Voltage accross parallel electric circuits.

[Anilkumar]

Why is the voltage accross the branches of parallel circuit, the same for all branches? Why dose a dead battery show its rated voltage accross its terminals? Thanks fellows, if you could kindly help me understand these things.

[Spyman]

Since all the parallel brances are connected to the same two points which the voltage is measured between, their voltages are all measured at the same location and that location can't have several voltages simultaneously. If the measurements are made at different locations then the voltage should differ, but if the wires connecting the branches to the parallel points have sufficient low resistance then the differences will be to small to pick up with an ordinary voltmeter.

 

A dead battery doesn't have to show any voltage at all across its terminals, it could very well be totally dead. However most used up batteries are only almost empty and as such there is likely enough energy left to get a reading from a voltmeter, but when it's connected to a load its voltage will drop because it gets drained.

Edited November 23, 2010 by Spyman

[Anilkumar]

Hi Spyman,

Thanks for the answer.

But I don't understand how voltage is location dependent? Primarily voltage forces the electrons to move. It is natural that resistors in series eat-up the voltage, and eventually it drops. But the same should happen with resistors connected in parellel. As, they too eat-up voltage. Resistors absorb energy. So they have to absorb energy, irrespective of how they are connected. Then why does the voltage not drop at each branch of the parallel connection?

Thanks again for caring to answer.

Regards.

[ewmon]

why does the voltage not drop at each branch of the parallel connection?

It does. In your circuit of parallel resistors, the voltage is supplied, and the current results from the voltage applied across the resistors. When applied across parallel resistors, the voltage produces currents in each branch equal to V/R, where R is the resistance for that branch. When the currents combine at either side of the resistors, the currents are additive (just like the waters of a river's branches splitting or combining).

 

Resistors absorb energy ... irrespective of how they are connected.

Yes, they absorb/dissipate the electrical power regardless of their configuration.

 

But I don't understand how voltage is location dependent?

In a detailed sense, wires have tiny (usually negligible) resistances. If you were to measure the voltage at a point down one of the branches toward its resistor instead of at the junction of the branches, the voltage would read slightly different due to the tiny voltage drop along the length of wire between the junction and the resistor.

 

It is natural that resistors in series eat-up the voltage, and eventually it drops. But the same should happen with resistors connected in parellel.

Regardless of the configuration, a resistor absorbs/dissipates an I²R amount of power, and the current depends upon the voltage and the resistance it's applied across. When a voltage is applied to a circuit, the current is a result of voltage working upon the circuit's overall resistance.

 

So, more accurately, the voltage drops across the combined resistances in series (that is, R1+R2+R3+...). The resulting current I=V/(R1+R2+R3+...) flows through all the resistors, and each resistor dissipates an I²R amount of power and an IR amount of voltage. Voltage drop across resistors is more accurately called IR drop. All the power dissipation and voltage drops add up, and everything is accounted for.

 

Voltages (or, more accurately, voltage drops) are additive with series resistors, and currents are additive with parallel resistors. These two facts are called Kirchhoff's Laws.

[Spyman]

Hi Spyman,

Thanks for the answer.

But I don't understand how voltage is location dependent? Primarily voltage forces the electrons to move. It is natural that resistors in series eat-up the voltage, and eventually it drops. But the same should happen with resistors connected in parellel. As, they too eat-up voltage. Resistors absorb energy. So they have to absorb energy, irrespective of how they are connected. Then why does the voltage not drop at each branch of the parallel connection?

Thanks again for caring to answer.

Regards.

A wire is nothing more than a long row of low level resistors in series and as such the location of the measurement determines how much of the voltage that are already used up.

 

Resistors in parallel eats up the voltage too, the difference is when they are in series they individually eat on different levels of the voltage, but when they are in parallel they all eat from the top.

 

How the resistors are connected and where they are placed in the circuit directly determines how much energy, of the avaible amount, there will be flowing through each of them.

 

Voltage is like a pressure forcing chains of electrons to move and resistors can be viewed as choke points restricting the amount of chains passing through them simultaneously. If the voltage is raised more chains can be forced to go through the choke point and if the voltage is lowered less chains will make it through.

 

When the resistors are placed in series all chains of electrons going through one of the resistors must also go through the rest of them, but since the resistors are placed in a row they get different levels of the voltage.

 

When the resistors are placed in parallel there will be different chains of electrons going through individual branches, but since all chains are going through the connection points together they will all share that voltage.

