Okay; I'm calm now. Lets try this again.
There must be a complete path from and back to the battery for current to flow through the circuit. That includes the positive battery cable, wires, fuses, connector terminals, switches, the load, (bulb, motor, etc.), More wires, then back to the battery's negative post. The ground you asked about is the part of that path that comes after the load. It includes wires that bolt to the body sheet metal, the negative battery cable, and it can include switches and connectors. Anything that breaks that circuit anywhere will cause current flow to stop. That could be an open, (turned-off) switch, a corroded connector terminal, a cut wire, blown fuse, or burned-out bulb.
A circuit that is protected by a ten-amp fuse can be expected to draw perhaps two to five amps. A brake light bulb has close to 12 ohms of resistance. If we use one of the twelve Ohm's Law formulas, 12 volts divided by 12 ohms equals one amp of current flow. To show that more clearly, imagine connecting the bulb right across the two battery posts. The bulb's 12 ohms is what limits current flow. The fact I impressed you with one of the Ohm's Law formulas is to make this description easier to follow. In practice, we never sit and calculate things when looking for a problem.
So now you have a 12-volt source, (the battery), and we use up that 12 volts across all of the loads. In this case the light bulb is a load, and it uses up all that 12 volts. To be technically correct, each of the connecting wires is also a load, sort of, but since they each have 0 ohms, none of the 12 volts is dropped across them. It's all used up across the bulb.
Now suppose we add another piece of wire in series. That's the same as making one of those wires longer. We still haven't added any resistance, and the bulb is full brightness. That wire has 0 ohms, but so does a short circuit. Instead of inserting that additional piece of wire, we inserted something that was shorted to the negative battery post, (ground). Now the entire circuit consists of the battery's positive post, a wire, the bulb, a wire, another wire that is touching the battery's negative post, and it is attached to that post. Everything in this circuit has 0 ohms of resistance, ... Except for the bulb. The bulb is the only thing limiting current flow to one amp.
Now lets move this to your vehicle. We still have the battery, the wire that goes to the bulb, (which is in the fuse socket), another wire, then to the unintentional short to ground, and then back to the battery's negative post. The bulb is the only thing with resistance, so it is the only thing that is limiting current flow to one amp. All of the battery's 12 volts is being used up, or "dropped" across the resistance, (the bulb).
If we were to remove the bulb and replace it with the fuse, that has no resistance, so there would be 0 ohms in the entire circuit. Ohm's Law says 12 volts divided by 0 ohms equals infinite amps. That would certainly cause a ten-amp fuse to blow! This is the condition you have now.
Keep in mind these numbers are perfect and only good for explaining theory. In actual practice, the wires, the fuse, even the battery, have a very tiny amount of resistance, but knowing that won't help my story.
If your circuit was working properly, there would be no short after the fuse. Current would have to travel through the desired load, then back to ground and the battery negative post. The intended load has the resistance that is supposed to limit current to a safe value. For the sake of this sad story, lets assume the wire after the fuse socket, (light bulb), has rubbed through and is touching ground. That short bypasses the desired load, which at this point appears to be the instrument cluster. Only the bulb limits current flow. Since it has the full 12 volts dropped across it, the bulb is full brightness.
During your troubleshooting, you find a connector at the firewall where the wire between the fuse socket and instrument cluster passes through. When that connector is unplugged, the test bulb turns off. That tells us there is now no path for current to get back to the battery. Whatever is shorted has to be beyond that connector. The clue here is the bulb went out completely.
Plug that connector back in, then continue on. Unplug the instrument cluster, and behold, the test lamp goes off again. The short is either in the cluster and something is creating a direct path to one of its ground wires, or the cluster is feeding a wire that is going to something else that is shorted to ground. At this point there are a couple of ways to approach this. The fastest is to install a replacement cluster, ... If you have one in your back pocket! Without that luxury, we need to look at the wiring diagram to see if we can find what else that circuit feeds, and identify some possible suspects. Suppose, for example, a wire leaves the cluster and feeds, ... Oh, ... The generator and voltage regulator. You find that wire is bare where it is rubbing on a sharp metal bracket on the engine. That is where current is able to bypass the intended load, the regulator's turn-on circuit, and take a shortcut back to the battery.
