# Batteries mounted in parallel. Amps wont add !

#### dl324

Joined Mar 30, 2015
13,820
3V 3mA or 1.5V 1.5mA when switched
Small detail... The OP is talking about putting batteries in parallel, not series.

#### sparky 1

Joined Nov 3, 2018
647
Why the current connecting one battery is the same as the current of two batteries in parallel. We need to examine 2 cases:
1.) 1k Ohm resistor 2.) 2.3 Ohm bulb

A math relationship was made by observation and measurement between battery and resistor load in a circuit.
The terminology for Ohms Law is not always apparent because of the inverse relationship. We convert between units
and read instruments without reviewing the qualitative each time. It is the nature of our electrical science to rely on numbers and instruments.
My attempt will be far less than perfect in comparison to an exact mathematical equation and it's terminology.

Both battery cathodes are connected to a load resistor.
Both battery anodes are connected to the opposite side of the load resistor.
Each battery will share whatever resistance load that is presented to the conductive path equally.

If you lower the resistance of the load the batteries will conduct a higher flow volume (The Current will increase).
The control of current is a result of the restrictive nature of the load in a conductive pathway.
Placing a tungsten filament in the conductive pathway presents only a small restrictive action resulting in
much higher current flow. The forces in the battery will change with the load also the forces in the conductor will evenly distribute the charge. The electrochemistry of the battery cell changes only slightly. Observation and measurement of a voltage drop across the load is important.
The voltage drop measurement was studied and an inverse relation was sculpted. The indirect math operation includes an inverse twist. We gather the values of voltage drop and resistance it describes the Current flow. When we say plug in the values also we perform an inverse function.

There are better explanations I am sure.

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#### DickCappels

Joined Aug 21, 2008
8,149
If you have one battery that can supply 1 amp when it is shorted, putting a second identical battery in parallel will supply 2 amps when shorted. In practice nothing is exact.

Incandescent light bulbs are terrible loads for this kind of test, particularly when measuring current. Since the resistance of the metal filament increases with temperature and higher resistance means less heating light bulbs operated at temperatures that make them glow have an in-built negative feedback and they act more like constant current loads than they do resistors.

#### Yuseph

Joined Jun 8, 2020
47
Thanks guys,

i never thought resistance could be a concern. In my books combining amps seemed as easy as combining volts.
You know today Ive run so many tests im starting to get the big picture. Its obvious that if I connect a 50w device instead of a 5w light bulb the result will be totally different.
But ive noticed another obvious condition. The voltage supply should always match the device voltage.
Here are all the results ive written down. If you re too bored to read it all ill summarize it for you. The only times i was successful at doubling the amps is when i had the multimeter connnected on the bigger amp side and the other side was half the value of the big amp side. This principle worked in all scenarios no matter what voltage i was applying and whether the supply voltage and the device voltage were matching or not.
Now ive seen that the voltage alignement is very important too. When i combined my adapter 12v with 6v of batteries the amp value jumped. But since it fell back when i moved to the 9v battery it is clear that too much amp gives the opposite effect. I think like some of you said that the device power matters. If 5w is its highest consumption then i think it eventually reaches a saturation point. At least thats the impression i get. There seems to be a saturation effect. Why it doesnt smoke i dont know.
All those test consisted of me connecting the second power supply to the light bulb directly. When connecting the 2nd power supply to the 1st supply power either it did nothing or i was getting a negative value.
Now I did some testing on my 2 solar panels and connecting the second panel to the device did not bring the expected result. I had to link the second panel to the 1st panel beforehand in order to get more amp. But it didnt double. But then since i only had a battery as the load i dont know it didnt drain much more amp from the 2nd panel.
Anyway i hope i have been clear enough.

See for yourself :

Incandescent 12v light bulb test :

- adapter 5v alone = 0.25A
- batteries 5v alone = 0.23A

- adapter 5v + batteries 5v = 0.16A

- adapter 6v + batteries 5v = 0.30A

- adapter 7.5v + batteries 5v = 0.64A

- adapter 9v + batteries 5v = 0.97A

- adapter 12v + batteries 5v = 1.64A
- adapter 12v+ batteries 6v = 1.80A
- adapter 12v + batteries 9v = 0.60A ???

