Thanks John! Very informative indeed. I like how you explain things.
So basically, they can be considered the same in terms of practical
application, right? But the first one will put more current through the
load, and the second one will split the current (if both the load and the
resistor in parallel are same resistance, it should split the current in
half, right?) so less current goes to the load. So why is it called a
voltage divider then? Shouldn't it be called a current divider? Or both
voltage and current divider? The reason I'm asking is that if I wanted to
divide the voltage only I could've used the first circuit. But if I wanted
to divide the voltage and the current (because the load draws less current)
then I'd use the second circuit. Am I close or far from the truth?

Somewhere in between...
Looking at the voltage divider first, classically it's configured like
this, which is the same as your first circuit but with the battery
replaced with E1 and 0V, which represent the output voltage from the
battery and the return to the battery.
E1


[R1]

+>E2

[R2]


0V
The reason it's called a voltage divider is because if E1 is sitting at,
say, 9V, then E2 must be somewhere between 9V and 0V, so the two
resistors are "dividing" the total voltage into two pieces; one from E2
to E1, and one from E2 to 0V.
It's the same as if you had a hose hooked up to a faucet and the
pressure at the faucet was at 9 PSI, and where the water was spouting
out of the other end of the hose the pressure was 0 PSI; If you took a
pressure gauge and ran it up and down the hose, the pressure would vary
from 0 PSI to 9 PSI depending on where along the hose it was. Having
two resistors is like having two hoses screwed together with a pressure
gauge where they're screwed together. If they're the same length and
the same diameter and the pressure at the faucet is 9 PSI with the water
running, then the pressure gauge between the hoses will read 4.5 PSI.
Same thing with the resistors. If E1 is 9 volts (9 PSI) and R1 is the
same resistance as R2, (two hoses with the same length and diameter)
then E2 will be 4.5 V (4.5 PSI)!
If you know E1 and you know the values of the two resistors or the
current you can find E2 using Ohm's law, and you'll find that if either
R1 or R2 vary, E2 will also vary. Sometimes this circuit is used to
drop the voltage from a source with too high a voltage to a voltage a
load can use, but it's usually not a good idea unless the load current
is small and the load always draws the same current no matter what and
is never disconnected from the circuit or connected to the circuit with
the power on. The reason for that is if it's disconnected the voltage
at the botton end of R1 will immediately go to whatever the source
voltage is, and then when it's reconnected the load will have that
voltage on it until it starts drawing enough current for the voltage to
drop down to where it's supposed to be. Problem is, it may never get
the chance to work properly if it gets connected to 9V and 5V is the
maximum it's ever supposed to see. This circuit is generally used to
get a voltage to use as a reference for something when the reference
needs to be somewhere between E1 and 0V and the device it's supplying
the reference voltage to (the load) has a resistance much, much, higher
than R2. Here's an example:
E1

[R1]

E2+<+
 
[R2] [R3]
 
+<+

0V
If we assume for a moment that R3 isn't connected, we can find the
voltage at E2 using Ohm's law, like we did before, or to make it easier,
we can say:
E2 = E1R2/R1+R2
Let's make E1 = 9V and R1 and R2 both = 1000 ohms. Then we can write:
E2 = (9V*1000R)/1000+1000 = 9000/2000 = 4.5 V
Which is just exactly what we said about the two equal hose segments!
Now, let's say that to be really sure of what E2 is we'll want to
measure it, and what we'll use to measure will be a digital voltmeter
(R3) with a resistance of 10 megohms. Since we know that the combined
resistances of R2 and R3 are going to be less than the resitance of R2
alone, and that if R2 isn't equal to R1 E2 won't be equal to 4.5V, we'll
need to find the value of 1000 ohms paralleled with 10 megohms and use
that to calculate the new R2. So:
Rt = R2R3/R2+R3 = (1000R*10000000R)/1000R+10000000R = 999.9 ohms
Now, using 999.9 ohms for R2 and plugging it into the voltage divider
equation equation we get:
E2 = E1R2/R1+R2 = (9V*999.9R)/1000R+999.9R ~ 4.4998V
so there's only a 200 microvolt error there, which is about 0.004%.
Not too bad...
Now, let's say that we measure the voltage with a crappy voltmeter that
has an input resistance of 1000 ohms. For the equivalent resistance
we'll get:
Rt = R2R3/R2+R3 = (1000R*1000R)/1000R+1000R = 500 ohms
and if we plug that into the voltage divider equation well get:
E2 = E1R2/R1+R2 = (5V*500R)/1000R+500R = 1.666V
a HUGE 56.7% error!
So, you can see that the uses for the circuit can vary from a gross way
to set the voltage across, or the current through, a load to a precise
way to generate a particular reference voltage.
But just _how_ to generate the reference, starting from scratch? Just
for fun, let's say that you have a 10V supply, you've decided to use a
10k ohm resistor for R1 and that you'd like to generate a 2.5V reference
for a comparator input. Going to the circuit:
E1


[R1]

+>E2

[R2]


0V
And rearranging
E2 = E1R2/R1+R2
to solve for R2, (since that's the only piece of the puzzle we don't
have) we get:
R2 = E2R1/E1E2
And plugging everything in:
R2 = (2.5V*10k)/10V2.5V = 3333.33... ohms
Now we have a problem because, if we're limited to standard 1% resistors
the closest we can get to 3333.33 ohms will be 3320 ohms on the low side
and 3400 ohms on the high side.
Just to see what we've got, plugging that 3320 ohms into the first
voltage divider equation will get us:
E2 = E1R2/R1+R2 = (10V*3.320k)/10k+3.32k ~ 2.4925V
which is only about 71/2 millivolts low. If you can live with that,
you're done. If you can't, you'll need to go back to the table of
resistances and see if you can find a "gear ratio" that works. Or use a
pot!^)
Phew...!