Why voltage regulators instead of voltage dividers for supplying power to loads?
No [it's not a duplicate of &quot;When would I use a voltage regulator vs voltage divider?&quot;] because I know why they are used differently and their application, I wanted to know the physics of why the resistors inside the [linear voltage regulator] component wouldn't burn in comparison to a voltage divider.

OK, I think I see the question you are asking, and the answer is fairly simple:

With a voltage divider, comprising only resistor components (which is typically what people mean when they talk about voltage dividers in this situation) the current for the whole load goes through the &quot;upper&quot; resistor. One of the effects of this (as well as poor regulation) is that the resistor has to be able to dissipate all the heat caused by passing that load current.
In this type of circuit, the resistors have to be comparatively low values, to reduce the effect of the load current on the voltage divider's &quot;output&quot; voltage. However using low resistor values increases the overall current flowing through the voltage divider to ground, and so increases the power dissipation in those resistors.

Using a linear voltage regulator IC, whether its feedback resistors are external or internal to the voltage regulator itself, the load current does not flow through those feedback resistors. Instead, the load current goes through what is called a &quot;pass element&quot; e.g. a transistor.


This difference means that the feedback resistors for a linear voltage regulator (and I'm addressing just your question above, about the resistors) only dissipate a small power since they only pass a tiny current, which is not related to the current required by the load. Those feedback resistors can be comparatively much higher in value, than the resistors in a &quot;simple resistor-only voltage divider&quot;.
For example, in page 1 of this datasheet for the old Signetics 7800 series, R19 and R20 are the feedback resistors (shown as 0.25kΩ + 5kΩ) so the current through them is just under 1mA at 5V output. The point is that this small current through those resistors stays approximately constant (and so does their power dissipation), no matter what the load current is.
(There is also this interesting webpage from Ken Shirriff, where he reverse-engineers a 7805 regulator. On that 7805 schematic, the feedback resistor divider is labelled R20 + R21.)
The pass element (e.g. BJT or FET) in a linear voltage regulator behaves like a variable resistor, under the control of an &quot;error amplifier&quot; (see below) and dissipates the same amount of power as the &quot;upper resistor&quot; in the equivalent voltage divider scenario.

Wouldn't the resistors [inside the linear voltage regulator] burn up due to the power dissipation?

No, it's the pass element (e.g. BJT or FET) which can dissipate significant power (and is designed for this, with heatsinking added by the circuit designer where necessary) - not the feedback resistors for the linear regulator, which don't dissipate enough power to &quot;burn up&quot;.
That pass element can be internal to a linear voltage regulator IC (typical these days), or external to it, or a combination of both, depending on the regulator IC and the circuit designer's choices.
In case it helps to see it, here is a block diagram of one type of linear voltage regulator. The load is connected to the VO terminals:

(Image source: From &quot;Figure 1 LDO block diagram&quot; of Linear Low Dropout Voltage Regulators, from Analog Devices ADALM1000 Active Learning Module)
The series pass element (in the diagram above, it's a P-Channel MOSFET) still dissipates a power related to the load current (P = (VI - VO)·IO approximately). The feedback resistors are termed &quot;Sampling Resistors&quot; in that diagram. As I explained, the load current IO does not flow through those sampling (feedback) resistors.
The &quot;Error Amplifier&quot; (measuring the difference between the reference voltage VR and VS which is the output voltage via the divider formed by sampling / feedback resistors R1 and R2) varies the effective resistance of the pass element, as the output voltage (and therefore VS) changes (whereas the reference voltage VREF and therefore VR, would be stable in an ideal regulator).
Does that explain what I think you are looking for in the question above, about why the resistors in a &quot;pure resistor voltage divider&quot; get hotter than the feedback resistors in a linear voltage regulator?

As the question has developed after I originally posted this answer, it's clear that a good approach to the whole problem is unlikely to involve a linear voltage regulator (or pure resistor voltage divider) at all. Instead, it may involve a buck-mode switching regulator (e.g. 12V to 5V) - perhaps several of them (e.g. one per RPi, or per several RPi boards).
There are advantages &amp; disadvantages of using one or more 12V PSUs (and additional buck regulators down to 5V) or using one or more 5V PSUs, depending on various factors (e.g. voltage drop over the DC power cabling). This has been explained in another answer.