Test on a 3v led :

- batteries 3v = 50mA

- adapter 4.5v + batteries 3v = 300mA
- adapter 4.5v + batteries 4.5v = 124mA ??

- adapter 5v + batteries 3v = 463mA
- adapter 5v + batteries 4.5v = 265mA ??
- adapter 5v + batteries 5v = 163mA ??

- adapter 6v + batteries 3v = 580mA
- adapter 6v + batteries 4.5v = 394mA ?
- adapter 6v + batteries 5v = 298mA ??
- adapter 6v + batteries 6v = 200mA ??

Test with 2 x 4 batteries
- 4 batteries 6v = 0.24A
- 4 batteries 6v + 4 batteries 5v = 0.18A ??
- 4 batteries 6v + 3 batteries 3.5v = 0.34A
- 4 batteries 6v + 2 batteries 2.5v = 0.65A

Until now i still consider that im not mastering the topic. I have no idea whats going on at a physical level. Those are things that are in total contradiction with what ive seen in my books or seen on youtube demos.

#### dl324

Joined Mar 30, 2015
13,820
i never thought resistance could be a concern. In my books combining amps seemed as easy as combining volts.
You aren't grasping the fundamentals.

According to Ohm's Law, I=V/R. If you don't change the voltage or resistance, current will remain constant. You could put 100 batteries in parallel. Assuming they're all the exact same voltage (which they won't be), and the load isn't changed, the current in the load will also be unchanged.

What *will* happen is that the current in the load will come from 100 batteries.

#### crutschow

Joined Mar 14, 2008
28,486
i never thought resistance could be a concern. In my books combining amps seemed as easy as combining volts.
Then either the books are wrong, or you need to read them more carefully.
Resistance is always involved with volts and amps.

#### sparky 1

Joined Nov 3, 2018
647
The Deduction of Ohms law is where the less than ideal is approached.
When R varies the law is not violated or when two uneven sources are paralleled the law is not violated.
However Ohms law by itself is not sufficient to explain it does not mean that the proportionality between I and V is directly related is false.
What Ohms law is and what it can do is misunderstood. I is still proportional to V.

#### Yuseph

Joined Jun 8, 2020
47
The situation now is im applying 12v to the bulb and i get 0.27A only, which means the light bulb is consuming 3.20w. so now the normal consequence should be, if i add another supply for the light bulb to reach its watt peak 5w, that amps go up to 0,40amps.
current situation : 3,20w = 11,85v * 0.27A
ideal situation : 5w = 12.20v * 0.41A

you guys are saying there is some resistance. So since i have 0,27A
11.85v / 0.27A = 43.88ohm
So because of 43 ohm of resistance adding another power supply will have no effect on the amps ?
guys what the hell.

#### Papabravo

Joined Feb 24, 2006
17,545
The situation now is im applying 12v to the bulb and i get 0.27A only, which means the light bulb is consuming 3.20w. so now the normal consequence should be, if i add another supply for the light bulb to reach its watt peak 5w, that amps go up to 0,40amps.
current situation : 3,20w = 11,85v * 0.27A
ideal situation : 5w = 12.20v * 0.41A

you guys are saying there is some resistance. So since i have 0,27A
11.85v / 0.27A = 43.88ohm
So because of 43 ohm of resistance adding another power supply will have no effect on the amps ?
guys what the hell.
No need to get vulgar. The two supplies will share the load according to their internal characteristics. Primarily the internal source resistance. Go back to the current hogging simulation that I showed in post#8. If I take one of those supplies away the lamp will draw the same amount of current as the sum of the two.

#### Yuseph

Joined Jun 8, 2020
47
No need to get vulgar. The two supplies will share the load according to their internal characteristics. Primarily the internal source resistance. Go back to the current hogging simulation that I showed in post#8. If I take one of those supplies away the lamp will draw the same amount of current as the sum of the two.
Nice schematic bro.
So im just wasting my time, right ? What double is only the capacity mAh, its never been the amps.
Im worried now because it means that even 10 solar panels will never get you 10A for oven and stuff to work properly. It just remains constant at one amp

#### dl324

Joined Mar 30, 2015
13,820
Im worried now because it means that even 10 solar panels will never get you 10A for oven and stuff to work properly. It just remains constant at one amp
You are failing to grasp the fundamentals.

