As in the above example you can test for the presence of voltage on the glow plug buss with a voltmeter, buy you probably won't know with certainty that the glow plugs are drawing current or not.
Let me give you an example. If you find 12 volts on the glow plug buss during the glow plug cycle, as controlled by the glow plug control timer and as switched onto the buss by the robust glow plug relay, how will you know if the glow plugs are all "open circuit" or not?
If the glow plugs are open circuit, they will not draw any current, or if only one glow plug is still within the resistance specification of about .4 ohms and the others are open how will you be able to determine that?
What do I mean by open circuit? I mean a circuit with infinite resistance to current. A glow plug is nothing more than a very low value resistor heating element. Somewhere in the less than 1 ohm of resistance. My Kioti manual say mine are .43 ohms each. and when placed in parallel the resistance is even less. So the glow plugs are a lot closer in value to a "dead short" than an open. An example of a "dead short" would be a chunk of wire. So a wire must be "shorted" by design and manufacture to function as a conductor of electricity. A lot of folks have a lot of trouble understanding the terms "open" and "shorted" Which are very loose terms indeed and can mean different things to different people.
But do wires and relay contacts have zero ohms of resistance?.. Um. not quite, they all exhibit some resistance to current and therefore have a resistance value although a very low one.
Now lets do a "thought experiment" where we don't actually do this, we just think about the results and apply logic and reason.
Take 1000 foot of 24 gauge wire, which is pretty small wire. and we look up in a table provided by the manufacture how many ohms of resistance it will have over that 1000 foot span. The table says 25 ohms. Oh wow! that is a lot more ohms than those glow plugs.
So we apply 12 volts to the battery end, and measure the voltage on the other end with a good voltmeter. What will it be? Will it have dropped by having to go thru that 25 ohms of resistance?
Well lets see now: E=IR so the voltage should be the current times the resistance right?. so...hey we don't know the current so how can we solve the equation for voltage drop if we don't know the current?.. hmmm . I did say the voltmeter was a good one, that will draw next to nothing in the way of current by making that measurement..... so I guess the current being 0 multiplied by the 25 ohms is still about zero huh?
So the voltage drop across a 25 ohm resistor in the form of 1000 foot of small gauge copper wire is near 0 volts of drop, so we would have the same 12 volts on the far end as we stuck in on the near (battery source end).
Now OK, lets try to draw some current thru that 1000 foot of small gauge wire.. Uh. oh... this is going to get dicey real quick. Lets start by putting a small load on this wire at first. Lets just put a 100 milliamp load on it. So a milliamp is 1/000 of an amp (Ampere) so 100 of those is 1/10 of an amp (Ampere) so far so good?
So lets do some quick calculations. E=IxR so we are solving for the E or the voltage drop in this circuit. E= .1 (amps) X 25 (ohms of resistance) so then .1x25=2.5 volts of drop with a pretty small load. So we could expect about 9.5 volts to be measured at the end of the wire with a very modest load on it. If we increase the load on this system, the voltage will continue to plummet all the way to zero volts.
What if we put one of our glow plugs on the end of the 1000 foot of 24 gauge wire and ground of the battery. Would we be able to power our glow plug up? After all we just got done measuring 12 volts on the end of the wire, so lets throw this glow plug on it and see what happens.
What will happen?
What if we increase the length of the wire to 2000 feet instead of 1000?. We will just drop twice as much voltage with a given load but the open circuit voltage will still measure 12 volts at the end of that 2000 foot of wire.
So the point to all of this "stuff" is that when you measure the voltage on the glow plug buss, you know the relay contacts are contacting the source battery to the buss, but you are still not sure the glow plugs are working. And by working I mean drawing massive amounts of current because of their extremely low resistance and making lots of heat in the process.. hmmmmm. Wait a minute you say...If they are a really big load and draw lots of current, shouldn't we get some voltage drop across the wiring and the relay contacts and even the internal resistance of the battery itself?... (light bulb comes on inside his head)
Yeah you should. Well, can we maybe see that voltage drop? Yeah we can. But the rub comes in in determining how much, and how to determine if ALL the glow plugs are working or just some of them.
Say your battery resting voltage is 12.7 volts. and you fire up the glow plug cycle, and you measure 12.65 or something on the glow plug buss. Are the glow plugs working? Probably not. Because the relay and the timer may drop voltages in the system a little bit like that, but remember these dang glow plugs are like dropping a crow bar across your battery.. it should knock the heck out of it voltage wise. And the resistance's in the wiring and relay, and the internal resistance of the battery itself should result in some pretty hefty voltage drops. maybe as much a 2 volts of drop. So if you measure maybe 10.5 volts on the glow plug buss in those few seconds it is operating, then likely they are working. BUT you can never be sure with this "estimating" method.
This is why you can use the infrared thermometer I discussed earlier or take off the glow plug buss and measure each plug's resistance to determine if they are "burnt out" (open) or within the resistance spec.
Probably enough to think about for now...I am sure you are probably tired of this game by now. Ask more questions if you want to go on.