# Solve Using Current Division Rule

Find current of resistors, use the current division rule.

Suppose that $R_1=2 \Omega$ , $R_2=4 \Omega$ , $R_3=1 \Omega$ , $I_S=5 A$ and $V_S=4 V$

Solution:
$R_2$ and $R_3$ are parallel. The current of $I_S$ is passing through them and it is actually divided between them. The branch with lower resistance has higher current because electrons can pass through that easier than the other branch. Using the current division rule, we get

$I_{R_2}=\frac{R_3}{R_2+R_3} \times I_S= \frac{1}{1+4} \times 5= 1 A$
$I_{R_3}=\frac{R_2}{R_2+R_3} \times I_S= \frac{4}{1+4} \times 5= 4 A$ .
Note that $I_{R_2} because $R_2>R_3$ .

What is the direction for $I_{R_2}$ and $I_{R_3}$ ? It is with the same direction as $I_S$ :

We have found $I_{R_2}$ and $I_{R_3}$ , now we need to find $I_{R_1}$ .

If you look at the circuit carefully, $R_1$ , $V_S$ and $I_S$ are all in series. The bottom node of $I_S$ is connected to the bottom node of $V_S$ and there is no other component connected there. And the other node of $V_S$ is connected to $R_1$ , again no other component here. This means that all current of $I_S$ should pass through $V_S$ and the same current must go through $R_1$ . So, the current of $R_1$ can be easily found:
$I_{R_1}=I_S= 5 A$ .

The direction is the same current as the current source direction:

Now, tell me what is the voltage drop across the current source?

## 3 thoughts on “Solve Using Current Division Rule”

$+V_{I_S}-R_3 \times I_{R_3}-R_1 \times I_{R_1}-V_S=0$
$V_{I_S}=R_3 \times I_{R_3}+R_1 \times I_{R_1}+V_S$
$V_{I_S}=1 \times 4+2 \times 5 +4=18 V$
If you label the voltage terminals of the current source the other way, you would get $-18 V$ .