# Potentiometers

by Kagetora
Tags: potentiometers
 P: 3 This isn't strictly physics, but I believe it relates enough to be relevant to this forum. I was wondering if anyone could explain to me how a potentiometer works? I'm a chem student trying to get the hang of electrochemistry, and I ran into a problem trying to understand how potentiometers relate to voltmeters. Specifically, I can't find any explanation anywhere that makes a lick of sense to me. My book explains it as follows: "When current flows through a wire, the frictional heating wastes some of the useful energy of the cell. A traditional voltmeter therefore measures a potential lower than the max cell potential. The key to determining maximum potential is to perform the measurement under conditions of zero current so that no energy is wasted. This measurement is accomplished by inserting a variable voltage device in opposition (italics--woo, that really helps) to the cell potential. The voltage on this instrument, called a potentiometer, is adjusted until no current flows into the cell circuit. Under such conditions the cell potential is equal in magnitude and opposite in sign to the voltage setting of the potentiometer. This value represents the max cell potential, since no energy is wasted heating the wire." Which leads me to ask: How does performing the measurement under zero current waste no energy in the measurement? How do you even measure a voltage with zero current? What does it mean when it says "in opposition"? I assume it's talking about a resistor. I suppose if I could only find a picture that shows this process step by step, I'd understand it. If anyone could help, I'd appreciate it.
 Emeritus Sci Advisor PF Gold P: 10,424 You can think of a potentiometer as a device which has an "input" and an "output." The input is some supply voltage, say 5V, referenced to ground. The output can vary between ground and 5V by turning a little dial. When the dial is turned all the way in one direction, the output is the same as ground. When the dial is turned all the way in the other direction, the output is the same as the supply. - Warren
 P: 3 I understand that you can vary output from a source with a potentiometer; I still don't understand how a potentiometer actually works.
Emeritus
PF Gold
P: 10,424

## Potentiometers

A potentiometer can be thought of as a length of resistive material (like graphite), and a movable "wiper" which contacts the resistive material.

You can move the wiper from the bottom to the top of the resistive material by turning a dial.

The resistance from the wiper to ground might be called R1, while the resistance between the wiper and the supply might be called R2. As you move the wiper, the ratio of R1 to R2 changes. R1 and R2 are simply a voltage divider.

- Warren
HW Helper
P: 2,886
 Quote by Kagetora This isn't strictly physics, but I believe it relates enough to be relevant to this forum. I was wondering if anyone could explain to me how a potentiometer works? I'm a chem student trying to get the hang of electrochemistry, and I ran into a problem trying to understand how potentiometers relate to voltmeters. Specifically, I can't find any explanation anywhere that makes a lick of sense to me. My book explains it as follows: "When current flows through a wire, the frictional heating wastes some of the useful energy of the cell. A traditional voltmeter therefore measures a potential lower than the max cell potential. The key to determining maximum potential is to perform the measurement under conditions of zero current so that no energy is wasted. This measurement is accomplished by inserting a variable voltage device in opposition (italics--woo, that really helps) to the cell potential. The voltage on this instrument, called a potentiometer, is adjusted until no current flows into the cell circuit. Under such conditions the cell potential is equal in magnitude and opposite in sign to the voltage setting of the potentiometer. This value represents the max cell potential, since no energy is wasted heating the wire." Which leads me to ask: How does performing the measurement under zero current waste no energy in the measurement? How do you even measure a voltage with zero current? What does it mean when it says "in opposition"? I assume it's talking about a resistor. I suppose if I could only find a picture that shows this process step by step, I'd understand it. If anyone could help, I'd appreciate it.
I actually thought that your book did a nice job

There is energy "wasted" or "dissipated" whenever there is Joule heating...power is dissipated through a resistor following Power= R I^2.
To avoid that, there must be no current through any resistor. Now, even if there is no external resistor connected to the battery, there is always some resistive effect inside a real battery so that as soon as some current flows, you lose energy and the potential difference across the terminals of the battery is affected.

