WEBVTT NOTE Power, and Why Things Get Hot 1 00:00:00.000 --> 00:00:09.482 The one you already know. P equals E times I Power in watts is voltage times current. 2 00:00:09.482 --> 00:00:32.015 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. 3 00:00:32.015 --> 00:00:55.775 Rearranged the useful way: I equals (P) divided by (E) 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. 4 00:00:55.775 --> 00:01:15.624 Substituting Ohm's law. Because E = I times R, you can substitute and get two more forms: P equals I to the power of 2 R P equals (E to the power of 2 ) divided by (R) These are not extra facts to memorise. 5 00:01:15.624 --> 00:01:27.861 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. 6 00:01:27.861 --> 00:01:42.976 The I-squared form is the one that matters. P equals I to the power of 2 R Power dissipated in a resistance goes up with the square of the current. Double the current and you quadruple the heat. 7 00:01:42.976 --> 00:01:57.389 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. 8 00:01:57.389 --> 00:02:11.725 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. 9 00:02:11.725 --> 00:02:27.385 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. 10 00:02:27.385 --> 00:02:42.344 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 2 to the power of 2 times 50 = 200 watts, all of it becoming heat. 11 00:02:42.344 --> 00:02:57.092 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. 12 00:02:57.092 --> 00:03:01.046 A battery stores energy; a radio consumes power. 13 00:03:01.046 --> 00:03:22.385 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. 14 00:03:22.385 --> 00:03:47.158 Worked examples. A transceiver draws 12 A at 13.8 V. What is its DC power consumption? P = E times I = 13.8 times 12 = 166 watts. How much current does a 60 W lamp draw from 120 V? I = P / E = 60 / 120 = 0.5 A. 15 00:03:47.158 --> 00:04:08.745 A 3 A current flows through a 25 ohm resistor. What power is dissipated? P = I to the power of 2R = 9 times 25 = 225 watts. That resistor needs to be physically large. Check yourself. 16 00:04:08.745 --> 00:04:32.574 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? 17 00:04:32.574 --> 00:04:40.596 Which formula would you reach for given only current and resistance? 18 00:04:46.596 --> 00:05:11.560 100 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. 19 00:05:11.560 --> 00:05:19.792 Heat follows the square of current. P = I to the power of 2R.