Which values for voltage divider
The best way to see the differences is to use the Thevenin equivalent for a resistor divider set up between two ideal (no source resistance of their own) voltage sources. Often, this is just some supply voltage and ground.

Let's look at the obvious case:







simulate this circuit &ndash; Schematic created using CircuitLab

The left side has a resistor divider between an ideal voltage source and ground and, without any load hanging off of \$V_\text{OUT}\$ (it's just open, as you can see), the voltage is easy to compute as \$V_\text{OUT}=V_\text{IN}\cdot\frac{R_2}{R_1+R_2}\$. However, what's missing from that simple calculation is the fact that \$V_\text{OUT}\$ is no longer ideal. It now has a source resistance that makes it non-ideal. That's because any current required by a load (currently not present) attached between \$V_\text{OUT}\$ and ground must cause an additional voltage drop across \$R_1\$ and that changes the voltage that the load experiences. So, again, \$V_\text{OUT}\$ is no longer ideal.

The effective, non-ideality of \$V_\text{OUT}\$ is expressed by first setting up a fictional \$V_\text{TH}\$ which is equal to the unloaded \$V_\text{OUT}\$ and then inserting a series resistor between this fictional \$V_\text{TH}\$ and \$V_\text{OUT}\$. This is shown on the right side, above. This resistor that represents the non-ideality of the voltage source is \$R_\text{TH}=\frac{R_1\cdot R_2}{R_1+R_2}\$.

The upshot of all this is that you now have a simpler way to view the resistor divider and you can easily see exactly how non-ideal it is by simply examining the value of \$R_\text{TH}\$. The closer this value is to zero, the more ideal is the voltage source. But the price you pay for getting closer to zero is a rapidly increasing power dissipation wasted in the resistor divider, itself.



Just to completely generalize the above, let's look at a resistor divider that sits between two different ideal voltage sources, where one is NOT zero volts. (That's just an arbitrary reference point, anyway.)







simulate this circuit

The only difference here is that now both voltages can be non-zero. In this case, the only new computation is the more general version: \$V_\text{TH}=\frac{V_\text{B}\cdot R_1+V_\text{A}\cdot R_2}{R_1+R_2}\$. That reduces to the equation I gave earlier, above, when \$V_\text{B}=0\:\text{V}\$.



The choice of resistor values will depend on the range of load impedances you want to allow attached to \$V_\text{OUT}\$ and how much voltage variation your loads can tolerate.

For example, suppose you have a power supply rail of \$5\:\text{V}\$ and want to use a voltage divider to create a voltage source at \$3.3\:\text{V}\$. Suppose also that the maximum current required by the device you'll attach to \$V_\text{OUT}\$ is \$10\:\text{mA}\$. Suppose that the device must not experience more than \$3.6\:\text{V}\$ nor less than \$3.1\:\text{V}\$ or else it won't work properly. And finally that the worst-case minimum current required by the device is \$100\:\mu\text{A}\$.

Given these specifications, we want a worst-case \$\Delta V=3.6\:\text{V}-3.1\:\text{V}=500\:\text{mV}\$ with a worst case current variation of \$\Delta I=10\:\text{mA}-100\:\mu\text{A}=9.9\:\text{mA}\$. This suggests an effective source impedance of \$R_\text{TH}=R_\text{SRC}=\frac{500\:\text{mV}}{9.9\:\text{mA}}\approx 50.5\:\Omega\$.

You now have two equations and two unknowns:

$$\begin{align*}
50.5\:\Omega &amp;= \frac{R_1\cdot R_2}{R_1+R_2}\\\\
5\:\text{V}\cdot\frac{R_2}{R_1+R_2} &amp;=3.6\:\text{V}+100\:\mu\text{A}\cdot 50.5\:\Omega
\end{align*}$$

Roughly speaking, you'd need \$R_1\approx 70\:\Omega\$ and \$R_2\approx 181\:\Omega\$. Note that just operating this divider requires \$\frac{\left(5\:\text{V}\right)^2}{70\:\Omega+181\:\Omega}\approx 100\:\text{mW}\$. (Also note that the output voltage (if the device didn't draw any current at all) might reach about \$5\frac12 \:\text{mV}\$ above the maximum \$3.6\:\text{mV}\$ spec. Which may be acceptable.