Just another dumb question,
If You have voltage drop thru. a wire does this affect the amps. output as well?
Joda.
Peter_n_Margaret said
11:16 PM Nov 16, 2018
No, but it does cost you power (watts), because Watts = Volts X Amps and if volts drop, so does watts (power), so you can't do as much with what is left.
eg... 12V x 10A = 120W.... 10V x 10A = 100W.
For those who wish to calculate voltage drop, here is the formulae. Note that cable length must include both wires, +ve and -ve.
VOTLAGE DROP = [cable length (in metres) X current (in amps) X 0.0164] divided by cable cross-section in mm.sq.
Cheers,
Peter
Joda said
08:46 AM Nov 17, 2018
Thanks for that info. P&M, if We use the water analogy then volts is volume, amps is pressure & watts is how much water is in the tank?
Cheers, John.
Peter_n_Margaret said
09:06 AM Nov 17, 2018
Joda wrote:
Thanks for that info. P&M, if We use the water analogy then volts is volume, amps is pressure & watts is how much water is in the tank? Cheers, John.
No.
Volts is pressure, Amps is flow rate.
How much is in the tank is watt hours.
Cheers,
Peter
Joda said
09:22 AM Nov 17, 2018
Ha! Watt HOURS, now I get it! Much appreciated,
Regards, John.
kgarnett said
10:43 AM Nov 17, 2018
As Peter has pointed out the voltage drop in the wire is caused by the current flowing through the resistance of the wire.
Does the voltage drop affect the amps output?It could be argued that the answer is Yes.
In the hypothetical case of a voltage source with zero internal resistance and wires with zero resistance, there would be no voltage drop in the wire and the current flow would be defined solely by the load resistance.
eg 12V source, wires with zero resistance and a 1ohm load, the current would be 12A. (12V/1ohm)
Now consider that the wires have 0.1 ohm resistance. In this case the total load seen by the voltage source is 1.1 ohms (ie 0.1 ohm wire resistance + 1 ohm load resistance). It then follows that the current through the load is reduced to 10.9A (12V/1.1ohm) and there would be 1.09V voltage drop across the wires.
If the connecting wire resistance is increased due to increased length or reduced CSA, then there would be a greater voltage drop in the wire and a greater reduction in current through the load.
The fact that there is voltage drop in the wire means that the voltage at the load is reduced by that same amount of voltage, and consequently the current through the load is reduced proportionally.
So it can be argued that if there is a voltage drop in the wire connecting the load, then the amps flowing through the load is affected (reduced).
Ken
Jaahn said
10:46 AM Nov 17, 2018
Hi Joda
Glad to see you got the light bulb moment with those answers from Peter
Just an additional comment from me. If you have too much resistance in the wires then usually you loose a bit of everything. Because the voltage is reduced, that means the pressure is less to push so much current through to the load. So it is a loose/ loose situation, and the wires overheat too. Bigger wires cost only a coupe of dollars more than skinny ones.
Jaahn
Aus-Kiwi said
10:55 AM Nov 17, 2018
Always proportional..
Whenarewethere said
11:20 AM Nov 17, 2018
Jaahn wrote:
Bigger wires cost only a coupe of dollars more than skinny ones.
Jaahn
Bigger wires may actually be cheaper because one will most likely not have to do the job a second time after months wasting time and stress trying to work out why nothing works as well as it should if it was done correctly in the first place!
Jaahn said
11:23 AM Nov 17, 2018
Whenarewethere wrote:
Jaahn wrote:
Bigger wires cost only a coupe of dollars more than skinny ones.
Jaahn
Bigger wires may actually be cheaper because one will most likely not have to do the job a second time after months wasting time and stress trying to work out why nothing works as well as it should if it was done correctly in the first place!
Jaahn
Joda said
12:09 PM Nov 17, 2018
Thanks guys, I,ve set up a stand alone system of 300ah battery bank & 400watts of panels.
I have a battery monitor in the system so I could see what is going on.
Wire sizes are o.k. , when I wired the car to run the van fridge I put a known load on increasing sizes of wire & measured the drops with a multimeter, round about way I know but soon figured out wire sizes.
My little system is only to run a 80lt compressor fridge at xmas on our family camp on the Murray, needed a bit of info to understand My monitor readings,
Thanks again,
John.
Whenarewethere said
12:29 PM Nov 17, 2018
Put extra insulation around the fridge.
Have a person designated as an official fridge door opener so the other person can get things in and out quickly, and there is a third hand available!
