Current Dividers

2 in Parallel §

Intuitively there’s twice as much current going to go through $R_1$ than $R_2$ because of half the resistance

The current has to add up to 6A, so the current flowing through $R_1$ has to be 4A because the current flowing through $R_2$ has to be half of $R_1$ and add up to 6A

4A + 2A = 6A

There’s a mathematical way to do this

$I_1=I_T[\frac{R_2}{R_1+R_2}]=6A[\frac{8}{4+8}]=4A$

$I_2=I_T[\frac{R_1}{R_1+R_2}]=6A[\frac{4}{4+8}]=2A$

3+ in Parallel §

$I_1=I_T[\frac{\frac{1}{R_1}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}]$