A very basic introduction to some simple maths that allow you to work out power, current and resistance.

In some of the guides I have included calculations for working out current, resistance etc. A basic understanding of how to do this will help when fault-finding or deciding on the correct cable size when adding accessories. If you have little or no electrical background it can get confusing, so hopefully this guide will make it clearer.

Lets look at the four key terms:

* Voltage* – measured in Volts, Symbol ‘V’

* Current* – measured in Amps, Symbol ‘I’

* Power* – measured in Watts, Symbol ‘W’

* Resistance* – measured in Ohms, Symbol Ω (Omega)

**So how are these four things related?**

You don’t need to know the theory behind the formulas, just remember these two triangles:-

Each triangle shows the three most used formulas that you will need. There are a few more and I’ve included a copy of the “Electric Wheel” at the end.* (My old Physics teacher told us an easy way to remember them “The Vatican is Roman Catholic” and “The Pope lives in the Vatican and he’s Catholic” – I meaning Current)*

As we usually know two parts of the equation, it’s easy to work out the third.

**Practical Examples**

Lets work out the current for a 12 volt television. The rated power for the television found from the manufacturers data is 65 Watts. We know the voltage is 12 volts so using P / V = I we can now work out the current:

P **65** Watts / V **12** Volts = I **5.41** Amps

If we were to measure the current using our multimeter, we probably wouldn’t get a reading of 5.41 Amps. The reason is it’s highly unlikely that the voltage is exactly 12 volts. If we are starting with a fully charged battery, the voltage might be 13.5 Volts, so working it out again for the corrected voltage we get:

P **65** Watts / V **13.5** Volts = I **4.81** Amps

We can see that the current drawn is less. What’s going on? Well manufacturers usually give a ‘nominal’ voltage rating, in this case 12 volts. That’s what they have based their calculations on. So that’s what we will use. When an Electrical Engineer designed the wiring for your caravan they will have used the ‘nominal’ voltage to choose the correct size cable and fuses for various circuits, with a little safety factor built-in of course.

Lets look at another example. The 12 volt accessory socket (cigarette lighter) has a fuse rated at 10 Amps – What wattage appliance can I plug in?

This time we will use V x I = P so:

V **12** Volts x I **10** Amps = P **120** Watts

A device with a maximum rating of 120 Watts can be used. Again we used the ‘nominal’ voltage for this calculation. If we used the actual fully charged battery voltage we would get 13.5 x 10 = 135 Watts which if we used as the battery voltage dropped would overload the circuit – P / V = I or 135 / 12 = 11.25 Amps so greater than the 10 Amp fuse.

Working anything out for mains appliances is the same. How much current does a 1.2Kw kettle draw? (1.2Kw = 1200 Watts)

P **1200** Watts / V **240** Volts = I **5** Amps

You are on a 10 Amp EHU – How many Watt’s is that?

V **240 **Volts x I **10** Amps = P **2400** Watts

**Resistance**

Your fridge has stopped working on 12 volts and you want to check the 12 volt heating element is OK by measuring the resistance. The information for the element says it is 12 volts, 170 Watts so we need to calculate the resistance in order to check it. First we need to work out the current:

P **170** Watts / V **12** Volts = I **14.1** Amps

Now we can work out the resistance by using:

V **12** Volts / I **14.1** Amps = R **0.85** Ω Ohms

Now that was long winded and required two calculations. If you look at the Electric Wheel below, you can se we could have used V² / P = R

V² (**12** x **12**) / P **170** Watts = R **0.85** Ω Ohms

So you can now measure the resistance of the heating element and you have a calculated figure as what to expect.

It’s as easy as that. However a word of caution. When you calculate wattage, resistance or current, these should be guides. In the real world things are not always to tolerance and you can get slight variations, so treat you calculated result as a guide and not an absolute.

**Electric ‘Wheel”**

There are various adaptations of this available on the internet, but this one shows all the basic formulas clearly…

*A simple way of having all the electrical formulas to hand (c) Unknown*

Copyright © 2011 – 2021 Simon P Barlow – All rights reserved

david bellamy

said:that was very helpful

Saittao

said:This is awesome I have always had a hard time learning and it takes a certin person for me to learn from and this was perfect for me! Thank you for much for sharing this Info!!!!!!

Ignatius Kabitchi

said:so helpful, thanks.

Paul Kelley

said:Why is Amps abbreviated to “I” instead of A and Watts abbreviated to “P” instead of W. Isn’t it a bit like calling an apple an orange? I’m sure there’s a reason, just wondering what it is, apart from making it harder to understand.

Paul

Simon Barlow

said:Hi Paul

Good question.

The use of the letter “I” originates from the French phrase intensité de courant, or in English current intensity. The “I” symbol was used by André-Marie Ampère after whom the unit of electric current is named. Some did not change from using C to I until 1896 as C was being used in formulas by physicists to denote the speed of light ( think it was used to denote the speed of light as a constant, hence “C”)

The watt unit is named after James Watt, the inventor of the steam engine and is a derived unit of power. One watt is defined as the energy consumption rate of one joule per second.

1W = 1J / 1s

One watt is also defined as the current flow of one ampere with voltage of one volt. 1W = 1V × 1A

Cheers

S

David Roberts

said:I have never been able to grasp this….so I ave up physics at O-level! So if you have a, say, 1kw something and an inverter….how many amps at 12v do you need?

Simon Barlow

said:Hi David

It’s quite simple, power is always the same no matter what the voltage. So 1Kw (1000w) at 230 volts = P (watts) divided by volts. 1000 / 230 = 4.347 Amps.

The same 1000 watts if supplied by a 12 volt system is 1000 / 12 = 83.33 Amps. The inverter power usage will probably add another 3 to 4 amps to this.

So to power a 1Kw running on 230 volts via an inverter from a 12 volt battery will mean you are drawing around 85 to 90 Amps.

David Roberts

said:Thanks…that helps me…so…if you need 90 amps what battery capacity should you have so you don’t b. them up? That’s assuming you do not want to pull 1kw for more than a few minutes..or is considering doing this off 12v a bit silly?

Simon Barlow

said:Hi David

Personally I wouldn’t look at a battery less than 120 Ah….. If physical size and weight is not an issue, I’d opt for two 100 Ah batteries connected in parallel. I would also keep the connecting leads from the batteries to the inverter as short as possible.

C Hodges

said:How do you test a 12 volt 6 prong relay?

Thank you.

Simon Barlow

said:Hi

The first thing you need to know are the pin-outs… or what pin is connected to what. A really good guide can be found here as it shows all the automotive types.. https://www.12voltplanet.co.uk/relay-guide.html

Once you know the correct pin connections using a 12 volt battery to energise the relay’s could you can test each contact to make sure it’s working as it should. If you see…

“C” marked, this is the ‘Common’ connection. “NO” means ‘Normally Open’

“NC” means ‘Normally Closed’

“Normally” is when the relay is not powered.

It is a simple matter now of using a multimeter to check the individual switch contact to make sure when the relay is not powered they are either closed or open and when the relay is powered up they change over correctly.

Jeffrey K Weaver

said:You would also need to check the resistance of the four non coil terminals to verify the relay is working properly, correct?

Simon Barlow

said:Hi Jeffrey

Generally if relay stops working it’s down to the coil either going OC or SC…. personally I’ve never come across a relay that has failed due to the contacts with one exception – I had a 40 amp relay that welded closed after carrying the starting current of a V6 4 x 4 Diesel.

Jeffrey K Weaver

said:Thank you for the reply. I’m trying to learn and came across this post which was very helpful.