Edited November 23, 2010 by Spyman

[TonyMcC]

Here is a little observation that may help if you think about it. Everything you connect to the mains in your home such as TV, fridge ,freezer etc. gets the same voltage (the mains voltage which is 240V in UK). This because everything in your home is connected to the mains in parallel. However the more items you plug in the bigger will be your electricity bill. This is because each item takes current and the mains supply has to provide the sum total of all of the currents.

[Anilkumar]

Thanks everybody for your kind attention. The allegory of chains & choke points given by Spyman is a little bit convincing. But I am still chewing it again and again to make it digestible. [Pardon me fellows, I am slow at digesting]. But [quite many 'but's here] what is the meaning of 'when they are in parallel they all eat from the top'.

 

The fact that the voltage when measured accross the terminal junctions of the parallel branches is equal to the rated voltage; is agreeable. But why is it, the same at the terminal ends of the individual resistors connected in parallel? Why does it not split like everything else [the electric current, water flowing through a branched pipe] does?

 

The little observation by TonyMcC is well known. I know it is so. But why? Is there an explanation?

 

And what I meant to say regarding the battery problem was; we know that voltage is joules/coulomb [energy spent per unit charge to displace it and it is proportional to the number of charged particles displaced.] Then when the battery goes on discharging; the number of charged particles reduces. But then why does the voltage accross its terminals, still show the rated voltage?

Why is it not reduced?

 

I think Ewmon's comparison 'of the waters of a river's branches splitting or combining' is finally getting me somewhere.

 

Thanks everybody for your time.

 

It would be great if you would carry this thing on, till that thing gets into my head.

 

Regards.

Edited November 24, 2010 by Anilkumar

[Spyman]

Water systems can be very good analogies for electrical circuits, were voltage is represented by pressure or altitude and current of electrons by the flow of water.

 

Let's try with a simple comparison of series and parallel branches, suppose you are standing on a roof pouring water down into a funnel.

 

From the funnel there are three pipes velved in series straight down and it is clear how the altitude changes from the funnel, through the first pipe, second pipe, third pipe and finally the ground.

 

When the three pipes are velved parallel they are all velved together to the outlet of the funnel or to a large pipe going horizontally across the roof and as such would not change the altitude of the water until it flows down into the pipes. All three pipes must reach all the way down to the ground or otherwise the circuit would be broken and not work. As we can notice the top altitude for all three pipes is the same as for the funnel and the bottom altitude is also at ground level for all three pipes, therefore the altitude difference over the length from top to ground for all three pipes are equal to each other and to the altitude of the funnel.

 

 

The battery is chemically charging the poles with voltage and even if is drained to such low level that the chemical energy is no longer able to support the ordinary load, it might still be able to output a small flow of electrons with rated voltage, and a voltmeter only need a small amount of electrons for a reading, while the ordinary load normally needs a huge flow.

 

The chemicals are lifting water droplets to the roof, in the analogy, and the ordinary load needs a large flow through the main pipe's outlet on a huge water wheel to make its axis turn around. While the voltmeter only needs a tiny flow of water to make its small low friction water wheel turn. In both cases the water flows from the same altitude (voltage) but the small current that is enough for the voltmeter doesn't suffice to even budge the larger water wheel.

 

I think it might need to be noted while the battery can be in its rated range it does still decrease when it gets drained. There is a flat slope of the output voltage from a battery, slowly dropping as the battery gets uncharged until a breaking point is reached and the slope sharply turns steep down, when the battery gets exhausted.

[Anilkumar]

Hello,

 

Great!

 

Thanks man.

 

For the painstaking explanation. As I had said earlier, I was chewing it the whole day. In the afternoon I did, like you said. Pondered over the water analogy of Ewmon; 'the waters of a river's branches splitting or combining' broke the ice. And now your detailed explanation confirmed it. If I have understood it rightly [i am doing a recheck, while you guys are arround, to correct me if I am wrong.] here in both these analogies the equivalent of EMF in electricity is GRAVITY, right?

 

I went through the books again. Got a clearer understanding now, of the thing called ELECTRIC POTENTIAL similar to GRAVITATIONAL POTENTIAL; both of which are PE per charge and mass respectively and are independent of charge and mass respectively [Here charge and mass in the fields of electricity and gravity respectively are analogical, if I am right again.] Voltage is Electric potential which is Joules per Coulomb. So till the last charged particle with PE exists in the battery, it gives the rated voltage reading. And when that too becomes 'neutralised' [tell me if this is the right word] the battery will not give any reading, I suppose.

