Chapter 19, Problem 85GP

(a)

To determine

The equivalent resistance of the given circuit.

Answer to Problem 85GP

The equivalent resistance of the circuit is $7.57\Omega$ .

Explanation of Solution

Given information:

Consider the given circuit.

Formula used:

Short the voltage source and open the current sources to determine the equivalent resistance of the circuit.

Calculation:

Draw the equivalent circuit.

In the circuit, $30\Omega$ and $18\Omega$ resistors are connected to common node. So, $30\Omega$ and $12\Omega$ resistors are connected in parallel and $4.5\Omega$ resistor is in series with the combination.Calculate the equivalent resistance of the circuit.

  $\begin{array}{c}{R}_{eq}=\left[\left(12\left|\right|30\right)+4.5\right]\left|\right|18\\ =\left[\frac{12\times 30}{12+30}+4.5\right]\left|\right|18\\ =\left(8.57+4.5\right)\left|\right|18\\ =13.07\left|\right|18\end{array}$

Further, simplify the equation.

  $\begin{array}{c}{R}_{eq}=\frac{13.07\times 18}{13.07+18}\\ =\frac{235.28}{31.07}\\ =7.57\Omega \end{array}$

Therefore, the equivalent resistance of the circuit is $7.57\Omega$ .

(b)

To determine

The current flowing through $18\Omega$ resistor.

Answer to Problem 85GP

The current flowing through $18\Omega$ resistor is $0.33\text{A}$ .

Explanation of Solution

Given information:

Consider the given circuit.

Formula used:

In the circuit, $30\Omega$ and $18\Omega$ resistors are connected to common node. So, the voltage across the branch of $18\Omega$ resistor is similar.

Write the expression to calculate the current using Ohm’s law.

  $V=I_{18}{R}_{18}$    … (1)

Where,

• $I_{18}$ is the current flowing through $18\Omega$resistor.

Calculation:

Substitute the values in equation (1).

  $\begin{array}{c}6=I_{18}\times 18\\ I_{18}=\frac{6}{18}\\ =0.33\text{A}\end{array}$

Therefore, the current flowing through $18\Omega$ resistor is $0.33\text{A}$ .

(c)

To determine

The current flowing through $12\Omega$ resistor.

Answer to Problem 85GP

The current flowing through $12\Omega$ resistor is $0.33\text{A}$ .

Explanation of Solution

Given information:

Consider the given circuit.

Formula used:

In the circuit, $30\Omega$ and $18\Omega$ resistors are connected to common node.

Calculation:

Calculate the equivalent resistance of $30\Omega$ , $18\Omega$ , and $4.5\Omega$ .

  $\begin{array}{c}{R}_1=\frac{30\times 12}{30+12}+4.5\\ =8.57+4.5\\ =13.07\Omega \end{array}$

Let the total current in the circuit is $I$ . Apply Ohm’s law to calculate the total current.

  $\begin{array}{c}I=\frac{6}{13.07}\\ =0.459\text{A}\end{array}$

Calculate the voltage across the node a using Ohm’s law.

  $V_a=I\times \left(12\left|\right|30\right)$

Substitute the values in the above expression.

  $\begin{array}{c}{V}_a=0.459\times \left(\frac{30\times 12}{30+12}\right)\\ =0.459\times 8.57\\ =3.93\text{V}\end{array}$

Since, $12\Omega$ resistor is in parallel across the node branch. Apply Ohm’s law to calculate the current flowing through $12\Omega$ resistor.

  $\begin{array}{c}{I}_{12}=\frac{3.93}{12}\\ =0.33\text{A}\end{array}$

Therefore, the current flowing through $12\Omega$ resistor is $0.33\text{A}$ .

(c)

To determine

The power dissipated in $4.5\Omega$ resistor.

Answer to Problem 85GP

The power dissipated in $4.5\Omega$ resistor is $0.948\text{W}$ .

Explanation of Solution

Given information:

From part (c), the current flowing through $4.5\Omega$ resistor is $0.459\text{A}$ .

Formula used:

Write the expression to calculate the power dissipated.

  $P_{4.5}=I^2{R}_{4.5}$

Calculation:

Substitute the values in the above expression.

  $\begin{array}{c}{P}_{4.5}={\left(0.459\right)}^2\times 4.5\\ =0.948\text{W}\end{array}$

Therefore, the power dissipated in $4.5\Omega$ resistor is $0.948\text{W}$ .