i have a question about this dummy load and DIY dummy load construction in general.

we have ten resistors in parallel in series with ten more resistors in parallel. how can their power dissipation be additive if all of the current flowing through the load has to pass through the first ten resistors before it passes through the second ten? meaning, wont the first ten resistors have to bear the full brunt of the power, anyway? what am i misunderstanding here?

let's say 72 watts going in to 8 ohms (and lets say these are all 40 ohm resistors to make it even and easy)

P = I^2 * R

72 = I^2 * 8

9 = I^2

I = 3A

that 3 amps splits ten ways as it enters the load (im ignoring the rectifier/fan cause my dummy load doesnt have one), so 300mA into each resistor of the first parallel combo:

P = I^2 * R

P = 0.3^2 * 40

P = 3.6 W each resistor

x10 = 36 W dissipated by the first group and the same by the second group

x2 = 72 W

now it makes sense to do it this way rather than just having one row of ten 80 ohms resistors, because in that case each resistor would be burning up 0.3^2 * 80 = 7.2 Watts instead of only 3.6 W.

the math makes sense to me. what doesnt make sense is that the series current (ALL of it) passes through one group and THEN the next. seems like it should ALL be dissipated in the first group seeing as it ALL passes through there. how does the current know there are ten more resistors to go? i guess the relationship here between voltage current resistance and power cannot be quite intuitive....

and say i built a 4 ohm dummy load with one row of ten 25 ohm 10 W resistors and one row of ten 15 ohms resistors and put those two rows in series (strictly hypothetical

) and i wanted to make this into an 8 ohm dummy load and i just added a 10 W 4 ohm resistor in series with both rows, right at the front or the end, that would be a bad idea:

let's say the same 72 watts, so 3A coming in -

that 3A passing through a row of ten 25 ohm resistors would only dissipate 0.3^2 * 25 = 2.25W per resistor, but when it passed through that added 4 ohm resistor it would dissipate 3^2 * 4 = 36W and bye bye resistor and dummy load and maybe OPT....

i guess after doing all this math i understand what's going on pretty well. the current itself only depends on total resistance, but the POWER dissipation depends on total current AND on how that current is divided/passes through the load.....