stephenkohnle53 said:
But since my voltmeter does not work, do you have any idea how I could calculate the resistance and length of resistor needed using that article. According to a video I watched a while ago I need 1 amp of current thus needing 9 ohms of resistance if I cut down a full pencil of HB from 175 millimeters to 82.89 millimeters would that work assuming my resistance and amperage is correct? I got that by dividing the 19 ohms of resistance the HB lead has by 9 to get 2.111111111 then divided 175mm by 2.111111111111.
You need to look at the data sheet for the battery you are trying to use.
https://d2ei442zrkqy2u.cloudfront.net/wp-content/uploads/2016/03/QU1604_US_UL1.pdfI get 82.89 as well. But...
You seem to be thinking that you can draw 1 amp from the battery at 9 volts. That would be 5 watts.
In looking at the data sheet, notice that the maximum power case shown (top right on page 2) is 1000mW, 1 watt. Under those conditions, the battery voltage drops rapidly and the battery dies in about 2 hours. If you have been experimenting with the same 9 volt battery for more than a few minutes, that battery is probably well on its way down that voltage curve.
So, for starters, you will need at least 5 fresh batteries wired in parallel. And in this case,10 batteries will last almost 3 times longer than 5.
According to that curve on the data sheet, your voltage is going to drop into the range of 8 to 8.5 very quickly. So you may want to adjust your resistance target accordingly.
Also notice that the internal resistance of the battery is 1.7 ohms. So if you are targeting a 8.3 ohm resister, about 17% of the heat will be expended by the batteries, not the pencil.
Looking at one of your links - that #2 pencil I was mentioning before is the same as your HB.