Batteries: Amp/Hour
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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.
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.
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tolipanebas,
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.
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.
lazy fairweather PPRuNer
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Navigator,
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.
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.
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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.
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.
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Hi again,
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!!
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!!
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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.
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.
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Hi Keith,
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 ]
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 ]
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Navigator,
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.
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.
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Hi,
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!
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!
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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.
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.