 

In the gravitational analogy, two objects placed at equal altitude have the same GP similar to two PD's accross two resistances connected in parallel, right? But when they are connected in series, it is like placing the two objects, one at greater altitude than the other, right?

 

One more question has proped up in my mind. [Can I go on spilling my doubts here? Hope so.] How do they make one battery more potent than the other? A stone or water can be lifted to higher altitudes to give them necessary PE. But how is this achieved in case of charged particles? By using one chemical more potent than the other to displace them, or what?

 

Thanks,

 

Regards.

[Spyman]

There are many different cinds of batteries with slightly different cell voltages in ranges from 1.2 to 3.6 Volts.

 

But more importantly several cells inside the battery can be connected in series to bring up the voltage.

 

Three 1.5 Volts batteries connected in series will together yield a voltage of 4.5 Volt and if they are connected in parallel they will instead be able to supply the tripple current of electrons.

 

If you take apart a standard 4.5 Volt battery you can see that it contains three 1.5 Volt cells in series.

(Be careful, batteries contains acids, which you don't want in your eyes.)

 

 

[EDIT]

I found this, showing six 1.5 Volts cells in series in a 9 Volt battery:

http://www.instructables.com/id/Making-a-45-volt-battery-pack-from-a-9V-battery/

Edited November 25, 2010 by Spyman

[dalemiller]

However most used up batteries are only almost empty and as such there is likely enough energy left to get a reading from a voltmeter, but when it's connected to a load its voltage will drop because it gets drained.

The immediate reduction of voltage across the battery terminals when a load has just been connected doesn't much occur from draining. As a battery weakens, its internal resistance increases. A voltage divider action places the open cell voltage across the battery's internal resistance in series with the load resistance shunted by the meter. IR drop across the internal resistance is a voltage that is subtracted from the open cell voltage. That reduced potential is all that remains across the load resistance.

[Anilkumar]

Well,

 

Thats it. Bulls-eye.

 

Thanks a lot.

 

You have not corrected me. It shows I have understood the theory correctly.

 

There are other doubts. But different matters. May be I should take-up a new thread.

 

Well, thanks again fellows, for helping out.

 

Regards.

[Spyman]

The immediate reduction of voltage across the battery terminals when a load has just been connected doesn't much occur from draining. As a battery weakens, its internal resistance increases. A voltage divider action places the open cell voltage across the battery's internal resistance in series with the load resistance shunted by the meter. IR drop across the internal resistance is a voltage that is subtracted from the open cell voltage. That reduced potential is all that remains across the load resistance.

The internal resistance in the battery increases due to a depletion of the chemical energy, which in turn reduces the final output voltage between the terminals.

 

¤ Before the load gets connected there is a buildup of a small charge that then rapidly gets depleted if the battery fails to sustain the voltage with the load.

 

¤ If the output current through the load brings the terminal voltage below the ratings, then a higher outflow than what the battery is able to support empties it.

 

¤ When the chemical reactions inside the battery no longer can keep up the rated voltage, then the chemical energy in the battery is getting exhausted.

 

I am not a native English speaker but the word "drained" seems correct to me.

[Anilkumar]

Hi Spyman,

 

Nice piece of explanation there,

 

But,

If the output current through the load brings the terminal voltage below the ratings, 'then' a higher outflow than what the battery is able to support 'empties' it.

this part I didn't understand.

 

Can a battery give a higher outflow than what the battery is able to support?

 

Does the word 'then' stand for 'it means/shows that' here and 'empties' stand for 'has emptied it'

 

Thanks.

Edited November 27, 2010 by Anilkumar

[ewmon]

I have repeatedly bit my tongue regarding Spyman's posts. I don't know if it's language skills getting in the way, but Spyman's posts sound like particular conditions cause particular results. However, this isn't true.

 

For example, 1) charges get depleted, 2) batteries will empty, and 3) batteries will get exhausted for many other reasons and not just the conditions stated in Spyman's last post. I don't know from what knowledge Spyman's speaks.

[Anilkumar]

Hi there,

 

I feel Spyman has good understanding of the subject.

 

Sometimes one can't express adequately. Which can be overcome by discussion.

[Spyman]

Hmm, sorry to cause confusion, I did not intend to say that only certain conditions can end up with exhausted batteries, I only tried to explain that the use of the word "drained" in my post #2 was not wrong, as opposed by dalemiller.

