Resistors are one of the most commonly used of electronic components.
In order to understand the theory and application of resistors it is necessary to understand Ohm’s Law.
You may cringe at the mention of mathematics but it really isn’t that complex. This page tries to explain it.
The Units of Measure
First of all we need to know about three units of measure. These are:
- Voltage, represented by the capital letter V
- Current, represented by the capital letter I
- Resistance in Ohms, represented by the capital letter R
What is Voltage?
It is a unit of potential, or push measured between two points in a circuit. The circuit may be test points on a circuit board, the terminals of a cell or a battery of cells, or even the potential between a storm cloud and the ground.
The potential represented by a Voltage can be thought of as being like the pressure of water or gas in a vessel. If a hole is made in a pressurised vessel, the rate at which the contents are forced out obviously depends upon the pressure (Voltage) and upon the resistance (size of the hole).
Current, measured in Amperes, is the rate of flow of electrons in a conductor, such as a length of copper wire.
Resistance, measured in Ohms, is quite simply the resistance to the flow of current in a conductor. Different materials have different resistance to the flow of electrical current. Good conductors, like copper or silver, have a low resistance relative to things usually used as insulators, like glass, epoxy or fibre-glass circuit boards or plastics.
Note that there is not really anything which is a true insulator. Apply a sufficient potential across any material and it will conduct. Clear air is an insulator to the level of potential present in everyday devices such as cells or batteries of cells, but a thunder storm will produce sparks of lightning between clouds, or from cloud to ground. This is because the potential difference between cloud and ground in a storm can reach many hundreds of thousands of Volts.
A Triplet of Equations
Ohm’s Law is comprised of three very simple equations, each containing the three units of measure detailed above:
- V = I * R (voltage equals current times resistance)
- I = V / R (Current = Voltage over resistance)
- R = V / I (Resistance equals Voltage over current)
From the above list of three equations it can be seen that given two of the three units of measure, it is possible to calculate the third.
There is a very simple way to remember the three equations, for which I need to describe a simple diagram.
Imagine a blank sheet of paper in front of you, upon which you have drawn a triangle, with the flat base at the bottom and the pointed apex of the triangle at the top.
Now imagine you have drawn a horizontal line across the inside of the triangle half-way down it’s height, which divides the triangle into a smaller triangle at the top and a rhombus at the bottom.
Now write in the letters which represent the three units of measure:
- Write a capital V in the top of the top smaller triangle, just under the apex.
- Write a capital I inside the lower rhombus, just inside it at what was the lower left point of the original triangle.
- Write a capital R at the lower right corner of the rhombus.
- Finally, write a small x half-way between the I and the R.
You can now imagine covering any one of the three units with the tip of one finger and ‘seeing’ the equation you need to calculate it.
Cover the V at the top, and you can ‘read off’ the equation:
V = I x R (Voltage = current times resistance)
Cover the I, and you have:
I = V / R (Current equals Voltage over resistance).
Cover the R and you have:
R = V / I (Resistance equals Voltage over current).
In the last two the line you drew to divide the original triangle represents the ‘over’ line.
If we were to place a 100 Ohm resistor across two points between which there is a potential (Voltage) of ten Volts, how much current will flow?
Here we want to calculate the current I, so imagine your fingertip placed over the I, and you are left with:
V / R (Voltage over resistance)
Replacing the letters with the known values:
10 / 100 (10 Volts over 100 Ohms)
Which obviously gives us 0.10 Amps.
Here we can use another interesting property of the calculation, if we give the resistance in Kilohms (a kilohm is 1000 Ohms) instead of Ohms, we will get the answer in milliamperes (one milliampere is one-thousandth of an Ampere):
10 / 0.1 (ten Volts over 0.10 kilohms)
Which gives 100 milliamps.
It is a Law
It is important to remember that Ohm’s Law is a law, that is given two things, the third MUST be true.
In our example above, given a 100 Ohm resistance across 10 Volts there MUST be 100 milliamps flowing in the circuit.
Another way of stating the above circuit conditions is to say that there MUST be a ‘Voltage drop’ of 10 Volts across the resistance because Ohm’s Law HAS to be obeyed by the circuit.
The above is a silly and very simple example, but imagine a resistor in a much more complex circuit. There absolutely MUST be a voltage-drop across any resistor in the circuit, and there MUST be a current flowing through a resistor given by the potential connected across it and it’s resistance.
It’s my experience that text books do not usually explain sufficiently the fact that this is a law, and that the rules of that law MUST be obeyed.
Some years ago I was in the habit of making very high Voltage power supplies to provide the anode Voltage to radio frequency power amplifiers using vacuum tubes.
It was desirable to provide a means of making the circuit cut-out if it was to draw too much current, for example in the event of a short circuit, which could cause damage to an expensive tube.
This was commonly achieved by putting a relay coil across the high voltage supply. The coil of the relay obviously has a resistance, and a voltage at which the actuator will ‘pull up’.
Thus it could be calculated what value of resistor to connect in parallel with the coil to cause there to be a voltage-drop across the coil connections in the event of the current flow reaching a certain level.
Remember, Ohm’s Law says that if current I is flowing across resistance R, there MUST be a Voltage (V) drop across the resistance. If that Voltage drop is higher than the level at which the relay actuates, then it will. In my protection circuit the relay contacts were used to cut the supply immediately and protect the tube from damage.