In many ways we can think of the flow of electricity through wires and lamps and switches etc. as being similar to the flow of water through the pipes and plumbing system of a house.
If we have a water pipe with some sort of a restriction in it, water will not flow through the pipe unless there is a pressure difference across the restriction.
The rate at which the water flows through the pipe depends upon the size and shape of the restriction and the pressure difference across it. For example water will pass through the pipe carrying the water from the cistern to the lavatory bowl when we flush it at more than two litres a second even though the water pressure is quite low, whereas a dripping tap might pass less than a litre of water an hour even though the water is at mains pressure.
Similarly the rate at which electricity flows through a resistor depends upon its resistance and the p.d. across it. We measure resistance in ohms, after Georg Ohm (1787 - 1854). The symbol for ohms is Ω, a capital omega, the last letter of the Greek alphabet. Not all browsers can show Greek letters properly, but if yours does not you can still see what a capital omega should look like by looking at some of the diagrams later on this Page.
As an aside, the first letter of the Greek alphabet is alpha, which is why we use alpha and omega for the first and the last. The second letter of the Greek alphabet is beta, hence our word alphabet (from alphabeta).
The unit of electrical charge is the coulomb, symbol C, after Charles Augustin de Coulomb (1736 - 1806). So we measure the flow of electricity (electric current) in coulombs per second. We call a current of one coulomb per second one ampere, symbol A, after André Ampère (1775 - 1836), although most non-French-speaking people usually miss out the accent. We often talk about amps rather than amperes although we should always write it as amperes (or A).
Many people have some difficulty with volts and amps. The key to mastering the difference between them is to mentally substitute potential difference for volts and coulombs per second for amps and think of water flowing through a restriction. This helps you to understand that volts is a pressure difference and amps a rate of flow.
The relationship between p.d., current and resistance is given by Ohm’s Law. We can write Ohm’s Law algebraically as
If a current of 3 A flows through a resistor with a resistance of 6 Ω the voltage across the resistor is 3 A × 6 Ω or 18 V
If there is a potential difference of 24 V across a resistor and the current through the resistor is 2 A then the resistance of the resistor is 24 V ÷ 2 A or 12 Ω
If there is a p.d. of 15 V across a 5 Ω resistor then the current through the resistor is 15 V ÷ 5 Ω or 3 A
There are some more worked examples on another Page.
Here we have a pipe with two restrictions, A and B in it
What goes in at one end of the pipe must come out of the other, that is, the flow through B must be the same as the flow through A. But although the flow rate is the same, the pressure difference needed to force the water through B at this rate is bigger that needed to force it through A.
Here we have a 3 Ω and a 5 Ω resistor in series. The current through both is 8 A. The p.d across the 3 Ω resistor is 3 Ω × 8 A or 24 V; similarly the p.d. across the 5 Ω resistor is 40 V.
The total p.d. across both resistors is therefore 24 V + 40 V or 64 V, and so the total resistance is 64 V ÷ 8 A or 8 Ω, that is, 5 Ω + 3 Ω.We have a 4 Ω resistor in series with a 12 V battery. Calculate the current.
The current is 12 V ÷ 4 Ω or 3 A.
Now we add a 8 Ω resistor in series. Calculate the new current.
The total resistance of the circuit now becomes 4 Ω + 8 Ω or 12 Ω so the new current is 12 V ÷ 12 Ω or 1 A.
These calculations, and many other calculations of this sort, assume that the resistance of a resistor does not change when the current through it changes. If this is true the graph of I against V is a straight line. Any device for which this is true is called a liniar device. Most resistors are liniar or nearly so but some other electrical components are not - there is more about these on another Page.
A buzzer is powered by a 12 V battery, and carries a current of 4 A. This makes it too loud. What value resistor must we put in series with the buzzer to reduce the current to 3 A?
