I'm rather new to the electrical space and Im starting out by trying to complete basic circuits in order to make sure I understand how all of the components work. After completing some basic labs with a 5v power source, an LED, and some resistors I decided to try replacing the resistors with a potentiometer. It's my understanding that by using terminals 1 and 2, you can turn the pot into a rheostat or adjustable resistor. So I put the pot in place of the resistor and sure enough, I could make the LED dim and brighten by adjusting the position of the knob on the pot. I also hooked up my multimeter in parallel to the LED and saw the voltage increase and decrease with the LED. This all made sense to me and was how I thought things should work. However, I then unplugged the LED from the breadboard just leaving the multimeter leads attached to the circuit. The reading returned immediately to 5v and stayed there regardless of where the pot was set. So now Im a little confused. Why would my multimeter only register a change in voltage when the LED was in parallel? Shouldnt the multimeter still be registering the changes in voltage based on where the pot is set?

I've found that if I configure the pot using all 3 terminals then it works as expected and I see the voltage changing on the multimeter even without the LED. In that configuration Im using terminal 1 as ground, 2 as output, and 3 as input. Im still confused why the initial experiment with using the pot as a rheostat with only two terminals wont register voltage change on my multimeter though.

  • \$\begingroup\$ In your second paragraph you describe using the 3 terminals of the potentiometer. In this configuration it is a voltage divider. Only using two terminals will make the potentiometer act as a variable resistor. So in this way, a potentiometer can be used both as a variable resistor, or a voltage divider depending on how you wire it up. \$\endgroup\$ – zack1544 Jan 5 '17 at 5:09
  • \$\begingroup\$ And an addition to the above comment I made, a voltage divider and a series resistor do completely different things. A series resistance, limits the current. A voltage divider, as the name suggests has a voltage output that is a fraction of the input. Hope this helps \$\endgroup\$ – zack1544 Jan 5 '17 at 5:15
  • \$\begingroup\$ Im worried that I confused the issue by bringing up the voltage divider use case I discovered. Should the initial test of using two terminals (pot as a variable resistor) have worked? Should I have been able to see variable voltage on my multimeter when adjusting the pot in that configuration? \$\endgroup\$ – jonlan Jan 5 '17 at 5:17
  • \$\begingroup\$ Please provide a schematic with the tool. \$\endgroup\$ – Voltage Spike Jan 5 '17 at 7:09
  • \$\begingroup\$ You'll have to excuse the awful diagram but Im not really clear on how to draw accurate circuit diagrams. I think this delivers the point though. The multimeter and the LED are in parallel and then the LED is removed leaving the multimeter attached. imgur.com/a/RMygG \$\endgroup\$ – jonlan Jan 5 '17 at 13:21

In your first test, using two terminals of the pot, the pot and the multimeter form a voltage divider. The input resistance of a multimeter is typically ~ 10 Megohms, and I expect that the resistance of your potentiometer is much less than that. With the pot and meter in series across the power supply, they form a votlage divider, and almost all of the supply voltage will be dropped across the meter, due to its very high resistance relative to the potentiometer.


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