12 volt vs. 120 volt

[dereckbc]

Get your glove ready.
No need for a circuit. The loss difference is a function of the square of the voltages: P = (V^2)/R

Don't see the 19,000%, more like 100x

And using the circuit:
100 ft, 50W, #12AWG, 104 deg F
120 volt: 0.148 volt drop, 0.17739 ohms => 0.12319 W
12 volt: 1.48 volt drop, 0.17739 ohms => 12.319 W

add: you said efficiency so:
50.123 W vs 62.319 W for 50 W
99.75% vs 80.23% = 19.52% more efficient

You have missed the point. Determine the resistance of the load into the circuit and work the voltage and power losses. I think you will find the 12 volt circuit will not even work. Yes I know about I^2 R losses, that is the point. FWIW 100 x = 10,000%

 

[mivey]

You have missed the point.

I wouldn't doubt that for a minute.

Determine the resistance of the load into the circuit and work the voltage and power losses. I think you will find the 12 volt circuit will not even work.

OK. I'll try again and see if I can follow your point.

Yes I know about I^2 R losses

I know you do. Sorry if it sounded like I was implying you didn't as I know you are well versed in this and more. Sometimes I just type things out as I am thinking along for clarity.

FWIW 100 x = 10,000%

Well I wasn't too clear here. Should have said I did not see a 19,000% efficiency change, but that the 12 volt power loss was 100 times greater than the 120 volt case. As I added, while the loss increased 100x, it was only a 19.5% change in efficiency.

 

[K8MHZ]

The trouble with 12v lights is that you need a transformer and they like to fail. Then you will find that unless you purchase the entire fixture, you won't be able to find a replacement transformer. A moot point, really, when you consider the transformer is usually covered with some sort of epoxy making it impossible to replace anyway.

 

[broadgage]

12 volt halogen lamps (mr 16 or similar) typicly have a greater effeciency and a longer life than similar size line voltage lamps.

A 12 volt 35 watt MR16 lamp gives more light than a 240 volt GU10 lamp (I presume that 120 volt GU10s are similar)
The 12 volt lamp will require a transformer which has losses, these losses reduce but dont eliminate the energy saving of 12 volt lamps.

A 12 volt 35 watt lamp worked from a transformer will draw from 38 to 40 watts from the line, still a useful saving.

The lamp life is variable, but the 12 volt lamps are usually longer lasting, and usually cheaper to buy than line voltage ones.

The transformers however cost money both initialy and for replacements, and the 12 volt lampholders sometimes fail.
Line voltage GU10 lampholders seem to last the life of the fixture.

LED or CFL would be far more efficient than either 12 volt or line voltage halogen.

 

[mivey]

You have missed the point. Determine the resistance of the load into the circuit and work the voltage and power losses. I think you will find the 12 volt circuit will not even work. Yes I know about I^2 R losses, that is the point.

I had the resistances wrong before, but I still don't know why you are saying it won't work unless you are saying the lamp won't work at 10.7 volts (a shorter run or bigger wire would make this moot). I still don't get your 19,000% efficiency change. Can you please explain, as I can't seem to get away from the following (maybe I'm stuck in a rut):

For 12 volts:
Load resistance = V^2/W = 144/50 = 2.88 ohms
L & N resistance = 0.355 ohms
Total resistance = 3.235 ohms
I = V/R = 12/3.235 = 3.7097 amps
VDrop = 3.7097 * 0.355 = 1.316 V
Vlamp = 10.684 V
Pin = V*I = 12 * 3.7097 = 44.516 W
Pout = Vlamp*I = 10.684 * 3.7097 = 39.634 W
efficiency = 89.03%

For 120 volts:
Load resistance = V^2/W = 14400/50 = 288 ohms
L & N resistance = 0.355 ohms
Total resistance = 288.355 ohms
I = V/R = 120/288.355 = 0.4162 amps
VDrop = 0.4162 * 288.355 = 0.1476 V
Vlamp = 119.852 V
Pin = V*I = 120 * 0.4162 = 49.938 W
Pout = Vlamp*I = 119.852 * 0.4162 = 49.877 W
efficiency = 99.88%

The 120 volt circuit is 10.8% more efficient than the 12 volt circuit.

