# Volts, Amps (current), resistance (Ohms) and powerUpdated a year ago

Let’s start with a look at the common terms used in electricity: Volts, Amps (current), resistance (Ohms) and power.

Everyone should have a ‘gut feel’ for what happens in a water sytem. In Fig 1 below you’ll agree that the higher I make the tank on the hill, the greater the pressure I would expect at the bottom of the pipe and as a result how much water will come out of the pipe into the bucket. Compare this with the tank on the roof of the house, assuming we use the same diameter pipe, you would rightly expect much less water to come out. The difference is the height of the tank or ‘head of water’ as it is also called. This pressure, corresponds nicely to the electrical term voltage (V). Note that many electrical units are named after real people so we give them capitals when the symbols are used.

Figure 1 Water pressure is like voltage

How do we compare currents? If you took a stop watch and timed how long it takes to fill the bucket in each case, the bucket near the hill will fill quicker. Why? Because more water (electrical current, amps) is flowing. So we can conclude that the higher the pressure (V) the more current (A also written I) will flow. This is very important to grasp. What else governs the time it takes to fill the bucket? Of course the diameter of the pipe. If we used thin tubing of 2mm it might take hours, but if we used a 50mm diameter pipe it might take seconds. This is what in electrical terms we would call resistance (R) and its unit is Ohms (Ω)

I came across this funny cartoon on the Internet showing Volt trying to push Amp through a constriction in the pipe being caused by Ohm. As you can see, the tighter Ohm squeezes his rope, the less of Amp will get through, also the harder Volt pushes, the more of Amp will get through! Spend some time pondering this This can be written as and is |

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**Ohm’s Law:**

**Ohm's law** states that the current through a conductor between two points is directly proportional to the potential difference across the two points. Where I is the current through the conductor in units of Amps, **V** is the potential difference measured *across* the conductor in units of Volts, and **R** is the resistance of the conductor in units of Ohms. More specifically, Ohm's law states that the **R** in this relation is constant, independent of the current.

Figure 2 Real electrical circuit

In this circuit we have a battery (1.5V) connected with wires of negligible resistance to a torch-globe of 2 Ohms. From the preceding Ohm’s law formula, we can calculate the current in the globe as

I = V/R or I = 1.5/2 = 0.75A. We can also express this in a more comfortable way for small currents as milliAmps (thousandths of an Amp) ie 750mA.

You might have picked up that our hill and tank diagram is not a true circuit, as the end does not go back to the beginning (that’s where the word circuit comes in) so lets close the loop as in Fig3.

This is looking more like a circuit now and the pump would be like a battery and the water wheel like the globe, ie its doing some work now. Note, to make it a real circuit with no breaks we would close the top of the tank and bleed out all the air.

**Figure 3 Pump is like voltage**

**Some worked examples:**

In the circuit above in fig. 2, let V = 12V and R = 100Ω we want to know I. In the triangle above put your finger over the one we want to calculate (I) and what’s left visible is V/R.

So, I = V/R = 12/100 = 0.12A (or 120mA)

Now lets find R assuming we only know V and I, once again put your finger over R and you get V/ I

So, R = V/ I = 12/0.12 = 100 Ω

Now lets find V assuming we only know R and I, once again put your finger over Vand you get I * R

So, V = I * R = 0.12 * 100 = 12V

OK, that was easy; now lets look at a real circuit where we have a long wire with some resistance due to the length of the wire and a resistance at the far end:

In the circuit at the left we have a battery (or PSU= Power Supply Unit) feeding a long roll of wire; the end of the wire is connected to a resistor. The V’s in circles are places where we will measure the voltage with a voltmeter. R |

The burning question is how much voltage are we going to have left at the end? If we tried to use our formula which value for R are we going to use? The answer is we first have to know the current in the whole circuit and to do that we must add up the total resistance.

So, R_{TOTAL} = R_{WIRE} + R_{L} = 0.1 + 10 = 10.1 Ω

Now we can find the current:

I = V/R = 12/10.1 = 1.19A Now, because we have the current (which of course has to be the same in both resistors) we can finally work out the voltage left across R_{L}

V = I *R = 1.19 * 10 = 11.9V

So you can see we have ‘lost’ a bit of voltage - where did it go? It’s warming the roll of wire!

