# Parallel Battery Power Consumption Understanding

## Main Question or Discussion Point

Let's say you have two batteries in parallel and at the same voltage. Well batteries fluctuate in voltage and when they are drained they will have a somewhat lower voltage than what they started with when fully charged. Because of this, when power (or current rather) is being drawn from a battery into a circuit (or load if you will), the battery with the slightly higher voltage will supply energy (current) to the circuit (load) until its voltage level is at an equal or lower level than the other battery voltage? From which the other battery will then be supplying the energy (current) to the circuit (load)? Is this correct?

I guess theoretically while one battery is at a higher voltage level there will be a small current going across the other battery, so a parallel setup would be considered less than ideal if given a choice between series, in general.

But I'm asking this because I'm come up with a dilemma regarding this setup in a solar regulating circuit and I want to make sure this is what is going on. Any replies or help in understanding is greatly appreciated.

I hope I was adequately clear.

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Dear Tach,
I think series combination of battery is efficient as compare to parallel.
We use parallel batteries to increase current, but if however one battery becomes weak as compare to other battery than more current will flow from the battery of higher potential. Further more if you connect two batteries in parallel, than these two batteries should have approximately equal voltage, otherwise charge will start to transfer from higher voltage battery to lower voltage battery. This transfer of charge from one battery to other is undesired.

Behavior of battery in parallel or series connection can be explain easily by simulation of it with an constant emf series with a variable resistor which the variation is related to its age and weak or strong conditions. When two batteries with different internal resistance are connected in parallel, a circulating current can be caused between them and share of each battery in load current can be determined by analyze of attached circuit.

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Creative thinking is breezy, Then think about your surrounding things and other thought products. http://electrical-riddles.com

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vk6kro
I'm not sure that is a good example.

Can you see that R1 and R2 are in parallel if e = e ?

Consider what would happen if you took the bottom end of R2 and attached it to the left voltage source.
The current has to be the same as it was before because e = e.

So if R1 and R2 are in parallel, a current does not flow outwards in R1 but inwards in R2. They both flow in the same direction, outwards from the power source if e is positive.

Dear Tach,
I think series combination of battery is efficient as compare to parallel.
We use parallel batteries to increase current, but if however one battery becomes weak as compare to other battery than more current will flow from the battery of higher potential. Further more if you connect two batteries in parallel, than these two batteries should have approximately equal voltage, otherwise charge will start to transfer from higher voltage battery to lower voltage battery. This transfer of charge from one battery to other is undesired.
Okay.

Thank you everyone for the replies. This clears everything up and hopefully someone else finds/found it useful too.

Connecting diodes on the terminats of your battery would prevent current flowing between the batteries. oviously you would charge the battery by connecting behind the diodes.

I'm not sure that is a good example.

.
Kirchhoff’s laws provide a set of procedures for determining voltages and currents in a
circuit. Other methods, such as Maxwell’s mesh equations and network analysis techniques,
are available to solve the same types of problems and, in many cases, are easier to
use when the circuits are complex. Please refer to attached example.

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vk6kro
I can't read your proof, but I did a simulation on the diagram:

[PLAIN]http://dl.dropbox.com/u/4222062/Batteries%20with%20same%20emf.PNG [Broken]

As you can see, there is no circulation between the batteries. This is always the case unless the emfs are different.

If you think about it, the voltage across the load has to be less than 30 volts ( because even the 1 ohm resistor drops some voltage), so the currents have to be from the EMFs to the load.

The 1, 2 and 3 ohm resistors are all in parallel with each other and in series with the 9 ohm load across 30 volts.
So, the current in the load is 30 volts / (0.5454 ohms + 9 ohms ) = 3.1428 amps.

This agrees with the simulation.

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VK6KRO - So per simulation 3.14 am flows through 9 ohm resistor. After that I wonder how it gets distributed in three, 30 volt batteries, which are in parallel [Does it happen automatically? Any law to support it?].

Second question is, what happens if I replace one 30 volt battery with 60 volt battery and keep all resistances equal [e.g. 1 ohm] and high power diode in front of battery so current doesn't back flow in battery.

I can't read your proof, but I did a simulation on the diagram:

[PLAIN]http://dl.dropbox.com/u/4222062/Batteries%20with%20same%20emf.PNG [Broken]

As you can see, there is no circulation between the batteries. This is always the case unless the emfs are different.

If you think about it, the voltage across the load has to be less than 30 volts ( because even the 1 ohm resistor drops some voltage), so the currents have to be from the EMFs to the load.

The 1, 2 and 3 ohm resistors are all in parallel with each other and in series with the 9 ohm load across 30 volts.
So, the current in the load is 30 volts / (0.5454 ohms + 9 ohms ) = 3.1428 amps.

This agrees with the simulation.

Last edited by a moderator:
vk6kro
VK6KRO - So per simulation 3.14 am flows through 9 ohm resistor. After that I wonder how it gets distributed in three, 30 volt batteries, which are in parallel [Does it happen automatically? Any law to support it?].

Second question is, what happens if I replace one 30 volt battery with 60 volt battery and keep all resistances equal [e.g. 1 ohm] and high power diode in front of battery so current doesn't back flow in battery.
Each battery sends out a current and gets the same current back. We take this for granted, but it is pretty amazing that each electron knows exactly where it has to end up. It happens automatically.
Kirchhoff's Law of currents and the Superposition Theorem cover it.

