Dear Johnvee:

I think I have a good explanation for the existence of the metal strip fuse in glow plug relays:

For a parallel circuit (e.g. 4 glow plugs, each with R1/2/3/4=0.6 Ohm)

The total equivalent resistance of this whole parallel circuit (consistent of 4 glow plugs) would be:

R = 1 / (1/R1 + 1/R2 + 1/R3 + 1/R4)

If we take R1=R2=R3=R4=0.6 Ohm, the total equivalent resistance is:

R= 1/6.67 = 0.15 Ohm

Since parallel glow plugs are designed to operate at a voltage of
11 V, the total amperage going through the whole parallel circuit
(all 4 glow plugs) is:

I = V / R = 11 / 0.15 = 73.33 A

(We may say that each glow plug gets about 73.33/4 = 18.33 A.)

Now we can clearly see why parallel pencil-type glow plugs are designed for 11 V, because if they are 12 V, the total amperage
will be I = 12 / 0.15 = 80 A. The glow plug relay's 80A metal strip fuse will blow up right away :-)

That's the reason why the metal strip fuse in the glow plug relay is rated at 80 A (about 6 A more than the maximum total amperage that will go through all 4 glow plugs at the VERY beginning of the pre-glow period, when the resistance of each individual glow plug is lowest.)

When the glow plugs become hotter and hotter, the resistance of the heating element in each glow plug will increase, i.e. R1 ~ R2 ~ R3 ~ R4 => higher, and the total equivalent resistance will become bigger (because of the above formula). Since V stays constant, the total amperage I will drop because R increases.

In many posts, people talked about the need to have an ammeter
that can handle 15-20 A. This means that the ammeter must be connected in series with EACH glow plug in the paralle circuit (where the amperage is about maximum 18.33 A assuming that
the COLD resistance of each individual glow plug is 0.6 Ohm, which is the value I used in my above calculations).

I need to borrow a good ammeter that can handle about 20 A. Most cheap junky multimeters can only handle a maximum 10 A.

Eric