View Full Version : Batteries: Amp/Hour
Two 12 volt/40 ampere hour batteries connected in parallel will produce:-
a. 12 volts for 80 hours
b. 12 volts for 40 hours
c. 24 volts for 80 hours
d. 24 volts for 40 hours
Can anyone help with the answer?
29th Nov 2001, 13:30
What current are you drawing ?, I would have thought
12v 80 hours at 1 amp
12v 40 hours at 2 amp
12v 20 hours at 4 amp
and so on.
29th Nov 2001, 13:57
under-exposed is correct in theory ,in practice a very high current will discharge the batteries in a shorter tlme than their rated capacity
29th Nov 2001, 14:37
A poorly worded question indeed! Assuming 1 amp steady current being drawn, option a is correct.
"Assumption - the mother of all f#ck-ups!"
29th Nov 2001, 22:23
The answer is A, but I'm not sure about the parallel configuration - doesn't make sense? :confused:
29th Nov 2001, 22:42
About the paralell config:
The tension measured between A and B is obviously 12 volts.
For the current, the answer is 40amp + 40amp, it should make 80amp.
Easy, isn't it.
Look through the window :rolleyes:
29th Nov 2001, 22:43
The question [as asked] is unanswerable. Methinks you have missed some info out, or got your units wrong. It would produce 12VA for 80 hours if thats any help.
30th Nov 2001, 00:05
It looks like this forum software ****ed my drawing...it was superb, I can assure you
Keep on looking through the window :rolleyes:
1st Dec 2001, 19:12
The comments regarding the poor quality of this question are all valid. But questions of this (poor) quality frequently appear in fedback from students taking the JAR exams.
So be prepared for badly worded questions if you're taking the exams. You get no points for picking none of the options!!!
2nd Dec 2001, 15:27
Examine this via a process of elmination:-
Parallel connection means any answer with 24 volts is wrong, so rule out c) and d) straight away. Now we've narrowed the choices by 50% just on simple logic!
A single battery rated at 12 V and 40Ah would "produce" 40 amps when discharged at 1 amp but we have two in parallel so answer b) is less "correct" than answer a).
By this process you arrive at answer a), even without the actual discharge rate being given.
Examiners who set such questions should themselves be subject to examination. I have sat many multiple-choice exams over the decades and even today these badly-worded questions still keep cropping up. Someone ought to write a book about them! (Not me, though, I've finished with exams of an academic nature now.)
4th Dec 2001, 19:38
Having just sat through a string of JAR modules I think I can answer you this one !
What you have to look out for when you are surfing for questions and answers is that you must take the worded questions with a pinch of salt. You'll find quite often a badly remembered question can lead you up the garden path. There is no substitution for getting stuck into the study books !
Enough waffle - 2 x 12v,40aH Batteries in parallel = 12v,80aH
5th Dec 2001, 01:03
We experiment quite often poor feedbacks from students and after check mostly of times it appears that, in fact, the question was correctly worded. To all students: be careful with "feedbacks" which have not been checked by an instructor. We are flooded by this kind of poor feedbacks. Don't lose your time trying to guess the questions, study your notes.
This feedback was unanswerable as worded.
5th Dec 2001, 02:00
This is nothing more then a very short question to check if you understand the basic 2 principles of electicity!
In this case they've used the parallel setup:
so tension remains unchanged
(answers c and d are thus eliminated)
If you suppose the consuption of electricity remains unchanged (very reasonable as no change is mentioned) we can assume the answer a is the only correct one.
I can make a similar question myself placing 2 12V/40hr batteries in serie.
What do you think is the correct answer now?
6th Dec 2001, 08:42
The UKCAA are adopting JAR 66 for licensing maintenance technicians and this involves transferring responsibility for Electrical Systems from the present avionics group to the present mechanical group. To qualify, existing "heavies" must pass a multi-guess question paper with questions like these.
Ye Gods! This is scary stuff indeed! :eek:
Imagine if they let us "fairies" lose on mechanical systems after examining us to the same level. :confused:
Or if they asked pilots questions like this on the CPL :p
To make the aeroplane climb the pilot must:
a. Push the control column forward
b. Pull the control column back
c. Move the control column to the left
d. move the control column to the right
Through difficulties to the cinema
7th Dec 2001, 00:37
Your logic sounds okay but lets just see where it takes us.
In parallel we have 12 volts producing a total current such that a 12AH battery lasts for 80 hours.
Each battery is therefore producing 1/2 amp. The total current is therefore 1amp.
Now if 12 volts produces 1amp then the resistance of our circuit is 12 ohms.
