Chapter 13, Problem 1SP

(a)

To determine

The equivalent resistance of the circuit.

Answer to Problem 1SP

The equivalent resistance is $4\Omega$

Explanation of Solution

Given info: The resistances are $6\Omega$ and $12\Omega$

Write the equation to find the equivalent resistance.

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

Here,

$R_{eq}$ is the equivalent resistance

$R_1$ is the first resistance

$R_2$ is the second resistance

Substitute $6\Omega$ $R_1$ and $12\Omega$ for $R_2$ to get $R_{eq}$

$\begin{array}{c}{R}_{eq}=\frac{6\Omega \times 12\Omega }{6\Omega +12\Omega }\\ =4\Omega \end{array}$

Conclusion:

Therefore, the equivalent resistance is $4\Omega$

(b)

To determine

The total current flowing through the circuit.

Answer to Problem 1SP

The total current is $0.125\text{A}$

Explanation of Solution

Here the equivalent resistance of $R_1$ and $R_2$ is in series with $R_3$

Write the equation to find the equivalent resistance of the circuit.

$R=R_{eq}+R_3$

Here,

$R$ is the equivalent resistance of the whole circuit

$R_{eq}$ is the equivalent resistance of parallel resistors

$R_3$ is the third resistance

Substitute $8\Omega$ for $R_3$ and $4\Omega$ for $R_{eq}$ to get $R$

$\begin{array}{c}R=8\Omega +4\Omega \\ =12\Omega \end{array}$

Write the equation to find the current.

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

Here,

$I$ is the current

$V$ is the voltage

$R$ is the resistance

Substitute $1.5\text{V}$ for $V$ and $12\Omega$ for $R$ to get $I$

$\begin{array}{c}I=\frac{1.5\text{V}}{12\Omega }\\ =0.125\text{A}\end{array}$

Conclusion:

Therefore, the current is $0.125\text{A}$

(c)

To determine

Current flowing through $6\Omega$ resistor.

Answer to Problem 1SP

Current is $0.0833\text{A}$

Explanation of Solution

Write the equation to find current.

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

Here,

$I$ is the current

$V$ is the voltage

$R$ is the resistance

Substitute $1.5\text{V}$ for $V$ and $6\Omega$ for $R$ to get $I$

$\begin{array}{c}I=\frac{1.5\text{V}}{6\Omega }\\ =0.0833\text{A}\end{array}$

Conclusion:

Therefore, the current flowing is $0.0833\text{A}$

(d)

To determine

Power dissipated in $8\Omega$ resistor.

Answer to Problem 1SP

Power dissipated is $125\text{mW}$

Explanation of Solution

Write the equation to find the current through $8\Omega$ resistor

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

Here,

$I$ is the current

$V$ is the voltage

$R$ is the resistance

Substitute $1.5\text{V}$ for $V$ and $8\Omega$ for $R$ to get $I$

$\begin{array}{c}I=\frac{1.5\text{V}}{8\Omega }\\ =0.1875\text{A}\end{array}$

Write the equation to find the power.

$P=I^{2}R$

Here,

$P$ is the power

$I$ is the current

$R$ is  the resistance

Substitute $0.1875\text{A}$ for $I$ and $8\Omega$ for $R$ to get $P$

$\begin{array}{c}P={\left(0.1875\text{A}\right)}^2\times 8\Omega \\ =125\text{mW}\end{array}$

Conclusion:

Therefore, the power consumed is $125\text{mW}$

(e)

To determine

Is the current through eight ohm resistor greater than six ohm resistor?

Answer to Problem 1SP

Yes, it is greater.

Explanation of Solution

The current through eight ohm resistor is greater than the current flowing through six ohm resistor since the eight ohm resistor is connected serially and the six ohm resistor is connected parallel.

Conclusion:

Yes it is greater.