 

A battery can not give a higher outflow than what it can support, but you can connect a load that would produce a higher outflow if the battery could support it, instead of giving this impossible outflow the battery instead chokes on internal resistance due to a depletion of chemical energy, which reduces the terminal voltage until the outflow matches what it can support.

 

Votlage, current and resistance follows Ohm's law, were: Voltage = Current * Resistance.

 

As such I meant that "then" is what will happen then if that condition happens and "empties" means that the battery is getting more and more empty because the battery's maximum outflow is lower than what the load is able to consume, which in turn very fast forces the voltage to drop lower and lower until the battery is no longer able to give noticeable outflow at all.

Edited November 29, 2010 by Spyman

[Anilkumar]

Hi Spyman,

 

Thanks for the trouble.

 

What makes a light bulb or any resistance, a 6V 1A, or 10V 2A, or xVolts yAmpere? Does it solely depend on the material, thickness, length and temperature of the coil?

 

What is regulated in a Fan and an Iron-Box: V, I, or R to increase/decrease the speed and heat respectively?

 

Thank you.

[Spyman]

Yes, the electrical resistance for a material depends on its properties, size and the environment.

 

There are different cinds of fans and therefore also different ways to regulate their speed, small fans for DC power, like in computers, usually have some cind of voltage regulation, but bigger industrial fans, like for ventilation, normally runs on AC power with frequency converters controlling their speeds.

 

Irons tools, electrical heating radiators and similar equipment have a rather simple adjustable thermostat that either turns the power on or off when the temperature is below or above desired value, but there exists voltage or current regulators for increased precision or smoother strain on the supply.

Edited November 30, 2010 by Spyman

[ewmon]

... A battery can not give a higher outflow than what it can support ...

... you can connect a load that would produce a higher outflow if the battery could support it ...

... instead of giving this impossible outflow, the battery instead chokes on internal resistance ...

... a depletion of chemical energy ... reduces the terminal voltage until the outflow matches what it can support.

... the battery's maximum outflow is lower than what the load is able to consume, which in turn very fast forces the voltage to drop lower and lower until the battery is no longer able to give noticeable outflow at all.

Spyman, your language skills are seriously getting in the way of describing the reality of electricity. I've studied and/or worked with electricity for decades, and you are not describing electricity properly or clearly. You repeatedly use the word "outflow", which I have never seen or heard used before. You still describe causes and effects improperly. For example, the statement below has very little scientific meaning.

 

the battery's maximum outflow is lower than what the load is able to consume

Any load that doesn't undergo a catastrophic failure is "able to consume" more of what you strangely call "outflow".

 

It seems that your source of information is not scientific.

[Spyman]

... A battery can not give a higher outflow than what it can support ...

... you can connect a load that would produce a higher outflow if the battery could support it ...

... instead of giving this impossible outflow, the battery instead chokes on internal resistance ...

... a depletion of chemical energy ... reduces the terminal voltage until the outflow matches what it can support.

... the battery's maximum outflow is lower than what the load is able to consume, which in turn very fast forces the voltage to drop lower and lower until the battery is no longer able to give noticeable outflow at all.

Spyman, your language skills are seriously getting in the way of describing the reality of electricity. I've studied and/or worked with electricity for decades, and you are not describing electricity properly or clearly. You repeatedly use the word "outflow", which I have never seen or heard used before. You still describe causes and effects improperly. For example, the statement below has very little scientific meaning.

 

the battery's maximum outflow is lower than what the load is able to consume

Any load that doesn't undergo a catastrophic failure is "able to consume" more of what you strangely call "outflow".

 

It seems that your source of information is not scientific.

Ok, ewmon let's see if we can come to an understanding...

 

First, here is the context:

Why dose a dead battery show its rated voltage accross its terminals?

However most used up batteries are only almost empty and as such there is likely enough energy left to get a reading from a voltmeter, but when it's connected to a load its voltage will drop because it gets drained.

 

With the background layed out here is another try to explain it more properly:

 

If Anilkumar connects a Voltmeter to an almost empty 1.5 Volt battery, the Voltmeter will load the battery with a resistance of around 10 Mega Ohm which will cause a current of ~0.15 mikro Amperes. This low strain will likely not exhaust the chemical energy left inside the battery during Anilkumar's measurement of the voltage.

 

In layman's terms: "the battery's maximum outflow is higher than what the load can consume".

 

With other words: the battery is able to keep the rated voltage while supporting this current and the load will have a current flow through it, with the rated voltage, that is lower than what the battery is able to sustain.

