Basic understanding of Voltage drop
What is the second/lower 15 k-ohm resistor for?
I understand why there is a 12 k-Ohm resistor in the top, that's for the V-out.

The arrangement shown gives \$ V_{OUT} = \frac {15}{12+15}9 = 5 \ \text V\$.

But why is there a 15 k-Ohm resistor in the lower right?

You wouldn't need it if the load required 5 V and was exactly 15 kΩ. This could be the case in some situations. In many cases you might not know the load impedance or it could change somewhat from device to device so it makes sense to make a divider using known resistor values to drive the following circuit.

Why would you need to make a voltage drop on the path back to the battery, isn't that just waste of energy?


If the 15 kΩ resistor was not there then VOUT would be 9 V as there is no current flowing so there is no voltage drop across the 12k resistor. When current is drawn from VOUT the voltage drop will vary with the current. This may not be desirable.
All current must return to the battery in any case.
Yes, energy is consumed and no, the circuit is (hopefully) performing a useful function so there is a benefit to the energy cost.


From the comments:

But why is Vout 9V without the 15 k-ohm resistor?




simulate this circuit &ndash; Schematic created using CircuitLab
Figure 1. With a very high input impedance voltmeter (10 MΩ, say) so little current will be drawn that there will be no voltage drop across R1 so the meter will read 9 V.

I thought that the Vout is only dependent on the 12 k-ohm resistor, as Vout lies above the 15 k-ohm resistor, and current flows from left to right and then down and back to battery? What you are writing is that, the Vout is also dependent on the lower resistor, even though Vout lies just above it. How can that be?

Perhaps the easiest way to imagine this is to consider a potentiometer.



simulate this circuit
Figure 2. Various configurations of voltage divider.

In Figure 2a we have added a GND point from which all voltage measurements will be taken. We have combined your potential divider into one resistor, R1: 12k + 15k = 27k.
It should be clear that the voltage reading at the bottom of R1 will be 0 V and at the top will be 9 V.
In Figure 2b we have replaced R1 with a potentiometer R2. I hope that it is clear that the voltages at the bottom and top are still 0 V and 9 V respectively. The wiper on the potentiometer is shown in mid-position. Q: What do you think the VM2 reading will be? A: 9 V / 2 = 4.5 V. As the wiper runs from bottom to top the VM2 reading will increase linearly from 0 to 9 V and will be 9 V times the ratio of the lower resistance divided by the total resistance. \$ V_{OUT} = \frac {R_L}{R_{Tot}} 9 \ \text V \$.
In Figure 2c we have replaced R2 with R3 and R4 which are in the ratio 12k:15k which is 4:5. This is equivalent to having the wiper of R2 at 5/9 of the way towards the top. From the previous paragraph we can see that this should give the result \$ V_{OUT} = \frac {R_L}{R_{Tot}} 9 = \frac {5}{9} 9 = 5 \ \text V \$.