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Let's say we have three 1V batteries and we're connecting them to a 1Ω resistor. Connected in series, the voltage is 3V and since the resistance is 1Ω, that means that from Ohm's law, the current should be 3A right? And then when we connect them in parallel, the voltage is 1V and the resistance is 1Ω so the current is 1A right?

So do I misunderstand the statement that batteries connected in parallel supply more current or do I misunderstand Ohm's law?

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    \$\begingroup\$ As others note "can" and "will" usually differ. Imagine each battery had a chemical to electrical conversion capability such that it COULD deliver up to 0.5A. If you connected a 1 Ohm load, Ohm's law would allow 1A IF the battery was able to supply it. But, as the battery was only able to supply 0.5 A max you'd see V = IR = 0.5 x 1 = 0.5 V across the resistor. ie the battery voltage would sag due to its limitations. Now use 3 similar capability batteries in parallel. Ability is now "up to 1.5 A" and actual with 1 Ohm load will be 1A, as expected. | \$\endgroup\$ – Russell McMahon Jun 26 at 6:16
  • \$\begingroup\$ Raphael, did you understand my answer? Someone didn't \$\endgroup\$ – Sunnyskyguy EE75 Jun 26 at 7:06
  • \$\begingroup\$ Well no to be honest. It's not you fault though. It's just that I'm still very much a beginner and you used a lot of words that I don't understand. \$\endgroup\$ – Raphael Jun 26 at 9:03
  • \$\begingroup\$ That’s OK, just know this. Learn to read schematics and Ohm’s Law can be applied to voltage sources like the equivalent batteries I showed, as everything has a effective series resistance (ESR) even diodes when saturated Rs, (when you move out of academia) and never be afraid to ask a question. That’s how I learned from the Japanese and I don’t even speak the language. FWIW @ElliotAnderson \$\endgroup\$ – Sunnyskyguy EE75 Jun 26 at 12:33
  • \$\begingroup\$ @Raphael he uses lots of words in ways that no one understands, including professionals. \$\endgroup\$ – hobbs Jun 26 at 16:15
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As others note "can" and "will" usually differ.

Imagine each battery had a chemical to electrical conversion capability such that it COULD deliver up to 0.5A.
If you connected a 1 Ohm load, Ohm's law would allow 1A IF the battery was able to supply it. But, as the battery was only able to supply 0.5 A max you'd see
V = IR = 0.5 x 1 = 0.5 V across the resistor.
ie the battery voltage would sag due to its limitations.

Now use 3 similar capability batteries in parallel.
Ability is now "up to 1.5 A" and
actual with 1 Ohm load will be 1A, as expected.

In situations like this which seem to not make sense, work through the problem and see how the resultant "needs" match the capabilities.
Here, with one cell, if it CAN provide 1A or more then all is well.
If the per cell capability is smaller than the need, work out what is impossible.
Here if cell can make 0.5A then.

V = IR = 0.5 max x 1 Ohm = 0.5V is false (if battery is at 1V) R = V/I = 1/.5 = 2 Ohms = false I = V/R = 1/1 = 1 A = false if max = 0.5A. Choose any 3 :-) - all false.
Now use 3 cells and try again. Ah!

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You're confusing unrelated concepts.

You have correctly calculated with Ohms Law how much current a particular circuit will draw. But this is not related to how much current your supply can deliver.

But how can the parallel batteries deliver more current?

Well first consider them in series, the current must pass through all of the batteries. So the total current can't exceed what a single battery is capable of providing. This is limited by the resistance of the battery. If a battery can deliver 1A, then no matter how many you put in series only 1A can come out.

If we put them in parallel however, there is a separate current path through each battery and their total available current can be added together.

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  • \$\begingroup\$ Did you understand my answer? \$\endgroup\$ – Sunnyskyguy EE75 Jun 26 at 7:13
  • \$\begingroup\$ hekete - I think you are still somewhat wrong. You appear to be confusing the amount of current a battery is capable of supplying with the current in the circuit. I'd suggest you re-read Sunnyskyguy answer again. Try experimenting with two batteries of different current supply capacity in series. \$\endgroup\$ – charmer Jun 26 at 14:04
  • \$\begingroup\$ @charmer what gives you that idea? \$\endgroup\$ – hekete Jun 26 at 14:09
  • \$\begingroup\$ @hekete just re-read your answer. The problem sentence is 'the total current can't exceed what a single battery is capable of providing.' I think you are only stating this for batteries with identical ratings, yes? On first read I had assumed it as a general statement: meaning that when batteries of different current capability were in series that the maximum current would then be limited by the battery with the lowest current capability. \$\endgroup\$ – charmer Jun 26 at 15:00
  • \$\begingroup\$ @charmer yes I'm assuming same battery specs, since you wouldn't want to put different batteries in parallel. \$\endgroup\$ – hekete Jun 27 at 4:42
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The parallel-connected batteries are capable of delivering more current than the series-connected batteries but the current actually delivered will depend on the applied voltage and load resistance.