 

 

When Anilkumar on the other hand connects a load with a low resistance, like a lightbulb of around 1 Ohm, then the load would be able to let through a current of 1.5 Amperes, if the battery is able to uphold the rated voltage while the current flows through the lamp making it shine bright.

 

But since the battery is almost empty, the chemical energy is nearly depleted, so it won't be able to uphold the rated voltage and ten millioooooon times higher current than with the Voltmeter. Instead the battery voltage drops to the highest value that gives the maximum current it is able to support, through the load in it's weakened state.

 

In layman's terms: "the battery's maximum outflow is lower than what the load is able to consume".

 

With other words: the battery is not able to keep the rated voltage while supporting this current and the load would have a current flow through it, with the rated voltage, that is higher than what the battery is able to sustain.

 

 

Is that scientific enough for you or is there still something nagging you?

 

Don't bite your tongue, please spit it out in the open for further discussion.

 

For instance, do you like dalemiller object to my use of the word "drained"?

[ewmon]

Is that scientific enough for you or is there still something nagging you?

To begin with, please define your word "outflow"?

 

Regardless, batteries generally do not maintain any voltage. The diagram below shows the discharge characteristics of a "1.5-volt" Duracell^® AA alkaline under various loads. Voltage drops with use, regardless of whether it's light or heavy. Look at the discharge curve for a battery, and you will see that the voltage almost always drops with usage. Most batteries have a nominal rating that's basically a "pre-use" rating. Once it undergoes any significant use, the voltage drops to another voltage.

 

 

Furthermore, most/all batteries operate for most of their operational life at much less than their "rated voltage", so talk about maintaining the rated voltage means nothing. For example, typical store-bought 1.5-volt batteries (regardless of whether they're regular, heavy duty, alkaline, rechargeable, etc or whether they're size D, C, A, AA, AAA, etc) will operate for most of their usable life at about 1.2 volts, not the "rated" 1.5 volts. Manufacturers of equipment know this, and they make sure their equipment can operate accordingly.

 

Lastly, a nearly depleted battery never regains it's rated voltage regardless of the load. In other words, usage draws down a battery pretty much permanently, so even when disconnected from the drain of a heavy load, the battery still cannot produce the "rated" voltage. This is why, when you use a voltmeter with a 10 megohm input impedance to measure the voltage of an almost drained battery, it reads something like 0.893 volts, and not the "rated" 1.500 volts.

 

Here's a good read.

Edited November 30, 2010 by ewmon

[Spyman]

To begin with, please define your word "outflow"?

Well, if a person who has "studied and/or worked with electricity for decades" is not able to understand my explanation in my post #21, then I simply suspect you don't want to either. From the tone of your posts and the following long windled explanations and arguments of details that is more or less beside the point of what Anilkumar asked for and my explanations of, I draw the conclusion that you for some unknown reason are trying to put me down instead of trying to be helpful.

 

Since your goal seems to be to only criticize and belittle me, I frankly don't see the point in wasting my time further on trying to "define" words for you.

 

But you might ponder this: If I would have used the expression: "Operating voltage is dictated by the state-of-discharge and the actual load imposed by the equipment", then I probably could have satisfied you but my goal was to help Anilkumar and I don't think that expression would have been very helpful.

 

 

Regardless, batteries generally do not maintain any voltage. The diagram below shows the discharge characteristics of a "1.5-volt" Duracell^® AA alkaline under various loads. Voltage drops with use, regardless of whether it's light or heavy. Look at the discharge curve for a battery, and you will see that the voltage almost always drops with usage. Most batteries have a nominal rating that's basically a "pre-use" rating. Once it undergoes any significant use, the voltage drops to another voltage.

Yes, the voltage always drops on usage and different batteries have different discharging curves.

 

But I don't see your point, I never opposed this or said something against it and already mentioned the discharge curve briefly in my post #8:

 

"I think it might need to be noted while the battery can be in its rated range it does still decrease when it gets drained. There is a flat slope of the output voltage from a battery, slowly dropping as the battery gets uncharged until a breaking point is reached and the slope sharply turns steep down, when the battery gets exhausted."

 

 

Furthermore, most/all batteries operate for most of their operational life at much less than their "rated voltage", so talk about maintaining the rated voltage means nothing. For example, typical store-bought 1.5-volt batteries (regardless of whether they're regular, heavy duty, alkaline, rechargeable, etc or whether they're size D, C, A, AA, AAA, etc) will operate for most of their usable life at about 1.2 volts, not the "rated" 1.5 volts. Manufacturers of equipment know this, and they make sure their equipment can operate accordingly.

