I have a potentiometer for the volume control in my electric guitar. My multimeter reads 485K ohms on a 500K pot putting my test leads on each outside lug, which is fine I believe. When I put a test lead on the wiper and one on an outside lug my multimeter just says 1 as I turn the knob and also switch to different ohm ranges on my meter. What does this mean?
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\$\begingroup\$ Set the wiper in the middle position and repeat your experiment. When it again does the switching, does it behave like it does when your leads are unconnected, or like it does when your leads are shorted together? \$\endgroup\$– Wouter van OoijenCommented Feb 26, 2014 at 19:50
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\$\begingroup\$ It behaves like the leads are unconnected. \$\endgroup\$– user37862Commented Feb 27, 2014 at 0:31
1 Answer
It probably indicates "overrange" on your meter. Presumably the most significant (leftmost) "1" comes on when the meter leads are disconnected (on ohms ranges) as on the below typical Asian multimeter.
It would appear to mean that the pot is faulty, since the resistance from the wiper to either end should vary from about 500K to a few ohms depending on the shaft position.
Here's what the pot looks like inside (viewed from the back, as is your pictorial diagram). When it's set to minimum resistance (clockwise from the front) A is shorted to W. When it's turned all the way counterclockwise from the front pin B (shield) is shorted to W the way your amplifier is wired.
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\$\begingroup\$ When I had the pot on the guitar, no sound was coming from one of my pickups. Is the result of a faulty pot that no sound would be output to my switch or that it would be output at max volumn? The neck volumn pot is the pot I'm referring to in the diagram seymourduncan.com/support/wiring-diagrams/… \$\endgroup\$ Commented Feb 26, 2014 at 20:11
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\$\begingroup\$ I would expect it to give no sound output from the relevant pickup if the pot is faulty. If you short the two pot connections you should get maximum volume from that pickup. That will pretty much confirm it unless there's some issue with your measurement (and that's pretty straightforward). \$\endgroup\$ Commented Feb 26, 2014 at 20:15
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\$\begingroup\$ I'm sorry, what do you mean by shorting the two pot connections? \$\endgroup\$ Commented Feb 26, 2014 at 20:24
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\$\begingroup\$ Connect them together in the amplifier, as if the pot was cranked all the way clockwise. \$\endgroup\$ Commented Feb 26, 2014 at 20:45
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\$\begingroup\$ So on your pot diagram, solder a wire from term A to B? My output is usually soldered to W. \$\endgroup\$ Commented Feb 26, 2014 at 23:13