Simple series circuits

Let's start with a series circuit consisting of three resistors and a single battery:

The first principle to understand about series circuits is that the amount of current is the same through any component in the circuit. This is because there is only one path for electrons to flow in a series circuit, and because free electrons flow through conductors like marbles in a tube, the rate of flow (marble speed) at any point in the circuit (tube) at any specific point in time must be equal.

From the way that the 9 volt battery is arranged, we can tell that the electrons in this circuit will flow in a counter-clockwise direction, from point 4 to 3 to 2 to 1 and back to 4. However, we have one source of voltage and three resistances. How do we use Ohm's Law here?

An important caveat to Ohm's Law is that all quantities (voltage, current, resistance, and power) must relate to each other in terms of the same two points in a circuit. For instance, with a single-battery, single-resistor circuit, we could easily calculate any quantity because they all applied to the same two points in the circuit:

Since points 1 and 2 are connected together with wire of negligible resistance, as are points 3 and 4, we can say that point 1 is electrically common to point 2, and that point 3 is electrically common to point 4. Since we know we have 9 volts of electromotive force between points 1 and 4 (directly across the battery), and since point 2 is common to point 1 and point 3 common to point 4, we must also have 9 volts between points 2 and 3 (directly across the resistor). Therefore, we can apply Ohm's Law (I = E/R) to the current through the resistor, because we know the voltage (E) across the resistor and the resistance (R) of that resistor. All terms (E, I, R) apply to the same two points in the circuit, to that same resistor, so we can use the Ohm's Law formula with no reservation.

However, in circuits containing more than one resistor, we must be careful in how we apply Ohm's Law. In the three-resistor example circuit below, we know that we have 9 volts between points 1 and 4, which is the amount of electromotive force trying to push electrons through the series combination of R₁, R₂, and R₃. However, we cannot take the value of 9 volts and divide it by 3k, 10k or 5k Ω to try to find a current value, because we don't know how much voltage is across any one of those resistors, individually.

The figure of 9 volts is a total quantity for the whole circuit, whereas the figures of 3k, 10k, and 5k Ω are individualquantities for individual resistors. If we were to plug a figure for total voltage into an Ohm's Law equation with a figure for individual resistance, the result would not relate accurately to any quantity in the real circuit.

For R₁, Ohm's Law will relate the amount of voltage across R₁ with the current through R₁, given R₁'s resistance, 3kΩ:

But, since we don't know the voltage across R₁ (only the total voltage supplied by the battery across the three-resistor series combination) and we don't know the current through R₁, we can't do any calculations with either formula. The same goes for R₂ and R₃: we can apply the Ohm's Law equations if and only if all terms are representative of their respective quantities between the same two points in the circuit.

So what can we do? We know the voltage of the source (9 volts) applied across the series combination of R₁, R₂, and R₃, and we know the resistances of each resistor, but since those quantities aren't in the same context, we can't use Ohm's Law to determine the circuit current. If only we knew what the total resistance was for the circuit: then we could calculate total current with our figure for total voltage (I=E/R).

This brings us to the second principle of series circuits: the total resistance of any series circuit is equal to the sum of the individual resistances. This should make intuitive sense: the more resistors in series that the electrons must flow through, the more difficult it will be for those electrons to flow. In the example problem, we had a 3 kΩ, 10 kΩ, and 5 kΩ resistor in series, giving us a total resistance of 18 kΩ:

In essence, we've calculated the equivalent resistance of R₁, R₂, and R₃ combined. Knowing this, we could re-draw the circuit with a single equivalent resistor representing the series combination of R₁, R₂, and R₃:

Now we have all the necessary information to calculate circuit current, because we have the voltage between points 1 and 4 (9 volts) and the resistance between points 1 and 4 (18 kΩ):

Knowing that current is equal through all components of a series circuit (and we just determined the current through the battery), we can go back to our original circuit schematic and note the current through each component:

Now that we know the amount of current through each resistor, we can use Ohm's Law to determine the voltage drop across each one (applying Ohm's Law in its proper context):

Notice the voltage drops across each resistor, and how the sum of the voltage drops (1.5 + 5 + 2.5) is equal to the battery (supply) voltage: 9 volts. This is the third principle of series circuits: that the supply voltage is equal to the sum of the individual voltage drops.