You understand Ohm's Law, but the "parallel batteries supply more current" statement should really be "parallel batteries CAN supply more current".

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  • \$\begingroup\$ Did you understand my answer? \$\endgroup\$ – Sunnyskyguy EE75 Jun 26 at 7:14
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  • This shows some 1.5V battery with some equivalent capacitance and ESR for each.
  • The parallel voltages are matched before putting in parallel.
  • The series batteries are fresh and have same capacity in mAh before loading.
  • Mismatch increases towards end of life so the weakest cell fails 1st.
  • The short circuit test , Isc is momentary.

schematic

simulate this circuit – Schematic created using CircuitLab

Conclusion

Yes, parallel batteries "can" supply twice the current when the load is less than the ESR of the battery. ( As shown above, for short circuit current, it is twice.)

But otherwise, when the load is equal to battery ESR, the current is the same.

With series cells it greater when the load R is higher than ESR, the higher V/R produces a higher current. OHm's Law always works when you can estimate the battery ESR= ratio for some >10% voltage drop for some rising current.

Load R should rarely go down to ESR of the batteries but often users underestimate the DCR start current on motors. Some efficient motors can be up to 12X rated current putting the DCR=Vbat/Idc close to the total battery ESR in a string.

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  • \$\begingroup\$ Yes, your answer makes perfect sense. I was going to post something similar, but I couldn't think of a simple way to put it. \$\endgroup\$ – hekete Jun 26 at 7:59
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    \$\begingroup\$ I was on the verge of downvoting but I won't. Instead I will explain why I think other people did a better job of answering the question. First, my impression of the OP is that they are very naive and know very little about circuits. Your answer begins with an acronym that you don't explain. You replace the batteries with an equivalent capacitor. I get it but I'm sure the OP was totally lost, and I don't see that the substitution improved the clarity of the answer. You talk about mAh and Isc without explaining these terms, and again I don't think they are relevant to the OP's question. \$\endgroup\$ – Elliot Alderson Jun 26 at 11:58
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    \$\begingroup\$ You talk about "ESR= ratio" and "some rising current". I couldn't parse that sentence. Finally, you start talking about motors and DCR, without explaining what "DCR" is, and I think this is probably way over the OP's head. The bottom line is that I don't think the OP could make sense of your answer. So, on another day I might have downvoted. I hope you will take this feedback in the spirit of constructive criticism. \$\endgroup\$ – Elliot Alderson Jun 26 at 12:02
  • \$\begingroup\$ @ElliotAlderson but answers aren't just for the OP. While I agree the OP probably didn't understand this answer (and is why I didn't put a similar thing in my answer, I thought about it but I couldn't think of a way in which to explain it well). Having a more technical answer is still valuable in the achieve sense for when others might search similar terms in the future. If it's down voted that gives the impression that it might be wrong? \$\endgroup\$ – hekete Jun 26 at 12:42
  • \$\begingroup\$ @hekete No, a downvote means "this answer is not useful". If the answer is not useful (i.e. understandable) by the OP, then what is the point? \$\endgroup\$ – Elliot Alderson Jun 26 at 13:34
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internal resistor can make a big difference

1- series connection enter image description here

total internal resistor rt= r1+r2+r3

let have an example of 0.5 Ω internal resistor and let the voltage each battery equal to 1v

the load is 1 Ω as your exemple

rt=0.5+0.5+0.5

rt=1.5 Ω

total resistor = R load+ rt

R= 1+1.5=2.5 Ω

I=Vt/R load+rt

Vt=1+1+1=3 V

I=3/ 1.5+1 =3/2.5= 1.2A

2- parallel connection enter image description here

the total resistance is the sume of all resistor

1/Rt=1/r1+1/r2+1/r3+1/Rload

1/Rt=1/0.5+1/0.5+1/0.5+1/1

Rt=1/7=0.14 Ω

I=vt/Rt

I= 1/0.14 = 7.14 A !!

so what do you think

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If an amp is flowing through a device with two electrodes, that means that one electrode is losing approximately 6.242×10^18 electrons per second worth of charge while the other is gaining a like amount. If two batteries are connected in parallel to a load, every electron's worth of charge that leaves the negative electrode of either battery will pass through the load before returning to the positive electrode of the same battery. If they are connected in series, each electron's worth of charge that passes through the load must pass through both batteries.

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Think of garden hoses. Think of water as current.

Attaching several hoses in a long line is like connecting in series. You can only force so much water down the single hose.

Tying several hoses together in parallel so that there are multiple hose ends sprouting water is analogous to batteries in parallel. Lots more water can come out the end.

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