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I'm using my digital multimeter to measure the resistance between the terminals of a 9 volt battery i had lying around.

Why does it read as 50 M Ω? Why is the resistance so high? I would think that the resistance in the battery would be 0 (zero) or close to it because if I ran a wire between the terminals to short them out, that WOULD short out the battery, which makes me think that there would be very little resistance between the terminals. I hope that makes sense?

And I would think that with such a high resistance - 50 M Ω - that even with a wire between the terminals, the battery wouldn't short.

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  • \$\begingroup\$ If the resistance between the two terminals was 0R, then you would never be able to store charge \$\endgroup\$ – JonRB Jan 15 '17 at 17:51
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A multimeter measures resistance by applying a voltage between its probes, and measuring the resulting current (or applies current and measures the voltage), so any reading you get when measuring across a power source, or across a component in a powered circuit, is meaningless.

NEVER try to measure resistance in a powered circuit.

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    \$\begingroup\$ Note that you might easily damage your multimeter that way. \$\endgroup\$ – Marcus Müller Jan 15 '17 at 17:56
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    \$\begingroup\$ @MarcusMüller You and Peter and Ryan have pointed out a scenario that any quality multimeter should be able to pass. Be very wary of any multimeter that include current scales on its main scale switch - its fuse should have easy access ;-) \$\endgroup\$ – glen_geek Jan 15 '17 at 18:15
  • \$\begingroup\$ @glen_geek indeed :) But anyway, it's pretty easy to damage even a 10A-fused current sensing path of a multimeter with household-grade AA batteries! \$\endgroup\$ – Marcus Müller Jan 15 '17 at 18:28

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