However, the method we just used to analyze this simple series circuit can be streamlined for better understanding. By using a table to list all voltages, currents, and resistances in the circuit, it becomes very easy to see which of those quantities can be properly related in any Ohm's Law equation:

The rule with such a table is to apply Ohm's Law only to the values within each vertical column. For instance, E_R1only with I_R1 and R₁; E_R2 only with I_R2 and R₂; etc. You begin your analysis by filling in those elements of the table that are given to you from the beginning:

As you can see from the arrangement of the data, we can't apply the 9 volts of E_T (total voltage) to any of the resistances (R₁, R₂, or R₃) in any Ohm's Law formula because they're in different columns. The 9 volts of battery voltage is not applied directly across R₁, R₂, or R₃. However, we can use our "rules" of series circuits to fill in blank spots on a horizontal row. In this case, we can use the series rule of resistances to determine a total resistance from the sum of individual resistances:

Now, with a value for total resistance inserted into the rightmost ("Total") column, we can apply Ohm's Law of I=E/R to total voltage and total resistance to arrive at a total current of 500 μA:

Then, knowing that the current is shared equally by all components of a series circuit (another "rule" of series circuits), we can fill in the currents for each resistor from the current figure just calculated:

Finally, we can use Ohm's Law to determine the voltage drop across each resistor, one column at a time:

Just for fun, we can use a computer to analyze this very same circuit automatically. It will be a good way to verify our calculations and also become more familiar with computer analysis. First, we have to describe the circuit to the computer in a format recognizable by the software. The SPICE program we'll be using requires that all electrically unique points in a circuit be numbered, and component placement is understood by which of those numbered points, or "nodes," they share. For clarity, I numbered the four corners of our example circuit 1 through 4. SPICE, however, demands that there be a node zero somewhere in the circuit, so I'll re-draw the circuit, changing the numbering scheme slightly:

All I've done here is re-numbered the lower-left corner of the circuit 0 instead of 4. Now, I can enter several lines of text into a computer file describing the circuit in terms SPICE will understand, complete with a couple of extra lines of code directing the program to display voltage and current data for our viewing pleasure. This computer file is known as the netlist in SPICE terminology:

series circuit
v1 1 0
r1 1 2 3k
r2 2 3 10k
r3 3 0 5k
.dc v1 9 9 1
.print dc v(1,2) v(2,3) v(3,0)
.end

Now, all I have to do is run the SPICE program to process the netlist and output the results:

v1            v(1,2)      v(2,3)      v(3)        i(v1)
9.000E+00     1.500E+00   5.000E+00   2.500E+00  -5.000E-04

This printout is telling us the battery voltage is 9 volts, and the voltage drops across R₁, R₂, and R₃ are 1.5 volts, 5 volts, and 2.5 volts, respectively. Voltage drops across any component in SPICE are referenced by the node numbers the component lies between, so v(1,2) is referencing the voltage between nodes 1 and 2 in the circuit, which are the points between which R₁ is located. The order of node numbers is important: when SPICE outputs a figure for v(1,2), it regards the polarity the same way as if we were holding a voltmeter with the red test lead on node 1 and the black test lead on node 2.

We also have a display showing current (albeit with a negative value) at 0.5 milliamps, or 500 microamps. So our mathematical analysis has been vindicated by the computer. This figure appears as a negative number in the SPICE analysis, due to a quirk in the way SPICE handles current calculations.

In summary, a series circuit is defined as having only one path for electrons to flow. From this definition, three rules of series circuits follow: all components share the same current; resistances add to equal a larger, total resistance; and voltage drops add to equal a larger, total voltage. All of these rules find root in the definition of a series circuit. If you understand that definition fully, then the rules are nothing more than footnotes to the definition.

Review

Components in a series circuit share the same current: ITotal = I1 = I2 = . . . In Total resistance in a series circuit is equal to the sum of the individual resistances: RTotal = R1+ R2 + . . . Rn Total voltage in a series circuit is equal to the sum of the individual voltage drops: ETotal = E1 + E2 + . . . En

Experiment: Parallel batteries

Parallel batteries

PARTS AND MATERIALS

• Four 6-volt batteries
• 12-volt light bulb, 25 or 50 watt
• Lamp socket

High-wattage 12-volt lamps may be purchased from recreational vehicle (RV) and boating supply stores. Common sizes are 25 watt and 50 watt. This lamp will be used as a "heavy" load for your batteries (heavy load = one that draws substantial current).

A regular household (120 volt) lamp socket will work just fine for these low-voltage "RV" lamps.