If you adjust the input to get rated voltage at the lamp, you get the same efficiencies.

If the length is limited to 43 ft, the volt drop for the 12 volt circuit is only about 5% and is only 4.98% less efficient than the 120 volt circuit.

 

[broadgage]

In a properly designed lighting circuit the losses should be only a few percent, no matter what voltage is used.
For 12 volt lamps this will require large and expensive cables if the 12 volt wires are more than a few feet long.
In practice though, allmost all 12 volt lamps are worked from a transformer with 120 volt or 277 volt input (240 in the UK) The line voltage wiring is sized in accordance with code, and good practice. The 12 volt wires are only at most a few feet, and often only inches in length thus allowing standard size wire.

 

[mivey]

In a properly designed lighting circuit the losses should be only a few percent, no matter what voltage is used.
For 12 volt lamps this will require large and expensive cables if the 12 volt wires are more than a few feet long.
In practice though, allmost all 12 volt lamps are worked from a transformer with 120 volt or 277 volt input (240 in the UK) The line voltage wiring is sized in accordance with code, and good practice. The 12 volt wires are only at most a few feet, and often only inches in length thus allowing standard size wire.

No doubt. The OP was about a fixture. I was trying to crunch the numbers using a 12V distribution so I could try to get closer to dereckbc's number of a 19,000% efficiebcy difference.

FWIW, a 100 ft 12 volt run with a 50 W load would probably need about an #8 wire, but that's another story.

 

[mivey]

For those who want the fixtures at the end of the wire (I'll ignore the lamp efficiency difference and the short run of wire at the fixture):

For 12 volt fixture:
Transformer efficiency = 94%
Load resistance = V^2/W = 14400/(50/eff%) = 270.720 ohms
L & N resistance = 0.355 ohms
Total resistance = 271.075 ohms
I = V/R = 120/271.075 = 0.4427 amps
VDrop = 0.4427 * 0.355 = 0.1571 V
Vtransformer = 119.843 V
Pin = V*I = 120 * 0.4427 = 53.122 W
Ptrans = Vlamp*I = 119.843 * 0.4427 = 53.052 W
Pout = Ptrans * transformer efficiency = 53.052*94% = 49.869 W
efficiency = 93.88%

For 120 volt fixture:
Transformer efficiency = 100%
Load resistance = V^2/W = 14400/(50/eff%) = 288.000 ohms
L & N resistance = 0.355 ohms
Total resistance = 288.355 ohms
I = V/R = 120/288.355 = 0.4162 amps
VDrop = 0.4162 * 0.355 = 0.1476 V
Vtransformer = 119.852 V
Pin = V*I = 120 * 0.4162 = 49.938 W
Ptrans = Vlamp*I = 119.852 * 0.4162 = 49.877 W
Pout = Ptrans * transformer efficiency = 49.877*100% = 49.877 W
efficiency = 99.88%

The 120 volt fixture is 6.00% more efficient than the 12 volt fixture.

To get to some real voltage drop, use a 4000 ft circuit to get:
For 120 volt fixture: 4.7% Vdrop, 95.3% efficiency
For 12 volt fixture: 4.98% Vdrop, 89.32% efficiency
The 120 volt fixture is 5.99% more efficient than the 12 volt fixture.


PS: FWIW I had a typo in post #25. In the 120 volt calcs,
"VDrop = 0.4162 * 288.355 = 0.1476 V" should have been
"VDrop = 0.4162 * 0.355 = 0.1476 V"

 

[dereckbc]

The 120 volt circuit is 10.8% more efficient than the 12 volt circuit.

If the length is limited to 43 ft, the volt drop for the 12 volt circuit is only about 5% and is only 4.98% less efficient than the 120 volt circuit.

I am looking at it differently than you are. Compare the power IIR losses on the wire.

12 volt power burned up on the wire is 4.885 watts
120 volt is .0615 watts

(4.885/.0615) x 100 = 7,943 % difference. OK I missed calculation earlier because I did not include the voltage divider effect of the wire resistance.

The point I am trying to make is you drop a lot less power on the conductors as the voltage goes higher. That is why POCO's use such high voltages. Every time you double the voltage, you get a 400% increase in efficiency given the same size length and diameter of wire.