Simple Circuit analysis of potentiometer problem

AI Thread Summary
A circuit with a power supply and a potentiometer can be analyzed as a voltage divider, with the potentiometer functioning as two resistors. This holds true if the wiper is connected to a high impedance; otherwise, the input impedance of the following circuit must be considered for accurate voltage calculations. It's important to avoid connecting a high impedance directly to the potentiometer wiper, as a minimum wiper current is necessary for reliable operation. The output voltage from the wiper corresponds to the voltage at the junction of the two resistors in the divider. Understanding these principles is essential for effective circuit design and analysis involving potentiometers.
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Really simple problem but:

If you have a circuit with a power supply, a potentiometer and an output voltage taken from midway down the potentiometer and just after it; would this act like a potential divider with the potentiometer acting as two resistors?
 
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Yes, It will act like a two resistors voltage divider.
But only without the load resistor.
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Thanks for the help
 
Keep in mind that is only true if the potentiometer wiper is connected to a high impedance. If there is a finite input impedance to the circuit that follows the wiper, then that impedance needs to be included in the circuit analysis to figure out the wiper voltage.

Also, it is not good design practice to connect a high impedance to a potentiometer wiper. There is a minimum wiper current needed to keep the potentiometer reliable. This is a good reference for using potentiometers:

http://www.google.com/url?q=http://www.bourns.com/pdfs/trmrpmr.pdf&sa=U&ei=vLOFT5nRKsSQiAK8vIH3BA&ved=0CBAQFjAA&usg=AFQjCNFDFeeCAZLuimBWRSHq48nFEuHt3Q

.
 
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Yes, this circuit can be analyzed as a potential divider with the potentiometer acting as two resistors. The potentiometer can be thought of as two resistors in series, with the adjustable wiper acting as a variable connection point between them. The output voltage taken from midway down the potentiometer would be equivalent to the voltage at the junction between the two resistors in a typical potential divider circuit. The ratio of the two resistors, determined by the position of the wiper, would determine the output voltage. This concept is commonly used in circuit analysis and can be applied to this simple potentiometer problem.
 
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