Joda said
12:38 PM Nov 17, 2018
You wanna tell My wife that? Go on I dare You!
Whenarewethere said
01:45 PM Nov 17, 2018
For the beer!
Peter_n_Margaret said
06:19 PM Nov 17, 2018
Jaahn wrote:
Hi Joda
Glad to see you got the light bulb moment with those answers from Peter
Just an additional comment from me. If you have too much resistance in the wires then usually you loose a bit of everything. Because the voltage is reduced, that means the pressure is less to push so much current through to the load. So it is a loose/ loose situation, and the wires overheat too. Bigger wires cost only a coupe of dollars more than skinny ones.
Jaahn
It depends... :)
If the load is constant, the volts will drop due to undersize cables and the amps will increase to maintain the power to feed the load.
Solar panels don't act like most other things either. They are a "constant power" source and with a PWM regulator (that has already had its losses counted compared with a MPPT), there is virtually no additional losses due to undersized wiring to the regulator, because the panel "pushes" harder and increases the amps. With an MPPT regulator, there will be losses with undersized cables. Don't ask me to explain it technically :)
Best solution is don't undersize cables.
Cheers,
Peter
Whenarewethere said
07:13 PM Nov 17, 2018
At the end of the day just do what you can do at the time of setting up your network. Put in cables which are over rated. If you are under rated run a second cable. Calculate distance for there and back (+ & -). Calculate losses for each individual input distance. Calculate for wiring distance from battery to fridge (I assume a fridge). Don't use controllers fixed to the back of solar panels. Install a quality controller near the battery. If you need extention cables to portable solar panels use 6awg cable and knock them up yourself so you know they are made correctly.
It is not that difficult to save a few percentage points here and there and it adds up pretty quickly.
PeterD said
08:10 AM Nov 18, 2018
Joda wrote:
Just another dumb question, If You have voltage drop thru. a wire does this affect the amps. output as well? Joda.
The short answer is yes. The change in current depends upon the load. If the load is resistive and the resistance does not change with heat then just add the resistance of the wire to the resistance of the load and recalculate (or measure with an ammeter.)
If you are charging a battery the current drops drastically as you increase the resistance. As the voltage drop increases in the wire the charge voltage at the battery drops so the current acceptance of the battery also drops. You will have great difficulty in calculating that one, you will need to measure it.
If you have a DC-DC charger or an inverter as the load the current will increase. These have circuitry that makes them work harder to supply the same output as the input voltage decreases. That means they have to draw more current to produce the same power.
oldtrack123 said
01:42 PM Nov 18, 2018
Joda wrote:
Just another dumb question, If You have voltage drop thru. a wire does this affect the amps. output as well? Joda.
Hi Joda
If you use the water analogy you will be able to work out many of those questions
Have a go Water pump,/ tank on a stand gives the pressure / volts. Piping is a restriction to flow rate /amps=cables resistance .A jet offers resistance = load ,
Higher pressure,from tank= more water [higher flow rate] = Amps & Volts at load
Bigger pipes /larger cables =less resistance to water/current flow ,less pressure /volts lost in piping /cables
That is just for starters
But the quick answer is: Voltage drop /lost in cables means less pressure/ volts at the load input. Load resistance/ jet size determines how much water /amps can be pushed into /through, that jet / load
-- Edited by oldtrack123 on Sunday 18th of November 2018 01:53:13 PM
Whenarewethere said
02:17 PM Nov 18, 2018
Same as air. I have 3 compressors to speed things up. 30sec 20-35PSI per corner 90Ssec 5-45PSI spare, I don't get anymore pressure but I do get volume, but I had to upgrade the air hoses to a larger ID so the air can get through efficiently.
Also run the engine to keep the voltage up. But I do have a 7% loss due to slightly under sized cable. But it is not a fridge and solar so am prepared to wear this loss.
eg... 12V x 10A = 120W.... 10V x 10A = 100W.
For those who wish to calculate voltage drop, here is the formulae. Note that cable length must include both wires, +ve and -ve.
VOTLAGE DROP = [cable length (in metres) X current (in amps) X 0.0164] divided by cable cross-section in mm.sq.
Cheers,
Peter
No.
Volts is pressure, Amps is flow rate.
How much is in the tank is watt hours.
Cheers,
Peter
As Peter has pointed out the voltage drop in the wire is caused by the current flowing through the resistance of the wire.