Yes, the batteries are not fixed on a constant voltage of exactly 1.500 Volts, depending on type and remaining charge the terminal voltage of the battery will differ, they have a known normal range for normal use.

 

I don't see your point here either, I never said something different and I have a hard time to understand why or how you fail to reach an understanding.

 

If I would have belived that you were genuine interested and trying to understand then I would have extended further on what I ment with "maintaining the rated voltage" but as I already have explained above I won't.

(Unless someone else is asking.)

 

 

Lastly, a nearly depleted battery never regains it's rated voltage regardless of the load. In other words, usage draws down a battery pretty much permanently, so even when disconnected from the drain of a heavy load, the battery still cannot produce the "rated" voltage. This is why, when you use a voltmeter with a 10 megohm input impedance to measure the voltage of an almost drained battery, it reads something like 0.893 volts, and not the "rated" 1.500 volts.

Here I strongly disagree with you, independent of where one might draw the limit of what a "nearly depleted battery" actually is and what the defined range of "rated" voltage are, a measurement of the battery voltage will still drop during a load and then regain some potential when the load is disconnected. Some batteries on the verge of their current use might even regain enough to run the load for another few minutes, if they get a longer duration of rest.

 

The main point here is that Anilkumar asked how a nearly empty battery would show a voltage in range with its ratings with a voltmeter and then still not function when the load gets connected, which is what I have tried to explain and seem to have succeded with, except for you even though you try to claim to already understand and have knowledge of this.

 

I agree that they never will regain their full voltage or get back to exactly 1.500 Volts, which I never said that they would either, but they are likely to return to a level within their normal range of rated voltage when the load gets disconnected and they still are only nearly empty.

 

 

I did test on three different depleted 1.5 Volt AA batteries which all had laid in a recycle bin for a longer duration, as load I used a 12 Volt 20 Watt lightbulb:

 

Battery 1

1.393 Volts before connecting Load

1.100 Volts after a few seconds with Load

1.310 Volts a few minutes after disconnecting load

0.900 Volts after more than 10 minutes with Load connected again

1.100 Volts 10 seconds after disconnecting Load again

1.150 Volts after 1 minute without Load

1.221 Volts after 10 minutes without Load

 

Note: This battery surprised me of having quite a lot of energy still left inside it, I guess it would suffice for usage in a computer mouse or a TV remote controller for some time yet and that it likely had been used in a high-drain load like a digital camera or something similar since someone has discarded it.

 

 

Battery 2

1.231 Volts before connecting Load

0.370 Volts directly after connecting Load

0.305 Volts after a few seconds with Load

0.985 Volts directly after disconnecting Load

1.175 Volts after 1 minute without the Load

1.218 Volts after 10 minutes without the Load

 

Note: This battery seems more like what I expected and what I thought of when explaining for Anilkumar.

 

 

Battery 3

0.912 Volts before connecting Load

0.008 Volts directly after connecting Load

0.005 Volts after a few seconds with Load

0.612 Volts directly after disconnecting Load

0.832 Volts after 1 minute without the Load

0.863 Volts after 10 minutes without the Load

 

Note: This was the most exhausted battery of them and is more like the "nearly depleted battery" ewmon is talking about, but even when it's this highly discharged the battery is still able to recover a large amount of its terminal voltage when the load is disconnected and almost return to the voltage it had before.

 

 

If you don't trust me then I suggest that you take a few depleted batteries and measure for yourself...

Edited December 1, 2010 by Spyman

[ewmon]

If anyone else here knows what the word "outflow" means, please define so the people who are learning electricity can relate it to the technical terminology that they find on the internet and in books.

[Spyman]

If anyone else here knows what the word "outflow" means, please define so the people who are learning electricity can relate it to the technical terminology that they find on the internet and in books.

LOL ewmon, what is most important for learning, the exactly correct technical terminology or the basic of the working principle for the phenomena in question?

 

Everything I have said are correct and IMHO helpful for people trying to understand electricity, although as I already have admitted there might be language problems and I have certainly used a lot of layman's terms instead of correct terminology because people starting to learn are not likely already well versed with it.

 

Anyone reading through this thread can see that your intentions with the last post is obviously not to be helpful at all, but instead solely made to flame me...

Edited December 1, 2010 by Spyman