Science Buddies: "Ask an Expert"

We are trying to do the pencil resistor experiment and are having trouble finding a miniature 9 volt light bulb. Any suggestions on where to get one?

lighteningbolt - An alternative might be to use a 6 volt lantern battery (or multiple battery 'pack' using 1.5 volt batteries) and a 6 volt, low current bulb. It will give you much more current for your experiment than a 9 volt battery as well.

Rick Marz

Hi:

I searched and found some options on Ebay and Amazon. If those don't work for you, you might be able to use a lower voltage bulb with a resistor. I found some LED replacements here: http://www.bulbtown.com/9_VOLT_LIGHT_BULBS_s/536.htm (don't know anything about "Bulb Town")

Best, Keith

OK. So we did this experiment using a 12.5 volt bulb and 8 1.5 D size batteries (12 Volts).
I'm not sure if our results make sense. We used a multimeter to measure the voltage and current. The current reading was -0.1 for all pencil resistors. The voltage for the 5.3 cm pencil was 0.48 Volts; the voltage for the 7.7 cm pencil resistor was 7.7 cm pencil was 0.62 Volts; the voltage for the 10.5 cm pencil was 0.8 Volts. and for the 14.5 cm pencil, the voltage was 1.o Volts. Therefore,using Ohms's Law, we calculated that the 14.5 cm pencil had a higher resistance of 10 Ohms, vs. 4.8 Ohms for the 5.3 cm. But shouldn't the voltage have remained constant and the current changed??

lighteningbolt - I'm assuming that you used the same piece of pencil lead, and just kept shortening it from 14.5 cm to 5.3 cm. Ohms law should have calculated the same resistance per centimeter of pencil length, but your readings, assuming the current measured was 0.1 amp for all readings, was correct. I think there are two factors that influence your measurements. The choice of using the 12 volt battery and bulb is a factor because the overall current of 0.1 Amperes in the circuit says the resistance of the bulb, plus the pencil lead is 120Ω in each case. That is likely incorrect. When deriving the resistance of the pencil lead using Ohm's law, your values should be as follows:

Length Voltage Drop Current Total Resistance of lead Resistance/cm (total resistance ÷ length)
5.3 cm 0.48 V 0.1A 4.8Ω 0.90Ω/cm
7.7 cm 0.62 V 0.1A 6.2Ω 0.81Ω/cm
10.0 cm 0.80 V 0.1A 8.0Ω 0.76Ω/cm
14.5 cm 1.00 V 0.1A 10.0Ω 0.69Ω/cm

The resistance/cm in the last column should be the same if it is the same pencil lead, and all distances and voltage and currents were measured precisely. (especially if you had the same current, as heating shouldn't have any effect). The choice of a high resistance bulb (about 110-115Ω) is high compared to the pencil lead (4.8-10.0Ω) making measurement difficult, but not impossible. I think you should have been able to get measurements that were more consistent. The current should not have been a uniform 0.1A in this case, and should have varied more in the range of the range of the pencil lead resistance (range of about 5Ω in 110Ω for example) Although it sounds like a modest difference, when added to any variance in voltage measurement you are getting about a 28% error (comparing your highest to lowest resistance/cm calculation) Might want to review your setup, technique and measurements.

Rick Marz