Heya
Basic electrical theory is that Power (P) = Voltage (V) x Current (I)
=> P = VI
By example, if the power of the device you are running is known (eg 90/130W head lamps), then max total power is 130 x 2 (lamps) = 260W.
So, 260 = VI
=> 260 = 12V x I
So, I = 260/12 = 21.7 Amps
So, in the case of a high beam headlamp fuse, you would run a 25A fuse minimum. I use heavy gauge wire in my installation, and use 30A fusing, which works fine, and will protect the wiring from a short.
As one of the other contributors mentioned, the fuses are there to protect the wiring, NOT the device. Shorts are caused most frequently through short circuits (eg wiring abraided and shorts to earth) OR through a device failing (eg fuel pump short circuits, effectively conducting power through the pump with no resistance in circuit). In these cases, if the fuse does not blow, the wiring will heat up and potentially cause a fire and/or damage the wiring. By blowing, the fuse protects the wiring from this outcome...the device is already stuffed, hence the fuse is not there to protect it.
Bottom line: The fuse needs to be 'big' enough to sustain the maximum current draw of the device when under load (including start up current), but small enough to blow BEFORE the wiring gets hot and fails. Hence, you need to know the power requirement of the device, and the power/current capacity of your wiring to design your fusing.
Cheers
Jamo
