Electronics > Projects, Designs, and Technical Stuff
Getting the correct wire guage for 12V/5A LED extension wires
cliffyk:
--- Quote from: DW1961 on July 17, 2020, 09:20:55 pm ---
--- Quote from: cliffyk on July 17, 2020, 09:10:53 pm ---
--- Quote from: DW1961 on July 17, 2020, 09:02:41 pm ---
--- Quote from: blueskull on July 17, 2020, 05:37:48 pm ---
--- Quote from: DW1961 on July 17, 2020, 05:14:16 pm ---When I input the metrics into this calculator, nothing changed except voltage percent of total volts. Voltage drop in volts stays the same.
--- End quote ---
Did you keep using the same current?
--- End quote ---
I did! 5V for both.
--- End quote ---
I assume you mean "5 A", which is of course why nothing but the voltage loss expressed as a percent of total voltage changed. The voltage loss is dependent on the current flow, not the voltage.
Ohm's law for dummies. Read it...
--- End quote ---
5 Amps is the "current flow?" That's how much the application needs, in this example we're using. Let's just say it needs 5 Amps to function. Maybe I am misunderstanding "current flow?" I'm understanding that as 5 Amps current.
--- End quote ---
That is exactly what it is. Amperes are not a measure of quantity like volts or ohms, but rather measurement of rate--the rate of flow of electrons in a circuit, that's why it's called current
DW1961:
--- Quote from: cliffyk on July 17, 2020, 10:35:58 pm ---
--- Quote from: DW1961 on July 17, 2020, 09:20:55 pm ---
--- Quote from: cliffyk on July 17, 2020, 09:10:53 pm ---
--- Quote from: DW1961 on July 17, 2020, 09:02:41 pm ---
--- Quote from: blueskull on July 17, 2020, 05:37:48 pm ---
--- Quote from: DW1961 on July 17, 2020, 05:14:16 pm ---When I input the metrics into this calculator, nothing changed except voltage percent of total volts. Voltage drop in volts stays the same.
--- End quote ---
Did you keep using the same current?
--- End quote ---
I did! 5V for both.
--- End quote ---
I assume you mean "5 A", which is of course why nothing but the voltage loss expressed as a percent of total voltage changed. The voltage loss is dependent on the current flow, not the voltage.
Ohm's law for dummies. Read it...
--- End quote ---
5 Amps is the "current flow?" That's how much the application needs, in this example we're using. Let's just say it needs 5 Amps to function. Maybe I am misunderstanding "current flow?" I'm understanding that as 5 Amps current.
--- End quote ---
That is exactly what it is. Amperes are not a measure of quantity like volts or ohms, but rather measurement of rate--the rate of flow of electrons in a circuit, that's why it's called current
--- End quote ---
OK, I think I see my malfunction. The calculator can't change the application itself in order to benefit from higher voltage, and it could only change % voltage drop? I remember that higher voltages can dramatically drop Amps and thus wire size to get the exact same power to the application. It always seems like it was getting something for nothing. You can't do that with water, for instance. You have either higher pressure or a larger pipe. Higher pressure means--at some point--thicker pipe, better fittings, etc.
cliffyk:
--- Quote from: DW1961 on July 17, 2020, 11:02:58 pm ---OK, I think I see my malfunction. The calculator can't change the application itself in order to benefit from higher voltage, and it could only change % voltage drop? I remember that higher voltages can dramatically drop Amps and thus wire size to get the exact same power to the application. It always seems like it was getting something for nothing. You can't do that with water, for instance. You have either higher pressure or a larger pipe. Higher pressure means--at some point--thicker pipe, better fittings, etc.
--- End quote ---
Same thing; pressure is voltage, pipe is resistance, flow rate is current. Increase the pressure (same as increasing voltage) or pipe size (same as reducing resistance) and you get more flow.
DW1961:
--- Quote from: cliffyk on July 17, 2020, 11:41:54 pm ---
--- Quote from: DW1961 on July 17, 2020, 11:02:58 pm ---OK, I think I see my malfunction. The calculator can't change the application itself in order to benefit from higher voltage, and it could only change % voltage drop? I remember that higher voltages can dramatically drop Amps and thus wire size to get the exact same power to the application. It always seems like it was getting something for nothing. You can't do that with water, for instance. You have either higher pressure or a larger pipe. Higher pressure means--at some point--thicker pipe, better fittings, etc.
--- End quote ---
Same thing; pressure is voltage, pipe is resistance, flow rate is current. Increase the pressure (same as increasing voltage) or pipe size (same as reducing resistance) and you get more flow.
--- End quote ---
Where the analogy breaks down for me is that you are getting more water pressure while using a thinner pipe, which doesn't work in hydraulics. The more pressure, the thicker the pipe. e.g., and I'm probably still not getting it, you can light a 120V 1000 watt bulb at 8.3 Amps. Using the same 10000 watt bulb with 240V, you only need to calculate for 4.1 Amps to supply the same power to the bulb.. So, you can use more pressure and a smaller pipe (wire) with electricity?
cliffyk:
--- Quote from: DW1961 on July 18, 2020, 06:19:54 pm ---
--- Quote from: cliffyk on July 17, 2020, 11:41:54 pm ---
--- Quote from: DW1961 on July 17, 2020, 11:02:58 pm ---OK, I think I see my malfunction. The calculator can't change the application itself in order to benefit from higher voltage, and it could only change % voltage drop? I remember that higher voltages can dramatically drop Amps and thus wire size to get the exact same power to the application. It always seems like it was getting something for nothing. You can't do that with water, for instance. You have either higher pressure or a larger pipe. Higher pressure means--at some point--thicker pipe, better fittings, etc.
--- End quote ---
Same thing; pressure is voltage, pipe is resistance, flow rate is current. Increase the pressure (same as increasing voltage) or pipe size (same as reducing resistance) and you get more flow.
--- End quote ---
Where the analogy breaks down for me is that you are getting more water pressure while using a thinner pipe, which doesn't work in hydraulics. The more pressure, the thicker the pipe. e.g., and I'm probably still not getting it, you can light a 120V 1000 watt bulb at 8.3 Amps. Using the same 10000 watt bulb with 240V, you only need to calculate for 4.1 Amps to supply the same power to the bulb.. So, you can use more pressure and a smaller pipe (wire) with electricity?
--- End quote ---
I believe you are confusing the "bursting" pressure of a physically thinner walled pipe ("thinner walled", not smaller I.D.) with pipe I.D.--if we define "smaller" pipe as just a smaller I.D. the analogy is spot on...
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