First of all thanks for all your replies.

Not enough information.

I cannot tell what the fuse should be because I don't know if the input is inductive, capacitive, or if the PSU only looks at the input peak voltage like into a bridge rectifier or what. So you are dependent on what the manufacture tells you. And because the input is spect for AC or DC I assume the input is a rectifier. If there is RFI filtering that may affect it.

I didn't get much of that. Sorry, very limited knowledge(just what I remember form highschool and a bit of wikipedia reaseach). You mean that due to lack of specification we can't get an accurate calculation so will be using the following rule of thumb.

Frankly I would probably take the input power, divide it by the input voltage, multiply that result by 2 or a little more like Steve suggests, and pick a slow-blow fuse of that value.

Input voltage=240V AC? Or should it be the RMS Volts?(and how would I calculate that? Would it be 240*0.707?)

Input power=?

Would we calculate it from:

Efficiency/100=EnergyOutput/Energy Input=>

Energy Input=100*Energy Output/Efficiency=100*Vdc*Idc/Efficiency=100*36*11/85=466W

?

Which would give us(assuming Input Voltage=240v): Input Power/Input Voltage=466/240=1.95A?

And then multiply by 2 or 3? (So Imax ac=4-6A?)

Why do we do the multiplication(by 2 or more -or between 2 and 10 as (*steve*) has suggested?

I don't know where you got your formula from but I work in electronic manufacturing and I do not use formulas That I cannot derive. Not even ohms laws. And your formula is just another reason why I don't use them. They often times tend to block understand and cause unnecessary confusion. I focus on a scientific analysis of capacitors, resistors, and inductors, and definitions. The rest is just math.

If you want a mathematical analysis of power factor PF I will be happy to do it.

I got the formula "Current (single phase): I = P / Vp×cos φ" from

this website and I found from

this document(from TDK-Lambda) that I

L = P

av / V

L×PF×Eff

where:

IL - RMS line current in Amps

Pav - average output power in Watts

VL - AC line voltage in Volts

PF - input power factor

Eff - efficiency of the supply

(so they add the Efficiency to the equation)

Finally one more thing.

Assuming we calculate the max input ac current to be for example 6 amps, would it be for some reason wrong to use 2 fuses in series, one slow-blow with a rating just above of the 6 amps(lets say 7 or 8) and a fast-blow wiath a 50 volt rating, so that the slow blow will cut off the supply in case we get a constant current(not a spike like the Inrush current) of above the 6A that is our limit whithin a few seconds and the fast blow will provide instant protection from classic short-circuits where we would get a very high current?

(yes, safety is very important for me as I am inexperienced and don't want to harm anyone)

I apologise for the far too many questions