I’ve fried a variable psu. there was short in one cable during test.
It’s fired just 1 of the 2 resistors inside, the biggest one. It was connected after the rectifier at the positive pins of 2 big caps 1000µf 25v. The other resistor it’s a 560ohm.
It’s completely fired than it’s impossible to know the value.
… you should draw the circuit with the used parts and all the power ratings of the PSU and post it.
Greats, Roger
The resistor in question forms part of a low pass filtger, in conjunction with the two electrolytic capacitors below it.
Most likely what has happened here is that when you shorted the supply, you exceeded the power rating (wattage :P) for the resistor. As this resistor is in series between the rectifier and the output, it took a whack.
Do like the rabbit says. Feed us more info.
Whatage?
;D
okokok! i will post more info and a circuit schematic to help, thankyou 
for now:
AC - DC Adaptor
Model: AL1000/10
Input: 230 ~50Hz
Power: 18W
Output: 1.5-3-4.5-6-7.5-9-12 V=== (the upper line it’s continuous, obviously).
Current: 1000mA 12VAmax.
What would help the most, would be a model number for the transformer.
From this, we can determine the voltage 
taps, from them, knowing the 560R parallel value, we can determine the current limiting/low pass filtering R(x) value.
Once we have done this, we (I) will suggest that you put in a higher power resistor, or do all your shorting out after a 78xx series regulator, as they have thermal cutout… :
