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We made a circuit with a power supply at 3V DC, a bulb in parallel with a test material (metal/insulator/salt solution) and measured the Amps in the combined circuit, and in the test material branch.

I expected that the Amps in the bulb branch would remain the same regardless, since $I=V/R$, and $V$ and $R$ are constant in this experiment, and the overall amount of Amps would rise with better conductors tested.

  • With plastic as the test substance the overall current was $0.3$ A and the test branch was $0$ A with a bright bulb.

  • When we used saline, the overall was $0.3$ A and test branch was $0.1$ A with a bright bulb.

  • Using a copper strip, the overall was $5.9$ A and test branch was $5.8$ A with a dim bulb.

This means that the bulb branch went from $0.3$ A to $0.2$ A to $0.1$ A as the resistance on the test branch decreased.

Is it just a sketchy experiment or is there some aspect of physics I'm missing? I'm familiar with resistances in parallel and series and with Ohm's law.

Any input appreciated!!!

Cheers,

S

(There's a "2" below the bottom ammeter; the photo cut if off!!)circuit

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I did not look into your data, but the resistance of the bulb increases with temperature (and, therefore, with the current flowing through the bulb). The resistance at low current may be an order of magnitude lower than at nominal current.

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  • $\begingroup$ Ah, yes, I remember the resistance would change as the bulb got hotter, but the bulb got dimmer as the resistance on the test branch decreased. I would have assumed with a constant Voltage on it's own branch the bulb would receive the same current and stay the same brightness, independent of what happens on the other branch? $\endgroup$ – William Nov 9 '16 at 22:17
  • $\begingroup$ @Weavermech: It is not obvious that the voltage is constant on the bulb branch: if the test branch resistor is low, the current in the circuit is high, so the voltage drop on the internal resistance of the battery is relatively high, so the voltage on the bulb branch is lower. $\endgroup$ – akhmeteli Nov 9 '16 at 23:08
  • $\begingroup$ This is interesting. I just edited my original to include we were using a 3V power supply, but are you saying that the pd over the bulb's branch can change depending on the the resistance of the material in the parallel branch?? $\endgroup$ – William Nov 9 '16 at 23:41
  • $\begingroup$ @Weavermech: Yes, this is what I am saying, and I explained why. $\endgroup$ – akhmeteli Nov 9 '16 at 23:45
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I'm a bit sceptical about the third test setup:

Using a copper strip overall was 5.9 and test branch was 5.8 with a dim bulb.

If you use a copper piece you have effectively removed all significant resistance - in other words, you've made a short circuit. The (total) current would increase and increase and increase until the power $P=VI$ on the copper wire is so large, that it melts.

During this process, there is no stable or steady current in the circuit. It is accelerating. Ohm's law only applies for steady currents.

On another note, mind your uncertainties:

This means the bulb branch went from 0.3A to 0.2A to 0.1A

Be careful with conclusions, when your uncertainty is at the only varying digit. If you really don't have more accuracy in your measurement devices than for 1 significant figure, these numbers might as well all be the same.

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  • $\begingroup$ "If you use a copper piece you have effectively removed all significant resistance - in other words, you've made a short circuit. The (total) current would increase and increase and increase until the power P=VI on the copper wire is so large, that it melts." Let me note that the battery has non-vanishing internal resistance. $\endgroup$ – akhmeteli Nov 9 '16 at 23:10
  • $\begingroup$ Sorry, I didn't mention it's not a battery. The power supply was set to 3V so I believe if the resistance in the copper strip was something like 0.5 Ohm the maximum current would be 6V? We did the copper a few times. $\endgroup$ – William Nov 9 '16 at 23:37
  • $\begingroup$ @Weavermech: Any power supply has non-zero internal resistance. And the maximum current cannot be measured in V. $\endgroup$ – akhmeteli Nov 9 '16 at 23:49

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