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 1, chapter 5: "Series and Parallel Circuits"

Lessons In Electric Circuits, Volume 1, chapter 11: "Batteries and Power Systems"

LEARNING OBJECTIVES

• Voltage source regulation
• Boosting current capacity through parallel connections

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

Begin this experiment by connecting one 6-volt battery to the lamp. The lamp, designed to operate on 12 volts, should glow dimly when powered by the 6-volt battery. Use your voltmeter to read voltage across the lamp like this:

The voltmeter should register a voltage lower than the usual voltage of the battery. If you use your voltmeter to read the voltage directly at the battery terminals, you will measure a low voltage there as well. Why is this? The large current drawn by the high-power lamp causes the voltage at the battery terminals to "sag" or "droop," due to voltage dropped across resistance internal to the battery.

Parallel Electrically common points

We may overcome this problem by connecting batteries in parallel with each other, so that each battery only has to supply a fraction of the total current demanded by the lamp. Parallel connections involve making all the positive (+) battery terminals electrically common to each other by connection through jumper wires, and all negative (-) terminals common to each other as well. Add one battery at a time in parallel, noting the lamp voltage with the addition of each new, parallel-connected battery:

There should also be a noticeable difference in light intensity as the voltage "sag" is improved.

Try measuring the current of one battery and comparing it to the total current (light bulb current). Shown here is the easiest way to measure single-battery current:

By breaking the circuit for just one battery, and inserting our ammeter within that break, we intercept the current of that one battery and are therefore able to measure it. Measuring total current involves a similar procedure: make a break somewhere in the path that total current must take, then insert the ammeter within than break:

Note the difference in current between the single-battery and total measurements.

Series-parallel

To obtain maximum brightness from the light bulb, a series-parallel connection is required. Two 6-volt batteries connected series-aiding will provide 12 volts. Connecting two of these series-connected battery pairs in parallel improves their current-sourcing ability for minimum voltage sag:

Experiment: Voltage divider

Voltage divider

PARTS AND MATERIALS

• Calculator (or pencil and paper for doing arithmetic)
• 6-volt battery
• Assortment of resistors between 1 KΩ and 100 kΩ in value

I'm purposely restricting the resistance values between 1 kΩ and 100 kΩ for the sake of obtaining accurate voltage and current readings with your meter. With very low resistance values, the internal resistance of the ammeter has a significant impact on measurement accuracy. Very high resistance values may cause problems for voltage measurement, the internal resistance of the voltmeter substantially changing circuit resistance when it is connected in parallel with a high-value resistor.

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 1, chapter 6: "Divider Circuits and Kirchhoff's Laws"

LEARNING OBJECTIVES

• Voltmeter use
• Ammeter use
• Ohmmeter use
• Use of Ohm's Law
• Use of Kirchhoff's Voltage Law ("KVL")
• Voltage divider design

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

Shown here are three different methods of circuit construction: on a breadboard, on a terminal strip, and "free-form." Try building the same circuit each way to familiarize yourself with the different construction techniques and their respective merits. The "free-form" method -- where all components are connected together with "alligator-" style jumper wires -- is the least professional, but appropriate for a simple experiment such as this. Breadboard construction is versatile and allows for high component density (many parts in a small space), but is quite temporary. Terminal strips offer a much more permanent form of construction at the cost of low component density.

Select three resistors from your resistor assortment and measure the resistance of each one with an ohmmeter. Note these resistance values with pen and paper, for reference in your circuit calculations.

Connect the three resistors in series, and to the 6-volt battery, as shown in the illustrations. Measure battery voltage with a voltmeter after the resistors have been connected to it, noting this voltage figure on paper as well. It is advisable to measure battery voltage while it's powering the resistor circuit because this voltage may differ slightly from a no-load condition. We saw this effect exaggerated in the "parallel battery" experiment while powering a high-wattage lamp: battery voltage tends to "sag" or "droop" under load. Although this three-resistor circuit should not present a heavy enough load (not enough current drawn) to cause significant voltage "sag," measuring battery voltage under load is a good scientific practice because it provides more realistic data.

Use Ohm's Law (I=E/R) to calculate circuit current, then verify this calculated value by measuring current with an ammeter like this ("terminal strip" version of the circuit shown as an arbitrary choice in construction method):

If your resistor values are indeed between 1 kΩ and 100 kΩ, and the battery voltage approximately 6 volts, the current should be a very small value, in the milliamp (mA) or microamp (μA) range. When you measure current with a digital meter, the meter may show the appropriate metric prefix symbol (m or μ) in some corner of the display. These metric prefix telltales are easy to overlook when reading the display of a digital meter, so pay close attention!