When the current is 4 A the resistance must be 12 V ÷ 4 A or 3 Ω. To reduce the current to 3 A the new total resistance must be 12 V ÷ 3 A or 4 Ω so we must add a 1 Ω resistor.
There are some more worked examples on another Page.
There are two special cases. The first is where one resistor is very much lower than all the others.
Here we have a 100 Ω resistor in series with a 0.01 Ω resistor connected to a 5 V power supply.
The total resistance is 100.01 Ω so the current is 5 V ÷ 100.01 Ω or 0.049995 A. The voltage drop across the 100 Ω resistor is 100 Ω × 0.049995 A or 4.9995 V, and that across the 0.01 Ω resistor is 0.01 Ω × 0.049995 A or 0.00049995 V. (These do not quite add up to 5 V because of rounding errors inside my calculator.)
In all the earlier examples we have ignored the resistance of the wires. This example shows that provided the wires have a very low resistance, for all but the most accurate work we are usually justified in doing so. (Excluding the people who supply us with our electricity and look after the cables which bring it to us of course!) The voltage drop along a wire is given by V = I × R so if R is very low then V is very low unless I is very high. This is what may happen if there is a short circuit: short circuits are discussed on the Page on Parallel Circuits (currentlu under construction).
In the second special case one resistor is very much bigger than all the others.
The total resistance is 100 100 Ω, so the total current is 5 V ÷ 100 100 Ω or 0.00004995 A. The voltage drop across the 100 Ω resistor is 0.00004995 A × 100 Ω or 0.004995 V, while that across the 100 000 Ω resistor is 0.00004995 A × 100 000 Ω or 4.995 V (Again these do not add up quite to 5 V because of rounding errors.)
If we change the 100 000 Ω resistor for a 1 000 000 Ω resistor the current will be even lower, the voltage drop across the 100 Ω resistor will be even lower, and the voltage drop across the 1 000 000 Ω resistor will be even nearer 5 V. Ultimately, if we have an infinitely large resistance the current will be infinitely small, and therefore the voltage drop across any finite resistance will be infinitely small. I don’t know about your calculator, but mine doesn’t have an “infinitely small” key, although I am told the zero key has the same effect. But the voltage drop when an infinitely small current flows through an infinitely large resistance is not infinitely small: 5 × n ÷ n = 5 even if n = 0.
These two special cases come together in a simple switch. We can think of a switch as something which has an infinitely low resistance when it is closed and an infinitely high resistance when it is open.
When the switch is closed it has an infinitely low resistance. So the total resistance is 100 Ω, the current is 5 V ÷ 100 Ω or 0.05 A, the voltage drop across the 100 Ω resistor is 0.05 A × 100 Ω or 5 V and the voltage drop across the switch is 0.05 A × 0 Ω or 0 V (Surprise!)
When the switch is open it has an infinitely high resistance. This means that the total resistance is infinitely high, so both the current through the 100 Ω resistor and the voltage drop across it are zero. But, as shown above, the p.d. across the switch is 5 V.
If you are one of the many people who have difficulty in understanding how we can have a voltage without a current think of, or better still actually take, a party balloon. Blow it up and tie a knot in it to stop the air escaping. Stick a piece of Sellotape® or similar onto it, and stick a pin through the Sellotape. Now stick a pin into the balloon anywhere else.
If you stick a pin into a balloon the rubber will usually rip and the balloon will explode with a loud bang. If however you stick a pin through the Sellotape the Sellotape will stop the balloon fabric from ripping, and you will just make a small hole. Because there is a pressure difference between the air inside the balloon and the air outside it air will come out of the hole. If you do not make a hole in the balloon no air will come out - but there is still a pressure difference between the inside and the outside.
Alternatively, look at this circuit for a light in your house.
When the switch is off there is no current in the live wire so no voltage drop along it. Every part of it is at 230 V. Similarly every part of the neutral wire and the wire from the switch to the lamp is at 0 V, so there is a p.d of 230 V across the switch and if you touch the wires inside it you may get a shock.
