Chapter 13, Problem 12E

(a)

To determine

The equivalent resistance of the combination.

Answer to Problem 12E

The equivalent resistance of the combination is $5\Omega$

Explanation of Solution

Given info: Resistances connected are three number of $15\Omega$ each.

Write the equation to find the equivalent resistance of resistors connected in parallel.

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

Here,

$R_{eq}$ is the equivalent resistance

$R_1$ is the first resistance

$R_2$ is the second resistance

$R_3$ is the third resistance

Substitute $15\Omega$ for $R_1$ , $15\Omega$ for $R_2$ and $15\Omega$ for $R_3$ in equation to get $R_{eq}$

$\begin{array}{c}{R}_{eq}=\frac{R_1{R}_2{R}_3}{R_2{R}_3+R_1{R}_3+R_1{R}_2}\\ =\frac{15\Omega \times 15\Omega \times 15\Omega }{15\Omega \times 15\Omega +15\Omega \times 15\Omega +15\Omega \times 15\Omega }\\ =5\Omega \end{array}$

Conclusion:

Therefore, the equivalent resistance of the combination is $5\Omega$

(b)

To determine

Find the total current through the circuit.

Answer to Problem 12E

The total current flowing through the circuit is $2\text{A}$

Explanation of Solution

Write the equation to find the current.

$I=\frac{V}{R_{eq}}$

Here,

$I$ is the current

$V$ is the voltage

$R_{eq}$ is the equivalent resistance

Substitute $5\Omega$ for $R_{eq}$ and $10\text{V}$ for $V$ in equation to get $I$

$\begin{array}{c}I=\frac{10\text{V}}{5\Omega }\\ =2\text{A}\end{array}$

Conclusion:

Therefore, the current flowing is $2\text{A}$

(c)

To determine

Find the current flowing through each resistor.

Answer to Problem 12E

A current of $0.67\text{A}$ flows through each of the resistor.

Explanation of Solution

Write the equation to find the current through first resistor.

$I_1=\frac{V}{R_1}$

Here,

$I_1$ is the current flowing through first resistor

$V$ is the voltage

$R_1$ is the first resistance

Substitute $10\text{V}$ for $V$ and $15\Omega$ for $R_1$ to get $I_1$

$\begin{array}{c}{I}_1=\frac{10\text{V}}{15\Omega }\\ =0.66\text{A}\end{array}$

Similarly write the equation to find the current through second resistor.

$I_2=\frac{V}{R_2}$

Here,

$I_2$ is the current flowing through second resistor

$V$ is the voltage

$R_2$ is the second resistance

Substitute $10\text{V}$ for $V$ and $15\Omega$ for $R_2$ to get $I_2$

$\begin{array}{c}{I}_2=\frac{10\text{V}}{15\Omega }\\ =0.66\text{A}\end{array}$

Write the equation to find the current through third resistor.

$I_3=\frac{V}{R_3}$

Here,

$I_3$ is the current flowing through third resistor

$V$ is the voltage

$R_3$ is the third resistance

Substitute $10\text{V}$ for $V$ and $15\Omega$ for $R_3$ to get $I_3$

$\begin{array}{c}{I}_3=\frac{10\text{V}}{15\Omega }\\ =0.67\text{A}\end{array}$

Conclusion:

Therefore, a current of $0.67\text{A}$ flows through all the three resistors.