# Batteries in parallel instead of series

1. Jan 16, 2008

### kmarinas86

Say I have three different circuits, each with three 9V batteries:

CIRCUIT ONE
Parallel

9v Battery 1 -> Coil 1 -> 9v Battery 1
9v Battery 2 -> Coil 1 -> 9v Battery 2
9v Battery 3 -> Coil 1 -> 9v Battery 3

CIRCUIT TWO
Parallel

9v Battery 1 -> Coil 2 -> 9v Battery 2
9v Battery 2 -> Coil 2 -> 9v Battery 3
9v Battery 3 -> Coil 2 -> 9v Battery 1

CIRCUIT THREE
Series

9v Battery 1 -> Wire 1 -> 9v Battery 2
9v Battery 2 -> Wire 2 -> 9v Battery 3
9v Battery 3 -> Coil 3 -> 9v Battery 1

Am I right in saying that the voltage of the first two circuits is 9V and the voltage of the circuit in series is 27V? Wire 1 and Wire 2 are connections between batteries which are not used to do work. Coils 1 and 2 are used to connect batteries 1, 2, and 3 in parallel. Coil 3 is just like Wires 1 and 2 except it is being used to perform work on a permanent magnet. Which circuit, then, is most suited to utilizing the greatest amount of electricity?

I would say its circuit 2 (though circuit 1 is almost exactly the same). Circuit three wastes the current that travels through Wires 1 and 2 unless these wires are used to do work. Some people say it makes little difference whether you use series or parallel, in that you will get the same energy. But to me, putting the ends of the battery together without doing anything with the current in between is the same as shorting the battery. I would expect the input emf between the batteries to be left unchanged whether they are used to produce work or not. What do you think?

2. Jan 16, 2008

### mgb_phys

I think we need a diagram or a much clearer explanation of the setup

3. Jan 16, 2008

### kmarinas86

Here are the attachments

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4. Jan 16, 2008

### NoTime

Are the coils in circuits 1 & 2 a single strand of wire with different color leads attached to the coil ends?
Or are the colored strands continuous and lost in the drawing?

Quiz question: What is the advantage of using parallel batteries?

5. Jan 16, 2008

### chroot

Staff Emeritus
The first two diagrams appear to be equivalent. They each deliver 9V across the load, with three times the current capability of a single battery alone.

The circuit in the third diagram would deliver 27V to the load, but would only be able to deliver the current of one battery by itself.

- Warren

6. Jan 17, 2008

### Odin42

Theoretically 1 and 3 could deliver the same energy(and power) and if I'm looking at two right(kind of confusing) you'd have a short.
But practically 3 is better to use than 1 because there is no guarantee that the three batteries have the same discharge properties and can cause problems.
see: http://en.wikipedia.org/wiki/Battery_(electricity)#Battery_packs

Last edited: Jan 17, 2008
7. Jan 17, 2008

### TheAnalogKid83

My guess would be that their internal ESRs drop the overall voltage source ESR by being in parallel, which gets the power supply closer to an ideal battery. Also, you get more power at the same voltage, since each additional battery can source additional current at the voltage.

8. Jan 17, 2008

### NoTime

The first sentence is correct.
The second need some work.

Note: If the colored strands are continuous thru the coil (rather than connected to the ends of the coil) then circuit 2 is equivalent to circuit 3 and not circuit 1.

9. Jan 17, 2008

### kmarinas86

With the last sentence, do you mean the actual current is three times less, or that the current is coming from a capacity that is three times less?

10. Jan 17, 2008

### kmarinas86

Which of the following would be more useful?
1) A 9V (150 mAh) battery connected behind a 1.5V (2500 mAh) battery in series
2) A 9V (150 mAh) battery connected in front of a 1.5V (2500 mAh) battery in series

I would guess the current from the 1.5V (2500 mAh) battery would exceed the current by the 9V battery, and therefore should be the leading battery. The lifetimes of the pair are limited by the same factor, which ever battery dies first. The 9V battery will probably die first. Therefore, I would expect that much more lifetime can be gained by putting the 1.5V (2500 mAh) battery in front. Higher current and same voltage would be expected, no?

The most extreme case would be putting a bunch of 9V batteries in series behind a lantern battery. It would be very dangerous if the load did not have a high impedance, I suspect. :)

11. Jan 17, 2008

### chroot

Staff Emeritus
The current depends only on the resistance of the load and the voltage applied to the load. If you connect multiple batteries in series, neither of those quantities change. All that changes is the total amount of current that can be supplied by the batteries. If one battery is capable of delivering 100 mA, then three batteries wired in parallel can deliver 300 mA. That doesn't mean the load would use all of that capability, though.

- Warren

12. Jan 18, 2008

### NoTime

I think you may be confused here.
The 150 mah designation is the amount of energy stored in the battery and not the maximum current you can obtain.
In the 9v case your load can draw 150 ma for one hour at which point the battery is dead and you need to get a new one.
Or if you put 6 of the 1.5 v 2500 mah batteries in series (to get 9v) your load could draw 150 ma for 16.6 hours.
Or if you put 3 9v batteries in parallel your load can draw 150 ma for three hours.

The max (short circuit) current a battery can deliver is dependent on its internal resistance and can be very much higher than the mah rating.