Voltage Divider Supply (Unloaded and Loaded)
A voltage divider is made from two resistors, and is used when we require a lower voltage than we have available currently. It can be used for power or signals, basically anything that needs to be reduced in amplitude.

 

Vin is a voltage across both resistors R1 and R2, and Vout is the voltage across R2 only.  

If we examine this circuit and use Ohms law, we know the following things:

The current through the resistors is:

I_divider = Vin / (R1 + R2)

If nothing is connected to Vout, then the same current that flows through R1 must flow through R2, so:

I_R2 = I_divider

Let's do an example with real values. Say we have:


Vin = 10V
R1 = 10&Omega;
R2 = 30&Omega;


From above, we have:

I_divider = 10V / (10&Omega; + 30&Omega;) = 0.25A

So with nothing connected to Vout, R2 has 0.25A flowing through it. Using Ohm's law, we can work out the voltage across it:

Vout = V_R2 = 0.25A * 30&Omega; = 7.5V

We can see easily that the voltage across R1 must be 2.5V, but just to make sure:

V_R1 = 0.25 * 10&Omega; = 2.5V

Loading

Now what happens when we connect a load to Vout? Let's say we connect a 100&Omega; load, now our circuit looks like this:



All we have done is put a resistor Rload in parallel with R2. From our knowledge of parallel resistors we can calculate the combined resistance of both, and replace then with a single effective resistor:

R_comb = 1 / ((1/R2) + (1/Rload)) = 1 / ((1/30) + (1/100)) = 23.07&Omega; (let's call it 23&Omega; for simplicity)

So now we are back with our simple voltage divider again, let's do the calculations:

I_divder = Vin / (R1 + R_comb) = 10V / (10&Omega; + 23&Omega;) = ~0.3A

We see the current has increased, due to the loading of the divider with Rload. You can think of this like a single pipe that has split into two, so more water flows (given the same pressure)
So what's Vout now? We use the same calculation as above:

Vout = Vcomb = I_divider * (R_comb) = 0.3A * 23&Omega; = 6.9V

So the current has increased, but the voltage at the divider output has dropped. Again , imagine the water flow (current) has increased, but the pressure (voltage) has dropped.

Our final result looks like this:



The greater the current drain, the greater is the change in terminal voltage with the application of the load

Let's have a look at what happens to Vout and I(Rload) as we change Rload from 1&Omega; up to 1k&Omega;:



The X axis represents the value of Rload. We can see that Vout changes little at the higher resistance values, then drops sharply as it approaches 1&Omega;. I(Rload) on the other hand does the opposite (the loading increases).
This is probably what your book is referring to.    

Hopefully this helps to see how the loading of a voltage divider works. If you have any questions, leave a comment and I will add some more detail.