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I apologize if this comes down to a lack of fundamental understanding of electrical flow, which it probably does. When I look at this:

I don't understand why Z2 is factored into Vout - as far as I can tell the current should branch out before getting to Z2. Does current go straight from input to ground, and then bound back and spread out?

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Think about the extreme case where Z2 = 0 (Vout shorted to ground) and you can see that Z2 matters. – Spehro Pefhany Jun 25 '14 at 10:51

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 – 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some of the current through \$R_1\$ 'branches out' through that path, the voltage divider equations derived above are no longer valid.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.

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Your reasoning is a problem with voltage dividers, but not for the reason you believe.

If we assume that the circuit connected to Vout has infinite impedance then all of the current from Vin flows through Z1 and consequently Z2. As the current flows through each of the resistors a voltage difference develops across them, proportional to the ratio of their resistances to the total (Ohm's Law). This voltage drop exists whether we measure it or not, and it holds Vout at a certain voltage above ground. This is the basic operation of a voltage divider.

The problem you almost touch upon is that Vout does not have infinite impedance. Some current does pass out of the node into the connected circuit, and this reduces the voltage drop across Z2. This is why voltage dividers cannot be used as voltage regulators for anything more complex than a FET or diode.

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A good example is the resistive voltage divider used to bias a transistor amplifier. This is a classical example for a loaded voltage divider. It is common practice to use a resistor niveau as low as possible (because of the input resistance of the whole circuit). The reason is as follows: The load (input impedance of the BJT at the base node, and in particular, the associated tolerances) should not have too much influence on the division ratio. – LvW Jun 25 '14 at 6:53

Using the water analogy, imagine you have 50 feet of half-inch diameter hose hooked to a hose bib on one end, that the other end is capped, and that there's a pressure gauge at the hose bib, another at the capped end of the hose, and another in the middle of the hose.

Now imagine that the hose bib is opened and that the gauge at the hose bib reads 50 PSI.

Since the other end of the hose is capped, there can be no water flow through the hose, and the other gauges will also read 50 PSI.

Likewise, if I connect two resistors in series, connect one end of the string to a 50 volt power supply and let the other end float, voltmeters connected to the power supply end of the string, to the junction of the resistors, and to the floating end of the string will all read 50 volts because there's no return to the supply and, therefore, no flow of charge through the string.

Now, uncap the hose.

Water will flow through it and since there's nothing holding the water back at the uncapped end, that gauge will read 0 PSI.

Assuming that there are no losses behind the bib, the pressure at the bib will stay at 50 PSI and then, with the other end of the hose at 0 PSI the pressure gauge at the middle of the hose must read 25 PSI.

With two equal valued resistors in the string, then, when the floating end of the string is returned to the supply charge will flow through the string.

Then, assuming no voltage drop at the loaded output of the supply, the voltmeter at that end of the string will read 50 volts, the one at the return end of the string will read zero volts, and the one at the junction of the resistors will read 25 volts.

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Beautiful example. – James M. Lay Sep 18 '15 at 5:22

First of all, you must realise that there is no bouncing back happening. Current due to DC voltages are always unidirectional and in the circuit you have given current flows from

$$V_{in} \rightarrow Z_1 \rightarrow Z_2 \rightarrow GND$$

Here the branch in \$V_{out}\$ is open (meaning infinite impedence). So as the circuit path is not complete in that way ideally no current will branch out.

This is exactly why voltage measuring devices (voltmeter have high impedences). Whenever you connect a high resistance device (general characteristics of loads) between \$V_{out}\$ and \$ground\$ the current sees a lower resistance path via \$Z_2\$ and chooses it.

Of course, some current also leaks out to load causing potential drop. To calculate the drop you must consider an equivalent resistance to \$Z_2\$ and \$Z_{load}\$ (load resistance) in parallel and replace the \$Z_2\$ in above circuit with the equivalent resistance calculated. Now you have a simple potential divider where \$V_{out}\$ can be calculated as

$$V_{out} = V_{in} \frac{Z_2}{Z_1 + Z_2}$$

Surely, \$V_{out}\$ without load will be greater than load with some finite resistance.

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