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I bought a multimeter with attached leads. I would think red is positive but when I take a reading it only works if I use black as positive. Was it assembled wrong? There's nothing in the manual. It's a greenlee dm-20.

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    \$\begingroup\$ Could you specify what exactly you're measuring and what results you get? \$\endgroup\$ – Keelan Jun 14 '13 at 21:32
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    \$\begingroup\$ You have to use both probes to take a reading. \$\endgroup\$ – Kaz Jun 15 '13 at 1:16
  • \$\begingroup\$ Were you trying to measure in-circuit resistance of a component with power still on the circuit or maybe with some residual power due to capacitors being still charged up? \$\endgroup\$ – Andy aka Jun 15 '13 at 12:26
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    \$\begingroup\$ I was at radio shack and brought in a battery to get tested and the salesperson connected the black to the + and I told him that was wrong? He said I should know I went to school for electronics... lol \$\endgroup\$ – user53947 Sep 24 '14 at 23:48
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Every dignified multimeter uses red for positive and black for negative.

In addition to that, when you're measuring volts, you can just switch the inputs. It won't go wrong, you'll just get a negative value.

If you suspect it might be assembled wrong, just check it with a voltage source of which you know which pole is the positive and which the negative. Connect the positive to the red lead and the negative to the black lead. If the reading is positive, red is for positive. Otherwise, it is indeed assembled incorrectly, and you should use red for negative.

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    \$\begingroup\$ It won't go wrong Be careful with this one in general. If the meter is analog, you can damage the needle with negative voltages in some cases. +1 for the rest. \$\endgroup\$ – AndrejaKo Jun 14 '13 at 21:59
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    \$\begingroup\$ +1 for the first sentence. Also, the sockets where the leads plug into the voltmeter are usually well labeled. In case of bannana plugs, its really how you decided to plug in the leads. Of course the only dignified way to do so is to plug the red into positive and black into negative. \$\endgroup\$ – Olin Lathrop Jun 14 '13 at 22:08
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    \$\begingroup\$ The Greenlee DM-20 specified by the OP is a digital multimeter with attached leads - the red lead should (must) be the positive lead. If you connect the lead "incorrectly" to a battery or other DC source, the meter should still work, but will show a "-" sign. It should work with either polarity when measuring AC. \$\endgroup\$ – Peter Bennett Jun 15 '13 at 1:04
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One observation- the POSITIVE output lead for the OHMS function is usually the BLACK lead on an analog multimeter, and usually the RED lead on a digital multimeter.

So if you measure the conductivity of a diode with an old-school analog multimeter you may have to swap the leads around to get it to conduct (or not conduct).

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    \$\begingroup\$ Thanks for bold mention of positive output lead ; because many people often tell for a device "it is the plus-lid and it is the minus lead". A plus may mean positive input or positive output (and so in negative also). Saying a casual plus or minus causes a lots of misunderstanding on basics. \$\endgroup\$ – Always Confused May 31 '16 at 17:19
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    \$\begingroup\$ I use an analog multimeter (Sanwa P3) , and yes on-which the black probe (to be attached on the common point) always acts as negative input terminus , and the red probe (to be attached on other points) always acts as positive input terminus ... irrespective of whether i'm measuring Volts or Ohms. I measure diodes, etc. accordingly. And black, the negative input terminus, is by default the positive output terminus. But digital multimeter's probes really change their behavior when measuring battery, diode, etc. according to the meter-set-up . That is interesting. Thanks very much. \$\endgroup\$ – Always Confused Jun 1 '16 at 13:54
  • \$\begingroup\$ [![Circuit diagram][1]][1] Not an answer, but i've attached the circuit diagram of that analog meter, given in the manual. I hope this will help those are seeking solution to the same question. i've put orange stars in the diagram which may take part in polarity. [1]: i.stack.imgur.com/CiLRQ.jpg \$\endgroup\$ – Always Confused Jun 4 '16 at 5:26

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