# Power

It is important for licensed  amateurs to know how much power they are allowed to use and how much power they  are actually using. In the UK the maximum amount of power permitted when  transmitting is 400w, but not on all bands. On some bands and parts of others  the maximum amount of power permitted is very much less. If you do not know how  much power you are using, you might be in breach of the terms and conditions of  your license, in which case you will lose your license. You have been warned.

Power is measured in Watts. This unit is named after a man called James Watt (1736 – 1819). 1  Watt is equal to 1 Joule per second. Joules are the units of energy, named after  James Prescott Joule who was a brewer (1818 – 1889).

If you are very old fashioned,  you might measure power in a different kind of unit = horse power. 1HP is the  same as 750 watts. I suppose this is how much power a horse is using when it  pulls a cart load of beer from the brewery to the pub.

Watts:

Power can be defined as “the rate of transferring energy“. Think about the light bulb in your bedside lamp; it might be 25w. This means that every second 25j of electrical energy are turned into light (that is only if it is 100% efficient, in fact some of the energy ends up as heat instead of light). When I am using maximum power on my motorcycle (well over the speed limit) 60,000 watts of energy are being used every second. Here are some typical values:

 25w the rating of the light bulb in a bedside lamp 100w the rating of the light bulb in my study 1Kw the power rating of my electric kettle 3Kw the power of a 3 bar electric fire 60Kw the power of my motorcycle Millions of watts the power supplied to a small town

There are a number of equations which can be used in calculating power, but the one you need to know about for the Radio Amateurs Exam (P = V x I) is explained below.

If you understood Ohm’s Law you will find this simple: there is a relationship between power, voltage and current shown in the equations below. If you know two of the values you can easily work out the third one.

 P = V x I P V = P / I I = P / V V I

The “P over V and I triangle” should help you to remember these three equations. If you know the voltage and current and want to calculate the power, you use the first equation. If you know the power and current and want to calculate the voltage, you use the second equation. Lastly, if you know the power and voltage and want to calculate the current, you use the third equation.

You can do some simple calculations for yourself using one of these equations. You know that your mains voltage is supposed to be 250v (that is a nominal value). You also know that your bedside lamp uses 25w. Using the third equation you will see that it uses 100 milliamps. What about your kettle? 1000w and 250v fed into the third equation gives 4amps.

You could go around your house and look on the back of various appliances to find their power ratings. There should be a little label on the back of your TV giving this information.

My rigs will only produce about 50w when I am transmitting by telephony. I would like to think that all the electrical energy fed into my antennas ends up as radio waves, but I know that 100% efficiency is just not possible: some of the energy will end up as heat. The current in my feeders (the co-axial cables feeding the antenna) should be about 1 amp and the voltage about 50v. (It is not really as simple as this because the currents in my feeders are r.f. (radio frequency) not ordinary direct currents.

Caveat: What happens in my radio antenna when I transmit is not quite so simple: it is not DC current. When I transmit some points on the antenna have a very high potential difference relative to earth and other points have a very high current. My radio amateurs licence allows me to use up to 400 watts on some frequencies. If I decided to purchase a more powerful transceiver (rig) you would be well advised to leave my antennas alone.