Current varies with voltage and resistance (load). If the voltage is constant and the load is constant, current will also be constant.

With solar cells, you would also use batteries so that power that isn't used isn't wasted and you can save power for use when the sun isn't shining. The batteries can also supplement daytime demand if it exceeds what the solar cells can provide.

#### WBahn

Joined Mar 31, 2012
26,398
Nice schematic bro.
So im just wasting my time, right ? What double is only the capacity mAh, its never been the amps.
Im worried now because it means that even 10 solar panels will never get you 10A for oven and stuff to work properly. It just remains constant at one amp
Let's say that your oven is designed to be powered by 20 V and, when supplied by 20 V, will draw 10 A.

Now let's say that you have a solar panel that can supply 32 V when it is open circuit (no load attached). If you then attach various loads what you will see is that as the current goes up, the voltage will go down. Let's say that you find a load that results in the output being 20 V and that the current in that load is 1 A. If you attach your oven to the panel, you would likely find that the voltage across it is, say, 3 V and that the current is 2 A. Your oven is too heavy a load for that panel to be able to power at its rated voltage and current. But if you put ten panels in parallel you would now find that, at 20 V you are able to deliver 10 A because, at that voltage, each of the ten panels is delivering 1 A and the total current is the sum of the current from all ten panels.

#### Papabravo

Joined Feb 24, 2006
17,545
What happens when a load has a lower resistance than a battery or a solar panel can provide is that the output voltage of the source drops. Here is where a second or additional source can be useful. It will allow the available current in each source to contribute to delivering power to the load. You still need to contend with the internal resistance and current hogging problem. It is not uncommon for parallel supplies to use a low forward drop (Schottky) diode to allow the sharing without allowing the supplies to fight each other. It is also how a battery backup at a lower voltage than the mains power supply can be used together,

Voltage, current, power, and batteries are kind of like horses and horsepower. You might be wondering how that can be. Well the standard mechanical definition of horsepower is lifting 550 lbs., 1 foot, in 1 second. Most draft horses have no problem with this task. However: there are no horses that I have ever seen or heard of that can lift 1 lb., 550 feet in 1 second. Do you see the problem?

#### sparky 1

Joined Nov 3, 2018
647
Nice schematic bro.
So im just wasting my time, right ? What double is only the capacity mAh, its never been the amps.
Im worried now because it means that even 10 solar panels will never get you 10A for oven and stuff to work properly. It just remains constant at one amp
Will people learn to use Watts ?
Every solar panel experimenter needs a Watt meter.

The conservation of energy yields to those maintaining the grid because we need to keep the power running. The smart grid should not interfere with off grid. The off grid needs to be aware of the Watt meter like connecting it to an Ethernet cable so they do a careful analysis on cloud.

#### ci139

Joined Jul 11, 2016
1,696
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#### sparky 1

Joined Nov 3, 2018
647
A smart battery might need sense resistors.
Most of the circuit's various conditions are better maintained with even charge distribution.

#### xox

Joined Sep 8, 2017
623
What double is only the capacity mAh, its never been the amps.

Im worried now because it means that even 10 solar panels will never get you 10A for oven and stuff to work properly. It just remains constant at one amp
Think of it this way: Current is not a variable to be manipulated. It's just the ratio between the voltage supplying a circuit divided by its total resistance.

So if the current goes up, it's ONLY because the voltage has increased, or the resistance decreased. Thus for any given voltage, a "larger" appliance simply has less internal resistance than a "smaller" one.

#### Yuseph

Joined Jun 8, 2020
47
In my books it says if you put a battery of higher voltage and a battery of lower voltage in parallel the battery with the highest voltage will discharge into the lower voltage battery.
I didnt notice that. But how could I in only a few seconds.

#### DickCappels

Joined Aug 21, 2008
8,149
I don't think the author meant it would discharge completely.

Maybe a better way to look at it is that the higher voltage battery will "try" to change the other battery to a higher voltage. What you have is two voltage sources connected to each other through their internal resistances. The voltage difference divided by the the sum of their internal resistances determines how much current flows.

Please note that the internal resistance of a battery is dependent upon its state of charge, its temperature, and probably a lot of other things.