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Probably a very basic question but I have no idea why this happens.

I have a 1K potentiometer. When I measure the resistance over its legs when it is not connected to anything, the results are as expected and alter expectedly when I move the knob.

However, when I connect the pot to a 9V battery and try to measure the pot's resistance, I don't get any readings.

What is the reason for this behavior?

Utku
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3 Answers3

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Because your multimeter can't measure resistance. So it applies a known current, measures the resulting voltage, and computes the resistance from that. 1

So when you're applying an external current to the potentiometer you are upsetting the meter's procedure, and the resulting voltage is probably outside the measurement range.

1 Unless it's really old. In which case it applies a fixed voltage, measures the current, and lets you read the resistance off an inverse scale.

enter image description here

  • Then how can I achieve the following: I want to find out the max current an LED can tolerate by making a simple circuit by connecting a battery, a pot and an LED, where the resistance of the pot starts at maximum. Then while measuring the current over the LED at all times, I will alter the resistance of the pot and when the LED burns out, the last current reading will give me the max current over the LED. However the problem is, as soon as I touch the probes to the legs of the LED, the LED goes off. What is the reason for this? And how can I determine the max current over an LED in this case? – Utku Dec 19 '15 at 17:29
  • In the Avometer, what was the utility in reversing the movement direction (I believe that is what the Rev MC button is for)? – tcrosley Dec 19 '15 at 17:34
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    @Utku For measuring current, you put the probes in series with the LED, not across it in parallel. – tcrosley Dec 19 '15 at 17:35
  • @tcrosley : "rev mc" can simplify bipolar measurements without replugging : for example, charge vs discharge currents for a battery. Nowadays there's a "-" sign right there on the display... –  Dec 19 '15 at 17:58
  • The led goes off because the current is flowing through your multimeterinstead – crasic Dec 19 '15 at 18:19
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Your meter measures resistance by injecting a small [voltage or current] and measures the resulting [current or voltage.]

That is fundamentally incompatible with having a voltage applied across the component by something else, such as your battery.

You could measure the current supplied by your 9V battery and deduce the resistance as R = V / I

Ecnerwal
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schematic

simulate this circuit – Schematic created using CircuitLab

Switch your meter to mA range and connect in series with the battery. I recommend that you add a series resistor to limit the current to protect the battery and meter. 33 Ω will limit the current to about 1/4 A with the pot turned to zero and the LED failed short-circuit.

Be careful with your potentiometer. A standard carbon-film pot might be rated at 0.25 W but that's the rating for the whole track. In the case of your 1 kΩ pot max current would be

$$R = \frac{0.25 W}{1000 Ω} = 0.25 mA$$

You can see from this that a 0.25 W potentiometer might not last too long.

Be careful with your multimeter. As others have pointed out, when measuring resistance your meter applies a small current from the internal battery to the resistor being tested. The voltmeter is usually on a sensitive range because the maximum voltage will be limited to a volt or so. If you were to connect the meter, on resistance range, to a resistor with more than a few volts across it you might damage the meter.

schematic

simulate this circuit

Note that the voltmeter has a very high internal resistance and 'all' of the current will go to Rtest. A little goes to the meter but in the case of a digital meter the input impedance is usually about 10 MΩ so it's tiny.

Transistor
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