If we now connect the batteries in series using the same load, we now have 24 volts across a 12 ohm resistnace. This gives a current of 2 amps.
Each battery is now applying 12 volts and passing 2 amps. So the batteries will now last for 40AH / 2A = 20 hours.
This option is not included in your question.
I suspect you have assumed that the batteries in series would produce the same 1 amp current as they did in parallel. This unfortunately is not how electricity works. And this was after all a question about electricity.
The original question should ideally have said something like
"when connected to a circuit a single 12V 40 AH battery lasts for 40 hours. If an identical battery is now connected in parallel with the first, how long will the batteries last if their internal resistances are ignored?
The correct answer would then be 80 hours.
7th Dec 2001, 02:05
May I point out a mistake in your comment on my alternative question?
<i>Now if 12 volts produces 1 ampere then the resistance of our (parallel) circuit is 12 ohms.</i>
This is totally correct, but note that this is the <b>total</b> resistance and as we have 2 identical resistances in parallel, each <b>individual</b> resistance in a parallel set-up equals 24 ohms!
Therefor, when you say,
<i>If we now connect the same 2 batteries in series, we now have 24 volts across a 12 ohm resistance. This gives a current of 2 amps.<i>
you make a mistake!
It should be:
<i>If we now connect the same batteries in series, we now have 24 volts across a 48 ohm resistance. This gives a current of 1/2 amps.</i>
You see, I didn't assume the current would stay 1 amp as you caim, however, it was you who overlooked the fact that the partition of the total resistance of the parallel set up between the 2 equal batteries would lead to a different total resistance when changing them to a setup in serie.
Needles to say that all this will now lead you straight to the correct answer on my alternative question (which is mentioned).
7th Dec 2001, 07:10
It's my first post here, but let me put my hand here!
The thing is actually a lot simpler than stated here in previous posts.
A battery does not produce anything! It just delivers energy that is stored in it in chemical form according to some demand. Demand here means a load that is connected across the battery terminals.
If we have a 40Ah 12V battery then it will deliver 12 V into a load at a current that depends on the load itself. If the load is 1 ohm then the current will be 12A. The battery will be able to supply this current for over 3 hours. It will supply 40A during one hour, 20A into 2 hours and so on. If the battery is subject to a short circuit then it will supply a very large current during a very short time if it isn’t damaged. In reality, the battery has some internal resistance which will cause the output voltage to drop and therefore the current will drop as well.
If 2 batteries are connected in parallel, then the current capacity is summed and the voltage remains the same. There are only 2 terminals in this configuration anyway. So 2 12V 40Ah batteries in parallel are equivalent to one 12V 80Ah battery.
If two batteries are connected in series, the result is a battery with a voltage that is the sum of the individual voltages and the current capacity remains the same. This is very similar to the 2 1.5V batteries that are used in some light torches in order to get 3V.
Going back to the original questions, I think that in my opinion, the wording is not the best. But answer A is the best one! For the series connection of batteries, the answer to tolipanebas ´s question should be C.
Hope this was clear!!! :)
7th Dec 2001, 10:41
Having recently sat through the Oz ATPL, I can assure half of the problem is understanding what CASA/CAA etc. are asking.
However, the key with these exams is making sure that you don’t choose the answers that are not least incorrect. :D
Capt Pit Bull
7th Dec 2001, 19:55
The Navigator and others.
I think you may have missed the point. In the given answers, the units used are amps rather than amp hours.
Thus the question as quoted is bogus. It either doesn't have enough information to answer it (if you assume the answer units are as intended), or the question is correct but the answers are all wrong since the units are wrong.
I'd put good money on assuming the units had all been mistyped.
I don't think anybody is really debating the methods for calculating capacities for different cell configurations.
This sort of thing really used to annoy me when I worked as a ground instructor. You spend ages trying to fill in the gaps in students basic scientific education, one of which is that the answer is meaningless unless the units are correct, then you find this kind of thing....
7th Dec 2001, 22:09
You appear to be assuming that the entire resistance in the circuit is due to battery internal resistances. Nothing in your qustions states that this is so.
More importantly it is unrealistic to suggest that we would have a circuit containing nothing but the batteries. Even when conducting capacity testing of batteries, a load is used to cotrol the discharge rate.
I suggest that it also rather improbable that a serviceable battery woould have a 24 ohm internal resistance.
I have (quite reasonably I think) assumed that the circuit included a load. I also ignored the internal resistance of the batteries because in most circuits they are a very small part of the total
The different ways in which we have interpreted this situation illustrates my original comment. If this is a real JAR question then it is a very poor one.
