# Help With Battery Symbols Please

Discussion in 'Homework Help' started by JDR04, May 22, 2012.

1. ### JDR04 Thread Starter Active Member

May 5, 2011
339
4
Hi Folks, I've attached a question and answer type table from my studies. I can't for the life of me work out how the answers were derived. You'll see my own scribblings in trying to work it out but just cannot get my head around it.

Lastly, if two 1.5V cells are connected in series then we have 3V...right. Now if another 1.5V cell is connected in Parallel to the first two which are in series, what do we get? I got 3 V with my meter but would like to know how it's worked out on paper.

I take it the longer line of a cell drawing is the positive and the shorter is the negative?

Thanks, your time and help is always appreciated-JDR04

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2. ### bretm Member

Feb 6, 2012
152
24
What a weird study problem. You'll never see backward batteries in any sensible schematic.

But anyway...long followed by short is +1.5V, short followed by long is battery inserted the opposite direction, for -1.5V. That will cancel out one of the +1.5V's.

1.5V in parallel with 3V is nothing to work out on paper--it's just a bad idea. The 3V will try to drive current into the 1.5V source. Bad things could happen depending on battery chemistry. You need protection diodes if you want to do that, such as having a backup battery for a circuit with another power source.

3. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,393
1,211
True, but when you introduce the human element, physical installations can result in those configurations. Luckily, the devices might not work giving the human a chance to double check their work.

4. ### JDR04 Thread Starter Active Member

May 5, 2011
339
4
Thanks Bretm for the quick reply.

I think what they are trying to establish is if I know the formula or understand how to work these things out.

The first two questions/diagrams tell me there will be no electron flow as it needs to flow from negative to positive but in these cases and as you rightly point out, some of the cells are reversed. I was just wondering if it all had some wierd thing to do with cells connected in parallel or parallel and series etc?? It's got me baffled.

The last one where the two cells are connected in series and then a single cell is connected in parallel to them has got me beat. I can't for the life of me find anything that will explain how to figure it out on paper when you need to. I agree with the protection diodes you mentioned but think they are trying to establish my understanding of this.

Thanks a lot and maybe somebody else could try and "clear these muddy waters" for me.

Thanks again -JDR04

5. ### JDR04 Thread Starter Active Member

May 5, 2011
339
4
Thanks Bretm for the quick reply.

I think what they are trying to establish is if I know the formula or understand how to work these things out.

The first question I understand OK. The next two questions/diagrams tell me there will be no electron flow as it needs to flow from negative to positive but in these cases and as you rightly point out, some of the cells are reversed. Where does the 3V come from in the second question/diagram? I was just wondering if it all had some wierd thing to do with cells connected in parallel or parallel and series etc?? It's got me baffled.

The last one where the two cells are connected in series and then a single cell is connected in parallel to them has got me beat. I can't for the life of me find anything that will explain how to figure it out on paper when you need to. I agree with the protection diodes you mentioned but think they are trying to establish my understanding of this.

Thanks a lot and maybe somebody else could try and "clear these muddy waters" for me.

Thanks again -JDR04

Last edited: May 22, 2012
6. ### bretm Member

Feb 6, 2012
152
24
That's because it's not solvable as stated. For ideal voltage sources, there is no solution. You have two equations, one that says v=1.5 and another that says v=3. Solve for v. Just doesn't make sense.

But reality isn't ideal. Batteries have internal resistance. Add a few ohms in series with each battery and see if you get anywhere.

7. ### JDR04 Thread Starter Active Member

May 5, 2011
339
4
Thanks Bretm, I think I'll get hold of my tutor and find out why such a problem is presented to us.

Once again,many thanks for sharing your time and knowledge. JDR04

8. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Voltages cancel,

A +1.5VDC + -1.5VDC = 0VDC

9. ### JDR04 Thread Starter Active Member

May 5, 2011
339
4
Thanks Bill.JDR04

10. ### WBahn Moderator

Mar 31, 2012
18,079
4,917
Technically, a battery is a collection of two or more cells, though it has come to be more generally used so that it includes single-cell "batteries", as well. But this heritage is still evident in that the symbol for a cell is a pair of parallel lines with one longer than the other while the symbol for a battery is at least two sets of these lines. Originally, the battery symbol used consisted of the number of cell symbols that actually made up the battery. Again, over time this technicallity has been softened and all but lost, but it still makes for a useful background to have in mind.

For a cell symbol, the longer line is the positive side of the cell. As you move through the cell, you increase the voltage by the cell voltage if you go from short to long and decrease it when you go from long to short. When you connect cells in series to form a battery, you have to remember that while moving through the cell changes the voltage, moving from one cell to the next has no effect on the voltage. To make this clear, you could draw the wires that connect one cell to another. That should help you figure out the overall voltages for the connections in your three problems.

As for the parallel connection case, as others have stated this is a bad idea unless the batteries being connected are at the same voltage. If they aren't, you will get current flow from the higher voltage source to the lower source that is limited only by the internal resistance of the sources. If these resistances are sufficiently low, you can quickly get enough current to generate sufficient heat to cause a fire or explosion. If you are lucky enough to avoid this outcome, you will end up with a final voltage that is probably somewhere between the two and will likely have one heavily discharged battery and one heavily overcharged battery. The exact balance that results and the consequences will depend on the types of batteries involved and what their initial state of charge was. It is not something that lends itself to any kind of "paper" calculation -- at least not unless you are intimately familiar with the electrochemistry and physical chemistry involved.

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