Chapter 19, Problem 81GP

(a)

To determine

The resistance of the resistor $R_2$ in the given circuit using.

Answer to Problem 81GP

The resistance of the resistor $R_2$ is $3.33\Omega$ .

Explanation of Solution

Given info:

Consider the given circuit.

  

In the circuit, the current flowing through resistor $R_1$ is $I$ .

Formula Used:

Apply Kirchhoff’s voltage law in the inner loop.

  $-12+10I+3=0$......... (1)

Calculation:

Solve equation (1).

  $\begin{array}{c}-12+10I+3=0\\ I=\frac{12-3}{10}\\ =0.9\text{A}\end{array}$

Apply Ohm’s law to calculate the value of the resistor $R_2$ .

  $V_T=I{R}_2$

Substitute the values in the above expression.

  $\begin{array}{c}3=0.9\times R_2\\ R_2=\frac{3}{0.9}\\ =3.33\Omega \end{array}$

Conclusion:

Therefore, the resistance of the resistor $R_2$ is $3.33\Omega$ .

(b)

To determine

The terminal voltage $V_T$ if a load with resistance $7\Omega$ is connected across 3 volt terminal.

Answer to Problem 81GP

The terminal voltage $V_T$ is $2.2\text{V}$ .

Explanation of Solution

Given info:

Consider the given circuit.

  

In the circuit, the load $R_3=7\Omega$ is connected across the 3-volt terminal.

Formula Used:

Write the expression to calculate the equivalent resistance.

  $R_{eq}=\frac{R_2\times R_3}{R_2+R_3}$......... (1)

Where,

• $R_{eq}$ is the equivalent resistance.

Calculation:

Substitute the values in equation (1).

  $\begin{array}{c}{R}_{eq}=\frac{3.33\times 7}{3.33+7}\\ =2.25\Omega \end{array}$

Apply Kirchhoff’s voltage law in the inner loop to determine the current.

  $-12+I{R}_1+I{R}_{eq}=0$

Substitute the values in the above expression.

  $\begin{array}{c}I=\frac{12}{10+2.25}\\ =0.97\text{A}\end{array}$

Apply Ohm’s law to calculate the terminal voltage across the load.

  $\begin{array}{c}{V}_T=0.97\times 2.25\\ =2.2\text{V}\end{array}$

Conclusion:

Therefore, the terminal voltage $V_T$ is $2.2\text{V}$ .