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I'm trying to understand simple things about electricity reading allaboutcircuits.com web. This chapter includes image which´s principle I don't understand. Here it is: enter image description here

When supplying constant voltage to circuit where resistance and current change, how is it possible that change of resistance affects voltage on resistor?

Please help.

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Would electronics.stackexchange.com be a better home for this question? –  Qmechanic Sep 8 '13 at 17:58
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@Qmechanic - I think it's simple enough to be just a school physics circuit question rather than a specifically electronic engineer one. –  Martin Beckett Sep 9 '13 at 2:49

3 Answers 3

up vote 1 down vote accepted

The potentiometer is not a normal resistor. They are often used simply as variable resistors, but in this example its use does its name justice -- "potentiometer" = voltage measure. The float is attached to a type of slider which contacts the resistor at some point in the middle. The result is that you can treat the potentiometer as two resistors (I'll call them R1 on top and R2 on bottom) in series, where R1 and R2 always add up to the same total resistance, and therefore where the potential drop across R1 and R2 always add up to the voltage of the source on the left.

As an example, if the float is at the top, you might find that R1 = 0 and R2 = 1000 Ohms. If the voltage source is 5V, then V1 (the voltage drop across R1) = 0 and V2 = 5V. If the float is exactly in the middle, R1 = R2 = 500 Ohms. Then V1 = V2 = 2.5V. If the float is at the bottom, then R1 would be 1000 Ohms and R2 would be 0, V1 would be 5V and V2 would be 0.

The movement doesn't change the voltage across the one whole resistor, but it changes how much of that voltage drop is above and how much is below the float. As drawn in your diagram, the voltmeter/level indicator measures V2, the voltage drop that is below the float.

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Then V1 = V2 = 5V Why both 5V? –  user35443 Sep 9 '13 at 12:59
    
Sorry, I had a typo. I just edited my answer to fix it. It should have read V1 = V2 = 2.5 V. That way V1 + V2 = 5 V which it has to for all values of the resistors. –  jdj081 Sep 10 '13 at 3:57
    
Ok , thank you! –  user35443 Sep 10 '13 at 4:37

The voltage across the entire resistor in the left hand circuit doesn't change. But the float moves the potentiometer arm along the resistor.
When it is at the top it gives nearly the +ve battery voltage, when it is near the bottom it gives nearly zero voltage - so forms a variable voltage supply to the right hand circuit.

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your circuit shows a voltmeter in parallel with the entire load resistor, so the variation of the floating conductor does not change the current in that resistor, only the EMF across the voltmeter. we can assume the total current flowing through the battery does not change because the voltmeter's resistance is several orders of magnitude higher than that of the load (there is almost no current flowing through the voltmeter despite having an EMF across it.

note the modified circuit below showing the voltmeter in parallel with the resistor and floater. now the current in the circuit varies with the position of the floater. in this case power dissipation varies according to water level height, which is inversely proportional to detected voltage on the meter. there are also some other concerns which would make this circuit infeasible...but that's for another day enter image description here

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