Chapter 13, Problem 4SP

(a)

To determine

The resistance of each of the two parallel connection of resistors.

Answer to Problem 4SP

The resistance of the first network is $1.5\Omega$ and $1\Omega$

Explanation of Solution

Given info: First network has two resistors of $3\Omega$ each , second network has three resistors of $3\Omega$ each.

Write the equation to find the equivalent resistance of the first network.

$R_{eq1}=\frac{R_1{R}_2}{R_1+R_2}$

Here,

$R_{eq1}$ is the equivalent resistance of first network.

$R_1$ is the first resistance

$R_2$ is the second resistance

Substitute $3\Omega$ for $R_1\text{and}{R}_2$ to get $R_{eq1}$

$\begin{array}{c}{R}_{eq1}=\frac{3\Omega \times 3\Omega }{3\Omega +3\Omega }\\ =3\Omega \end{array}$

Write the equation to find the equivalent resistance of the second network.

$R_{eq2}=\frac{R_1{R}_2{R}_3}{R_1{R}_2+R_2{R}_3+R_1{R}_3}$

Here,

$R_{eq2}$ is the equivalent resistance of second network.

$R_1$ is the first resistance

$R_2$ is the second resistance

$R_3$ is the third resistance

Substitute $3\Omega$ for $R_1\text{and}{R}_2$ and $R_3$ to get $R_{eq2}$

$\begin{array}{c}{R}_{eq2}=\frac{\left(3\Omega \right)\left(3\Omega \right)\left(3\Omega \right)}{\left(3\Omega \times 3\Omega \right)+\left(3\Omega \times 3\Omega \right)+\left(3\Omega \times 3\Omega \right)}\\ =1\Omega \end{array}$

Conclusion:

Therefore, the resistance of the first network is $1.5\Omega$ and $1\Omega$

(b)

To determine

The total equivalent resistance between A and B.

Answer to Problem 4SP

The total equivalent resistance is $5.5\Omega$

Explanation of Solution

The parallel connection of first network and the second network is in series with the single resistance connected in the middle. Therefore the total equivalent resistance will be the sum of them.

$R_{tot}=R_{eq1}+R_{eq2}+R_s$

Here,

$R_{tot}$ is the total resistance

$R_{eq1}$ is the equivalent resistance of the first network

$R_{eq2}$ is the equivalent resistance of second network

$R_s$ is the resistance in series

Substitute $1\Omega$ for $R_{eq2}$ , $3\Omega$ for $R_{eq1}$ and $3\Omega$ for $R_s$ in equation to find $R_{tot}$

$\begin{array}{c}{R}_{tot}=1\Omega +3\Omega +3\Omega \\ =7\Omega \end{array}$

Conclusion:

Therefore, The total equivalent resistance is $7\Omega$

(c)

To determine

The current flowing through the entire connection.

Answer to Problem 4SP

The current flowing is $2.14\text{A}$

Explanation of Solution

Given info: The voltage applied is $15\text{V}$

Write the equation to find the current through the entire circuit

$I=\frac{V}{R}$

Here,

$I$ is the current

$V$ is the voltage

$R$ is the resistance

Substitute $15\text{V}$ for $V$ and $7\Omega$ for $R$ to get $I$

$\begin{array}{c}I=\frac{15\text{V}}{7\Omega }\\ =2.14\text{A}\end{array}$

Conclusion:

Therefore, the current flowing through the entire circuit is $2.14\text{A}$

(d)

To determine

Current flowing through each of the three resistor parallel network.

Answer to Problem 4SP

The current is $5\text{A}$

Explanation of Solution

Write the equation to find the current.

$I=\frac{V}{R}$

Here,

$I$ is the current

$V$ is the voltage

$R$ is the resistance

The equivalent resistance of the three resistor network is $1\Omega$

Substitute $15\text{V}$ for $V$ and $1\Omega$ for $R$ to get $I$

$\begin{array}{c}I=\frac{15\text{V}}{1\Omega }\\ =15\text{A}\end{array}$

Since it is a parallel connection the same current flows through each resistor.

Therefore, current through each resistor can be found by dividing the current by three.

$\begin{array}{c}I=\frac{15\text{A}}{3}\\ =5\text{A}\end{array}$

Conclusion:

Therefore, the current is $5\text{A}$