Now let’s look at a practical application of what we just learned, but first there is one more term to understand and that’s Power with the symbol P and it’s measured in Watts with symbol W. Power in a circuit is simply the Volts times the Amps ie P = V * I. In the above example the power in the resistor is:

P = V * I = 11.9 * 1.19 = 14.16W

But how much power is being wasted in the roll of wire? To find that we need to know the voltage across the roll, which is simply the difference between the input and the output voltages, ie

(12 – 11.9) = 0.1V

So, the wasted power heating up the roll of wire would be P = V * I = 0.1 * 1.19 = 0.12W. It doesn’t seem much, but remember it’s ‘stealing’ power that was meant for the load.

Now let’s look at a practical application of this:

Here we have almost the same circuit but with a LED strip as the resistor. In this real case we don’t know the resitance of either the roll or the LED’s but we are going to work it out. What we know, is the LED strip is rated at 9.6W/m and is 30cm (0.3m) long, and the roll is 100m long. |

We start by measuring the voltage at the start of the roll (at the PSU) and it’s 12V, then we measure the voltage at the end of the roll and it’s 10.5V in other words we have ‘lost’ 1.5V in the roll.

Using the power formula we can find the current in the circuit: if the LED is 9.6W/m and we have 1/3 of a meter the power used by the LEDs is 9.6/3 = 3.2W

We measured the voltage across it to be 10.5V so the current will be I = P/V = 3.2/10.5 = 0.31A

Because we know the voltage at the start and the end of the roll of wire we know the voltage across it = (12 – 10.5) = 1.5V. We also know the current through it (see above) is 0.31A

So using the Ohm’s law formula we have the resistance of the roll: R = V/ I = 1.5/0.31 = 4.8 Ω (for 100m)

We are now in a powerful position to predict how much dimming in any circuit if we use this (rather poor) wire – probably rip cord!

If 100m has a resitance of 4.8 Ω the Ohms per meter is 4.8/100 = 0.048 Ω/m

Now lets apply this: Assume we use 20m of this wire to light a LED spotlight that is rated at 5A at 12V (this would be a power of 12*5 = 50W). The resistance of our piece of wire of 20m is: 20*0.048 = 0.96 Ω

If we now try draw 5A through the wire the voltage across the wire will be V= I *R = 5*0.96 = 4.8V

This now leaves you with only (12-4.8) = 7.2V to light the spotlight!!!!!!!

So you can see the huge impact the wire size and consequently its Ohms/meter has. The key message is: USE THICKER WIRE!!!!

Lets have a look at other power examples and why we can use thin wire for the 230V mains input but not for the DC output. Assuming we have a 300W, 230V to 12V PSU fully loaded (ignoring the 20% under-rating for the moment) power going in is: P=V * I, so I = P/V = 300/230 = 1.3A However the power going out is the same because you can’t get more power out than you put in, so now V is 12 and therefore I = P/V = 300/12 = 25A Big surprise! Note that V* I is the same in both cases ie 300W.

You should see now the big difference in wire sizes required between input and output. The key message is THE HIGHER THE CURRENT THE THICKER THE WIRE required. In a typical installation, assuming 10m of wire needed, the incoming mains could perhaps use wire of 1.5mm^{2} cross-section area but the outgoing 12V would need 5mm^{2} (see wire tables later on).

What happens if we put voltages in series or parallel? Fig 4 below shows that in parallel, the pressure (V) remains the same but in series they add up.

Figure 4 Voltages in series or parallel

These 2 waterfalls clearly show a comparison between a high voltage circuit with not much current flowing and a low voltage one with plenty of current flowing. Why? The resistance (the gap ) is much less in the second case.

Figure 5 Voltage and current

The last analogy to consider is power. We are not talking about power in the source here (as in a 300W power supply), but the power being used up in the load. In Fig 3 above, the water wheel is turning and could be coupled to a grinder or generator. The power it would be able to exert will be due to 2 things; the pressure (V) across it, and the flow (A or I) through it. It’s easy to calculate: P=V* I. In our circuit in fig 2 that would mean P = 1.5*0.75 which is 1.13W (W is Watts – already introduced)