Power supply calculation and voltage drop

Hello,

If someone could guide me how to calcute next problem,

I have a 50 loads that can be powerd from 6-12V and current consumption is 80mA connected in paralel with a distance between each other 10m. I have taken cabel resistance 20ohm per 100 meters.

Total cable lenght to connect all this loads is 500m which gives me 10kOhm over pozitive wire and 10kOhm resistance over negative wire. Total current that must go over circuit is 4Amps. I want to use car battery 12V 30ah.


Does this 12V 30Ah means that battery is ready to provide 30A over one hour but voltage wont drop ?
Will it be sufficent power supply 12V 4Ah ?
Do i must connect multiple battery packs ?

Goal is to secure sufficent voltage over network 12V for max voltage drop to 6V and supplying 80mA to all loads.

Regards

cipi-cips said:

Total current that must go over circuit is 4Amps. I want to use car battery 12V 30ah.


Does this 12V 30Ah means that battery is ready to provide 30A over one hour but voltage wont drop ?

Click to expand...

A 12V lead-acid battery is about 12.6V full charge resting. It gradually drops to about 10.5V when fully discharged.
In simple math, a 30Ah bat sends 1A for 30 hours. Or 30A for 1 hr but you'll see immediate voltage drop and you may only get 1/2 hour of satisfactory performance.

Will it be sufficent power supply 12V 4Ah ?

Click to expand...

At 4A discharge you should get 6 or 7 hours from your 30 Ah bat.

cipi-cips said:

I have taken cabel resistance 20ohm per 100 meters.

Total cable lenght to connect all this loads is 500m which gives me 10kOhm over pozitive wire and 10kOhm resistance over negative wire. Total current that must go over circuit is 4Amps. I want to use car battery 12V 30ah.

Click to expand...

It will be 10 Ohm (not 10k; must be a typo) for 500m. For the return path you will have another 10 Ohm. That makes a total of 20 Ohm lead resistance.

A 12V battery cannot deliver 1A because of the lead resistance. Your load is 50 number of 80mA (in parallel); total load current will be 50x80mA=4A.

Your 12V battery is insufficient for this job; you need to get fat cables with resistance of 1-2 Ohm /100m. You see, you are using low voltage high current circuits and line drops will be a concern.

If your loads are not all at the same place, they will see different voltages. That may be a problem.

c_mitra said:

It will be 10 Ohm (not 10k; must be a typo) for 500m. For the return path you will have another 10 Ohm. That makes a total of 20 Ohm lead resistance.

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Can you explain me how it will be 10 Ohm ? Probably type error I see I have written 10k but it will be 100Ohm's per positive and 100 Ohms per negative wire.

c_mitra said:

A 12V battery cannot deliver 1A because of the lead resistance. Your load is 50 number of 80mA (in parallel); total load current will be 50x80mA=4A.

Your 12V battery is insufficient for this job; you need to get fat cables with resistance of 1-2 Ohm /100m. You see, you are using low voltage high current circuits and line drops will be a concern.

If your loads are not all at the same place, they will see different voltages. That may be a problem.

Click to expand...

Is it possible to use multiple batteries connected in parallel ?

Regards