Batteries: Amp/Hour
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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?
Many thanks!
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?
Many thanks!
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Hey chaps,
About the paralell config:
_____ 12v_____
A___/ \___B
\_____ 12v_____/
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
About the paralell config:
_____ 12v_____
A___/ \___B
\_____ 12v_____/
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
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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!!!
So be prepared for badly worded questions if you're taking the exams. You get no points for picking none of the options!!!
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.)
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.)
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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
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
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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.
This feedback was unanswerable as worded.
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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.
a:12V/40hr
b:12V/80hr
c:24V/40hr
D:24V/80hr
What do you think is the correct answer now?
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.
a:12V/40hr
b:12V/80hr
c:24V/40hr
D:24V/80hr
What do you think is the correct answer now?
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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!
Imagine if they let us "fairies" lose on mechanical systems after examining us to the same level.
Or if they asked pilots questions like this on the CPL
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
Ye Gods! This is scary stuff indeed!
Imagine if they let us "fairies" lose on mechanical systems after examining us to the same level.
Or if they asked pilots questions like this on the CPL
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
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tolipanebas,
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.
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.
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Hi Keith,
May I point out a mistake in your comment on my alternative question?
You say:
<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).
May I point out a mistake in your comment on my alternative question?
You say:
<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).
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Hi there!
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!!!
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!!!
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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.
However, the key with these exams is making sure that you don’t choose the answers that are not least incorrect.
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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....
CPB
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....
CPB
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tolipanebas,
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.
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.