7th Dec 2001, 22:35
Ha it's a trick question.
The correct answer is that you shouldn't really connect two batteries in parallel unless you know a bit more about them.
Why? Well what would limit the current if one battery was was really 11.5V and the other 12.5V.
It wouldn't be wise to connect two NiCad batteries together unless the cells were carefully matched and the charge states were close to identical - even then you might get high currents flowing that could damage one of the batteries.
7th Dec 2001, 23:11
Having thought about this for a moment I think your interpretaion poses an even greater problem than simply not being very realistic.
If we assume that the entire circuit resistance is within the batteries that means the two terminals must be conected together by a zero resistance cable. Now that's not really possible, but if we use a very thick short metal bar then we will come pretty close to it.
To simplify the experiment let's first imagine a single 12 volt battery with a 24 ohm internal resistance. We connect its two terminals together with a two centimeter thick brass bar. The current flowing through the bar will be about 12V / 24 ohms = 1/2 amp.
We now need to know the voltage in the circuit so we connect a voltmeter across the terminals. But the voltmeter is also across the (almost) zero resistance bar, so it will read zero volts. This is because the entire voltage is being dropped within the battery.
Now it really makes no difference how many batteries you use or how you connect them. If the entire circuit resistance is within the batteries, then the entire voltage drop will be within the batteries. So as long as the circuit is connected and the current is flowing, the (external) voltage across the battery terminals will be zero.
7th Dec 2001, 23:24
Correct (as sure you know). Aircraft batteries are rated in ampere/hours. This means they are garanteed to deliver the advertised amount of amps for one hour i.e. a 40 ampere/hour battery will deliver 40 amps for one hourand divisions there-of.
Its all to do with total power failure back-up requirements.
8th Dec 2001, 01:56
When Batteries are connected in Parallel the voltage remains the same, the available current is added together.
In answer to your question a 80ah battery will discharge 1 amp for eighty hours or 80 amp for 1 hour or any combination in between.
There should be minimal voltage drop.
8th Dec 2001, 06:20
Capt Pit Bull: The question clearly states ampere hour. So my reasoning remains correct.
Keith Williams: I strongly suggest that you revise a bit of basic theory of electrical circuits. You insist in including resistances in your arguments although there is no information whatsoever about that in the question. You can’t talk about a load until you know what load it is. Of course, a battery is of no use if you do not have a load but when you do, you know how much current is being drawn because you can measure it. In addition, in the capacity tests you mentioned, the load is known.
I agree with you when you say that the value of the internal resistance may be neglected for most of the cases. But I doubt that it is 24ohm as you suggested. Just think that 40 A are being drawn from a battery (due to a load equivalent to 0.3ohm) . If the internal resistance of the battery was about 20ohm then, a voltage of 20 x 40 = 800 V would develop across the internal resistance. As this is not possible, the output voltage of the battery would just drop to zero and the current would stop. Therefore, the battery could not meet the specifications! The internal resistance is very low and much less then 1 ohm! Ideally, it should be zero but unfortunately, that is not the case.
Cwatters: You may connect two batteries in parallel provided that both are serviceable. If there is a difference in the output voltage of the two batteries, then a current will flow from one to the other and charge it until both are at the same voltage. This is what happens when you charge a battery or jump-start a car with a dead battery. Also, an “alternator” charges a battery with a higher voltage. The process of charging a battery always involves the connection of a higher voltage source to the battery. “Intelligent” chargers may control electronically the amount of current delivered to extend battery life.
This is not a very difficult subject but it is important that we discuss it until everybody agrees!!
8th Dec 2001, 16:28
Navigator, I think you need to read my last post and the few preceding it again.
I did not suggest that a battery would have an internal resistance of 24 ohm. I actually argued that this is an unrealistic figure.
If you look at the posts from tolipanebas, you will see that it is he who suggested a 24 ohm internal resistance. Indeed his entire argument was based upon that assumption.
You are also correct that no load is specified. But my original estimate of a load of 12 ohms was in response to toilpanebas's assumption that the current was 1 amp.
To summarise my comments:
The question is flawed in that its asks for the life of a 24 ampere hour battery but does not indicate what current is flowing.
The assumption by tolipanebas that the batteries have a 24 ohm resistnace and that this is the only resistance in the circuit, is unfounded.
If tolipanebas is correct however, then when the batteries are connected and a current flows the voltage measured across the terminal will be zero.
I think you will find that my electrical theory is quite sound but you need to read the entire string to following the arguments.
Capt Pit Bull
8th Dec 2001, 16:37
Hate to appear picky, but its Amp Hours, not Amp/Hours, which is quite different. Like I said, units are vital.
