Cool, this is going to be fun...a debate. The german physicist Georg Ohm's law more correctly is I = V * nAd/T. Where nAd is effectively the measure of free electrons in a wire segment and T is the time it takes for those electrons to move through the segment. Super simplifying this becomes basically I = V/R where R is the resistance of a DC circuit. This translates into the more common and well known form of Ohm's law: V = I * R.
An analogy when thinking of this is that V is like a water pump, R is like a valve, and I is like the water flowing through the pipe. For any given setting of the valve (R), the higher the water pump pressure (V) the more water will flow through the valve and the pipe in a given amount of time. Conversely, closing the valve somewhat (increasing its resistance) will decrease the amount of water flowing through the pipe.
So, for a fixed resistance in a simple DC circuit (which is what a light bulb is), the higher the voltage the more current and therefore the BRIGHTER the bulb. This is what you are seeing as Power. Power is therefore consumed by a circuit while voltage and current is conserved. The equation P = V*I is representative of that but is not Ohm's law. Bottom line is that you cannot control P directly, you must do so by adjusting V or I. You can control V by adjusting the battery (or regulator output) voltage or using a voltage divider circuit to control the voltage the bulb sees (this would apply more of a load to the battery which is why I didn't suggest it). You can control I by adjusting the resistance in SERIES with the load (bulb) in the circuit. When you select a bulb by wattage you are basically selecting between various internal resistances. A fuse has ZERO internal resistance by design so putting a fuse in line with something will only protect that circuit from going above the current value limited by the fuse. It will no lower the current flowing in the circuit. To test this theory put a resistor (it's more fun with a potentiometer) in series with a light bulb in a dc circuit. As you adjust it you will see the brightness of the bulb go up and down. By putting different value fuses in, you will see no change (unless the fuse value is too small and the current pops it).
The comments about a 120V 100wt bulb drawing half the current when 240V is applied is correct. However, it is for the following reason. You cannot take a bulb designed for 120V use and simply apply 240V to it or you will fry it. The effective Resistance of a 120V 100wt bulb is approx 145 ohms. Applying 120V volts to the bulb causes is to draw approx .83A. If you simply applied 240V to it you would now draw approx 1.6A which would surely blow it. A 240V 100wt bulb is designed to consume 100 wts and to accomplish this the internal resistance is approx 571 ohms. The point is that they're not the same two bulbs.
Also the comment about putting a resistor in SERIES (not PARALLEL) with the bulb supposedly drawing more current from the battery is completely wrong. If a resistor is put in series with the bulb, the resistor value will be effectively added to the resistance of the bulb increasing the overall resistance of the circuit therefore lowering the current for a fixed voltage (remember the correct ohms law V = I * R). If you put the resistor in parallel with the bulb, then it would cause the load on the circuit to be increased. This is because series resistances simply add but parallel resistances actually lower the effective resistance a circuit sees (to get the effective resistance you must invert the value of each resistance, add them all, and then reinvert them to get the new value).
Anyway, that's how I learned it when I got my electrical engineering degree, but that has been a few years...
BTW, I put a DVM on my bike tonite and at 3000rpm the gauge looked like 15V (halfway between the 14 and 16) while the meter was reading 14.3V. Given the comment about the gauge being logarithmic, that could make what I was seeing on the gauge closer to accurate. Also, I checked with someone and found that most automotive bulbs will handle at least 20V, so 15 or 16 probably wouldn't hurt them afterall...
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Topic: In-dash voltmeter reading seems high (Read 8980 times)