The measured value of current should agree closely with your Ohm's Law calculation. Now, take that calculated value for current and multiply it by the respective resistances of each resistor to predict their voltage drops (E=IR). Switch you multimeter to the "voltage" mode and measure the voltage dropped across each resistor, verifying the accuracy of your predictions. Again, there should be close agreement between the calculated and measured voltage figures.

Voltage divider Divider, voltage

Each resistor voltage drop will be some fraction or percentage of the total voltage, hence the name voltage dividergiven to this circuit. This fractional value is determined by the resistance of the particular resistor and the total resistance. If a resistor drops 50% of the total battery voltage in a voltage divider circuit, that proportion of 50% will remain the same as long as the resistor values are not altered. So, if the total voltage is 6 volts, the voltage across that resistor will be 50% of 6, or 3 volts. If the total voltage is 20 volts, that resistor will drop 10 volts, or 50% of 20 volts.

Kirchhoff's Voltage Law KVL

The next part of this experiment is a validation of Kirchhoff's Voltage Law. For this, you need to identify each unique point in the circuit with a number. Points that are electrically common (directly connected to each other with insignificant resistance between) must bear the same number. An example using the numbers 0 through 3 is shown here in both illustrative and schematic form. In the illustration, I show how points in the circuit may be labeled with small pieces of tape, numbers written on the tape:

Using a digital voltmeter (this is important!), measure voltage drops around the loop formed by the points 0-1-2-3-0. Write on paper each of these voltages, along with its respective sign as indicated by the meter. In other words, if the voltmeter registers a negative voltage such as -1.325 volts, you should write that figure as a negative number. Do not reverse the meter probe connections with the circuit to make the number read "correctly." Mathematical sign is very significant in this phase of the experiment! Here is a sequence of illustrations showing how to "step around" the circuit loop, starting and ending at point 0:

Using the voltmeter to "step" around the circuit in this manner yields three positive voltage figures and one negative:

These figures, algebraically added ("algebraically" = respecting the signs of the numbers), should equal zero. This is the fundamental principle of Kirchhoff's Voltage Law: that the algebraic sum of all voltage drops in a "loop" add to zero.

It is important to realize that the "loop" stepped around does not have to be the same path that current takes in the circuit, or even a legitimate current path at all. The loop in which we tally voltage drops can be any collection of points, so long as it begins and ends with the same point. For example, we may measure and add the voltages in the loop 1-2-3-1, and they will form a sum of zero as well:

Try stepping between any set of points, in any order, around your circuit and see for yourself that the algebraic sum always equals zero. This Law holds true no matter what the configuration of the circuit: series, parallel, series-parallel, or even an irreducible network.

Kirchhoff's Voltage Law is a powerful concept, allowing us to predict the magnitude and polarity of voltages in a circuit by developing mathematical equations for analysis based on the truth of all voltages in a loop adding up to zero. This experiment is intended to give empirical evidence for and a deep understanding of Kirchhoff's Voltage Law as a general principle.

COMPUTER SIMULATION

Netlist (make a text file containing the following text, verbatim):

Voltage divider
v1 3 0
r1 3 2 5k
r2 2 1 3k
r3 1 0 2k
.dc v1 6 6 1
* Voltages around 0-1-2-3-0 loop algebraically add to zero:
.print dc v(1,0) v(2,1) v(3,2) v(0,3)
* Voltages around 1-2-3-1 loop algebraically add to zero:
.print dc v(2,1) v(3,2) v(1,3)
.end

SPICE Computer simulation Simulation, computer

This computer simulation is based on the point numbers shown in the previous diagrams for illustrating Kirchhoff's Voltage Law (points 0 through 3). Resistor values were chosen to provide 50%, 30%, and 20% proportions of total voltage across R₁, R₂, and R₃, respectively. Feel free to modify the voltage source value (in the ".dc" line, shown here as 6 volts), and/or the resistor values.

When run, SPICE will print a line of text containing four voltage figures, then another line of text containing three voltage figures, along with lots of other text lines describing the analysis process. Add the voltage figures in each line to see that the sum is zero.

Experiment: Current divider

Current divider

PARTS AND MATERIALS

• Calculator (or pencil and paper for doing arithmetic)
• 6-volt battery
• Assortment of resistors between 1 KΩ and 100 kΩ in value

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 1, chapter 6: "Divider Circuits and Kirchhoff's Laws"

LEARNING OBJECTIVES

• Voltmeter use
• Ammeter use
• Ohmmeter use
• Use of Ohm's Law
• Use of Kirchhoff's Current Law (KCL)
• Current divider design

SCHEMATIC DIAGRAM

ILLUSTRATION

Normally, it is considered improper to secure more than two wires under a single terminal strip screw. In this illustration, I show three wires joining at the top screw of the rightmost lug used on this strip. This is done for the ease of proving a concept (of current summing at a circuit node), and does not represent professional assembly technique.