The potentiometer is connected to apply a voltage opposite to the one of the voltage you are trying to measure (the positive terminal to the positive terminal). If you adjust the voltage of the potentiometer until there is no current at all, you know that the true voltage of the battery is equal to the voltage of the potentiometer...all that with no current flowing and so no waste of energy.

Hope this makes sense.

Pat
 Emeritus Sci Advisor PF Gold P: 10,424 The problem you're trying to solve is this: your chemical cell has a relatively high output impedance. This means that as you draw current from it, its voltage droops. What you want to do is measure the voltage of the cell without actually pulling any current from it -- a so-called "no load" condition. You can achieve this by connecting the output of the cell to a potentiometer, and then varying the potentiometer until no current flows into or out of the cell. Then, disconnect the cell and measure the voltage produced by the potentiometer. In the real world, you'd probably use a lab power supply, which have very vey low output impedances, to perform the same kind of experiments. You'd just have to vary the supply's output voltage until the current through the circuit is zero. - Warren
 Sci Advisor HW Helper PF Gold P: 1,382
 P: 3 I suppose it would help if I emphasized that I have almost no understanding of the physics of electricity whatsoever beyond the very basic concepts explained in my chemistry book, which is why I can't just look at a wikipedia explanation and go, "Oh, so that's how it works!" :P This: "The potentiometer is connected to apply a voltage opposite to the one of the voltage you are trying to measure (the positive terminal to the positive terminal). If you adjust the voltage of the potentiometer until there is no current at all, you know that the true voltage of the battery is equal to the voltage of the potentiometer...all that with no current flowing and so no waste of energy." and this: "You can achieve this by connecting the output of the cell to a potentiometer, and then varying the potentiometer until no current flows into or out of the cell. Then, disconnect the cell and measure the voltage produced by the potentiometer." ... sort of make sense to me. How, though, can you tell that there is no current flowing in or out of the cell if you're already applying a resistive voltage in the opposite direction? Does the current of the cell itself sort of negate the current of the potentiometer until it is balanced both ways?
 Emeritus Sci Advisor PF Gold P: 10,424 If you apply two equal voltages to both ends of a wire, no current will flow through the wire. You can use an ammeter (current meter) to measure the current through the wire. - Warren
 Sci Advisor HW Helper PF Gold P: 1,382 When using a potentiometer as an instrument for measurement of electrical potential, there are several things you have to do: 1) connect a resistance to a power source, meaning tie the leads from either end of a slidewire to the terminals of a battery, DC power supply, or the silver and iron forks you stuck into a raw frog leg (we'll forget AC measurements for the moment); 2) connect one terminal of the unknown voltage source to the corresponding terminal of the potentiometer power supply (+ to +, or - to -, whatever is convenient, or specified in the potentiometer instructions); 3) connect the other terminal of the unknown to one terminal of an ammeter, for high sensitivity, a galvanometer); 4) connect the other terminal of the galvanometer to the potentiometer "tap" which makes contact with the potentiometer resistance at any point between the two power supply connections; 5) adjust the "tap" until the galvanometer indicates zero current passing from (or to) the tap through the galvanometer to (or from) the unknown voltage source; 6) the "tap" voltage is then equal to the unknown source voltage, and zero current is being drawn.Help any? No one's specifically said to you that the two voltage sources (potentiometer and unknown) have to be tied together at a common point, and then the voltage at the other unknown terminal and the "tap" voltage are compared with the galvanometer. Thought I'd go ahead and spell it out for you.
P: 92
 Quote by Bystander When using a potentiometer as an instrument for measurement of electrical potential, there are several things you have to do: 1) connect a resistance to a power source, meaning tie the leads from either end of a slidewire to the terminals of a battery, DC power supply, or the silver and iron forks you stuck into a raw frog leg (we'll forget AC measurements for the moment); 2) connect one terminal of the unknown voltage source to the corresponding terminal of the potentiometer power supply (+ to +, or - to -, whatever is convenient, or specified in the potentiometer instructions); 3) connect the other terminal of the unknown to one terminal of an ammeter, for high sensitivity, a galvanometer); 4) connect the other terminal of the galvanometer to the potentiometer "tap" which makes contact with the potentiometer resistance at any point between the two power supply connections; 5) adjust the "tap" until the galvanometer indicates zero current passing from (or to) the tap through the galvanometer to (or from) the unknown voltage source; 6) the "tap" voltage is then equal to the unknown source voltage, and zero current is being drawn.Help any? No one's specifically said to you that the two voltage sources (potentiometer and unknown) have to be tied together at a common point, and then the voltage at the other unknown terminal and the "tap" voltage are compared with the galvanometer. Thought I'd go ahead and spell it out for you.
let me add this for the benefit of the non-electrical engineer who asked the original question:

