I assume it's because we don't know RL of the circuit below.



If we use buck converter, we can get constant voltage regardless of RL, right?

 

R/L is seldom a fixed Value.
A Voltage-Regulator will supply the same Voltage regardless of changes in R//L.
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Yes, as well as save power.

In your circuit, R2 is redundant unless it is way smaller than any RL. And then it wastes even more power.

If RL is actually a fixed resistance, then get rid of R2 and you have a voltage divider composed of R1 and RL.

 

Even if \( \mathsf{R_L} \) is known and fixed, a voltage divider is not the equivalent of a buck converter. While they both drop the voltage, the voltage divider does it by converting the unwanted part of the current to heat, using the came power as if all of the current were being used productively but wasting the part that isn't needed.

The buck converter, on the other hand, only provides the desired power to the load by integrating over time making it far more efficient.

 

The buck converter, on the other hand, only provides the desired power to the load by integrating over time making it far more efficient.

This is true, but the way I look at it is that the buck converter drops the voltage across an inductor rather than a resistor. An inductor drops voltage without dissipating any power, whereas a resistor does dissipate power.

 

Why use Buck Converter instead of potentiometer?

A voltage divider will only work if it's made very stiff. The rule-of-thumb for dividers is that divider current should be at least 10 times the load current. Do the math and see how much power will be wasted by trying to use such a simplistic approach.

If we use buck converter, we can get constant voltage regardless of RL, right?

In theory. If the load is sufficiently large, the regulator will stop regulating.