I am a grade 12 student, working with electromagnets for a physics project. My equation requires me to measure the current of a 20-25 cm bare copper wire 18 gauge. The power supply would be an aaa battery (1.5v) and a rechargeable aaa battery (1.2v). How much current can I expect from this battery (completely new, haven't even taken it out the box)with neodymium magnets (which conduct) on both the positive terminal and the negative (one on each side of the battery basically). I haven't really dealt like this with anything before but I want an approx value so that I can test this equation out.

Thank you.

Edit: The circuit would be 20-25cm long

  • 2
    \$\begingroup\$ Why not just measure it? Anyway, it'll probably be close to the short-circuit current of the battery, which depends on a lot of factors. \$\endgroup\$ – Hearth Dec 2 '18 at 23:24
  • \$\begingroup\$ do you mean that it would short-circuit? \$\endgroup\$ – Himanshu Singh Dec 2 '18 at 23:54
  • \$\begingroup\$ What are the magnets for? \$\endgroup\$ – jonathanjo Dec 2 '18 at 23:55
  • \$\begingroup\$ @jonathanjo The magnets presumably are a way to connect the wires to the battery without using a battery holder \$\endgroup\$ – C_Elegans Dec 3 '18 at 0:38
  • \$\begingroup\$ The 1.5V aaa battery will not supply much current. A few amps, I would say. The rechargeable battery may supply 10 Amps or so. I would guess that when you complete the circuit with the rechargeable battery you will see a large spark (much larger than with the other battery) and the wire will heat up rapidly. Getting a minor burn is not out of the question if you hold onto the wire. Suggest that you plan to avoid holding the wire after making the connection, and also, wear some kind of glasses just in case particles are ejected. Not trying to exaggerate the danger. It is not that bad. \$\endgroup\$ – mkeith Dec 3 '18 at 1:34

As they will always tell you: don't short batteries out, they can explode.

  • First you look up the resistivity of the wire: Wikipedia lists 18 AWG wire as having approx 20 mOhm/metre
  • You have 0.25 metre, so you have 5 mOhm
  • IF the voltage stayed at 1.5 V you'd have V/R = I, 1.5/0.005 = 300 A
  • 300 A is an awful lot -- it would melt 18 AWG cable in a few seconds, for example
  • However there is no way to get 300 A out of a small battery
  • The typical short-circuit behaviour of a battery is still dangerous: a high current (12 A), a very rapid (3 seconds) rise in temperature of the battery -- over boiling, perhaps 120 C and a very rapid decrease in current as the energy is used, followed by a slower decrease in temperature. (Taken from battery manufacturer graph )

Measuring that would be tricky and require good safety work.

I'd suggest it's not a suitable experiment for grade 12 without considerable supervision on safety, and indeed how to capture the measurement quickly.


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