If you set the current regulator to 5A and load the power supply with less than 5A then the current regulator is saturated and does not get hot.
This situation the heat sink size
But if you have a load that tries to draw more than 5A (maybe the output is accidently shorted) then the current regulator will have its max current (which might drop to only 1A) and it will have the max voltage across it (maybe 36V) and it will be extremely hot.
This situation the heat sink size and how design size of heat sink
Your voltage regulator with the four current-boosting transistors has the base-emitter voltage of the transistors (maybe 0.8V) plus the voltage across the 0.1 ohm resistors (0.125V) for a total of 0.925V. The 4.7 ohm resistor with a voltage across it of 0.925V has a current of 197mA which is also the current of the regulator. If the input voltage is 36V, the output voltage is 5V and the current is 5A then the regulator didssipates 6.1W and the transistors each dissipate 37.2W.
how the regulator didssipates 6.1W
V = 5V and I = 197mA then P = 0.985W = ?
and how the transistors each dissipate 37.2W
V = ? and I = 5A then p = ?