-1
\$\begingroup\$

enter image description hereenter image description hereI have a power cord that is supposed to supply an appliance with 12V D/C power. The power cord has positive and negative leads. One lead has the typical writing on it which could identify it as the positive lead but I don't want to assume that because the appliance that is being powered is very expensive to replace, so I am hoping I can identify which lead is which with a multimeter however I am confused about whether or not this can be done if there is no current running through the power cord. In other words, I have the appliance which the power cord is plugged into and the the other end of the power cord is cut. I tried touching the exposed wires with the multimeter leads but there is no reading at all. It does read accurately when I test various batteries so I know the multimeter works fine but that is what made me think I need to run some kind of battery power through the cord in order to test the polarity of the wires??

\$\endgroup\$

closed as off-topic by PeterJ, Voltage Spike, Dmitry Grigoryev, DoxyLover, Brian Carlton Aug 9 '17 at 23:44

This question appears to be off-topic. The users who voted to close gave these specific reasons:

  • "Questions on the use of electronic devices are off-topic as this site is intended specifically for questions on electronics design." – PeterJ, Voltage Spike, Brian Carlton
  • "Questions on the repair of consumer electronics, appliances, or other devices must involve specific troubleshooting steps and demonstrate a good understanding of the underlying design of the device being repaired. See also: Is asking on how to fix a faulty circuit on topic?" – Dmitry Grigoryev, DoxyLover
If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ I don't understand very well what you are saying. Do you have only one side of the cord exposed? Is the other side plugged in the appliance and there is no way to remove it from there? \$\endgroup\$ – nickagian Aug 3 '17 at 6:23
  • \$\begingroup\$ It is not clear what difficulties you are facing. If you know how to measure DC voltages with a multimeter (as it seems, since you measured batteries), you should be able to discover the polarity of the power cord, if the power supply is connected to mains. Please, explain your problem better. You may also embed a photo of the thing. \$\endgroup\$ – Lorenzo Donati Aug 3 '17 at 8:14
  • 1
    \$\begingroup\$ Checking for polarity involves voltage measurement. You don't need current running through the cord to do that. \$\endgroup\$ – Dmitry Grigoryev Aug 3 '17 at 8:40
  • \$\begingroup\$ Impedance mode on a multimeter \$\endgroup\$ – Voltage Spike Aug 3 '17 at 15:33
2
\$\begingroup\$

To find out which of the two leads go where at the jack connector on the other side is quite easy and you don't need any external power source for that.

Just use your multi-meter to measure the resistance between the leads and the two sides of the connector (internal and external). So put one probe of the multi-meter on one lead and then check with the second probe which of the connector sides gives you around 0Ω.

But this of course doesn't tell you where you should connect the positive and where the negative potential. For this to know for sure you have to find the appliance specifications / data sheet.

All the devices I have seen until today that use such a connector, they always use the outer part of the connector for the negative potential and the inner part for the positive one.

That should be the standard way of wiring it, but as I said you have to find information about your device to be able to know that without guessing.

\$\endgroup\$
  • \$\begingroup\$ Ok so just checking to see if this is it. I touched what I think is the positive wire (exposed) with the black multimeter lead and the red multimeter lead I touch first the outside of the jack (connector shown in the image on the right side). The mutlimeter reads around 0.2-0.8. Then I place the red multimeter lead on the inside (center) of the jack and the reading is "0.L" which I believe is zero? This then confirms that the wire I am touching with the black multimeter lead is the positive wire? \$\endgroup\$ – astrodoc71 Aug 3 '17 at 11:12
  • \$\begingroup\$ First of all the reading "0.L" means that the measured resistance is bigger than the range the multi-meter can display. In other words this means that there is no connection between the two leads. The 0.2-0.8 reading is what you need. This tells you there is a conducting path between the two leads in this case. So, the wire where you have the red lead seems to be connected with the outside of the jack. And this corresponds to what I have expected. \$\endgroup\$ – nickagian Aug 3 '17 at 11:25
  • \$\begingroup\$ Have in mind that normally those kind of cables have the negative wire marked (with white dashes for the '-') and not the positive one. So, if it is as I guess it is, and you have chosen the wire with the dashes on it as the one you think is the positive wire, then your assumption is wrong! As I said, normally the outside of the jack is the negative potential. But again, although you have found which wire goes where on the connector, only by studying the datasheet of the device you can be sure where to connect the positive potential and where the negative one. \$\endgroup\$ – nickagian Aug 3 '17 at 11:29
1
\$\begingroup\$

It only took me five minutes to find the manual for your ATIK Camera, then another minute to find the page that references the power jack.

(Do some research, if I had a $900.00 camera I would do the research my self)

Go to this link for the manual for your camera;

http://3ainmfntxe31vi9qd1pxgpd1-wpengine.netdna-ssl.com/wp-content/uploads/2015/07/AtikSeries4Manual.pdf

Go to page 5, section 3.1 it states that the power connection is "center positive"

Page 5 of user manual

\$\endgroup\$
0
\$\begingroup\$

If you now want to know which lead or wire is connected to the center post of the power jack, simply run a continuity test;

  1. Place the selector of your Volt-Meter (VM) on the continuity test mode setting

  2. Place the VM's red or black lead on the center post of the jack

  3. Place the VM's other lead on one of the two wires connected to the jack

  4. If the result of the test is a tone or the display of the VM indicates a positive test (0 resistance reading indicating a closed circuit), then label the wire and its polarity with a piece of tape

  5. If the test indicated an open connection then place the lead on the other wire and record the result, if this is the correct wire label it with a piece of tape.

  6. if neither test were positive test the VM's functionality by shorting the leads of the VM together. If the VM functionality test is positive but neither of the power jack assemblies wire's tests were positive - the power jack assembly could be damaged.

VoltMeter-Continuity Mode

\$\endgroup\$
-1
\$\begingroup\$

There should be no problem measuring the polarity of a conductor. However there does need to be some sort of current source on the wire. ie. A battery, or power supply. You say that the "cord" you have is supposed to supply 12v DC, so assuming that your "cord" is actually a AC/DC transformer or "power brick" as it's sometimes called. You should be able to plug in the cord, taking care not to allow the ends of the cord touch each other, and then take your multimeter probes to the two wires. If the voltage readout is something like (12.48) then that means the wire you have the red probe on is positive and the wire you have the black probe on is negative. If however it shows something like (-12.48) then that means the polarity is reversed, and the wire you have the black probe attached to is actually positive and the one with the red probe is actually negative. They're just switched in this scenario.

Please let me know if this explanation helps you, I can comment instead of post on my own posts since my rep is below 50.

\$\endgroup\$
  • \$\begingroup\$ @Harry Svensson You're right, but I never said there needed to be a load. Just a current source. I think you must have thought since I said current that I meant current flow. \$\endgroup\$ – Bradyn Claycomb Aug 3 '17 at 4:13

Not the answer you're looking for? Browse other questions tagged or ask your own question.