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I need to measure current in a small electromagnet powered off 3 x AAA batteries in series. This is the manual for my multimeter. When I run this multimeter in series and try to find the current, the electromagnet doesn't work anymore. I suspect the internal resistance of the multimeter to be too high compared to my electromagnet (just a few coils of 20AWG magnet wire) so the electromagnet isn't getting enough voltage anymore. Is there a cheap/free solution for me to find the current in the electromagnet?

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    \$\begingroup\$ AAA batteries still pack quite a punch, so if you practically short them, I'd still expect a couple Ampere flowing. Maybe you just broke your multimeter's fuse? \$\endgroup\$ – Marcus Müller Jun 11 '16 at 13:31
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    \$\begingroup\$ there's a fuse in line only in current, not in voltage measurement mode. So this is not an indication of functionality. \$\endgroup\$ – Marcus Müller Jun 11 '16 at 13:44
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    \$\begingroup\$ @PeterZhu why should it? In voltmeter mode, the device has an enormous inner resistance, so no fuse will be needed. \$\endgroup\$ – Marcus Müller Jun 11 '16 at 13:51
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    \$\begingroup\$ Then it is possible you have blown the 10A fuse too. \$\endgroup\$ – user_1818839 Jun 11 '16 at 13:59
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    \$\begingroup\$ well, \$\frac{4.5\,\text V}{0.2\,\Omega}=22.5\,\text A\$. Happily, the batteries won't allow this, because of their source impedance, which is what I asked you to measure by comparing voltage with and without load (see comments under answer). Also, 0.2 Ohm is something that requires very good contact when measuring. I'd guess that it should be lower if you had a proper measurement device and proper connectors for resistances of this order of magnitude. \$\endgroup\$ – Marcus Müller Jun 11 '16 at 14:06
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You should use the 10A range. Set the meter to 10A and plug the red lead into the 10A jack. There are two fuses, one 10A fuse for the 10A range and one 250mA fuse for the lower current ranges. You may have blown the fuse for the lower ranges. The fuse replacement instructions are on the last page.

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  • \$\begingroup\$ :) thanks for picking up on this! Definitely, "just a few windings of 20AWG" will easily blow a fuse (which pretty certainly is "thinner" than 20AWG) rated 250mA. \$\endgroup\$ – Marcus Müller Jun 11 '16 at 13:50
  • \$\begingroup\$ So I still have a few questions. Does the voltmeter not go through the fuse because I can still measure voltage just fine on that socket? Will the batteries really output more than 0.25A? \$\endgroup\$ – Peter Zhu Jun 11 '16 at 13:54
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    \$\begingroup\$ @PeterZhu you've gotten answers to that in the comments. If you don't believe what we write, why are you asking? \$\endgroup\$ – Marcus Müller Jun 11 '16 at 13:55
  • \$\begingroup\$ No, usually, the voltmeter setting will not be fused. That would be nonsensical, because a voltmeter has very high internal resistance. \$\endgroup\$ – Marcus Müller Jun 11 '16 at 13:56
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    \$\begingroup\$ It's broken. when nearly shorting batteries across the fuse, no current flows. What better test can there be??? \$\endgroup\$ – Marcus Müller Jun 11 '16 at 14:22

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