The non-professional nature of the "free-form" construction method merits no further comment.

INSTRUCTIONS

Once again, I show different methods of constructing the same circuit: breadboard, terminal strip, and "free-form." Experiment with all these construction formats and become familiar with their respective advantages and disadvantages.

Select three resistors from your resistor assortment and measure the resistance of each one with an ohmmeter. Note these resistance values with pen and paper, for reference in your circuit calculations.

Connect the three resistors in parallel to and each other, and with the 6-volt battery, as shown in the illustrations. Measure battery voltage with a voltmeter after the resistors have been connected to it, noting this voltage figure on paper as well. It is advisable to measure battery voltage while it's powering the resistor circuit because this voltage may differ slightly from a no-load condition.

Measure voltage across each of the three resistors. What do you notice? In a series circuit, current is equal through all components at any given time. In a parallel circuit, voltage is the common variable between all components.

Use Ohm's Law (I=E/R) to calculate current through each resistor, then verify this calculated value by measuring current with a digital ammeter. Place the red probe of the ammeter at the point where the positive (+) ends of the resistors connect to each other and lift one resistor wire at a time, connecting the meter's black probe to the lifted wire. In this manner, measure each resistor current, noting both the magnitude of the current and the polarity. In these illustrations, I show an ammeter used to measure the current through R₁:

Measure current for each of the three resistors, comparing with the current figures calculated previously. With the digital ammeter connected as shown, all three indications should be positive, not negative.

Now, measure total circuit current, keeping the ammeter's red probe on the same point of the circuit, but disconnecting the wire leading to the positive (+) side of the battery and touching the black probe to it:

Note both the magnitude and the sign of the current as indicated by the ammeter. Add this figure (algebraically) to the three resistor currents. What do you notice about the result that is similar to the Kirchhoff's Voltage Law experiment? Kirchhoff's Current Law is to currents "summing" at a point (node) in a circuit, just as Kirchhoff's Voltage Law is to voltages adding in a series loop: in both cases, the algebraic sum is equal to zero.

This Law is also very useful in the mathematical analysis of circuits. Along with Kirchhoff's Voltage Law, it allows us to generate equations describing several variables in a circuit, which may then be solved using a variety of mathematical techniques.

Current divider Divider, current

Now consider the four current measurements as all positive numbers: the first three representing the current through each resistor, and the fourth representing total circuit current as a positive sum of the three "branch" currents. Each resistor (branch) current is a fraction, or percentage, of the total current. This is why a parallel resistor circuit is often called a current divider.

Disconnect the battery from the rest of the circuit, and measure resistance across the parallel resistors. You may read total resistance across any of the individual resistors' terminals and obtain the same indication: it will be a value less than any of the individual resistor values. This is often surprising to new students of electricity, that you read the exact same (total) resistance figure when connecting an ohmmeter across any one of a set of parallel-connected resistors. It makes sense, though, if you consider the points in a parallel circuit in terms of electrical commonality. All parallel components are connected between two sets of electrically common points. Since the meter cannot distinguish between points common to each other by way of direct connection, to read resistance across one resistor is to read the resistance of them all. The same is true for voltage, which is why battery voltage could be read across any one of the resistors as easily as it could be read across the battery terminals directly.

If you divide the battery voltage (previously measured) by this total resistance figure, you should obtain a figure for total current (I=E/R) closely matching the measured figure.

The ratio of resistor current to total current is the same as the ratio of total resistance to individual resistance. For example, if a 10 kΩ resistor is part of a current divider circuit with a total resistance of 1 kΩ, that resistor will conduct 1/10 of the total current, whatever value that current total happens to be.

COMPUTER SIMULATION

Schematic with SPICE node numbers:

Ammeters in SPICE simulations are actually zero-voltage sources inserted in the paths of electron flow. You will notice the voltage sources V_ir1, V_ir2, and V_ir3 are set to 0 volts in the netlist. When electrons enter the negative side of one of these "dummy" batteries and out the positive, the battery's current indication will be a positive number. In other words, these 0-volt sources are to be regarded as ammeters with the red probe on the long-line side of the battery symbol and the black probe on the short-line side.

