The drop over ten metres

You need

• From the drawer: a two-conductor cord ten metres long (12 V extension, lamp cord), with the AWG gauge legible on the jacket, in copper and not copper-clad aluminium (cheap speaker wire often is, and resists one and a half times more); a 12 V source that can hold 2 A (12 V battery, bench power supply, car battery as a last resort); the multimeter; four alligator clips.
• To buy: a 12 V 21 W automotive bulb (P21W, the reversing-light bulb, 3 $) and its socket, or a 12 V 20 W halogen bulb.
• Optional: a second cord of another gauge, to compare.

Do

BeforeTwelve volts do not hurt. What hurts is the short circuit: a car battery delivers hundreds of amperes into a clip that slips, and a thin wire melts in smoke. Put a 5 A fuse in the positive lead if the source has none, and connect the load before the source. The ammeter goes in series, on the 10 A range, and never across the source: the meter’s fuse blows, at best. The bulb burns fingers after a minute.

1. Read the gauge on the jacket and note the resistance per metre from the table: AWG 16, 13.2 mΩ; AWG 18, 21.0 mΩ; AWG 14, 8.29 mΩ. Ten metres of cord is twenty metres of copper, out and back.

   expected0.26 Ω for the loop in AWG 16, 0.42 Ω in AWG 18

2. Lay the cord in a U on the bench, both ends within reach. Set up: the source, the cord, the bulb at the end; the ammeter in series near the source, 10 A range.

   expectedthe bulb lights brightly, 1.5 to 1.8 A

3. Measure the drop along one conductor, not the difference of two readings at 12 V: the voltmeter on the 2 V range, one probe on the bare copper of the positive conductor at the source end, the other on the same conductor at the bulb end, under load. Then the same on the negative conductor. The clips and the ammeter stay out of the measurement.

   expected0.20 to 0.25 V per conductor in AWG 16, 0.35 to 0.40 V in AWG 18; the two conductors within 5 % of each other

4. Calculate the cord’s resistance: the sum of the two drops divided by the current.

   expected0.25 to 0.28 Ω in AWG 16, within five percent of the table; one and a half times above, it is copper-clad aluminium, not copper

5. Calculate the power lost in the cord, the sum of the drops times the current.

   expected0.8 W in AWG 16, spread over twenty metres; the cord does not even get warm, but the bulb at the end receives 20 W instead of 21

6. Calculate the drop as a percentage of the source.

   expected3.8 % in AWG 16, 6 % in AWG 18; the rule of thumb at low voltage is three percent, and lamp cord already exceeds it

7. Redo the calculation for the same 21 W bulb fed at 48 V: the current, the drop, the heat.

   expected0.44 A, 0.11 V, 0.05 W; sixteen times less heat in the same copper, which is the reason for 48 and 56 V packs