Chapter 8, Problem 1P

For the network of Fig. 8.103:

a. Find the currents I₁ and I₂.

b. Determine the voltage V_s.

Fig. 8.103

To determine

(a)

Currents $I_1$ and $I_2$

Answer to Problem 1P

  $\begin{array}{l}{I}_1=4.8A\\ I_2=1.2A\end{array}$

Explanation of Solution

Given:

The given electric network is:

  

Calculation:

Use current divider rule:

  $\begin{array}{l}{I}_1=6A\left(\frac{R_2}{R_1+R_2}\right)\\ I_1=6A\left(\frac{8\Omega }{8\Omega +2\Omega }\right)\\ I_1=4.8A\end{array}$

Apply KCL:

  $\begin{array}{l}{I}_2=6-I_1\\ I_2=6-4.8\\ I_2=1.2A\end{array}$

To determine

(b)

Supply voltage $V_S$

Answer to Problem 1P

  $V_S=9.6\text{V}$

Explanation of Solution

Given:

The given electric network is:

  

Calculation:

As we can see, the resistors and current source are connected in parallel. Therefore, the voltage across the source is:

  $\begin{array}{l}{V}_S=I_1{R}_1\\ V_S=\left(4.8\text{A}\right)\left(2\Omega \right)\\ V_S=9.6\text{V}\end{array}$