9th Dec 2001, 04:07
I suspect I got confused with the posts between you and tolipanebas! I apologise! I just found it a bit strange the arguments were going that way! A bit confusing in my opinion!
May be what you are trying to say is that the question should be about the equivalent capacity of the various configurations of bateries or alternatively they should have defined a load for the given configurations.
Capt Pit Bull is right about being picky! We should be! Units are fundamental in aviation (and evrywhere else I guess)! There were already a few incidents due to the lack of knowledge of units...
PS - I need to correct something I said on my earlier post! Answers A & B are both correct!
[ 09 December 2001: Message edited by: TheNavigator ]
9th Dec 2001, 17:11
You are correct in saying that the arguments were going in a strange direction, but this was simply in response to the suggestion by tolipanebas that the entire circuit resistance would be due to a 48 ohm internal resistance for each battery.
You are also correct in stating that both A and B are correct. This is precisely why I stated that the question was defective.
Your earlier suggestion that I had erroneously introduced a load is however more problematic.
The questions specifies two 12 volt, 40 Ampere hour batteries connected in parallel. The options offered are combinations of 24v, 12V, 40 hours and 80 hours. We have all agreed that when connected in parallel the circuit voltage will be 12V. But the two choices of 40 hours and 80 hours imply two possible circuit resistances.
The batteries are connected in parallel, so each provides half of the total current. So if the life is 40 hours then we have 40 Ampere hours divided by 40 hours = 1 Amp from each battery. This gives 2a for the entire circuit. If a circuit voltage of 12 V results in a current of 2A then the resistance must be 12 volts divided by 2 amp, which is 6 ohms. So if the correct answer is 40 hours then the total circuit resistance must be 6 ohms. By applying the same logic it can be shown that for a life of 80 hours, the total circuit resistance must be 12 ohms. So total resistances of 6 and 12 ohms are implicit in the options offered in the question.
The problem with the question is that the total resistance is not stated, nor is the matter of whether or not internal resistances should be ignored.
Those members who have suggested that 80 hour is the more correct of the two options, appear to be assuming that the original condition (when the 40 Ah capacity was measured) was 1 Amp. But there is no justification for this assumption.
The fact that the batteries are connected in parallel certainly implies that each will provide half of the circuit current. This in turn will have an effect on their life in the circuit. This effect is however not as obvious as it might at first appear. The critical question to consider is "what was the original condition?".
To illustrate this let's just assume that the total resistance is a 12 ohm resistor and the internal resistances of the batteries are sufficiently small to be ignored.
a. A single battery would provide a circuit voltage of 12V and a current of 1A. This would give a life of 40 Ah divided by 1A which is 40 hours.
b. Two batteries connected in series would provide a circuit voltage of 24V and a current of 2A. This would give a life of 40 Ah divided by 2A which is 20 hours.
c. Two batteries connected in parallel would give a circuit voltage of 12V and a current of 1A. But each battery would provide only 1/2A. So the life of each battery would be 40Ah divided by 1/2A which is 80 hours.
If we use the alternative assumption that the total resistance is 6 Ohms then the results will be as follows:
a. A single battery would provide a circuit voltage of 12V and a current of 2A. This would give a life of 40 Ah divided by 2A which is 20 hours.
b. Two batteries connected in series would provide a circuit voltage of 24V and a current of 4A. This would give a life of 40 Ah divided by 4A which is 10 hours.
c. Two batteries connected in parallel would give a circuit voltage of 12V and a current of 2A. But each battery would provide only 1A. So the life of each battery would be 40Ah divided by 1A which is 40 hours.
So both the 40 hour and 80 hour life are equally justifiable.
For those preferring to skip the analysis and simply assume that "putting two batteries in parallel reduces the current from each by half, so the life is doubled", the above figures might be surprising. Comparing a single battery with two in parallel does indeed double the life. But comparing two in series with two in parallel gives 4 times the life.
10th Dec 2001, 03:56
As a final remark!
May be those questions are designed to provide more chances for the candidates! :) I can't believe that the model answers assume only one of those options to be the right one! If that is the case, there is a serious problem within the system!!!
Thanks for your posts Keith!
19th Dec 2001, 02:03
If the question was stated as given then the correct answer is "12 volts untill well after the best before date". If it was given this way on an exam then the person who set it would be better employed in agriculture or road maintenance at a rather junior level and never allowed near an airplane. It reminds me of one on an RCAF tradesman's test I took.
What are the advantages of an electric current?
The obvious answer, "None at all to someone awaiting execution in Florida" wasn't in the list either.