The one you already know
Power in watts is voltage times current. A mobile radio pulling 10 amperes from a 13.8 volt supply is consuming 138 watts — which, incidentally, is why a 5 amp “12 volt” wall adapter cannot run it, no matter how confidently the box is labelled.
Rearranged the useful way:
A 100 watt load on a 13.8 volt supply draws 100 / 13.8 = 7.2 amperes. Sizing a power supply, a fuse, or a length of wire is this calculation and nothing else.
Substituting Ohm’s law
Because E = I × R, you can substitute and get two more forms:
These are not extra facts to memorise. They are the same law with Ohm’s law pushed into it, and either can be re-derived in ten seconds if you forget the exact shape.
The I-squared form is the one that matters
Power dissipated in a resistance goes up with the square of the current. Double the current and you quadruple the heat. That single relationship explains most of what goes wrong physically in a radio station:
- A corroded ground connection has a small extra resistance. At receive currents, nothing happens. At transmit currents, the same resistance dissipates enough power to get warm, which accelerates the corrosion, which raises the resistance further. Bad connections fail progressively.
- Undersized DC power wire is not a performance problem first — it is a heating problem first. The voltage drop that makes the radio unhappy and the heat that melts the insulation are the same phenomenon.
- A dummy load rated for 200 watts is rated for the heat it can shed, not the RF it can absorb. 2 amperes into a 50 ohm load is 2² × 50 = 200 watts, all of it becoming heat.
Power is a rate
Watch the wording in exam questions. Power is the rate at which electrical energy is used. Energy is measured in joules or watt-hours; power is joules per second. A battery stores energy; a radio consumes power.
The distinction matters when you size a battery for field operation: a 20 amp hour battery running a radio that draws 1 amp average will last roughly 20 hours, but that same battery says nothing about whether it can deliver the 20 amp peak an amplifier wants.
Worked examples
A transceiver draws 12 A at 13.8 V. What is its DC power consumption? P = E × I = 13.8 × 12 = 166 watts.
How much current does a 60 W lamp draw from 120 V? I = P / E = 60 / 120 = 0.5 A.
A 3 A current flows through a 25 ohm resistor. What power is dissipated? P = I²R = 9 × 25 = 225 watts. That resistor needs to be physically large.
Check yourself
- A 100 W HF radio is about 50% efficient on transmit. Roughly what DC current does it draw from a 13.8 V supply at full output?
- You halve the current through a fixed resistance. What happens to the heat?
- Which formula would you reach for given only current and resistance?
Answers
- 100 W out at 50% efficiency means about 200 W in. I = 200 / 13.8 = roughly 14.5 A — which is why 100 W radios come with heavy DC leads and a 25 or 30 amp supply.
- It drops to one quarter. Heat follows the square of current.
- P = I²R.