Once you lift up that grounded wire, current has to go through the voltage regulator, but the test bulb is also still in the circuit. There are two loads that each now drop part of the battery's 12 volts. If the regulator drops four volts, the bulb will drop the other eight volts, so it will be dim. THAT is the indication you're looking for. A dim bulb proves there is current flowing through it, so the rest of the circuit is not broken, and a less-than-full-brightness bulb proves there is no short further down the line.
Finally, ... To illustrate how this works in real life, when I worked at the dealership, I had a right rear tail lamp harness from a Dodge Viper. It had three sockets for a total of five filaments, so I could plug in as many bulbs as the circuit needed. The individual wires were tied together, then soldered to a blown fuse. That was just so it could be plugged in quickly to save time fiddling with jumper leads.
I had a full-size Dodge van come in with a blown fuse for the tail lights. The customer said they had been replacing fuses, then they'd blow again in a minute or a day, but not always right away. I plugged in the test harness, then drove around the parking lot until the bulb went full brightness. Luckily it stayed that way while I carefully drove into the shop. It did flicker a few times. So I knew this wasn't a solid, or dead short. I ran around the van with a rubber hammer, and when I got near the left rear tail light, the test light flickered quite a bit. When I pushed on that lens just right, the test light went dim, and at the same time I could see a faint glow from the tail lamp bulb.
My test bulb was limiting current to one amp. When the short was gone, the current also had to go through the resistance of all the other bulbs. Since it was dividing up among all the van's bulbs that were in "parallel", only some of that current went through each bulb. The low current through each one of those bulbs is why they were so dim. Since there were so many bulbs on that circuit, their total resistance in the circuit was very low. Relatively speaking, the test bulb presented more resistance, so more of the 12 volts was dropped across it, and it was still fairly bright. The change between "fairly bright" and full brightness was enough to show when I was doing something that made the short go away.
Upon removing the lens, I found that someone had spliced in a trailer wiring harness. They caught the wire under a screw when they put it back together. Simply cutting off the damaged part, and reconnecting the wire solved the short.
If I had not learned this light bulb trick at a tv servicing class in the '70s, how many fuses would I have had to blow before I figured out there must be some better way to diagnose this?
Getting back to your test light that you want to use in place of the fuse, that bulb is much too small and has too much resistance. Most test lights draw less than a quarter amp. Given their high resistance in relation to the circuit you're testing, most of the 12 volts will be dropped across it, even when there is no short further down in the circuit. The bulb will be near full brightness when the circuit has no defect, so you won't get the desired indication of when a defect is removed.
Even my brake light bulb is too small for working in a high-current circuit like for a heater fan motor or head light circuit. This is where I plugged more bulbs into my Viper harness. All of them together would allow just over four amps to flow. If you need more than that, a low-beam head lamp bulb will pass five amps, and a high-beam is good for six amps. If you had a short in a heater fan circuit, for example, that motor might draw ten amps at a mid speed. A brake light bulb in place of the fuse would limit current to one amp. When the short is removed, the expected clue is the motor will run very slowly, but that tiny bulb limits current too much for that to happen. You need the head light bulb in place of the fuse. That will pass enough current for the motor to run fast enough for you to know it's running. If the short occurs again, the five amps the bulb will pass is not nearly high enough to overheat any wire in a car.
Keep in mind that the information I just posted took about six weeks to get to in my eight-week-long Automotive Electrical class. By that time we had covered basic electrical theory over and over until everyone got it. This is a very hard subject to learn for people who learn best by taking things apart. People who are good at visualizing abstract ideas learn this material easier, so don't sit in a corner and whimper if it doesn't make sense.
Okay, my fingertips are sore from pounding on this miserable keyboard. My next step, while waiting for you to try out this wondrous trick, is to look at the wiring diagram to see what else is involved with the instrument cluster. I'll be back tomorrow to see how you're doing.
Thursday, December 29th, 2016 AT 6:12 PM