OK guys, time for me to be an engineering geek! Sorry...
Anyone familiar with Ohm's Law? The long and the short of it is: Voltage / resistance = amperage. Or, V/R=I. You can swap that around however you like. In the case of our dirty roadster fuse block, we have some added resistance. Let's put some numbers into the equation. Voltage stays the same at 12V. If we assume a resistance of 1 ohm in the circuit, we have this: 12Volts/1 ohm = 12 Amps. The consumer in the circuit (the lights in this case) can draw 12 Amps.
Now, let's get a nasty dirty fuse block. Our circuit now has a resistance of 4 ohms. Plug in the numbers and you get: 12V/4 ohms = 3 amps. The total power our bulbs can pull is only 3 amps. That's why they are dim. Resistance goes up, current goes down and the lights suffer.
So if the resistor (in this case a bad fuse block with lots of corrosion) is preventing the bulbs from drawing enough current, where is all that energy going? It goes to heat. The fact that the fuse heated up is a dead giveaway that the problem is in the fuse block. Had the excessive resistance been in a connector with corroded terminals, the connector would have heated up, not the fuse.
Hopes this helps.