picture a meter-stick with a thin wire stretched along the length of it, over the numerical markings, 1-100cm.
a power supply is connected to the wire: one end to each end of the meter stick. the power supply is turned up a little, so that there's some voltage difference from one end of the meter stick to the other. you've got a pretty accurate voltmeter handy, so you can set the voltage across the total length to be, say, one volt. with a lot of decimal places to the right of the "1".

now, you have your electrochemical cell, and it's got some unknown voltage appearing across its terminals, and as bystander well-put it, if you draw any current from it, the internal losses will lower the voltage at its external terminals, so you won't know the REAL voltage inside.

thus, the trick is to find the voltage without drawing any current from the cell.

this takes two more steps: first, connect one side of the cell to one end of the meter stick. preferably the "minus side" connection of the power supply, but you can change ends if the next step doesn't work....

next, get yourself a hugely sensitive current meter, called a galvanometer. this baby will deflect its needle if the tiniest difference in voltage appears across its terminals.

connect one side of the galvanometer to the side of your e-chem celll that's not connected to the meter stick.

now, and this is the cool part... take the other connection from the galvanometer and run a wire over to the meter stick, and slide this wire up or down the meter stick until the galvanometer reads exactly zero!

for example, if the power supply connected to the meter-stick is set to exactly [in quotes] 1.000 volts, each millimeter of the meter stick will equal exactly [again in quotes] .001 volts.

thus, the "distance up the meter stick" is proportional to the voltage of the electrochemical cell when the galvanometer reads exactly zero.

try drawing this as a diagram. it might help, too.

hope this helped. if not, email me and i'll draw it out and send you a picture.
+af
 HW Helper P: 1,446 A potentiometer is used to mention the electrical potential $\xi$ of an unknown source. This can only be achieved if one measures the source while drawing no current from it, since if the source do have an internal resistance the measured voltage over its terminals will drop below its potential. The potentiometer measures the potential of such a source by connecting it to a variable known source (they are connected so that the two sources oppose each other, the source with the larger output voltage wiil then control the flow of current in the circuit). If there is a difference in output volts between the two sources then there will be current flowing in the circuit. When the two sources have the same output (by adjusting the known source) no current will flow in the circuit (a galvanometer in the circuit is used to monitor the flow of current) and we have the potential of the unknown source by looking at the setting of the known source.
P: 92
 Quote by andrevdh A potentiometer is used to mention the electrical potential $\xi$ of an unknown source. This can only be achieved if one measures the source while drawing no current from it, since if the source do have an internal resistance the measured voltage over its terminals will drop below its potential. The potentiometer measures the potential of such a source by connecting it to a variable known source (they are connected so that the two sources oppose each other, the source with the larger output voltage wiil then control the flow of current in the circuit). If there is a difference in output volts between the two sources then there will be current flowing in the circuit. When the two sources have the same output (by adjusting the known source) no current will flow in the circuit (a galvanometer in the circuit is used to monitor the flow of current) and we have the potential of the unknown source by looking at the setting of the known source.
---- in physics and chemistry.

many of the web references or wikipedia entries or google results will include "potentiometers" which are used as voltage dividers, as in any and all analog volume controls in audio equipment. in those situations, current does flow and resistance/impedance of the load matters, but not at all like the galvanometer-balancing situations described below.
+af

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