Chapter 17, Problem 89E

(a)

To determine

The current $I_2$ in the given circuit.

Answer to Problem 89E

The current $I_2$ in the given circuit is $\underset{\_}{1\text{A}}$.

Explanation of Solution

The circuit is given in Figure 17.41.

Apply Kirchhoff’s current rule in the junction where the three currents $I_1$, $I_2$, and $I_3$ meet. In the junction the current $I_1$ is entering the junction and $I_2$, and $I_3$ are leaving the junction. Thus, according to Kirchhoff’s current rule it can be written that,

  $I_1=I_2+I_3$        (I)

Solve equation (I) for $I_2$.

  $I_2=I_1-I_3$        (II)

Conclusion:

Substitute $4\text{A}$ for $I_1$, $3\text{A}$ for $I_3$ in equation (II) to find $I_2$.

  $\begin{array}{c}{I}_2=4\text{A}-3\text{A}\\ =1\text{A}\end{array}$

Therefore, the current $I_2$ in the given circuit is $\underset{\_}{1\text{A}}$.

(b)

To determine

The resistance $r$ in the given circuit.

Answer to Problem 89E

The resistance $r$ in the given circuit is $\underset{\_}{10\text{ohm}}$.

Explanation of Solution

The circuit is given in Figure 17.41. Apply Kirchhoff’s voltage rule in the loop involving the resistors $5\text{ohm}$ and $r$.

  $\left(10\text{V}\right)+I_{2}r+\left(5\text{ohm}\right)\left(4\text{A}\right)-\left(40\text{V}\right)=0$        (III)

Solve equation (III) for $r$.

  $r=\frac{\left(40\text{V}\right)-\left(10\text{V}\right)-\left(5\text{ohm}\right)\left(4\text{A}\right)}{I_2}$        (IV)

Conclusion:

Substitute $1\text{A}$ for $I_2$ in equation (IV) to find $r$.

  $\begin{array}{c}r=\frac{\left(40\text{V}\right)-\left(10\text{V}\right)-\left(5\text{ohm}\right)\left(4\text{A}\right)}{1\text{A}}\\ =10\text{ohm}\end{array}$

Therefore, the resistance $r$ in the given circuit is $\underset{\_}{10\text{ohm}}$.

(c)

To determine

The EMF $\epsilon$ in the given circuit.

Answer to Problem 89E

The EMF $\epsilon$ in the given circuit is $\underset{\_}{14\text{V}}$.

Explanation of Solution

The circuit is given in Figure 17.41. Apply Kirchhoff’s voltage rule in the loop involving the resistors $2\text{ohm}$ and $r$.

  $\epsilon +\left(2\text{ohm}\right)\left(3\text{A}\right)-I_{2}r-\left(10\text{V}\right)=0$        (V)

Solve equation (V) for $\epsilon$.

  $\epsilon =\left(10\text{V}\right)-\left(2\text{ohm}\right)\left(3\text{A}\right)+I_{2}r$        (VI)

Conclusion:

Substitute $1\text{A}$ for $I_2$, and $10\text{ohm}$ for $r$ in equation (VI) to find $\epsilon$.

  $\begin{array}{c}\epsilon =\left(10\text{V}\right)-\left(2\text{ohm}\right)\left(3\text{A}\right)+\left(1\text{A}\right)\left(10\text{ohm}\right)\\ =20\text{V}-6\text{V}\\ =14\text{V}\end{array}$

Therefore, the EMF $\epsilon$ in the given circuit is $\underset{\_}{14\text{V}}$.