How exactly does a series voltage divider work?
How exactly does a series voltage divider work?


I'm afraid that the circuit, as drawn, is confusing to those just learning the fundamentals.  Although it's not apparent, the two impedances are series connected which means that all of the current through \$Z_1\$ is through \$Z_2\$ - there is no 'branching out' before getting to \$Z_2\$.  This is crucial to understanding the voltage divider equation.

Let's redraw the circuit to emphasize the series connection:







simulate this circuit &ndash; Schematic created using CircuitLab

Clearly, this is a series connected circuit and the series current is just

$$I = \frac{V_{in}}{R_1 + R_2}$$

By Ohm's law, the voltage across a resistor is product of the current through and the resistance.  Thus, the voltage across each resistor is given by

$$V_{R1} = I R_1 =  V_{in}\frac{R_1}{R_1 + R_2}$$

$$V_{R2} = I R_2 =  V_{in}\frac{R_2}{R_1 + R_2}$$

This is voltage division and this result crucially depends on the fact that the resistors are series connected so that they have identical current through.

Now, if the output of our circuit is the voltage across \$R_2\$, we might label the node where the two resistors connect as \$V_{out}\$ as in the schematic in your question.

And, we might not draw the input voltage source explicitly as in your schematic.

One last thing, if we attach another circuit in parallel with \$R_2\$ such that some of the current through \$R_1\$ 'branches out' through that path, the voltage divider equations derived above are no longer valid.

However, we could account for this external circuit by replacing \$R_2\$ in the equations above with \$R_{EQ}\$ where

$$R_{EQ} = R_2 || R_L$$

and \$R_L\$ is the equivalent resistance of the external circuit.