Series and parallel pairs turn up so often that both have a shortcut worth knowing cold. They look almost identical on the page, and one of them crosses over. That crossing is this lesson.

The voltage divider: two resistors in series across a supply, and you want the voltage at the point between them. $V_{out} = V_{in} \dfrac{R_2}{R_1 + R_2}$, read aloud V-out equals V-in times R-two over R-one plus R-two. Fix the subscript convention now, because everything depends on it: $R_1$ is the upper resistor, from the supply down to the tap; $R_2$ is the lower one, from the tap down to ground; $V_{in}$ is the supply in volts and $V_{out}$ is the voltage at the tap, measured to ground. The same current runs through both, so the volts split in proportion to the ohms — and the resistor you measure across is the one on top of the fraction. One condition: it holds only while whatever you connect to the tap draws no appreciable current.

The current divider: two resistors in parallel, and you want the current in one branch. $I_1 = I_t \dfrac{R_2}{R_1 + R_2}$ — I-one equals I-total times R-two over R-one plus R-two — where $I_t$ is the total current arriving at the node in amperes, $I_1$ is the current in branch 1, and $R_1$ and $R_2$ are the two branch resistances in ohms. Look at the numerator: to find the current in branch 1 you put branch 2's resistance on top. The OTHER branch's resistance.

That crossing is the marquee trap of the chapter, and it is also completely sensible. Current takes the easy road, so a branch's share is set by how hard the OTHER road is. If the two look the same to you on the page, test them at an extreme: make $R_1$ enormous, and branch 1 should carry almost nothing. Only the crossed version does that.