Part 1: Fundamental Concepts: Investigating Resistance

In the previous section, electrical resistance was identified as a property of materials.  If you had a rod of resistive material, its ends could be connected to a power source in the same way that a lightbulb or resistor is connected.  You could then use a voltmeter, a device that measures the difference in voltage between its two leads, and touch the rod at any point to determine the voltage that is present where you touched it.  This is shown in the figure below.  The circle with the letter “V” in it, stands for a voltmeter.  In this case, the voltmeter is shown with a red (upper) and black (lower) probe touching the circuit where the measurement is to be made.  The voltmeter shown is not really part of the circuit.  It is there to indicate the voltage present across the points the probes touch.

Don’t be confused between the symbol for a voltage source and a voltmeter.  A voltmeter (used to measure a voltage) has a V in the circle and a voltage source (which provides potential like a battery) has a “+” and “-“ in the circle.  A voltmeter indicates the voltage present on its red or positive (+) lead in reference to its negative (-) or black lead.  It is convention to use wire with red insulation to indicate the conductor is carrying a positive voltage and black insulation to indicate the wire is carrying a negative voltage. Figure 1. Resistive rod and voltmeter.

Figuring out what the voltmeter will read when its red lead is at one end or the other (at point “a” or point “c”) is easy to work out.  The voltmeter reads the potential, or voltage, between its two leads.  When the red lead is at point “a” and the black lead is at point “c” then the voltmeter is connected to the conductors attached to the voltage source’s terminals.  Given that we are treating conductors as perfect, you expect to read the value of the voltage source, which is 10V.  When the red lead is moved to point “c” at the bottom of the rod, both of the voltmeter’s leads are connected to the same place so there is no potential difference between them and the voltmeter reads 0V.

When the red voltmeter lead is moved to the middle of the rod with the black on at the bottom, you can guess what the voltage is.  Assuming that the rod is made of the same material throughout, if one end measures 10V and the other 0V, it is safe to assume that when it is measured in the middle, the voltage is 5V.  The voltmeter lead is half way down the rod so you would expect to measure half of the voltage.

Let’s use Ohm's law to mathematically figure this out.  To do this, the problem can be rearranged to make it easier to sort things out.  Refer to Figure 2, shown below, to help follow along with the discussion.  We can take our resistive rod and split it, half way down the length, and reattach the ends with a conductor.  The overall length of rod is still the same so the overall resistance hasn’t changed.  Given a conductor is used to rejoin the ends, there is no change in electrical behavior because conductors don’t electrically alter the circuit.  With that done, the circuit can be redrawn using resistor symbols.  In the far right schematic shown in Figure 2, notice that the direction of the current flow is shown and labeled.  In addition, the point where the voltage is being measured is shown with an outward pointing arrow labeled Vout.

The only tricky bit in all of this is figuring out what the values of the resistors are.  The actual resistance value of the rod isn’t known, and as you will see why in a bit, the actual value isn’t important, only the ratio is.  Assuming the rod is homogeneous (it has consistent properties over its length), then breaking it in half would mean there are two identical pieces.  Assigning the resistance of the original rod a value of R, it can reasonably be assumed that two pieces each have a value of R/2.  The actual resistance value isn’t important — what is important is knowing the ratio of the resistance values. Figure 2. Simplifying the rod problem.

The potential driving the circuit (10V) is known, and the total resistance in the circuit is R.  With these two pieces of information, it is possible to find the current, I, flowing through the circuit in relationship to the resistance R.  This is done using Ohm's law as follows:

V = I * R   ;Ohm's law

10 = I * R  ;Substitute the value for V

I = 10/R    ;Rearrange to find I

Vout is measured across the lower resistor.  Ohm's law tells us how much current flows for a given resistance with a particular voltage across it, or how much voltage appears across a resistor when there is a particular amount of current flowing through it.  Applying Ohm's law again, this time for just the lower resistor, tells us what the voltmeter reads.  Using the value found above for the current (I=10/R) the voltage across the lower resistor can be found as follows:

V = I * R                  ;Ohm's law

V = I * (R/2)            ;Put in lower resistor value

V = (10/R) * (R/2)    ;Substitute in I, found previously

V = 5V                     ;Solve for V (the Rs cancel leaving 10/2)

Our suspicions were correct, the voltage at Vout is 5V.  The two equal values resistors caused the voltage at their center junction to be ½ that applied across them.