\left(\frac{{10}^3\text{Ω}}{1\text{kΩ}}\right)}\\ I=5.478\times {10}^{-3}\text{A}\end{array}$

Substitute $5.478\times {10}^{-3}\text{A}$ for $I$ and $2.0\text{kΩ}$ for $R_A$ in the Equation (V).

$\begin{array}{l}{V}_A=\left(5.478\times {10}^{-3}\text{A}\right)\left(2.0\text{kΩ}\right)\left(\frac{{10}^3\Omega }{1\text{kΩ}}\right)\\ V_A=10.956\text{V}\end{array}$

Interpretation Introduction

(d)

Interpretation:

The voltage drop across the resistance $R_D$ is to be stated.

Concept introduction:

The resistive electrical circuit is the circuit which contains the voltage source and the resistor. The analysis of the circuit is done to evaluate the equivalent resistance, current through the circuit and power consumed by the circuit.

The voltage drop across the resistor is defined as $V=IR$, where $I$ is the current through the resistor and the $R$ is the resistance of the resistor.

Answer to Problem 2.6QAP

The voltage drop across the resistor $R_D$ is $3.707\text{V}$.

Explanation of Solution

Write the expression for the equivalent resistance between the point 1 and 2.

$R_{12}=\frac{R_B{R}_C}{R_B+R_C}$ ....... (I)

Write the expression for the equivalent resistance between the point 2 and 3.

$R_{23}=\frac{R_D{R}_E}{R_D+R_E}$ ....... (II)

Write the expression for the total equivalent resistance.

$R=R_A+R_{12}+R_{23}$ ....... (III)

The voltage drop across the resistor $R_D$ is the voltage drop across the resistor $R_{23}$.

Write the expression for the voltage drop across the resistor $R_D$.

$V_D=\left(\frac{R_{23}}{R}\right)V$ ....... (IV)

Substitute $3.0\text{kΩ}$ for $R_B$ and $4.0\text{kΩ}$ for $R_C$ in the Equation (I).

$\begin{array}{l}{R}_{12}=\frac{\left(3.0\text{kΩ}\right)\left(4.0\text{kΩ}\right)}{\left(3.0\text{kΩ}\right)+\left(4.0\text{kΩ}\right)}\\ R_{12}=1.714\text{kΩ}\end{array}$

Substitute $2.0\text{kΩ}$ for $R_D$ and $1.0\text{kΩ}$ for $R_E$ in the Equation (II).

$\begin{array}{l}{R}_{23}=\frac{\left(2.0\text{kΩ}\right)\left(1.0\text{kΩ}\right)}{\left(2.0\text{kΩ}\right)+\left(1.0\text{kΩ}\right)}\\ R_{23}=0.667\text{kΩ}\end{array}$

Substitute $1.714\text{kΩ}$ for $R_{12}$, $0.667\text{kΩ}$ for $R_{23}$ and $2.0\text{kΩ}$ for $R_A$ in the Equation (III).

$\begin{array}{l}R=\left(2.0\text{kΩ}\right)+\left(1.714\text{kΩ}\right)+\left(0.667\text{kΩ}\right)\\ R=4.381\text{kΩ}\end{array}$

Substitute $0.667\text{kΩ}$ for $R_{23}$, $4.381\text{kΩ}$ for $R$ and $24\text{V}$ for $V$ in the Equation (IV).

$\begin{array}{l}{V}_D=\left(\frac{0.667\text{kΩ}}{4.318\text{kΩ}}\right)\left(24\text{V}\right)\\ V_D=3.707\text{V}\end{array}$

Interpretation Introduction

(e)

Interpretation:

The potential difference between the points 5 and 4.

Concept introduction:

The resistive electrical circuit is the circuit which contains the voltage source and the resistor. The analysis of the circuit is done to evaluate the equivalent resistance, current through the circuit and power consumed by the circuit.

The voltage drop across the resistor is defined as $V=IR$, where $I$ is the current through the resistor and the $R$ is the resistance of the resistor.

Answer to Problem 2.6QAP

The potential difference across the points 5 and 4 is $13.044\text{V}$.

Explanation of Solution

Write the expression for the equivalent resistance between the point 1 and 2.

$R_{12}=\frac{R_B{R}_C}{R_B+R_C}$ ....... (I)

Write the expression for the equivalent resistance between the point 2 and 3.

$R_{23}=\frac{R_D{R}_E}{R_D+R_E}$ ....... (II)

Write the expression for the total equivalent resistance.

$R=R_A+R_{12}+R_{23}$ ....... (III)

Write the expression for the current flowing through the circuit.

$I=\frac{V}{R}$ ....... (IV)

Here, $V$ is the voltage of the voltage source.

Write the expression for the voltage drop across the resistor $R_A$.

$V_A=I{R}_A$ ....... (V)

Write the expression for the potential difference across the points 5 and 4.

$V_{54}=V_{12}+V_{23}$ ....... (VI)

Write the expression for the total voltage.

$V=V_A+V_{12}+V_{23}$ ....... (VII)

Substitute $3.0\text{kΩ}$ for $R_B$ and $4.0\text{kΩ}$ for $R_C$ in the Equation (I).

$\begin{array}{l}{R}_{12}=\frac{\left(3.0\text{kΩ}\right)\left(4.0\text{kΩ}\right)}{\left(3.0\text{kΩ}\right)+\left(4.0\text{kΩ}\right)}\\ R_{12}=1.714\text{kΩ}\end{array}$

Substitute $2.0\text{kΩ}$ for $R_D$ and $1.0\text{kΩ}$ for $R_E$ in the Equation (II).

$\begin{array}{l}{R}_{23}=\frac{\left(2.0\text{kΩ}\right)\left(1.0\text{kΩ}\right)}{\left(2.0\text{kΩ}\right)+\left(1.0\text{kΩ}\right)}\\ R_{23}=0.667\text{kΩ}\end{array}$

Substitute $1.714\text{kΩ}$ for $R_{12}$, $0.667\text{kΩ}$ for $R_{23}$ and $2.0\text{kΩ}$ for $R_A$ in the Equation (III).

$\begin{array}{l}R=\left(2.0\text{kΩ}\right)+\left(1.714\text{kΩ}\right)+\left(0.667\text{kΩ}\right)\\ R=4.381\text{kΩ}\end{array}$

Substitute $4.381\text{kΩ}$ for $R$ and $24\text{V}$ for the $V$ in the Equation (IV).

$\begin{array}{l}I=\frac{24\text{V}}{\left(4.381\text{kΩ}\right)\left(\frac{{10}^3\text{Ω}}{1\text{kΩ}}\right)}\\ I=5.478\times {10}^{-3}\text{A}\end{array}$

Substitute $5.478\times {10}^{-3}\text{A}$ for $I$ and $2.0\text{kΩ}$ for $R_A$ in the Equation (V).

$\begin{array}{l}{V}_A=\left(5.478\times {10}^{-3}\text{A}\right)\left(2.0\text{kΩ}\right)\left(\frac{{10}^3\Omega }{1\text{kΩ}}\right)\\ V_A=10.956\text{V}\end{array}$

Substitute $V_{12}+V_{23}$ for $V_{54}$ in the Equation (VII).

$\begin{array}{l}V=V_A+V_{54}\\ V_{54}=V-V_A\end{array}$

Substitute $10.956\text{V}$ for $V_A$ and $24\text{V}$ for $V$.

$\begin{array}{l}{V}_{54}=\left(24\text{V}\right)-\left(10.956\text{V}\right)\\ V_{54}=13.044\text{V}\end{array}$