The second question is easier. If one of the emfs is 60 volts, the load may have more than 30 volts on it, (depending on the load and the internal resistances) so the diodes will be reverse biased and no current will flow.

Each battery sends out a current and gets the same current back. We take this for granted, but it is pretty amazing that each electron knows exactly where it has to end up. It happens automatically.
Kirchhoff's Law of currents and the Superposition Theorem cover it.

The second question is easier. If one of the emfs is 60 volts, the load may have more than 30 volts on it, (depending on the load and the internal resistances) so the diodes will be reverse biased and no current will flow.

That's interesting to know that exact current will flow through each battery automatically.

Let's consider that instead of three batteries we have only 2 batteries connected in parallel [e.g. in your diagram batteries with resistors numbered 1 & 2]. Out of which one is 60 v and other is 30 v. Now if I put diodes on both batteries in series with the resistors number 1 and 2 so not reverse current flows in each battery from + end side. assume that each battery have 1 ohm internal resistance. What would be the current flowing from 9 ohm resistor and what will be voltage across it?
Note: The diodes can withstand against voltage of 300 v.

vk6kro
That's interesting to know that exact current will flow through each battery automatically.

Let's consider that instead of three batteries we have only 2 batteries connected in parallel [e.g. in your diagram batteries with resistors numbered 1 & 2]. Out of which one is 60 v and other is 30 v. Now if I put diodes on both batteries in series with the resistors number 1 and 2 so not reverse current flows in each battery from + end side. assume that each battery have 1 ohm internal resistance. What would be the current flowing from 9 ohm resistor and what will be voltage across it?
Note: The diodes can withstand against voltage of 300 v.
The 30 V battery will not supply any current because of the reverse biased diode.

The 60 V battery has a load of 1 ohm in series with 9 ohms, so 10 ohms.

Allowing 0.6 V forward drop for the diode, the net voltage will be 60 Volts minus 0.6 Volts or 59.4 Volts.
So the current will be 59.4 volts / 10 ohms or 5.94 amps.

There will be 53.46 volts across the 9 ohm resistor (5.94 amps * 9 ohms) so you can see why the diode on the 30 volt battery is reverse biased.

The 30 V battery will not supply any current because of the reverse biased diode.

The 60 V battery has a load of 1 ohm in series with 9 ohms, so 10 ohms.

Allowing 0.6 V forward drop for the diode, the net voltage will be 60 Volts minus 0.6 Volts or 59.4 Volts.
So the current will be 59.4 volts / 10 ohms or 5.94 amps.

There will be 53.46 volts across the 9 ohm resistor (5.94 amps * 9 ohms) so you can see why the diode on the 30 volt battery is reverse biased.
So is there any way / arrangement so we can get current from 30 v battery as well, which will add up and will flow through 9 ohm resistor?

vk6kro
You could put the batteries in series.

Or, you could put a 30 volt to 60 volt DC to DC converter after the 30 volt battery. This would then contribute to the output power.

But, no, not really anything practical.

You could put the batteries in series.

Or, you could put a 30 volt to 60 volt DC to DC converter after the 30 volt battery. This would then contribute to the output power.

But, no, not really anything practical.

Would it be helpful if 60 v battery produces 0.5 amp and 30 v battery produces 1 amp. Can we couple them as I suggested previously? Here their VI is same and lets assume they have same internal resistance and have 9 ohm as load as shown in your figure with no additional resistors?

So is there any way / arrangement so we can get current from 30 v battery as well, which will add up and will flow through 9 ohm resistor?
Could you not be bothered to put your figures into my graphical method to answer your question?

vk6kro
Would it be helpful if 60 v battery produces 0.5 amp and 30 v battery produces 1 amp. Can we couple them as I suggested previously? Here their VI is same and lets assume they have same internal resistance and have 9 ohm as load as shown in your figure with no additional resistors?
The current from each battery will depend on its load, not what it is capable of.

Unfortunately, if you put the batteries in parallel, with protective diodes in series with each battery, the higher voltage one will deliver all the current and the lower voltage one is not used at all.

This does depend on the internal resistances and the load resistance. For example, if the internal resistance of the 60 volt battery was more than 9 ohms, then the lower voltage battery would not see a reverse biased diode and it could deliver some current.

Each battery sends out a current and gets the same current back. We take this for granted, but it is pretty amazing that each electron knows exactly where it has to end up. It happens automatically.
Kirchhoff's Law of currents and the Superposition Theorem cover it.

The second question is easier. If one of the emfs is 60 volts, the load may have more than 30 volts on it, (depending on the load and the internal resistances) so the diodes will be reverse biased and no current will flow.
It is no more amazing than "What comes up must come down." Electrons are not cognitive agents capable of "knowing" anything in any meaningful sense of the word, it's just basic physics.

When an electron gets pushed out of a power source (such as a battery), it leaves a void behind it. This void pulls an electron in from behind to fill it. Repeat this process billions upon billions of times, and you have an electric current.

An object thrown upward does not "know" that it "should" be falling, its just minimizing it's gravitational potential energy. Same concept with current.

vk6kro