Chapter 19, Problem 29P

To determine

The magnitudes and direction of current across each resistor in the given circuit.

Answer to Problem 29P

The magnitudes and direction of current across each resistor in the given circuitare $I_1=0.1335\text{A}$ , $I_2=0.3145\text{A}$ , and $I_3=0.1810\text{A}$ .

Explanation of Solution

Given information:

Consider the given circuit.

Formula used:

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

Calculation:

Apply Kirchhoff’s voltage law in the upper loop.

  $\begin{array}{c}-25I_1-18I_2+9=0\\ 25I_1+18I_2=-9\end{array}$    … (1)

Apply Kirchhoff’s voltage law in the lower loop.

  $\begin{array}{c}18I_2+35I_3-12=0\\ 18I_2+35I_3=12\end{array}$    … (2)

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

  $I_2=I_1+I_3$    … (3)

Substitute the value from equation (3) to equation (1).

  $\begin{array}{c}18I_2+35\left(I_2-I_1\right)=12\\ 18I_2+35I_2-35I_1=12\\ -35I_1+53I_2=12\end{array}$    … (4)

On solving equation (1) and (4),

  $\begin{array}{l}{I}_1=0.1335\text{A}\\ I_2=0.3145\text{A}\end{array}$

Substitute the values in equation (3).

  $\begin{array}{c}0.3145=0.1335+I_3\\ I_3=0.1810\text{A}\end{array}$

Therefore, themagnitudes and direction of current across each resistor in the given circuit are $I_1=0.1335\text{A}$ , $I_2=0.3145\text{A}$ , and $I_3=0.1810\text{A}$ .