Netlist (make a text file containing the following text, verbatim):

Current divider
v1 1 0
r1 3 0 2k
r2 4 0 3k
r3 5 0 5k
vitotal 2 1 dc 0
vir1 2 3 dc 0
vir2 2 4 dc 0
vir3 2 5 dc 0
.dc v1 6 6 1
.print dc i(vitotal) i(vir1) i(vir2) i(vir3)
.end

When run, SPICE will print a line of text containing four current figures, the first current representing the total as a negative quantity, and the other three representing currents for resistors R₁, R₂, and R₃. When algebraically added, the one negative figure and the three positive figures will form a sum of zero, as described by Kirchhoff's Current Law.

Experiment: Potentiometer as voltage divider

Potentiometer as a Voltage Divider

PARTS AND MATERIALS

• Two 6-volt batteries
• Carbon pencil "lead" for a mechanical-style pencil
• Potentiometer, single turn, 5 kΩ to 50 kΩ, linear taper (Radio Shack catalog # 271-1714 through 271-1716)
• Potentiometer, multi turn, 1 kΩ to 20 kΩ, (Radio Shack catalog # 271-342, 271-343, 900-8583, or 900-8587 through 900-8590)

Potentiometer

Potentiometers are variable voltage dividers with a shaft or slide control for setting the division ratio. They are manufactured in panel-mount as well as breadboard (printed-circuit board) mount versions. Any style of potentiometer will suffice for this experiment.

Potentiometer, linear vs. audio taper Audio taper potentiometer Linear taper potentiometer

If you salvage a potentiometer from an old radio or other audio device, you will likely be getting what is called an audio taper potentiometer. These potentiometers exhibit a logarithmic relationship between division ratio and shaft position. By contrast, a linear potentiometer exhibits a direct correlation between shaft position and voltage division ratio. I highly recommend a linear potentiometer for this experiment, and for most experiments in general.

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 1, chapter 6: "Divider Circuits and Kirchhoff's Laws"

LEARNING OBJECTIVES

• Voltmeter use
• Ohmmeter use
• Voltage divider design and function
• How voltages add in series

SCHEMATIC DIAGRAM

ILLUSTRATION

INSTRUCTIONS

Begin this experiment with the pencil "lead" circuit. Pencils use a rod made of a graphite-clay mixture, not lead (the metal), to make black marks on paper. Graphite, being a mediocre electrical conductor, acts as a resistor connected across the battery by the two alligator-clip jumper wires. Connect the voltmeter as shown and touch the red test probe to the graphite rod. Move the red probe along the length of the rod and notice the voltmeter's indication change. What probe position gives the greatest voltage indication?

Essentially, the rod acts as a pair of resistors, the ratio between the two resistances established by the position of the red test probe along the rod's length:

Now, change the voltmeter connection to the circuit so as to measure voltage across the "upper resistor" of the pencil lead, like this:

Move the black test probe position along the length of the rod, noting the voltmeter indication. Which position gives the greatest voltage drop for the meter to measure? Does this differ from the previous arrangement? Why?

Manufactured potentiometers enclose a resistive strip inside a metal or plastic housing, and provide some kind of mechanism for moving a "wiper" across the length of that resistive strip. Here is an illustration of a rotary potentiometer's construction:

Some rotary potentiometers have a spiral resistive strip, and a wiper that moves axially as it rotates, so as to require multiple turns of the shaft to drive the wiper from one end of the potentiometer's range to the other. Multi-turn potentiometers are used in applications where precise setting is important.

Linear potentiometers also contain a resistive strip, the only difference being the wiper's direction of travel. Some linear potentiometers use a slide mechanism to move the wiper, while others a screw, to facilitate multiple-turn operation:

It should be noted that not all linear potentiometers have the same pin assignments. On some, the middle pin is the wiper.

Set up a circuit using a manufactured potentiometer, not the "home-made" one made from a pencil lead. You may use any form of construction that is convenient.

Measure battery voltage while powering the potentiometer, and make note of this voltage figure on paper. Measure voltage between the wiper and the potentiometer end connected to the negative (-) side of the battery. Adjust the potentiometer mechanism until the voltmeter registers exactly 1/3 of total voltage. For a 6-volt battery, this will be approximately 2 volts.

Now, connect two batteries in a series-aiding configuration, to provide approximately 12 volts across the potentiometer. Measure the total battery voltage, and then measure the voltage between the same two points on the potentiometer (wiper and negative side). Divide the potentiometer's measured output voltage by the measured total voltage. The quotient should be 1/3, the same voltage division ratio as was set previously: