The load is simply a 96ohm resistor.
What I wanted to know was what the current in the 24v input would be.
As nothing is 100% efficient, it would be nice to know how to determine it exactly.
Best Wishes
Tony Hunt
Tony; here is a hint. The output current from the 317 goes to two parallel branches; one is the 96Ω load resistor, and the other is the voltage divider that makes 1.25V out of 2.4V. Since you do not show those, you will have to calculate the current that flows through the divider. It should be ~5mA, so the total current coming out of the pin is 30mA.
Now if you were to look at the 317's data sheet, it would tell you that the Adujustment Pin current that flows out of that pin is ~50uA, which is tiny compared to the load current
If you consider Kirchoff's current law, it basically says, sum of current in = current out, which tells us that the current flowing into the 317 is 30mA.
Now consider this: if we think about the power supply, it is delivering 24V*30mA = 0.72W.
Where is the Power going? Well, the power dissipated in the load resistor is only 2.4V*25mA = 0.06W. Where is the rest (0.72-0.06 = 0.66W) going?
You guessed it, into the regulator, and the voltage divider resistors. The voltage divider resistors dissipate 2.4V*5mA = 0.012W (insignificant), leaving 0.648W for the regulator itself.
Now, I'm here to tell you that if you ask a LM317 in a TO220 metal tab package (without an add-on heatsink) to to dissipate 0.65W, it will get hot to the touch, but be short of burning your fingers. If you ask a TO-92 style 317 to dissipate 0.65W, it will get stinking hot to the point where it will shut itself down, reducing the output voltage to near zero, thereby reducing the current through it, to protect itself, and your next post to the forum would be "
how come my 317 doesn't put out the voltage I expected"