Does the voltage drop affect the amps output? It could be argued that the answer is Yes.
In the hypothetical case of a voltage source with zero internal resistance and wires with zero resistance, there would be no voltage drop in the wire and the current flow would be defined solely by the load resistance.
eg 12V source, wires with zero resistance and a 1ohm load, the current would be 12A. (12V/1ohm)
Now consider that the wires have 0.1 ohm resistance. In this case the total load seen by the voltage source is 1.1 ohms (ie 0.1 ohm wire resistance + 1 ohm load resistance). It then follows that the current through the load is reduced to 10.9A (12V/1.1ohm) and there would be 1.09V voltage drop across the wires.
If the connecting wire resistance is increased due to increased length or reduced CSA, then there would be a greater voltage drop in the wire and a greater reduction in current through the load.
The fact that there is voltage drop in the wire means that the voltage at the load is reduced by that same amount of voltage, and consequently the current through the load is reduced proportionally.
So it can be argued that if there is a voltage drop in the wire connecting the load, then the amps flowing through the load is affected (reduced).
Ken
Hi Joda
Glad to see you got the light bulb moment with those answers from Peter
Just an additional comment from me. If you have too much resistance in the wires then usually you loose a bit of everything. Because the voltage is reduced, that means the pressure is less to push so much current through to the load. So it is a loose/ loose situation, and the wires overheat too.
Bigger wires cost only a coupe of dollars more than skinny ones.
Jaahn
Bigger wires may actually be cheaper because one will most likely not have to do the job a second time after months wasting time and stress trying to work out why nothing works as well as it should if it was done correctly in the first place!
Jaahn
Put extra insulation around the fridge.
Have a person designated as an official fridge door opener so the other person can get things in and out quickly, and there is a third hand available!
For the beer!
It depends... :)
If the load is constant, the volts will drop due to undersize cables and the amps will increase to maintain the power to feed the load.
Solar panels don't act like most other things either. They are a "constant power" source and with a PWM regulator (that has already had its losses counted compared with a MPPT), there is virtually no additional losses due to undersized wiring to the regulator, because the panel "pushes" harder and increases the amps. With an MPPT regulator, there will be losses with undersized cables. Don't ask me to explain it technically :)
Best solution is don't undersize cables.
Cheers,
Peter
At the end of the day just do what you can do at the time of setting up your network. Put in cables which are over rated. If you are under rated run a second cable. Calculate distance for there and back (+ & -). Calculate losses for each individual input distance. Calculate for wiring distance from battery to fridge (I assume a fridge). Don't use controllers fixed to the back of solar panels. Install a quality controller near the battery. If you need extention cables to portable solar panels use 6awg cable and knock them up yourself so you know they are made correctly.
It is not that difficult to save a few percentage points here and there and it adds up pretty quickly.
The short answer is yes. The change in current depends upon the load. If the load is resistive and the resistance does not change with heat then just add the resistance of the wire to the resistance of the load and recalculate (or measure with an ammeter.)
If you are charging a battery the current drops drastically as you increase the resistance. As the voltage drop increases in the wire the charge voltage at the battery drops so the current acceptance of the battery also drops. You will have great difficulty in calculating that one, you will need to measure it.
If you have a DC-DC charger or an inverter as the load the current will increase. These have circuitry that makes them work harder to supply the same output as the input voltage decreases. That means they have to draw more current to produce the same power.
Hi Joda
If you use the water analogy you will be able to work out many of those questions
Have a go Water pump,/ tank on a stand gives the pressure / volts. Piping is a restriction to flow rate /amps=cables resistance .A jet offers resistance = load ,
Higher pressure,from tank= more water [higher flow rate] = Amps & Volts at load
Bigger pipes /larger cables =less resistance to water/current flow ,less pressure /volts lost in piping /cables
That is just for starters
But the quick answer is: Voltage drop /lost in cables means less pressure/ volts at the load input. Load resistance/ jet size determines how much water /amps can be pushed into /through, that jet / load
-- Edited by oldtrack123 on Sunday 18th of November 2018 01:53:13 PM
Same as air. I have 3 compressors to speed things up. 30sec 20-35PSI per corner 90Ssec 5-45PSI spare, I don't get anymore pressure but I do get volume, but I had to upgrade the air hoses to a larger ID so the air can get through efficiently.
Also run the engine to keep the voltage up. But I do have a 7% loss due to slightly under sized cable. But it is not a fridge and solar so am prepared to wear this loss.