Chapter 19, Problem 32P

(a)

To determine

The magnitude of the current $I_1$ , $I_2$ , and $I_3$ in the circuit.

Answer to Problem 32P

The magnitude of the current is $I_1=-0.7687\text{A}$ , $I_2=-0.7141\text{A}$ , and $I_3=-0.0545\text{A}$ .

Explanation of Solution

Given information:

Consider the given circuit.

The value of the internal resistance is $1\Omega$ .

Formula used:

Apply Kirchhoff’s voltage law in the circuit to determine the currents.

Calculation:

Apply Kirchhoff’s voltage law in the upper loop.

  $\begin{array}{c}-8I_1-1\cdot I_1-12I_1-10I_2-1\cdot I_2-12-12=0\\ -21I_1-11I_2-24=0\\ 21I_1+11I_2=-24\end{array}$  … (1)

Apply Kirchhoff’s voltage law in the lower loop.

  $\begin{array}{c}-15I_3+10I_2+1\cdot I_2-18I_3-I_3-6+12=0\\ 11I_2-34I_3+6=0\\ 11I_2-34I_3=-6\end{array}$  … (2)

At node (a), apply Kirchhoff’s current law.

  $I_1=I_2+I_3$    … (3)

Substitute the value of $I_3$ from equation (3) to equation (2).

  $\begin{array}{c}11I_2-34\left(I_1-I_2\right)=-6\\ -34I_1+45I_2=-6\end{array}$    … (4)

On solving equations (1) and (4),

  $\begin{array}{c}{I}_1=-0.7687\text{A}\\ I_2=-0.7141\text{A}\end{array}$

Substitute the values in equation (3).

  $\begin{array}{c}{I}_3=-0.7687-\left(-0.7141\right)\\ =-0.0545\text{A}\end{array}$

Therefore, themagnitude of the current is $I_1=-0.7687\text{A}$ , $I_2=-0.7141\text{A}$ , and $I_3=-0.0545\text{A}$ .

(b)

To determine

The terminal voltage of the 6-volt battery.

Answer to Problem 32P

The terminal voltage of the 6-volt battery is $5.94$ volt.

Explanation of Solution

Given information:

Consider the given circuit.

Formula used:

From part (a), the magnitude of the current is $I_1=-0.7687\text{A}$ , $I_2=-0.7141\text{A}$ , and $I_3=-0.0545\text{A}$ .

Calculate the terminal voltage of 6-volt battery.

  $\begin{array}{c}V=6-1\cdot \left(0.0545\right)\\ =5.94\text{V}\end{array}$

Therefore, the terminal voltage of the 6-volt battery is $5.94$ volt.