Electronic – Consumer Batteries Needed to Ignite a 10ohm Resistor

DO NOT destructively test resistors or any other devices without the DIRECT SUPERVISION OF AN ADULT QUALIFIED TO INSURE SAFETY

It depends on the wattage rating of the resistor, airflow, atmospheric pressure and composition, what the leads are connected to internally and externally, the temperature coefficient of the resistor, and the degree to which its materials are flame retardant.

DO NOT destructively test resistors or any other devices without the DIRECT SUPERVISION OF AN ADULT QUALIFIED TO INSURE SAFETY

You can find out if you are exceeding the manufacturer's rating by comparing the square of voltage, divided by resistance, to the rated power limit in watts. For example, 9*9/10 or 8.1 watts is drastically more than the 1/4 watt rating of most small through hole resistors so it's clear that the component is being abused and there is a substantial risk that something bad and potentially dangerous will happen. (1/2 and 1/8 watt units are also common, though you can get higher rated ones).

DO NOT destructively test resistors or any other devices without the DIRECT SUPERVISION OF AN ADULT QUALIFIED TO INSURE SAFETY

And don't assume the resistor will just burn; it could also explode. Either one can be hazardous to your health and surrounding property. Also consider that it could be hot enough ignite other materials or produce nasty gases even if the resistor itself does not burst into flame. You could also overheat whatever is sourcing power to the resistor.

DO NOT destructively test resistors or any other devices without the DIRECT SUPERVISION OF AN ADULT QUALIFIED TO INSURE SAFETY

No resistor at all. Once again, questions should stick to what you want to know or accomplish, not how you think it should be done.

Your basic question is apparently how to power this "speaker" (clearly more than just a speaker) from the power source you supply rather than the two AAA batteries it is designed for. You have available some sort of lithium battery and a regulated 5 V supply generated from that somehow.

First, you need to find out whether the batteries in your speaker unit are ground-referenced. If they are, you can proceed. If not, then this is beyond your level at this time and you either need to find a different speaker unit or a altogether different approach. Run the speaker normally with a fairly strong signal into it. With a voltmeter, measure between the negative terminal of the combined AAA battery pack and the outer ring of the 3.5 mm plug. There should be 0 V, both when measuring AC and DC. Of course exactly 0 will never happen, so in this case anything over about 10 mV means the two points aren't really connected. If they are connected, then the battery is ground-referenced and you can proceed.

If the lithium battery voltage is around 3 V, then use it directly. If this battery is a single cell, this might just work. Basically, if the lithium battery voltage is below the regulated 5 V output, try connecting the battery to the + side of where the AAA pair would go, and ground to the - side.

If the lithium battery voltage is higher than 5 V, then it would be best to to use that directly to make some sort of regulated 3 V to drive the speaker unit with. A linear 3.3 V regulator is a quick and simple answer, but might get warm when the speaker is producing loud sound. Try it and see. If that is not acceptable or the lithium battery voltage is substantially higher than 5 V, then use a switching regulator instead. There are many switching regulator chips out there that can do this with a few external parts. You can even use one that has a fixed 3.3 V output.

Added:

You now say the lithium battery puts out 7.4 V and the link to the speaker unit rates it as 1/2 W, but it's not clear if that is input power or power to the speaker. Just to see where you're at, .5W / 3V = 170 mA. We can't really tell from the sparse information in the link, but lets say top current draw of the speaker unit is 200 mA at 3 V. With just a linear regulator, the regulator would dissipate (7.4V - 3V) * 200mA = 880 mW. That's rather wasteful and something like a TO-220 package will get hot but probably OK with a modest heat sink. You can try a 7803 regulator.

The other thing to try is to power the speaker unit from your existing 5 V source. I don't know what a "BEC" is, so can't tell if this is a linear or switching regulator and how much current it can support. The speaker will draw more current at 5 V than at 3 V. If a lot more, it may get damaged. After all, it's meant to run from 3 V. 5 V may be OK, but you're a test pilot then and you can't complain if it vanishes into a greasy puff of black smoke.